# The singular set in the Stefan problem

Alessio Figalli  ETH Zürich, Department of Mathematics, Rämistrasse 101, 8092
Zürich, Switzerland<EMAIL_ADDRESS>,  Xavier Ros-Oton  ICREA,
Pg. Lluís Companys 23, 08010 Barcelona, Spain & Universitat de Barcelona,
Departament de Matemàtiques i Informàtica, Gran Via de les Corts Catalanes
585, 08007 Barcelona, Spain<EMAIL_ADDRESS>and  Joaquim Serra  ETH Zürich,
Department of Mathematics, Rämistrasse 101, 8092 Zürich, Switzerland
<EMAIL_ADDRESS>
###### Abstract.

In this paper we analyze the singular set in the Stefan problem and prove the
following results:

  * •

The singular set has parabolic Hausdorff dimension at most $n-1$.

  * •

The solution admits a $C^{\infty}$-expansion at all singular points, up to a
set of parabolic Hausdorff dimension at most $n-2$.

  * •

In $\mathbb{R}^{3}$, the free boundary is smooth for almost every time $t$,
and the set of singular times $\mathcal{S}\subset\mathbb{R}$ has Hausdorff
dimension at most $1/2$.

These results provide us with a refined understanding of the Stefan problem’s
singularities and answer some long-standing open questions in the field.

###### Key words and phrases:

Stefan problem; free boundary

###### 2010 Mathematics Subject Classification:

35R35; 35B65.

##  1\. Introduction

The Stefan problem, dating back to the XIXth century [St, LC], aims to
describe phase transitions, such as ice melting to water, and it is among the
most classical and well-known free boundary problems. In its simplest
formulation, this problem consists in finding the evolution of the temperature
$\theta(x,t)$ of the water when a block of ice is submerged inside. Then, the
function $\theta\geq 0$ satisfies $\partial_{t}\theta-\Delta\theta=0$ in the
region $\\{\theta>0\\}$, while the evolution of the free boundary
$\partial\\{\theta>0\\}$ (the interphase ice/water) is dictated by the Stefan
condition $\partial_{t}\theta=|\nabla_{x}\theta|^{2}$ on
$\partial\\{\theta>0\\}$.

Let $\chi_{A}$ denotes the characteristic function of a set $A$. After the
transformation $u(x,t):=\int_{0}^{t}\theta(x,\tau)d\tau$, one can note that
$\\{u>0\\}=\\{\theta>0\\}$. Also, as explained in [Duv73, Bai74] (see also
[Fig18a]), the Stefan problem becomes locally equivalent to the so-called
“parabolic obstacle problem”:

$\left\\{\begin{split}\partial_{t}u-\Delta u&=-\chi_{\\{u>0\\}}\\\ u&\geq 0\\\
\partial_{t}u&\geq 0\\\ \partial_{t}u&>0\quad{\rm
in}\,\\{u>0\\}\end{split}\right.\qquad\qquad\quad\textrm{in}\quad\Omega\times(0,T)\subset\mathbb{R}^{n}\times\mathbb{R}.$
(1.1)

The regularity of the free boundary for the Stefan problem (1.1) was developed
in the groundbreaking paper [Caf77]. The main result therein establishes that
the free boundary (i.e., the interface $\partial\\{u>0\\}$) is $C^{\infty}$ in
space and time, outside some closed set $\Sigma\subset\Omega\times(0,T)$ of
_singular points_ at which the contact set $\\{u=0\\}$ has zero density.

###  1.1. The singular set

The fine understanding of singularities is a central research topic in several
areas related to nonlinear PDEs and Geometric Analysis. A major question in
such context is to establish estimates for the _size_ of the singular set —
see [Sim68, CKN82, Whi00, Alm00] for some famous examples.

For the Stefan problem (1.1), a variant of the techniques used in the study of
the elliptic obstacle problem yields the following result: for every $t$, let
$\Sigma_{t}$ denote the singular points at time $t$ (so that
$\Sigma=\cup_{t\in(0,T)}\Sigma_{t}\times\\{t\\}$). Then
$\Sigma_{t}\subset\mathbb{R}^{n}$ is locally contained in a
$(n-1)$-dimensional $C^{1}$ manifold [LM15, Bla06]. Furthermore, the whole
singular set $\Sigma\subset\mathbb{R}^{n}\times\mathbb{R}$ is contained in a
$(n-1)$-dimensional manifold which is $C^{1}$ in space and $C^{0,1/2}$ in
time; see [LM15] for more details. This result is optimal in space, in the
sense that the singular set could be $(n-1)$-dimensional for a fixed time
$t=t_{0}$. However, it is not clear what the size of the singular set _in
time_ should be.

The first natural question in this direction is to estimate the parabolic
Hausdorff dimension111 The parabolic Hausdorff dimension is, by definition,
the Hausdorff dimension associated with the “parabolic distance” $d_{\rm
par}\bigl{(}(x_{1},t_{1}),(x_{2},t_{2})\bigr{)}=\sqrt{|x_{1}-x_{2}|^{2}+|t_{1}-t_{2}|}$.
Notice that, if we denote by ${\rm dim}_{\mathcal{H}}(E)$ the standard
Hausdorff dimension of a set
$E\subset\mathbb{R}^{n+1}=\mathbb{R}^{n}\times\mathbb{R}$, then ${\rm
dim}_{\mathcal{H}}(E)\leq{\rm dim}_{\rm par}(E)$. On the other hand, the time
axis has parabolic Hausdorff dimension 2, while it has standard Hausdorff
dimension 1.  of the singular set $\Sigma$, denoted ${\rm dim}_{\rm
par}(\Sigma)$. The results in [LM15, Bla06] imply that ${\rm dim}_{\rm
par}(\Sigma)\leq n+\frac{1}{2}$, and no better estimate was known. Here, by
refining our understanding of singular points, we establish the following:

######  Theorem 1.1.

Let $\Omega\subset\mathbb{R}^{n}$ be an open set, let $u\in
L^{\infty}(\Omega\times(0,T))$ solve the Stefan problem (1.1), and let
$\Sigma\subset\Omega\times(0,T)$ be the set of singular points. Then

${\rm dim}_{\rm par}(\Sigma)\leq n-1.$

As mentioned above, the singular set could be $(n-1)$-dimensional in space for
a fixed time $t=t_{0}$. Hence our result gives the optimal bound for the
parabolic Hausdorff dimension of $\Sigma$. Also, since the time axis has
parabolic Hausdorff dimension 2, Theorem 1.1 implies that, in
$\mathbb{R}^{2}$, the free boundary is smooth for almost every time
$t\in(0,T)$. Therefore, it is natural to ask ourselves if a similar result
holds in the physical space $\mathbb{R}^{3}$ and, more in general, “how often”
singular points may appear.

In [Caf78], Caffarelli showed that any $C^{1}$ curve contained inside $\Sigma$
cannot evolve with time (i.e., it must be contained in a fixed time slice
$\\{t=t_{0}\\}$). However, apart from this result, nothing else was known
concerning this question. Here, we prove the following estimate on the size of
singular times in the physical space $\mathbb{R}^{3}$:

######  Theorem 1.2.

Let $\Omega\subset\mathbb{R}^{3}$, let $u\in L^{\infty}(\Omega\times(0,T))$
solve the Stefan problem (1.1), and let

$\mathcal{S}:=\big{\\{}t\in(0,T)\ :\ \exists\,(x,t)\in\Sigma\big{\\}}$ (1.2)

denote the set of “singular times”. Then,

$\dim_{\mathcal{H}}(\mathcal{S})\leq\frac{1}{2}.$

In particular, for almost every time $t\in(0,T)$, the free boundary is a
$(n-1)$-dimensional submanifold of $\mathbb{R}^{n}$ of class $C^{\infty}$.

This is the first result on the size of the set of singular times for the
Stefan problem. In particular, prior to this result, it was not even known if
solutions to the Stefan problem (1.1) could have singularities for _all_ times
$t\in(0,T)$ (not even when $n=2$).

We do not know if the dimensional bound $1/2$ is sharp for the Stefan problem
in $\mathbb{R}^{3}$, but this bound is definitely critical222The reader
familiar with fluid equations may note that $1/2$ is the same bound currently
known for the dimension of singular times in the Navier-Stokes equation; see
the classical result of Caffarelli, Kohn, and Nirenberg [CKN82]. We do not see
any connection between the fact that the numbers are the same. Actually, while
for Navier-Stokes many people would like to prove that singular points do not
exist, in the Stefan problem singular points exist and can be rather large.
Also, given the analysis performed in our paper, the estimate $1/2$ is sharp
in many points (see Remark 1.5)., and we would not be surprised if our bound
turned out to be optimal.

Our Theorem 1.2 above follows from more general result valid in arbitrary
dimension $\mathbb{R}^{n}$. As explained in more detail below, we prove that
the singular set is $C^{\infty}$, outside a small subset of parabolic
dimension at most $n-2$.

######  Theorem 1.3.

Let $\Omega\subset\mathbb{R}^{n}$, and let $u\in
L^{\infty}(\Omega\times(0,T))$ solve the Stefan problem (1.1). Then there
exists $\Sigma^{\infty}\subset\Sigma$ such that

${\rm dim}_{\rm par}(\Sigma\setminus\Sigma^{\infty})\leq
n-2,\qquad\dim_{\mathcal{H}}\big{(}\\{t\in(0,T)\ :\
\exists\,(x,t)\in\Sigma^{\infty}\\}\big{)}=0,$

and $\Sigma^{\infty}\subset\Omega\times(0,T)$ can be covered by countably many
$(n-1)$-dimensional submanifolds in $\mathbb{R}^{n+1}$ of class
$C^{\infty}$.333Here, the submanifolds that cover $\Sigma^{\infty}$ are of
class $C^{\infty}$ as subset of $\mathbb{R}^{n+1}$ with the usual Euclidean
distance, not with respect the parabolic distance. So, our statement is much
stronger than the previously known results (for instance, [LM15] proved
$C^{1}$ regularity of $\Sigma$ with respect to the parabolic distance, which
implies only $C^{1/2}$ regularity in time).

In a sense, this result says that the singular set can be split into two
separate pieces: one which is very smooth and extremely rare in time (the set
$\Sigma^{\infty}$), and one which is small (of dimension at most $n-2$).

As a consequence, we also deduce the following corollary in 2 dimensions:

######  Corollary 1.4.

Let $\Omega\subset\mathbb{R}^{2}$, and let $u\in
L^{\infty}(\Omega\times(0,T))$ solve the Stefan problem (1.1). Let
$\mathcal{S}$ be as in (1.2). Then

$\dim_{\mathcal{H}}(\mathcal{S})=0.$

In the next section, we briefly explain the general strategy behind the proofs
of our results.

###  1.2. Ideas of the proof

To prove Theorems 1.1, 1.2, and 1.3 we need to introduce a variety of new
ideas with respect to the existing literature, combining tools from geometric
measure theory (GMT), PDE estimates, dimension reduction-type arguments, and
new monotonicity formulas. Let us give a quick overview of the main steps in
the proofs.

Let $(x_{\circ},t_{\circ})$ be a singular point. It is well known that

$u(x_{\circ}+x,t_{\circ}+t)=p_{2,x_{\circ},t_{\circ}}(x)+o(|x|^{2}+|t|),$
(1.3)

where $p_{2,x_{\circ},t_{\circ}}(x)$ is a quadratic polynomial of the form
$\frac{1}{2}x^{T}Ax$, with $A\geq 0$ and ${\rm tr}(A)=1$. In particular, the
set of singular points can be descomposed as
$\Sigma=\cup_{m=0}^{n-1}\Sigma_{m}$, where

$\Sigma_{m}:=\big{\\{}(x_{\circ},t_{\circ})\in\Sigma\,:\,{\rm
dim}(\\{p_{2,x_{\circ},t_{\circ}}=0\\})=m\big{\\}},\qquad m=0,\ldots,n-1.$

Moreover, for each $m$, the set $\Sigma_{m}\cap\\{t=t_{\circ}\\}$ can be
covered by a $C^{1}$ manifold of dimension $m$. Unfortunately, the previous
expansion implies only $C^{1/2}$ regularity in time for the covering
manifolds. In particular, because of this, (1.3) implies a non-sharp bound on
the parabolic Hausdorff dimension of $\Sigma_{m}$.

To prove Theorem 1.1, our strategy is to refine (1.3) in the time variable, by
developing a parabolic version of [FS17]. This is the content of the first
part of the paper. A key tool for this is a truncated version of the parabolic
frequency function

$\phi(r,w):=\frac{r^{2}\int_{\\{t=-r^{2}\\}}|\nabla
w|^{2}G\,dx}{\int_{\\{t=-r^{2}\\}}w^{2}G\,dx},$

where $G$ is the time-reversed heat kernel. We will see that (a truncated
version of) $\phi(r,w)$ is essentially monotone in $r$ for

$w=u(x_{\circ}+\cdot,t_{\circ}+\cdot)-p_{2,x_{\circ},t_{\circ}}.$

Thanks to this fact, assuming with no loss of generality that
$(x_{\circ},t_{\circ})=(0,0)$, we can prove that

$\frac{(u-p_{2})(rx,r^{2}t)}{\|u-p_{2}\|_{r}}\longrightarrow
q(x,t)\qquad\text{as }r\to 0$ (1.4)

along subsequences, where $q$ is a parabolically homogeneous function.

We then show the following:

  * (i)

If $(0,0)\in\Sigma_{m}$ with $m\leq n-2$, then the function $q$ is a
_quadratic_ caloric polynomial. This means that the expansion (1.3) cannot be
improved at these points! Hence, to get an improved dimensional bound on
$\Sigma_{m}$ we employ a barrier argument in the spirit of [FRS20]. More
precisely, since $m\leq n-2$, we can exploit the fact that $\Sigma_{m}$ has
zero capacity to build a refined barrier and show that

$\qquad\qquad\qquad\qquad{\rm dim}_{\rm par}(\Sigma_{m})\leq m,\qquad 0\leq
m\leq n-2.$

  * (ii)

If $(0,0)\in\Sigma_{n-1}$, then $q$ is a homogeneous solution of the parabolic
thin obstacle problem. We denote by $\Sigma_{n-1}^{<3}$ the subset at which
the homogeneity is less than $3$.

    * (a)

If $(0,0)\in\Sigma_{n-1}^{<3}$, we show that $\partial_{t}q\not\equiv 0$ and
that $q$ is _convex_ in all directions that are tangential to $\\{p_{2}=0\\}$.
This allows us to perform a dimension reduction that, combined with a barrier
argument, implies

${\rm dim}_{\rm par}(\Sigma_{n-1}^{<3})\leq n-2.$

    * (b)

If $(0,0)\in\Sigma_{n-1}\setminus\Sigma_{n-1}^{<3}$ we show that $q$ is always
3-homogeneous, hence

$u(x_{\circ}+\cdot,t_{\circ}+\cdot)=p_{2,x_{\circ},t_{\circ}}+O(|x|^{3}+|t|^{3/2}),$
(1.5)

and the same barrier argument as in (ii)-(a) implies that

${\rm dim}_{\rm par}(\Sigma_{n-1}\setminus\Sigma_{n-1}^{<3})\leq n-1.$

Combining these estimates, Theorem 1.1 follows.

We wish to remark that, although these proofs require a series of delicate new
estimates, the strategy behind this first result is a generalization of the
ideas in [FS17, FRS20] to the parabolic setting. However, to prove Theorems
1.2 and 1.3, completely new ideas are needed.

Indeed, in the elliptic setting one, can show that, outside a small set of
dimension $n-2$, the “second blow-up” $q$ is a cubic harmonic polynomial and
the frequency function is still (almost) monotone for the function
$w=u-p_{2}-q$. This fact was crucially used in [FRS20] and it allowed us to
prove that, in the elliptic setting, an expansion of the form

$u=p_{2}+p_{3}+p_{4}+o(|x|^{5-\varepsilon})$

holds, up to a set of dimension $n-2$.

Unfortunately, these methods completely break down in the parabolic setting,
as the “second blow-up” $q(x,t)$ is never a cubic caloric polynomial (as we
shall see, it is typically of the form $t|x_{n}|+\frac{1}{6}|x_{n}|^{3}$), and
the frequency function $\phi(r,w)$ is _never (almost) monotone_ for
$w=u-p_{2}-q$.444This issue is already present in dimension $1$, and it can be
understood as follows: since
$u\sim\frac{1}{2}x_{n}^{2}+t|x_{n}|+\frac{1}{6}|x_{n}|^{3}$, in a parabolic
cylinder $\mathscr{C}_{r}:=B_{r}\times[-r^{2},0]$ the set $\\{u=0\\}$ has
volume $r^{n+3}$ (roughly, it behaves as the set
$\mathscr{C}_{r}\cap\\{|x_{n}|\lesssim|t|\\}$). This error becomes critical
when considering a frequency formula of order at least 3, and a completely
different strategy needs to be found. Thus, if we want to improve (1.5), we
need to develop completely new methods.

One of the main difficulties in the present paper is actually to pass from a
cubic expansion (1.5) to an expansion of order $3+\beta$ for some $\beta>0$.
More precisely, the goal is to prove that

$u(x,t)={\textstyle\frac{1}{2}}x_{n}^{2}+a|x_{n}|(t+{\textstyle\frac{1}{6}}x_{n}^{2})+p_{3}^{odd}(x,t)+O\big{(}(|x|+|t|^{1/2})^{3+\beta}\big{)},$
(1.6)

for some $a>0$ and $p_{3}^{odd}$ an odd cubic caloric polynomial. One of our
key results here (which is probably the most delicate part of the whole paper)
is that, up to a change of coordinates, (1.6) holds up to an
$(n-2)$-dimensional set.

This is a new paradigm of dimension reduction, where we do _not_ have a
frequency or similar powerful monotonicity formula, and it is based on
barriers, maximum principle, compactness arguments, parabolic regularity
estimates, non-homogeneous blow-ups, and new GMT covering arguments. This is
developed in the second part of the paper, see Sections 10-12, and in
particular Propositions 11.3 and 12.2.

Once we have proven (1.6), we need to push the expansion to higher order. For
this, we develop a barrier argument to show that the set $\\{u>0\\}$ splits
into two separate connected components inside the set
$\Omega^{\beta}:=\\{|x|^{2+\beta}<-t\\}$.

Note that, under the parabolic scaling $(x,t)\to(rx,r^{2}t)$, the set
$\Omega^{\beta}$ converges to $\mathbb{R}^{n}\times(-\infty,0)$ as $r\to 0.$
In other words, we have “broken the parabolic scaling”. This means that we may
expect the solution $u$ to behave, inside $\Omega^{\beta}$, as two (almost)
independent solutions of the Stefan problem inside each connected component of
$\\{u>0\\}$. In other words, $u$ should behave as the sum of two solutions of
the Stefan problem for which $(0,0)$ is a regular point!

The last step, which is carried out in the third part of the paper, is to
prove a $C^{\infty}$ regularity result near _regular_ points, which is robust
enough to be applied in our setting. More precisely, we want to show that if
$\bar{u}$ is a solution of the Stefan problem such that $\\{\bar{u}=0\\}$ is
sufficiently close to $\\{x_{n}\leq 0\\}$ inside $\Omega^{\beta}$, then we
have a $C^{\infty}$ expansion for $\bar{u}$ at $(0,0)$. We then apply this
result to our solution $u$ restricted to each connected component of
$\\{u>0\\}\cap\Omega^{\beta}$, and we obtain a $C^{\infty}$-type regularity
for $u$.

As a corollary of this $C^{\infty}$ expansion we are able to prove that,
outside a $(n-2)$-dimensional set, if $(x_{\circ},t_{\circ})$ and
$(x_{1},t_{1})$ are singular points then

$|t_{\circ}-t_{1}|=o(|x_{\circ}-x_{1}|^{k})\quad\textrm{for every}\ k\gg 1.$

This implies Theorem 1.3, and finally Theorem 1.2 and Corollary 1.4 follow as
immediate consequences.

######  Remark 1.5.

In view of our proof strategy, the dimensional bound $1/2$ in Theorem 1.2 for
the set of singular times in $\mathbb{R}^{3}$ is _critical_ in at least two
ways.

On the one hand, in the lower strata $\Sigma_{m}$ with $m\leq n-2$, both (1.3)
and the bound ${\rm dim}_{\rm par}(\Sigma_{m})\leq m$ are optimal. In
particular, in $\mathbb{R}^{3}$, the projection of $\Sigma_{1}$ on the $t$
axis has dimension at most $1/2$, and it looks unlikely to us that one can
improve this bound.

On the other hand, the dimension bound $1/2$ is _also_ critical when we look
at $\Sigma_{2}$. Indeed, the set of points at which (1.6) does _not_ hold has
parabolic dimension at most $n-2(=1)$. So, also for this set, its projection
on the $t$ axis is expected to have dimension $1/2$. Whether this bound can be
improved is very unclear to us, and if possible, it would require a completely
new understanding of this type of points.

###  1.3. Organization of the paper

In Section 2 we introduce some notation that will be used throughout the
paper. Then, in Section 3, we first recall some classic results on the Stefan
problem and then establish some new basic properties of solutions. In Sections
4 and 5 we introduce some new parabolic functionals and monotonicity formulas
of Weiss and Almgren-type. This allows us, in Section 6, to consider the
second blow-up at singular points (see (1.4)), and then prove in Section 7
that this blow-up is 3-homogeneous at “most” singular points. In Section 8 we
construct some appropriate barriers, adapted to each type of singular point,
which allow us to prove Theorem 1.1. We then study cubic blow-ups in Section
9. As explained before, a key and difficult part in the proof of our results
is to pass from the cubic blow-ups to an estimate of order $3+\beta$, with
$\beta>0$ (see (1.6)). This is accomplished in Sections 10, 11, and 12.
Finally, we show in Section 13 that this implies a $C^{\infty}$ estimate at
“most” singular points, and in Section 14 we finally give the proof of Theorem
1.3 and its consequences.

###  1.4. Acknowledgments

AF and JS have received funding from the European Research Council (ERC) under
the Grant Agreement No 721675. XR has been supported by the European Research
Council (ERC) under the Grant Agreement No 801867, and by the Swiss NSF. JS
has been supported by Swiss NSF Ambizione Grant PZ00P2 180042 and by the
European Research Council (ERC) under the Grant Agreement No 948029.

##  2\. Notation used throughout the paper

In this section we introduce the notation and the mathematical objects that
will be used consistently throughout the paper. In particular, the letters $G$
and $\zeta$ always refer in the sequel to the Gaussian and the spatial cut-off
introduced below, and — since they appear often in the paper — we will not
recall this every time.

###  2.1. Operators

We define the following useful operators that will be used in all the paper.

