# Bowl Soliton Asymptotics and Applications

Sathya Rengaswami, José Torres Santaella

###### Abstract

In this paper, we obtain the asymptotic expansion for the analogue of the
bowl-soliton for a large ‘nondegenerate’ class of fully nonlinear curvature
flows. We use this to show the uniqueness of these bowl-type solitons in their
asymptotic class. We also give examples to illustrate the situation for
‘degenerate’ speeds and how they different they can be. Finally, we show how
to construct ‘wing-like’ solitons for these flows, which are complete,
connected translators that are not graphical, entire or convex. We also obtain
asymptotic expansions for them to show the variety of solutions that one can
obtain depending on the choice of speed function.

##  1 Introduction

Geometric evolution equations for hypersurfaces have seen significant
development over the last few decades. We have witnessed a remarkable growth
in this field, leading to the emergence of intriguing nonlinear partial
differential equations. These equations have played a crucial role in
addressing fundamental questions within both mathematics and physics.

In this research, our focus lies on a particular type of evolving
hypersurfaces known as ”translators.” These hypersurfaces in
$\mathbb{R}^{n+1}$ undergo evolution by translation along a fixed unit
direction when subjected to an extrinsic curvature flow, i.e. the normal
velocity at each point of the hypersurface is a 1-homogeneous smooth symmetric
function of their principal curvatures. It’s worth noting that translators
represent a significant class of second-order elliptic partial differential
equations.

More precisely, given an immersed hypersurface
$\Sigma_{0}=F_{0}(\Sigma)\subset\mathbb{R}^{n+1}$, a solution to an extrinsic
curvature flow, or $f$-flow for short, with initial data $\Sigma_{0}$
corresponds to a $1$-parameter family of immersions
$F:\Sigma\times\mathbb{R}\to\mathbb{R}^{n+1}$, that solves

$\displaystyle\begin{cases}\dfrac{\partial F}{\partial
t}(x,t)=f(\lambda(x,t))\nu(x,t),\>(x,t)\mbox{ in }\Sigma\times(0,T),\\\
F(x,0)=F_{0}(x),\end{cases}$ (1)

where $\nu(x,t)$ is the inward unit normal vector of the hypersurface
$\Sigma_{t}=F(\Sigma,t)$ in $\mathbb{R}^{n+1}$,
$\lambda(x,t)=(\lambda_{1}(x,t),\ldots,\lambda_{n}(x,t))$ are the principal
curvatures of $\Sigma_{t}$ with respect to to $\nu(x,t)$. Here,
$f:\Gamma\to\mathbb{R}$ is a smooth function of the principal curvatures of
$\Sigma_{t}$ with the following properties:

  1. a)

$\Gamma\subset\mathbb{R}^{n}$ is an open symmetric cone that contains the
positive cone
$\Gamma_{+}\coloneqq\left\\{\lambda\in\mathbb{R}^{n}:\lambda_{i}>0\right\\}$.

  2. b)

$f$ is positive and symmetric, i.e.:
$f(\sigma(\lambda_{1}),\ldots,\sigma(\lambda_{n}))=f(\lambda_{1},\ldots,\lambda_{n})$
for every permutation $\sigma\in S_{n}$.

  3. c)

$f$ is strictly increasing in each variable, i.e.: $\dfrac{\partial
f}{\partial\lambda_{i}}(\lambda)>0$ holds for every $\lambda\in\Gamma$ and
$i=1,\ldots,n$.

  4. d)

$f$ is $1$-homogeneous, i.e.: $f(c\lambda)=cf(\lambda)$ for every $c>0$.

  5. e)

$f$ vanishes at boundary of $\Gamma$, i.e.: there exist a continous function
$\tilde{f}:\overline{\Gamma}\to\mathbb{R}$ such that
${\left.\kern-1.2pt\tilde{f}\vphantom{\big{|}}\right|_{\Gamma}}=f$ and
${\left.\kern-1.2pt\tilde{f}\vphantom{\big{|}}\right|_{\partial\Gamma}}=0$.

Given a speed function $f$, a translating solution to Equation (1), or a _f-
translator_ for short, is a solution of the form

$\displaystyle F(x,t)=F_{0}(x)+e_{n+1}t.$

(up to tangential reparametrizations.) Note that by hypotheses (b)) and (c)),
the $f$-flow is invariant under the isometries of ambient space and parabolic
rescalings. Thus there is no loss of generality in fixing the translation
direction to be $e_{n+1}=(0,\ldots,0,1)\in\mathbb{R}^{n+1}$.

Importantly, translating solutions can be studied by the parabolic and by the
elliptic PDE points of view, since every time slice $\Sigma_{t}$ satisfies the
Equation

$\displaystyle f(\lambda)=-\left<{\nu,e_{n+1}}\right>,$ (2)

recall that $\nu$ is the inward pointing unit normal of $\Sigma_{t}$ in
$\mathbb{R}^{n+1}$. In fact, in local coordinates $\Sigma_{t}$ can be seen as
a graph of a function for which Equation (2) correspond to a nonlinear
elliptic PDE (quasilinear when $f=H$ and fully nonlinear when $f\neq H$.).
Moreover, from the parabolic point of view, $f$-translators without boundary
are examples of noncompact _eternal solutions_ of Equation (1), i.e.:
solutions that are defined for all $t\in(-\infty,\infty)$, see [10] for
details.

It is worth mentioning that $f$-translators have been widely studied when
$f=H$, see for instance [4] for a complete survey about $H$-translators, model
of singularities, and minimal surfaces theory. In addition, the reader will be
referred to [11] for existence and properties of
$\sqrt[\alpha]{S_{n}}$-translators for $\alpha>0$.

In a different work, discussed in [7], the first author explored the
existence, uniqueness, regularity, and asymptotic geometry of ”bowl”-type
solutions. These solutions are constructed for fairly general speed functions
that are $\alpha$-homogeneous with $\alpha>0$. To be precise, a “bowl”-type
solution of (1) is a complete, strictly convex, rotationally symmetric111This
solution is unique among strictly convex rotationally symmetric translating
graphs. translating graph in $\mathbb{R}^{n+1}$ which may be defined in a ball
of finite radius or all of $\mathbb{R}^{n}$.

This dichotomy is characterized by the value of $f(0,1,\ldots,1)$ and the
asymptotic behavior of the implicit solution of $f(x,y,\ldots,y)=1$ as
$y\to\infty$, i.e. the behavior of the speed function near the boundary of the
positive cone. In addition, when $f(0,1,\ldots,1)>0$, the “bowl”-type solution
is always entire and behaves like a paraboloid at infinity, i.e:

$\displaystyle\dfrac{|x|^{2}}{2f(0,1,\ldots,1)}+o(|x|^{2}),\mbox{ as
}|x|\to\infty.$

It is an interesting question what the lower order terms are. When $f$ is the
mean curvature, the authors in [1] showed that the bowl soliton is smoothly
asymptotic to

$\frac{|x|^{2}}{2(n-1)}-\ln(|x|)+O(|x|^{-1}),\mbox{ as }|x|\to\infty.$ (3)

We extend the above result to a large class of speeds which we define below.
For convenience, we denote
$\mathbf{e}\coloneqq(1,\ldots,1)\in\mathbb{R}^{n-1}$, the $(n-1)-$tuple of
1’s.

######  Theorem 1.1.

Assume that $f:\Gamma\to\mathbb{[}0,\infty)$ satisfies properties a)-d) and is
nondegenerate. Then, the entire “bowl”-type solution is smoothly asymptotics
to

$\dfrac{|x|^{2}}{2f(0,\mathbf{e})}-{\left.\kern-1.2pt\dfrac{\partial
f}{\partial\lambda_{1}}\vphantom{\big{|}}\right|_{(0,\mathbf{e})}}\ln(|x|)+O(|x|^{-1}),\mbox{
as }|x|\to\infty.$

Furthermore, by applying the same techniques employed in [5], we show that
that the bowl-type soliton is essentially unique in the asymptotic class of
$O(|x|^{2})$ solutions via the following theorem:

######  Theorem 1.2.

Let $\Sigma\subset\mathbb{R}^{n+1}$ be a strictly convex complete
$f$-translator with a single end smoothly asymptotic to the “bowl”-type
solution. Then, if $f$ satisfies properties a)-e) and is nondegenerate,
$\Sigma$ is the “bowl”-type solution up to vertical translations.

To show that Theorem 1.1 does not apply to all speed functions, we discuss the
degenerate speed function $f=\sqrt[n]{S_{n}}$ (the $n^{th}$ root of the Gauss
curvature) and show that the bowl soliton does not have quadratic asymptotics
for any $n\geq 2$.

######  Theorem 1.3.

Let $r\coloneqq|x|$ be the Euclidean norm of an $n$-tuple. The “bowl”-type
solution for the speed function $f=\sqrt[n]{S_{n}}$ is smoothly asymptotic to

$\displaystyle\begin{cases}\int\limits_{0}^{r}e^{\frac{s^{2}}{2}}ds+O\left(\int\limits_{0}^{r}\sqrt{e^{\frac{s^{2}}{2}}-1}ds\right),&\mbox{
for }n=2,\\\ \dfrac{r^{4}}{12}+O(r),&\mbox{ for }n=3,\\\
\dfrac{(n-2)^{\frac{n-1}{n-2}}}{2(n-1)^{\frac{1}{n-2}}}r^{\frac{2(n-1)}{n-2}}+O\left(r^{\frac{2}{n-2}}\right),&\mbox{
for }n\geq 4\end{cases},\mbox{ as }r\to\infty.$

Then, we discuss a special kind of translator known as the _wing-like_
solution firstly studied in the context of mean curvature flows in [1].

######  Theorem 1.4.

For every $R>0$, there exist a non-convex complete rotationally symmetric
$f$-translator $W_{R}$ with respect to $x_{n+1}$-axis $f=\sqrt[k]{S_{k}}$ and
$f=\dfrac{S_{k}}{S_{k-1}}$ such that:

  1. 1.

For $f=\sqrt[k]{S_{k}}$, we distinguish:

    1. (a)

When $k$ is even: $W_{R}\setminus B_{R_{1}}(0)$ with $R_{1}>R$ posses two
graphical branches $W_{R}^{+},W_{R}^{-}:\mathbb{R}^{n}\setminus
B_{R}(0)\to\mathbb{R}$ smoothly asymptotic to

$\displaystyle
W_{R}^{\pm}(x)=\pm\left(\dfrac{|x|^{2}}{2f(0,\mathbf{e})}-{\left.\kern-1.2pt\dfrac{\partial
f}{\partial\lambda_{1}}\vphantom{\big{|}}\right|_{(0,\mathbf{e})}}\ln(|x|)+O(|x|^{-1})+C^{\pm}\right),\mbox{
as }|x|\to\infty.$

    2. (b)

When $k$ is odd: $W_{R}$ posse a $\mathbb{S}^{n-1}$ boundary component and
$W_{R}\setminus B_{R_{1}}(0)$ with $R_{1}>R$ is given by a vertical graph
smoothly asymptotic to

$\displaystyle
W_{R}(x)=\dfrac{|x|^{2}}{2f(0,\mathbf{e})}-{\left.\kern-1.2pt\dfrac{\partial
f}{\partial\lambda_{1}}\vphantom{\big{|}}\right|_{(0,\mathbf{e})}}\ln(|x|)+O(|x|^{-1})+C,\mbox{
as }|x|\to\infty.$

  2. 2.

For $f=\dfrac{S_{k}}{S_{k-1}}$ we have that $W_{R}\setminus B_{R_{1}}(0)$ with
$R_{1}>R$ posses two graphical branches $W_{R}^{+},W_{R}^{-}$ such that

$\displaystyle
W_{R}^{+}(x)=\dfrac{|x|^{2}}{2f(0,\mathbf{e})}-{\left.\kern-1.2pt\dfrac{\partial
f}{\partial\lambda_{1}}\vphantom{\big{|}}\right|_{(0,\mathbf{e})}}\ln(|x|)+O(|x|^{-1})+C^{+},\mbox{
as }|x|\to\infty,$ $\displaystyle\lim\limits_{|x|\to\infty}|\nabla
W_{R}^{-}(x)|=0.$

(a) $S_{k}/S_{k-1}$

(b) $\sqrt[k]{S_{k}}$ for $k$ even

(c) $\sqrt[k]{S_{k}}$ for $k$ odd

Figure 1: Winglike translators for various speed functions

Finally, we provide some applications of the above theorems, namely an
asymptotic growth estimate:

######  Theorem 1.5.

Let $\Sigma=\left\\{(x,u(x)):x\in\mathbb{R}^{n}\right\\}$ be an entire convex
translating solution of (1) such that $f$ is nondegenerate. Assume furhter,
that there exist $a,b,C_{1},C_{2},R>0$ such that

$\displaystyle C_{1}|x|^{a}\leq u(x)\leq C_{2}|x|^{b},\mbox{ for }|x|\geq R,$
(4)

then, $a\leq 2\leq b$. In addition, if $a=b=2$, then $u(x)$ agrees with the
“bowl”-type solution up to vertical translations.

The organization of this article goes as follows: In Section 2, we discuss the
preliminaries of the differential geometry of axially symmetric translators
and the ODE theory needed for the analysis of the ODE that the translator
solves. In Section 3, we derive the asymptotic expansion up to $o(|x|^{-2}$)
of the slope field of the translator. We use this estimate in Section 4 to
prove the uniqueness result in Theorem 1.2. Section 5 discusses the degenerate
example $\sqrt[n]{S_{n}}.$ Section 6 concerns the discussion of winglike
translators. Finally, in Section 7 we give a proof of Theorems 1.5.

Acknowledgment: We would like to acknowledge the support of our advisors Drs.
Mat Langford and Theodora Bourni, and Franciso Martín, Miguel Sanchez and
Mariel Saez, both in terms of technical discussions as well as travel funding.
The second author was partially supported by the project PID2020-116126GB-I00
funded by MCIN/ AEI /10.13039/501100011033, by the project PY20-01391 (PAIDI
2020) funded by Junta de Andalucía FEDER and by the framework of IMAG-Mar ía
de Maeztu grant CEX2020- 001105-M funded by MCIN/AEI/ 10.13039/50110001103.

##  2 Preliminaries

###  2.1 The rotational translator ODE

For a real-valued $C^{2}$-function $u$ of a single real variable we consider
$\Sigma$ to be the graph of $y=u(r)$, where $r=|x|$ and $x\in\mathbb{R}^{n}$.

Then, the inward unit normal of $\Sigma$ at a point $(x,u(r))$ is given by

$\displaystyle\vec{N}=\left(\frac{u^{\prime}}{\sqrt{1+u^{\prime
2}}}\frac{x}{r},\frac{-1}{\sqrt{1+u^{\prime
2}}}\right)\in\mathbb{R}^{n}\times\mathbb{R}$

Moreover, the principal curvatures of $\Sigma$ are given by

$\displaystyle\lambda_{1}=\frac{u^{\prime\prime}}{(1+u^{\prime
2})^{3/2}}\mbox{ and }\lambda_{i}=\frac{u^{\prime}}{r\sqrt{1+u^{\prime
2}}},\mbox{ for }i=2,\ldots,n.$ (5)

######  Definition 1.

A speed function is a function $f:\Gamma\to\mathbb{R}$ that satisfies
properties a)-d). A speed function $f$ is said to be _nondegenerate_ if
$f(0,\mathbf{e})>0$, where $\mathbf{e}=(1,\ldots,1)\in\mathbb{R}^{n-1}$.

