# A game-theoretic approach to the asymptotic behavior of solutions to an
obstacle problem for the mean curvature flow equation

Kuniyasu Misu

###### Abstract

We consider the asymptotic behavior of solutions to an obstacle problem for
the mean curvature flow equation by using a game-theoretic approximation, to
which we extend that of Kohn and Serfaty [14]. The paper [14] gives a
deterministic two-person zero-sum game whose value functions approximate the
solution to the level set mean curvature flow equation without obstacle
functions. We prove that moving curves governed by the mean curvature flow
converge in time to the boundary of the convex hull of obstacles under some
assumptions on the initial curves and obstacles. Convexity of the initial set,
as well as smoothness of the initial curves and obstacles, are not needed. In
these proofs, we utilize properties of the game trajectories given by very
elementary game strategies and consider reachability of each player. Also,
when the equation has a driving force term, we present several examples of the
asymptotic behavior, including a problem dealt in [9].

email<EMAIL_ADDRESS>
Keywords— Viscosity solutions, Mean curvature flow equation, Asymptotic shape,
Large time behavior, Deterministic game

Mathematics Subject Classification 2020— 35B40, 35D40, 91A50

##  1 Introduction

#### Obstacle problem for the mean curvature flow equation.

We consider the following obstacle problem for the mean curvature flow
equation:

$\begin{cases}V=-\kappa\ \mbox{on $\partial D_{t}$},\\\ O_{-}\subset
D_{t},\end{cases}$ (1.1)

where $\\{D_{t}\\}_{t>0}$ is the unknown family of open sets in
$\mathbb{R}^{d}$, $V$ is the velocity of a point in $\partial D_{t}$ in the
direction of its outward normal vector, $\kappa$ is the mean curvature of
$\partial D_{t}$ at the point and $O_{-}$ is a fixed open set in
$\mathbb{R}^{d}$. Our main goal is to investigate the asymptotic behavior of
solutions to (1.1) by the level set equation and its game-theoretic
approximation. We mainly deal with the case $d=2$ in this manuscript.

The mean curvature flow equation has been attracting much attention. In the
early stages, the smoothness of the initial surface was naturally assumed and
the surface evolution was considered as long as singularities do not occur. In
paticular when $d=2$, the mean curvature flow equation is often called the
curve shortening problem and the curve evolution was analysed in e.g. [6, 11].

The level set method for surface evolution equations was first rigorously
analyzed in [3, 5]. The basic idea of this method is to represent moving
surfaces as level sets of auxiliary functions and to rewrite surface evolution
equations by level set equations, whose unknown functions are the auxiliary
functions. A great advantage is the point that viscosity solutions of level
set equations follow the long time behavior of the moving surfaces even after
topological change of surfaces. The level set method is applied to various
surface evolution equations including the mean curvature flow equation. See
also [7] in detail.

Recently obstacle problems for the mean curvature flow equation have been
considered in [10, 13, 21]. Obstacle problems are problems that have regions
called obstacles which the solutions cannot exceed.

According to the unpublished paper [20] by Mercier, the level set method is
still valid for (1.1). The corresponding level set equation to (1.1) is the
following:

$\begin{cases}u_{t}(x,t)+F(Du(x),D^{2}u(x))=0&\quad\mbox{in}\
\mathbb{R}^{d}\times(0,\infty),\\\ u(x,0)=u_{0}(x)&\quad\mbox{in}\
\mathbb{R}^{d},\\\ \Psi_{-}(x)\leq u(x,t)&\quad\mbox{in}\
\mathbb{R}^{d}\times(0,\infty),\end{cases}$ (1.2)

where $\Psi_{-}\in Lip(\mathbb{R}^{d})$ is a given obstacle function that
satisfies $O_{-}=\\{x\in\mathbb{R}^{d}\mid\Psi_{-}(x)>0\\}$. The function
$u_{0}\in C(\mathbb{R}^{d})$ is an initial datum and $F$ is given by

$F(Du,D^{2}u)=-\lvert Du\rvert\mathrm{div}\left(\frac{Du}{\lvert
Du\rvert}\right).$

Namely $F$ is the level-set mean curvature flow operator defined as

$F(p,X)=-\mathrm{Tr}\left(\left(I-\frac{p\otimes p}{\lvert
p\rvert^{2}}\right)X\right),\quad p\in\mathbb{R}^{d}\setminus\\{0\\},\
X\in\mathbb{S}^{d},$

where $p\otimes p=(p_{i}p_{j})_{i,j=1}^{d}$ for a vector
$p=(p_{1},\dots,p_{d})\in\mathbb{R}^{d}$ and $\mathbb{S}^{d}$ is the set of
$d\times d$ real symmetric matrices. For a comparison principle to (1.2), see
also [13].

Throughout this paper we follow the 0 level set of the solution $u$. Together
with it, we assume on the initial data $u_{0}$ as follows:

For some $a<0$ and $R>0$, $u_{0}=a$ in $B^{c}(0,R)$. (1.3)

#### Intuitive observation.

$\Longrightarrow$$\Longrightarrow$$\Longrightarrow$$t\to\infty$$t\to\infty$$t\to\infty$obstacle
$O_{-}$initial curve $\partial D_{0}$ Figure 1: Conjecutures on the asymptotic
shapes

For the solution to (1.1) with $d=2$ and without obstacles, it is known that
$D_{t}$ becomes convex at some time, the moving curve $\partial D_{t}$
converges to a single point and then vanishes, provided $\partial D_{0}$ is a
smooth closed curve ([6, 11]). On the other hand, for our problem (1.1), it is
obvious that the solution does not converge to any single point. Also it is
not clear whether $D_{t}$ becomes convex at some time. However it is natural
to expect that in many cases $D_{t}$ converges to the convex hull of $O_{-}$,
which we hereafter denote by $Co(O_{-})$, as $t\to\infty$ because of the curve
shortening property of the solution and the smoothing effect of the curvature
flow as we draw some examples in Figure 1. As shown in Figure 1, even for the
same obstacles, different initial curves may converge to different limits.
Thus we shall assume at least that one connected component of $D_{0}$ contains
the whole $O_{-}$ and expect that the asymptotic shape is $Co(O_{-})$ under
this assumption. Our main theorem (Theorem 3.2) is intended to justify this
expectation as much as possible.

#### Game interpretation.

Our first result is the extension of [14] to problems including (1.2). First,
let us briefly explain the game rule for (1.2) with $d=2$ and without obstacle
function by following [14, Section 1.6]. The game is a deterministic two-
person zero-sum game. For convenience, we name the first player Paul and the
second player Carol. Let $\epsilon>0$. Also, let $x_{0}=x\in\mathbb{R}^{2}$ be
the initial position of this game and $t>0$ be the terminal time. At the
$i$-th round of this game, Paul chooses directions $v_{i}\in\mathbb{R}^{2}$
with $\lvert v_{i}\rvert=1$ and Carol chooses a number $b_{i}=\pm 1$ after
Paul’s choice. Then the game position that we henceforth regard as Paul’s
position conveniently moves from $x_{i-1}$ to the next place $x_{i}$ depending
on their choice as follows:

$x_{i}=x_{i-1}+\sqrt{2}\epsilon b_{i}v_{i}$ (1.4)

After the $N$-th round, where $N\sim t\epsilon^{-2}$, the game ends and Carol
pays the terminal cost $u_{0}(x_{N})$ to Paul. Paul’s goal is maximizing the
cost while Carol’s goal is minimizing it. The value function
$u^{\epsilon}(x,t)$ is defined as the cost optimized by both the players, that
is,

$u^{\epsilon}(x,t)=\max_{v_{1}}\min_{b_{1}}\dots\max_{v_{N}}\min_{b_{N}}u_{0}(x_{N}).$

This value function approximates the viscosity solution $u$ of (1.2) with
$d=2$ and without obstacle function. In fact the convergence $u^{\epsilon}\to
u$ is shown in [14].

In order to handle (1.2) that has the obstacle function $\Psi_{-}$, we modify
the game rule as follows. At each $i$-th round, we suppose that Paul has the
right to quit the game. If Paul quits the game, the game cost is given by
$\Psi_{-}(x_{i})$. By doing this modification, the value function
$u^{\epsilon}$ is restricted to $\Psi_{-}\leq u^{\epsilon}$. Such an
interpretation of parameters of PDEs is well understood for first order
equations; see [1]. The cost $\Psi_{-}(x_{i})$ is called stopping cost and an
optimal control problem with stopping cost is called optimal stopping time
problem. For second order equations, see e.g. [18, 4], which deal with the
optimal stopping time problem corresponding to the obstacle problem for the
infinity Laplacian equation and p-Laplacian equation respectively.

The value function $u^{\epsilon}(x,t)$ satisfies the following Dynamic
Programming Principle:

$u^{\epsilon}(x,t)=\max\\{\Psi_{-}(x),\max_{\lvert v\rvert=1}\min_{b=\pm
1}u^{\epsilon}(x+\sqrt{2}\epsilon bv,t-\epsilon^{2})\\}$

for $t>0$. This is a key equation in the proof of the convergence result.

The paper [14] also mention the game interpretation for higher dimensional
case. Based on this, we can generalize our game to the case $d\geq 3$. In the
game, Paul chooses $d-1$ orthogonal unit vectors
$v_{i}^{j}(j=1,2,\cdots,d-1)$, Carol chooses $d-1$ values $b_{i}^{j}\in\\{\pm
1\\}(j=1,2,\cdots,d-1)$, and the state equation is
$x_{i}=x_{i-1}+\sqrt{2}\epsilon\sum_{j=1}^{d-1}b_{i}^{j}v_{i}^{j}$ instead of
(1.4).

The precise statement of the convergence of the value functions is described
by the half relaxed limits of the value functions, which are defined as
follows:

$\overline{u}(x,t):=\varlimsup_{\begin{subarray}{c}(y,s)\to(x,t)\\\
\epsilon\searrow 0\end{subarray}}u^{\epsilon}(y,s),\
\underline{u}(x,t):=\varliminf_{\begin{subarray}{c}(y,s)\to(x,t)\\\
\epsilon\searrow 0\end{subarray}}u^{\epsilon}(y,s).$

As a consequence of Proposition A.3, we present the convergence result for
(1.2) at the moment. We describe a game interpretation and the same type of
convergence result for more general PDEs than (1.2) in Appendix A.

######  Proposition 1.1.

The functions $\overline{u}$ and $\underline{u}$ are respectively viscosity
sub- and supersolution of (1.2). Moreover
$\overline{u}(x,0)=\underline{u}(x,0)=u_{0}(x)$ for $x\in\mathbb{R}^{d}$.

#### Asymptotic behavior.

$\partial D_{0}$$O_{-}$ Figure 2: Example of $D_{0}$ and $O_{-}$

$x$$v$ Figure 3: Strategy

We study the asymptotic behavior of solutions to (1.1). To explain an outline
of the proof of the main theorem (Theorem 3.2), at the moment, we identify
$u^{\epsilon}$ with $u$ and consider a specific figure (Figure 3). Since we
consider 0 level set of solutions to (1.2), our concern is whether the game
cost is positive or negative. Thus, from Paul’s point of view, the victory
condition is that the game cost becomes positive. Namely, from Carol’s point
of view, the victory condition is that the game cost becomes negative. There
is no need to give optimal strategies. Hereafter, even if a strategy taken by
the players is not optimal, we often use present tense such as ”Paul takes
some strategy when he is in some domain”. To show that the asymptotic shape is
$Co(O_{-})$, we have to prove that Paul wins if he starts from $Co(O_{-})$ and
Carol wins if Paul starts from $\overline{Co(O_{-})}^{c}$. (We avoid the
argument on the boundary of $Co(O_{-})$.) Furthermore, by the rule of the game
explained above, we see that the victory condition of Paul is whether he
reaches $O_{-}$ at some round or $D_{0}$ at the final round.

The easiest situation for Paul is that the initial game position $x\in O_{-}$.
In this case, it suffices for Paul to quit the game at the first round and
gain the stopping cost $\Psi_{-}(x)>0$. If we take $x$ as shown in Figure 3, a
strategy for Paul to win is the following: He keeps taking $v$ parallel to the
dotted line segment as in Figure 3 until he reaches the domain inside the
dotted circle. Once he gets there, he quits the game and gains the positive
stopping cost. Even if he does not get there, he can gain the positive
terminal cost at the final round of the game because the dotted line segment
in Figure 3 is contained in $D_{0}$.

$C$$x$$z$$\tau=\tau_{1}$$0<\tau_{1}<\tau_{2}$$\tau=\tau_{2}$ Figure 4:
Strategy for $x\in Co(O_{-})$

For the other $x\in Co(O_{-})$, we consider a strategy for Paul to reach the
domain from where we above overview that he could win if he started. To
construct it, we prepare a type of strategies of the game called  concentric
strategy, which is also introduced in [14] and [16, Lemma 2.5 2.6]. See
Definition 2.5. If Paul takes a concentric strategy, he can choose his
favorable point $z\in\mathbb{R}^{2}$ and can control the distance from $z$ to
game positions regardless of Carol’s choices as follows:

$\lvert x_{n}-z\rvert=\sqrt{\lvert x_{0}-z\rvert^{2}+2n\epsilon^{2}}.$

In paticular $\lvert x_{n}-z\rvert$ is monotonically increasing with respect
to $n$ and, denoting the game time $n\epsilon^{2}$ by $\tau$, it goes to
infinity as $\tau\to\infty$. Figure 4 shows an example of $x\in Co(O_{-})$ and
an appropriate concentric strategy. In Figure 4 the center of the arc $C$ is
$z$. Paul’s strategy is to choose this $z$ and keep taking the concentric
strategy until he reaches a neighborhood of the bold curve in $D_{0}$. Since
the domain enclosed by the arc $C$ and the bold curve is bounded, he indeed
reaches a neighborhood of the bold curve. Therefore he wins if he starts at
this initial position $x$.

In the main theorem we state a condition on $D_{0}$ and $O_{-}$ that we are
able to apply above technique. To indicate above bounded domain, we construct
an appropriate Jordan closed curve in the proof and Appendix D and then use
the Jordan curve theorem.