$\begin{array}[]{lll}\text{\small\bf H}:=(\Delta-\partial_{t})&&(\mbox{that
is, }\text{\small\bf H}f:=\Delta f-\partial_{t}f)\vspace{5pt}\\\
Z:=(x\cdot\nabla+2t\partial_{t})&&(\mbox{that is, }Zf:=x\cdot\nabla
f+2t\partial_{t}f).\end{array}$

###  2.2. Gaussian kernel

We introduce the “reversed heat kernel” in $\mathbb{R}^{n}\times(-\infty,0)$:

$G(x,t):=\frac{1}{(4\pi t)^{n/2}}\exp\left(\frac{|x|^{2}}{4t}\right).$

###  2.3. A bilinear form

Given $r\in(0,1]$, we denote by $\langle\,,\,\rangle_{r}$ the following
bilinear form defined for pairs of functions
$f,g:\mathbb{R}^{n}\times(-1,0)\to\mathbb{R}^{k}$:

$\langle f,g\rangle_{r}:=\int_{\mathbb{R}^{n}}(f\cdot
g)(x,-r^{2})\,G(x,-r^{2})\,dx.$

In the special case $r=1$, we sometimes use the notation
$\langle\,,\,\rangle:=\langle\,,\,\rangle_{1}$

###  2.4. Functionals $D$ and $H$

Throughout the paper we shall use the following dimensionless quantities
defined for all functions $w:\mathbb{R}^{n}\times(-1,0)\to\mathbb{R}$
sufficiently regular (in our application, $w$ will be at least $C^{1,1}$ in
space and $C^{0,1}$ in time):

$D(r,w):=2r^{2}\langle\nabla w,\nabla
w\rangle_{r}=2r^{2}\int_{\\{t=-r^{2}\\}}|\nabla w|^{2}G,\qquad H(r,w):=\langle
w,w\rangle_{r}=\int_{\\{t=-r^{2}\\}}w^{2}G.$

###  2.5. Parabolic rescaling

We introduce here a useful notation for parabolic rescaling and normalization.
Given a function $w:\mathbb{R}^{n}\times(-1,0)\to\mathbb{R}$, for $r>0$ we
define $w_{r}$ and $\tilde{w}_{r}$ as

$w_{r}(x,t):=w(rx,r^{2}t)\qquad\mbox{and}\qquad\tilde{w}_{r}(x):=\frac{w_{r}}{H(1,w_{r})^{\frac{1}{2}}}=\frac{w(r\,\cdot\,,r^{2}\,\cdot\,)}{H(r,w)^{\frac{1}{2}}}.$
(2.1)

Note that $H(1,\tilde{w}_{r})=1$. Also, the following commutation properties
of the parabolic rescaling with H and $Z$ hold:

$\text{\small\bf H}(f_{r})=r^{2}(\text{\small\bf H}f)_{r}\quad\mbox{and}\quad
Z(f_{r})=(Zf)_{r}.$

Furthermore,

$\langle f,g\rangle_{r}=\langle f_{r},g_{r}\rangle,\qquad\mbox{where
}\langle\,,\,\rangle=\langle\,,\,\rangle_{1}.$

###  2.6. Homogeneous functions

We say that a function $f(x,t)$ is (parabolically) $\lambda$-homogeneous
whenever $f(rx,r^{2}t)=r^{\lambda}f(x,t)$ for all $r>0$. Equivalently, $f$ is
$\lambda$-homogeneous whenever $Zf=\lambda f$.

###  2.7. Spatial cut-off

Throughout the paper, we will save the letter $\zeta$ to denote a spatial cut-
off as follows:

$\mbox{ $\zeta=\zeta(x)\quad$ with $\zeta\in C^{\infty}_{c}(B_{1/2})$, \
$\zeta\geq 0$, \ and \ $\zeta\equiv 1$ in $B_{1/4}$}.$ (2.2)

This cut-off function will be crucial, since the functionals $D$ and $H$ that
appear in monotonicity formulae need functions defined in the whole space.
Hence, we shall multiply our solution and its derivatives (which are defined
only in a bounded domain) by $\zeta$ in order to define functions defined in
the whole space (of course, by doing so we will introduce errors in the
equation but these will be exponentially small).

###  2.8. Parabolic cylinders

Given $r>0$, we define the parabolic cylinder $\mathscr{C}_{r}$ as

$\mathscr{C}_{r}:=B_{r}\times(-r^{2},0).$

###  2.9. Spatial projection and projection onto time axis

We use the notation $\pi_{x}:\mathbb{R}^{n}\times\mathbb{R}\to\mathbb{R}^{n}$
and $\pi_{t}:\mathbb{R}^{n}\times\mathbb{R}\to\mathbb{R}$ to refer to the
canonical projections:

$\pi_{x}(x,t):=x\qquad\mbox{and}\qquad\pi_{t}(x,t):=t.$ (2.3)

##  3\. Classical and new facts about solutions to the Stefan problem

It is well known, by the results of Caffarelli [Caf77], that solutions to the
Stefan problem are $C^{1,1}$ in space and $C^{0,1}$ in time. So, we consider
here $u\in C^{1,1}_{x}\cap C^{0,1}_{t}(B_{1}\times(-1,1))$ a solution of the
parabolic obstacle problem

$\begin{cases}\text{\small\bf H}u=\chi_{\\{u>0\\}}\\\ u\geq 0\\\
\partial_{t}u>0\quad\mbox{in }\\{u>0\\}\end{cases}\qquad\textrm{in}\quad
B_{1}\times(-1,1).$ (3.1)

Note that, since $u$ is nonnegative, (3.1) implies that $u$ is nondecreasing
in time inside $B_{1}\times(-1,1)$.

By Caffarelli [Caf77], any solution $u$ of (3.1) with
$(0,0)\in\partial\\{u=0\\}$ satisfies

$\sup_{B_{1/2}\times(-1/2,0)}|D^{2}u|+|\partial_{t}u|\leq
C\|u(\,\cdot\,,0)\|_{L^{\infty}(B_{1})}\qquad\mbox{and}\qquad\sup_{B_{r}(0)}u(\,\cdot\,,0)\geq
cr^{2}\quad\forall\,r\in(0,1),$ (3.2)

where $C,c$ are positive dimensional constants.

Notice that, if $u$ is a solution of (3.1), then

$u^{x_{\circ},t_{\circ},r}:=r^{-2}u(x_{\circ}+r\,\cdot\,,t_{\circ}+r^{2}\,\cdot\,)$

is also a solution (in the corresponding rescaled and translated domain).

The classical theory defines the following two type of points on the free
boundary $\partial\\{u>0\\}$:

  * •

Regular points: $(x_{\circ},t_{\circ})\in\partial\\{u>0\\}$ is a regular point
if

$\lim_{r\to
0}u^{x_{\circ},t_{\circ},r}(t,x)=\frac{1}{2}\big{(}\max\\{0,\boldsymbol{e}\cdot
x\\}\big{)}^{2}$

for some $\boldsymbol{e}\in\mathbb{S}^{n-1}$.

  * •

Singular points: $x_{\circ}\in\partial\\{u>0\\}$ is a singular point if

$\lim_{r\to 0}u^{x_{\circ},t_{\circ},r}(t,x)\to p_{2,x_{\circ},t_{\circ}}(x),$

where $p_{2,x_{\circ},t_{\circ}}$ is a convex quadratic $2$-homogeneous
polynomial which does not depend on time.

When $(x_{\circ},t_{\circ})=(0,0)$ is a singular point, we simplify the
notation by writing $p_{2}$ instead of $p_{2,0,0}$. Throughout the paper we
will denote

$\Sigma:=\\{\mbox{singular points in
}B_{1}\times(-1,1)\\}\subset\partial\\{u>0\\}.$

The following result follows from [Caf77], from [KN77] for the higher
regularity near regular points, and from [Bla06] for the uniqueness of blow-up
at singular points:

######  Theorem 3.1.

Let $u\in C^{1,1}_{x}\cap C^{0,1}_{t}(B_{1}\times(-1,1))$ solve (3.1). Then,
every free boundary point is either regular or singular. Moreover, the set of
regular points is relatively open inside the free boundary, and it is a
$C^{\infty}$ $(n-1)$-dimensional manifold inside $B_{1}\times(-1,1)$.

As a consequence of the parabolic Monneau-type monotonicity formula proved in
[Bla06, LM15], we have the following result for the singular set.

######  Proposition 3.2.

Let $u\in C^{1,1}_{x}\cap C^{0,1}_{t}(B_{1}\times(-1,1))$ solve (3.1), and let
$\Sigma\subset B_{1}\times(-1,1)$ denote the set of singular points. Then,
$\pi_{x}(\Sigma)\subset B_{1}$ can be locally covered by a $(n-1)$-dimensional
$C^{1}$ manifold.

###### Proof.

This follows from the more general result of $C^{1}$ regularity in space and
$C^{0,1/2}$ in time given in [LM15, Theorem 1.9]. ∎

We will also need the following estimate, whose proof is contained in [Caf77,
CF79].

######  Proposition 3.3.

Let $u\in C^{1,1}_{x}\cap C^{0,1}_{t}(B_{1}\times(-1,1))$ solve (3.1), and let
$(0,0)$ be a free boundary point. Then

$|\partial_{t}u|+(D^{2}u)_{-}\leq C|\log r|^{-\varepsilon}\quad\mbox{ in
}B_{r}\times(-r^{2},r^{2}),$

where $C,\varepsilon>0$ depend only on $n$.

After this summary of some well-known properties, we now prove some new
results that will be crucial in our analysis. We begin with the following
local semiconvexity property in time.

######  Proposition 3.4.

Let $u\in C^{1,1}_{x}\cap C^{0,1}_{t}(B_{1}\times(-1,1))$ solve (3.1). Then,

$\partial_{tt}u\geq-C\quad\mbox{ in
}B_{1/2}\times(-{\textstyle\frac{1}{2}},{\textstyle\frac{1}{2}}),$

where $C$ depends only on $n$ and $\|u\|_{L^{\infty}}$.

To prove the proposition above, we need the parabolic version of the standard
$L^{1}$ to $L^{1}_{weak}$ estimate of Calderón and Zygmund for elliptic PDE
[Jon64]:

######  Theorem 3.5.

Let $w:\mathscr{C}_{1}\to\mathbb{R}$ be any solution of $\text{\small\bf
H}w=f$ in $\mathscr{C}_{1}$, with $f\in L^{1}$. Then

$\sup_{\theta>0}\theta\big{|}\\{|D^{2}w|+|\partial_{t}w|>\theta\\}\cap\mathscr{C}_{1/2}\big{|}\leq
C(\|f\|_{L^{1}(\mathscr{C}_{1})}+\|w\|_{L^{1}(\mathscr{C}_{1})}),$

for some constant $C$ depending only on $n$.

We will also need the following result (see e.g. [Wa92, Theorem 4.16]):

######  Lemma 3.6.

Let $w:\mathscr{C}_{1}\to R$ satisfy $\text{\small\bf H}w\leq 0$. Then

$\sup_{\mathscr{C}_{1/2}}w\leq
C\bigg{(}\int_{\mathscr{C}_{1}}(w_{+})^{\varepsilon}\bigg{)}^{\frac{1}{\varepsilon}},$

for some $\varepsilon>0$ and $C$ depending only on $n$.

We can now prove the desired semiconvexity in time:

###### Proof of Proposition 3.4.

Fix $r_{1},r_{2}$ such that $1/2<r_{1}<r_{2}<1$, and for $r\in(0,1)$ define
the domains $\Omega_{r}:=B_{r}\times(-r,r)$. For $h\in\mathbb{R}$ and given a
function $w$, we define the discrete time derivative as
$\delta_{t,h}w(x,t):=h^{-1}\big{(}w(x,t+h)-w(x)\big{)}$.

Now, consider $v^{h}:=\delta_{t,h}\partial_{t}u$. By Proposition 3.3 the
function $v^{h}$ is continuous. Also, by (3.1), $v^{h}\geq 0$ on $\\{u=0\\}$.
Notice in addition that $\text{\small\bf H}v^{h}\leq 0$ in $\\{u>0\\}$.
Indeed, inside the open set $\\{u>0\\}$ we have $\text{\small\bf
H}(\partial_{t}\delta_{t,h}u)=0$ and hence we have (in the sense of
distributions)

$\text{\small\bf
H}(\partial_{t}\delta_{t,h}u)(\,\cdot\,,\,\cdot\,)=\frac{\partial_{t}(\chi_{\\{u>0\\}})(\,\cdot\,,\,\cdot\,+h)}{h}\geq
0,$

where the right hand side is a nonnegative measure since
$\partial_{t}(\chi_{\\{u>0\\}})$ is a nonnegative measure by the monotonicity
property of $u$.

This proves that $(v^{h})_{-}$ is subcaloric, and using Lemma 3.6 we get

${\rm ess\,sup}_{\Omega_{1/2}}(v^{h})_{-}\ \leq\
\sup_{\theta>0}\theta\big{|}\\{(v^{h})_{-}>\theta\\}\cap\Omega_{r_{1}}\big{|}=C\sup_{\theta>0}\theta|\\{|\partial_{t}\hat{w}|>\theta\\}\cap\Omega_{r_{1}}\big{|},\quad\mbox{
where }\hat{w}:=\delta_{t,h}u.$

Note that the function $\hat{w}$ satisfies $\text{\small\bf
H}\hat{w}=\delta_{t,h}(\chi_{\\{u>0\\}})\geq 0$ and $\hat{w}\geq 0$, since $u$
and $\\{u>0\\}$ are nondecreasing in time. Moreover, by the parabolic
Calderón-Zygmund type estimate (Theorem 3.5), we have

$\sup_{\theta>0}\theta|\\{|\partial_{t}\hat{w}|>\theta\\}\cap\Omega_{r_{1}}\big{|}\leq
C(\|\text{\small\bf
H}\hat{w}\|_{L^{1}(\Omega_{r_{2}})}+\|\hat{w}\|_{L^{1}(\Omega_{r_{2}})}).$

Now, fix $\xi=\xi(x,t)$ a smooth nonnegative cut-off, compactly supported in
$\Omega_{1}$ and such that $\xi\equiv 1$ in $\Omega_{r_{2}}$. For
$h\ll(1-r_{2})$, we have

$\|\text{\small\bf
H}\hat{w}\|_{L^{1}(\Omega_{r_{2}})}\leq\int_{\Omega_{1}}\text{\small\bf
H}\hat{w}\,\xi=\int_{\Omega_{1}}(\text{\small\bf
H}\delta_{t,h}u)\,\xi=\int_{\Omega_{1}}u\,\delta_{t,-h}(\Delta\xi+\partial_{t}\xi)\leq
C\|u\|_{L^{1}(\Omega_{1})},$

and similarly

$\|\hat{w}\|_{L^{1}(\Omega_{r_{2}})}\leq\int_{\Omega_{1}}\hat{w}\,\xi=\int_{\Omega_{1}}\delta_{t,h}u\,\xi=\int_{\Omega_{1}}u\,\delta_{t,-h}\xi\leq
C\|u\|_{L^{1}(\Omega_{1})}.$

Hence, combining all these estimates together, we have shown that

${\rm ess\,sup}_{\Omega_{1/2}}(v^{h})_{-}\leq C\|u\|_{L^{1}(\Omega_{1})}.$

Letting $h\downarrow 0$, we conclude the proof. ∎

We will also need the following relation between the derivatives of $u$.

######  Proposition 3.7.

Let $u\in C^{1,1}_{x}\cap C^{0,1}_{t}(B_{1}\times(-1,1))$ solve (3.1). Then
there exists a constant $C>0$ such that

$(D^{2}u)_{-}\leq C\,\partial_{t}u\quad\mbox{ in }B_{1/2}\times(-1/2,1/2).$

###### Proof.

Given $(x_{\circ},t_{\circ})\in B_{1/2}\times(-1/2,1/2)$, and fixed
$e\in\mathbb{S}^{n-1}$, we consider

$w_{x_{\circ},t_{\circ}}:=\partial_{t}u+c_{1}\left(\textstyle\frac{|x-x_{\circ}|^{2}}{4n}-\frac{(t-t_{\circ})}{2}-u\right)-c_{2}(\partial_{ee}u)_{-},$

where $c_{1}$ and $c_{2}$ are small positive constants to be chosen. Note that
$w_{x_{\circ},t_{\circ}}$ is nonnegative on $\partial\\{u>0\\}$.

Let $\Omega:=B_{1/4}(x_{\circ})\times(t_{\circ}-1/4,t_{\circ})$ and denote by
$\partial_{par}\Omega$ its parabolic boundary. Since $\partial_{t}u>0$ in
$\\{u>0\\}$, it follows by the first bound in (3.2) that

$w_{x_{\circ},t_{\circ}}\geq
c_{1}\big{(}{\textstyle\frac{1}{64n}}-C_{0}\varrho^{2}\big{)}-c_{2}C_{0}\quad\mbox{in
}\partial_{par}\Omega\cap\big{(}\\{u=0\\}+(B_{\rho}\times(0,\rho^{2}))\big{)}$

where $C_{0}$ depends only on $n$ and $\|u\|_{L^{\infty}}$, and
$\\{u=0\\}+(B_{\rho}\times(0,\rho^{2}))$ denotes the Minkowski sum of sets.
Hence, choosing $\varrho>0$ sufficiently small and $c_{1}\geq CC_{0}c_{2}$, we
will have $w_{x_{\circ},t_{\circ}}>0$ on this piece of boundary.

On the other hand, since by assumption $\partial_{t}u>0$ in $\\{u>0\\}$, we
have

$w_{x_{\circ},t_{\circ}}\geq\partial_{t}u-C(c_{1}+c_{2})\geq
c_{u}-C(c_{1}+c_{2})\quad\mbox{in
}\Omega\setminus\big{(}\\{u=0\\}+(B_{\rho}\times(0,\rho^{2}))\big{)},$

where $c_{u}>0$ is a constant (depending on $u$).

Hence, choosing $c_{1}$ and $c_{2}$ sufficiently small so that
$c_{u}-C(c_{1}+c_{2})>0$ and $c_{1}\geq CC_{0}c_{2}$, it follows by the
maximum principle that

$w_{x_{\circ},t_{\circ}}\geq 0\quad\mbox{in }\Omega.$

In particular, evaluating at $(x_{\circ},t_{\circ})$ we obtain

$\partial_{t}u(x_{\circ},t_{\circ})-c_{1}u(x_{\circ},t_{\circ})-c_{2}(\partial_{ee}u)_{-}(x_{\circ},t_{\circ})>0.$

Since $u(x_{\circ},t_{\circ})\geq 0$, and both $(x_{\circ},t_{\circ})\in
B_{1/2}\times(-{\textstyle\frac{1}{2}},{\textstyle\frac{1}{2}})$ and
$e\in\mathbb{S}^{n-1}$ can be chosen arbitrarily, the lemma follows. ∎

Finally, we shall also need the following result.

######  Lemma 3.8.

Let $u\in C^{1,1}_{x}\cap C^{0,1}_{t}(B_{1}\times(-1,1))$ solve (3.1). There
exists a dimensional constant $\varepsilon_{\circ}>0$ such that the following
holds: Assume that

$\big{|}u-{\textstyle\frac{1}{2}}(x\cdot
e)^{2}\big{|}\leq\varepsilon_{\circ}\qquad\text{and}\qquad\partial_{ee}u\geq-\varepsilon_{\circ}$

for some $e\in\mathbb{S}^{n-1}$. Then

$\partial_{ee}u\geq 0\quad\mbox{in
}B_{1/2}\times(-{\textstyle\frac{1}{2}},{\textstyle\frac{1}{2}}).$

###### Proof.

We may assume $e=\boldsymbol{e}_{n}$. Write
$x=(x^{\prime},x_{n})\in\mathbb{R}^{n-1}\times\mathbb{R}$, and set
$v^{h}(x^{\prime},x_{n},t):=h^{-2}\big{(}u(x^{\prime},x_{n}+h,t)+u(x^{\prime},x_{n}-h,t)-2u(x^{\prime},x_{n},t)\big{)}$.
We note that

$\text{\small\bf H}v^{h}\leq 0\quad\mbox{in }\\{u>0\\}\qquad\mbox{ and }\qquad
v^{h}\geq 0\quad\mbox{in }\\{u=0\\}.$

Now, fix $(x_{\circ},t_{\circ})\in B_{1/2}\times(-1/2,1/2)$ and consider the
function

$w^{h}_{x_{\circ},t_{\circ}}:=v^{h}+256\,n\,\varepsilon_{\circ}\left(\textstyle\frac{|x-x_{\circ}|^{2}}{4n}-\frac{(t-t_{\circ})}{2}-u\right)\quad\text{inside
}U_{x_{\circ},t_{\circ}}:=\\{|x-x_{\circ}|\leq 1/4,-1/4\leq t-t_{\circ}\leq
0\\}.$

Note that $w_{x_{\circ},t_{\circ}}$ is supercaloric in
$U_{x_{\circ},t_{\circ}}\cap\\{u>0\\}$. Therefore, if we show that —for
$\varepsilon_{\circ}$ small— $w_{x_{\circ},t_{\circ}}$ is nonnegative on
$\partial_{par}U_{x_{\circ},t_{\circ}}\cup(U_{x_{\circ},t_{\circ}}\cap\\{u=0\\})$,
then it will follow by the maximum principle that
$w^{h}_{x_{\circ},t_{\circ}}\geq 0$ inside $U_{x_{\circ},t_{\circ}}$, and in
particular

$v^{h}(x_{\circ},t_{\circ})\geq
w^{h}_{x_{\circ},t_{\circ}}(x_{\circ},t_{\circ})\geq 0.$

Since $(x_{\circ},t_{\circ})\in B_{1/2}\times(-1/2,1/2)$ is arbitrary, and
$v^{h}\to\partial_{nn}u$ as $h\downarrow 0$, the lemma will follow. So, we are
left with proving that $w_{x_{\circ},t_{\circ}}\geq 0$ on
$\partial_{par}U_{x_{\circ},t_{\circ}}\cup(U_{x_{\circ},t_{\circ}}\cap\\{u=0\\})$

Since $v^{h}\geq 0$ on $\\{u=0\\}$, it follows that
$w^{h}_{x_{\circ},t_{\circ}}\geq 0$ on
$U_{x_{\circ},t_{\circ}}\cap\\{u=0\\})$. Hence, it only remains to show that
$w^{h}_{x_{\circ},t_{\circ}}\geq 0$ on $\partial_{par}U_{x_{\circ},t_{\circ}}$
—provided $\varepsilon_{\circ}$ is chosen sufficiently small. Note that, on
$\partial_{par}U_{x_{\circ},t_{\circ}}$, we have
$\frac{|x-x_{\circ}|^{2}}{4n}-\frac{(t-t_{\circ})}{2}>\frac{1}{64n}$. Let us
divide the parabolic boundary in two pieces: the piece (I) where
$\\{u\leq\frac{1}{128n}\\}$, and (II) where $\\{u\geq\frac{1}{128n}\\}$. On
the piece (I), by the assumption $\partial_{nn}u\geq-\varepsilon_{\circ}$ we
obtain

$w^{h}_{x_{\circ},t_{\circ}}=v^{h}+256\,n\,\varepsilon_{\circ}\left(\textstyle\frac{|x^{\prime}-x_{\circ}^{\prime}|^{2}}{4(n-1)}-\frac{(t-t_{\circ})}{2}-u\right)\geq-2\varepsilon_{\circ}+256\,n\,\varepsilon_{\circ}\left(\frac{1}{64n}-\frac{1}{128n}\right)\geq
0.$

To estimate $w^{h}_{x_{\circ},t_{\circ}}$ on the piece (II) we note that,
since
$\big{|}u-{\textstyle\frac{1}{2}}x_{n}^{2}\big{|}\leq\varepsilon_{\circ}$, it
follows that

$\\{u>\frac{1}{128n}\\}\subset\left\\{|x_{n}|>\frac{1}{10n^{1/2}}\right\\},\qquad\\{u>0\\}\supset\left\\{|x_{n}|>\frac{1}{20n^{1/2}}\right\\},\qquad\|u\|_{L^{\infty}(U_{x_{\circ},t_{\circ}})}\leq
1,$

provided $\varepsilon_{\circ}$ is taken sufficiently small. Thus, using
standard interior regularity for the heat equation and interpolation555Note
that $u-\frac{1}{2}x_{n}^{2}$ is caloric inside
$\left\\{|x_{n}|>\frac{1}{20n^{1/2}}\right\\}$ and it has small $L^{\infty}$
norm. we obtain that $|\partial_{nn}(u-\frac{1}{2}x_{n}^{2})|\leq
C_{n}\varepsilon_{\circ}$ inside
$\left\\{|x_{n}|>\frac{1}{10n^{1/2}}\right\\}$. In particular
$v^{h}\geq\frac{1}{2}$ in $\left\\{|x_{n}|>\frac{1}{10n^{1/2}}\right\\}$ for
$\varepsilon_{\circ}$ sufficiently small. Therefore, recalling that $|u|\leq
1$ inside $U_{x_{\circ},t_{\circ}}$, it follows that

$w^{h}_{x_{\circ},t_{\circ}}=v^{h}+256\,n\,\varepsilon_{\circ}\left(\textstyle\frac{|x^{\prime}-x_{\circ}^{\prime}|^{2}}{4(n-1)}-\frac{(t-t_{\circ})}{2}-u\right)\geq\frac{1}{2}-C_{n}\varepsilon_{\circ}\geq
0\qquad\text{on (II)},$

provided we choose $\varepsilon_{\circ}$ small. ∎

##  4\. Parabolic functionals and useful formulae

In this section we establish formulae which are valid for arbitrary functions
$\mathbb{R}^{n}\times(-1,1)\rightarrow\mathbb{R}$ having $C^{1,1}_{x}\cap
C^{0,1}_{t}$ regularity. Later on we will apply these formulae for example to
$w(x,t)=(u-p_{2})(x,t)\zeta(x)$. Recall that $Z$, $G$, $H$, $D$, and $\zeta$,
were defined in Section 2.