######  Remark 1.

We emphasize that being _nondegenerate_ is equivalent to requiring that the
cylinder $S^{n-1}\times\mathbb{R}$ is not a stationary solution to the
$f$-flow (1).

######  Example 1.

The class of speed functions $f:\Gamma\to\mathbb{R}$ that additionally
satisfies Property e) is vast and includes:

  * •

The mean curvature $H=\lambda_{1}+\ldots+\lambda_{n}$, supported in
$\Gamma_{1}=\left\\{\lambda\in\mathbb{R}^{n}:H>0\right\\}$.

  * •

The $k$-th roots of the symmetric elemental polynomial $\sqrt[k]{S_{k}}$,
where

$\displaystyle S_{k}(\lambda)=\sum_{1\leq i_{1}<\ldots<i_{k}\leq
n}\lambda_{i_{1}}\ldots\lambda_{i_{k}},$

supported in the g$\accentset{\circ}{a}$rdin cone
$\Gamma_{k}:=\left\\{\lambda\in\mathbb{R}^{n}:S_{l}(\lambda)>0,\>l=1,\ldots,k\right\\}$.

  * •

The inverse of the $k$-th harmonic sum $\left(\sum\limits_{1\leq
i_{1}<\ldots<i_{k}\leq
n}\dfrac{1}{\lambda_{i_{1}}+\ldots+\lambda_{i_{k}}}\right)^{-1}$ supported in
$\Gamma=\left\\{\lambda\in\mathbb{R}^{n}:\lambda_{1}+\ldots+\lambda_{k}>0\right\\}$,
where $\lambda_{1}\leq\ldots\leq\lambda_{n}$.

  * •

Any $1$-homogeneous symmetric combination of the above functions.

It is worth to mention that by removing hypothesis e), the Hessian quotients
functions $Q_{k,l}=\left(\dfrac{S_{k}}{S_{l}}\right)^{\frac{1}{k-l}}$
supported in $\Gamma_{k}$ can be included in this class of functions.

######  Remark 2.

Due to axial symmetry of $\Sigma$, $f$ only depends on two variables (because
there are only two distinct principal curvatures), and hence we sometimes use
$f(x,y)$ instead of $f(x,y\mathbf{e}).$

Therefore, by Remark 2 equation

$\displaystyle f(\lambda)=\left<{\nu,e_{n+1}}\right>$

for a rotationally symmetric graph $(x,u(r))$ with respect to $x_{n+1}$-axis
is given by

$\displaystyle f\left(\frac{u^{\prime\prime}}{(1+u^{\prime
2})^{3/2}},\frac{u^{\prime}}{r\sqrt{1+u^{\prime
2}}}\right)=\frac{1}{\sqrt{1+u^{\prime 2}}}\,.$

We can reduced the above equation to a first order ODE by setting
$v=u^{\prime}$, and with the $1$-homogeneity of $f$ we may write it by

$f\left(\frac{v^{\prime}}{1+v^{2}},\frac{v}{r}\right)=1.$ (6)

######  Remark 3.

Geometrically, $v$ is the gradient of the profile curve
$(r,u(r))\in\mathbb{R}^{2}$, and the solution is unique up to vertical
translations.

In addition, since $f$ is strictly monotone in each argument, we may apply the
Implicit Function theorem to obtain a unique function $x=g(y,z)$ in the sense
that

$\displaystyle f(g(y,z),y)=z.$ (7)

Note that in Eq. (6), we have $z=1$, and in this case, we will suppress the
second argument and refer to $g(y,1)$ as simply $g(y)$. We will return to
using $g(y,z)$ in Section 6.

Moreover, since $f$ is $C^{1}$ and has a non-singular derivative w.r.t. $x$,
we have that $g$ is of class $C^{1}$ as well.

Consequently, the slope function $v$ of a rotationally symmetric
$f$-translators satisfies

$\displaystyle\begin{cases}v^{\prime}(r)=\left(1+v^{2}(r)\right)g\left(\dfrac{v(r)}{r}\right),r\geq
0,\\\ v(0)=0.\end{cases},$ (8)

and we can recover $u(r)$ from $v(r)$ via an integration procedure.

######  Remark 4.

It is important to mention that even though Equation (8) appears to have a
singularity for the initial condition $v(0)=0$, this is only a coordinate
singularity.
We refer the interested reader to [7] for a study of general
$\alpha-$homogeneous speeds, where the questions of existence, regularity,
uniqueness, and convexity of this solution are addressed. A complete
classification of speeds based on whether the resulting solution is
asymptotically cylindrical is also presented therein.

######  Example 2.

The following examples are the expressions of the function $f(x,y)$ and $g(y)$
for the following speed functions:

  1. 1.

The mean curvature, $H$:

$\displaystyle f(x,y)=x+(n-1)y\mbox{ and }g(y)=1-(n-1)y.$

  2. 2.

The $k$th-root of the symmetric elemental polynomials, $\sqrt[k]{S_{k}}$:

$\displaystyle
f(x,y)=\sqrt[k]{\binom{n-1}{k}y^{k}+\binom{n-1}{k-1}xy^{k-1}}\mbox{ and
}g(y)=\binom{n-1}{k-1}^{-1}y^{1-k}-\frac{n-k}{k}y.$

  3. 3.

The quotients of the symmetric elemental polynomials,
$Q_{k+1,k}=\dfrac{S_{k+1}}{S_{k}}$:

$\displaystyle
f(x,y)=\dfrac{\binom{n-1}{k}xy^{k}+\binom{n-1}{k+1}y^{k+1}}{\binom{n-1}{k-1}xy^{k-1}+\binom{n-1}{k}y^{k}}\mbox{
and }g(y)=\dfrac{n-k}{k+1}y\dfrac{(k+1)-(n-k-1)y}{(n-k)y-k}.$

###  2.2 Differentiability properties of _nondegenerate_ $f$

We are interested in speed function $f$ that satisfies $f(0,1)>0$, in this
section and beyond, we normalize $f$ so that $f(0,1)=1$.

In addition, since $f(x,y)$ is $1$-homogeneous, we have the identity

$f(x,y)=f_{x}x+f_{y}y.$

where $f_{x}=\frac{\partial f}{\partial x}$ and $f_{y}=\frac{\partial
f}{\partial x}$.
We note that the partial derivatives $f_{x}$ and $f_{y}$ are $0$-homogeneous
functions, and by our normalization, we have

$\displaystyle f_{y}(0,1)=1$ (9)

Now, we outline an estimation trick that we will use repeatedly in this paper.
Due to $1$-homogeneity of $f$, we have $f(x,y)=yf(x/y,1)$:

  * •

When $f$ is $C^{1}$, using the mean value theorem, we may write

$\displaystyle f(x,y)=y(1+f_{x}(\xi,1))\dfrac{x}{y}=y+f_{x}(\xi,1)x$ (10)

for some $0\leq\xi\leq\dfrac{x}{y}$.

  * •

When $f$ is $C^{2}$, Taylor’s remainder theorem yields

$f(x,y)=y\left(1+f_{x}(0,1)\dfrac{x}{y}+\frac{1}{2}f_{xx}(\xi,1)\dfrac{x^{2}}{y^{2}}\right)=y+f_{x}(0,1)x+\frac{1}{2y}f_{xx}(\xi,1)x^{2}$
(11)

Therefore, by the normalization of $f$, we will assume in most calculations in
this paper that

$\displaystyle 0<x\leq y\Leftrightarrow 0<\xi\leq 1,$

and consequently, $|f_{x}(\xi,1)|,|f_{xx}(\xi,1)|$ are bounded by compactness
of $[0,1]$ and the continuity of these functions.

###  2.3 Differentiability properties of $g$

Recall that there exist a unique $\mathcal{C}^{1}$ function $g$ that satisfies
$f(g(y),y)=1$.

Then, due to our normalization $f(0,1)=1$, we have $g(1)=0$. Moreover, by
chain rule,

$f_{x}(g(y),y)g_{y}(y)+f_{y}(g(y),y)=0$

which, by suppressing the arguments, we get

$f_{x}g_{y}+f_{y}=0.$

Differentiating once more, we get

$f_{xx}g_{y}^{2}+2f_{xy}g_{y}+f_{yy}+f_{x}g_{yy}=0.$

Next, we note that whenever $f_{x}\neq 0$ it holds

$g_{y}=-\frac{f_{y}}{f_{x}}.$

In particular, $g_{y}\leq 0$, indicating that $g(y)$ is decreasing.

######  Remark 5.

In particular, when $f_{xx},f_{xy},f_{yy}$ are defined at $(x,y)=(g(y),y)$, we
have

$g_{yy}=-\frac{f_{xx}g_{y}^{2}+2f_{xy}g_{y}+f_{yy}}{f_{x}}.$

Therefore, it is interesting that $g$ will be convex when $f$ is concave and
vice versa. However, since our results do not rely on the convexity properties
of $f$, we won’t be using this fact.

###  2.4 ODE Theory

Throughout the paper we will use the technique of super-solutions and sub-
solutions for an ODE of the form

$x^{\prime}(t)=f(t,x(t)).$ (12)

We refer the reader to [9] for the following classic ODE results.

A differentiable function $x_{+}(t)$ satisfying

$x_{+}^{\prime}(t)>f(t,x_{+}(t))$

is called a _super-solution_ to (12). Similarly, a differentiable function
$x_{-}(t)$ satisfying

$x_{-}^{\prime}(t)<f(t,x_{-}(t))$

is called a _sub-solution_ to (12).

######  Lemma 2.1.

Let $x_{+}(t)$, $x_{-}(t)$ be super, sub-solutions of the differential
equation $x^{\prime}=f(t,x)$ on $[t_{0},T)$ respectively. For every solution
$x(t)$ on $[t_{0},T)$ we have

$x(t)<x_{+}(t),t\in[t_{0},T)\text{ whenever }x(t_{0})\leq x_{+}(t_{0})$

respectively

$x_{-}(t)<x(t),t\in[t_{0},T)\text{ whenever }x_{-}(t_{0})\leq x(t_{0})$

######  Remark 6.

If one replaces strong inequality by weak inequality in the definitions of sub
and super-solutions, one gets weak inequalities instead of strong ones in the
above lemma.

##  3 Asymptotic of bowl-type solutions

Recall from the introduction that a bowl-type soliton is a complete strictly
convex smooth solution of (8) with $u^{\prime}=v$, which is defined in $[0,R)$
where $R\in\left\\{\frac{1}{f(1,1)},\infty\right\\}$.

Furthermore, by the _nondegenerate_ property together with the normalization
$f(0,1)=1$, we have that $R=\infty$ and

$\displaystyle u(r)=r^{2}+o(r^{2}),\mbox{ as }r\to\infty.$

In this section we will prove Theorem 1.1 which state:

######  Theorem 3.1.

Assume that $f$ is a normalized _nondegenerte_ speed function. Then, the
corresponding bowl-type soliton for this flow has the asymptotics

$u(r)=\frac{r^{2}}{2}-c\log r+o(r^{-1}),\mbox{ as }r\to\infty,$

where $c=f_{x}(0,1)$.

We will prove the above theorem by showing that $v$ satisfies

$\displaystyle v(r)=r-\frac{c}{r}+o(r^{-2}),\mbox{ as }r\to\infty,$

whence the claim of the theorem follows at once by integration.

######  Remark 7.

Intuitively, a differentiable solution $v$ to (8) has the symmetries of an odd
function, and hence one does not expect even powers in the asymptotic
expansion, and indeed one can formally plug in a Laurent series for $v$, and
without much effort, one can determine the coefficients of the asymptotic
expansion. In what follows, we provide rigorous proofs for the correctness of
these coefficients.

The method that we will follow consists in a bootstrapping approach to
progressively refine the asymptotic terms. For this, we will use the ODE’s
theory of sub- and super-solutions denoted by $w_{\pm,\varepsilon}(r)$, where
the $\varepsilon$ will be used in asymptotic little-oh notation.

On the other hand, from Section 2.4, we recall that $w$ is a sub-solution
(super-solution resp.) to Eq. (8) if

$w^{\prime}\leq(1+w^{2})g(w/r),\>(\geq\mbox{ resp.}).$

However, in some cases it may be more convenient to check the following
equivalent condition

$\displaystyle f\left(\frac{w^{\prime}}{1+w^{2}},\frac{w}{r}\right)\leq
1,\>(\geq\mbox{ resp.}).$ (13)

######  Proposition 3.2.

The functions

$\displaystyle w_{+}(r)=r\mbox{ and }w_{-,\varepsilon}(r)=(1-\varepsilon)r$

satisfy the following properties:

  1. 1.

$w_{+}(r)$ is a super-solution to Eq. (8) $r>0$.

  2. 2.

For every $\epsilon\in(0,1)$, $w_{-,\varepsilon}(r)$ is a sub-solution to Eq.
(8) for sufficiently large $r$. Moreover, given any $r_{0}>0$, there exists
$r_{1}>r_{0}$ such that $v(r_{1})\geq r_{1}$. Thus, $v\geq(1-\varepsilon)r$
for sufficiently large $r$.

###### Proof.

The proofs are all direct computations.

  1. 1.

Firstly, we note that $f(x,y)$ is increasing in each variable, then it holds

$\displaystyle
f\left(\frac{w_{+}^{\prime}}{1+w_{+}^{2}},\frac{w_{+}}{r}\right)=f\left(\frac{1}{1+r^{2}},1\right)\geq
f\left(0,1\right)=1.$

Therefore, $w_{+}(r)$ is a super-solution to Eq. (8) $r>0$.

  2. 2.

Next, by evaluating $w_{-,\varepsilon}$ in (6) and taking the limit as
$r\to\infty$, we see that

$\displaystyle\displaystyle\lim_{r\to\infty}f\left(\frac{w_{-,\varepsilon}^{\prime}}{1+w_{-,\varepsilon}^{2}},\frac{w_{-,\varepsilon}}{r}\right)$
$\displaystyle=\lim_{r\to\infty}f\left(\frac{1-\epsilon}{1+((1-\varepsilon)r)^{2}},1-\varepsilon\right)$
$\displaystyle=f\left(0,1-\varepsilon\right)$ $\displaystyle<f(0,1)=1.$

Therefore, the claim is true for sufficiently large $r$.

On the other hand, we will prove the following part by contradiction. Let
$r_{0}>0$ and assume that $v(r)<(1-\epsilon)r$ for all $r>r_{0}$. Then, since
$g(y)$ is decreasing in $y$, it follows that

$\displaystyle v^{\prime}$ $\displaystyle=(1+v^{2})g\left(\frac{v}{r}\right)$
$\displaystyle>(1+v^{2})g\left(1-\varepsilon\right)$
$\displaystyle=C(1+v^{2})$

for $r>r_{0}$ and some $C>0$. Consequently, this inequality implies that $v$
blows up at some finite $r_{1}>r_{0}$ contradicting that $v(r)$ exist for all
$r\geq 0$.