One also define a concentric strategy of Carol (Definition 2.5) that has
similar effect to that of Paul. Namely if Carol chooses a point $z$ and takes
the concentric strategy, she can force the distance $\lvert x_{n}-z\rvert$ to
be monotonically increasing with respect to $n$ and go to infinity as
$\tau\to\infty$. For $x\in\overline{Co(O_{-})}^{c}$ we can take an open ball
$B$ such that $\overline{Co(O_{-})}\subset B$ and $x\in B^{c}$ by the
hyperplane separation theorem and the boundedness of $\overline{Co(O_{-})}$.
If Carol chooses $z$ and takes the concentric strategy, she wins for
sufficiently large $\tau$ owe to the boundedness of $D_{0}$. If we assume a
kind of strict convexity on the obstacle $O_{-}$, we can take above open ball
$B$ for $x\in\overline{Co(O_{-})}^{c}$ whose radius does not depend on $x$.
This means that the moving surface sticks to the obstacle in finite time
(Theorem 3.7).

#### Literature.

We give some other related works on the asymptotic behavior of solutions to
obstacle problems for the mean curvature flow equation. Spadaro considers
(1.1) to characterize the mean-convex hull set in his unpublished paper [22].
He considers (1.1) by a variational discrete scheme, which is different from
our approach, but is guaranteed to approximate the viscosity solution to (1.2)
by [20]. According to [22], the part of the limit hypersurface that does not
touch the obstacle is a minimal surface. (This result enhances the
plausibility of our expectation.) Compared to our result, his result works in
higher dimensional case $d\leq 7$, while it needs to assume at least that the
initial set $D_{0}$ is convex when $d=2$. For $d\geq 8$, [23, Proposition 4.2]
implies that the limit hypersurface may have non empty boundary. [21] proves
the convergence of moving surfaces in a situation that both initial surface
and obstacles are given as periodic graphs. For problems with driving force,
[10] proves the solution $u(x,t)$ to the problem (2.1) with $f=0$ (We will
introduce it later.) converges as $t\to\infty$ to the stationary solution.
They also give the result concerning to the shape of the stationary solution.
However it is limited to the case where the initial data and the obstacle
function are radially symmetric.

Concerning to an approach other than the level set method, Takasao [24]
considers an obstacle problem for the mean curvature flow equation in the
sense of Brakke’s mean curvature flow ([2]). He proves the global existence of
the weak solution by using the Allen-Cahn equation with forcing term.

While [13, 24] and this paper consider given obstacle problems, they arise
from many different situations. In [9] an obstacle problem naturally appears
in the motion of the top and the bottom of the solution of birth and spread
type equations though the equations have no obstacle functions. [17] shows
that large exponent limit of power mean curvature flow equation is formulated
by an obstacle problem involving 1-Laplacian.

#### Organization.

This paper is organized as follows. Section 2 contains definitions, notations
and lemmas that are needed to prove our results. In Section 3 we prove the
theorems on the asymptotic shape of the solution to (1.1). Section 4 is
devoted to compute several examples of asymptotic shapes of solutions to
problems with driving force. The convergence of the value functions of the
game and some arguments to complement the proof of the main theorem are
presented in appendices.

##  2 Preliminary result

###  2.1 Definitions and notations

In this subsection we introduce the notion of viscosity solution to the
following obstacle problem, which is the most general form in this manuscript.
Moreover we remark some known results.

$\begin{cases}u_{t}(x,t)-\nu\lvert
Du(x,t)\rvert+F(Du(x,t),D^{2}u(x,t))=f(x)&\quad\mbox{in}\
\mathbb{R}^{d}\times(0,\infty),\\\ u(x,0)=u_{0}(x)&\quad\mbox{in}\
\mathbb{R}^{d},\\\ \Psi_{-}(x)\leq u(x,t)\leq\Psi_{+}(x)&\quad\mbox{in}\
\mathbb{R}^{d}\times(0,\infty),\end{cases}$ (2.1)

where $\Psi_{+},\Psi_{-}\in Lip(\mathbb{R}^{d})$ are obstacle functions which
satisfy $\Psi_{-}\leq\Psi_{+}$ in $\mathbb{R}^{d}$. A real number $\nu$ is a
constant and $f$ is a locally bounded function. This equation without
obstacles is a birth and spread type equation introduced in [9]. Though the
source term $f$ is not considered in the proof of the main theorem, we take it
into consideration in Appendix A to prepare for the forthcoming paper
”Asymptotic shape of solutions to the mean curvature flow equation with
discontinuous source terms” by Hamamuki and the author. We do so all the more
because it is natural extension in light of optimal control theory.

We denote the upper and lower semicontinous envelope of $u$ by $u^{\ast}$ and
$u_{\ast}$ respectively.

######  Definition 2.1 (Viscosity solution).

  1. 1.

A function $u$ is a viscosity subsolution of (2.1) if it satisfies the
following conditions.

    1. (a)

$\Psi_{-}(x)\leq u^{\ast}(x,t)\leq\Psi_{+}(x)$ for all
$(x,t)\in\mathbb{R}^{d}\times(0,\infty)$.

    2. (b)

$u^{\ast}(x,0)\leq u_{0}(x)$ for all $x\in\mathbb{R}^{d}$.

    3. (c)

Whenever $\phi(x,t)$ is smooth, $u^{\ast}-\phi$ has a local maximum at
$(x_{0},t_{0})\in\mathbb{R}^{d}\times(0,\infty)$ and
$u^{\ast}(x_{0},t_{0})-\Psi_{-}(x_{0})>0$, we have

$\phi_{t}(x_{0},t_{0})-\nu\lvert
D\phi(x_{0},t_{0})\rvert+F_{\ast}(D\phi(x_{0},t_{0}),D^{2}\phi(x_{0},t_{0}))\leq
f^{\ast}(x_{0}).$

  2. 2.

A function $u$ is a viscosity supersolution of (2.1) if it satisfies the
following conditions.

    1. (a)

$\Psi_{-}(x)\leq u_{\ast}(x,t)\leq\Psi_{+}(x)$ for all
$(x,t)\in\mathbb{R}^{d}\times(0,\infty)$.

    2. (b)

$u_{\ast}(x,0)\geq u_{0}(x)$ for all $x\in\mathbb{R}^{d}$.

    3. (c)

Whenever $\phi(x,t)$ is smooth, $u_{\ast}-\phi$ has a local minimum at
$(x_{0},t_{0})\in\mathbb{R}^{d}\times(0,\infty)$ and
$u_{\ast}(x_{0},t_{0})<\Psi_{+}(x_{0})$, we have

$\phi_{t}(x_{0},t_{0})-\nu\lvert
D\phi(x_{0},t_{0})\rvert+F^{\ast}(D\phi(x_{0},t_{0}),D^{2}\phi(x_{0},t_{0}))\geq
f_{\ast}(x_{0}).$

  3. 3.

A function $u$ is a viscosity solution of (2.1) if it is a viscosity
subsolution and a viscosity supersolution of (2.1).

We now give the definition of solutions of the following surface evolution
equation:

$\begin{cases}V=-\kappa+\nu,\ \mbox{on $\partial D_{t}$},\\\ O_{-}\subset
D_{t}\subset O_{+},\end{cases}$ (2.2)

where $\\{D_{t}\\}_{t>0}$ is the unknown family of open sets in
$\mathbb{R}^{d}$. Furthermore $O_{-}$ and $O_{+}$ are fixed open sets in
$\mathbb{R}^{d}$. We also introduce the closed version of (2.2):

$\begin{cases}V=-\kappa+\nu,\ \mbox{on $\partial E_{t}$}\\\ C_{-}\subset
E_{t}\subset C_{+},\end{cases}$ (2.3)

where $\\{E_{t}\\}_{t>0}$ is the unknown family of closed sets in
$\mathbb{R}^{d}$. $C_{-}$ and $C_{+}$ are fixed closed sets in
$\mathbb{R}^{d}$. The PDE (2.1) with $f=0$ is the level set equation for these
surface evolution equations. Since we only consider bounded initial surfaces
in this manuscript, we employ the following class of solutions:

$\displaystyle K_{a}(\mathbb{R}^{d}\times[0,\infty)):=$ $\displaystyle\\{u\in
C(\mathbb{R}^{d}\times[0,\infty))\mid\forall T>0\ \exists R>0\ \mbox{s.t.
$u=a$ in}\ B_{R}^{c}(0)\times[0,T]\\}.$

######  Definition 2.2.

  1. 1.

Let $D_{0}$, $O_{-}$ and $O_{+}$ be open sets in $\mathbb{R}^{d}$. A family of
open sets $\\{D_{t}\\}_{t\geq 0}$ is called an open evolution of (2.2) with
$D_{0}$, $O_{-}$ and $O_{+}$ if there exist $\Psi_{-},\Psi_{+}\in
Lip(\mathbb{R}^{d})$, $u_{0}\in C(\mathbb{R}^{d})$ and a solution $u\in
K_{a}(\mathbb{R}^{d}\times[0,\infty))$ of (2.1) with $\Psi_{-}$, $\Psi_{+}$,
$u_{0}$ and $f=0$ such that $O_{-}=\\{x\in\mathbb{R}^{d}\mid\Psi_{-}(x)>0\\}$,
$O_{+}=\\{x\in\mathbb{R}^{d}\mid\Psi_{+}(x)>0\\}$ and
$D_{t}=\\{x\in\mathbb{R}^{d}\mid u(x,t)>0\\}$ for $t\geq 0$.

  2. 2.

Let $E_{0}$, $C_{-}$ and $C_{+}$ be closed sets in $\mathbb{R}^{d}$. A family
of closed sets $\\{D_{t}\\}_{t\geq 0}$ is called an closed evolution of (2.3)
with $E_{0}$, $C_{-}$ and $C_{+}$ if there exist $\Psi_{-},\Psi_{+}\in
Lip(\mathbb{R}^{d})$, $u_{0}\in C(\mathbb{R}^{d})$ and a solution $u\in
K_{a}(\mathbb{R}^{d}\times[0,\infty))$ of (2.1) with $\Psi_{-}$, $\Psi_{+}$,
$u_{0}$ and $f=0$ such that $C_{-}=\\{x\in\mathbb{R}^{d}\mid\Psi_{-}(x)\geq
0\\}$, $C_{+}=\\{x\in\mathbb{R}^{d}\mid\Psi_{+}(x)\geq 0\\}$ and
$E_{t}=\\{x\in\mathbb{R}^{d}\mid u(x,t)\geq 0\\}$ for $t\geq 0$.

######  Remark 2.3.

The open evolutions and the closed evolutions uniquely exist ([20]).

######  Remark 2.4.

Our main equation in this manuscript is (1.1), which has an obstacle on one
side. We interpret problems that have only $O_{-}$ as (2.2) with
$O_{+}=\mathbb{R}^{d}$ and problems that have only $O_{+}$ as (2.2) with
$O_{-}=\emptyset$.

Whenever we consider (2.2) in this manuscript, we simultaniouly consider the
solution $\\{E_{t}\\}_{t\geq 0}$ to (2.3) with $E_{0}=\overline{D_{0}}$,
$C_{-}=\overline{O_{-}}$ and $C_{+}=\overline{O_{+}}$. Throughout the
manuscript, we denote an open evolution by $\\{D_{t}\\}_{t\geq 0}$ and a
closed evolution by $\\{E_{t}\\}_{t\geq 0}$. As explained above, we assume
that $D_{0}$ is bounded.

#### Notations.

For a point $z\in\mathbb{R}^{d}$, we denote the set
$\\{x\in\mathbb{R}^{d}\mid\lvert x-z\rvert<r\\}$ by $B_{r}(z)$ or sometimes
$B(z,r)$. For a set $S\subset\mathbb{R}^{d}$, we denote the set
$\\{x\in\mathbb{R}^{d}\mid dist(x,S)<r\\}$ by $B_{r}(S)$. We denote by
$S^{d-1}$ the set of unit vectors in $\mathbb{R}^{d}$. The line segment with
end points $x$ and $y$ will be denoted by $l_{x,y}$. When two lines $l_{1}$
and $l_{2}$ are parallel, we will write $l_{1}\parallel l_{2}$. For a set $A$,
we denote the convex hull of $A$ by $Co(A)$. For a family of sets
$\\{D_{t}\\}_{t\geq 0}$, we define

$\varlimsup_{t\to\infty}D_{t}:=\bigcap_{\tau>0}\bigcup_{t>\tau}D_{t},\
\varliminf_{t\to\infty}D_{t}:=\bigcup_{\tau>0}\bigcap_{t>\tau}D_{t}.$

If $\varlimsup_{t\to\infty}D_{t}=\varliminf_{t\to\infty}D_{t}$, we will write

$\lim_{t\to\infty}D_{t}:=\varlimsup_{t\to\infty}D_{t}=\varliminf_{t\to\infty}D_{t}.$

###  2.2 Basic strategy of the game

We prepare special strategies of both players that we explained in Section 1.

######  Definition 2.5 (Concentric strategy).

Let $\epsilon>0$ and $z\in\mathbb{R}^{d}$. Let $x\in\mathbb{R}^{d}$ be the
current position of the game.

  1. 1.

A set of $d-1$ orthogonal unit vectors $v^{j}\in S^{d-1}(j=1,2,\cdots d-1)$
chosen by Paul is called a $z$ concentric strategy (by Paul) if $\langle
v^{j},x-z\rangle=0$ for all $j$, where $\langle\cdot,\cdot\rangle$ stands for
the Euclidean inner product.

  2. 2.

Let $\\{v^{1},v^{2},\cdots,v^{d-1}\\}$ be a choice by Paul in the same round.
A choice $(b^{1},b^{2},\cdots,b^{d-1})\in\\{\pm 1\\}^{d-1}$ by Carol is called
a $z$ concentric strategy (by Carol) if $\langle b^{j}v^{j},x-z\rangle\geq 0$
for all $j$.