######  Lemma 4.1.

Let $f\in C^{1,1}_{x}\cap C^{0,1}_{t}$. Then

$\frac{d}{dr}\bigg{|}_{r=1}\langle f,G\rangle_{r}=\langle Zf,G\rangle.$

###### Proof.

Recall that

$\langle f,G\rangle_{r}=\int_{\\{t=-r^{2}\\}}fG,\qquad\langle
Zf,G\rangle=\int_{\\{t=-1\\}}Zf\,G,\qquad Z=x\cdot\nabla+2t\cdot\partial_{t}.$

Also, thanks to the scaling property $G(rx,r^{2}t)r^{n}=G(x,t)$, we have

$\int_{\\{t=-r^{2}\\}}f(x,t)G(x,t)dx=\int_{\\{t=-1\\}}f(rx,rt)G(rx,r^{2}t)r^{n}\,dx=\int_{\\{t=-1\\}}f(rx,r^{2}t)G(x,t)\,dx.$

Then, the result follows by differentiating with respect to $r$ the expression
above, and evaluating at $r=1$. ∎

######  Lemma 4.2.

Let $f,g\in C^{1,1}_{x}\cap C^{0,1}_{t}$. Then

$2\langle\nabla f,\nabla g\rangle=\langle Zf,g\rangle-2\langle\text{\small\bf
H}f,g\rangle=\langle f,Zg\rangle-2\langle f,\text{\small\bf H}g\rangle.$ (4.1)

###### Proof.

Noting that $\nabla G=-\frac{x}{2}G$ and $Z=x\cdot\nabla-2\cdot\partial_{t}$
on $\\{t=-1\\}$, using integration by parts we obtain

$\begin{split}2\langle\nabla f,\nabla g\rangle&=2\int_{\\{t=-1\\}}\nabla
f\cdot\nabla g\,G=-2\int_{\\{t=-1\\}}f{\rm div}(\nabla g\,G)\\\
&=-2\int_{\\{t=-1\\}}\big{(}f\Delta g\,G+f\nabla g\cdot\nabla
G\big{)}=-\int_{\\{t=-1\\}}\big{(}2f\Delta g\,G-fx\cdot\nabla g\,G\big{)}\\\
&=-\int_{\\{t=-1\\}}f\big{(}2\Delta
g-2\partial_{t}g-(-2\partial_{t}g+x\cdot\nabla
g)\big{)}G=-\int_{\\{t=-1\\}}f\big{(}2\text{\small\bf H}g-Zg\big{)}G\\\
&=-2\langle f,\text{\small\bf H}g\rangle+\langle f,Zg\rangle.\end{split}$

By symmetry, also the other identity $2\langle\nabla f,\nabla
g\rangle=-2\langle\text{\small\bf H}f,g\rangle+\langle Zf,g\rangle$ holds. ∎

######  Lemma 4.3.

Let $w\in C^{1,1}_{x}\cap C^{0,1}_{t}$. Then

$H^{\prime}(1,w)=2\langle w,Zw\rangle=2D(1,w)+4\langle w,\text{\small\bf
H}w\rangle.$ (4.2)

###### Proof.

The first equality is an immediate application of Lemma 4.1 with $f=w^{2}$. To
prove the second equality, we use (4.1). ∎

######  Remark 4.4.

As a consequence of Lemma 4.3 we obtain the following identity:

$D(r,w)=\langle w,Zw\rangle-2\langle w,\text{\small\bf H}w\rangle.$ (4.3)

######  Lemma 4.5.

Let $w\in C^{1,1}_{x}\cap C^{0,1}_{t}$. Then

$D^{\prime}(1,w)=2\langle Zw,Zw\rangle-4\langle Zw,\text{\small\bf
H}w\rangle.$

###### Proof.

Using Lemma 4.1 with $f=|\nabla w|^{2}$ we obtain

$\begin{split}\frac{d}{dr}D(r,w)\bigg{|}_{r=1}&=\frac{d}{dr}\bigg{|}_{r=1}2\bigg{(}r^{2}\int_{\\{t=-r^{2}\\}}|\nabla
w|^{2}G\bigg{)}=\sum_{i}4\langle
Z\partial_{i}w,\partial_{i}w\rangle+2D(1,w)\\\ &=\sum_{i}4\langle
Z\partial_{i}w+\partial_{i}w,\partial_{i}w\rangle=\sum_{i}4\langle\partial_{i}Zw,\partial_{i}w\rangle=4\langle\nabla
Zw,\nabla w\rangle\\\ &=2\langle Zw,Zw\rangle-4\langle Zw,\text{\small\bf
H}w\rangle,\end{split}$

where the last equality follows from Lemma 4.2. ∎

We now introduce the frequency functions

$\phi(r,w):=\frac{D(r,w)}{H(r,w)}\qquad\textrm{and}\qquad\phi^{\gamma}(r,w):=\frac{D(r,w)+\gamma
r^{2\gamma}}{H(r,w)+r^{2\gamma}}.$

######  Lemma 4.6.

Let $w\in C^{1,1}_{x}\cap C^{0,1}_{t}$. Then

$\frac{d}{dr}\phi^{\gamma}(r,w)\geq\frac{2}{r}\,\frac{\big{(}\langle
Zw,Zw\rangle_{r}\langle w,w\rangle_{r}-\langle
w,Zw\rangle_{r}^{2}\big{)}+\big{(}2r^{2}\langle w,\text{\small\bf
H}w\rangle_{r}\big{)}^{2}+E^{\gamma}(r,w)}{\big{(}H(r,w)+r^{2\gamma}\big{)}^{2}},$

where

$E^{\gamma}(r,w):=2r^{2}\langle w,\text{\small\bf
H}w\rangle_{r}\big{(}D(r,w)+\gamma r^{2\gamma}\big{)}-2r^{2}\langle
Zw,\text{\small\bf H}w\rangle_{r}\big{(}H(r,w)+r^{2\gamma}\big{)}.$ (4.4)

###### Proof.

By scaling it is enough to compute, for $a>0$,

$\frac{d}{dr}\phi^{\gamma,a}(r,w)\bigg{|}_{r=1}\qquad\mbox{where}\quad\phi^{\gamma,a}(r,w):=\frac{D(r,w)+\gamma(ar)^{2\gamma}}{H(r,w)+(ar)^{2\gamma}}.$

By Lemmas 4.1, 4.3, and 4.5, and by (4.3), we have

$A_{1}:=\frac{d}{dr}\bigg{|}_{r=1}\big{(}D(r,w)+\gamma(ar)^{2\gamma}\big{)}=2\langle
Zw,Zw\rangle-4\langle Zw,\text{\small\bf H}w\rangle+2\gamma^{2}a^{2\gamma},$
$D(1,w)+\gamma a^{2\gamma}=\langle w,Zw\rangle-2\langle w,\text{\small\bf
H}w\rangle+\gamma a^{2\gamma},$
$A_{2}:=\frac{d}{dr}\bigg{|}_{r=1}\big{(}H(r,w)+(ar)^{2\gamma}\big{)}=2\langle
w,Zw\rangle+2\gamma a^{2\gamma},$ $H(1,w)+(ar)^{2\gamma}=\langle
w,w\rangle+a^{2\gamma}.$

Therefore,

$\begin{split}\frac{d}{dr}\bigg{|}_{r=1}\phi^{\gamma,a}(r,w)&=\frac{A_{1}\big{(}H(1,w)+a^{2\gamma}\big{)}-A_{2}\big{(}D(1,w)+\gamma
a^{2\gamma}\big{)}}{\big{(}H(1,w)+a^{2\gamma}\big{)}^{2}}\\\
&=2\,\frac{X^{2}+2\langle w,\text{\small\bf H}w\rangle\big{(}\langle
Zw,w\rangle+\gamma a^{2\gamma}\big{)}-2\langle Zw,\text{\small\bf
H}w\rangle\big{(}\langle
w,w\rangle+a^{2\gamma}\big{)}}{\big{(}H(1,w)+a^{2\gamma}\big{)}^{2}},\end{split}$

where

$\begin{split}X^{2}:=&\,\big{(}\langle
Zw,Zw\rangle+\gamma^{2}a^{2\gamma}\big{)}\big{(}\langle
w,w\rangle+a^{2\gamma}\big{)}-\big{(}\langle w,Zw\rangle+\gamma
a^{2\gamma}\big{)}^{2}\\\ \geq&\,\langle Zw,Zw\rangle\langle
w,w\rangle-\langle w,Zw\rangle^{2}\geq 0.\end{split}$

Using again (4.3), this gives

$\begin{split}\frac{d}{dr}\bigg{|}_{r=1}\phi^{\gamma,a}(r,w)&\geq
2\,\frac{\big{(}\langle Zw,Zw\rangle\langle w,w\rangle-\langle
w,Zw\rangle^{2}\big{)}+4\big{(}\langle w,\text{\small\bf
H}w\rangle\big{)}^{2}}{\big{(}H(1,w)+a^{2\gamma}\big{)}^{2}}\\\
&\qquad+2\,\frac{2\langle w,\text{\small\bf H}w\rangle\big{(}D(1,w)+\gamma
a^{2\gamma}\big{)}-2\langle Zw,\text{\small\bf
H}w\rangle\big{(}H(1,w)+a^{2\gamma}\big{)}}{\big{(}H(1,w)+a^{2\gamma}\big{)}^{2}}.\end{split}$

Applying the previous equality with $w$ replaced by
$w_{r}=w(r\,\cdot\,,r^{2}\,\cdot\,)$ and $a$ replaced by $r$, the lemma
follows. ∎

##  5\. Monotonicity formulae for the Stefan problem

Let $\mathcal{P}$ denote the set of nonnegative $2$-homogeneous polynomials in
the $x$ variable —i.e. $p=p(x)$— which satisfy $\Delta p=1$. In particular,
such polynomials always satisfy the same equation as the parabolic obstacle
problem.

######  Remark 5.1.

Let $p\in\mathcal{P}$. Then, since $\text{\small\bf H}u=\chi_{\\{u>0\\}}$ and
$\text{\small\bf H}p\equiv\Delta p\equiv 1$, we have

$(u-p)\text{\small\bf H}(u-p)=p\chi_{\\{u=0\\}}\geq 0,\qquad
Z(u-p)\,\text{\small\bf H}(u-p)=Zp\,\chi_{\\{u=0\\}}=2p\,\chi_{\\{u=0\\}}\geq
0.$ (5.1)

Note that the solution $u(x,t)$ that we are considering is defined only for
$x\in B_{1}$. However the formulae given in Section 4 are for function $w$
defined for all $x$ all of $\mathbb{R}^{n}$. Morally, we would like to apply
the formulae in Section 4 with $w$ replaced by $u-p$. Since this is not
possible, we will instead use make use of the cut-off function $\zeta$ in
(2.2) and set $w:=(u-p)\zeta$, with $w=0$ outside $B_{1/2}$. However, such $w$
is not a exact solution and errors coming from this truncation need to be
taken into account.

Note that if $u$ is a solution of the parabolic obstacle problem in
$\mathscr{C}_{1}=B_{1}\times(-1,0)$ and $p\in\mathcal{P}$, then by Remark 5.1
we have666From now on, given a function $v:\mathscr{C}_{1}\to\mathbb{R}$, the
product $\zeta\,v$ will always be seen as a function defined in all
$\mathbb{R}^{n}\times(-1,0)$, where we extended the product by $0$ outside of
$B_{1}$.

$\langle\zeta\,Z(u-p),\zeta\,\text{\small\bf
H}(u-p)\rangle_{r}=2\langle\zeta\,(u-p),\zeta\,\text{\small\bf
H}(u-p)\rangle_{r}\geq 0.$ (5.2)

The following lemma will be useful to control truncation errors, which are
exponentially small as $r\downarrow 0$.

######  Lemma 5.2.

Let $v\in C^{1,1}_{x}\cap C^{0,1}_{t}(\mathscr{C}_{1})$. Then, for all
$r\in(0,1/2)$ we have

$\big{|}\langle\zeta\,v,\text{\small\bf H}(\zeta
v)\rangle_{r}-\langle\zeta\,v,\zeta\,\text{\small\bf
H}v\rangle_{r}\big{|}+\big{|}\langle Z(\zeta v),\text{\small\bf H}(\zeta
v)\rangle_{r}-\langle\zeta\,Zw,\zeta\,\text{\small\bf
H}v\rangle_{r}\big{|}\leq
CM_{v}^{2}\,{\textstyle\exp\left(-\frac{1}{(8r)^{2}}\right)},$

where

$M_{v}:=\sup_{B_{1/2}\times(-1/2,0)}|v|+|\nabla v|+|\partial_{t}v|+|\Delta
v|,$

and $C$ depends only on $\zeta$.

###### Proof.

We note that, since $\zeta=\zeta(x)$,

$\text{\small\bf H}(\zeta v)=\zeta\,\text{\small\bf
H}v+\Delta\zeta\,v+2\nabla\zeta\cdot\nabla v\qquad\text{and}\qquad Z(\zeta
v)=\zeta\,Zv+(x\cdot\nabla\zeta)v.$

Thus

$\begin{split}\langle\zeta\,v,\text{\small\bf H}(\zeta
v)\rangle_{r}=\int_{\\{t=-r^{2}\\}}\zeta\,v\,\text{\small\bf H}(\zeta
v)\,G=\int_{\\{t=-r^{2}\\}}\zeta\,v\,\zeta\,\text{\small\bf
H}v\,G+\int_{\\{t=-r^{2}\\}}\zeta\,v\,(\Delta\zeta\,v+2\nabla
v\cdot\nabla\zeta)\,G,\end{split}$

and similarly

$\begin{split}\langle Z(\zeta v),\text{\small\bf H}(\zeta
v)\rangle_{r}&=\int_{\\{t=-r^{2}\\}}Z(\zeta v)\,\text{\small\bf H}(\zeta
v)\,G\\\ &=\int_{\\{t=-r^{2}\\}}\zeta\,Zv\,\zeta\,\text{\small\bf
H}v\,G+\int_{\\{t=-r^{2}\\}}\big{(}\zeta\,Zv(\Delta\zeta\,v+2\nabla
v\cdot\nabla\zeta)+(x\cdot\nabla\zeta)\,v\,\text{\small\bf
H}v\big{)}\,G.\end{split}$

Since $\Delta\zeta$, $\nabla\zeta$, and $Z\zeta$ vanish in $B_{1/4}$, and

$G(x,-r^{2})=\frac{1}{(4\pi)^{n/2}r^{n}}\exp\left({\textstyle-\frac{|x|^{2}}{4r^{2}}}\right)\leq
C_{n}\exp\left({\textstyle-\frac{1}{(8r)^{2}}}\right)\quad\mbox{ for
}|x|\geq\frac{1}{4},$

the lemma follows. ∎

We next prove the following Weiss’ type monotonicity formula (as well as a
useful consequence of it).

######  Lemma 5.3 (Weiss’-type formula).

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), and
$(0,0)$ a singular point. Given $p\in\mathcal{P}$, set $w:=(u-p)\zeta$. Let

$W_{\lambda}(r,w):=r^{-2\lambda}\big{(}D(r,w)-\lambda H(r,w)\big{)}.$

Then:

(a) For all $\lambda\geq 2$

$\hskip 71.13188pt\frac{d}{dr}W_{\lambda}(r,w)\geq 2r^{-2\lambda-1}\langle
Zw-\lambda w,Zw-\lambda
w\rangle_{r}^{2}-Ce^{-\frac{1}{r}},\qquad\forall\,r\in(0,{\textstyle\frac{1}{2}}),$

where $C>0$ depends only on $n$, $\|u(\,\cdot\,,0)\|_{L^{\infty}(B_{1})}$, and
$\lambda$.

(b) Moreover, $W_{2}(0^{+},w)=0$ and

$\hskip 71.13188ptD(r,w)-2H(r,w)\geq-
Ce^{-\frac{1}{r}},\qquad\forall\,r\in(0,{\textstyle\frac{1}{2}}),$ (5.3)

where $C>0$ depend only on $n$ and $\|u(\,\cdot\,,0)\|_{L^{\infty}(B_{1})}$.

###### Proof.

As we shall see, (b) is a rather direct consequence of (a). We begin with the
first part.

$\bullet$ Proof of (a). To compute $W^{\prime}_{\lambda}(r,w)$ we use Lemmas
4.3 and 4.5, as well as (4.3) (rescaled), to obtain

$\begin{split}W^{\prime}_{\lambda}(r,w)&=\frac{1}{r^{2\lambda}}\left(D^{\prime}(r,w)-\lambda
H^{\prime}(r,w)\big{)}-\frac{2\lambda}{r}D(r,w)+\frac{2\lambda^{2}}{r}H(r,w)\right)\\\
&=\frac{2}{r^{2\lambda+1}}\big{(}\langle Zw,Zw\rangle_{r}-2r^{2}\langle
Zw,\text{\small\bf H}w\rangle_{r}-\lambda\langle w,Zw\rangle_{r}-\lambda
D(r,w)+\lambda^{2}H(r,w)\big{)}\\\ &=\frac{2}{r^{2\lambda+1}}\big{(}\langle
Zw,Zw\rangle-2r^{2}\langle Zw,\text{\small\bf H}w\rangle_{r}-\lambda\langle
w,Zw\rangle_{r}-\lambda(\langle w,Zw\rangle_{r}-2r^{2}\langle
w,\text{\small\bf H}w\rangle_{r})+\lambda^{2}\langle w,w\rangle_{r}\big{)}\\\
&=\frac{2}{r^{2\lambda+1}}\big{(}\langle Zw-\lambda w,Zw-\lambda
w\rangle_{r}+2r^{2}\big{\\{}\lambda\langle w,\text{\small\bf
H}w\rangle-r^{2}\langle Zw,\text{\small\bf
H}w\rangle\big{\\}}\big{)}.\end{split}$

Set $\tilde{w}:=u-p_{2}$, so that $w=\zeta\tilde{w}.$ Using (5.2) (see also
Remark 5.1) with $p=p_{2}$, and by Lemma 5.2 applied to $\tilde{w}$, we have

$\lambda\langle w,\text{\small\bf H}w\rangle_{r}-\langle Zw,\text{\small\bf
H}w\rangle_{r}=\lambda\langle\zeta\tilde{w},\text{\small\bf
H}(\zeta\tilde{w})\rangle_{r}-\langle Z(\zeta\tilde{w}),\text{\small\bf
H}(\zeta\tilde{w})\rangle_{r}\\\
\geq\lambda\langle\zeta\tilde{w},\zeta\,\text{\small\bf
H}\tilde{w}\rangle_{r}-\langle\zeta\,Z\tilde{w},\text{\small\bf
H}\tilde{w}\rangle_{r}-C_{\zeta}\big{(}M_{\tilde{w}}\big{)}^{2}e^{-\frac{1}{(8r)^{2}}}=(\lambda-2)\langle\zeta\tilde{w},\zeta\,\text{\small\bf
H}\tilde{w}\rangle_{r}-Ce^{-\frac{2}{r}},$ (5.4)

where $M_{\tilde{w}}$ is defined as in (5.2). Recalling (3.2) and that the
cut-off $\zeta$ is fixed, it follows that $M_{\tilde{w}}$ (and therefore also
the constant $C$ above) depends only on $n$ and
$\|u(\,\cdot\,,0)\|_{L^{\infty}(B_{1})}$. Since $\lambda\geq 2$,
$-r^{-2\lambda+1}e^{-\frac{2}{r}}\geq-Ce^{-\frac{1}{r}}$, and
$\langle\zeta\tilde{w},\zeta\,\text{\small\bf H}\tilde{w}\rangle_{r}\geq 0$
(see Remark 5.1), the conclusion follows.

$\bullet$ Proof of (b). Since $(0,0)$ is a singular point we have that
$w_{r}(x)=(u-p)(rx,r^{2}t)=(p_{2}-p)(rx)+o(r^{2})$, and therefore

$W_{2}(0^{+},w)=\lim_{r\downarrow 0}W(1,r^{-2}w_{r})=W_{2}(1,p_{2}-p)=0.$

As a consequence, (5.3) follows by integrating $(a)$ with $\lambda=2$ between
$0$ and $r$. ∎

We can now prove our approximate frequency formula.