∎

######  Remark 8.

Proposition 3.2 implies that the solution $v(r)=r+o(r)$ as $r\to\infty$.

######  Proposition 3.3.

The function $w_{-,\varepsilon}(r)=r-\varepsilon$ satisfies:

  1. 1.

For every $\varepsilon\in(0,1)$, $w_{-,\varepsilon}(r)$ is a sub-solution to
Eq. (6) for sufficiently large $r$.

  2. 2.

Given any $r_{0}>0$, there exists $r_{1}>r_{0}$ such that $v(r_{1})\geq
r_{1}$. Thus, $v\geq r-\epsilon$ for sufficiently large $r$.

###### Proof.

  1. 1.

We will show that $w_{-\varepsilon}$ verifies Eq. (13). Indeed, by Eq. (10),
we may write

$\displaystyle
f\left(\frac{w_{-,\varepsilon}^{\prime}}{1+w_{-,\varepsilon}^{2}},\frac{w_{-,\varepsilon}}{r}\right)$
$\displaystyle=f\left(\frac{1}{1+(r-\varepsilon)^{2}},1-\frac{\epsilon}{r}\right)$
$\displaystyle=1-\frac{\varepsilon}{r}+\frac{f_{x}(\xi,1)}{1+(r-\varepsilon)^{2}}.$

Then, since

$\displaystyle
0\leq\xi\leq\dfrac{1}{\left(1-\dfrac{\varepsilon}{r}\right)(1+(r-\varepsilon)^{2})}<1$

for $r$ sufficiently large, we obtain

$\displaystyle
f\left(\frac{w_{-,\varepsilon}^{\prime}}{1+w_{-,\varepsilon}^{2}},\frac{w_{-,\varepsilon}}{r}\right)$
$\displaystyle=1-\frac{\varepsilon}{r}+o(r^{-1})<1.$

for sufficiently large $r$.

  2. 2.

Next, by arguing by contradiction, we fix $r_{0}>0$ and assume that
$v(r)<r-\varepsilon$ holds for all $r>r_{0}$ and $\varepsilon\in(0,1)$.
Then, by part 2 of Proposition 3.2, we have that $v\geq\dfrac{r}{2}$ for
sufficiently large $r$ and

$\displaystyle v^{\prime}$ $\displaystyle=(1+v^{2})g\left(\frac{v}{r}\right)$
$\displaystyle>\left(1+\dfrac{r^{2}}{4}\right)g\left(1-\dfrac{\varepsilon}{r}\right).$

Finally, recall that $g(1)=0$. Then, by the mean value theorem, we have

$\displaystyle v^{\prime}$
$\displaystyle\geq\left(1+\dfrac{r^{2}}{4}\right)\left(g(1)-g_{y}\left(\xi\right)\dfrac{\varepsilon}{r}\right),$

for some $\xi\in\left(1-\dfrac{\varepsilon}{r},1\right)$. Consequently, since
$g$ is decreasing, there is $C>0$ depending on $\varepsilon$ such that

$\displaystyle v^{\prime}$ $\displaystyle\geq Cr,\mbox{ for all }r>r_{0}.$

However, this fact contradicts $v(r)<r-\varepsilon$ for all $r>r_{0}$.

∎

######  Remark 9.

Proposition 3.3 implies that $v(r)=r+o(1)$ as $r\to\infty$.

######  Proposition 3.4.

Let $c=f_{x}(0,1)$ and consider

$\displaystyle w_{+,\varepsilon}(r)=r-\frac{c-\varepsilon}{r}\mbox{ and
}w_{-,\varepsilon}(r)=r-\frac{c+\epsilon}{r}.$

  1. 1.

For every $\varepsilon\in(0,c)$, $w_{+,\varepsilon}(r)$ is a super-solution to
Eq. (6) for sufficiently large $r$. In addition, given any $r_{0}>0$, there
exists $r_{1}>r_{0}$ such that $v(r_{1})\leq
r_{1}-\frac{c-\varepsilon}{r_{1}}$. Thus, $v\leq r-\frac{c-\varepsilon}{r}$
for sufficiently large $r$.

  2. 2.

For every $\epsilon>0$, $w_{-,\varepsilon}(r)$ is a sub-solution to Eq. (6)
for sufficiently large $r$. Moreover, given any $r_{0}>0$, there exists
$r_{1}>r_{0}$ such that $v(r_{1})\geq r_{1}$. Thus, $v\geq
r-\frac{c+\varepsilon}{r}$ for sufficiently large $r$.

###### Proof.

  1. 1.

Taking sufficiently large $r$ as in part 2 of Proposition 3.3, but using Eq.
11 instead, we have

$\displaystyle
f\left(\frac{w_{+,\varepsilon}^{\prime}}{1+w_{+,\varepsilon}^{2}},\frac{w_{+,\varepsilon}}{r}\right)$
$\displaystyle=f\left(\frac{1+\frac{c-\varepsilon}{r^{2}}}{1+(r-\frac{c-\varepsilon}{r})^{2}},1-\frac{c-\varepsilon}{r^{2}}\right)$
$\displaystyle\geq
f\left(\frac{1}{1+r^{2}},1-\frac{c-\varepsilon}{r^{2}}\right)$
$\displaystyle\geq
1-\frac{c-\varepsilon}{r^{2}}+\frac{f_{x}(0,1)}{1+r^{2}}+\frac{1}{2(1-\frac{c-\varepsilon}{r^{2}})}f_{xx}(\xi,1)\frac{1}{(1+r^{2})^{2}}$
$\displaystyle=1+\frac{\varepsilon}{r^{2}}-\frac{c}{r^{2}(1+r^{2})}+\frac{1}{2(1-\frac{c-\varepsilon}{r^{2}})}\frac{f_{xx}(\xi,1)}{(1+r^{2})^{2}}$
$\displaystyle\geq 1+\frac{\varepsilon}{r^{2}}-O(r^{-4})$
$\displaystyle>1,\mbox{ for sufficiently large }r.$

For the next part, let $r_{0}>0$ and we assume

$\displaystyle v(r)>r-\frac{c-\epsilon}{r},\mbox{ for all }r>r_{0}.$

Then, by Proposition 3.2, we have

$\displaystyle v^{\prime}$ $\displaystyle=(1+v^{2})g\left(\frac{v}{r}\right)$
$\displaystyle\leq(1+r^{2})g\left(1-\frac{c-\varepsilon}{r^{2}}\right),$

since $g$ is decreasing.
We note that the second-order expansion of $g$ allows us to write

$\displaystyle v$
$\displaystyle\leq(1+r^{2})\left(g(1)-g_{y}(1)\frac{c-\epsilon}{r^{2}}+\frac{g_{yy}(\xi)}{2}\dfrac{c^{2}}{r^{4}}\right).$

Recall (9), $c=f_{x}(0,1)$ and $g(1)=0$, then we have $g_{y}(1)=-c^{-1}$ and

$\displaystyle v^{\prime}$
$\displaystyle\leq(1+r^{2})\left(\dfrac{c-\varepsilon}{cr^{2}}+\dfrac{g_{yy}(\xi)}{2}\dfrac{c^{2}}{r^{4}}\right)$
$\displaystyle=(1+r^{2})\dfrac{c-\varepsilon}{cr^{2}}+O(r^{-2})$
$\displaystyle\leq\left(1-\dfrac{\varepsilon}{c}\right)+O(r^{-2}).$

Therefore, we have obtained $v$ grows no more rapidly than
$\left(1-\dfrac{\epsilon}{c}\right)r$, contradicting that
$v(r)>r-\dfrac{c-\varepsilon}{r}$ for large enough $r>r_{0}$.

  2. 2.

Arguing as in the previous part, for sufficiently large $r$, we have

$\displaystyle
f\left(\frac{w_{-,\varepsilon}^{\prime}}{1+w_{-,\varepsilon}^{2}},\frac{w_{-,\varepsilon}}{r}\right)$
$\displaystyle=f\left(\frac{1-\frac{c+\varepsilon}{r^{2}}}{1+(r-\frac{c+\varepsilon}{r})^{2}},1-\frac{c+\varepsilon}{r^{2}}\right)$
$\displaystyle\leq f\left(\frac{1}{1+r^{2}},1-\frac{c-\epsilon}{r^{2}}\right)$
$\displaystyle=1-\frac{c+\varepsilon}{r^{2}}+\frac{f_{x}(0,1)}{1+r^{2}}+\frac{f_{xx}(\xi,1)}{2\left(1-\frac{c+\varepsilon}{r^{2}}\right)}\frac{1}{(1+r^{2})^{2}}$
$\displaystyle=1-\frac{\varepsilon}{r^{2}}-\frac{c}{r^{2}(1+r^{2})}+O(r^{-4})$
$\displaystyle=1-\frac{\varepsilon}{r^{2}}+O(r^{-4})$ $\displaystyle<1.$

For the next part, we fix $r_{0}>0$ and assume

$\displaystyle v(r)<r-\frac{c+\varepsilon}{r},\mbox{ for all }r>r_{0}.$

Then, by part (2) of Proposition 3.3 for sufficiently large $r$, we have
$v\geq r-1$ and hence,

$\displaystyle v^{\prime}$ $\displaystyle=(1+v^{2})g\left(\frac{v}{r}\right)$
$\displaystyle\geq(1+(r-1)^{2})g\left(1-\frac{c+\epsilon}{r^{2}}\right)$
$\displaystyle=(1+(r-1)^{2})\left(g(1)-g_{y}\left(1\right)\frac{c+\varepsilon}{r^{2}}+\dfrac{g_{yy}(\xi)}{2}\left(\frac{c+\varepsilon}{r^{2}}\right)^{2}\right)$
$\displaystyle\geq(1+(r-1)^{2})\left(\frac{c+\varepsilon}{cr^{2}}-O(r^{-4})\right)$
$\displaystyle\geq 1+\frac{\varepsilon}{c}+O(r^{-2}).$

Therefore, $v$ grows faster than
$\left(1+\dfrac{\varepsilon}{c}\right)r+O(r^{-1})$, contradicting
$v(r)<r-\dfrac{c+\epsilon}{r}$ for large enough $r>r_{0}$.

∎

######  Remark 10.

Proposition 3.4 implies $v(r)=r-\dfrac{c}{r}+o(r^{-1})$ as $r\to\infty$.

######  Remark 11.

The results of Proposition 3.4 were obtained independently in [2].

######  Proposition 3.5.

Let $c=f_{x}(0,1)$ and consider

$\displaystyle
w_{+,\varepsilon}(r)=r-\frac{c}{r}+\dfrac{\varepsilon}{r^{2}}\mbox{ and
}w_{-,\varepsilon}(r)=r-\frac{c}{r}-\dfrac{\varepsilon}{r^{2}}.$

  1. 1.

For every $\varepsilon\in(0,c)$, $w_{+,\varepsilon}(r)$ is a super-solution to
Eq. (6) for sufficiently large $r$. In addition, given any $r_{0}>0$, there
exists $r_{1}>r_{0}$ such that $v(r_{1})\leq
r_{1}-\dfrac{c}{r_{1}}+\dfrac{\varepsilon}{r_{1}^{2}}$. Thus, $v\leq
r-\dfrac{c}{r}+\dfrac{\varepsilon}{r^{2}}$ for sufficiently large $r$.

  2. 2.

For every $\varepsilon>0$, $w_{-,\varepsilon}(r)$ is a sub-solution to Eq. (6)
for sufficiently large $r$. Moreover, given any $r_{0}>0$, there exists
$r_{1}>r_{0}$ such that $v(r_{1})\geq
r_{1}-\dfrac{c}{r_{1}}-\dfrac{\varepsilon}{r_{1}^{2}}$. Thus, $v\geq
r-\dfrac{c}{r}-\dfrac{\varepsilon}{r^{2}}$ for sufficiently large $r$.

###### Proof.

  1. 1.

Indeed, as in the previous proof for sufficiently large $r$, we have

$\displaystyle
f\left(\frac{w_{+,\varepsilon}^{\prime}}{1+w_{+,\varepsilon}^{2}},\frac{w_{+,\varepsilon}}{r}\right)$
$\displaystyle=f\left(\frac{1+\frac{c}{r^{2}}-\dfrac{2\varepsilon}{r^{3}}}{1+\left(r-\dfrac{c}{r}+\dfrac{\varepsilon}{r^{3}}\right)^{2}},1-\dfrac{c}{r^{2}}+\dfrac{\varepsilon}{r^{3}}\right)$
$\displaystyle\geq
f\left(\dfrac{1+\dfrac{c}{r^{2}}-\dfrac{2\varepsilon}{r^{3}}}{1+r^{2}},1-\dfrac{c}{r^{2}}+\dfrac{\varepsilon}{r^{3}}\right)$
$\displaystyle\geq
1-\dfrac{c}{r^{2}}+\dfrac{\varepsilon}{r^{3}}+\dfrac{c}{1+r^{2}}\left(1+\dfrac{c}{r^{2}}-\dfrac{2\varepsilon}{r^{3}}\right)+\dfrac{f_{xx}(\xi,1)}{2}\dfrac{\left(1+\dfrac{c}{r^{2}}-\dfrac{2\varepsilon}{r^{3}}\right)^{2}}{\left(1+\dfrac{c}{r^{2}}-\dfrac{\varepsilon}{r^{3}}\right)(1+r^{2})^{2}}$
$\displaystyle=1+\frac{\varepsilon}{r^{3}}-O(r^{-4})$ $\displaystyle\geq 1.$

For the next part, fix $r_{0}>0$ and suppose
$v(r)>r-\dfrac{c}{r}+\dfrac{\varepsilon}{r^{2}}$ for all $r>r_{0}$. Then, by
recalling that $g(1)=0$, $g_{y}(1)=-c^{-1}$, we obtain

$\displaystyle v^{\prime}$ $\displaystyle=(1+v^{2})g\left(\frac{v}{r}\right)$
$\displaystyle\leq(1+r^{2})g\left(1-\frac{c}{r^{2}}+\frac{\varepsilon}{r^{3}}\right)$
$\displaystyle\leq(1+r)^{2}\left(g(1)-g_{y}(1)\left(\frac{c}{r^{2}}-\frac{\varepsilon}{r^{3}}\right)+\dfrac{g_{yy}(\xi)}{2}\left(\frac{c}{r^{2}}-\frac{\varepsilon}{r^{3}}\right)^{2}\right)$
$\displaystyle\leq 1-\frac{\varepsilon}{cr}+O(r^{-2}).$

However, this contradicts $v(r)>r-\dfrac{c}{r}+\dfrac{\varepsilon}{r^{2}}$ for
all $r>r_{0}$.

  2. 2.