One can easily understand the behaviors of trajectories given when one player
takes above strategies. Let $d_{n}=\lvert x_{n}-z\rvert$ for fixed
$z\in\mathbb{R}^{d}$. If Paul takes a $z$ concentric strategy through the
game, then we see that the sequence $\\{d_{n}\\}$ satisfies

$R_{n+1}=\sqrt{R_{n}^{2}+2(d-1)\epsilon^{2}}$ (2.4)

by the Pythagorean theorem regardless of Carol’s choices. The solution
$\\{R_{n}\\}$ of (2.4) is explicitly obtained by

$R_{n}=\sqrt{R_{0}^{2}+2(d-1)n\epsilon^{2}}.$ (2.5)

On the other hand, if Carol takes a $z$ concentric strategy through the game,
we have

$\displaystyle\left\lvert x+\sum_{j}(\sqrt{2}\epsilon
b^{j}v^{j})-z\right\rvert^{2}$ $\displaystyle=\lvert
x-z\rvert^{2}+2(d-1)\epsilon^{2}+\sum_{j}\sqrt{2}\epsilon\langle
b^{j}v^{j},x-z\rangle$ $\displaystyle\geq\lvert
x-z\rvert^{2}+2(d-1)\epsilon^{2},$

which implies $d_{n+1}\geq\sqrt{d_{n}^{2}+2(d-1)\epsilon^{2}}$. Therefore we
obtain $d_{n}\geq\sqrt{d_{0}^{2}+2(d-1)n\epsilon^{2}}$.

######  Remark 2.6.

When Carol takes a $z$ concentric strategy, she can control the distance
$|x_{n}-z|$. However we notice that she can not control the moves in
tangential direction. For instance, let $d=2$, $z=(0,0)$ and the current game
position $x_{n}=(0,1)$. If Paul chooses $v=(1,0)$ at this point, both $b=1$
and $b=-1$ are $(0,0)$ concentric strategies by Carol. One may think that if
Carol makes the further decision that for Paul’s choice $v$ tangential to the
circle centered at the origin passing through $x_{n}$, she chooses $b$ so that
$bv$ becomes clockwise, then she could control the trajectory of the game to
be clockwise. However this is not true. Carol’s greedy attempt to move as she
pleases in the tangential direction will be thwarted by Paul. Indeed when
Carol makes above decision, Paul can move counterclockwise while the distance
$|x_{n}-z|$ meets almost the condition (2.5) by slightly leaning the vector
$v$ from tangential one. (e.g. $v=(\cos{-\epsilon^{2}},\sin{-\epsilon^{2}})$
at $x_{n}=(0,1)$)

##  3 Asymptotic behavior of solutions

Throughout this section we consider the main equation (1.1).

In the following lemma, we estimate the asymptotic shape from above by
considering Carol’s strategies.

######  Lemma 3.1.

$\varlimsup_{t\to\infty}E_{t}\subset Co(C_{-}).$

###### Proof.

Let $u$ be the unique solution to (1.2) with $u_{0}$ and $\Psi_{-}$ that are
as in Definition 2.2. We notice that the conclusion holds if and only if for
$x\in Co(C_{-})^{c}$, there exists $\tau>0$ such that $u(x,t)<0$ for $t>\tau$.
To prove $u^{\epsilon}<0$, it is sufficient to give a Carol’s strategy that
makes the game cost negative but is not necessarily optimal one. For $x\in
Co(C_{-})^{c}$ we can take an open ball $B$ such that $C_{-}\subset B$ and
$x\in\partial B$ by the hyperplane separation theorem and the boundedness of
$Co(C_{-})$. Let $z$ be the center of $B$ and $r=\lvert z-x\rvert$. If Carol
takes a $z$ concentric strategy, then we see that regardless of Paul’s choice,
the game trajectory $\\{x_{n}\\}$ satisfies $\lvert
x_{n}-z\rvert\geq\sqrt{r^{2}+2n(d-1)\epsilon^{2}}$, where
$\sqrt{r^{2}+2n(d-1)\epsilon^{2}}$ is the solution of (2.4) with $R_{0}=r$.
Letting $\tau=2R(R+r)/(d-1)$, we have

$\lvert x_{N}-z\rvert\geq r+2R$ (3.1)

for the last position $x_{N}$ of the game, where $R>0$ is a constant taken in
(1.3). The inequality (3.1) and $Co(C_{-})\subset B$ imply
$dist(x_{N},Co(C_{-}))\geq dist(x_{N},B)\geq 2R$. Also $C_{-}\subset
E_{0}\subset B_{R}(0)$ implies $Co(C_{-})\subset B_{R}(0)$. Hence we have
$x_{N}\notin B_{R}(0)$, which means $u_{0}(x_{N})=a<0$ by (1.3). If Paul quits
the game on the way, the stopping cost is at most

$\sup_{b\in B^{c}}\Psi_{-}(b)<0.$

The comparison principle ([13]) for (1.2) and the convergence results in
Appendix A imply that $u^{\epsilon}(x,t)$ converges to $u(x,t)$ locally
uniformly in $(x,t)$. See also [1, Chapter V Lemma 1.9] if necessary. We also
notice that the uniform boundedness of $u^{\epsilon}$ is satisfied owe to the
rule of the game and the boundedness of $u_{0}$ and $\Psi_{-}$. Since both
upper bound of the terminal cost $a$ and that of the stopping cost $\sup_{b\in
B^{c}}\Psi_{-}(b)$ do not depend on $\epsilon$, we conclude that $u(x,t)<0$
for $t>\tau$ and $x\in Co(C_{-})^{c}$. ∎

$\Downarrow$$\Downarrow$The graph $G$ is connected.The graph $G$ is connected.
Figure 5: $D_{0}$ and $O_{-}$ that satisfy the assumption of Theorem 3.2 (In
this figure the loops of $G$ are ignored.) The graph $G$ is not connected.The
graph $G$ is not connected.$\Downarrow$$\Downarrow$ Figure 6: $D_{0}$ and
$O_{-}$ that do not satisfy the assumption of Theorem 3.2
$C$$x$$a$$b$$\hat{a}$$\hat{b}$ Figure 7: strategy in the case 3)

For an obstacle $O_{-}$ and an initial set $D_{0}$, we define the graph
$G=(V,E)$ as follows:

$V:=\\{O\subset\mathbb{R}^{2}\mid O\ \mbox{ is a connected component of}\
O_{-}\\},$ $E:=\\{\langle O,P\rangle\mid O,P\in V\ \mbox{and $l_{x,y}\subset
D_{0}$ for some $x\in O$ and $y\in P$}\\}.$

See Appendix C for definitions of terms in graph theory.

######  Theorem 3.2.

Assume that $d=2$ and the graph $G$ is connected. Then

$Co(O_{-})\subset\varliminf_{t\to\infty}D_{t}\subset\varlimsup_{t\to\infty}D_{t}\subset\overline{Co(O_{-})}$

and

$Co(O_{-})\subset\varliminf_{t\to\infty}E_{t}\subset\varlimsup_{t\to\infty}E_{t}\subset\overline{Co(O_{-})}.$

######  Remark 3.3.

Figure 5 (resp. Figure 6) shows examples of $D_{0}$ and $O_{-}$ that satisfy
(resp. do not satisfy) the assumption of Theorem 3.2.

###### Proof.

For $D_{0}$ and $E_{0}$, we take $u_{0}$ as in Definition 2.2. Indeed it
suffices to let

$u_{0}(x)=\begin{cases}dist(x,\partial D_{0}),\ x\in D_{0}\\\
\max\\{a,-dist(x,\partial D_{0})\\},\ x\in D_{0}^{c}.\end{cases}$ (3.2)

Similarly it suffices to let

$\Psi_{-}(x)=\begin{cases}dist(x,\partial O_{-}),\ x\in O_{-}\\\
\max\\{a,-dist(x,\partial O_{-})\\},\ x\in O_{-}^{c}.\end{cases}$ (3.3)

By Lemma 3.1 and $D_{t}\subset E_{t}$, it suffices to prove
$Co(O_{-})\subset\varliminf_{t\to\infty}D_{t}$. Namely our goal is to prove
that for $x\in Co(O_{-})$, there exists $\tau>0$ such that $u(x,t)>0$ for
$t>\tau$. To prove $u^{\epsilon}>0$, it is sufficient to give a Paul’s
strategy, which makes the game cost positive and is not necessarily optimal
one. Let $x\in Co(O_{-})$. It would be convenient to introduce the set

$L:=\\{z\in l_{x,y}\mid x,y\in O_{-},l_{x,y}\subset D_{0}\\}$

in doing case analysis for $x\in Co(O_{-})$.

$1)~{}x\in O_{-}.$ In this case, it suffices for Paul to quit the game at the
first round and gain the stopping cost $\Psi_{-}(x)>0$. Recall that
$u^{\epsilon}(x,t)$ converges to $u(x,t)$ locally uniformly in $(x,t)$. Thus
we obtain $u(x,t)>0$ for any $t>0$.

$2)~{}x\in L\setminus O_{-}.$ Let $z,w\in O_{-}$ satisfy $x\in l_{z,w}$ and
$l_{z,w}\subset D_{0}$. We take $\delta>0$ to satisfy
$\overline{B_{\delta}(z)}\subset O_{-}$ and $\overline{B_{\delta}(w)}\subset
O_{-}$. For the initial position $x$, Paul’s strategy is to keep taking
$v=\frac{z-x}{\lvert z-x\rvert}$ until he reaches $B_{\delta}(z)\cup
B_{\delta}(w)$. If he reaches $B_{\delta}(z)\cup B_{\delta}(w)$, then he quits
the game. By doing this, Paul gains positive game cost in either case he quits
the game or not. See Figure 3. More precisely, Paul gains at least

$\min\left\\{\min_{y\in\overline{B_{\delta}(z)}\cup\overline{B_{\delta}(w)}}\Psi_{-}(y),\min_{y\in
l_{z,w}}u_{0}(y)\right\\}>0$

regardless of $\epsilon\in(0,\delta/\sqrt{2})$, where $\epsilon$ is taken
small enough for Paul not to stride over $B_{\delta}(z)$ or $B_{\delta}(w)$.
Hence, as in the case 1), we obtain $u(x,t)>0$ for any $t>0$.

$3)~{}x\in Co(O_{-})\setminus L.$ Henceforth we give a strategy by Paul that
includes a $z$ concentric strategy and makes the game cost positive. To do so,
we are going to construct a closed curve that consists of an arc $C$ with its
center at $z$ and a path $\hat{\Gamma}$ in $L$. Since $O_{-}\subset L\subset
Co(O_{-})$, we have $Co(O_{-})=Co(L)$. By Lemma B.2 in Appendix B, we can take
$a,b\in L$ such that $x\in l_{a,b}$. We only show the case $a,b\notin O_{-}$,
since otherwise we would prove it in a simpler manner.

We first explain how to construct a path $\Gamma\subset L$ that contains $a$
and $b$. We take a specific path rather than just a path. By doing so, we are
able to indicate a region that includes final positions of the games to
guarantee that $u^{\epsilon}$ is uniformly positive. Since $a\in L$, there is
a line segment that is in $D_{0}$, contains $a$, and has endpoints in some
connected components $A$ and $B$ of $O_{-}$ respectively. Similarly there is a
line segment that is in $D_{0}$, contains $b$, and has endpoints in some
connected components $C$ and $D$ of $O_{-}$ respectively. Notice that $\langle
A,B\rangle,\langle C,D\rangle\in E$, recalling $E$ is the set of unordered
pairs of the graph $G$ defined above. From Proposition C.2 there is a path
$P=(V^{\prime},E^{\prime})$ of the graph $G$ such that $A,B,C,D\in V^{\prime}$
and $\langle A,B\rangle,\langle C,D\rangle\in E^{\prime}$. Writing
$V^{\prime}=\\{O_{0},O_{1},\cdots,O_{n}\\}$ and $E^{\prime}=\\{\langle
O_{i},O_{i+1}\rangle\mid i=0,1,\cdots,n-1\\}$, we see that for some points
$y_{i},\tilde{y}_{i}\in O_{i}$ $(i=0,1,\cdots,n-1)$, there are line segments
$l_{\tilde{y}_{i},y_{i+1}}$ $(i=0,1,\cdots,n-1)$ such that $a\in
l_{\tilde{y}_{0},y_{1}}$, $b\in l_{\tilde{y}_{n-1},y_{n}}$ and
$l_{\tilde{y}_{i},y_{i+1}}\subset D_{0}$ $(i=0,1,\cdots,n-1)$. Let
$\Gamma_{i}$ be a polygonal line in $O_{i}$ with endpoints $y_{i}$ and
$\tilde{y}_{i}$. Now we define

$\Gamma:=l_{\tilde{y}_{0},y_{1}}\cup\Gamma_{1}\cup
l_{\tilde{y}_{1},y_{2}}\cup\Gamma_{2}\cup
l_{\tilde{y}_{2},y_{3}}\cup\cdots\cup
l_{\tilde{y}_{n-2},y_{n-1}}\cup\Gamma_{n-1}\cup l_{\tilde{y}_{n-1},y_{n}}.$

We may assume $l_{a,b}\nparallel l_{\tilde{y}_{0},y_{1}}$ and
$l_{a,b}\nparallel l_{\tilde{y}_{n-1},y_{n}}$, since otherwise we would retake
either $a$ or $b$ as an element of $O_{-}$. Without loss of generality we can
also assume that $\Gamma$ and $l_{a,b}$ do not cross each other except at $a$
and $b$. By Lemma B.1 in Appendix B we take $\delta>0$ small enough to satisfy
$B_{3\delta}(\Gamma)\subset L$, noticing that $L$ is an open set and $\Gamma$
is a compact set. Let $w_{0},w_{1},w_{2}$ be unit vectors in $\mathbb{R}^{2}$
such that $w_{0}\parallel l_{a,b}$, $w_{1}\parallel l_{\tilde{y}_{0},y_{1}}$,
$w_{2}\parallel l_{\tilde{y}_{n-1},y_{n}}$ and $(w_{0}\cdot w_{1})(w_{0}\cdot
w_{2})\geq 0$. Let $\hat{a}=a+\delta w_{1}$ and $\hat{b}=b+\delta w_{2}$. We
temporarily define $C$ as the arc passing through $\hat{a}$, $\hat{b}$ and
$x$. Combining $\Gamma$ and $C$, we make a closed curve $C\cup\hat{\Gamma}$,
where

$\hat{\Gamma}:=l_{\hat{a},y_{1}}\cup\Gamma_{1}\cup
l_{\tilde{y}_{1},y_{2}}\cup\Gamma_{2}\cup
l_{\tilde{y}_{2},y_{3}}\cup\cdots\cup
l_{\tilde{y}_{n-2},y_{n-1}}\cup\Gamma_{n-1}\cup l_{\tilde{y}_{n-1},\hat{b}}.$

$x$$x$$\Longrightarrow$ Figure 8: make a Jordan closed curve
$C$$x$$a$$b$$\hat{a}$$\hat{b}$$x$$a$$b$$\hat{a}$$\hat{b}$$\Longrightarrow$$C$
Figure 9: retake $C$ and $\Omega$

For the closed curve $C\cup\hat{\Gamma}$ and $x$ in it, there is a Jordan
closed curve $\hat{C}$ such that $x\in\hat{C}$ and $\hat{C}\subset
C\cup\hat{\Gamma}$ (Figure 8). See Appendix D in detail. Thus, based on the
Jordan curve theorem, we let $\Omega$ be the bounded domain that satifies
$\partial\Omega=\hat{C}$. If $\Omega$ touches the arc $C$ from inside, then we
retake the other pair of $(\hat{a},\hat{b})$ and, together with it, retake $C$
and $\Omega$ so that $\Omega$ touches the arc $C$ from outside (Figure 9). We
notice that the domain enclosed by $l_{\tilde{y}_{0},y_{1}}$,
$l_{\tilde{y}_{n-1},y_{n}}$ and the two arcs shown in Figure 9 is bounded, and
hence, so is the new $\Omega$. We further notice that we can take $C$ so that
$C$ and $\hat{\Gamma}$ intersect only at $\hat{a}$ and $\hat{b}$ by taking
$\delta$ smaller if necessary.