######  Proposition 5.4 (Frequency formula).

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), and
$(0,0)$ a singular point. Given $p\in\mathcal{P}$, set $w:=(u-p)\zeta$. Then,
for all $\gamma\in(2,\infty)$ we have

$\frac{d}{dr}\phi^{\gamma}(r,w)\geq\frac{2}{r}\,\frac{\big{(}\langle
Zw,Zw\rangle_{r}\langle w,w\rangle_{r}-\langle
w,Zw\rangle_{r}^{2}\big{)}+\big{(}2r^{2}\langle w,\text{\small\bf
H}w\rangle_{r}\big{)}^{2}}{\bigl{(}H(r,w)+r^{2\gamma}\bigr{)}^{2}}-Ce^{-\frac{1}{r}},\quad\forall\,r\in(0,1),$

and

$r^{2}\langle w,\text{\small\bf H}w\rangle_{r}\geq-
Ce^{-\frac{1}{r}},\quad\forall\,r\in(0,1),$

where $C>0$ depends only on $n$, $\|u(\,\cdot\,,0)\|_{L^{\infty}(B_{1})}$, and
$\gamma$.

###### Proof.

Since $-r^{-4\gamma}e^{-\frac{2}{r}}\geq-Ce^{-\frac{1}{r}}$, thanks to Lemma
4.6 it suffices to show that $E^{\gamma}(r,w)\geq-Ce^{-\frac{2}{r}}$. In
particular, it is enough to prove that

$\langle w,\text{\small\bf H}w\rangle_{r}\big{(}D(r,w)+\gamma
r^{2\gamma}\big{)}-\langle Zw,\text{\small\bf
H}w\rangle_{r}\big{(}H(r,w)+r^{2\gamma}\big{)}\geq-Ce^{-\frac{2}{r}}.$

On the one hand, by Lemma 5.3, for $\gamma>2$ and $r$ sufficiently small we
have

$0\leq\frac{1}{2}D(r,w)+\frac{\gamma}{2}r^{2\gamma}-H(r,w)-2r^{2\gamma}\leq
C.$

Thus, using (5.2) and Lemma 5.2 (as in the proof of Lemma 5.3) we obtain

$\langle w,\text{\small\bf H}w\rangle_{r}\big{(}D(r,w)+\gamma
r^{2\gamma}-2H(r,w)-2r^{2\gamma}\big{)}\geq-Ce^{-\frac{2}{r}}.$

On the other hand, we similarly have $0\leq H(r,w)+r^{2\gamma}\leq C$, so by
using (5.2) and Lemma 5.2 we obtain

$\big{(}2\langle w,\text{\small\bf H}w\rangle_{r}-\langle Zw,\text{\small\bf
H}w\rangle_{r}\big{)}\big{(}H(r,w)+r^{2\gamma}\big{)}\geq-Ce^{-\frac{2}{r}}.$

This finishes the proof. ∎

We now begin to discuss a series of consequences from the previous bounds. The
first observation is that, as an immediate consequence of (5.3), the following
holds:

######  Lemma 5.5.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), and
$(0,0)$ is a singular point. Given $p\in\mathcal{P}$, set $w:=(u-p)\zeta$.
Then, for all $\gamma>2$, it holds $\phi^{\gamma}(0^{+},w)\geq 2$.

We also have the following:

######  Lemma 5.6.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), and
$(0,0)$ a singular point. Given $p\in\mathcal{P}$, set $w:=(u-p)\zeta$, and
let $\gamma>2$.

  1. (a)

We have

$\frac{d}{dr}\frac{H(r,w)+r^{2\gamma}}{r^{4}}\geq-Ce^{-\frac{1}{2r}}.$

  2. (b)

Assume $\phi^{\gamma}(0^{+},w)=\lambda\leq 3$, and that there exist $\delta>0$
and $R\in(0,1)$ such that $\phi^{\gamma}(r,w)\leq\lambda+\frac{\delta}{2}$ for
all $r\in(0,R)$. Then

$c\left(\frac{R}{r}\right)^{2\lambda}\leq\frac{H(R,w)+R^{2\gamma}}{H(r,w)+r^{2\gamma}}\leq
C_{\delta}\left(\frac{R}{r}\right)^{2\lambda+\delta},$

where $c,C_{\delta}>0$ depend only on $n$, $\|u\|_{L^{\infty}}$, $\gamma$, and
(only in the case of $C_{\delta}$) $\delta$.

###### Proof.

It is a minor modification of the proof of [FRS20, Lemma 4.1]. We define

$F(r):=\frac{2r^{2}\langle w,\text{\small\bf
H}w\rangle_{r}}{H(r,w)+r^{2\gamma}},$

and use that, by Proposition 5.4,

$\frac{d}{dr}\phi^{\gamma}(r,w)\geq\frac{2}{r}\big{(}F(r)\big{)}^{2}-Ce^{-\frac{1}{r}}.$

In particular, thanks to Lemma 5.5, it follows that

$\phi^{\gamma}(r,w)\geq 2-Ce^{-\frac{1}{r}}.$ (5.5)

In particular, from Lemma 4.3 we have

$\begin{split}\frac{\frac{d}{dr}(H(r,w)+r^{2\gamma})}{(H(r,w)+r^{2\gamma})}&=\frac{2}{r}\,\frac{D(r,w)+\gamma
r^{2\gamma}+2r^{2}\langle w,\text{\small\bf
H}w\rangle_{r}}{H(r,w)+r^{2\gamma}}=\frac{2}{r}\phi^{\gamma}(r,w)+\frac{2}{r}F(r)\\\
&\geq\frac{2}{r}\phi^{\gamma}(r,w)-\frac{C}{r}e^{-\frac{1}{r}}\geq\frac{4}{r}-Ce^{-\frac{1}{2r}},\end{split}$
(5.6)

where the first inequality follows from Proposition 5.4, while the second one
from (5.5).

Then, (a) is a direct consequence of (5.6). On the other hand, (b) follows
exactly as in the proof of [FRS20, Lemma 4.1], that is, integrating the first
line of (5.6) and using the Cauchy-Schwartz inequality to control
$\left|\int_{r}^{R}\frac{1}{\rho}F(\rho)d\rho\right|$. ∎

It is well known (see [Bla06]) that the singular set $\Sigma$ can be split
into the following sets:

$\Sigma_{m}:=\big{\\{}(x_{\circ},t_{\circ})\mbox{ singular point with }{\rm
dim}\big{(}\\{p_{2,x_{\circ},t_{\circ}}=0\\}\big{)}=m\big{\\}}\,,\quad 0\leq
m\leq n-1.$ (5.7)

Recall that, as an immediate consequence of [LM15, Theorem 1.9], we have the
following.

######  Proposition 5.7.

Let $u\in C^{1,1}_{x}\cap C^{0,1}_{t}(B_{1}\times(-1,1))$ solve (3.1). Then
$\pi_{x}(\Sigma_{m})\subset B_{1}$ can be locally covered by an
$m$-dimensional $C^{1}$ manifold.

We next show the following:

######  Lemma 5.8.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), and
$(0,0)\in\Sigma_{m}$. Given $p\in\mathcal{P}$, set $w:=(u-p_{2})\zeta$.

  * (a)

If $0\leq m\leq n-2$, then $\phi^{\gamma}(0^{+},w)=2$ for all $\gamma>2$.

  * (b)

If $m=n-1$, then $\phi^{\gamma}(0^{+},w)\leq 3$ for all $\gamma>3$.

###### Proof.

As we shall see, the proofs of the two results are very similar.

$\bullet$ Proof of (a). By Lemma 5.6(b) it is enough to show that if
$(0,0)\in\Sigma_{m}$, with $m\leq n-2$, then for each $\epsilon>0$ we have

$\liminf_{r\downarrow 0}\frac{{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{B_{r}}|u(\cdot,-r^{2})-p_{2}|}{r^{2+\epsilon}}>0.$
(5.8)

We will use a subsolution to show that, since $(0,0)$ is a singular point ant
$\partial_{t}u>0$ outside of the contact set $\\{u=0\\}$, then

${\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\partial
B_{r}}\partial_{t}u(x,t)\,dx\geq
c_{\epsilon}r^{\epsilon}\qquad\forall\,t\in(-r^{2},0).$ (5.9)

Then using that both $u$ and $\partial_{t}u$ are nonnegative, (5.8) follows
immediately by integrating (5.9) with respect to $t$.

To show (5.9), set $L:=\\{p_{2}=0\\}$ and define the “parabolic cone”

$\mathcal{C}_{\delta}:=\left\\{(x,t)\ :\ x\in\mathbb{R}^{n},t<0,\ {\rm
dist}(x,L)\geq\delta(|x|+|t|^{1/2})\right\\}.$

Note that, since $r^{-2}u_{r}-p_{2}\to 0$ locally uniformly, for all
$\delta>0$ there exists $r_{\delta}>0$ such that

$\mathcal{C}_{\delta/2}\cap\mathscr{C}_{r_{\delta}}\subset\\{u>0\\}$

(recall that
$\mathscr{C}_{r_{\delta}}=B_{r_{\delta}}\times(-r_{\delta}^{2},0)$). We now
consider the following eigenvalue problem and make the following claim:

Claim: For any $\epsilon>0$ there exists $\delta$ such that the following
holds: There exist $N\in(0,\epsilon)$ and a positive $N$-homogenous function
$\Phi:\mathbb{R}^{n}\times(-\infty,0)\to(0,+\infty)$ (i.e.,
$\Phi(rx,r^{2}t)=r^{N}\Phi(x,t)$ for all $r>0$) satisfying

$\text{\small\bf H}\Phi=0$ in $\mathcal{C}_{\delta}$ and  $\text{\small\bf
H}\Phi=0$ on $\partial{\mathcal{C}_{\delta}}$.

To prove this, we look for $\Phi$ of the form
$\Phi(x,t)=t^{N/2}\phi(x/t^{1/2})$, where
$\phi:\mathbb{R}^{n}\to\mathbb{R}_{+}$ solves the following eigenvalue problem
for the Ornstein-Uhlenbeck operator in
$\widetilde{\mathcal{C}_{\delta}}:=\\{x_{n}\geq\delta(|x|+1)\\}$:

$\begin{cases}\mathcal{L}_{OU}\phi+\frac{N_{\delta}}{2}\phi=0\quad&\mbox{in
}\widetilde{\mathcal{C}_{\delta}}\\\ \phi(x)=0&\mbox{on
}\partial\widetilde{\mathcal{C}_{\delta}},\end{cases}$

where

$\mathcal{L}_{OU}\phi(x):=\Delta\phi(x)-\frac{x}{2}\cdot\nabla\phi(x)=e^{|x|^{2}/4}{\rm
div}(e^{-|x|^{2}/4}\nabla\phi)$ (5.10)

is the Ornstein-Uhlenbeck operator.

Note that, as $\delta\to 0$, the first eigenvalue $N_{\delta}$ small is very
close to the first eigenvalue $N_{0}$ of $\mathcal{L}_{OU}$ in the limiting
domain $\mathbb{R}^{n}\setminus L$. Also, since ${\rm dim}(L)\leq n-2$, $L$
has zero harmonic capacity. Therefore, it follows by the Raleigh quotient
characterization

$\frac{N_{0}}{2}=\inf_{\phi\in C^{1}_{c}(\mathbb{R}^{n}\setminus
L)}\frac{\int|\nabla\phi|^{2}\,e^{-|x|^{2}/4}}{\int\phi^{2}\,e^{-|x|^{2}/4}}$
(5.11)

that $N_{0}=0$, hence $N_{\delta}$ can be made arbitrarily small provided we
choose $\delta$ small enough. This proves the claim.

We now note that

$\overline{\mathcal{C}_{\delta}}\cap\partial_{par}\mathscr{C}_{r_{\delta}}\subset\subset\\{u>0\\},$

and that $\partial_{t}u$ is positive and supercaloric inside $\\{u>0\\}$.
Hence, we can use $\Phi$ as lower barrier to show that

$\partial_{t}u\geq c\Phi\qquad\mbox{in }\mathscr{C}_{r_{\delta}},$

for some $c>0$, and (5.9) follows.

$\bullet$ Proof of (b). This part is similar to the argument used for (a).
However in this case, since ${\rm dim}(L)=n-1$, $L$ is a set of positive
harmonic capacity and therefore $N_{0}>0$. By Talenti’s rearrangement (with
Gaussian weights), one can easily show that the infimum in (5.11) is attained
by the function $|x_{n}|$, and thus $N_{0}=1$. Therefore, in this case we can
find a $N$-homogeneous function $\Phi$ as above, but with $N\leq 1+\epsilon$.
Using this $\Phi$ as lower barrier for $\partial_{t}u$, we conclude that

${\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\partial
B_{r}}|\partial_{t}u|(x,t)\,dx\geq
c_{\epsilon}r^{1+\epsilon}\qquad\Rightarrow\qquad\liminf_{r\downarrow
0}\frac{{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{B_{r}}|u(\cdot,-r^{2})-p_{2}|}{r^{3+\epsilon}}>0$

(cp. (5.9)-(5.8)), which leads to $\phi(0^{+},u-p_{2})\leq 3$. ∎

Choosing $\gamma>3,$ as a direct consequence of Lemmas 5.8 and 5.6 we obtain
the following:

######  Corollary 5.9.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), and
$(0,0)$ a singular point. Then, for any $\delta>0,$

$H\big{(}r,(u-p_{2})\zeta\big{)}\gg r^{3+\delta}$ as $r\downarrow 0$.

Moreover, the limit $\lim_{r\downarrow 0}\phi\big{(}r,(u-p_{2})\zeta\big{)}$
exists and equals $\phi^{\gamma}(r,(u-p_{2})\zeta)$ for any $\gamma>3$.
Finally, given $K>1$, there exists $C_{K}>1$ such that the following holds:

$\frac{1}{C_{K}}\leq\frac{H(r,(u-p_{2})\zeta)}{H(\theta r,(u-p_{2})\zeta)}\leq
C_{K}\qquad\forall\,r\in(0,1),\,\theta\in[K^{-1},K].$

##  6\. The 2nd blow-up

After the preliminary results from the previous sections, we now start
investigating the structure of “2nd blow-ups” at singular points, namely,
blow-ups of $u-p_{2}$.

The following lemma gives estimates for the difference of two solutions. Note
that both $u$ and $p_{2}$ satisfy the same equation $\text{\small\bf
H}v=\chi_{\\{v>0\\}}$. Although we will first apply the following lemma to $u$
and $p_{2}$, we give a more general version (for difference of approximate
solutions) which will be very useful later on in the paper. Recall that
$\mathscr{C}_{r}=B_{r}\times(-r^{2},0)$.

######  Lemma 6.1.

Let $u_{i}:\mathscr{C}_{2}\to\mathbb{R}$, $i=1,2$, solve

$\begin{cases}\text{\small\bf
H}u_{i}=\chi_{\\{u_{i}>0\\}}(1+\varepsilon_{i}(x,t))\quad\textrm{in}\quad\mathscr{C}_{2}\\\
u_{i}\geq 0\\\ \partial_{t}u_{i}\geq 0,\end{cases}$ (6.1)

with $|\varepsilon_{i}(x,t)|\leq\bar{\varepsilon}<\frac{1}{100}$, and denote
$w:=u_{1}-u_{2}$. Then

$\|w\|_{L^{\infty}(\mathscr{C}_{1})}\leq
C\big{(}\|w\|_{L^{2}(\mathscr{C}_{2})}+\bar{\varepsilon}\big{)}$ (6.2)

and

$\left(\int_{\mathscr{C}_{1}}|\nabla w|^{2}+|w\,\text{\small\bf
H}w|+|\partial_{t}w|^{2}\right)^{\frac{1}{2}}\leq
C\big{(}\|w\|_{L^{2}(\mathscr{C}_{2})}+\bar{\varepsilon}\big{)},$ (6.3)

where $C$ is a dimensional constant.

###### Proof.

We first prove (6.2). On the one hand, we note that inside $\\{u_{1}=0\\}$ it
holds

$\text{\small\bf H}w=1+\varepsilon_{1}-\text{\small\bf H}u_{2}\geq
1-\bar{\varepsilon}-(1+\varepsilon_{2})\chi_{\\{u_{2}>0\\}}\geq-2\bar{\varepsilon},\qquad
w=-u_{2}\leq 0,$ (6.4)

therefore $w_{+}=\max(0,w)$ satisfies $\text{\small\bf
H}w_{+}\geq-2\bar{\varepsilon}$ in $\mathscr{C}_{2}$. Since
$(u_{2}-u_{1})_{+}=w_{-}$, by symmetry we also have $\text{\small\bf
H}w_{-}\geq-2\bar{\varepsilon}$. Thus,

$(\Delta-\partial_{t})|w|\geq-2\bar{\varepsilon}\qquad\mbox{ in
}\mathscr{C}_{2},$

so the parabolic Harnack inequality yields

$\sup_{B_{1}\times(-1,0)}|w|\leq
C_{n}\bigg{(}\int_{B_{2}\times(-2,0)}|w|+\bar{\varepsilon}\bigg{)}\leq
C_{n}\big{(}\|w\|_{L^{2}(B_{2}\times(-2,0))}+\bar{\varepsilon}\big{)},$

which proves (6.2).

We next prove (6.3). Notice that $w\,\text{\small\bf
H}w\geq-2\bar{\varepsilon}|w|$, since

$(u_{1}-u_{2})\,\text{\small\bf
H}(u_{1}-u_{2})=\begin{cases}-|u_{1}-u_{2}|(|\varepsilon_{1}|+|\varepsilon_{2}|)\geq-2\bar{\varepsilon}|u_{1}-u_{2}|\quad&\mbox{if
}u_{1}>0,\ u_{2}>0\\\ u_{1}(1+\varepsilon_{1}-0)\geq 0&\mbox{if }u_{1}>0,\
u_{2}=0\\\ -u_{2}(0-1-\varepsilon_{2})\geq 0&\mbox{if }u_{1}=0,\ u_{2}>0\\\
0&\mbox{if }u_{1}=0,\ u_{2}=0.\end{cases}$

Similarly $\partial_{t}w\,\text{\small\bf
H}w\geq-2\bar{\varepsilon}|\partial_{t}w|$, since

$\partial_{t}(u_{1}-u_{2})\,\text{\small\bf
H}(u_{1}-u_{2})=\begin{cases}-|\partial_{t}(u_{1}-u_{2})|(|\varepsilon_{1}|+|\varepsilon_{2}|)\geq-2\bar{\varepsilon}|\partial_{t}(u_{1}-u_{2})|\quad&\mbox{if
}u_{1}>0,\ u_{2}>0\\\ \partial_{t}u_{1}(1+\varepsilon_{1}-0)\geq 0&\mbox{if
}u_{1}>0,\ u_{2}=0\\\ -\partial_{t}u_{2}(0-1-\bar{\varepsilon}_{2})\geq
0&\mbox{if }u_{1}=0,\ u_{2}>0\\\ 0&\mbox{if }u_{1}=0,\ u_{2}=0.\end{cases}$

On the one hand, choosing

$\tilde{\eta}=\tilde{\eta}(x)\in C^{\infty}_{c}(B_{5/3})\quad\mbox{nonnegative
and such that $\tilde{\eta}\equiv 1$ in $\overline{B}_{4/3}$},$ (6.5)

we obtain

$\frac{d}{dt}\int_{B_{2}\times\\{t\\}}w^{2}\tilde{\eta}+2\int_{B_{2}\times\\{t\\}}(w\,\text{\small\bf
H}w+|\nabla
w|^{2})\tilde{\eta}=\int_{B_{2}\times\\{t\\}}\Delta(w^{2})\tilde{\eta}=\int_{B_{2}\times\\{t\\}}w^{2}\Delta\tilde{\eta}\leq
C\int_{B_{5/3}\times\\{t\\}}w^{2}.$ (6.6)

Integrating in $t$ the last inequality, and recalling that $w\,\text{\small\bf
H}w\geq-2\bar{\varepsilon}|w|$, using (6.5) we obtain

$\int_{-3/2}^{0}\biggl{(}\int_{B_{4/3}\times\\{t\\}}\big{(}|w\,\text{\small\bf
H}w|+|\nabla w|^{2}\big{)}\biggr{)}\,dt\leq
C_{n}\sup_{t\in(-3/2,0)}\int_{B_{5/3}\times\\{t\\}}\big{(}|w|^{2}+\bar{\varepsilon}|w|\big{)}.$

Now from the $L^{\infty}$ estimate in Step 1 (plus a standard covering
argument, to obtain a control in $\mathscr{C}_{3/2}$ instead of
$\mathscr{C}_{1}$), we obtain that

$\sup_{t\in(-3/2,0)}\int_{B_{5/3}\times\\{t\\}}\big{(}|w|^{2}+\bar{\varepsilon}|w|\big{)}\leq
C_{n}\big{(}\|w\|_{L^{2}(\mathscr{C}_{2})}+\bar{\varepsilon}\big{)}^{2},$

and therefore

$\int_{-3/2}^{0}\biggl{(}\int_{B_{4/3}\times\\{t\\}}\big{(}|w\,\text{\small\bf
H}w|+|\nabla w|^{2}\big{)}\biggr{)}\,dt\leq
C_{n}\big{(}\|w\|_{L^{2}(\mathscr{C}_{2})}+\bar{\varepsilon}\big{)}^{2}.$
(6.7)

On the other hand, choosing now $\eta=\eta(x)\in C^{\infty}_{c}(B_{4/3})$ such
that $\eta\equiv 1$ in $\overline{B}_{1}$, and using $\text{\small\bf
H}w(\partial_{t}w)\geq-2\bar{\varepsilon}|\partial_{t}w|$, for any $M>0$ we
have

$\begin{split}2\int_{B_{2}\times\\{t\\}}(\partial_{t}w)^{2}\eta^{2}+&\frac{d}{dt}\int_{B_{2}\times\\{t\\}}|\nabla
w|^{2}\eta^{2}=2\int_{B_{2}\times\\{t\\}}(\partial_{t}w)^{2}\eta^{2}+\int_{B_{2}\times\\{t\\}}2\nabla
w\cdot(\nabla\partial_{t}w)\eta^{2}\\\
&=\int_{B_{2}\times\\{t\\}}\Big{(}2(\partial_{t}w-\Delta
w)(\partial_{t}w)\eta^{2}-4\partial_{t}w\nabla
w\cdot\nabla\eta\,\eta\Big{)}\\\
&\leq\int_{B_{2}\times\\{t\\}}\Big{(}4\bar{\varepsilon}|\partial_{t}w|\eta^{2}+2M|\nabla
w|^{2}|\nabla\eta|^{2}+\frac{2}{M}(\partial_{t}w)^{2}\eta^{2}\Big{)}.\end{split}$

Since $4\bar{\varepsilon}|\partial_{t}w|\leq
C\bar{\varepsilon}^{2}+\frac{1}{2}|\partial_{t}w|^{2}$, choosing $M=4$ and
using that $\eta$ is supported in $B_{4/3}$, we obtain

$\int_{B_{1}\times\\{t\\}}(\partial_{t}w)^{2}\eta^{2}+\frac{d}{dt}\int_{B_{2}\times\\{t\\}}|\nabla
w|^{2}\eta^{2}\leq C_{n}\bigg{(}\int_{B_{4/3}\times\\{t\\}}|\nabla
w|^{2}+\bar{\varepsilon}^{2}\bigg{)}.$

Multiplying this inequality by a nonnegative smooth function $\psi\in
C_{c}^{\infty}((-3/2,0])$ such that $\psi|_{[-1,0]}\equiv 1$, and integrating
over $[-3/2,0]$, we get

$\displaystyle\int_{\mathscr{C}_{1}}(\partial_{t}w)^{2}\eta^{2}$
$\displaystyle\leq\int_{0}^{3/2}\biggl{(}\int_{B_{1}}(\partial_{t}w)^{2}\eta^{2}\biggr{)}\psi(t)\,dt$
$\displaystyle\leq\int_{0}^{3/2}\biggl{(}\int_{B_{2}\times\\{t\\}}|\nabla
w|^{2}\eta^{2}\biggr{)}\psi^{\prime}(t)\,dt+C_{n}\bigg{[}\int_{0}^{3/2}\biggl{(}\int_{B_{4/3}}|\nabla
w|^{2}\biggr{)}\psi(t)\,dt+\bar{\varepsilon}^{2}\bigg{]}$ $\displaystyle\leq
C_{n}\bigg{[}\int_{0}^{3/2}\biggl{(}\int_{B_{4/3}}|\nabla
w|^{2}\biggr{)}\,dt+\bar{\varepsilon}^{2}\bigg{]}.$

Combining this bound with (6.7), the result follows. ∎

As a consequence, we find:

######  Corollary 6.2.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), $(0,0)$
a singular point, and let $w:=u-p_{2}$. Then, for all $r\in(0,1)$,

$\|w_{r}\|_{L^{\infty}(\mathscr{C}_{1})}+\|\partial_{t}w_{r}\|_{L^{\infty}(\mathscr{C}_{1})}+\|\nabla
w_{r}\|_{L^{2}(\mathscr{C}_{1})}+\|\nabla\partial_{t}w_{r}\|_{L^{2}(\mathscr{C}_{1})}\leq
C\|w_{r}\|_{L^{2}(\mathscr{C}_{2})}.$

Also, for all $e\in\\{p_{2}=0\\}\cap\mathbb{S}^{n-1}$, we have

$\|\nabla\partial_{e}w_{r}\|_{L^{2}(\mathscr{C}_{1})}\leq
C\|w_{r}\|_{L^{2}(\mathscr{C}_{2})}.$

The constant $C$ depends only on $n$ and $\|u\|_{L^{\infty}}$.