As in part one for sufficiently large $r$, we have

$\displaystyle
f\left(\frac{w_{-,\varepsilon}^{\prime}}{1+w_{-,\varepsilon}^{2}},\frac{w_{-,\varepsilon}}{r}\right)$
$\displaystyle=f\left(\frac{1+\dfrac{c}{r^{2}}+\dfrac{2\varepsilon}{r^{3}}}{1+\left(r-\dfrac{c}{r}-\dfrac{\varepsilon}{r^{2}}\right)^{2}},1-\frac{c}{r^{2}}-\frac{\varepsilon}{r^{3}}\right)$
$\displaystyle\leq 1-\frac{\varepsilon}{r^{3}}+O(r^{-4})$ $\displaystyle\leq
1.$

On the other hand, fix $r_{0}>0$ and we assume that
$v(r)<r-\dfrac{c}{r}-\dfrac{\varepsilon}{r^{2}}$ for all $r>r_{0}$. Then, by
Part 2 of Proposition 3.4 for sufficiently large $r$, we have
$v>r-\dfrac{2c}{r}$ and hence,

$\displaystyle v^{\prime}$
$\displaystyle=\left(1+v^{2}\right)g\left(\frac{v}{r}\right)$ $\displaystyle
v^{\prime}$
$\displaystyle\geq\left(1+\left(r-\frac{2c}{r}\right)^{2}\right)g\left(1-\frac{c}{r^{2}}-\frac{\varepsilon}{r^{3}}\right)$
$\displaystyle\geq 1+\frac{\varepsilon}{cr}-O(r^{-2}).$

However, this contradics $v(r)<r-\dfrac{c}{r}-\dfrac{\varepsilon}{r^{2}}$ for
large enough $r>r_{0}$.

∎

######  Remark 12.

In the above proposition we have proven Theorem 1.1,
$v(r)=r-\dfrac{c}{r}+o(r^{-2})$.

######  Remark 13.

By following this process further with the speed functions
$f=Q_{k+1,k}=\dfrac{S_{k+1}}{S_{k}}$, we obtained the following expression for
the “bowl”-type solution

$\displaystyle
v(r)=r-\dfrac{c}{r}+\dfrac{n^{2}(n-k-1)(n-k-4)}{(k+1)^{3}(n-k)^{2}}\dfrac{1}{r^{3}}+O(r^{-5}).$

Note that when $k=0$, $Q_{k+1,k}=H$ and this expression coincide with the one
obtained in [1].

##  4 Uniqueness of entire solutions

With the fine asymptotic information about the bowl-type solitons that we have
obtained so far, we can now demonstrate the uniqueness result for entire
strictly convex solutions of (2) that are smoothly asymptotic to the
“bowl”-type solution for nondegenerate speed $f$.

To this end, we will recall a tangential principle for $f$-translators in
$\mathbb{R}^{n+1}$ developed by the second author in [10].

######  Theorem 4.1.

Let $\Sigma_{1},\Sigma_{2}\subset\mathbb{R}^{n+1}$ be two complete, embedded,
connected $f$-translators such that

  1. 1.

$f:\Gamma\to(0,\infty)$ satisfies properties a)-d).

  2. 2.

$\Sigma_{1}$ is strictly convex, i.e: the principal curvatures
$\lambda\in\Gamma_{+}$.

  3. 3.

$\Sigma_{2}$ is convex, i.e: the principal curvatures
$\lambda\in\overline{\Gamma}_{+}$ .

Then,

  1. 1.

Assume that there exists an interior point $p\in\Sigma_{1}\cap\Sigma_{2}$ such
that the tangent spaces coincide at $p$. If $\Sigma_{1}$ lies at one side of
$\Sigma_{2}$, then both hypersurfaces coincide.

  2. 2.

Assume that the boundaries $\partial\Sigma_{i}$ lie in the same hyperplane
$\Pi$ and the intersection of $\Sigma_{i}$ with $\Pi$ is transversal. If
$\Sigma_{1}$ lies at one side of $\Sigma_{2}$ and there exist
$p\in\partial\Sigma_{1}\cap\partial\Sigma_{2}$ such that the tangent spaces to
$\Sigma_{i}$ and $\partial\Sigma_{i}$ coincide, then both hypersurfaces
coincide.

###### Proof.

See [10] Theorem 1.4 for a proof and an additional remark. ∎

######  Remark 14.

We would like to highlight that the hypotheses of the Tangency Principle can
be modified by the followings:

  1. 1.

$\Gamma\subset\mathbb{R}^{n}$ is the connected component of
$\left\\{\lambda\in\mathbb{R}^{n}:\gamma(\lambda)>0\right\\}$ containig
$\Gamma_{+}$ and it is a convex cone.

  2. 2.

$f:\Gamma\to\mathbb{R}$ is smooth and symmetric and satisfies properties
c)-d).

  3. 3.

The principal curvatures of $\Sigma_{1}$ lies in $\Gamma$ and the principal
curvatures of $\Sigma_{2}$ lies in $\partial\Gamma\cap\left\\{\frac{\partial
f}{\partial\lambda_{i}}>0\right\\}$.

The reason of this is that the proof is based on a convex combination argument
of the local graphs near a tangency point of the hypersurfaces. An in
particular, by shrinking the domain if it necessary, the principal curvatures
of the convex combination is an admissible family for the functional
$f(\lambda)-\left<{\nu,e_{n+1}}\right>$, where a Hopf’s maximum principle
holds. We refer the reader to Thm. 1.1 in [3] for details.

######  Definition 2.

Let $\Sigma$ be an entire translating graphs that satisfies Equation (1).
Then, the end of $\Sigma$ is _smoothly asymptotic_ to the “bowl”-type soliton
if $\Sigma$ can be expressed outside a ball as a vertical graph of a function
$u_{\Sigma}$ such that

$\displaystyle
u_{\Sigma}(x)=\dfrac{|x|^{2}}{2}-\frac{c}{2}\ln(|x|^{2})+O\left(|x|^{-1}\right),\mbox{
as }|x|\to\infty.$ (14)

Note that the normalization $f(0,1)=1$ is used and $c\coloneqq\dfrac{\partial
f}{\partial\lambda_{1}}(0,1)$.

######  Remark 15.

As is noted in [10] (see sec. 4), Equation (2) is invariant under rotation
fields that fix the $x_{n+1}$-axis in $\mathbb{R}^{n+1}$. Consequently, under
the hypothesis of Theorem 1.2, it is enough to show that $\Sigma$ is symmetric
along the plane $\left\\{x_{1}=0\right\\}$ to obtain that $\Sigma$ is
rotationally symmetric.

We will now use Alexandrov’s moving plane method to establish uniqueness. To
this end, we recall the following definitions used in [5] and [6]:

  * •

$\Pi_{t}\coloneqq\left\\{x\in\mathbb{R}^{n+1}:\textbf{p}(x)=t\right\\}$, where
$\textbf{p}(x_{1},\ldots,x_{n+1})=x_{1}$ is the projection onto the first
coordinate. In addition, $\Pi\coloneqq\Pi_{0}$.

  * •

$Z_{t}\coloneqq\left\\{x_{n+1}>t\right\\}$.

  * •

Let $A\subset\mathbb{R}^{n+1}$ be an arbitrary subset. Then we set
$A_{+}(t)\coloneqq\left\\{x\in A:\textbf{p}(x)\geq t\right\\}$,
$A_{-}(t)\coloneqq\left\\{x\in A:\textbf{p}(x)\leq t\right\\}$ and
$\delta_{t}(A):=A\cap\Pi_{t}$. Note that $A_{+}(t)$ and $A_{-}(t)$ are the
right hand side and the left hand side, respectively, along $\Pi_{t}$ of $A$
(see Definition 3 given below).

  * •

The $1$-parameter families of right (respectively, left) reflections of $A$,
respectively, along the hyperplane $\Pi_{t}$ are given by
$A_{+}^{*}(t):=\left\\{(2t-x_{1},x_{2},\ldots,x_{n+1})\in\mathbb{R}^{n+1}:(x_{1},\ldots,x_{n+1})\in
A_{+}(t)\right\\}$ (respectively,
$A_{-}^{*}(t):=\left\\{(2t-x_{1},x_{2},\ldots,x_{n+1})\in\mathbb{R}^{n+1}:(x_{1},\ldots,x_{n+1})\in
A_{-}(t)\right\\}$).

  * •

We denote by $\pi:\mathbb{R}^{n+1}\to\Pi_{0}\subset\mathbb{R}^{n+1}$ the
orthogonal projection given by
$\pi(x_{1},\ldots,x_{n+1})=(0,x_{2},\ldots,x_{n+1})$.

######  Definition 3.

Let $A,B$ be two subsets of $\mathbb{R}^{n+1}$. Then, it will be said “$A$ is
on the right side of $B$” (denoted by $B\leq A$) if, and only if, for every
$x\in\Pi=\left\\{x_{1}=0\right\\}$ such that

$\displaystyle\pi^{-1}(\left\\{x\right\\})\cap A\neq\emptyset\mbox{ and
}\pi^{-1}(\left\\{x\right\\})\cap B\neq\emptyset,$

we have that

$\displaystyle\sup\left\\{\mathbf{p}(p):p\in\pi^{-1}(\left\\{x\right\\})\cap
B\right\\}\leq\inf\left\\{\mathbf{p}(p):p\in\pi^{-1}(\left\\{x\right\\})\cap
A\right\\}.$ (15)

######  Remark 16.

The proof of Theorem 1.2 uses the method of moving planes of Alexandrov in the
spirit of [8], [5] and [6]. This method specifically requires three properties
to be applied:

  * •

A family of hyperplanes as translating solutions of Equation (2). To
accomplish this, the Property e) on $f$ is needed.

  * •

Tangential principles in the interior and at the boundary, since we will need
to understand how the translators intersect tangentially with their
reflections along $\Pi_{t}$.

  * •

“Enough space” to start the method. This means that we can find $t>0$ large
enough such that the reflection of $\Sigma$ along $\Pi_{t}$ is a graph over
$\Pi$ and $\Sigma_{-}(t)\leq\Sigma^{*}_{+}(t)$. To compare the horizontal
distance between $\Sigma_{+}^{*}$ and $\Sigma_{-}$, we will need the
asymptotic behavior of the “bowl”-type solution at infinity.

Next, we will follow the arguments used in [6] theorem 6, for $f$-translators
which are smoothly asymptotic to the “bowl”-type solution.

######  Lemma 4.2.

There exists $r_{0}>R$ such that $\Sigma_{+}(t)$ is a graph over $\Pi$ for
every $t>r_{0}$.

###### Proof.

Firstly, we note that for every $t>R$, $\Sigma_{+}(t)$ possess only one
unbounded connected component. If this were false, one could choose a compact
component $\Sigma^{\prime}\subset\Sigma_{+}(t)$ and a $t$ large enough such
that $\Sigma\cap\Pi_{t}=\emptyset$. Then, by translating $\Pi_{t}$ until it
touches $\Sigma^{\prime}$ at a first order contact point222First, this is
possible because the first-order contact point coincides with the point at
which the two hypersurfaces first touch. In addition, the first order contact
point means that the graph and gradient coincide. In particular, the tangent
planes coincide because the unit normal vectors of the two hypersurfaces
coincide at this point., say
$\Sigma^{\prime}\cap\Pi_{t^{\prime}}=\left\\{p\right\\}$ for some
$0<t^{\prime}<t$, Theorem 4.1, applied to $\Sigma^{\prime}$ with
$\Pi_{t^{\prime}}$, implies that $\Sigma^{\prime}$ is totally geodesic
contradicting that $\Sigma^{\prime}$ is strictly convex.

Next, by Equation (14), we have

$\displaystyle
d_{x}u_{\Sigma}(e_{1})\geq\left(1-\dfrac{C}{|x|^{2}}\right)\left<{x,e_{1}}\right>,$

for some constant $C>0$ and $|x|\geq R$. Therefore, by choosing $r_{0}>R$
large enough such that $1-\dfrac{C}{|x|^{2}}\geq\varepsilon>0$, it follows
that $d_{x}u_{\Sigma}(e_{1})>0$ whenever $\left<{x,e_{1}}\right>\geq r_{0}$.
Finally, Lemma 4.2 follows since $\Sigma$ is properly embedded and
$\Sigma_{+}(r_{1})\cup\pi\left(\Sigma_{+}(r_{1})\right)$ bounds a domain in
$\mathbb{R}^{n+1}$. ∎

From Lemma 4.2, it follows that $\Sigma^{*}_{+}(t)\cap Z_{R}$ is a vertical
graph of a function satisfying

$\displaystyle u^{*}_{\Sigma}(x)=u_{\Sigma}(2t-x_{1},x_{2},\ldots,x_{n})$

for any $t>r_{0}$ .

######  Lemma 4.3.

Let $a>0$. There exists $r_{1}>r_{0}$ such that for $|x|\geq r_{1}$ and
$t>a+x_{1}$ we have

$\displaystyle u^{*}_{\Sigma}(x)-u_{\Sigma}(x)\geq\epsilon,$

for some $\epsilon>0$.

###### Proof.

The proof follows the same method as in [5] Step 3 Theorem A. ∎

From Lemma 4.3 we have that for every $a>0$ and $t\geq r_{1}$

$\displaystyle\Sigma_{-}(t+a)\cap\left\\{x_{1}\leq
t\right\\}\leq\Sigma_{+}^{*}(t+a)\cap\left\\{x_{1}\leq t\right\\}.$

Moreover, since $\Sigma_{+}(t)$ is a graph over $\Pi$ for $t\geq r_{1}$, it
follows that

$\displaystyle\Sigma_{-}(t+a)\cap\left\\{t\leq x_{1}\leq
t+a\right\\}\leq\Sigma_{+}^{*}(t+a)\cap\left\\{t\leq x_{1}\leq t+a\right\\}.$

In conclusion, by setting

$\displaystyle\mathcal{A}\coloneqq\left\\{t\in[0,\infty):\Sigma_{+}(t)\mbox{
is a graph over }\Pi\mbox{ and }\Sigma_{-}(t)\leq\Sigma_{+}^{*}(t)\right\\},$

we see that $\mathcal{A}\neq\emptyset$ since $t+a\in\mathcal{A}$ for $t\geq
r_{1}$.

###### Proof of Theorem 1.2.

Firstly, by the Remark 15 it is only necessary to prove that $\Sigma$ is
symmetrical about the hyperplane $\Pi=\left\\{x_{1}=0\right\\}$. Moreover, by
the lemmas 4.2 and 4.3, the set

$\displaystyle\mathcal{A}=\left\\{t\in[0,\infty):\Sigma_{+}(t)\mbox{ is a
graph over }\Pi\mbox{ and }\Sigma_{-}(t)\leq\Sigma_{+}^{*}(t)\right\\},$

is not empty. Then, we apply the same arguments as in [5] to obtain
$\mathcal{A}$ is closed and open subset of $[0,\infty)$. Consequently,
$\mathcal{A}=[0,\infty)$, which means $0\in\mathcal{A}$. In fact, we have
obtained $\Sigma_{-}(0)\leq\Sigma_{+}^{*}(0)$, and by analogous arguments, it
can be shown that $\Sigma_{-}^{*}(0)\leq\Sigma_{+}(0)$. Note that the
combination of these two properties implies that $\Sigma$ is symmetric with
respect the hyperplane $\Pi$. ∎

##  5 A Degenerate Example: $\sqrt[n]{S_{n}}$.

As seen in the previous section, the asymptotic behavior of the “bowl”-type
solution is that of a paraboloid when the velocity function is nondegenerate.