We now give a strategy by Paul for the initial position $x$. Paul first takes
a $z$ concentric strategy, where $z$ is the center of the arc $C$. If Paul
enters $B_{\delta}(\Gamma_{i})$ for some $i$, then he quits the game at this
point. Once he enters $B_{\delta}(l_{\tilde{y}_{i},y_{i+1}})$ for some $i$, he
takes a similar strategy to that in the case 2). To see that it attains
positive game cost, we notice two propeties of game trajectories $\\{x_{n}\\}$
given when Paul keeps taking a $z$ concentric strategy by some round. One is
that $x_{n}\in B\left(z,\sqrt{\lvert
x_{0}-z\rvert^{2}+2n\epsilon^{2}}\right)^{c}$. The other is that
$x_{n}\in(C\cup\Omega)\setminus N_{\delta}$ implies $x_{n+1}\in\Omega$ for
$\sqrt{2}\epsilon<\delta$, where we denote
$(\bigcup_{i}B_{\delta}(\Gamma_{i}))\cup(\bigcup_{i}B_{\delta}(l_{\tilde{y}_{i},y_{i+1}}))$
by $N_{\delta}$. Thus it takes at most finite time $\tau$ (satisfying
$\Omega\subset B(z,\sqrt{\lvert x_{0}-z\rvert^{2}+2\tau})$) for Paul to reach
$N_{\delta}$. We see that the game ends at some point in $N_{2\delta}$ and
Paul gains at least

$\min\left\\{\inf_{y\in\bigcup_{i}B_{2\delta}(\Gamma_{i})}\Psi_{-}(y),\inf_{y\in\bigcup_{i}B_{\delta}(l_{\tilde{y}_{i},y_{i+1}})}u_{0}(y)\right\\}>0,$

regardless of $\epsilon\in(0,\delta/\sqrt{2})$. Therefore we conclude that
$u(x,t)>0$ for any $t>\tau$, noticing that $\tau$ may depend on $x$, but does
not depend on $\epsilon$. ∎

######  Remark 3.4.

In general, fattening of the level set may occur under the assumption of
Theorem 3.2. i.e., $\overline{D_{t}}=E_{t}$ may fail at some $t>0$. Theorem
3.2 states that even if the curve evolutions are not unique, they have the
same limit.

We give some sufficient conditions to the assumption of Theorem 3.2.

######  Corollary 3.5.

Assume that $Co(O_{-})\subset D_{0}$. Then the same conclusion as that of
Theorem 3.2 holds.

Figure 10: $O_{-}$ is connected

$x$$z$ Figure 11: Paul’s strategy to achieve positive game cost for the
initial position $x$

######  Corollary 3.6.

Assume that $O_{-}$ is connected (Figure 11 and 11). Then the same conclusion
as that of Theorem 3.2 holds.

Under the following assumption, the moving surface sticks to the obstacle in
finite time:

$\exists r>0\ \forall w\in\partial O_{-}\ \exists z\in B_{r}(w),\ \mbox{s.t.}\
O_{-}\subset B_{\lvert z-w\rvert}(z).$ (3.4)

In the following theorem, there is no need to asuume $d=2$.

######  Theorem 3.7.

Assume (3.4). Then

$\lim_{t\to\infty}D_{t}=O_{-}$

and

$\lim_{t\to\infty}E_{t}=C_{-}.$

Moreover there exists $\tau>0$ such that $D_{t}=D_{\tau}=O_{-}$ and
$E_{t}=E_{\tau}=C_{-}$ for $t\geq\tau$.

###### Proof.

We notice that the condition (3.4) implies that $O_{-}$ is convex (Proposition
B.4). It is now clear that $O_{-}\subset D_{t}$ for any $t>0$ and hence
$O_{-}\subset\bigcap_{t>0}D_{t}\subset\varliminf_{t\to\infty}D_{t}$. We prove
that there exists $\tau>0$ such that
$\bigcup_{t\geq\tau}E_{t}\subset\overline{O_{-}}$. Namely we show that there
exists $\tau>0$ such that $u(x,t)<0$ for $t\geq\tau$ and
$x\in(\overline{O_{-}})^{c}$. The difference from Lemma 3.1 is that we now
have to take $\tau$ independent of $x$. Indeed, for
$x\in(\overline{O_{-}})^{c}$, we can take a ball $B_{\lvert z-w\rvert}(z)$ in
(3.4) such that $x\notin B_{\lvert z-w\rvert}(z)$ and it suffices for Carol to
keep taking a $z$ concentric strategy until the game ends. The value $r$ in
(3.1) is now taken independent of $x$ and then so is $\tau$. Therefore we
obtain

$O_{-}\subset\varliminf_{t\to\infty}D_{t}\subset\varlimsup_{t\to\infty}D_{t}\subset\bigcup_{t\geq\tau}D_{t}\subset\bigcup_{t\geq\tau}E_{t}\subset\overline{O_{-}}.$

Since $\bigcup_{t\geq\tau}D_{t}$ is open, we have
$O_{-}=\bigcup_{t\geq\tau}D_{t}$, which means
$\varliminf_{t\to\infty}D_{t}=\varlimsup_{t\to\infty}D_{t}$ and moreover
$D_{t}=D_{\tau}=O_{-}$ for $t\geq\tau$.

We also have

$O_{-}\subset\bigcap_{t>0}D_{t}\subset\bigcap_{t>0}E_{t}\subset\varliminf_{t\to\infty}E_{t}\subset\varlimsup_{t\to\infty}E_{t}\subset\overline{O_{-}}.$

Since $\bigcap_{t>0}E_{t}$ is closed, we similarly have
$\overline{O_{-}}=\bigcap_{t>0}E_{t}$, which means
$\varliminf_{t\to\infty}E_{t}=\varlimsup_{t\to\infty}E_{t}$ and moreover
$E_{t}=E_{0}=\overline{O_{-}}$ for $t\geq 0$. ∎

######  Remark 3.8.

The hair-clip solution, which can be regarded as an explicit solution to the
curve shortening problem with the Dirichlet condition (See e.g. [19]), implies
that the solution of our obstacle problem can be apart from the asymptotic
shape at any time. i.e., there are $D_{0}$ and $O_{-}$ such that
$Co(O_{-})\subsetneq D_{t}$ for any $t>0$.

##  4 With driving force

In this section we consider the following surface evolution equations in the
plane that have obstacles on one side:

$\begin{cases}V=-\kappa+\nu,\ \mbox{on $\partial D_{t}$},\\\ O_{-}\subset
D_{t},\end{cases}$ (4.1)

or

$\begin{cases}V=-\kappa+\nu,\ \mbox{on $\partial D_{t}$},\\\ D_{t}\subset
O_{+},\end{cases}$ (4.2)

where $D_{t}$, $O_{-}$ and $O_{+}$ are open sets in $\mathbb{R}^{2}$ and
$\nu>0$ is a constant. These equations are specific cases of (2.2).

While Theorem 3.2 is invarient with respect to similarity transformation, the
behaviors of moving curves governed by (4.1) or (4.2) depend not only on shape
of the obstacles and the initial curve but also on size of them. It is easily
understood by considering the case $\partial D_{0}$ is a circle. For
$D_{0}=B_{R}((0,0))$, the circle $\partial D_{0}$ shrinks if $R<\nu^{-1}$, and
it spreads if $R>\nu^{-1}$ as time goes. So it is difficult to present a
concise result as Theorem 3.2. We give several examples of computations of the
asymptotic shapes here.

###  4.1 Basic strategy of the game

Concerning to the game interpretation for (4.1), the difference from the game
in Section 1 is that Paul has the right to choose $w_{i}\in S^{1}$ at each
round $i$ and the control system is

$x_{i}=x_{i-1}+\sqrt{2}\epsilon b_{i}v_{i}+\nu\epsilon^{2}w_{i},$

instead of (1.4). In the game for (4.2), not Paul but Carol has the right to
quit the game. If Carol quits the game at round $i$, then the cost is given by
$\Psi_{+}(x_{i})$. See Appendix A in detail. As explained later in the proof
of Lemma 3.1, the value functions $u^{\epsilon}$ locally uniformly converge to
the solution $u$ of the corresponding level set equation.

We now prepare several types of game strategies and give the properties of the
game trajectories when they are used. The first one is similar to the one in
Definition 2.5.

######  Definition 4.1 (Concentric strategy).

Let $\nu>0$, $\epsilon>0$ and $z\in\mathbb{R}^{2}$. Let $x\in\mathbb{R}^{2}$
be the current position of the game.

  1. 1.

A choice $(v,w)\in S^{1}\times S^{1}$ by Paul is called a $z$ concentric
strategy (by Paul) if

$w=\frac{z-x}{\lvert z-x\rvert}~{}\mbox{and}~{}v\perp w.$

When $x=z$, any $(v,w)\in S^{1}\times S^{1}$ satisfying $v\perp w$ is called a
$z$ concentric strategy.

  2. 2.

Let $(v,w)\in S^{1}\times S^{1}$ be a choice by Paul in the same round. A
choice $b\in\\{\pm 1\\}$ by Carol is called a $z$ concentric strategy (by
Carol) if

$\langle bv,x+\nu\epsilon^{2}w-z\rangle\geq 0.$

As in Section 2.2, let $d_{n}=\lvert x_{n}-z\rvert$ for fixed
$z\in\mathbb{R}^{2}$. If Paul takes a $z$ concentric strategy through the
game, then the sequence $\\{d_{n}\\}$ satisfies

$R_{n+1}=\sqrt{(R_{n}-\nu\epsilon^{2})^{2}+2\epsilon^{2}}.$ (4.3)

If Carol takes a $z$ concentric strategy through the game, we have $d_{n}\geq
R_{n}$, where $\\{R_{n}\\}$ is the solution to (4.3) with $R_{0}=d_{0}$. In
the following lemmas, we give basic propeties of the behaviors of the
solutions to (4.3).

######  Lemma 4.2.

Fix $\nu>0$. Let $\epsilon>0$ and $\\{R_{n}\\}$ be a sequence satisfying the
condition (4.3). Then the following properties hold.

  1. 1.

If $R_{0}=\nu^{-1}+\frac{\nu}{2}\epsilon^{2}$, then
$R_{n}=\nu^{-1}+\frac{\nu}{2}\epsilon^{2}$ for all $n$. If
$R_{0}>\nu^{-1}+\frac{\nu}{2}\epsilon^{2}$, then
$R_{n}>\nu^{-1}+\frac{\nu}{2}\epsilon^{2}$ for all $n$ and $\\{R_{n}\\}$ is
decreasing for $\epsilon\leq\sqrt{2}\nu^{-1}$. If $\nu\epsilon^{2}\leq
R_{0}<\nu^{-1}+\frac{\nu}{2}\epsilon^{2}$, then
$R_{n}<\nu^{-1}+\frac{\nu}{2}\epsilon^{2}$ for all $n$ and $\\{R_{n}\\}$ is
increasing.

  2. 2.

$\lim_{n\to\infty}R_{n}=\nu^{-1}+\frac{\nu}{2}\epsilon^{2}.$

  3. 3.

$R_{n+1}-R_{n}\leq\frac{\epsilon^{2}}{R_{n}}-\nu\epsilon^{2}+\frac{\nu^{2}\epsilon^{4}}{2R_{n}}.$
(4.4)

If $R_{n+1}\geq R_{n}$, then

$\frac{\epsilon^{2}}{R_{n+1}}-\nu\epsilon^{2}+\frac{\nu^{2}\epsilon^{4}}{2R_{n+1}}\leq
R_{n+1}-R_{n}.$ (4.5)

###### Proof.

  1. 1.

We first notice that $R_{n+1}\leq R_{n}$ is equivalent to
$R_{n}\geq\frac{\nu}{2}\epsilon^{2}+\nu^{-1}$. Also $R_{n+1}\geq R_{n}$ is
equivalent to $R_{n}\leq\frac{\nu}{2}\epsilon^{2}+\nu^{-1}$. These facts imply
the first assertion.

Assume that $R_{k}>\frac{\nu}{2}\epsilon^{2}+\nu^{-1}$ for some $k$. Since
$(R_{k}-\nu\epsilon^{2})^{2}>\left(\nu^{-1}-\frac{\nu}{2}\epsilon^{2}\right)^{2}$
for $\epsilon\leq\sqrt{2}\nu^{-1}$, we have

$\displaystyle R_{k+1}$
$\displaystyle=\sqrt{(R_{k}-\nu\epsilon^{2})^{2}+2\epsilon^{2}}>\sqrt{\left(\nu^{-1}-\frac{\nu}{2}\epsilon^{2}\right)^{2}+2\epsilon^{2}}=\frac{\nu}{2}\epsilon^{2}+\nu^{-1}.$

Hence, if $R_{0}>\frac{\nu}{2}\epsilon^{2}+\nu^{-1}$, we have by induction
that $R_{n}>\frac{\nu}{2}\epsilon^{2}+\nu^{-1}$ for all $n$. The second
assertion follows from this.