###### Proof.

We apply Lemma 6.1 with $u_{1}=r^{-2}u_{r}$, $u_{2}=p_{2}$, and
$\bar{\varepsilon}=0$, to obtain

$\|w_{r}\|_{L^{\infty}(\mathscr{C}_{1})}+\|\partial_{t}w_{r}\|_{L^{2}(\mathscr{C}_{1})}+\|\nabla
w_{r}\|_{L^{2}(\mathscr{C}_{1})}\leq C\|w_{r}\|_{L^{2}(\mathscr{C}_{2})}.$

Now, to prove the $L^{\infty}$ bound on $\partial_{t}w_{r}$ we observe that
$\partial_{t}(u-p_{2})=\partial_{t}u\geq 0$ and $\text{\small\bf
H}\partial_{t}(u-p_{2})=\partial_{t}\chi_{\\{u>0\\}}\geq 0$. Hence
$\partial_{t}w_{r}$ is nonnegative and subcaloric, and therefore the weak
Harnack inequality implies that

$\|\partial_{t}w_{r}\|_{L^{\infty}(\mathscr{C}_{1/2})}\leq
C_{n}\|\partial_{t}w_{r}\|_{L^{1}(\mathscr{C}_{1})}\leq
C_{n}\|\partial_{t}w_{r}\|_{L^{2}(\mathscr{C}_{1})}.$

Also, since $(\partial_{t}w_{r})\,\text{\small\bf H}(\partial_{t}w_{r})\geq
0$, similarly to (6.7) (replacing $w$ by $\partial_{t}w_{r}$) we obtain

$\|\nabla\partial_{t}w_{r}\|_{L^{2}(\mathscr{C}_{1/2})}\leq
C\|\partial_{t}w_{r}\|_{L^{2}(\mathscr{C}_{1})}\leq
C\|\partial_{t}w_{r}\|_{L^{\infty}(\mathscr{C}_{1})}\leq
C\|w_{r}\|_{L^{2}(\mathscr{C}_{2})}.$

By a covering argument we can replace $\mathscr{C}_{1/2}$ by
$\mathscr{C}_{1}$, the first part of the corollary follows.

Finally, given $e\in\\{p_{2}=0\\}$, since $\partial_{e}p_{2}\equiv 0$ we
obtain that $\partial_{e}w_{r}=0$ on $\\{u=0\\}$ and that $\text{\small\bf
H}\partial_{e}w_{r}=0$ in $\\{u>0\\}$. This yields
$\partial_{e}w_{r}\text{\small\bf H}\partial_{e}w_{r}=0$, and arguing as
before we obtain

$\|\nabla\partial_{e}w_{r}\|_{L^{2}(\mathscr{C}_{1/2})}\leq
C\|\partial_{e}w_{r}\|_{L^{2}(\mathscr{C}_{1})}\leq C\|\nabla
w_{r}\|_{L^{2}(\mathscr{C}_{1})}\leq C\|w_{r}\|_{L^{2}(\mathscr{C}_{2})}.$

Again by a covering argument we can replace $\mathscr{C}_{1/2}$ by
$\mathscr{C}_{1}$, which concludes the proof. ∎

The next lemma will be very useful in the sequel. We recall that $\zeta$
denotes the cut-off function defined in Section 2.7.

######  Lemma 6.3.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), $(0,0)$
a singular point, and let $w:=u-p_{2}$. Then, for all $r\in(0,1)$ we have

$\frac{1}{C}H(r,w\zeta)^{1/2}\leq\|w_{r}\|_{L^{2}(\mathscr{C}_{1})}\leq
CH(r,w\zeta)^{1/2},$

for some constant $C>1$ depending only on $n$ and $\|u\|_{L^{\infty}}$.

To prove Lemma 6.3, we will use the following simple consequence of the
Gaussian log-Sobolev inequality.

######  Lemma 6.4.

Let $f:\mathbb{R}^{n}\to\mathbb{R}$ satisfy

$\int_{\mathbb{R}^{n}}f^{2}dm=1\quad\mbox{and}\quad\int_{\mathbb{R}^{n}}|\nabla
f|^{2}dm\leq 4,$

where
$dm=G(x,-1)dx=\frac{1}{(4\pi)^{n/2}}\exp\left(-\frac{|x|^{2}}{4}\right)dx$ is
the Gaussian measure. Then, for some dimensional $R_{\circ}>0$, we have

$\int_{B_{R_{\circ}}}f^{2}dm\geq\frac{1}{2}.$

###### Proof.

Given $\lambda$ large (to be fixed later), choose $R_{\circ}$ so that

$\lambda\int_{\mathbb{R}^{n}\setminus B_{R_{\circ}}}dm\leq\frac{1}{4}.$

By assumption

$\int_{B_{R_{\circ}}}f^{2}dm+\int_{\mathbb{R}^{n}\setminus
B_{R_{\circ}}}f^{2}\chi_{\\{f^{2}\leq\lambda\\}}dm+\int_{\mathbb{R}^{n}\setminus
B_{R_{\circ}}}f^{2}\chi_{\\{f^{2}>\lambda\\}}dm=1.$

Suppose now, by contradiction, that $\int_{B_{R_{\circ}}}f^{2}dm<1/2$. Since

$\int_{\mathbb{R}^{n}\setminus
B_{R_{\circ}}}f^{2}\chi_{\\{f^{2}\leq\lambda\\}}dm\leq\lambda\int_{\mathbb{R}^{n}\setminus
B_{R_{\circ}}}dm\leq\frac{1}{4},$

we have

$\int_{\mathbb{R}^{n}\setminus
B_{R_{\circ}}}f^{2}\chi_{\\{f^{2}>\lambda\\}}dm\geq\frac{1}{4}.$ (6.8)

Recall the Gaussian log-Sobolev inequality:

$\int_{\mathbb{R}^{n}}F^{2}\log F^{2}dm\leq\int_{\mathbb{R}^{n}}|\nabla
F|^{2}dm+\biggl{(}\int_{\mathbb{R}^{n}}F^{2}dm\biggr{)}\,\log\biggl{(}\int_{\mathbb{R}^{n}}F^{2}dm\biggr{)}.$

To reach a contradiction, we apply this inequality to
$F:=|f|\chi_{\\{|f|>e^{1/2}\\}}$. Indeed, since $|\nabla F|\leq|\nabla f|$,
$\log(F^{2})\geq 1$, $F^{2}\leq e+f^{2}$, and $F=f$ inside
$\\{f^{2}\geq\lambda\\}$, we get

$\displaystyle\log\lambda\int_{\mathbb{R}^{n}\setminus
B_{R_{\circ}}}f^{2}\chi_{\\{f^{2}>\lambda\\}}dm$
$\displaystyle\leq\int_{\mathbb{R}^{n}\setminus
B_{R_{\circ}}}F^{2}\log(F^{2})\chi_{\\{f^{2}>\lambda\\}}dm\leq\int_{\mathbb{R}^{n}}F^{2}\log
F^{2}dm$ $\displaystyle\leq\int_{\mathbb{R}^{n}}|\nabla
F|^{2}dm+\biggl{(}\int_{\mathbb{R}^{n}}F^{2}dm\biggr{)}\,\log\biggl{(}\int_{\mathbb{R}^{n}}F^{2}dm\biggr{)}$
$\displaystyle\leq\int_{\mathbb{R}^{n}}|\nabla
f|^{2}dm+\biggl{(}e+\int_{\mathbb{R}^{n}}f^{2}dm\biggr{)}\,\log\biggl{(}e+\int_{\mathbb{R}^{n}}f^{2}dm\biggr{)}$
$\displaystyle\leq 4+(e+1)\log(e+1).$

Combining this bound with (6.8), we get

$\log\lambda\leq 16+4(e+1)\log(e+1),$

a contradiction if $\lambda$ is chosen large enough. ∎

We can now prove Lemma 6.3.

###### Proof of Lemma 6.3.

We divide the proof into three steps.

$\bullet$ Step 1. We first show that

$\frac{1}{C}H(r,w\zeta)^{1/2}\leq\|w_{r}\|_{L^{2}(\mathscr{C}_{2R_{\circ}})},$
(6.9)

where $R_{\circ}$ is the dimensional constant from Lemma 6.4.

Let $\tilde{w}_{r}:=\frac{(w\zeta)_{r}}{H(r,w\zeta)^{1/2}}$. Since
$\phi(r,w\zeta)\leq 4$ for $r$ small, we have

$\int_{\\{t=-1\\}}(\tilde{w}_{r})^{2}G=1\quad\mbox{ and
}\quad\int_{\\{t=-1\\}}|\nabla\tilde{w}_{r}|^{2}G\leq 4.$

Then, using Lemma 6.4 we obtain

$\int_{\\{|x|\leq R_{\circ},\ t=-1\\}}(\tilde{w}_{r})^{2}G\geq\frac{1}{2}.$

Replacing $r$ by $\theta^{1/2}r$ with $\theta\in(1,4)$, by scaling we obtain

$\int_{\\{|x|\leq 2R_{\circ},\
t=-\theta\\}}(\tilde{w}_{r})^{2}G\geq\frac{1}{2}.$

Hence, after integrating with respect to $\theta\in(1,4)$, we find

$\int_{\\{|x|\leq 2R_{\circ},\ -4\leq
t\leq-1\\}}(\tilde{w}_{r})^{2}G\geq\frac{1}{C}.$

Since $\\{|x|\leq 2R_{\circ},\ -4\leq
t\leq-1\\}\subset\mathscr{C}_{2R_{\circ}}$, (6.9) follows.

$\bullet$ Step 2. We now prove that, for any given $M>1$, there exists a
constant $C_{M}\geq 1,$ depending only on $n$ and $M$, so that the following
holds for all in $r\in(0,\frac{1}{4})$:

$\|w_{r}\|_{L^{2}(\mathscr{C}_{1})}\geq
C_{M}H(r,w\zeta)^{1/2}\qquad\Rightarrow\qquad\|w_{2r}\|_{L^{2}(\mathscr{C}_{1})}\geq
M\|w_{r}\|_{L^{2}(\mathscr{C}_{1})}.$ (6.10)

To prove this, note that the $L^{\infty}$ estimate from Corollary 6.2 gives

$\|w_{r}\|_{L^{\infty}(\mathscr{C}_{1})}\leq
C_{\circ}\|w_{r}\|_{L^{2}(\mathscr{C}_{2})},$

where $C_{\circ}$ is dimensional. Therefore, whenever
$\|w_{2r}\|_{L^{2}(\mathscr{C}_{1})}\leq M\|w_{r}\|_{L^{2}(\mathscr{C}_{1})}$,
we have

$\|w_{r}\|_{L^{\infty}(\mathscr{C}_{1})}\leq
C_{\circ}\|w_{r}\|_{L^{2}(\mathscr{C}_{2})}=2^{n+2}C_{\circ}\|w_{2r}\|_{L^{2}(\mathscr{C}_{1})}\leq
2^{n+2}C_{\circ}M\|w_{r}\|_{L^{2}(\mathscr{C}_{1})}.$

In particular, for any $\tau>0$ small, we have

$\int_{\\{x\in B_{1},-1<t<-\tau\\}}w_{r}^{2}\geq\int_{\\{x\in
B_{1},-1<t<0\\}}w_{r}^{2}-\tau\|w_{r}\|^{2}_{L^{\infty}(\mathscr{C}_{1})}\geq(1-\tau
2^{2n+4}C_{\circ}^{2}M^{2})\|w_{r}\|^{2}_{L^{2}(\mathscr{C}_{1})}.$

Choosing $\tau:=(2^{2n+5}C_{\circ}^{2}M^{2})^{-1}$ and using that $G\geq
c_{\tau}>0$ inside $\\{x\in B_{1},-1<t<-\tau\\}$, we get

$\frac{1}{2}\|w_{r}\|^{2}_{L^{2}(\mathscr{C}_{1})}\leq\int_{\\{x\in
B_{1},-1<t<-\tau\\}}w_{r}^{2}\leq C_{1}\int_{\\{x\in
B_{1},-1<t<-\tau\\}}w_{r}^{2}G\leq
C_{2}\int_{\tau}^{1}H(\theta^{2}r,w\zeta)d\theta.$

Since the last term can be bounded by $C_{3}H(r,w\zeta)$ (thanks to Corollary
5.9), this proves the contrapositive of (6.10) with $C_{M}=(2C_{3})^{1/2}$.

$\bullet$ Step 3. We now conclude the proof of the lemma combining Steps 1 and
2.

First, (6.9) rescaled yields

$\frac{1}{C}H(c_{\circ}r,w\zeta)^{1/2}\leq\|w_{r}\|_{L^{2}(\mathscr{C}_{1})},$

where $c_{\circ}=\frac{1}{2R_{\circ}}$, that combined with
$H(c_{\circ}r,w\zeta)^{1/2}\geq\frac{1}{C}H(r,w\zeta)^{1/2}$ (see Corollary
5.9) gives

$\frac{1}{C}H(r,w\zeta)^{1/2}\leq\|w_{r}\|_{L^{2}(\mathscr{C}_{1})}.$

This proves the first inequality in the statement.

To prove the second inequality, assume by contradiction that
$\|w_{r}\|_{L^{2}(\mathscr{C}_{1})}\geq C_{M}H(r,w\zeta)^{1/2}$, with
$C_{M}\geq 1$ as in (6.10) and $M$ large enough to be chosen. Then, by (6.10),

$\|w_{2r}\|_{L^{2}(\mathscr{C}_{1})}\geq
M\|w_{r}\|_{L^{2}(\mathscr{C}_{1})}\geq MC_{M}H(r,w\zeta)^{1/2}.$

Let $N>1$ be a large constant to be fixed. Since $H(r,w\zeta)^{1/2}\geq
C_{4}^{-1}H(2r,w\zeta)^{1/2}$ (by Corollary 5.9), choosing $M=NC_{4}$ we
obtain

$\|w_{2r}\|_{L^{2}(\mathscr{C}_{1})}\geq NC_{M}H(2r,w\zeta)^{1/2}.$

This allows us to apply again (6.10) (with the same $M$) and with $r$ replaced
by $2r$, and we obtain

$\|w_{4r}\|_{L^{2}(\mathscr{C}_{1})}\geq
M\|w_{2r}\|_{L^{2}(\mathscr{C}_{1})}\geq MNC_{M}H(2r,w\zeta)^{1/2}\geq
MNC_{M}C_{4}^{-1}H(4r,w\zeta)^{1/2}=N^{2}C_{M}H(4r,w\zeta)^{1/2}.$

Iterating this argument $\ell$ times, this yields

$\|w_{2^{\ell}r}\|_{L^{2}(\mathscr{C}_{1})}\geq
N^{\ell}C_{M}H(2^{\ell}r,w\zeta)^{1/2}$

Choosing $\ell$ such that $\frac{1}{8}\leq 2^{\ell}r\leq\frac{1}{4}$, and
using that at scales of order $1$ both quantities
$\|w_{2^{\ell}r}\|_{L^{2}(\mathscr{C}_{1})}$ and $H(2^{\ell}r,w\zeta)^{1/2}$
are comparable to $1$, we obtain

$C\geq N^{\ell}C_{M}\geq N^{\ell},$

a contradiction if $N$ is chosen large enough. ∎

The following Lipschitz estimate for the rescaled difference $u-p_{2}$ will be
useful in the sequel. We recall that $w_{r}$ has been defined in (2.1), while
$\Sigma_{n-1}$ is defined as in (5.7).

######  Lemma 6.5.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1), and
$(0,0)\in\Sigma_{n-1}$. Then, given $R\geq 1$, for all
$r\in\left(0,\frac{1}{10R}\right)$ and
$\boldsymbol{e}\in\mathbb{S}^{n-1}\cap\\{p_{2}=0\\}$, we have

$\sup_{\mathscr{C}_{1}}\big{(}(\partial_{\boldsymbol{e}\boldsymbol{e}}w_{r})_{-}+|\partial_{t}w_{r}|+|\nabla
w_{r}|\big{)}\leq C\|w_{r}\|_{L^{2}(\mathscr{C}_{2})},$

for some constant $C$ depending only on $n$,
$\|u(\,\cdot\,,0)\|_{L^{\infty}(B_{1})}$, and $R$.

###### Proof.

Given a function $f:\mathbb{R}^{n}\to\mathbb{R}$, a vector
$\boldsymbol{e}\in\mathbb{S}^{n-1}$, and $h\in(0,1)$, let

$\delta^{2}_{\boldsymbol{e},h}f:=\frac{f(\,\cdot\,+h\boldsymbol{e},\,\cdot\,)+f(\,\cdot\,-h\boldsymbol{e},\,\cdot\,)-2f}{h^{2}}\qquad\mbox{and}\qquad\delta_{t,h}f:=\frac{f\big{(}\,\cdot\,,\cdot+h\big{)}-f}{h}.$

Note that, for $\boldsymbol{e}\in\\{p=0\\}\cap\mathbb{S}^{n-1}$, we have
$\delta^{2}_{\boldsymbol{e},h}p\equiv 0$. Thus, since $\text{\small\bf H}u=1$
outside of $\\{u=0\\}$ and $\text{\small\bf H}u\leq 1$ in $B_{1}\times(-1,0)$
we have

$\text{\small\bf
H}\big{(}\delta^{2}_{\boldsymbol{e},h}w\big{)}=\frac{\text{\small\bf
H}u\big{(}\,\cdot\,+h\boldsymbol{e},\,\cdot\,\big{)}+\text{\small\bf
H}u\big{(}\,\cdot\,-h\boldsymbol{e},\,\cdot\,\big{)}-2\text{\small\bf
H}u}{h^{2}}\leq 0\qquad\mbox{ in }\\{u>0\\}.$

Similarly,

$\text{\small\bf H}\big{(}\delta_{t,h}w\big{)}\leq
0\qquad\mbox{and}\qquad\text{\small\bf H}\big{(}\delta_{t,-h}w\big{)}\leq
0\qquad\mbox{ in }\\{u>0\\}.$

On the other hand, since $u\geq 0$, we have

$\delta^{2}_{\boldsymbol{e},h}w=\delta^{2}_{\boldsymbol{e},h}u(\,\cdot\,)\geq
0\quad\mbox{and}\quad\delta_{t,\pm h}w=\delta_{t,\pm h}u(\,\cdot\,)\geq
0\quad\mbox{ inside }\\{u=0\\}.$

As a consequence, the negative part of the second order incremental quotient
$(\delta^{2}_{\boldsymbol{e},h}w_{r})_{-}$ is subcaloric, and so is its limit
$(\partial_{\boldsymbol{e}\boldsymbol{e}}^{2}w_{r})_{-}$ (recall that $u\in
C^{1,1}_{x}\cap C^{0,1}_{t}$ , and thus
$(\delta^{2}_{\boldsymbol{e},h}w_{r})_{-}\to(\partial_{\boldsymbol{e}\boldsymbol{e}}^{2}w_{r})_{-}$
and a.e. as $h\to 0$). Similarly, both $(\delta_{t}w_{r})_{-}$ and
$(-\delta_{t}w_{r})_{-}$ are subcaloric, and thus so is $|\partial_{t}w_{r}|$.