In this section we will show that the situation can be very different for
nondegenerate speeds, taking as an example the function $\sqrt[n]{S_{n}}$.

######  Theorem 5.1.

The slope of the “bowl”-type $\sqrt[n]{S_{n}}$-translator is given by

$\displaystyle
v(r)=\begin{cases}e^{\frac{r^{2}}{2}}+O\left(\sqrt{e^{\frac{r^{2}}{2}}-1}\right),&\mbox{
for }n=2,\\\ \dfrac{r^{3}}{3}+O(1),&\mbox{ for }n=3,\\\
\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+O\left(\dfrac{1}{r^{\frac{n}{n-2}}}\right),&\mbox{
for }n\geq 4.\end{cases},\mbox{ as }r\to\infty$

The proof of this theorem will be split into the following propositions.
First, recall that the slope of the “bowl”-type $\sqrt[n]{S_{n}}$-translator
satisfies the ODE

$\displaystyle\begin{cases}v^{\prime}=(1+v^{2})\left(\dfrac{r}{v}\right)^{n-1},\\\
v(0)=0.\end{cases}$ (16)

We start with the case of $n=2$.

######  Proposition 5.2.

For $n=2$ the smooth solution to (16) satisfies

$\displaystyle
v(r)=e^{\frac{r^{2}}{2}}+O\left(\sqrt{e^{\frac{r^{2}}{2}}-1}\right),\mbox{ as
}r\to\infty.$

###### Proof.

Firstly, for $n=2$, we note that Equation (16) has the form

$\displaystyle v^{\prime}=(1+v^{2})\dfrac{r}{v}.$ (17)

Therefore, $v=\sqrt{e^{\frac{r^{2}}{2}}-1}$ for $r>0$. Then, Proposition 5.2
holds by the explicit expression of the solution. ∎

######  Proposition 5.3.

For $n=3$ the smooth solution to (16) satisfies

$\displaystyle v(r)=\dfrac{r^{3}}{3}+O(1),\mbox{ as }r\to\infty.$

###### Proof.

For $n\geq 3$, Equation (16) can be solved implicitly by

$\displaystyle\dfrac{v^{n-2}}{n-2}-\int\limits_{0}^{v}\dfrac{t^{n-3}}{1+t^{2}}dt=\dfrac{r^{n}}{n}.$
(18)

Particularly, for $n=3$, we have the implicit equation

$\displaystyle v-\arctan(v)=\dfrac{r^{3}}{3}.$

On the other hand, the function $f(x)=x-\arctan(x)$ has the following
asymptotic expansion

$\displaystyle f(x)=x-\dfrac{\pi}{2}+O(x^{-1}),\mbox{ as }x\to\infty.$

Then, an easy calculation reveals that $f^{-1}(x)$ has the following
asymptotic given by

$\displaystyle f^{-1}(x)=x+\dfrac{\pi}{2}-O(x^{-1}),\mbox{ as }x\to\infty.$

Consequently, we obtain that the asymptotic expansion of $v$ is given by

$\displaystyle v(r)=\dfrac{r^{3}}{3}+\dfrac{\pi}{2}-O(r^{-3}),\mbox{ as
}r\to\infty.$

∎

######  Proposition 5.4.

For $n>3$, the asymptotic behavior of the solution to (16) is given by

$\displaystyle
v(r)=\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+O\left(\dfrac{1}{r^{\frac{n}{n-2}}}\right),\mbox{
as }r\to\infty.$

###### Proof.

The proof is divided in several steps.

######  Step 5.1.

$v(r)\geq\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}$ for
$r>0$.

Firstly, we are going to show that the function
$w(r)=\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}$ is a sub-
solution to Eq. (16) for $r\geq
R(n):=\left(\dfrac{n-2}{2}\right)^{\frac{n-2}{2}}$. Indeed, we note that

$\displaystyle w^{\prime}-(1+w^{2})\left(\dfrac{r}{w}\right)^{n-1}$
$\displaystyle=\left(\dfrac{n}{n-2}\right)^{\frac{n-3}{n-2}}r^{\frac{2}{n-1}}-\left(1+\left(\dfrac{n-2}{n}\right)^{\frac{2}{n-2}}r^{\frac{2n}{n-2}}\right)\left(\frac{n-2}{n}\right)^{-\frac{(n-1)}{n-2}}r^{\frac{-2(n-1)}{n-2}}$
$\displaystyle=-\dfrac{1}{\left(\dfrac{n-2}{n}\right)^{\frac{n-1}{n-2}}r^{\frac{2(n-1)}{n-2}}}<0.$

Since $w(0)=v(0)=0$, we get $v(r)\geq w(r)$ for all $r\geq R(n)$.

######  Step 5.2.

For $n>3$,
$v(r)=\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+o\left(r^{\frac{n}{n-2}}\right)$
as $r\to\infty$.

We argue by contradiction which means there exists $\varepsilon>0$ such that
for all $r_{0}\gg 1$ it holds

$\displaystyle(1+\varepsilon)w(r)\leq v(r),\mbox{ for all }r\geq r_{0}.$

In particular,

$\displaystyle
v^{\prime}=(1+v^{2})\left(\dfrac{r}{v}\right)^{n-1}\leq\dfrac{r^{n-1}w^{-(n-3)}}{(1+\varepsilon)^{n-1}}=\dfrac{1}{(1+\varepsilon)^{n-1}}\left(\dfrac{n-2}{n}\right)^{-\frac{n-3}{n-2}}r^{\frac{2}{n-2}}.$

Then, by the previous step, we have

$\displaystyle\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}\leq
v\leq\dfrac{\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}}{(1+\varepsilon)^{n-1}}r^{\frac{n}{n-2}}.$

Note that this is impossible since $\varepsilon>0$. Consequently,
$v(r)=w(r)+o\left(r^{\frac{n}{n-2}}\right)$ for $r\to\infty$.

Next, since
$v(r)=\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+o(r^{\frac{n}{n-2}})$
as $r\to\infty$, we can find a non-negative function $\varphi(r)$ with the
following property: for every $C>0$ and all $r\gg 1$ it holds
$|\varphi(r)|\leq Cr^{\frac{n}{n-2}}$ and

$\displaystyle
v(r)=\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+\varphi(r),\mbox{
for }r\geq r_{0}.$

In addition, $\varphi$ satisfies the following equation

$\displaystyle\varphi^{\prime}$
$\displaystyle=v^{\prime}-\left(\dfrac{n}{n-2}\right)^{\frac{n-3}{n-2}}r^{\frac{2}{n-2}}$
$\displaystyle=\left(1+\left(\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+\varphi\right)^{2}\right)\dfrac{r^{n-1}}{\left(\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+\varphi\right)^{n-1}}-\left(\frac{n}{n-2}\right)^{\frac{n-3}{n-2}}r^{\frac{2}{n-2}}.$

######  Step 5.3.

$\varphi\to 0$ as $r\to\infty$.

We argue by contradictions, this means there exist $\varepsilon>0$ such that
for all $r_{0}\gg 1$ and every $C>0$ it holds

$\displaystyle\varepsilon\leq\varphi(r)\leq Cr^{\frac{n}{n-2}},\mbox{ for all
}r\geq r_{0}.$

In particular, by letting $C\to 0$ in

$\displaystyle\varphi^{\prime}=$
$\displaystyle\left(1+\left(\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+\varphi\right)^{2}\right)\dfrac{r^{n-1}}{\left(\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+\varphi\right)^{n-1}}-\left(\frac{n}{n-2}\right)^{\frac{n-3}{n-2}}r^{\frac{2}{n-2}}$
$\displaystyle\leq$
$\displaystyle\left(1+r^{\frac{2n}{n-2}}\left(\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}+C\right)^{2}\right)\dfrac{r^{n-1}}{\left(\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+\epsilon\right)^{n-1}}-\left(\frac{n}{n-2}\right)^{\frac{n-3}{n-2}}r^{\frac{2}{n-2}}$
$\displaystyle=$
$\displaystyle\left(\left(\dfrac{n}{n-2}\right)^{\frac{n-1}{n-2}}\left(\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}+C\right)^{2}-\left(\dfrac{n}{n-2}\right)^{\frac{n-3}{n-2}}\right)r^{\frac{2}{n-2}}$
$\displaystyle-\dfrac{(n-1)\left(\dfrac{n}{n-2}\right)^{\frac{n}{n-2}}\left(\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}+C\right)^{2}\epsilon}{r}+O\left(\dfrac{1}{r^{\frac{(n-1)}{(n-2)}}}\right),$

we observe that

$\displaystyle\varphi^{\prime}\leq-(n-1)\left(\dfrac{n}{n-2}\right)\dfrac{\varepsilon}{r}+O\left(\dfrac{1}{r^{\frac{(n-1)}{(n-2)}}}\right)\leq
0.$

Therefore, $\varphi$ is strictly decreasing but this contradicts that
$\varepsilon\leq\varphi$.

######  Step 5.4.

$\varphi\geq\left(\dfrac{n}{n-2}\right)^{\frac{1}{n-2}}\dfrac{1}{r^{\frac{n}{n-2}}}$
for large enough $r$.

Let $r_{1}\gg 1$ and $C>0$ such that

$\displaystyle 0\leq\varphi(r)\leq Cr^{\frac{n}{n-2}},\mbox{ for all }r\geq
r_{1}.$

Then, we may estimate $\varphi$ by letting $C\to 0$ in the following
expression

$\displaystyle\varphi^{\prime}=$
$\displaystyle\left(1+\left(\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+\varphi\right)^{2}\right)\dfrac{r^{n-1}}{\left(\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+\varphi\right)^{n-1}}-\left(\frac{n}{n-2}\right)^{\frac{n-3}{n-2}}r^{\frac{2}{n-2}}$
$\displaystyle\geq$
$\displaystyle\dfrac{1+r^{\frac{2n}{n-2}}\left(\frac{n-2}{n}\right)^{\frac{2}{n-2}}}{r^{\frac{2(n-1)}{n-2}}\left(\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}+C\right)^{n-1}}-\left(\frac{n}{n-2}\right)^{\frac{n-3}{n-2}}r^{\frac{2}{n-2}}.$
$\displaystyle=$
$\displaystyle\left(\dfrac{\left(\frac{n-2}{n}\right)^{\frac{2}{n-2}}}{\left(\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}+C\right)^{n-1}}-\left(\dfrac{n}{n-2}\right)^{\frac{n-3}{n-2}}\right)r^{\frac{2}{n-2}}+\dfrac{1}{\left(\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}+C\right)^{n-1}r^{\frac{2(n-1)}{n-2}}}.$

Therefore, we have that

$\displaystyle\varphi^{\prime}\geq\left(\dfrac{n}{n-2}\right)^{\frac{n-1}{n-2}}\dfrac{1}{r^{\frac{2(n-1)}{n-2}}},\mbox{
as }r\to\infty.$

Then, by integrating form $r$ to $\infty$ we finally obtain

$\displaystyle\varphi\geq\left(\dfrac{n}{n-2}\right)^{\frac{1}{n-2}}\dfrac{1}{r^{\frac{n}{n-2}}}.$

######  Step 5.5.

$\varphi=O\left(\dfrac{1}{r^{\frac{n}{n-2}}}\right)$.

We note that for $r_{0}\gg 1$ it holds

$\displaystyle\left(\dfrac{n}{n-2}\right)^{\frac{1}{n-2}}\dfrac{1}{r^{\frac{n}{n-2}}}\leq\varphi(r)\leq
Cr^{\frac{n}{n-2}},\mbox{ for all }r\geq r_{0}.$

Next, by letting $C\to 0$ in

$\displaystyle\varphi^{\prime}=$
$\displaystyle\left(1+\left(\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+\varphi\right)^{2}\right)\dfrac{r^{n-1}}{\left(\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{n}{n-2}}+\varphi\right)^{n-1}}-\left(\frac{n}{n-2}\right)^{\frac{n-3}{n-2}}r^{\frac{2}{n-2}}$
$\displaystyle\leq$
$\displaystyle\left(1+r^{\frac{2n}{n-2}}\left(\left(\frac{n-2}{n}\right)^{\frac{1}{n-2}}+C\right)^{2}\right)\dfrac{r^{\frac{2(n-1)^{2}}{n-2}}}{\left(\left(\dfrac{n-2}{n}\right)^{\frac{1}{n-2}}r^{\frac{2n}{n-2}}+\left(\dfrac{n}{n-2}\right)^{\frac{1}{n-2}}\right)^{n-1}}$
$\displaystyle-\left(\frac{n}{n-2}\right)^{\frac{n-3}{n-2}}r^{\frac{2}{n-2}},$

we deduce that

$\displaystyle\varphi^{\prime}\leq-\dfrac{(n-2)\left(\dfrac{n}{n-2}\right)^{\frac{n-1}{n-2}}}{r^{\frac{2(n-1)}{n-2}}}+O\left(\dfrac{1}{r^{\frac{4(n-1)}{n-2}}}\right).$

Finally, by integrating over $r$ to $\infty$, we obtain

$\displaystyle\varphi\leq\dfrac{(n-2)\left(\dfrac{n}{n-2}\right)^{\frac{1}{n-2}}}{r^{\frac{n}{n-2}}}.$

∎

This concludes the proof of Proposition 16.

######  Remark 17.

We note that the proof of the uniqueness theorem presented in section 4 also
applies for entire strictly convex solutions smoothly asymptotic to the
“bowl”-type translators of the $\sqrt[n]{S_{n}}$-flow. To see this, we choose
$R>0$. Then, the entire $\sqrt[n]{S_{n}}$-translator can be written as a
vertical graph $u_{\Sigma}:\mathbb{R}^{n}\setminus B_{R}(0)\to\mathbb{R}$ such
that

$\displaystyle u_{\Sigma}=h_{1}(r^{2})+O\left(h_{2}(r^{2})\right),\mbox{ as
}r\to\infty,$

where $|x|=r$ and

$\displaystyle h_{1}(r^{2})=\begin{cases}\int e^{\frac{r^{2}}{2}}dr,&\mbox{
for }n=2,\\\ \frac{1}{12}r^{4},&\mbox{ for }n=3,\\\
\dfrac{(n-2)^{\frac{n-1}{n-2}}}{2(n-1)n^{\frac{1}{n-2}}}r^{\frac{2(n-1)}{n-2}},&\mbox{
for }n\geq
4.\end{cases},\>h_{2}(r^{2})=\begin{cases}\int\sqrt{e^{\frac{r^{2}}{2}}-1}dr,&\mbox{
for }n=2,\\\ r,&\mbox{ for }n=3,\\\ \frac{1}{r^{\frac{2}{n-2}}},&\mbox{ for
}n\geq 4.\end{cases}$

Then, Lemma 4.2 holds for $f=\sqrt[n]{S_{n}}$, since

$\displaystyle
d_{x}u_{\Sigma}(e_{1})\geq\left(h_{1}^{\prime}(|x|^{2})-Ch_{2}(|x|^{2})\right)\left<{x,e_{1}}\right>,$

and

$\displaystyle
0<h_{1}^{\prime}(|x|^{2})-Ch_{2}^{\prime}(|x|^{2})=\begin{cases}e^{\frac{|x|^{2}}{2}}-C\sqrt{e^{\frac{|x|^{2}}{2}}-1},&\mbox{
for}n=2\\\ \dfrac{|x|^{3}}{3}-C,&\mbox{ for }n=3\\\
\dfrac{(n-1)}{(n-2)^{\frac{n-3}{n-2}}}|x|^{\frac{2}{n-2}}-\dfrac{C}{|x|^{\frac{n}{n-2}}},&\mbox{
for }n\geq 4.\end{cases},$ (19)

for large enough $r\geq R$ and some positive constants $C$.