Assume that $\nu\epsilon^{2}\leq R_{k}<\frac{\nu}{2}\epsilon^{2}+\nu^{-1}$.
Since
$(R_{k}-\nu\epsilon^{2})^{2}<\left(\nu^{-1}-\frac{\nu}{2}\epsilon^{2}\right)^{2}$
for $\epsilon\in\left(0,\sqrt{2}\nu^{-1}\right)$, we get
$R_{k+1}<\frac{\nu}{2}\epsilon^{2}+\nu^{-1}.$ In the same way,
$\nu\epsilon^{2}\leq R_{0}<\frac{\nu}{2}\epsilon^{2}+\nu^{-1}$ gives
$R_{n}<\frac{\nu}{2}\epsilon^{2}+\nu^{-1}$. Therefore the third assertion is
obtained.

  2. 2.

Since $\\{R_{n}\\}$ is monotone and bounded, it is convergent. Its limit value
is given by taking limit for both sides of (4.3) and solving the limit
equation.

  3. 3.

The inequality (4.4) is given by the following computation.

$R_{n+1}-R_{n}=\frac{R_{n+1}^{2}-R_{n}^{2}}{R_{n+1}+R_{n}}\leq\frac{R_{n+1}^{2}-R_{n}^{2}}{2R_{n}}=\frac{\epsilon^{2}}{R_{n}}-\nu\epsilon^{2}+\frac{\nu^{2}\epsilon^{4}}{2R_{n}}.$

Similarly (4.5) is obtained by

$R_{n+1}-R_{n}=\frac{R_{n+1}^{2}-R_{n}^{2}}{R_{n+1}+R_{n}}\geq\frac{R_{n+1}^{2}-R_{n}^{2}}{2R_{n+1}}\geq\frac{\epsilon^{2}}{R_{n+1}}-\nu\epsilon^{2}+\frac{\nu^{2}\epsilon^{4}}{2R_{n+1}}.$

∎

It is sometimes convenient to describe the behavior of the trajectory by an
operator. For fixed $\nu>0$, we define the operator
$T_{h}:\mathbb{R}\to\mathbb{R}$ as

$T_{h}(R):=\sqrt{(R-\nu h)^{2}+2h}.$

We denote $n$ times composition of $T_{h}$ by $T_{h}^{n}$. The solution
$\\{R_{n}\\}$ to (4.3) with $R_{0}=R$ is described by
$R_{n}=T_{\epsilon^{2}}^{n}(R)$.

######  Lemma 4.3.

  1. 1.

$T_{h}(R)<T_{h}(R^{\prime})$, provided $\nu h\leq R<R^{\prime}$.

  2. 2.

$T_{h}(R)>T_{h/2}^{2}(R)$.

###### Proof.

The proofs are done by direct computation, so are omitted. ∎

We also prepare a notation that represents the time to pass through the set
$B_{b}(z)\setminus B_{a}(z)$ when Paul takes a $z$ concentric strategy. For
$a,b\geq 0$ and $\epsilon>0$ satisfying $a\leq
b\leq\nu^{-1}+\frac{\nu}{2}\epsilon^{2}$, we define

$t_{\epsilon}(a,b):=\epsilon^{2}\lvert\\{n\in\mathbb{N}\mid
T_{\epsilon^{2}}^{n}(a)<b\\}\rvert,$ (4.6)

where we denote the set $\\{0,1,2,\cdots\\}$ by $\mathbb{N}$.

######  Lemma 4.4.

Let $t>0$ and $0<R<\nu^{-1}$. Define $N=\lceil t\epsilon^{-2}\rceil$. Then
$T_{\epsilon^{2}}^{N}(R)<\nu^{-1}$ for sufficiently small $\epsilon>0$.

###### Proof.

We prove that

$t_{\epsilon}\left(\nu^{-1}-\epsilon,\nu^{-1}\right)\geq
C\log_{2}\epsilon^{-1}$

for some constant $C>0$.

First let us explain the property of $t_{\epsilon}(a,b)$. Let $0<a\leq b\leq
c\leq\nu^{-1}+\frac{\nu}{2}\epsilon^{2}$. By the definition of $t_{\epsilon}$,
we see

$t_{\epsilon}(a,b)\leq t_{\epsilon}(a,c).$

By Lemma 4.3 1, we have

$t_{\epsilon}(b,c)\leq t_{\epsilon}(a,c).$

Let $n^{\ast}=t_{\epsilon}(a,b)\epsilon^{-2}$ and
$d=T_{\epsilon^{2}}^{n^{\ast}-1}(a)$. Then we have

$t_{\epsilon}(a,c)=t_{\epsilon}(a,b)+t_{\epsilon}(d,c)-\epsilon^{2}.$

Since $d<b$, we deduce that

$t_{\epsilon}(a,c)\geq t_{\epsilon}(a,b)+t_{\epsilon}(b,c)-\epsilon^{2}.$
(4.7)

We next estimate $t_{\epsilon}\left(\nu^{-1}-\epsilon,\nu^{-1}\right)$. When
$\nu^{-1}-2^{-m}\epsilon\leq R_{n}\leq\nu^{-1}-2^{-m-1}\epsilon$ for
$m,n\in\mathbb{N}$, the inequality (4.4) implies

$\displaystyle
R_{n+1}-R_{n}\leq\frac{\epsilon^{2}}{\nu^{-1}-2^{-m}\epsilon}-\nu\epsilon^{2}+\frac{\nu^{2}\epsilon^{4}}{2\nu^{-1}-2^{-m+1}\epsilon}=\frac{2^{-m}\nu\epsilon^{3}+\nu^{2}\epsilon^{4}}{2\nu^{-1}-2^{-m+1}\epsilon},$

and in particular, if $\epsilon\leq\nu^{-1}$,

$R_{n+1}-R_{n}\leq\frac{2^{-m}\nu\epsilon^{3}+\nu^{2}\epsilon^{4}}{2\nu^{-1}-2^{-m+1}\epsilon}\leq\frac{2^{-m}\nu\epsilon^{3}+\nu^{2}\epsilon^{4}}{\nu^{-1}}=\nu^{2}\epsilon^{3}\left(2^{-m}+\nu\epsilon\right).$

Thus
$t_{\epsilon}\left(\nu^{-1}-2^{-m}\epsilon,\nu^{-1}-2^{-m-1}\epsilon\right)$
can be compared to the exit time for sequences with the constant speed of
$\nu^{2}\epsilon^{3}\left(2^{-m}+\nu\epsilon\right)$ per round. Then we have

$\displaystyle
t_{\epsilon}\left(\nu^{-1}-2^{-m}\epsilon,\nu^{-1}-2^{-m-1}\epsilon\right)$
$\displaystyle\geq\left(\left(\nu^{-1}-2^{-m-1}\epsilon\right)-\left(\nu^{-1}-2^{-m}\epsilon\right)\right)\frac{\epsilon^{2}}{\nu^{2}\epsilon^{3}\left(2^{-m}+\nu\epsilon\right)}=\frac{1}{2\nu^{2}+2^{m+1}\nu^{3}\epsilon}.$

In particular, if $2^{m}\leq\epsilon^{-1}$, we get

$\frac{1}{2\nu^{2}+2^{m+1}\nu^{3}\epsilon}\geq\frac{1}{2\nu^{2}+2\nu^{3}}=\frac{1}{2\nu^{2}(\nu+1)}.$

Let $m^{\ast}$ be the minimal integer $m$ satisfying $2^{m}>\epsilon^{-1}$.
Using (4.7), we obtain

$\displaystyle t_{\epsilon}\left(\nu^{-1}-\epsilon,\nu^{-1}\right)$
$\displaystyle\geq\sum_{k=0}^{m^{\ast}-1}\left(t_{\epsilon}\left(\nu^{-1}-2^{-k}\epsilon,\nu^{-1}-2^{-k-1}\epsilon\right)-\epsilon^{2}\right)$
$\displaystyle\geq\sum_{k=0}^{m^{\ast}-1}\frac{1}{3\nu^{2}(\nu+1)}=\frac{\log_{2}\epsilon^{-1}}{3\nu^{2}(\nu+1)},$

which is the conclusion. ∎

######  Lemma 4.5.

For $\nu>0$ and $\nu^{-1}\geq\delta>0$, there exist $\epsilon_{0}>0$ and $M>0$
such that

$t_{\epsilon}(a,\nu^{-1}-\delta)\leq M$

for all $0<\epsilon<\epsilon_{0}$ and $0\leq a\leq\nu^{-1}-\delta$.

###### Proof.

Let $n_{\epsilon}=t_{\epsilon}(a,\nu^{-1}-\delta)\epsilon^{-2}$. Namely
$T_{\epsilon^{2}}^{n_{\epsilon}}(a)<\nu^{-1}-\delta$ and
$T_{\epsilon^{2}}^{n_{\epsilon}+1}(a)\geq\nu^{-1}-\delta$ are satisfied. Since
$T_{\epsilon^{2}}(R)-R$ is monotonically decreasing with respect to $R$ and
$T_{\epsilon^{2}}^{n_{\epsilon}+1}(a)\leq\nu^{-1}-\delta/2$ for sufficiently
small $\epsilon>0$, we have by (4.5) in Lemma 4.2

$\displaystyle T_{\epsilon^{2}}^{n+1}(a)-T_{\epsilon^{2}}^{n}(a)\geq
T_{\epsilon^{2}}^{n_{\epsilon}+1}(a)-T_{\epsilon^{2}}^{n_{\epsilon}}(a)\geq\frac{\epsilon^{2}}{T_{\epsilon^{2}}^{n_{\epsilon}+1}(a)}-\nu\epsilon^{2}+\frac{\nu^{2}\epsilon^{4}}{2T_{\epsilon^{2}}^{n_{\epsilon}+1}(a)}\geq\frac{\delta\nu^{2}}{2-\delta\nu}\epsilon^{2}$

for $n=0,1,\cdots,n_{\epsilon}$. This means

$n_{\epsilon}\leq(\nu^{-1}-\delta-a)\frac{2-\delta\nu}{\delta\nu^{2}\epsilon^{2}}$

and hence

$t_{\epsilon}(a,\nu^{-1}-\delta)\leq(\nu^{-1}-\delta-a)\frac{2-\delta\nu}{\delta\nu^{2}\epsilon^{2}}\epsilon^{2}\leq(\nu^{-1}-\delta)\frac{2-\delta\nu}{\delta\nu^{2}},$

which is the conclusion. ∎

To study the asymptotic behavior of solutions, the following types of
strategies of the game are also useful.

######  Definition 4.6 (Push by moving circle).

For a sequence $\\{z_{n}\\}\subset\mathbb{R}^{2}$ such that $\lvert
z_{n+1}-z_{n}\rvert\leq C$ for some constant $C>0$, a strategy for Paul (resp.
for Carol) to keep taking a $z_{n}$ concentric strategy at each round $n$ is
called a push by moving circle strategy by Paul (resp. by Carol).

######  Lemma 4.7 (Properties of push by moving circle strategies).

Let $\delta>0$. Let $\epsilon>0$ be sufficiently small. i.e., we assume that
$0<\epsilon<\epsilon_{0}(\nu,\delta)$ for some $\epsilon_{0}(\nu,\delta)>0$.

  1. 1.

If either Paul or Carol takes a push by moving circle strategy with
$C=\frac{\delta}{2}\nu^{2}\epsilon^{2}$ and
$x_{0}\in\overline{B(z_{0},\nu^{-1}-\delta)}^{c}$, then
$x_{n}\in\overline{B(z_{n},\nu^{-1}-\delta)}^{c}$ for all round $n$.

  2. 2.

If Paul takes a push by moving circle strategy with
$C=\frac{\nu^{2}\delta}{2(1+\nu\delta)}\epsilon^{2}$ and
$x_{0}\in\overline{B(z_{0},\nu^{-1}+\delta)}$. then
$x_{n}\in\overline{B(z_{n},\nu^{-1}+\delta)}$ for all round $n$.

###### Proof.

  1. 1.

We notice that $\lvert x_{n+1}-z_{n}\rvert\geq T_{\epsilon^{2}}(\lvert
x_{n}-z_{n}\rvert)$ whichever player takes the push by moving circle strategy.
We prove that $x_{n}\in\overline{B(z_{n},\nu^{-1}-\delta)}^{c}$ implies
$x_{n+1}\in\overline{B(z_{n+1},\nu^{-1}-\delta)}^{c}$. If $\lvert
x_{n}-z_{n}\rvert\geq\nu^{-1}$, we see from Lemma 4.2 that $\lvert
x_{n+1}-z_{n}\rvert\geq\nu^{-1}$ and hence $\lvert
x_{n+1}-z_{n+1}\rvert\geq\nu^{-1}-\frac{\delta}{2}\nu^{2}\epsilon^{2}\geq\nu^{-1}-\delta$
for sufficiently small $\epsilon$. Thus we asuume $\lvert
x_{n}-z_{n}\rvert<\nu^{-1}$ hereafter.

It suffices to show
$|x_{n+1}-z_{n}|>\nu^{-1}-\delta+\frac{\delta}{2}\nu^{2}\epsilon^{2}$. Indeed
this inequality is obtained by

$\lvert x_{n+1}-z_{n}\rvert\geq T_{\epsilon^{2}}(\lvert
x_{n}-z_{n}\rvert)>T_{\epsilon^{2}}(\nu^{-1}-\delta)\geq\nu^{-1}-\delta+\frac{\delta}{2}\nu^{2}\epsilon^{2}.$

The second inequality is derived from the monotonicity property of
$T_{\epsilon^{2}}$ stated in Lemma 4.3. The third inequality is computed by
(4.5) in Lemma 4.2. We notice that
$T_{\epsilon^{2}}(\nu^{-1}-\delta)\leq\nu^{-1}-\delta/2$ for sufficiently
small $\epsilon$. Letting $R_{n}=\nu^{-1}-\delta$ and hence
$R_{n+1}=T_{\epsilon^{2}}(\nu^{-1}-\delta)$, we indeed have

$\displaystyle
R_{n+1}-R_{n}\geq\frac{\epsilon^{2}}{R_{n+1}}-\nu\epsilon^{2}+\frac{\nu^{2}\epsilon^{4}}{2R_{n+1}}\geq\frac{\epsilon^{2}}{\nu^{-1}-\delta/2}-\nu\epsilon^{2}+\frac{\nu^{2}\epsilon^{4}}{2\nu^{-1}-\delta}\geq\frac{\delta}{2}\nu^{2}\epsilon^{2}.$

Thus the proof is complete.