Therefore, by weak Harnack inequality (see Lemma 3.6), for any $\varepsilon>0$
we have

$\sup_{\mathscr{C}_{4/3}}\big{(}|\partial_{t}w_{r}|+(\partial_{\boldsymbol{e}\boldsymbol{e}}^{2}w_{r})_{-}\big{)}\leq
C_{n}\biggl{(}\int_{\mathscr{C}_{5/4}}(|\partial_{t}w_{r}|+|\partial_{\boldsymbol{e}\boldsymbol{e}}w_{r}|)^{\varepsilon}\biggr{)}^{1/\varepsilon}.$

Also, by standard interpolation inequalities, the $L^{\varepsilon}$ norm with
$\varepsilon<1$ can be controlled by the weak $L^{1}$ norm, namely

$\biggl{(}\int_{\mathscr{C}_{5/4}}(|\partial_{t}w_{r}|+|\partial_{\boldsymbol{e}\boldsymbol{e}}w_{r}|)^{\varepsilon}\biggr{)}^{1/\varepsilon}\leq
C(n,R)\,\sup_{\theta>0}\,\theta\bigl{|}\bigl{\\{}(|\partial_{t}w_{r}|+|\partial_{\boldsymbol{e}\boldsymbol{e}}w_{r}|)>\theta\bigr{\\}}\cap\mathscr{C}_{5/4}\bigr{|}.$

Furthermore, by the parabolic Calderón-Zygmund theory (see Theorem 3.5), the
right hand side above is controlled by

$\|\text{\small\bf
H}w_{r}\|_{L^{1}(\mathscr{C}_{5/3})}+\|w_{r}\|_{L^{1}(\mathscr{C}_{5/3})}.$

In addition, since $\text{\small\bf H}w_{r}\leq 0$ in $\mathscr{C}_{5/3}$, the
norm $\|\text{\small\bf H}w_{r}\|_{L^{1}(\mathscr{C}_{5/3})}$ is controlled by
the $L^{1}$ norm of $w_{r}$ inside $\mathscr{C}_{2}$: indeed, if $\chi$ is a
smooth nonnegative cut-off function that is equal to $1$ in
$\mathscr{C}_{5/3}$ and vanishes outside $\mathscr{C}_{2}$, then integration
by parts gives

$\|\text{\small\bf
H}w_{r}\|_{L^{1}(\mathscr{C}_{5/3})}\leq-\int_{\mathscr{C}_{2}}\chi\,\text{\small\bf
H}w_{r}=-\int_{\mathscr{C}_{2}}(\Delta+\partial_{t})\chi\,w_{r}\leq
C_{n}\int_{\mathscr{C}_{2}}|w_{r}|\leq\|w_{r}\|_{L^{2}{(\mathscr{C}_{2})}}.$
(6.11)

This proves that

$\sup_{\mathscr{C}_{4/3}}\big{(}|\partial_{t}w_{r}|+(\partial_{\boldsymbol{e}\boldsymbol{e}}^{2}w_{r})_{-}\big{)}\leq
C_{n}\|w_{r}\|_{L^{2}{(\mathscr{C}_{2})}}.$ (6.12)

Finally, since $\\{p_{2}=0\\}$ is $(n-1)$-dimensional (recall
$(0,0)\in\Sigma_{n-1}$), as a consequence of (6.12) and
$(\Delta-\partial_{t})w_{r}\leq 0$ we deduce that

$\partial_{\boldsymbol{e}^{\prime}\boldsymbol{e}^{\prime}}w_{r}\leq
C\|w_{r}\|_{L^{2}{(\mathscr{C}_{2})}}\quad\mbox{in
}\mathscr{C}_{4/3},\quad\mbox{ where
}\boldsymbol{e}^{\prime}\in\\{p=0\\}^{\perp}\mbox{ with
}|\boldsymbol{e}^{\prime}|=1.$

The above semiconcavity estimate, combined with the semiconvexity bound in
(6.12), implies the desired uniform bound on $\|\nabla
w_{r}\|_{L^{\infty}(\mathscr{C}_{1})}$. ∎

As a consequence of the previous lemma, we get the following result (recall
that, as a consequence of Lemma 5.8 and Corollary 5.9, we know that
$\phi(0^{+},w\zeta)\in[2,3]$).

######  Proposition 6.6.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1),
$(0,0)\in\Sigma_{n-1}$, and set $w:=u-p_{2}$ and
$\lambda^{2nd}:=\phi(0^{+},w\zeta)\in[2,3]$. Then

$\\{u(\cdot\,,t)=0\\}\cap B_{r}\subset\big{\\{}x\,:\,{\rm
dist}(x,\\{p_{2}=0\\})\leq Cr^{\lambda^{2nd}-1}\big{\\}}$ (6.13)

for all $r\in(0,1/2)$ and $t\geq-r^{2}$. In addition, the constant $C$ depends
only on $n$ and $\|u(\,\cdot\,,0)\|_{L^{\infty}(B_{1})}$.

###### Proof.

As a consequence of Lemmas 5.6 and 6.5 we obtain

$\|\nabla u-\nabla p_{2}\|_{L^{\infty}(B_{r}\times\\{-r\\}))}\leq
Cr^{\lambda^{2nd}-1}.$

Thus, on $\\{u=0\\}\cap\bigl{(}B_{r}\times\\{-r^{2}\\}\bigr{)}$ we have
$|\nabla p_{2}(x)|=\big{|}{\rm dist}(x,\\{p_{2}=0\\})\big{|}\leq
Cr^{\lambda^{2nd}-1}$. Since $\\{u=0\\}$ shrinks with time, we obtain (6.13).
∎

Using the previous results, we can now prove the following:

######  Proposition 6.7.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a solution of (3.1), and $(0,0)$ be
a singular point. Let $w:=u-p_{2}$, $m:={\rm
dim}(\\{p_{2}=0\\})\in\\{0,1,2,\dots n-1\\}$,
$\lambda^{2nd}:=\phi(0^{+},w\zeta)$, and define

$\tilde{w}_{r}:=\frac{(u-p_{2})_{r}}{H\big{(}r,(u-p_{2})\zeta\big{)}}.$ (6.14)

Then, for every sequence $r_{k}\downarrow 0$ there is a subsequence
$r_{k_{\ell}}$ such that

$\tilde{w}_{r_{k_{\ell}}}\rightarrow
q\qquad\mbox{and}\qquad\nabla\tilde{w}_{r_{k_{\ell}}}\rightharpoonup\nabla
q\qquad\mbox{in }L^{2}_{\rm loc}(\mathbb{R}^{n}\times(-\infty,0])$

as $\ell\to\infty$, where $q\not\equiv 0$ is a $\lambda^{2nd}$-homogeneous
function. Moreover, for any $e$ unit vector tangent to $\\{p_{2}=0\\}$, the
following “growth estimates” hold:

$\sup_{\mathscr{C}_{R}}\big{(}|W|+R^{2}|\partial_{t}W|\big{)}+\bigg{(}{\mathchoice{\kern
4.30554pt\vbox{\hrule height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern
4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\mathscr{C}_{R}}R^{2}|\nabla
W|^{2}+R^{4}|\nabla\partial_{e}W|^{2}+R^{6}|\nabla\partial_{t}W|^{2}\bigg{)}^{\frac{1}{2}}\leq
C_{\delta}R^{\lambda^{2nd}+\delta}$ (6.15)

with $W=\tilde{w}_{r_{k_{\ell}}}$ (for all $R$ satisfying $1\leq R\ll
r_{k_{\ell}}^{-1}$) and for $W=q$ (for all $R\geq 1$).

In addition, we have:

  1. (a)

If $0\leq m\leq n-2$ then $\lambda^{2nd}=2$, and —in some appropriate
coordinates—

$\qquad
p_{2}(x)=\frac{1}{2}\sum_{i=m+1}^{n}\mu_{i}x_{i}^{2}\qquad\mbox{and}\qquad
q(x)=At+\nu\sum_{i=m+1}^{n}x_{i}^{2}-\sum_{j=1}^{m}\nu_{j}x_{j}^{2}\,,$ (6.16)

where $\mu_{i}>0$, $\nu$ and $A$ are nonnegative, and
$A-2(n-m)\nu+2\sum_{j=1}^{m}\nu_{j}=0$.

  2. (b)

If $m=n-1$ then $\tilde{w}_{r_{k_{\ell}}}\rightarrow q$ in $C^{0}_{\rm
loc}(\mathbb{R}^{n}\times(-\infty,0])$, $\lambda^{2nd}\in[2+\alpha_{\circ},3]$
for some dimensional constant $\alpha_{\circ}>0$, and $q$ solves the parabolic
thin obstacle problem

$\begin{cases}\text{\small\bf H}q\leq 0\quad\text{and}\quad q\text{\small\bf
H}q=0\quad&\mbox{in }\mathbb{R}^{n}\times(-\infty,0)\\\ \text{\small\bf
H}q=0&\mbox{in }\mathbb{R}^{n}\times(-\infty,0)\setminus\\{p_{2}=0\\}\\\
\,q\geq 0&\mbox{on }\\{p_{2}=0\\}\\\ \partial_{t}q\geq 0&\mbox{in
}\,\mathbb{R}^{n}\times(-\infty,0).\end{cases}$ (6.17)

###### Proof.

Given $r_{k}\downarrow 0$, by the $H^{1}$ space-time estimates of Corollary
6.2 we obtain (up to subsequence)

$\tilde{w}_{r_{k}}\rightarrow q\quad\mbox{and
}\quad(\nabla,\partial_{t})\tilde{w}_{r_{k}}\rightharpoonup(\nabla,\partial_{t})q\quad\mbox{in
}L^{2}_{loc}\big{(}\mathbb{R}^{n}\times(-\infty,0]\big{)},$

for some $q\in L^{2}_{loc}((-\infty,0],H^{1}_{loc}(\mathbb{R}^{n}))$.
Moreover, since $\phi(r_{k},w\zeta)\leq C$ we have
$\int_{\\{t=-1\\}}\big{|}\nabla(\widetilde{w\zeta})_{r_{k}}\big{|}^{2}G\leq
C$, from which it follows that $\int_{\\{t=-1\\}}|\nabla q|^{2}G\leq C$.

Furthermore, by Lemma 5.6(b) applied with $\gamma=4$, Lemma 6.3 combined with
Corollary 6.2 implies the validity of the “growth estimates” (6.15) for
$W=\tilde{w}_{r_{k}}$, for all $R$ satisfying $1\leq R\ll r_{k}^{-1}$. Taking
the limit as $k\to\infty$, since all the seminorms in the estimates are lower
semicontinuous, we obtain that (6.15) holds also $W=q$ (for all $R\geq 1$). It
is also easy to see, using the monotonicity of the frequency in Proposition
5.4, that $q$ is $\lambda^{2nd}$-homogeneous.

Let us show that, in addition, $q$ satisfies

$\int_{\\{t=-1\\}}(p_{2}-p)q\,G\,dx\geq 0\qquad\textrm{for all}\quad
p\in\mathcal{P}.$ (6.18)

Indeed, for any fixed $p\in\mathcal{P}$, Lemma 5.6(a) gives

$\frac{1}{r^{4}}(H\big{(}r,(u-p)\zeta\big{)}+r^{8}\big{)}+Ce^{-c/r}\geq\lim_{r\downarrow
0}\frac{1}{r^{4}}(H\big{(}r,(u-p)\zeta\big{)}\big{)},$

and therefore

$\int_{\\{t=-1\\}}\big{(}r^{-2}u_{r}-p)\zeta_{r}\big{)}^{2}G+Cr^{4}\geq\int_{\\{t=-1\\}}(p_{2}-p)^{2}G.$

This allows us to proceed exactly as in the proof of [FS17, Lemma 2.11] (with
obvious modifications), and (6.18) follows.

Now in order to conclude the proof, we consider two cases.

(a) If $m\leq n-2$ then $w_{r_{k}}\in H^{1}_{\rm loc}$ it is caloric outside
of an infinitesimal neighborhood of $\\{p_{2}=0\\}\times(-\infty,0)$. Since
${\rm dim}(\\{p_{2}=0\\})\leq n-2$ which has zero capacity, we deduce that
$\text{\small\bf H}q\equiv 0$ in all of $\mathbb{R}^{n}\times(-\infty,0)$.
Also, by Lemma 5.8(a) we know that $\lambda^{2nd}=2$. Then, using that
$\partial_{t}q\geq 0$ and (6.18), we easily conclude that $q$ must be of the
form (6.16).

(b) If $m=n-1$, then by Lemma 5.8(b) we have $\lambda^{2nd}\in[2,3]$.
Moreover, using the Lipschitz estimate in Lemma 6.5, we deduce that (up to
subsequence) $\tilde{w}_{r_{k}}\to q$ locally uniformly in
$\mathbb{R}^{n}\times(-\infty,0)$. Clearly $\partial_{t}q\geq 0$ (since
$\partial_{t}\tilde{w}_{r_{k}}\geq 0$). Moreover, $q$ is a solution of (6.17)
because $\text{\small\bf H}w_{r_{k}}\leq 0$ implies $\text{\small\bf
H}w_{r_{k}}\rightharpoonup^{*}\ \text{\small\bf H}q\leq 0$ as measures, and
thus $w_{r_{k}}\text{\small\bf H}w_{r_{k}}\geq 0$ implies $q\text{\small\bf
H}q\geq 0$. Also, $\text{\small\bf H}q\leq 0$ is supported on $\\{p_{2}=0\\}$
where $q\geq 0$ (since on $\\{p_{2}=0\\}$ we have $w_{r_{k}}=u_{r_{k}}\geq
0$).

Now, we notice that there are no $2$-homogeneous solutions $q$ to (6.17) and
satisfying (6.18) (see [FS17, Proposition 2.10] for a very similar argument),
and therefore it must be $\lambda^{2nd}>2$. Finally, the fact that
$\lambda^{2nd}\geq 2+\alpha_{\circ}$ can be proved either by compactness, or
using [Shi20, Proposition 9] to deduce that there are no $\lambda$-homogeneous
solutions to (6.17) for $\lambda\in(2,2+\alpha_{\circ})$. ∎

##  7\. The set $\Sigma_{n-1}^{<3}$

Let us define the sets

$\Sigma_{n-1}^{<3}:=\big{\\{}(x_{\circ},t_{\circ})\in\Sigma_{n-1}\ :\
\phi\big{(}0^{+},u(x_{\circ}+\,\cdot\,,t_{\circ}+\,\cdot\,)-p_{2,x_{\circ},t_{\circ}}\big{)}<3\big{\\}}$
(7.1)

and

$\Sigma_{n-1}^{=3}:=\Sigma_{n-1}\setminus\Sigma_{n-1}^{<3}.$ (7.2)

In this section we investigate the structure of the possible blow-ups $q$ at
points of $\Sigma_{n-1}^{<3}$. We start proving the following .

######  Lemma 7.1.

Let $(0,0)\in\Sigma_{n-1}^{<3}$ with $p_{2}=\frac{1}{2}x_{n}^{2}$. Let
$\tilde{w}_{r}$ be defined by (6.14) and suppose that
$\tilde{w}_{r_{k}}\rightharpoonup q$ in $W^{1,2}_{\rm
loc}(\mathbb{R}^{n}\times(-\infty,0])$, for some sequence $r_{k}\downarrow 0$.
Then:

  * (i)

$q$ is $\lambda^{2nd}$-homogeneous, where $\lambda^{2nd}\in(2,3)$, and
$\int_{\\{t=-1\\}}q^{2}G=1$;

  * (ii)

$q$ satisfies the growth estimates in (6.15);

  * (iii)

$q$ is a solution of (6.17);

  * (iv)

$\partial_{t}q\not\equiv 0$;

  * (v)

$\partial_{ee}q\geq 0$ in $\mathbb{R}^{n}$ for all $e\in\\{x_{n}=0\\}$ and
$\partial_{tt}q\geq 0$ in $\mathbb{R}^{n}$.

In the proof of Lemma 7.1(iv) we will use the following result, which gives a
classification of all convex solutions to the Signorini problem. (Notice that,
previously, this was only known for homogeneities $\lambda\leq 2$.)

######  Lemma 7.2.

Let $q=q(x)$ be a $\lambda$-homogeneous solution of the elliptic Signorini
problem, namely

$\Delta q\leq 0$ and $q\Delta q=0$ in $\mathbb{R}^{n}$, $\Delta q=0$ on
$\\{x_{n}\neq 0\\}$, $q\geq 0$ on $\\{x_{n}=0\\}$.

Assume in addition that $q$ is convex with respect to all directions parallel
to $\\{x_{n}=0\\}$. Then, either $\lambda=3/2$, or $q$ is a harmonic
polynomial and $\lambda$ is an integer.

###### Proof.

Since $q$ is homogeneous and convex with respect to the first $n-1$ variables,
this implies that the set $K:=\\{q=0\\}\subset\\{x_{n}=0\\}$ is convex. Now,
if $K$ has empty interior as a subset of $\\{x_{n}=0\\}$, then it will be
contained in a $(n-2)$-dimensional subspace, so it follows form a standard
capacity argument that $q$ must be harmonic in all of $\mathbb{R}^{n}$. Hence,
$q$ is a homogenous harmonic polynomial and $\lambda$ will be an even integer.
On the other hand, if $K=\\{x_{n}=0\\}$, then $q$ would be an odd harmonic
polynomial and $\lambda$ will be an odd integer.

Consider now the case when $K\neq\\{x_{n}=0\\}$ has non-empty interior inside
$\\{x_{n}=0\\}$. Up to replacing $q(x^{\prime},x_{n})$ by
$q(x^{\prime},x_{n})+q(x^{\prime},-x_{n})$, we may assume that $q$ is even.
Pick a direction $e\in\mathbb{S}^{n-1}\cap\\{x_{n}=0\\}$ such that $-e\in{\rm
int}(K)$. Then, for any point $z^{\prime}\in\mathbb{R}^{n-1}$ we have
$(z^{\prime},0)-et\in K$ for $t>0$ large enough. We claim that this implies
that $\partial_{e}q\geq 0$ in all $\mathbb{R}^{n}$. Indeed, if it was
$\partial_{e}q(z)<0$ at some point $z=(z^{\prime},z_{n})$ with $z_{n}>0$ then,
by convexity of $q$ in the direction $e=(e^{\prime},0)$, we would have

$q(z^{\prime}-se^{\prime},z_{n})\geq
q(z^{\prime},z_{n})-\partial_{e}q(z^{\prime},z_{n})s\quad\mbox{for all }s>0.$

Hence, $q(z^{\prime}-s_{\circ}e^{\prime},z_{n})>0$ for $s=s_{\circ}$
sufficiently large. However, since $(z-s_{\circ}e^{\prime},0)\in K$, then
(recall that $q$ is even)

$q(z-s_{\circ}e^{\prime},0)=0\quad\mbox{and}\quad\partial_{n}q(z-s_{\circ}e^{\prime},0^{+})\leq
0.$ (7.3)

Since $\Delta q\leq 0$ and $\partial_{ii}q\geq 0$ for all $i\leq n-1$ we have
that $q$ is concave in the variable $x_{n}$, a contradiction.

Once we have shown that $\partial_{e}q\geq 0$ in all $\mathbb{R}^{n}$ for all
$e\in-{\rm int}(K)$, we can use the argument from [CRS17, RS17] in order to
classify blow-ups. Indeed, we look at two different directions in the
monotonicity cone, $e$ and $e^{\prime}$, and observe that
$\partial_{e}q|_{\mathbb{S}^{n-1}}$ and
$\partial_{e^{\prime}}q|_{\mathbb{S}^{n-1}}$ are both positive harmonic
functions in the same cone, and they both vanishing on the boundary.
Therefore, by uniqueness of positive harmonic functions in cones, one must be
a multiple of the other. This argument, applied to $n-1$ independent
directions of the monotonicity cone, leads to the fact that $q$ depends only
on two euclidean variables, namely, $x_{n}$ and a linear combination of
$x_{1},\dots,x_{n-1}$. One then reduces the classification problem to the
well-known situation of $\mathbb{R}^{2}$, in which all homogeneous solutions
are explicit and the only convex ones have homogeneity $3/2$. This concludes
the proof. ∎

We can now prove Lemma 7.1.

###### Proof of Lemma 7.1.

Note that (i)-(iii) follow from Proposition 6.7.

To show (iv), we recall that, by Proposition 3.7, we have

$(D^{2}u)_{-}\leq C\,\partial_{t}u.$

Hence, if $e\in\\{x_{n}=0\\}$,

$\big{(}\partial_{ee}(u-p_{2})\big{)}_{-}=(\partial_{ee}u)_{-}\leq
C\,\partial_{t}u=C\,\partial_{t}(u-p_{2}),$

and, as a consequence,

$(\partial_{ee}\tilde{w}_{r})_{-}\leq C\,\partial_{t}\tilde{w}_{r},\quad\mbox{
and thus }(\partial_{ee}q)_{-}\leq C\,\partial_{t}q.$ (7.4)

Assume now by contradiction that $\partial_{t}q\equiv 0$. Then $q=q(x)$ is
convex with respect to all directions in $\\{x_{n}=0\\}$. However, using Lemma
7.2, we obtain a contradiction with the fact that, by assumption, $q$ is
homogeneous of degree $\lambda\in(2,3)$.

We now prove (v). By differentiating (6.17) with respect to $t$, and using
that $q$ is homogeneous, it follows that $\varphi_{1}:=\partial_{t}q(x,-1)$
solves

$\mathcal{L}_{OU}\varphi_{1}=\lambda_{1}\varphi_{1}\quad\mbox{in
}\mathbb{R}^{n}\setminus\big{(}\\{x_{n}=0\\}\cap\\{q=0\\}\big{)},\qquad\varphi_{1}=0\quad\mbox{on
}\\{x_{n}=0\\}\cap\\{q=0\\},$ (7.5)

where $\lambda_{1}:=\frac{\lambda^{2nd}-2}{2}$, and $\mathcal{L}_{OU}$ is as
in (5.10). Let $dm=G(x,-1)dx$ be the Gaussian measure. Since $\varphi_{1}$
belongs to $H^{1}(\mathbb{R}^{n},dm)$ (as a consequence of the estimates in
(6.15)), then $\varphi_{1}$ must be the first eigenfunction of
$\mathcal{L}_{OU}$, and $\lambda_{1}$ the first eigenvalue.

Also, as shown in the beginning of the proof of Lemma 6.5, the function
$\big{(}\partial_{ee}(u-p_{2})\big{)}_{-}\geq 0$ is subcaloric. Therefore,
$f_{r}:=(\partial_{ee}\tilde{w}_{r})_{-}$ satisfies $f_{r}\,\text{\small\bf
H}f_{r}\geq 0$, and we obtain (cp. (6.7))

$R^{2}{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\mathscr{C}_{R}}|\nabla
f_{r}|^{2}\leq C_{n}{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\mathscr{C}_{2R}}f_{r}^{2}.$

Since $f_{r}\leq C\,\partial_{t}\tilde{w}_{r}$ (by (7.4)), the estimates in
(6.15) yield

$\bigg{(}R^{2}{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\mathscr{C}_{R}}|\nabla(\partial_{ee}\tilde{w}_{r})_{-}|^{2}\bigg{)}^{\frac{1}{2}}\leq
C\bigg{(}{\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{\mathscr{C}_{2R}}|\partial_{t}\tilde{w}_{r}|^{2}\bigg{)}^{\frac{1}{2}}\leq
C_{\delta}R^{\lambda^{2nd}+\delta-2}.$ (7.6)

Set $\psi:=\big{(}\partial_{ee}q(x,-1)\big{)}_{-}$. Then, $\psi$ is a
$(\lambda^{2nd}-2)$-homogeneous nonnegative subsolution of (7.5). Also, since
$\tilde{w}_{r_{k}}\to q$ in $L^{2}_{\rm loc}$, it follows by (7.6) that $\psi$
belongs to $H^{1}(\mathbb{R}^{n},dm)$. By uniqueness of the first
eigenfunction, it follows that $\psi$ is a nonnegative multiple of
$\varphi_{1}$.