In addition, we have that Lemma 4.3 also holds, since by the mean value
theorem, it follows that

$\displaystyle u^{*}_{\Sigma}(x)-u_{\Sigma}(x)$
$\displaystyle=2(t-x_{1})\dfrac{\partial u_{\Sigma}}{\partial
x_{1}}(\xi,x_{2},\ldots,x_{n}),$

for some $\xi\in(x_{1},2t-x_{1})$ with $|x|>r_{0}$.

Then, by writing $(\xi,x^{\prime})=(\xi,x_{2},\ldots,x_{n})$, the asymptotic
expression of $u_{\Sigma}$ gives us

$\displaystyle 2(t-x_{1})\dfrac{\partial u_{\Sigma}}{\partial
x_{1}}(\xi,x_{2},\ldots,x_{n})$ $\displaystyle\geq
4(t-x_{1})\xi\left(h_{1}^{\prime}(|(\xi,x^{\prime})|^{2})-Ch_{2}^{\prime}(|(\xi,x^{\prime})|^{2})\right)$
$\displaystyle\geq
4ar_{0}(h_{1}(|(\xi,x^{\prime})|^{2})^{\prime}-h_{2}^{\prime}(|(\xi,x^{\prime})|^{2})).$

Finally, by choosing $r_{1}\geq r_{0}$ as in (19), the difference
$h_{1}(|(\xi,x^{\prime})|^{2})^{\prime}-h_{2}^{\prime}(|(\xi,x^{\prime})|^{2})$
is uniformly bounded from below. Consequently, the proof of Theorem 1.2 holds.

##  6 Wing-like translators

An exotic translating solution to the mean curvature is the _wing-like_
translator discussed in [1]. This solution is connected but not graphical, and
is rather the union of two graphs each defined on the complement of a ball
$\mathbb{R}^{n}-B_{R}$. The upper and lower halves are each asymptotic to
paraboloids.

In this section, we will discuss the existence and the asymptotic behavior of
_wing-like_ translators for _nondegenerate_ speeds (see definition 1). In
particular, we will see that the upper and lower branches need not always have
the same asymptotics.

More precisely, a _wing-like_ solution is a non-convex rotationally symmetric
hypersurface in $\mathbb{R}^{n+1}$ with compact boundary (possible empty)
along the $x_{n+1}$-axis which is at distance $R>0$ from the origin that
satisfies Eq. (2).
Furthermore, by removing a large enough ball from a _wing-like_ solution we
have at possible two branches given by vertical graphs:

  * •

We are going to show that the upper branch is always asymptotic to the
“bowl”-type solution.

  * •

Meanwhile, the lower branch has shown different behaviors at infinity with
respect to the speed function $f$.

In particular, we will show that for the function $f=\sqrt[k]{S_{k}}$ with $k$
even the lower branch is asymptotic to symmetric reflection of “bowl” type
solution, for $k$ odd, the lower posses a boundary component, and for the
function $f=Q_{k+1,k}$, the lower branch whose gradient vanishes at infinity.

######  Remark 18.

We note that the ODE satisfying a rotationally symmetric
$Q_{k+1,k}$-translator has as solution $v=0$, but recall that the horizontal
hyperplanes are not vertical translators and that $Q_{k+1,k}$ is not well
defined at $0$.

######  Remark 19.

It is an open problem to characterize the dichotomy of the lower branch of
wing-like solutions for any _nondegenerate_ speed $f$.

Now we will follow the construction given in [1]. Firstly, we will start with
the a small non-convex portion of the _wing-like_ solution.

######  Step 6.1.

Construction of a small portion of the _wing-like_ $f$-translator as a graph
over the $x_{n+1}$-axis.

At the point where the tangent space is not orthogonal to $e_{n+1}$, the
translator can be represented locally as a graph of a function
$r:(a,b)\to(0,\infty)$ over the $x_{n+1}$-axis by

$\displaystyle\bigcup_{x_{n+1}}r(x_{n+1})\cdot\mathbb{S}^{n-1}\times\\{x_{n+1}\\}.$

Note that $r(x_{n+1})$ represents the radius of the hypersurface given its
last coordinate.

Next, we are going to find the ODE the $r$ satisfies. Firstly, at points where
the tangent space is not parallel to $e_{n+1}$, the _wing-like_ $f$-translator
can be described as a graph of a rotationally symmetric function $u(r)$ over
the hyperplane $\left\\{x_{n+1}=0\right\\}$.

Note that by construction $u\circ r(x_{n+1})=x_{n+1}$, and by the chain rule,
we have the following equations:

$\displaystyle u^{\prime}(r)=\dfrac{1}{r^{\prime}}\mbox{ and
}u^{\prime\prime}(r)=-\dfrac{r^{\prime\prime}}{(r^{\prime})^{3}}.$

Consequently, since a _wing-like_ solution is not convex, Eq. (2) has the form
(7). Then, since $u$ satisfies equation (8), we obtain

$\displaystyle
r^{\prime\prime}=-(1+(r^{\prime})^{2})g\left(r^{-1},r^{\prime}\right),$

where
$g\left(r^{-1},r^{\prime}\right)=r^{\prime}g\left((rr^{\prime})^{-1},1\right)$
since $g(y,z)$ is $1$-homogeneous. We note that in previous sections we denote
$g(y)$ instead of $g(y,1)$ to economize notation.

In addition, at points of the _wing-like_ $f$-translator where the tangent
space is vertical, we may argue in the same way as before by considering the
branches separately.

Finally, we claim that there exists $\varepsilon>0$ and a strictly convex
solution to the problem

$\displaystyle\begin{cases}r^{\prime\prime}=-(1+(r^{\prime})^{2})g\left(r^{-1},r^{\prime}\right),\>x_{n+1}\in(h_{0}-\varepsilon,h_{0}+\varepsilon),\\\
r(h_{0})=R,\>r^{\prime}(h_{0})=0.\end{cases},$ (20)

where $h_{0}\in\mathbb{R}$ and $R>0$. Note that a different choice of $h_{0}$
corresponds to a vertical translation of the _wing-like_ $f$-translator.
Therefore, we will assume that $h_{0}=0$.

Indeed, since Eq. (20) is not degenerate and the he right-hand side is at
least of class $\mathcal{C}^{2}$, the classical theory of ODEs holds. This
means that there exist $\varepsilon>0$ and a solution to Eq. (20) in
$(-\varepsilon,\varepsilon)$.

Then, by shrinking $\varepsilon>0$ if necessary, we may assume that this
solution is strictly convex. This is due to continuity and the fact that

$\displaystyle r^{\prime\prime}(0)=-g\left(R^{-1},0\right)>0.$

To see this, we note that by taking derivatives with respect $z$ in Eq.(7), we
have$f_{x}g_{z}=1$. Then, since $f$ is strictly increasing, $g_{z}(y,z)>0$.
Therefore, by recalling that

$g(1,1)=0\mbox{ and }g(R^{-1},0)<0$

the claim holds. Completing the construction of the small portion of the
“wing”-like $f$-translator.

######  Remark 20.

In the particular case of $f=\sqrt[k]{S_{k}}$, the ODE that $r$ satisfies is

$\displaystyle
r^{\prime\prime}=-(1+(r^{\prime})^{2})\left(\dfrac{(r^{\prime})^{k}r^{k-1}}{\binom{n-1}{k-1}}-\dfrac{(n-k)}{kr}\right).$

Therefore, when $k$ is an even number, we have that $\tilde{r}=r(-x_{n+1})$ is
also a solution to Eq. (20).

######  Proposition 6.1.

The principal curvatures of the upper half small portion of the _wing-like_
$f$-translator belong into the cone

$\left\\{f\mbox{ satisy properties }\ref{a}-\ref{e}\mbox{ and
}f(0,1)>0\right\\}.$

###### Proof.

Firstly, we note that Eq. (5) implies that

$\displaystyle\lambda_{1}=\frac{-r^{\prime\prime}}{(1+r^{\prime
2})^{3/2}}\mbox{ and }\lambda_{i}=\frac{1}{r\sqrt{1+r^{\prime 2}}}.$

are the principal curvatures of the small portion of the _wing-like_
$f$-translator.

Then, by differentiating them

$\displaystyle\lambda_{1}^{\prime}=\frac{-r^{\prime\prime\prime}}{(1+r^{\prime
2})^{3/2}}+\frac{3(r^{\prime\prime})^{2}r^{\prime}}{(1+r^{\prime
2})^{5/2}}\mbox{ and
}\lambda_{i}^{\prime}=\frac{-r^{\prime}}{r^{2}\sqrt{1+r^{\prime
2}}}-\frac{r^{\prime}r^{\prime\prime}}{r(1+r^{\prime 2})^{3/2}}$

Next, we will be evaluate the principal curvatres at $u=0$333 $u>0$ is the
upper part of the winglike solution and $u<0$ is the lower.. Differentiating
$r^{\prime\prime}$, we have

$\displaystyle
r^{\prime\prime\prime}=-2r^{\prime}r^{\prime\prime}g-(1+r^{\prime
2})g_{y}\frac{r^{\prime}}{r^{2}}-(1+r^{\prime 2})g_{z}r^{\prime\prime},$

where $g,g_{y},g_{z}$ are all evaluated at the point
$\left(r^{-1},r^{\prime}\right)$. Therefore, we obtain

$\displaystyle
r^{\prime\prime\prime}(0)=-g_{z}\left(R^{-1},0\right)r^{\prime\prime}(0)=g_{z}\left(R^{-1},0\right)g\left(R^{-1},0\right).$

Finally, by recalling $r(0)=R$, $r^{\prime}(0)=0$ and
$r^{\prime\prime}(0)=-g(R^{-1},0)<0$, we see that

$\displaystyle\lambda_{1}^{\prime}(0)=-g_{z}\left(R^{-1},0\right)g\left(R^{-1},0\right)>0,\mbox{
and }\lambda_{i}^{\prime}(0)=0.$

Therefore, since in the upper half of the small portion
$\left\\{u>0\right\\}$, $r$ is strictly decreasing. We obtain that
$\lambda_{i}^{\prime}>0$ for all $i=1,\ldots,n$ finalizing the proposition. ∎

######  Step 6.2.

Construction of the upper branch of the _wing-like_ $f$-translator.

We have a solution on some interval $u\in[0,\varepsilon)$. To continue the
solution, we may revert to the standard representation

$\displaystyle\begin{cases}v^{\prime}=(1+v^{2})g\left(\dfrac{v}{r}\right)\\\
v(r_{0})=v_{0}\end{cases}$

where we may choose $r_{0}\in(R,R+\varepsilon)$ arbitrarily. Then, since
$v\to\infty$ as $r\to R^{+}$, we may assume $\dfrac{v_{0}}{r_{0}}\geq 1$, by
choosing $r_{0}$ sufficiently close to $R.$ By standard ODE theory we have
existence and uniqueness as long as the right hand side is well-defined.

If $\dfrac{v_{0}}{r_{0}}>1$, the right hand side of the ODE is initially
negative, hence the function decreases until $g\left(\dfrac{v}{r}\right)=0$
(i.e.: $\dfrac{v}{r}=1$). So we only need to discuss the case
$\dfrac{v_{0}}{r_{0}}=1$. If this is the case, then the function $v_{+}=r$ is
a supersolution to the ODE, and $v_{-}=\dfrac{r}{f(1,1)}$ is a subsolution, as
can be seen in [7] Section 7.1. The proof that $\dfrac{v}{r}\to 1$ is just a
special case of Proposition 2 of [7].

On the other hand, for the lower branch of the _wing-like_ $f$-translator, we
have a solution on some interval $(-\varepsilon,0]$. As with the upper branch,
we may revert to the standard representation

$\displaystyle\begin{cases}v^{\prime}=(1+v^{2})g\left(\dfrac{v}{r}\right),\\\
v(r_{0})=v_{0}.\end{cases},$

where $r_{0}\in(R,R+\varepsilon)$. Then, Since $v\to-\infty$ as $r\to R^{+}$,
we may assume $\dfrac{v_{0}}{r_{0}}\leq-1$, by choosing $r_{0}$ sufficiently
close to $R.$ By standard ODE theory we have existence and uniqueness as long
as the right hand side is well-defined.

######  Step 6.3.

Construction and asymptotic behavior of the lower branch of the _wing-like_
$\sqrt[k]{S_{k}}$-translator for $k$ even.

Firstly, by construction the principal curvatures of the small portion of
$W_{R}$ are given by

$\displaystyle\lambda_{1}=\dfrac{g(r^{-1},r^{\prime})}{\sqrt{1+(r^{\prime})^{2}}}=\dfrac{\dfrac{(r^{\prime})^{k}r^{k-1}}{\binom{n-1}{k-1}}-\dfrac{(n-k)}{kr}}{\sqrt{1+(r^{\prime})^{2}}}\mbox{
and }\lambda_{i}=\dfrac{1}{r\sqrt{1+(r^{\prime})^{2}}}.$

Then, since $\lambda\in\Gamma_{k}$ if, and only if,

$\displaystyle\binom{n-1}{l}\lambda_{i}^{l}+\binom{n-1}{l-1}\lambda_{1}\lambda_{i}^{l-1}>0,\mbox{
for }l=1,\ldots,k,$

or equivalently, we have

$\displaystyle
0<\dfrac{\binom{n-1}{l-1}}{r^{l}(1+(r^{\prime})^{2})^{\frac{l}{2}}}\left(rg(r^{-1},r^{\prime})+\dfrac{n-l}{l}\right)\Leftrightarrow
0<\dfrac{(r^{\prime}r)^{k}}{\binom{n-1}{k-1}}+\dfrac{n(k-l)}{kl}.$

Consequently, since $r^{\prime}$ only vanish at the origin, it follows that
the principal curvatures of the small portion satisfy
$\lambda(0)\in\partial\Gamma_{k}\cap\Gamma_{k-1}$, or equivalently

$\lambda(0)\in\partial\Gamma_{k}\cap\left\\{\dfrac{\partial\sqrt[k]{S_{k}}}{\partial\lambda_{i}}>0\right\\},$

and $\lambda(p)\in\Gamma_{k}$ for $p\in W_{R}\setminus\left\\{0\right\\}$.