  2. 2.

We prove that $x_{n}\in\overline{B(z_{n},\nu^{-1}+\delta)}$ implies
$x_{n+1}\in\overline{B(z_{n+1},\nu^{-1}+\delta)}$. If $\lvert
x_{n}-z_{n}\rvert<\nu\epsilon^{2}$, it is clear that $\lvert
x_{n+1}-z_{n+1}\rvert\leq\nu^{-1}+\delta$ for sufficiently small $\epsilon$.
If not, we have

$\lvert x_{n+1}-z_{n+1}\rvert\leq
T_{\epsilon^{2}}(\nu^{-1}+\delta)+\frac{\nu^{2}\delta}{2(1+\nu\delta)}\epsilon^{2}.$

Letting $R_{n}=\nu^{-1}+\delta$ and hence
$R_{n+1}=T_{\epsilon^{2}}(\nu^{-1}+\delta)$, we have from (4.4) in Lemma 4.2

$R_{n}-R_{n+1}\geq-\frac{\epsilon^{2}}{R_{n}}+\nu\epsilon^{2}-\frac{\nu^{2}\epsilon^{4}}{2R_{n}}=\frac{\nu^{2}\delta}{1+\nu\delta}\epsilon^{2}+o(\epsilon^{2}).$

Therefore we obtain

$\lvert x_{n+1}-z_{n+1}\rvert\leq\nu^{-1}+\delta$

for sufficiently small $\epsilon$.

∎

We set

$\mathcal{L}_{\nu}:=\\{\gamma([0,1])\mid\gamma\in
C^{2}([0,1];\mathbb{R}^{2}),\ \mbox{$\gamma$ is regular and}\
\lvert\kappa_{\gamma(t)}\rvert\leq\nu\ \mbox{for all}\ t\in(0,1)\\},$

where we denote the curvature of $\gamma([0,1])$ at $\gamma(t)$ by
$\kappa_{\gamma(t)}$.

$x$$\hat{x}$$\Gamma$a circle with the radius of
$\nu^{-1}$$\hat{x}+\frac{\hat{x}-x}{\nu\lvert\hat{x}-x\rvert}$ Figure 12:
$\Gamma$ tube strategy

######  Definition 4.8 ($\Gamma$ tube strategy).

Let $\nu>0$ and $\Gamma\in\mathcal{L}_{\nu}$. Let $x\in\mathbb{R}^{2}$ be the
current position of the game. Let $\hat{x}$ satisfy $\min_{y\in\Gamma}\lvert
x-y\rvert=\lvert x-\hat{x}\rvert$. A
$\hat{x}+\frac{\hat{x}-x}{\nu\lvert\hat{x}-x\rvert}$ concentric strategy is
called a $\Gamma$ tube strategy (Figure 12).

######  Lemma 4.9 (Property of $\Gamma$ tube strategies).

Let $\Gamma=\gamma([0,1])\in\mathcal{L}_{\nu}$. Let $\delta>0$ and
$\epsilon\ll\delta$. If Paul takes a $\Gamma$ tube strategy at round $n$ and
$x_{n}\in B_{\delta}(\Gamma)\setminus(B_{\delta}(\gamma(0))\cup
B_{\delta}(\gamma(1)))$, then $x_{n+1}\in B_{\delta}(\Gamma)$.

###### Proof.

It is clear that the statement holds if $x_{n}\in B_{\delta/2}(\Gamma)$. Thus,
there is no loss of generality to assume that $\hat{x}_{n}=(0,0)$ and
$x_{n}\in\\{(0,q)\mid\delta/2\leq q<\delta\\}$, where $\hat{x}_{n}$ is a
minimizer taken in Definition 4.8. There are a graph
$f:\mathbb{R}\to\mathbb{R}$ and $\delta_{0}>0$ such that
$\hat{\Gamma}:=\\{(p,q)\mid q=f(p),-\delta_{0}<p<\delta_{0}\\}\subset\Gamma$.
Let $C=\partial B_{\nu^{-1}}((0,-\nu^{-1}))$. If Paul takes a $\Gamma$ tube
strategy, we see that $dist(x_{n+1},C)<dist(x_{n},C)<\delta$. Since
$f(p)\geq-\nu^{-1}+\sqrt{\nu^{-2}-p^{2}}$ for $-\delta_{0}<p<\delta_{0}$, we
have

$dist(x_{n+1},\Gamma)\leq dist(x_{n+1},\hat{\Gamma})\leq
dist(x_{n+1},C)<dist(x_{n},C)<\delta,$

which is the conclusion. ∎

###  4.2 Examples

We first observe the behavior of the solution to (4.1) by considering two
examples of $O_{-}$. In both of the problems we assume

  * (A1)

$D_{0}\subset B_{R}(z)$ for some $z\in\mathbb{R}^{2}$ and $R<\nu^{-1}$.

Set

$A:=\bigcap\left\\{\overline{B_{\nu^{-1}}(z)}\mid
O_{-}\subset\overline{B_{\nu^{-1}}(z)},\ z\in\mathbb{R}^{2}\right\\}$

for $O_{-}\subset\mathbb{R}^{2}$. Under the assumption (A1), this $A$ is a
major candidate of the asymptotic shape. Indeed we can show at least that the
asymptotic shape is bounded by $A$ from above for general $O_{-}$ satisfying
(A1).

######  Lemma 4.10.

Let $O\subset\mathbb{R}^{d}$ and $r>0$. Then the set
$\\{y\in\mathbb{R}^{d}\mid O\subset\overline{B_{r}(y)}\\}$ is convex.

###### Proof.

Let $y_{1},y_{2}\in\mathbb{R}^{d}$ satisfy $O\subset\overline{B_{r}(y_{1})}$
and $O\subset\overline{B_{r}(y_{2})}$. Then, for $x\in O$, we have $\lvert
y_{i}-x\rvert\leq r$ ($i=1,2$). This implies that

$\left\lvert\frac{y_{1}+y_{2}}{2}-x\right\rvert=\frac{\lvert
y_{1}-x+y_{2}-x\rvert}{2}\leq r$

for all $x\in O$ and the lemma follows. ∎

We hereafter take $u_{0}$ and $\Psi_{-}$ as (3.2) and (3.3) respectively. We
set $O_{\delta_{1}}:=\\{x\in\mathbb{R}^{2}\mid\Psi_{-}(x)>-\delta_{1}\\}$ and
$C_{\delta_{1},\delta_{2}}:=\\{z\in\mathbb{R}^{2}\mid
O_{\delta_{1}}\subset\overline{B(z,\nu^{-1}-\delta_{2})}\\}$ for
$\delta_{1},\delta_{2}>0$.

######  Lemma 4.11.

Assume $\mathrm{(A1)}$. For $x\in A^{c}$, there exist
$\delta_{1},\delta_{2}>0$ and $\hat{z}\in\mathbb{R}^{2}$ such that $\hat{z}\in
C_{\delta_{1},\delta_{2}}$ and $x\in\overline{B_{\nu^{-1}}(\hat{z})}^{c}$.

###### Proof.

Step 1. We first show that $B_{\delta_{1}+\delta_{2}}(z)\subset C_{0,0}$
implies $z\in C_{\delta_{1},\delta_{2}}$. Indeed
$B_{\delta_{1}+\delta_{2}}(z)\subset C_{0,0}$ implies

$O_{-}\subset\bigcap\\{\overline{B_{\nu^{-1}}(\tilde{z})}\mid\tilde{z}\in
B_{\delta_{1}+\delta_{2}}(z)\\}=\overline{B(z,\nu^{-1}-(\delta_{1}+\delta_{2}))}.$

By taking $\delta_{1}$ neighborhood of both sides, we have

$O_{\delta_{1}}\subset\overline{B(z,\nu^{-1}-\delta_{2})},$

which means $z\in C_{\delta_{1},\delta_{2}}$.

Step 2. Let $C^{x}_{0,0}:=\\{z\in\mathbb{R}\mid
O_{-}\subset\overline{B_{\nu^{-1}}(z)},x\in\overline{B_{\nu^{-1}}(z)}^{c}\\}$.
We next show $(C_{0,0}^{x})^{int}\neq\emptyset$. By the assumption (A1), we
see that $\overline{B(z,\nu^{-1}-R)}\subset C_{0,0}$ for some
$z\in\mathbb{R}^{2}$. We also notice that
$C^{x}_{0,0}=C_{0,0}\setminus\overline{B_{\nu^{-1}}(x)}$. If
$\overline{B(z,\nu^{-1}-R)}\setminus\overline{B_{\nu^{-1}}(x)}\neq\emptyset$,
then it has the interior and so does $C^{x}_{0,0}$. Since $x\in A^{c}$, we
have $C^{x}_{0,0}\neq\emptyset$ and let $\tilde{z}\in C^{x}_{0,0}$. Since
$C_{0,0}$ is convex (Lemma 4.10), we have
$Co\left(\\{\tilde{z}\\}\cup\overline{B(z,\nu^{-1}-R)}\right)\subset C_{0,0}$.
Hence, even if $\overline{B(z,\nu^{-1}-R)}\subset\overline{B_{\nu^{-1}}(x)}$,
the set
$Co\left(\\{\tilde{z}\\}\cup\overline{B(z,\nu^{-1}-R)}\right)\setminus\overline{B_{\nu^{-1}}(x)}$
has the interior and so does $C^{x}_{0,0}$.

Therefore, for small $\delta_{1}>0$ and $\delta_{2}>0$, there exists
$\hat{z}\in C_{\delta_{1},\delta_{2}}$ and
$x\in\overline{B_{\nu^{-1}}(\hat{z})}^{c}$. ∎

######  Lemma 4.12.

Assume $\mathrm{(A1)}$. Then

$\varlimsup_{t\to\infty}E_{t}\subset A.$

###### Proof.

We give appropriate strategies of Carol as in the proof of Lemma 3.1. For
$x\in A^{c}$, there exist $\delta_{1},\delta_{2}>0$ and
$z_{0}\in\mathbb{R}^{2}$ such that $R<\nu^{-1}-\delta_{2}$, $z_{0}\in
C_{\delta_{1},\delta_{2}}$ and
$x\notin\overline{B(z_{0},\nu^{-1}-\delta_{2})}$ (Lemma 4.11). We define the
sequence $\\{z_{n}\\}$ by

$z_{n}:=z_{0}+\min\left\\{\frac{\delta_{2}}{2}n\nu^{2}\epsilon^{2},\lvert
z-z_{0}\rvert\right\\}\frac{z-z_{0}}{\lvert z-z_{0}\rvert},$

where $z$ is a point taken in the assumption (A1). Carol’s strategy is to take
a push by moving circle strategy with this $\\{z_{n}\\}$.

If she does so, then $z_{n}\in C_{\delta_{1},\delta_{2}}$ for all $n$ because
$z_{0},z\in C_{\delta_{1},\delta_{2}}$ and $C_{\delta_{1},\delta_{2}}$ is
convex (Lemma 4.10). By 1 in Lemma 4.7 we see that if Paul quits the game at
round $i$, the game position $x_{i}$ is not in $O_{\delta_{1}}$. Thus the
stopping cost is at most $-\delta_{1}$ regardless of $\epsilon$. If Paul does
not quit the game, the last position $x_{N}$ of the game is not in
$\overline{B(z,\nu^{-1}-\delta_{2})}$ for sufficiently large $t$. Thus the
terminal cost is at most $R-\nu^{-1}+\delta_{2}<0$ for large $t$ regardless of
$\epsilon$. Therefore we conclude that for $x\in A^{c}$, there exists $\tau>0$
such that $u(x,t)<0$ for $t>\tau$. ∎

The first example of $O_{-}$ is the following:

$O_{-}:=B_{R^{\prime}}((0,0))\setminus\\{(p,q)\mid q\geq\lvert p\rvert\\},$

where $R^{\prime}<\nu^{-1}$ (Figure 13). Notice that
$A^{int}=B_{R^{\prime}}((0,0))\setminus\left\\{(p,q)\mid
q\geq\sqrt{\nu^{-2}-p^{2}}+R^{\prime}-\sqrt{\nu^{-2}-R^{\prime 2}}\right\\}$
for this obstacle, where we denote the interior of $A$ by $A^{int}$.

$\Longrightarrow$$t\to\infty$ Figure 13: Small Pac-Man

######  Proposition 4.13 (Small Pac-Man).

Assume $\mathrm{(A1)}$. Then

$A^{int}\subset\varliminf_{t\to\infty}D_{t}\subset\varlimsup_{t\to\infty}D_{t}\subset
A$

and

$A^{int}\subset\varliminf_{t\to\infty}E_{t}\subset\varlimsup_{t\to\infty}E_{t}\subset
A.$

###### Proof.

By Lemma 4.12, it suffices to show
$A^{int}\subset\varliminf_{t\to\infty}D_{t}$. As in the proof of Theorem 3.2,
we do case analysis for the initial game position $x\in A^{int}$ and give an
appropriate strategy of Paul.

$1)~{}x\in O_{-}.$ In this case, it suffices for Paul to quit the game at the
first round and gain the stopping cost $\Psi_{-}(x)>0$ as in the proof of
Theorem 3.2.