We claim that $\psi\equiv 0$. Indeed, since $\varphi_{1}$ is the first
eigenfuction, it is positive everywhere (except on the boundary). Hence, if by
contradiction $\psi$ is not zero everywhere, then we get
$(\partial_{ee}q(x,-1)\big{)}_{-}>0$ in
$\mathbb{R}^{n}\setminus\big{(}\\{x_{n}=0\\}\cap\\{q(x,-1)=0\\}\big{)}$, and
in particular $\partial_{ee}q<0$ on $\\{x_{n}=0\\}\setminus\\{q(x,-1)=0\\}$.
However, since $q=|\nabla q|=0$ on $\\{x_{n}=0\\}\cap\partial\\{q(x,-1)=0\\}$,
this contradicts the fact that $q\geq 0$ on $\\{x_{n}=0\\}$. Hence $\psi\equiv
0$, or equivalently $\partial_{ee}q\geq 0$, as desired.

Finally, to show that $\partial_{tt}q\geq 0$, we recall that, thanks to
Proposition 3.4, we have

$\partial_{tt}(u-p_{2})=\partial_{tt}u\geq-C.$

After scaling, we obtain

$\partial_{tt}\tilde{w}_{r}\geq-C\frac{r^{4}}{H\big{(}r,(u-p_{2})\zeta\big{)}^{1/2}}=o(1)\qquad\mbox{as
}r\downarrow 0,$

and letting $r\to 0$ the result follows. ∎

We can now prove a “dimension reduction” lemma for the set
$\Sigma_{n-1}^{<3}$. We say that a set of points $\\{y_{i}\\}_{i\in I}$ has
rank at most $m$ if, for any finite subset $\\{y_{i_{j}}\\}_{1\leq j\leq N}$,
it holds ${\rm span}(y_{i_{1}},\ldots,y_{i_{N}})\leq m$.

######  Lemma 7.3.

Let $(0,0)\in\Sigma_{n-1}^{<3}$ with $p_{2}=\frac{1}{2}x_{n}^{2}$, let
$\tilde{w}_{r}$ be defined as in (6.14), and suppose that
$\tilde{w}_{r_{k}}\to q$ in $L^{2}_{\rm
loc}(\mathbb{R}^{n}\times(-\infty,0])$, for some sequence $r_{k}\downarrow 0$.
Suppose in addition that $(x_{k}^{(j)},t_{k}^{(j)})$ are sequences of singular
points with $t_{k}^{(j)}\geq 0$ such that
$y_{k}^{(j)}:=\frac{x_{k}^{(j)}}{r_{k}}\to y_{\infty}^{(j)}\neq 0$. Then
$y_{\infty}^{(j)}\in\\{x_{n}=0\\}$, and the set
$\big{\\{}y_{\infty}^{(j)}\big{\\}}_{j}$ has rank at most $n-2$.

Before giving the proof of Lemma 7.4 we need a couple of auxiliary lemmas.

######  Lemma 7.4.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1). Then:

(a) The singular set is closed —more precisely
$\Sigma\cap\overline{B}_{\varrho}$ is closed for any $\varrho<1$. Moreover,

$\Sigma\cap\overline{B}_{\varrho}\ni(x_{k},t_{k})\to(x_{\infty},t_{\infty})\qquad\Rightarrow\qquad
p_{2,x_{k},t_{k}}\to p_{2,x_{\infty},t_{\infty}}.$

(b) The function

$\Sigma\ni(x_{\circ},t_{\circ})\mapsto\phi(0^{+},u(x_{\circ}+\,\cdot\,\,,t_{\circ})-p_{2,x_{\circ},t_{\circ}})$

is upper semi-continuous.

###### Proof of Lemma 7.4.

(a) In [Caf77], Caffarelli proved that the regular set is relatively open
inside the free boundary, hence the singular set is closed. Moreover, it
follows from Lemma 5.6(a) that $p_{2,x_{k},t_{k}}\to
p_{2,x_{\infty},t_{\infty}}$ whenever
$(x_{k},t_{k})\to(x_{\infty},t_{\infty})$ (cp. [Fig18b, Theorem 8.1]).

(b) Fixed $\gamma>3$, this follows from the (almost) monotonicity of the
frequency $\phi^{\gamma}$ and Corollary 5.9. ∎

######  Lemma 7.5.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a bounded solution of (3.1),
$(x_{k},t_{k})\in\Sigma$, and assume that $(x_{k},t_{k})\to(0,0)\in\Sigma$ .
Also, assume that $t_{k}\geq-C^{\prime}|x_{k}|^{2}$ for some $C^{\prime}>0$.
Then

$\textstyle{\rm dist}\left(\frac{x_{k}}{|x_{k}|},\\{p_{2}=0\\}\right)\to
0\qquad\mbox{as $k\to\infty$}.$

###### Proof.

Let $r_{k}:=|x_{k}|$, $y_{k}:=x_{k}/r_{k}$, and $s_{k}:=t_{k}/r_{k}^{2}$ . We
know that $U_{k}:=r_{k}^{-2}u_{r_{k}}$ converges locally uniformly to $p_{2}$
as $k\to\infty$. Moreover, since $U_{k}$ has a singular point at
$(s_{k},t_{k})$ and $U_{k}$ is increasing in time, it follows that
$U_{k}(y_{k},s)=0$ for all $s\leq s_{k}$. In particular, thanks to our
assumption, $U_{k}(y_{k},-C^{\prime})=0$. This implies that, for any
accumulation point $y_{\infty}$ of $\\{y_{k}\\}$, we must have
$p_{2}(y_{\infty})=0$. The lemma follows. ∎

We can now prove Lemma 7.3.

###### Proof of Lemma 7.3.

The fact that $y_{\infty}^{(j)}\in\\{x_{n}=0\\}$ follows from Lemma 7.5. Also,
since $(x_{k}^{(j)},t_{k}^{(j)})$ is a singular point, we have
$u(x_{k}^{(j)},t_{k}^{(j)})=0$. Hence, since $t_{k}\geq 0$ and $u$ is
increasing in time, we deduce that $(u-p_{2})(x_{k}^{(j)},t)\leq 0$ for all
$t\leq 0$, and therefore

$\tilde{w}_{r_{k}}(y_{k}^{(j)},t)\leq 0\quad\mbox{for all }t\leq 0.$

Since $\tilde{w}_{r_{k}}\to q$ locally uniformly, it follows that
$q(y_{\infty}^{(j)},t)\leq 0$, but since $y_{\infty}^{(j)}\in\\{x_{n}=0\\}$
(where $q$ is nonnegative, see (6.17)), it must be

$q(y_{\infty}^{(j)},t)\equiv 0\quad\mbox{for all }t\leq 0.$

Now, the fact that $q$ is homogeneous, monotone in $t$, and convex with
respect to the first $n-1$ variables (see Lemma 7.1), implies that the cone
$K\subset\\{x_{n}=0\\}$ generated by $y_{\infty}^{(j)}$ is contained inside
$\\{q=0\\}\cap\\{x_{n}=0\\}$.

Assume now by contradiction that the rank of
$\big{\\{}y_{\infty}^{(j)}\big{\\}}_{j}$ is $n-1$. Then $K$ has non-empty
interior inside $\\{x_{n}=0\\}$. In particular, if we pick a direction
$e\in\mathbb{S}^{n-1}\cap\\{x_{n}=0\\}$ such that $-e\in{\rm int}(K)$, then
for any point $z^{\prime}\in\mathbb{R}^{n-1}$ we have $(z^{\prime},0)-se\in K$
for $s>0$ large enough.

Set $\hat{q}(x):=q(x,-1)$. We claim that

$\partial_{e}\hat{q}\geq 0\qquad\textrm{inside}\quad\mathbb{R}^{n}$

for any such direction $e$. Indeed, suppose by contradiction that
$\partial_{e}\hat{q}(z,-1)<0$ at some point $z=(z^{\prime},z_{n})$ with (with
no loss of generality) $z_{n}>0$. Then, by convexity of $q$ in the direction
$e=(e^{\prime},0)$, we have

$\hat{q}(z^{\prime}-se^{\prime},z_{n})\geq\hat{q}(z^{\prime},z_{n})-\partial_{e}\hat{q}(z^{\prime},z_{n})s.$

Hence for all $s>s_{\circ}$ (with $s_{\circ}$ large enough depending on $z$)
we have

$\hat{q}(z^{\prime}-se^{\prime},z_{n})\geq c_{\circ}s\qquad\mbox{where
}c_{\circ}>0,$ (7.7)

and

$(z^{\prime}-se^{\prime},0)\in K\subset\\{\hat{q}=0\\}\cap\\{x_{n}=0\\}.$
(7.8)

However, since $q$ is convex in the first $n-1$ variables, we have

$\partial_{nn}q\leq\Delta q\leq\partial_{t}q\leq
C_{\delta}R^{\lambda^{2nd}-2+\delta}\quad\mbox{in }B_{R}\times(-R^{2},0).$
(7.9)

Therefore, since $\hat{q}(z^{\prime}-se^{\prime},0)=0$ and
$\partial_{n}\hat{q}(z^{\prime}-se^{\prime},0^{+})\geq 0$ (by (7.8) and
(6.17)), integrating (7.9) along the segment joining
$(z^{\prime}-se^{\prime},0)$ and $(z^{\prime}-se^{\prime},z_{n})$ we obtain

$\hat{q}(z^{\prime}-se^{\prime},z_{n})\leq
C_{\delta}R^{\lambda^{2nd}-2+\delta}\,\frac{|z_{n}|^{2}}{2}\qquad\mbox{whenever}\quad\big{|}(z^{\prime}-se^{\prime},z_{n})\big{|}\leq
R.$

This implies that $\hat{q}(z^{\prime}-se^{\prime},z_{n})\leq
C_{\delta}s^{\lambda^{2nd}-2+\delta}|z_{n}|^{2}$ for $s$ sufficiently large.
However, since $\lambda^{2nd}<3$, choosing $\delta>0$ such that
$\lambda^{2nd}-2+\delta<1$, this contradicts (7.7).

Thus, we have shown that $\partial_{e}\hat{q}\geq 0$. Then, differentiating
(6.17) with respect to $e$ and using that $q$ is homogeneous, we deduce that
$\psi:=\partial_{e}\hat{q}$ is a nonnegative solution of

$\mathcal{L}_{OU}\psi=\lambda\psi\quad\mbox{in
}\mathbb{R}^{n}\setminus\big{(}\\{x_{n}=0\\}\cap\\{\hat{q}=0\\}\big{)},\qquad\psi=0\quad\mbox{on
}\\{x_{n}=0\\}\cap\\{\hat{q}=0\\}.$

Also, since $\partial_{e}q$ is $(\lambda^{2nd}-1)$-homogeneous, its associated
eigenvalue is $\lambda=\frac{\lambda^{2nd}-1}{2}$. On the other hand, we
showed in the proof of Lemma 7.1(v) that the positive function
$\varphi_{1}(x):=\partial_{t}q(x,-1)$ is the first eigenfunction of this
problem and has eigenvalue $\frac{\lambda^{2nd}-2}{2}$. Since the first
eigenfunction is the only one which does not change sign, this provides the
desired contradiction. ∎

We can now prove that $\pi_{x}\big{(}\Sigma_{n-1}^{<3}\big{)}$ has Hausdorff
dimension at most $n-2$ (recall (2.3)). Note that this is not standard, since
Lemma 7.3 only shows that, around any point
$(x_{\circ},t_{\circ})\in\Sigma_{n-1}^{<3}$, the set
$\pi_{x}(\Sigma_{n-1}^{<3}\cap\\{t\geq t_{\circ}\\}\big{)}\cap
B_{r}(x_{\circ})$ is contained in a $\varepsilon$-neighborhood of some
$(n-2)$-plane. The fact that we can only control points “in the future”
$\\{t\geq t_{\circ}\\}$ forces us to prove some new appropriate GMT result
that incorporates this feature. We state it as an abstract result.

######  Proposition 7.6.

Let $E\subset\mathbb{R}^{n}\times\mathbb{R}$, and suppose that for all
$(x,t)\in E$ and $\varepsilon>0$ there exists $\varrho_{x,t,\varepsilon}>0$
such that, for all $r\in(0,\varrho_{x,t,\varepsilon})$,

$\pi_{x}\big{(}E\cap(\overline{B_{r}}(x)\times[t,\infty])\big{)}\subset(x+L+\overline{B_{r\varepsilon}})\quad\mbox{for
some $L\subset\mathbb{R}^{n}$ linear space, with ${\rm dim}(L)=m$}.$ (7.10)

Then

${\rm dim}_{\mathcal{H}}\big{(}\pi_{x}(E)\big{)}\leq m.$

To prove this proposition, we need the following classical GMT lemma, whose
proof can be found for instance in [Ros21, Lemma 9].

######  Lemma 7.7.

Given $\alpha>0$ there exists $\varepsilon>0$, depending only on $n$ and
$\alpha$, such that the following holds: Let $F\subset\mathbb{R}^{n}$, and
suppose that there exists $\varrho>0$ such that, for all $x\in F$ and
$r\in(0,\varrho)$,

$F\cap\overline{B_{r}}(x)\subset(x+L+\overline{B_{r\varepsilon}})\quad\mbox{for
some $L\subset\mathbb{R}^{n}$ linear space, with ${\rm dim}(L)=m$}.$

Then $\mathcal{H}^{m+\alpha}(F)=0$.

To do this, for any arbitrarily small $\delta>0$ we will construct a cover of
$\pi_{x}(E)$ by countably many balls $B_{r_{i}}(x_{i})$ of diameter
$\leq\delta$ such that $\sum_{i}r_{i}^{m+\alpha}$ can be made arbitrarily
small. This implies that $\mathcal{H}^{m+\alpha}(\pi_{x}(E))=0$ for all
$\alpha>0$ and hence ${\rm dim}_{\mathcal{H}}\big{(}\pi_{x}(E)\big{)}\leq m$.

We can now prove Proposition 7.6.

###### Proof of Proposition 7.6.

Up to taking countable unions, we may assume that
$E\subset\overline{B_{1/2}}(z)\times[-1,1]$. Also, since

$E=\bigcup_{\ell}E_{\ell},\qquad\textrm{where}\quad E_{\ell}=\big{\\{}(x,t)\in
E:\rho_{x,t,\varepsilon}>{\textstyle\frac{1}{\ell}}\big{\\}},$

it suffices to prove that ${\rm
dim}_{\mathcal{H}}\big{(}\pi_{x}(E_{\ell})\big{)}\leq m$ for each $\ell$. To
this aim, given $\alpha>0$ arbitrarily small, we shall prove that
$\mathcal{H}^{m+\alpha}\big{(}\pi_{x}(E_{\ell})\big{)}=0$.

Fix $\varepsilon>0$ (depending only on $n$ and $\alpha$, to be chosen later).
By assumption, for all $(x,t)\in E_{\ell}$ and for all $r\in(0,1/\ell]$,

$\pi_{x}\big{(}E_{\ell}\cap(\overline{B_{r}}(x)\times[t,\infty])\big{)}\subset(x+L_{x,t,r}+\overline{B_{r\varepsilon}})\quad\mbox{for
some $L_{x,t,r}$ linear and $m$-dimensional}.$ (7.11)

Note that (7.11) holds also for all $(\bar{x},\bar{t})$ in the closure
$\overline{E_{\ell}}$ of $E_{\ell}$, provided we define
$L_{\bar{x},\bar{t},r}$ as an arbitrary limit of the hyperplanes
$L_{x,t,\varepsilon}$ as $(x,t)\in E_{\ell}$ converges to $(\bar{x},\bar{t})$.

Now, given $(x,t)\in E_{\ell}$ and $r\leq 1/\ell$, by compactness there exists
$(\bar{x},\bar{t})\in\overline{B_{r}}(x)\times[-1,1]\cap\overline{E_{\ell}}$
such that
$\bar{t}\leq\pi_{t}\big{(}\overline{B_{r}}(x)\times[-1,1]\cap\overline{E_{\ell}}\big{)}\subset\mathbb{R}$.
Thus, applying (7.11) at the point $(\bar{x},\bar{t})$, we get

$\pi_{x}\big{(}E_{\ell}\big{)}\cap
B_{r}(x)\subset\pi_{x}\left(\overline{E_{\ell}}\cap(\overline{B_{r}}(\bar{x})\times[\bar{t},\infty))\right)\subset\bar{x}+L_{\bar{x},\bar{t},r}+\overline{B_{r\varepsilon}}.$

Thus, choosing $\varepsilon$ sufficiently small, Lemma 7.7 applied with
$F=\pi_{x}(E_{\ell})$ implies that
$\mathcal{H}^{m+\alpha}\big{(}\pi_{x}(E_{\ell})\big{)}=0$, as desired. ∎

Finally, as an immediate consequence of Lemma 7.3 and Proposition 7.6, we
deduce the following:

######  Proposition 7.8.

The set $\Sigma_{n-1}^{<3}$ satisfies ${\rm
dim}_{\mathcal{H}}\big{(}\pi_{x}(\Sigma_{n-1}^{<3})\big{)}\leq n-2$.

##  8\. Quadratic cleaning of the singular set and proof of Theorem 1.1

In this section we prove that, for any given $\varepsilon>0$, there exists
$C_{\varepsilon}$ such that, for any singular point
$(x_{\circ},t_{\circ})\in\Sigma$, it holds

$\Sigma\cap\big{\\{}|x-x_{\circ}|\leq r,\ t\geq
t_{\circ}+Cr^{2-\varepsilon}\big{\\}}=\emptyset.$ (8.1)

As we shall see, this estimate will allow us to prove Theorem 1.1.

We begin with the following:

######  Lemma 8.1.

Let $u:B_{1}\times(-1,1)\to[0,\infty)$ be a solution of (3.1) and $(0,0)$ a
singular point. Assume that $H(r,u-p_{2})^{1/2}\leq\omega(r)$, where
$\omega(r)=o(r^{2})$ as $r\downarrow 0$. Suppose in addition that
$\boldsymbol{e}_{n}$ is an eigenvector $D^{2}p_{2}$ with maximal eigenvalue,
and that there exists $c>0$ such that

${\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{B_{r}\cap\\{|x_{n}|\geq\frac{r}{10}\\}}\partial_{t}u(\cdot,-r^{2})\geq
c\,r^{\beta}\quad\mbox{ for all $r\in(0,1)$, for some $\beta\in(0,1]$}.$ (8.2)

Then there exists $C>0$ such that

$\\{u=0\\}\cap\big{(}B_{r/2}\times[C\omega(r)r^{-\beta}+r^{2},1)\big{)}=\emptyset\qquad\forall\,r\in(0,1).$

###### Proof.

Since $\text{\small\bf H}(u-p_{2})=-\chi_{\\{u=0\\}}\leq 0$, the function
$u-p_{2}$ is supercaloric. Also, since $\partial_{t}u\geq 0$, then $u-p_{2}$
is nondecreasing in time. Thus, thanks to the bound
$H(r,u-p_{2})^{1/2}\leq\omega(r)$ we get

$u\geq p_{2}-C_{1}\omega(r)\quad\textrm{in}\quad B_{r}\times[-r^{2}/2,1).$
(8.3)

In particular, since $\omega(r)=o(r^{2})$ and $\boldsymbol{e}_{n}$ is an
eigenvector $D^{2}p_{2}$ with maximal eigenvalue, for any fixed $\delta>0$
small, we obtain

$\\{u=0\\}\cap\big{(}B_{r}\times[-r^{2}/2,1)\big{)}\subset\\{|x_{n}|\leq
r\delta^{2}\\}\qquad\forall\,r\ll 1.$ (8.4)

Thus $\text{\small\bf H}(\partial_{t}u)=0$ inside
$\big{(}B_{r}\cap\\{|x_{n}|>r\delta^{2}\\}\big{)}\times[-r^{2}/2,1)$, and
therefore (8.2) and Harnack inequality imply that
$\partial_{t}u(\cdot,-r^{2}/4)\geq 2c_{2}r^{\beta}$ inside
$B_{r}\cap\\{|x_{n}|>r\delta\\}$, for some $c_{2}=c_{2}(n,\delta)>0$.

Combining this bound with the estimate $\partial_{tt}u\geq-C$ (see Proposition
3.4), we get

$\partial_{t}u\geq c_{2}r^{\beta}\quad\mbox{in
}\big{(}B_{r}\cap\\{|x_{n}|>r\delta\\}\big{)}\times[-r^{2}/4,c_{3}r],$ (8.5)

for some $c_{3}>0$ (recall that $\beta\leq 1$). In particular, combining (8.3)
and (8.5), we obtain

$u(\cdot,-r^{2}/4+h)\geq p_{2}-C_{1}\omega(r)+c_{2}r^{\beta}h\qquad\mbox{in
}B_{r}\cap\\{|x_{n}|>r\delta\\},$ (8.6)

for all $h\in[0,c_{3}r]$. Choosing $h\geq
r^{2}/4+2C_{1}c_{2}^{-1}\omega(r)r^{-\beta}$, and using again (8.3), since $u$
is nondecreasing we obtain

$u\geq
p_{2}+C_{1}\omega(r)\bigl{(}-1+2\chi_{\\{|x_{n}|>r\delta\\}}\bigr{)}\qquad\forall\,(x,t)\in
B_{r}\times[2C_{1}c_{2}^{-1}\omega(r)r^{-\beta},1).$ (8.7)

Now, let $h^{\delta}$ be the solution to

$\begin{cases}\text{\small\bf H}h^{\delta}=0\quad&\mbox{in
}B_{1}\times(0,\infty)\\\ h^{\delta}=2&\mbox{on }\big{(}\partial
B_{1}\cap\\{|x_{n}|>\delta\\}\big{)}\times[0,\infty)\\\ h^{\delta}=0&\mbox{on
}\big{(}\partial B_{1}\cap\\{|x_{n}|<\delta\\}\big{)}\times[0,\infty)\\\
h^{\delta}=0&\mbox{at }t=0.\end{cases}$

Since $h^{\delta}\to 2$ as $\delta\to 0$, it follows that
$h^{\delta}\geq\frac{3}{2}$ inside $B_{1/2}$ for all $t\geq 1$, provided
$\delta$ is small enough. Now we can observe that

$\psi(x,t):=p_{2}(x)+C_{1}\omega(r)\bigg{(}-1+h^{\delta}\Big{(}\frac{x}{r},\frac{t-2C_{1}c_{2}^{-1}\omega(r)r^{-\beta}}{r^{2}}\Big{)}\bigg{)}$

satisfies $\text{\small\bf H}\psi=1$ in
$B_{r}\times[2C_{1}c_{2}^{-1}\omega(r)r^{-\beta},\infty)$ and, by (8.7), we
have $u\geq\psi$ on the parabolic boundary
$\partial_{par}\bigl{(}B_{r}\times[2C_{1}c_{2}^{-1}\omega(r)r^{-\beta},1)\bigr{)}$.
Hence, by the maximum principle,

$u\geq\psi\quad\mbox{in }B_{r},\mbox{ for }t\geq
2C_{1}c_{2}^{-1}\omega(r)r^{-\beta}.$

Evaluating at $t=2C_{1}c_{2}^{-1}\omega(r)r^{-\beta}+r^{2}$ (and using that
$h^{\delta}\geq\frac{3}{2}$ in $B_{1/2}$ for all $t\geq 1$) we obtain

$u\geq\psi=p_{2}+\frac{C_{1}}{2}\omega(r)>0\qquad\mbox{in }B_{r/2},\mbox{ for
}t\geq 2C_{1}c_{2}^{-1}\omega(r)r^{-\beta}+r^{2},$

and the result follows. ∎

The (almost) quadratic cleaning (8.1) will be proved by applying Lemma 8.1,
with different $\omega$ and $\beta$, in each of the following cases (recall
(5.7), (7.1), and (7.2)):

  * (i)

If $(x_{\circ},t_{\circ})\in\Sigma_{m}$ for some $m\leq n-2$, we will use
$\omega(r)=o(r^{2})$ and $\beta=\varepsilon$;

  * (ii)

if $(x_{\circ},t_{\circ})\in\Sigma_{n-1}^{<3}$, we will use
$\omega(r)=r^{\lambda^{2nd}}$ and $\beta=\lambda^{2nd}-2+\varepsilon$;

  * (iii)

if $(x_{\circ},t_{\circ})\in\Sigma_{n-1}^{=3}$, we will use $\omega(r)=r^{3}$
and $\beta=1$ (in this case we can prove the exact quadratic cleaning without
loosing $\varepsilon$ in the exponent)

We start with the easiest case (i).