Next, the ODE for $f=\sqrt[k]{S_{k}}$ is given by

$\displaystyle\begin{cases}v^{\prime}=(1+v^{2})\left(\dfrac{1}{\binom{n-1}{k-1}}\left(\dfrac{r}{v}\right)^{k-1}-\dfrac{(n-k)v}{kr}\right),\>r\geq
r_{0},\\\ v(r_{0})=v_{0}.\end{cases}$ (21)

We note that when $k$ is an even number, then the function $-v$, where $v$
denotes the upper half solution, is also a solution to (21)444Recall Remark 20
and the construction of the _wing-like_ solution is by extending the small
piece $r$ as the initial data of Eq. (21) in terms of $u$ with
$v^{\prime}=u$..
In particular, the right hand side of Eq. (21) is always negative and finite.
This means the solution exists for all $r\geq r_{0}$, and the asymptotic
expression of $v$ for $k$ even is given by

$\displaystyle
v(r)=-\dfrac{r}{\sqrt[k]{\binom{n-1}{k}}}+\dfrac{\binom{n-1}{k}\binom{n-1}{k-1}}{k\sqrt[k]{\binom{n-1}{k}}}\dfrac{1}{r}+O(|x|^{-2}),\mbox{
as }r\to\infty.$

Consequently, the principal curvatures of $W_{R}^{-}$ do not belong
$\Gamma_{k}$ for any $k$ as $r\to\infty$.

(a) $Q_{k+1,k}$

(b) $\sqrt[k]{S_{k}}$ for $k$ even

(c) $\sqrt[k]{S_{k}}$ for $k$ odd

Figure 2: Profile curves for winglike translators for various speed functions

######  Remark 21.

It is important to note that when $k$ is odd, the right hand side of Eq. (21)
has a singularity when $v=0$. This is because the translator equation 6 can
never be satisfied by any axially symmetric graph that possesses a point where
$u^{\prime}=0$ and $r>0$. Thus, if $u^{\prime}\to 0$ at some point
$r_{1}>r_{0}$, $u^{\prime}$ can never be extended past $r_{1}$, i.e. such
graph will possess a boundary $S^{n-1}$.

######  Step 6.4.

Construction and asymptotic behavior of the lower branch of the _wing-like_
$Q_{k+1,k}$-translator.

Firstly, the ODE that $r$ satisfies is

$\displaystyle
r^{\prime\prime}=-(1+(r^{\prime})^{2})\dfrac{(n-k)}{(k+1)r}\left(\dfrac{(k+1)r^{\prime}r-(n-k-1)}{(n-k)-krr^{\prime}}\right).$

Therefore, the principal curvatures of the small portion of $W_{R}$ are given
by

$\displaystyle\lambda_{1}=\dfrac{g(r^{-1},r^{\prime})}{\sqrt{1+(r^{\prime})^{2}}}=\dfrac{(n-k)}{(k+1)}\dfrac{\dfrac{(k+1)r^{\prime}r-(n-k-1)}{(n-k)-krr^{\prime}}}{r\sqrt{1+(r^{\prime})^{2}}}\mbox{
and }\lambda_{i}=\dfrac{1}{r\sqrt{1+(r^{\prime})^{2}}}.$

Then, we will have that $\lambda\in\Gamma_{k+1}$ if, and only if,

$\displaystyle
0<\dfrac{(n-k)}{k+1}\dfrac{(k+1)rr^{\prime}-(n-k-1)}{(n-k)-krr^{\prime}}+\dfrac{n-l}{l},\>l=1,\ldots,k+1.$

Consequently, since

$\displaystyle 0<\dfrac{n-l}{l}-\dfrac{n-k-1}{k+1}\Leftrightarrow 0<k+1-l,$

we obtain that $\lambda(0)\in\partial\Gamma_{k+1}\cap\Gamma_{k}$, or
equivalently

$\lambda(0)\in\partial\Gamma_{k+1}\cap\left\\{\dfrac{\partial
Q_{k+1,k}}{\partial\lambda_{i}}>0\right\\},$

and $\lambda(p)\in\Gamma_{k}$ for $p\in W_{R}\setminus\left\\{0\right\\}$.

Next, the ODE for $f=Q_{k+1,k}$ is given by

$\displaystyle\begin{cases}v^{\prime}=\dfrac{n-k}{k+1}(1+v^{2})\dfrac{v}{r}\dfrac{(k+1)-(n-k-1)\dfrac{v}{r}}{(n-k)\dfrac{v}{r}-k},\>r\geq
r_{0},\\\ v(r_{0})=v_{0}.\end{cases},$ (22)

Firstly, since $v\to-\infty$ as $r\to R^{+}$, we may take $r_{0}$ to be
sufficiently close to $R$ so that $v_{0}<0$. Note that the right hand side of
Eq. (22) is positive when $v<0$. Thus the solution is increasing as long as
this is the case. On the other hand, $v_{+}=0$ is a solution to the Eq. (22),
therefore $v$ and $v_{+}$ cannot coincide which means that $v$ remains
negative for all $r\geq r_{0}$.

Now we show that $v\to 0$ as $r\to\infty$. Firstly, since $v$ is increasing
and bounded above by $0$, $\displaystyle L\coloneqq\lim_{r\to\infty}v$ exists
in $[v_{0},0]$. If $L\neq 0$, then there exists $\varepsilon>0$ such that

$\displaystyle\dfrac{v}{r}<\dfrac{-\varepsilon}{r}.$

Then, since $k<n-1$ and the function
$g(y)=\dfrac{n-k}{k+1}y\dfrac{(k+1)-(n-k-1)y}{(n-k)y-k}$ is decreasing, we
have

$\displaystyle
v^{\prime}\geq(1+v^{2})g\left(\dfrac{-\varepsilon}{r}\right)\geq(1+v^{2})\dfrac{(n-k)\varepsilon}{kr}.$

Therefore, $\tan\left(\dfrac{(n-k)\varepsilon}{k}\ln(r)\right)=O(v)$, but this
contradicts that $v$ is bounded.

######  Remark 22.

We remark that the asymptotic behavior of the lower branch of the
$Q_{k+1,k}$-translator cannot decay to $0$ faster than
$\dfrac{-1}{r^{\frac{n-k}{k}(1+\varepsilon)}}$ for every $\varepsilon>0$ and
$1\leq k$.
To see this, let $C,\varepsilon>0$ and consider
$w_{C,\varepsilon}(r)=-Cr^{-(1+\varepsilon)a}$ where $a=\dfrac{n-k}{k}$. Then,
$w_{C,\varepsilon}$ is a super-solution to Eq. (22) with $v(r_{0})<0$. In
fact,

$\displaystyle
w^{\prime}(r)=\dfrac{(1+\varepsilon)aC}{r^{(1+\varepsilon)a+1}}$

and the RHS of Eq. (22) is

$\displaystyle
aC(1+w^{2})\dfrac{1}{r^{(1+\varepsilon)a+1}}\dfrac{(k+1)r^{(1+\varepsilon)a+1}+(n-k-1)}{(n-k)+kr^{(1+\varepsilon)a+1}}$
$\displaystyle\leq aC(1+w^{2})\dfrac{1}{r^{(1+\varepsilon)a+1}}$
$\displaystyle\leq w^{\prime},$

for $r>r_{0}$, where $r_{0}$ is sufficiently large. Then, for each
$\varepsilon>0$, we may fix $r_{1}>r_{0}$ and then choose C small enough so
that $v(r_{1})<w_{C,\varepsilon}(r_{1})$. Consequently, we have that
$w_{C,\varepsilon}>v$ for all $r\geq r_{1}$.

######  Remark 23.

By the above remark, we have that the asymptotic behavior of the principal
curvatures of the lower branch at infinity behave as $O(-r^{-a-1})$, where
$a=\dfrac{n-k}{k}$. Then, it follows that $\lambda\in\Gamma_{k+1}$ if, and
only if,

$\displaystyle
0<\binom{n-1}{l}\dfrac{(-1)^{l-1}}{r^{l(a+1)}}\dfrac{n(k-l)}{kl}\mbox{, holds
for }l=1,\ldots,k+1.$

Therefore, $\lambda\not\in\Gamma_{k+1}$ for any $k$ as $r\to\infty$ of the
lower branch.

######  Step 6.5.

Finally, the construction of the _wing-like_ solution.

This will be done by extending the small piece
$r:(-\varepsilon,\varepsilon)\to\mathbb{R}$ in the two branches by plugging
the initial conditions

$\displaystyle\begin{cases}u^{\prime}\left(r(-\varepsilon)\right)=\dfrac{1}{r^{\prime}(-\varepsilon)}\\\
u(r(-\varepsilon))=-\varepsilon.\end{cases}\begin{cases}u^{\prime}\left(r(\varepsilon)\right)=\dfrac{1}{r^{\prime}(\varepsilon)}\\\
u(r(\varepsilon))=\varepsilon.\end{cases}$

in Equations (21) and (22), respectively.

##  7 Application: Growth estimate

In this section we will prove the theorem 1.5, which we repeat here for the
convenience of the reader.

######  Theorem 7.1.

Let $\Sigma=\left\\{(x,u(x)):x\in\mathbb{R}^{n}\right\\}$ be an entire convex
$f$-translator for some nondegenerate speed $f$. Assume further, that there
exist $a,b,C_{1},C_{2},R>0$ such that

$\displaystyle C_{1}|x|^{a}\leq u(c)\leq C_{2}|x|^{b},\mbox{ for }|x|\geq R,$

then, $a\leq 2\leq b$.

In addition, if $a=b=2$, then $u(x)$ agrees with the “bowl”-type solution up
to vertical translations.

###### Proof.

Let assume first that $a>2$, and let $P$ be the “bowl”-type $f$-translator in
$\mathbb{R}^{n+1}$. Recall that $P$ is an entire strictly convex rotationally
symmetric graph smoothly asymptotic to

$\displaystyle\dfrac{|x|^{2}}{f(0,1)}-{\left.\kern-1.2pt\dfrac{\partial
f}{\partial\lambda_{1}}\vphantom{\big{|}}\right|_{\lambda=(0,1)}}\ln(|x|)+O(|x|^{-1}).$

Next, by translating suitably $P$ over $\Sigma$, we can find a $t_{0}>0$ such
that $P-te_{n+1}$ lies strictly below from $\Sigma$ for $t\geq t_{0}$. Note
that this is possible since $a>2$. Now, we may translate $P-te_{n+1}$ upward
until touches $\Sigma$ for the first time. Finally, by the interior tangential
principle Theorem 4.1, we obtain $\Sigma=P$, but this contradicts $a>2$.

The case $b<2$ is analogous, the only change is to place the “bowl”-type
soliton above $\Sigma$ and move it down until it touches $\Sigma$.

Finally, when $a=b=2$, then $\Sigma$ is smoothly asymptotic to $P$, and by
Theorem 1.2, $\Sigma=P$ up to a vertical translation. ∎

######  Remark 24.

Theorem 1.5 also holds when the hypotheses of $f$ are changed by the one given
in Remark 14. This means also holds for entire $f$-translators such that the
principal curvatures of the graph belongs to $\Gamma$, and $\Gamma$ is a
convex cone of $\left\\{f>0\right\\}$ that contains the point
$(1,\mathbf{e})$.

## References

  * [1] Julie Clutterbuck, Oliver C. Schnürer and Felix Schulze  “Stability of translating solutions to mean curvature flow”  In _Calc. Var. Partial Differential Equations_ 29.3, 2007, pp. 281–293  DOI: 10.1007/s00526-006-0033-1
  * [2] A. Cogo, S. Lynch and O. Martínez  “Rotational symmetry of ancient solutions to fully nonlinear curvature flows”, 2023  arXiv:2310.08301 [math.DG]
  * [3] Francisco Fontenele and Sérgio L Silva  “A tangency principle and applications”  In _Illinois Journal of Mathematics_ 45.1 Duke University Press, 2001, pp. 213–228
  * [4] David Hoffman, Tom Ilmanen, Francisco Martín and Brian White  “Notes on translating solitons for mean curvature flow”  In _Minimal Surfaces: Integrable Systems and Visualisation: m: iv Workshops, 2016–19_ , 2021, pp. 147–168  Springer
  * [5] Francisco Martín, Andreas Savas-Halilaj and Knut Smoczyk  “On the topology of translating solitons of the mean curvature flow”  In _Calc. Var. Partial Differential Equations_ 54.3, 2015, pp. 2853–2882  DOI: 10.1007/s00526-015-0886-2
  * [6] Antonio Martínez and AL Martínez-Triviño  “Equilibrium of surfaces in a vertical force field”  In _Mediterranean Journal of Mathematics_ 19.1 Springer, 2022, pp. 3
  * [7] Sathyanarayanan Rengaswami  “Rotationally symmetric translating solutions to extrinsic geometric flows”  In _arXiv preprint arXiv:2109.10456_ , 2021
  * [8] Richard M Schoen  “Uniqueness, symmetry, and embeddedness of minimal surfaces”  In _Journal of Differential Geometry_ 18.4 Lehigh University, 1983, pp. 791–809
  * [9] Gerald Teschl  “Ordinary differential equations and dynamical systems”  American Mathematical Soc., 2012
  * [10] José Torres-Santaella  “Maximum Principles and Consequences for $\gamma$-translators in $\mathbb{R}^{n+1}$”  In _arxiv preprint arXiv:2306.03649_ , 2023
  * [11] John Urbas  “Complete noncompact self-similar solutions of Gauss curvature flows. I. Positive powers”  In _Math. Ann._ 311.2, 1998, pp. 251–274  DOI: 10.1007/s002080050187


# Improving Application of Bayesian Neural Networks to Discriminate Neutrino
Events from Backgrounds in Reactor Neutrino Experiments

Ye Xua Corresponding author, e-mail address<EMAIL_ADDRESS>WeiWei Xua
YiXiong Menga Bin Wua

###### Abstract

The application of Bayesian Neural Networks(BNN) to discriminate neutrino
events from backgrounds in reactor neutrino experiments has been described in
Ref.[1]. In the paper, BNN are also used to identify neutrino events in
reactor neutrino experiments, but the numbers of photoelectrons received by
PMTs are used as inputs to BNN in the paper, not the reconstructed energy and
position of events. The samples of neutrino events and three major backgrounds
from the Monte-Carlo simulation of a toy detector are generated in the signal
region. Compared to the BNN method in Ref.[1], more 8He/9Li background and
uncorrelated background in the signal region can be rejected by the BNN method
in the paper, but more fast neutron background events in the signal region are
unidentified using the BNN method in the paper. The uncorrelated background to
signal ratio and the 8He/9Li background to signal ratio are significantly
improved using the BNN method in the paper in comparison with the BNN method
in Ref.[1]. But the fast neutron background to signal ratio in the signal
region is a bit larger than the one in Ref.[1].

###### keywords:

Bayesian neural networks, neutrino oscillation, identification

aDepartment of Physics, Nankai University, Tianjin 300071, China

PACS numbers: 07.05.Mh, 29.85.Fj, 14.60.Pq

##  1 Introduction

The main goals of reactor neutrino experiments are to detect
$\bar{\nu_{e}}\rightarrow\bar{\nu_{x}}$ oscillation and precisely measure the
mixing angle of neutrino oscillation $\theta_{13}$. The experiment is designed
to detect reactor $\bar{\nu_{e}}$’s via the inverse $\beta$-decay reaction

$\bar{\nu_{e}}+p\rightarrow e^{+}+n$.