$z_{0}$$z_{1}$$z_{2}$$x$$\partial B(z_{0},\nu^{-1}+\delta_{2})$ Figure 14: A
strategy of Paul

$2)~{}x\in A^{int}\setminus O_{-}.$ We set
$O^{\delta_{1}}:=\\{x\in\mathbb{R}^{2}\mid\Psi_{-}(x)>\delta_{1}\\}$. We see
that for $x\in A^{int}\setminus O_{-}$, there exist $\delta_{1},\delta_{2}>0$
and $z_{0}\in\\{(0,q)\mid q\leq x\cdot(0,1)\\}$ such that
$x\in\overline{B(z_{0},\nu^{-1}+\delta_{2})}$ and $O^{\delta_{1}}\cap\partial
B(z_{0},\nu^{-1}+\delta_{2})\cap\\{(p,q)\mid q>0\\}\neq\emptyset$ (Figure 14).
Define the sequence $\\{z_{n}\\}$ so that

$z_{n+1}=z_{n}+\left(0,-\frac{\nu^{2}\delta_{2}}{2(1+\nu\delta_{2})}\epsilon^{2}\right).$

Paul’s strategy is to keep taking a push by moving circle strategy with this
$\\{z_{n}\\}$ until he reaches $O^{\delta_{1}}$, where he quits the game.

By doing this strategy, Paul actually gains positive game cost. Set

$\Omega_{n}=B_{\delta_{1}}(\\{(p,q)\mid q\geq\lvert p\rvert\\})\cap
B(z_{n},\nu^{-1}+\delta_{2})$

for $n=0,1,\cdots$. By 2 in Lemma 4.7 we see that if $x_{n}\in\Omega_{n}$,
then $x_{n+1}\in\Omega_{n+1}$ or Paul quits the game. If $n\epsilon^{2}$ is
sufficiently large, then $\Omega_{n}=\emptyset$. That means Paul definitely
quits the game in finite time and the stopping cost is at least

$\inf_{y\in O^{\delta_{1}}}\Psi_{-}(y)=\delta_{1}>0$

regardless of $\epsilon$. Hence, together with the case 1), it turns out that
for $x\in A^{int}$, there exists $\tau>0$ such that $u(x,t)>0$ for $t>\tau$. ∎

arc with curvature $\nu$a circle with the radius of
$\nu^{-1}$$\Longrightarrow$$\Gamma\in\mathcal{L}_{\nu}$$t\to\infty$ Figure 15:
Two small balls

We next consider (4.1) with $O_{-}=B_{r}((d,0))\cup B_{r}((-d,0))$. In this
problem we further assume

  * (A2)

There exists a function $f:[a,b]\to\mathbb{R}$ such that $\Gamma:=\\{(p,q)\mid
q=f(p),a\leq p\leq b\\}\in\mathcal{L}_{\nu}$, $(a,f(a))\in B_{r}((-d,0))$,
$(b,f(b))\in B_{r}((d,0))$ and $\Gamma\subset D_{0}$.

######  Proposition 4.14 (Two small balls).

Assume $\mathrm{(A1)}$ and $\mathrm{(A2)}$. Then

$A^{int}\subset\varliminf_{t\to\infty}D_{t}\subset\varlimsup_{t\to\infty}D_{t}\subset
A$

and

$A^{int}\subset\varliminf_{t\to\infty}E_{t}\subset\varlimsup_{t\to\infty}E_{t}\subset
A.$

###### Proof.

We take $\delta>0$ small enough to satisfy $B_{2\delta}(\Gamma)\subset D_{0}$
and $B_{2\delta}((a,f(a)))\cup B_{2\delta}((b,f(b)))\subset O_{-}$.

$1)~{}x\in O_{-}.$ The proof for this case is the same as before.

$x$$\Gamma$$z_{0}$$z_{1}$$z_{2}$$\partial B(z_{0},\nu^{-1}+\delta_{2})$ Figure
16: A strategy of Paul

$2)~{}x\in B_{\delta}(\Gamma)\setminus O_{-}.$ Paul keeps taking a $\Gamma$
tube strategy until he reaches $B_{\delta}((a,f(a)))\cup
B_{\delta}((b,f(b)))$. Once he reaches $B_{\delta}((a,f(a)))\cup
B_{\delta}((b,f(b)))$, he quits the game.

By his doing this strategy, the game trajectory $\\{x_{n}\\}$ is restricted to
$B_{\delta}(\Gamma)$ as shown in Lemma 4.9. Thus, whether he quits the game or
not, he gains at least $\delta>0$.

$3)~{}x\in A^{int}\setminus(B_{\delta}(\Gamma)\cup O_{-}).$ We extend $f$ as
follows:

$\tilde{f}(p):=\begin{cases}\max\\{f(p),\sqrt{(r-\delta_{1})^{2}-(p+d)^{2}}\\},&-d-r+\delta_{1}\leq
p\leq-d+r-\delta_{1}\\\
\max\\{f(p),\sqrt{(r-\delta_{1})^{2}-(p-d)^{2}}\\},&d-r+\delta_{1}\leq p\leq
d+r-\delta_{1}\\\ f(p),&-d+r-\delta_{1}<p<d-r+\delta_{1}.\end{cases}$

Without loss of generality we can assume $x\in\\{(p,q)\mid
q\geq\tilde{f}(p),-d-r+\delta_{1}\leq p\leq d+r-\delta_{1}\\}$. We take
$\delta_{1},\delta_{2}>0$ and $\\{z_{n}\\}$ as in the case 2) in the proof of
Proposition 4.13. See also Figure 16.

Paul’s strategy is to keep taking a push by moving circle strategy with this
$\\{z_{n}\\}$ until he reaches $O^{\delta_{1}}\cup B_{\delta}(\Gamma)$. Once
he reaches $O^{\delta_{1}}$, he quits the game. Once he reaches
$B_{\delta}(\Gamma)$, he takes the same strategy as in the case 2).

By adopting this strategy, Paul actually gains positive game cost. Set

$\Omega_{n}=\\{(p,q)\mid q\geq\tilde{f}(p),-d-r+\delta_{1}\leq p\leq
d+r-\delta_{1}\\}\cap B(z_{n},\nu^{-1}+\delta_{2})\setminus
B_{\delta}(\Gamma)$

for $n=0,1,\cdots$. By 2 in Lemma 4.7 we see that if $x_{n}\in\Omega_{n}$,
then $x_{n+1}\in\Omega_{n+1}$ or Paul reaches $O^{\delta_{1}}\cup
B_{\delta}(\Gamma)$ (Figure 16). If $n\epsilon^{2}$ is sufficiently large,
then $\Omega_{n}=\emptyset$. That means Paul definitely reaches
$O^{\delta_{1}}\cup B_{\delta}(\Gamma)$ in finite time. Since the stopping
cost is at least $\delta_{1}>0$ and the terminal cost is at least $\delta>0$,
it turns out that for $x\in A^{int}$, there exists $\tau>0$ such that
$u(x,t)>0$ for $t>\tau$. ∎

######  Remark 4.15.

We expect that the same conclusion holds for more general $O_{-}$. As an
analogue of Theorem 3.2, we define the graph $G=(V,E)$ as follows:

$V:=\\{O\subset\mathbb{R}^{2}\mid O\ \mbox{is a connected component of}\
O_{-}\\},$ $E:=\\{\langle O,P\rangle\mid\mbox{There exist $x\in O$, $y\in P$
and $\Gamma_{x,y}\in\mathscr{L}_{\nu}$ such that $\Gamma_{x,y}\subset
D_{0}$}\\},$

where we denote a curve $\Gamma=\gamma([0,1])\in\mathscr{L}_{\nu}$ satisfying
$\gamma(0)=x$ and $\gamma(1)=y$ by $\Gamma_{x,y}$. It seems that if $O_{-}$
satisfies (A1) and the graph $G$ is connected, then the same conclusion holds.

$t\to\infty$$\Longrightarrow$$t\to\infty$$\Longrightarrow$obstacle
$O_{+}^{c}$$\partial
D_{0}$$\Longrightarrow$$\Longrightarrow$$t\to\infty$$R=\sqrt{2}\nu^{-1}$$R<\sqrt{2}\nu^{-1}$$R>\sqrt{2}\nu^{-1}$
Figure 17: Square boxes

Finally we give an example of computation of the asymptotic shape of solutions
to (4.2). We can deal with the problem that remains in [9, Section 6], where
$O_{+}=\mathbb{R}^{2}\setminus\\{(p,q)\in\mathbb{R}^{2}\mid\lvert
p\rvert+\lvert q\rvert\leq R\\}$ with $R=\sqrt{2}\nu^{-1}$. We also give a
game theoritic proof for the case $R\neq\sqrt{2}\nu^{-1}$, which is considered
in a different way in [9, Section 6].

We set

$A:=\bigcup\left\\{B_{\nu^{-1}}(z)\mid B_{\nu^{-1}}(z)\subset O_{+},\
z\in\mathbb{R}^{2}\right\\}.$

We notice that $A=B_{\nu^{-1}}((0,0))$ if $R=\sqrt{2}\nu^{-1}$ and
$A=\emptyset$ if $R<\sqrt{2}\nu^{-1}$. Figure 17 shows the result of the
asymptotic shapes. In Figure 17, dotted circles are circles with the radius of
$\nu^{-1}$.

######  Proposition 4.16.

Assume either of the following:

  1. 1.

$R=\sqrt{2}\nu^{-1}$ and $B_{\nu^{-1}}((0,0))\subset D_{0}$,

  2. 2.

$R<\sqrt{2}\nu^{-1}$,

  3. 3.

$R>\sqrt{2}\nu^{-1}$ and $B(\hat{z},\nu^{-1}+\delta)\subset D_{0}$ for some
$\hat{z}\in\mathbb{R}^{2}$ and $\delta>0$.

Then

$A\subset\varliminf_{t\to\infty}D_{t}\subset\varlimsup_{t\to\infty}D_{t}\subset\overline{A}$

and

$A\subset\varliminf_{t\to\infty}E_{t}\subset\varlimsup_{t\to\infty}E_{t}\subset\overline{A}.$

###### Proof.

It suffices to take $u_{0}$ as (3.2). Similarly it suffices to let

$\Psi_{+}(x)=\begin{cases}dist(x,\partial O_{+}),\ x\in O_{+}\\\
\max\\{a,-dist(x,\partial O_{+})\\},\ x\in O_{+}^{c}.\end{cases}$

  1. 1.

$1)~{}x\in B_{\nu^{-1}}((0,0)).$ Paul’s strategy is to keep taking a (0,0)
concentric strategy. We denote by $V^{\epsilon}(x,t)$ the total cost when Paul
takes this strategy with the game variables $(x,t,\epsilon)$ and Carol does
not quit the game on the way. Notice that $V^{\epsilon}(x,t)$ does not depend
on Carol’s choices $\\{b_{n}\\}$ because $u_{0}$ is radially symmetric. Since
$u_{0}(x_{n})$ is monotonically decreasing with respect to $n$ and
$u_{0}(x_{n})\leq\Psi_{+}(x_{n})$, Carol’s optimal strategy for the strategy
of Paul is not to quit the game on the way. Thus the inequality
$u^{\epsilon}(x,t)\geq V^{\epsilon}(x,t)$ is satisfied. Fix $x\in
B_{\nu^{-1}}((0,0))$ and $t>0$. By Lemma 4.4 we have $V^{\epsilon}(x,t)>0$ for
sufficiently small $\epsilon>0$. Lemma 4.3 2 implies that for any subsequence
$\\{\epsilon_{n}\\}$ satisfying $\epsilon_{n}=2^{-n}\epsilon_{0}$,
$V^{\epsilon_{n}}(x,t)$ is monotonically increasing with respect to $n$.
Therefore we obtain $u(x,t)>0$.

$z_{0}$$\nu^{-1}-\delta_{1}$$x$$z_{1}$$z_{2}$ Figure 18: Carol’s strategy

$2)~{}x\in\overline{O_{+}}^{c}.$ Carol’s strategy is to quit the game at the
first round.

$3)~{}x=(p,q)\in\overline{A}^{c}\setminus\overline{O_{+}}^{c}.$ We may assume
$q\geq\lvert p\rvert$. We can take $z_{0}\in\mathbb{R}^{2}$, $\delta_{1}>0$
and $\delta_{2}>0$ so that $\lvert x-z_{0}\rvert\geq\nu^{-1}-\delta_{1}$,
$z_{0}\in\\{(0,y)\mid y>0\\}$ and $\partial
B(z_{0},\nu^{-1}-\delta_{1})\cap\overline{B_{\delta_{2}}(O_{+})}^{c}\neq\emptyset$.
Carol’s strategy is to take a push by moving circle strategy with
$\\{z_{n}\\}\subset\\{(0,q)\mid q>0\\}$ satisfying
$z_{n+1}=z_{n}+(0,\frac{\delta_{1}}{2}\nu^{2}\epsilon^{2})$ for all $n$ until
Paul reaches $\overline{B_{\delta_{2}}(O_{+})}^{c}$. Once he reaches there,
she quits the game. See Figure 18.

By doing above strategy, Carol actually pays nagative game cost. To show it,
set

$\Omega_{n}=\\{(p,q)\mid q\geq\lvert
p\rvert\\}\cap\left(\overline{B_{\delta_{2}}(O_{+})}\setminus
B(z_{n},\nu^{-1}-\delta_{1})\right)$

for $n=0,1,\cdots$. We see that if $x_{n}\in\Omega_{n}$, then
$x_{n+1}\in\Omega_{n+1}$ or Carol quits the game at round $n+1$. If
$n\epsilon^{2}$ is sufficiently large, then $\Omega_{n}=\emptyset$. That means
Carol definitely quits the game in finite time and the stopping cost is at
most

$\sup_{y\in\overline{B_{\delta_{2}}(O_{+})}}\Psi_{+}(y)=-\delta_{2}<0$

regardless of $\epsilon$. Therefore we obtain $u(x,t)<0$ for
$x\in\overline{B_{\nu^{-1}}((0,0))}^{c}$ and sufficiently large $t>0$.

  2. 2.

We take $\delta_{1}>0$ to satisfy
$B((0,0),\nu^{-1}-2\delta_{1})\cap\overline{O_{+}}^{c}\neq\emptyset$.

$1)~{}x\in\overline{O_{+}}^{c}.$ Carol’s strategy is to quit the game at the
first round.

$2)~{}x\in\overline{O_{+}}\setminus B((0,0),\nu^{-1}-\delta_{1}).$ Carol’s
strategy is similar to that in the case 3) in $1$.