######  Proposition 8.2.

Let $(0,0)\in\Sigma_{m}$. Then, for any $\varepsilon>0$ there exists
$r_{\varepsilon}>0$ such that

$\\{u=0\\}\cap\bigl{(}B_{r}\times[r^{2-\varepsilon},1)\bigr{)}=\emptyset\qquad\forall\,r\in(0,r_{\varepsilon}).$

###### Proof.

As shown in the proof of Lemma 5.8(a) (see (5.9)), we have ${\mathchoice{\kern
4.30554pt\vbox{\hrule height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern
4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{B_{r}}\partial_{t}u(\cdot,-r^{2})\geq
c_{\varepsilon}r^{\varepsilon}$ in $B_{r}$, for all $r$ sufficiently small.
Then, the result follows using Lemma 8.1 with $\omega(r)=o(r^{2})$ and
$\beta=\varepsilon$. ∎

We now consider the case (iii), which is also easier than (ii) .

######  Proposition 8.3.

Let $\varrho>0$, and let
$(x_{\circ},t_{\circ})\in\Sigma_{n-1}^{=3}\cap\bigl{(}B_{1-\varrho}\times(-1+\varrho^{2},1)\bigr{)}$.
Then

$\\{u=0\\}\cap\big{(}B_{r}(x_{\circ})\times[t_{\circ}+Cr^{2},1)\big{)}=\emptyset\qquad\forall\,r\in(0,\varrho),$

where $C$ is independent of $(x_{\circ},t_{\circ})$.

To prove Proposition 8.3, we will need the following result.

######  Lemma 8.4.

Let $(x_{\circ},t_{\circ})\in\Sigma_{n-1}^{=3}\cap
B_{1-\varrho}\times(-1+\varrho^{2},1)$, and assume (with no loss of
generality, up to a rotation in space) that
$p_{2,x_{\circ},t_{\circ}}=\frac{1}{2}x_{n}^{2}$. There exist positive
$\hat{c}$ and $\hat{r}$, independent of $(x_{\circ},t_{\circ})$, such that

${\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{B_{r}\cap\\{|x_{n}|\geq\frac{r}{10}\\}}\partial_{t}u(x_{\circ}+\cdot,t_{\circ}-r^{2})\geq\hat{c}\,r\qquad\forall\,r\in(0,\hat{r}).$
(8.8)

###### Proof.

Proposition 6.6 applied with $\lambda^{2nd}=3$ to the function
$\varrho^{-2}u(x_{\circ}+\varrho\,\cdot\,,t_{\circ}+\varrho^{2}\,\cdot\,,)$
implies that

$\\{u(x_{\circ}+\,\cdot\,,t_{\circ}+\,\cdot\,)=0\\}\cap
B_{\varrho}\subset\bigl{\\{}|x_{n}|\leq
C_{1}(|x^{\prime}|^{2}-t)\bigr{\\}}\qquad\forall\,t\in[-\varrho^{2},0].$ (8.9)

where $C_{1}$ depends only on $n$, $\|u\|_{L^{\infty}}$, and $\varrho$. We now
consider a barrier of the form

$\phi(x,t):=g\big{(}x_{n}-C_{1}(|x^{\prime}|^{2}-t)\big{)}.$

Note that

$\text{\small\bf
H}\phi=\big{(}1+4C_{1}^{2}|x^{\prime}|^{2}\big{)}g^{\prime\prime}-C_{1}(2n-1)g^{\prime}.$

Hence, choosing
$g(s)=\left(e^{As}-1\right)\chi_{(0,1)}(s)+\left(e^{A}-1\right)\chi_{(1,\infty)}(s)$
with $A$ large (as in the standard barrier from Hopf’s Lemma), we see that
$\phi\geq 0$ is subcaloric inside the domain
$D:=\\{x_{n}-C_{1}(|x^{\prime}|^{2}-t)<1\\}.$ Hence, since $\partial_{t}u$ is
positive and caloric inside $\\{u>0\\}$, it follows by (8.9) and the maximum
principle that

$\partial_{t}u\geq\bar{c}\,\phi\qquad\text{in
}B_{\hat{r}}(x_{\circ})\times[t_{\circ}-\hat{r}^{2},t_{\circ}],$

where $\bar{c},\hat{r}>0$ may depend on $u$ and $\varrho$, but may be chosen
independently of $(x_{\circ},t_{\circ})$. Using the explicit formula for
$\phi$, (8.8) follows easily. ∎

As a consequence, we get:

######  Corollary 8.5.

Let $(x_{\circ},t_{\circ})\in\Sigma_{n-1}^{=3}\cap
B_{1-\varrho}\times(-1+\varrho,1)$. Then

$\frac{r^{3}}{C}\leq
H\big{(}r,(u(x_{\circ}+\,\cdot\,,t_{\circ}+\,\cdot\,)-p_{2})\big{)}^{1/2}\leq
Cr^{3}$

for all $r\in(0,\varrho)$, where $C>0$ may depend on $u$ and $\varrho$, but is
independent of $(x_{\circ},t_{\circ})$.

###### Proof.

Choosing $\gamma=4$, Corollary 5.9 and Lemma 5.6(b) applied with $R=\varrho$
imply that

$H(r,u(x_{\circ}+\,\cdot\,,t_{\circ}+\,\cdot\,)-p_{2,x_{\circ},t_{\circ}})\leq
Cr^{6}\qquad\forall\,r\in(0,\varrho).$

Viceversa, the opposite inequality follows easily by integrating in time the
estimate in Lemma 8.4, using that $\partial_{t}(u-p_{2})=\partial_{t}u\geq
0$.∎

We can now prove Proposition 8.3.

###### Proof of Proposition 8.3.

In view of Lemma 8.4, the result follows from Lemma 8.1 with
$\omega(r)=Cr^{3}$ and $\beta=1$. ∎

Finally, we consider the remaining case (ii).

######  Proposition 8.6.

Let $(0,0)\in\Sigma_{n-1}^{<3}$. Then, for any $\varepsilon>0$ there exists
$r_{\varepsilon}>0$ such that

$\\{u=0\\}\cap\big{(}B_{r}\times[r^{2-\varepsilon},1)\big{)}=\emptyset\qquad\textrm{for
all}\quad r\in(0,r_{\varepsilon}).$

To prove Proposition 8.6 we will need the following:

######  Lemma 8.7.

Let $(0,0)\in\Sigma_{n-1}^{<3}$, and assume that $p_{2}=\frac{1}{2}x_{n}^{2}$.
Then, for any $\varepsilon>0$ there exists $r_{\varepsilon}>0$ such that

${\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{B_{r}\cap\\{|x_{n}|\geq\frac{r}{10}\\}}\partial_{t}u(\,\cdot\,,-r^{2})\geq
r^{\lambda^{2nd}-2+\varepsilon}\quad\mbox{ in }B_{r},\qquad\textrm{for
all}\quad r\in(0,r_{\varepsilon}).$

###### Proof.

Let $w=u-p_{2}$. For any $r_{k}\downarrow 0$ there exists a subsequence
$r_{k_{\ell}}\downarrow 0$ such that

$\tilde{w}_{r_{k_{\ell}}}=\frac{w_{r_{k_{\ell}}}}{H(r_{k_{\ell}},w\zeta)^{1/2}}\rightarrow
q,$

where $q$ satisfies the properties stated in Lemma 7.1. In particular, for any
such limit $q$, we have $\partial_{t}q\geq 0$ and $\partial_{t}q\not\equiv 0$.
This implies the existence of a constant $c>0$ such that

$\int_{B_{1}\times[-2,-3/2]\cap\\{|x_{n}|\geq
1/5\\}}\frac{\partial_{t}w_{r_{k_{\ell}}}}{H(r_{k_{\ell}},w\zeta)^{1/2}}\geq
c.$

Also, by a compactness argument, the constant $c$ can be chosen to be
independent of any subsequence, therefore

$\int_{B_{1}\times[-2,-3/2]\cap\\{|x_{n}|\geq
1/5\\}}\frac{\partial_{t}w_{r}}{H(r,w\zeta)^{1/2}}\geq c>0\qquad\forall\,r>0.$

Hence, since $w_{r}$ is caloric in
$\big{(}B_{2}\times(-2,0)\big{)}\cap\big{\\{}|x_{n}|\geq{\textstyle\frac{1}{20}}\big{\\}}$
for $r\ll 1$, the classical Harnack inequality for the heat equation implies
that

${\mathchoice{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern 4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{(B_{r}\cap\\{|x_{n}|\geq\frac{r}{10}\\})\times\\{-r^{2}\\}}\frac{\partial_{t}u}{H(r,w\zeta)^{1/2}}={\mathchoice{\kern
4.30554pt\vbox{\hrule height=0.4pt,width=6.0pt,depth=0.0pt}\kern-9.7pt}{\kern
4.30554pt\vbox{\hrule
height=0.4pt,width=4.3pt,depth=0.0pt}\kern-7.0pt}{}{}}\int_{(B_{1}\cap\\{|x_{n}|\geq\frac{1}{10}\\})\times\\{-1\\}}\frac{\partial_{t}w_{r}}{H(r,w\zeta)^{1/2}}\geq
c>0.$

Recalling that $H(r,w\zeta)^{1/2}\geq
c_{\varepsilon}r^{\lambda^{2nd}+\varepsilon/2}$ (this follows from Lemma
5.6(b) with $\gamma>3$), the result follows. ∎

We can now prove Proposition 8.6.

###### Proof of Proposition 8.6.

Arguing as in the proof of Lemma 8.5, we get $H(r,u-p_{2})\leq
Cr^{2\lambda^{2nd}}$. Hence, in view of Lemma 8.7, the result follows from
Lemma 8.1 with $\omega(r)=Cr^{\lambda^{2nd}}$ and
$\beta=\lambda^{2nd}-2+\varepsilon$. ∎

Combining Propositions 8.2, 8.3, and 8.6, we immediately deduce the following:

######  Corollary 8.8.

For any $(x_{\circ},t_{\circ})\in\Sigma$ and $\varepsilon>0$, there exists
$\rho=\rho(x_{\circ},t_{\circ},\varepsilon)\in(0,1)$ such that

$\\{u=0\\}\cap\big{(}B_{r}(x_{\circ})\times[t_{\circ}+r^{2-\varepsilon},1)\big{)}=\emptyset\qquad\textrm{for
all}\quad r\in(0,\rho).$

Finally, to prove Theorem 1.1, we will also need the following simple GMT
lemma:

######  Lemma 8.9.

Let $E\subset\mathbb{R}^{n}\times(-1,1)$ with

$\dim_{\mathcal{H}}\big{(}\pi_{x}(E)\big{)}\leq\beta.$

Assume that for any $\varepsilon>0$ and $(x_{\circ},t_{\circ})\in E$ there
exists $\rho=\rho(\varepsilon,x_{\circ},t_{\circ})>0$ such that

$\big{\\{}(x,t)\in B_{\rho}(x_{\circ})\times(-1,1)\ :\
t-t_{\circ}>|x-x_{\circ}|^{2-\varepsilon}\big{\\}}\cap E=\emptyset.$

Then $\dim_{\rm par}(E)\leq\beta$.

###### Proof.

Fix $\beta^{\prime}>\beta$. We need to show that, for any given $\delta>0$,
the set $E$ can be covered by countably many of cylinders
$B_{r_{i}}(x_{i})\times(t_{i}-r_{i}^{2},t_{i}+r_{i}^{2})$ so that

$\sum_{i}r_{i}^{\beta^{\prime}}\leq\delta.$ (8.10)

Choose $\varepsilon:=(\beta^{\prime}-\beta)/3$,
$\beta^{\prime\prime}:=\beta+\varepsilon$, and decompose

$E=\cup_{\ell\geq 1}E_{\ell},\quad\mbox{where}\quad
E_{\ell}:=\big{\\{}(x,t)\in E\ :\
2^{-\ell+1}\geq\rho(\varepsilon,x,t)>2^{-\ell}\big{\\}}.$

Now, given a pair of points $(x,t)$ and $(x^{\prime},t^{\prime})$ belonging to
$E_{\ell}$, with $|x-x^{\prime}|<2^{-\ell}$, by applying Corollary 8.8 at both
points we deduce that $|t-t^{\prime}|\leq|x-x^{\prime}|^{2-\varepsilon}$. This
proves that, for all $(x,t)\in E_{\ell}$ and $r\in(0,2^{-\ell})$, we have

$E_{\ell}\cap\pi^{-1}_{x}(B_{r}(x))\subset
B_{r}\times(t-r^{2-\varepsilon},t+r^{2-\varepsilon}).$ (8.11)

Now, the assumption $\dim_{\mathcal{H}}\big{(}\pi_{x}(E)\big{)}\leq\beta$
implies $H^{\beta^{\prime\prime}}(\pi_{x}(E_{\ell}))=0$. Thus, for any given
$\ell$, there exists a family of balls $\\{B_{r_{k}}(x_{k})\\}_{k}$, with
$r_{k}\in(0,2^{-\ell})$, such that

$\pi_{x}(E_{\ell})\subset\bigcup_{k}B_{r_{k}}(x_{k})\quad\mbox{ and
}\quad\sum_{k}r_{k}^{\beta^{\prime\prime}}\leq\delta 2^{-\ell}.$

Noting that $(-r^{2-\varepsilon},r^{2+\varepsilon})$ can be covered by
$r^{-\varepsilon}$ many intervals of length $2r^{2}$, (8.11) implies that
$E_{\ell}\cap\pi^{-1}_{x}(B_{r_{k}}(x_{k}))$ can be covered by cylinders of
the form
$\\{B_{r_{k}}(x_{k})\times(t_{j,k}-r_{k}^{2},t_{j,k}+r_{k}^{2})\\}_{j\in
I_{k}}$, with $\\#I_{k}\leq r_{k}^{-\varepsilon}$. This gives a covering of
$E_{\ell}$ such that

$\sum_{k}\sum_{j\in I_{k}}r_{k}^{\beta^{\prime}}\leq
r_{k}^{-\varepsilon}\sum_{k}r_{k}^{\beta^{\prime}}=\sum_{k}r_{k}^{\beta^{\prime\prime}}\leq\delta
2^{-\ell}.$

Taking the union over $\ell\geq 1$ of all these coverings, since
$\sum_{\ell\geq 1}\delta 2^{-\ell}=\delta$, we obtain a covering for $E$
satisfying (8.10). ∎

We are now in position to prove one of our main results.

###### Proof of Theorem 1.1.

The result follows from Proposition 3.2, Corollary 8.8, and Lemma 8.9. ∎

##  9\. Cubic blow-ups

The following lemma classifies possible 2nd blow-ups at points of
$\Sigma_{n-1}^{=3}$.

######  Lemma 9.1.

Let $q$ be a 3-homogeneous solution of (6.17), with
$\\{p_{2}=0\\}=\\{x_{n}=0\\}$. Then

$q(x,t)=a|x_{n}|\bigg{(}x_{n}^{2}+6bt-3\sum_{\alpha,\beta=1}^{n-1}b_{\alpha\beta}x_{\alpha}x_{\beta}\bigg{)}+\bar{a}x_{n}\bigg{(}x_{n}^{2}+6\bar{b}t-3\sum_{\alpha,\beta=1}^{n-1}\bar{b}_{\alpha\beta}x_{\alpha}x_{\beta}\bigg{)}\,,$
(9.1)

where $a\geq 0$, $(b_{\alpha\beta})\in\mathbb{R}^{(n-1)\times(n-1)}$ is
nonnegative definite, and $b={\rm trace}(b_{\alpha\beta})$, $\bar{b}={\rm
trace}(\bar{b}_{\alpha\beta})$.

As we shall see below, Lemma 9.1 follows easily from the following result.

######  Lemma 9.2.

Let $q:\mathbb{R}^{n}\times(-\infty,0)\rightarrow\mathbb{R}$ be a continuous
and $\lambda$-homogeneous function with polynomial growth, such that
$q|_{\\{x_{n}=0\\}}\geq 0$ and $\text{\small\bf H}q$ is a locally bounded
signed measure concentrated on $\\{x_{n}=0\\}$. If $\lambda>0$ is an odd
integer, then $q\equiv 0$ on $\\{x_{n}=0\\}$.

###### Proof.

For simplicity we give the proof in the case $\lambda=3$, which anyhow is the
only one relevant for our purposes. The interested reader will notice that the
proof for $\lambda=5,7,9...$ is identical, but using the functions $Q$
constructed in [FRS20, Lemma B.5].

Let $x=(x^{\prime},x_{n})$ and define

$Q(x):=|x_{n}|\big{(}3|x^{\prime}|^{2}-(n-1)x_{n}^{2}\big{)}.$

Note that, since $\text{\small\bf H}q=\mu$ and $q$ is $\lambda$-homonegenous,
$q(x,-1)$ satisfies

$\mathcal{L}_{OU}q+\frac{\lambda}{2}q=\mu\quad\mbox{in }\mathbb{R}^{n}$

with $\lambda=3$, where $\mathcal{L}_{OU}$ is the operator defined in (5.10).
On the other hand, an explicit computation shows that

$\mathcal{L}_{OU}Q+\frac{\lambda}{2}Q=6|x^{\prime}|^{2}\mathcal{H}^{n-1}|_{\\{x_{n}=0\\}}\quad\mbox{in
}\mathbb{R}^{n}.$

Hence, since $\mu$ is concentrated on $\\{x_{n}=0\\}$, using integration by
parts (which is justified by the polynomial growth of $q$ and exponential
decay of the Gaussian kernel) and denoting $dm=G(x,-1)dx$ the Gaussian
density, we obtain

$\displaystyle
6\int_{\\{x_{n}=0\\}}q|x^{\prime}|^{2}G(x,-1)\,d\mathcal{H}^{n-1}=\int_{\mathbb{R}^{n}}q\,\big{(}\mathcal{L}_{OU}+{\textstyle\frac{\lambda}{2}}\big{)}Q\,dm=\int_{\mathbb{R}^{n}}\big{(}\mathcal{L}_{OU}+{\textstyle\frac{\lambda}{2}}\big{)}q\,Qdm=\int_{\mathbb{R}^{n}}G(x,-1)\,Q\,d\mu=0,$

where in the last equality we used that $\mu$ is supported on $\\{x_{n}=0\\}$,
where $Q$ vanishes.

Thus $\int_{\\{x_{n}=0\\}}q|x^{\prime}|^{2}G(x,-1)\,d\mathcal{H}^{n-1}=0$, and
since $q\geq 0$ on $\\{x_{n}=0\\}$, this forces $q\equiv 0$ on
$\\{x_{n}=0\\}$, as wanted. ∎

We can now prove Lemma 9.1.

###### Proof of Lemma 9.1.

Thanks to Lemma 9.2, we know that $q$ vanishes on $\\{x_{n}=0\\}$. Thus
$q|_{\\{x_{n}>0\\}}$ and $q|_{\\{x_{n}<0\\}}$ are 3-homogenous caloric
functions vanishing on $\\{x_{n}=0\\}$ and therefore, by Liouville Theorem,
$q$ must be of the form (9.1) satisfying $b={\rm trace}(b_{\alpha\beta})$ and
$\bar{b}={\rm trace}(\bar{b}_{\alpha\beta})$. Recalling that $q$ solves
(6.17), and hence it is a supercaloric function, we obtain the extra
conditions that $a\geq 0$ and that $(b_{\alpha\beta})$ is nonnegative
definite. ∎

Our next result is the following monotonicity formula.

######  Lemma 9.3.

Let $(0,0)\in\Sigma_{n-1}^{=3}$ with $p_{2}=\frac{1}{2}(x_{n})^{2}$. Let $Q$
be a 3-homogeneous solution of (6.17). Then

$\frac{d}{dr}\left(\frac{1}{r^{6}}\int_{\\{t=-r^{2}\\}}(u-p_{2})\zeta\,Q\,G\right)\geq-C\|Q\|_{L^{2}(\mathscr{C}_{1})}\,,$
(9.2)

where $C$ depends only on $n$ and $\|u\|_{L^{\infty}}$.

###### Proof.

After scaling we have

$\frac{1}{r^{6}}\int_{\\{t=-r^{2}\\}}(u-p_{2})\zeta\,Q\,G=\frac{1}{r^{3}}\int_{\\{t=-1\\}}\big{(}(u-p_{2})\zeta\big{)}_{r}QG,$

therefore

$\frac{d}{dr}\left(\frac{1}{r^{6}}\int_{\\{t=-r^{2}\\}}(u-p_{2})\zeta\,Q\,G\right)=\frac{1}{r^{4}}\int_{\\{t=-1\\}}Z\big{(}(u-p_{2})\zeta\big{)}_{r}\,Q\,G-\frac{3}{r^{4}}\int_{\\{t=-1\\}}\big{(}(u-p_{2})\zeta\big{)}_{r}\,Q\,G,$

where $Z$ is defined in Section 2.1. Recall the integration by parts
identities from Lemma 4.2,

$\int_{\\{t=-1\\}}Zf\,g\,G=\int_{\\{t=-1\\}}f\,Zg\,G+2\int_{\\{t=-1\\}}(\text{\small\bf
H}f\,g-f\,\text{\small\bf H}g)\,G.$

Since $ZQ=3Q$, $\text{\small\bf
H}Q=2At\,\mathcal{H}^{n-1}|_{\\{x_{n}=0\\}}\leq 0$, $u-p_{2}=u\geq 0$ on
$\\{x_{n}=0\\}$, and $\text{\small\bf H}(u-p_{2})=\chi_{\\{u=0\\}}$, this
yields