The signature is a delayed coincidence between $e^{+}$ and the neutron
captured signals. In the paper, only three important sources of backgrounds
are taken into account and they are the uncorrelated background from natural
radioactivity and the correlated backgrounds from fast neutrons and 8He/9Li.
The backgrounds like the neutrino events consist of two signals, a fast signal
and a delay signal. It is vital to separate neutrino events from backgrounds
accurately in the reactor neutrino experiments. Bayesian neural networks
(BNN)[2] are algorithms of the neural networks trained by Bayesian statistics.
They are not only non-linear functions as neural networks, but also controls
model complexity. So their flexibility makes them possible to discover more
general relationships in data than traditional statistical methods and their
preferring simple models make them possible to solve the over-fitting problem
better than the general neural networks[3]. BNN have been used to particle
identification and event reconstruction in the experiments of the high energy
physics, such as Ref.[1, 4, 5, 6]. The application of BNN to discriminate
neutrino events from backgrounds in reactor neutrino experiments has been
described in Ref.[1]. In the paper, BNN are also used to identify neutrino
events in the signal region[1] in reactor neutrino experiments, but the
numbers of photoelectron received by PMTs are used as inputs to BNN, not the
reconstructed energy and position of events.

##  2 The Classification with BNN[1, 2, 6]

The idea of BNN is to regard the process of training a neural network as a
Bayesian inference. Bayes’ theorem is used to assign a posterior density to
each point, $\bar{\theta}$, in the parameter space of the neural networks.
Each point $\bar{\theta}$ denotes a neural network. In the methods of BNN, one
performs a weighted average over all points in the parameter space of the
neural network, that is, all neural networks. The methods are described in
detail in Ref.[1, 2, 6]. The posterior density assigned to the point
$\bar{\theta}$, that is, to a neural network, is given by Bayes’ theorem

$p\left(\bar{\theta}\mid x,t\right)\propto\mathit{p\left(t\mid
x,\bar{\theta}\right)p\left(\bar{\theta}\right)}$ (1)

Where $x$ is a set of input data which corresponds to a set of target $t$. The
likelihood $p\left(t\mid x,\bar{\theta}\right)$ can be obtained by using a
training sample. And a Gaussian prior is specified for each weight using the
Bayesian neural networks package of Radford Neal111R. M. Neal, _Software for
Flexible Bayesian Modeling and Markov Chain Sampling_ ,
http://www.cs.utoronto.ca/~radford/fbm.software.html. Given an event with data
$x^{\prime}$, an estimate of the probability that it belongs to the signal is
given by the weighted average

$\bar{y}\left(x^{\prime}|x,t\right)=\int
y\left(x^{\prime},\bar{\theta}\right)p\left(\bar{\theta}\mid
x,t\right)d\bar{\theta}$ (2)

Currently, the only way to perform the high dimensional integral in Eq. (2) is
to sample the density $p\left(\bar{\theta}\mid x,t\right)$ with Markov Chain
Marlo Carlo (MCMC) methods[2, 7, 8, 9]. In MCMC methods, one steps through the
$\bar{\theta}$ parameter space in such a way that points are visited with a
probability proportional to the posterior density, $p\left(\bar{\theta}\mid
x,t\right)$. Points where $p\left(\bar{\theta}\mid x,t\right)$ is large will
be visited more often than points where $p\left(\bar{\theta}\mid x,t\right)$
is small.

Eq. (2) approximates the integral using the average

$\bar{y}\left(x^{\prime}\mid
x,t\right)\approx\frac{1}{L}\sum_{i=1}^{L}y\left(x^{\prime},\bar{\theta_{i}}\right)$
(3)

where $L$ is the number of points $\bar{\theta}$ sampled from
$p\left(\bar{\theta}\mid x,t\right)$. Each point $\bar{\theta}$ corresponds to
a different neural network with the same structure. So the average is an
average over neural networks, and the probability of the data $x^{\prime}$
belongs to the signal. The average is closer to the real value of
$\bar{y}\left(x^{\prime}\mid x,t\right)$, when $L$ is sufficiently large.

##  3 Toy Detector and Monte-Carlo Simulation[5]

In the paper, a toy detector is used to simulate central detectors in the
reactor neutrino experiments, such as Daya Bay experiment[10] and Double Chooz
experiment[11], with CERN GEANT4 package[12]. The toy detector is the same as
Ref.[5]. A total of 366 PMTs are arranged in 8 rings of 30 PMTs on the lateral
surface of the oil region, and in 5 rings of 24, 18, 12, 6, 3 PMTs on the top
and bottom caps.

The responses of neutrino events and backgrounds deposited in the toy detector
are simulated with GEANT4. Although the physical properties of the
scintillator and the oil (their optical attenuation length, refractive index
and so on) are wave-length dependent, only averages[13] (such as the optical
attenuation length of Gd-LS with a uniform value is 8 meter and the one of LS
is 20 meter) are used in the detector simulation.

According to the anti-neutrino interaction in detectors of the reactor
neutrino experiments[14], a neutrino event is uniformly generated throughout
Gd-LS region (see Fig. 1). A uncorrelated background event is generated in
such a way that a $\gamma$ event generated on the base of the energy
distribute of the natural radioactivity in the proposal of the Day Bay
experiment[10] is regarded as the fast signal, a neutron event of the single
signal is regarded as the delay signal, its delay time is uniformly generated
from 2 $\mu$s to 100 $\mu$s and the positions of the fast signal and the delay
signal are uniformly generated throughout Gd-LS region. A fast neutron event
is uniformly generated throughout Gd-LS region and its energy are uniformly
generated from 0 MeV to 50 MeV, therein an event of two signals are regarded
as a fast neutron background event. Since the behaviors of 8He/9Li decay
events in detectors couldn’t be simulated by the Geant4 package, a 8He/9Li
event is generated in such a way that the neutron signal from a fast neutron
event is regarded as its delay signal, an electron event generated at the same
position as the fast neutron event on the base of the energy distribute of
8He/9Li events in the proposal of the Day Bay experiment[10] is regarded as
its fast signal in the paper.

Energies and positions of neutrino events and backgrounds are reconstructed by
the method in Ref.[5]. The signal region is determined by using the
reconstructed energies and positions, as well as the neutron delay
time(described in Ref.[1]).

##  4 Neutrino Discrimination with BNN

Choosing inputs to BNN is vital to identify neutrino events . The
reconstructed energies, the distance between reconstructed the positions of
neutron and positron and the neutron delay time were used as inputs to the BNN
method in Ref.[1], but the energies and the distance are both the
reconstructed physics variables, and they make BNN discriminations worse
because of their reconstruction uncertainties. So we try to use raw data as
inputs to BNN. Obviously, the numbers of photoelectrons received by 366 PMTs
are rawer than the reconstructed variables. An event consists of two signals
(a fast signal and a delay signal), so if the numbers of photoelectron
received by PMTs will be directly used as inputs to BNN, BNN will have 732
inputs at least. It will take too much time to run a BNN program in a general
computer because of such many inputs. The method of reducing inputs to BNN in
the paper is that the photoelectrons received by several neighboring PMTs are
added up. That is several neighboring PMTs incorporate a PMT patch. In the
paper, a PMT patch is a 3(azimuth direction)$\times$4(z direction) PMTs array
on the detector lateral surface or a 120∘ sector(including 21 PMTs) on the
detector top and bottom caps. The delay time between two signals is very
important to discriminate neutrino events from the uncorrelated background, so
the number of photoelectrons received by a patch is multiply by the delay
time, and the result is used as the inputs to all neural networks, which have
the same structure. Then all the networks have a input layer of 52 inputs, the
single hidden layer of fifteen nodes and a output layer of a single output
which is just the probability that an event belongs to the neutrino event.
Discriminating neutrino events from backgrounds is actually a binary response
problem, that is the target is ’1’ or ’0’. Neutrino events are labeled by t=1,
and background events are labeled by t=0. So the output of BNN has to be a
number between 0 and 1. If the output is less than 0.5, the event is regarded
as a background event, and If the output is larger than 0.5, the event is
regarded as a neutrino event.

A Markov chain of neural networks is generated using the Bayesian neural
networks package of Radford Neal, with a training sample consisting of
neutrino events and background events. One thousand iterations, of twenty MCMC
steps each, are used. The neural network parameters are stored after each
iteration, since the correlation between adjacent steps is very high. That is,
the points in neural network parameter space are saved to lessen the
correlation after twenty steps here. It is also necessary to discard the
initial part of the Markov chain because the correlation between the initial
point of the chain and the point of the part is very high. The initial three
hundred iterations are discarded here. It takes about 120 hours to run 1000
iterations on a computer with two 3.4GHz Intel Pentium D processors (only one
of which are used).

Neutrino identification efficiencies are defined by the ratios between the
number of the events in neutrino test sample regarded as neutrinos and the
number of neutrino test sample. Background identification efficiencies are
defined by the ratios between the numbers of the events in background test
samples regarded as background events and the numbers of background test
samples. The identification efficiencies are measured with the test sample
which is different from the training sample. Other 3000 events each of the
neutrino and the three backgrounds are used to test the identification
capability of the trained BNN. In the paper, BNN are trained by the different
training samples, which consist of neutrinos and three backgrounds at
different rates, since the different identification efficiencies are obtained
using those BNN.

##  5 Results and Discussion

As Tab. 1 shows, most neutrino events, uncorrelated background events and
8He/9Li background events in the signal region can be identified using the BNN
method in the paper, but only a small part of fast neutron background events
can be identified using the BNN method in the paper. Since most fast neutron
events can’t be discriminate from neutrino events using the BNN method in the
paper, neutrino discriminations are concerned with neutrino rates in training
samples, as well as ratios of neutrino events and fast neutron events in
training samples. The neutrino discrimination in the signal region increases
from 90.5% to 93.7% with the increase of the neutrino rate from 50.0% to 57.1%
in the training sample using the BNN method in the paper. And the neutrino
discrimination also increases from 90.5% to 94.1% with the increase of the
ratio of neutrino events and fast neutron events from 2:1 to 3:1 in the
training sample. The different background to signal ratios in the signal
region are obtained using the BNN trained by the training samples consisting
of neutrino events and background events at different rates in the reactor
neutrino experiments.

Neutrino events are discriminated from fast neutrons and 8He/9Li events via
their fast signals identification, that is positron signals from neutrino
events are separated from recoil proton signals from fast neutrons and
electron signals from 8He/9Li events. $\gamma$ signals induced by positrons
and recoil protons are closer to point sources, but $\gamma$ signals induced
by electrons are closer to line sources. There is an effect on the
distribution of photoelectrons over all the PMTs in the detector due to the
difference between a point source and a line source. The effect can be
extracted from the inputs by the BNN method in the paper. So neutrino events
can be better discriminated from 8He/9Li events using the BNN method in the
paper, but distinguishing between neutrinos events and fast neutrons becomes
worse using the BNN method in the paper.

The events in the signal region can be identified using BNN one by one, once
those BNN are trained by training samples. If the BNN method in the paper is
used to the reactor neutrino experiments, the background to signal ratios will
be changed. We only roughly estimate the changes here. We assume that the
uncorrelated background fraction in the signal region is $A/N$, the fast
neutrons background fraction in the signal region is $F/N$, and the 8He/9Li
background fraction in the signal region is $L/N$ in the reactor neutrino
experiments. Those background fractions are very low (for example, they are
<0.2%, 0.1%, 0.3% in one of the near detector claimed by the proposal of the
Daya Bay experiment[10], respectively). If neutrino events are discriminated
from background events using the BNN method in Ref.[1], the background to
signal ratios can reach 0.2*(A/N), 0.68*(F/N) and 0.66*(L/N), respectively. If
the efficiencies of the first column in Tab. 1 are use to the estimation, we
get the result of the identification using the BNN method in the paper:

Uncorrelated background/Signal=(A/N)*(1-0.983)/0.941=0.018*(A/N)

Fast neutrons Background/Signal=(F/N)*(1-0.293)/0.941=0.751*(F/N)

8He/9Li Background/Signal=(L/N)*(1-0.913)/0.941=0.092*(L/N)

As the above equations show, the uncorrelated background to signal ratio and
8He/9Li background to signal ratio in the signal region are significantly
improved using the BNN method in the paper in comparison with the BNN method
in Ref.[1]. And the fast neutron background to signal ratio is a bit larger
than the one in Ref.[1]. But the fast neutron fraction in the signal region is
lower than the ones of the uncorrelated background and 8He/9Li background, so
the total background to signal ratio using the BNN method in the paper is much
lower than the one in Ref.[1]. In a word, the BNN method in the paper can be
applied to discriminate neutrino events from background events better than the
BNN method in Ref.[1] and the method based on the cuts in reactor neutrino
experiments.

##  6 Acknowledgements

This work is supported in part by the National Natural Science Foundation of
China (NSFC) under the contract No. 10605014, the national undergraduate
innovative plan of China under the contract No.081005517 and the physical base
of Nankai University under the contract No. J0730315.

## References

  * [1] Y. Xu, Y. X. Meng, and W. W. Xu, _Applying bayesian neural networks to separate neutrino events from backgrounds in reactor neutrino experiments_ , Journal of Instrumentation, 3, P08005 (2008), arXiv: 0808.0240
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Table 1: The different identification efficiencies are obtained with the BNNs trained by the different training samples, which consist of the neutrino and three backgrounds at different rates. The term after $\pm$ is the statistical error of the identification efficiencies. The numbers of the train samples are 24000, respectively. The 3000 events each of the uncorrelated background, fast neutron and 8He/9Li are regarded as the test sample. neutrino rate (%) | 50.0 | 50.0 | 54.5 | 57.1
---|---|---|---|---
uncorrelated background rate (%) | 16.7 | 12.5 | 9.1 | 9.5
fast neutron rate (%) | 16.7 | 25.0 | 27.3 | 23.8
8He/9Li rate (%) | 16.7 | 12.5 | 9.1 | 9.5
neutrino eff.(%) | 94.1$\pm$0.43 | 90.5$\pm$0.54 | 92.6$\pm$0.48 | 93.7$\pm$0.44
uncorrelated background eff.(%) | 98.3$\pm$0.24 | 98.1$\pm$0.25 | 96.4$\pm$0.34 | 96.7$\pm$0.33
fast neutrons eff.(%) | 29.3$\pm$0.83 | 35.8$\pm$0.88 | 34.6$\pm$0.87 | 32.8$\pm$0.86
8He/9Li eff.(%) | 91.3$\pm$0.51 | 90.6$\pm$0.53 | 87.7$\pm$0.60 | 87.5$\pm$0.60
Figure 1: The neutrino events for the Monte-Carlo simulation of the toy
detector are uniformly generated throughout Gd-LS region. (a) is the
distribution of the positron energy; (b) is the distribution of the energy of
the neutron captured by Gd; (c) is the distribution of the distance between
the positron and neutron positions; (d) is the distribution of the delay time
of the neutron signal.