$3)~{}x\in\overline{O_{+}}\cap B((0,0),\nu^{-1}-\delta_{1}).$ Carol keeps
taking a $(0,0)$ concentric strategy until Paul is forced to reach
$B((0,0),\nu^{-1}-\delta_{1})^{c}\cap B_{\delta_{1}}(O_{+})^{c}$. By Lemma 4.5
it takes at most finite time for Paul to reach there. Once Paul reaches
$B_{\delta_{1}}(O_{+})^{c}$, Carol quits the game. Once Paul reaches
$\overline{O_{+}}\setminus B((0,0),\nu^{-1}-\delta_{1})$, Carol’s strategy is
as in the case 2).

Therefore, for $x\in\mathbb{R}^{2}$, we obtain $u(x,t)<0$ for sufficiently
large $t>0$.

  3. 3.

$1)~{}x\in\overline{O_{+}}^{c}.$ Carol’s strategy is to quit the game at the
first round.

$2)~{}x\in\overline{O_{+}}\setminus A.$ Carol’s strategy is similar to that in
the case 3) in $1$.

$3)~{}x\in A.$ We set
$O^{\delta_{1}}:=\\{x\in\mathbb{R}^{2}\mid\Psi_{+}(x)>\delta_{1}\\}$ and
$C_{\delta_{1},\delta_{2}}:=\\{z\in\mathbb{R}^{2}\mid
B(z,\nu^{-1}+\delta_{2})\subset O^{\delta_{1}}\\}$ for
$\delta_{1},\delta_{2}>0$. For $x\in A$, there exist $\delta_{1},\delta_{2}>0$
such that

$x\in\bigcup\left\\{B(z,\nu^{-1}+\delta_{2})\mid z\in
C_{\delta_{1},\delta_{2}}\right\\}$

and $C_{0,\delta}\subset C_{\delta_{1},\delta_{2}}$. Let
$z_{0}\in\mathbb{R}^{2}$ be a point satisfying $x\in
B(z_{0},\nu^{-1}+\delta_{2})$ and $z_{0}\in C_{\delta_{1},\delta_{2}}$. We
define the sequence $\\{z_{n}\\}$ by

$z_{n}:=z_{0}+\min\left\\{\frac{\delta_{2}}{2}n\nu^{2}\epsilon^{2},\lvert\hat{z}-z_{0}\rvert\right\\}\frac{\hat{z}-z_{0}}{\lvert\hat{z}-z_{0}\rvert}.$

Since $C_{\delta_{1},\delta_{2}}$ is now convex, we have
$l_{z_{0},\hat{z}}\subset C_{\delta_{1},\delta_{2}}$.

Paul’s strategy is to take a push by moving circle strategy with this
$\\{z_{n}\\}$. Indeed if Carol quits the game on the way, the stopping cost is
at least $\delta_{1}>0$ because the game trajectory $\\{x_{n}\\}$ is contained
in $O^{\delta_{1}}$. If Carol does not quit the game, the terminal cost is at
least $\delta-\delta_{1}$, which is positive. That is because $x_{N}\in
B(\hat{z},\nu^{-1}+\delta_{1})\subset B(\hat{z},\nu^{-1}+\delta)\subset D_{0}$
for any last position $x_{N}$ of the game.

∎

##  Appendix A Game interpretation and convergence of value functions

In this appendix we give a game whose value functions converge to the
viscosity solution to (2.1). We introduce the rule of the game corresponding
to (2.1) with $\nu\geq 0$ and $d=2$ and give the proof of the convergence
result for the case. We also remark on the other $\nu\in\mathbb{R}$ and $d$.

The game is almost the same as explained in Section 1. We define the total
number $N$ of rounds by $N=\lceil t\epsilon^{-2}\rceil$, where $\lceil
r\rceil$ stands for the minimum integer that is no less than $r$. The actions
of both players in each round $i$ ($i=1,2,\cdots,N$) are modified as follows:

  1. 1.

Paul decides whether to quit the game.

  2. 2.

Carol decides whether to quit the game.

  3. 3.

Paul chooses $v_{i},w_{i}\in S^{1}$. ($S^{1}$ is the set of unit vectors in
$\mathbb{R}^{2}$.)

  4. 4.

Carol chooses $b_{i}\in\\{\pm 1\\}$ after Paul’s choice.

  5. 5.

Determine the next states as follows.

$\displaystyle x_{i}=x_{i-1}+\sqrt{2}\epsilon
b_{i}v_{i}+\nu\epsilon^{2}w_{i}.$ (A.1)

The total cost is also modified as follows. If Paul quits the game at round
$i$, the total cost is given by $\Psi_{-}(x_{i})$. If Carol quits the game at
round $i$, it is given by $\Psi_{+}(x_{i})$. If both players go throughout $N$
rounds of the game, it is given by
$u_{0}(x_{N})+\sum_{i=0}^{N-1}\epsilon^{2}f(x_{i})$. The value function
$u^{\epsilon}(x,t)$ is defined inductively based on the following Dynamic
Programming Principle and the initial condition:

$\begin{split}&u^{\epsilon}(x,t)=\\\
&\max\\{\Psi_{-}(x),\min\\{\Psi_{+}(x),\sup_{v,w\in S^{1}}\min_{b=\pm
1}[u^{\epsilon}(x+\sqrt{2}\epsilon bv+\nu
w\epsilon^{2},t-\epsilon^{2})+\epsilon^{2}f(x)]\\}\\}\end{split}$ (A.2)

for $t>0$.

$u^{\epsilon}(x,t)=u_{0}(x)$ (A.3)

for $t\leq 0$.

These value functions mean the total cost optimized by both players.

######  Remark A.1.

As explained in Section 1, we can generalize our game to the case $d\geq 3$.
In the game corresponding to (2.1) with $\nu\geq 0$, Paul chooses a unit
vector $w_{i}$ and $d-1$ orthogonal unit vectors $v_{i}^{j}(j=1,2,\cdots
d-1)$. Carol chooses $d-1$ values $b_{i}^{j}\in\\{\pm 1\\}(j=1,2,\cdots d-1)$.
The state equation is $x_{i}=x_{i-1}+\nu
w_{i}\epsilon^{2}+\sqrt{2}\epsilon\sum_{j=1}^{d-1}b_{i}^{j}v_{i}^{j}$ instead
of (A.1).

######  Remark A.2.

The Dynamic Programming Principle corresponding to (2.1) with $\nu<0$ is given
by

$u^{\epsilon}(x,t)=\max\\{\Psi_{-}(x),\min\\{\Psi_{+}(x),\sup_{v\in
S^{1}}\inf_{\begin{subarray}{c}w\in S^{1}\\\ b=\pm
1\end{subarray}}[u^{\epsilon}(x+\sqrt{2}\epsilon bv+\nu
w\epsilon^{2},t-\epsilon^{2})+\epsilon^{2}f(x)]\\}\\}.$

Namely, not Paul but Carol has the right to choose $w_{i}\in S^{1}$.

For these value functions, the same type of result as Proposition 1.1 holds.

######  Proposition A.3.

the functions $\overline{u}$ and $\underline{u}$ are respectively viscosity
sub- and supersolution of (2.1). Moreover
$\overline{u}(x,0)=\underline{u}(x,0)=u_{0}(x)$ for $x\in\mathbb{R}^{d}$.

######  Remark A.4.

As explained before, (2.1) with $f=0$ is a level set equation. By choosing
$\Psi_{+}$ so that $\Psi_{+}>\|u_{0}\|_{\infty}$ for $O_{+}=\mathbb{R}^{d}$,
we can ignore Carol’s stopping cost $\Psi_{+}$ when we consider obstacle
problems that have an obstacle on one side such as (1.1). Similarly, by
choosing $\Psi_{-}$ so that $\Psi_{-}<-\|u_{0}\|_{\infty}$ for
$O_{-}=\emptyset$, we can ignore Paul’s stopping cost $\Psi_{-}$.

We especially show the proof of Proposition A.3 with $d=2$ and $\nu\geq 0$
because the other case is similar. Our proof is based directly on the game as
in [14], whereas those in [12, 15] are based on the properties of the operator
whose fixed point is the solution of the Dynamic Programming Principle. Also
since the proof in [14] is local argument, roughly speaking, all we have to do
is to do the local argument in $\\{(x,t)\mid\Psi_{-}(x)<\overline{u}(x,t)\\}$
or in $\\{(x,t)\mid\Psi_{+}(x)>\underline{u}(x,t)\\}$. However we need to care
about the point that $\\{(x,t)\mid\Psi_{-}(x)<\overline{u}(x,t)\\}$ or
$\\{(x,t)\mid\Psi_{+}(x)>\underline{u}(x,t)\\}$ may not be open.

The proof consists of three steps. We show that the limits of the value
functions satisfy the conditions (a) and (c) in Definition 2.1 in the first
two propositions, (Proposition A.5 and A.7) and they satisfy the initial
condition (b) in the last one. (Proposition A.11) We mention that the initial
data $u_{0}$ is assumed to be just continuous, not to be Lipschitz continuous
as in [12] or bounded uniformly continuous as in [15]. Regarding the last
proposition, the idea of the proof is similar to that of [8, Proposition 3.1]
though the situation is different.

To visualize choices of players of the game, we give another description of
the level-set mean curvature flow operator $F$:

$F(Du,D^{2}u)=-\left\langle D^{2}u\frac{D^{\perp}u}{\lvert
Du\rvert},\frac{D^{\perp}u}{\lvert Du\rvert}\right\rangle$

for $Du\neq 0$. Here we denote by $D^{\perp}u$ a vector field satisfying
$Du\cdot D^{\perp}u=0$ and $\lvert Du\rvert=\lvert D^{\perp}u\rvert$ in
$\mathbb{R}^{2}$.

######  Proposition A.5.

The function $\underline{u}$ is a viscosity supersolution of (2.1) in
$\mathbb{R}^{2}\times(0,\infty)$.

###### Proof.

As for Definition 2.1-2(a), we directly have $\Psi_{-}(x)\leq
u^{\epsilon}(x,t)\leq\Psi_{+}(x)$ by the Dynamic Programing Principle (A.2).
Thus we obtain $\Psi_{-}(x)\leq\underline{u}(x,t)\leq\Psi_{+}(x)$ since
$\Psi_{+}$ and $\Psi_{-}$ are continuous. To prove the viscosity inequality,
we argue by contradiction. For a smooth function
$\phi:\mathbb{R}^{2}\times(0,\infty)\to\mathbb{R}$, a positive constant
$\theta_{0}>0$ and $(x,t)\in\mathbb{R}^{2}\times(0,\infty)$, we consider the
following condition (C):

$\partial_{t}\phi(x,t)-\nu\lvert D\phi(x,t)\rvert-\left\langle
D^{2}\phi(x,t)\frac{D^{\perp}\phi(x,t)}{\lvert
D\phi(x,t)\rvert},\frac{D^{\perp}\phi(x,t)}{\lvert
D\phi(x,t)\rvert}\right\rangle-f_{\ast}(x)\leq-\theta_{0}<0$ (A.4)

if $D\phi(x,t)\neq 0$, and

$\partial_{t}\phi(x,t)-\nu\lvert
D\phi(x,t)\rvert-\inf_{\lvert\zeta\rvert=1}\left\langle
D^{2}\phi(x,t)\zeta,\zeta\right\rangle-f_{\ast}(x)\leq-\theta_{0}<0$ (A.5)

if $D\phi(x,t)=0$.

We assume that there exist a smooth function $\phi$ and $(\hat{x},\hat{t})$
such that $(\hat{x},\hat{t})$ is a strict local minimum of
$\underline{u}-\phi$, $\underline{u}<\Psi_{+}$ at $(\hat{x},\hat{t})$ and the
condition (C) is satisfied at $(\hat{x},\hat{t})$ with $\phi$ and some
$\theta_{0}>0$. Then we can take a $\delta$ neighborhood of
$(\hat{x},\hat{t})$ where $\underline{u}-\phi$ attains its unique minimum at
$(\hat{x},\hat{t})$ and the condition (C) holds, retaking smaller
$\theta_{0}>0$ if necessary. For technical reasons, we take such $\delta$
neighborhood as
$N_{\delta}((\hat{x},\hat{t})):=\\{(x,t)\in\mathbb{R}^{2}\times[0,\infty)\mid\lvert
x-\hat{x}\rvert+\lvert t-\hat{t}\rvert<\delta\\}$ and $\delta>0$ small enough
to satisfy $\delta\leq\frac{a}{2\max\\{L,M\\}}$, where
$a:=\Psi_{+}(\hat{x})-\underline{u}(\hat{x},\hat{t})$, $L$ is the Lipschitz
constant of $\Psi_{+}$ and $M=\sup_{y\in B_{1}(\hat{x})}\lvert f(y)\rvert$.

From the definition of $\underline{u}$, there are sequence
$\\{\epsilon_{n}\\}$, $\\{x_{\epsilon_{n}}^{0}\\}$, and
$\\{t_{\epsilon_{n}}^{0}\\}$ satisfying

$\epsilon_{n}\searrow
0,~{}~{}(x_{\epsilon_{n}}^{0},t_{\epsilon_{n}}^{0})\to(\hat{x},\hat{t}),~{}~{}u^{\epsilon_{n}}(x_{\epsilon_{n}}^{0},t_{\epsilon_{n}}^{0})\to\underline{u}(\hat{x},\hat{t}).$

We may substitute $\epsilon$ for $\epsilon_{n}$ hereafter. We take $\epsilon$
small enough to satisfy $(x_{\epsilon}^{0},t_{\epsilon}^{0})\in
N_{\delta/2}((\hat{x},\hat{t}))$ and
$\Psi_{+}(x_{\epsilon}^{0})-u^{\epsilon}(x_{\epsilon}^{0},t_{\epsilon}^{0})\geq
a/2$. For any $\epsilon$, we construct the sequence $\\{X_{k}\\}$ satisfying

$X_{0}=(x_{\epsilon}^{0},t_{\epsilon}^{0}),$ $X_{k+1}=X_{k}+(\sqrt{2}\epsilon
b_{k}\eta_{k}^{\perp}+\nu\eta_{k}\epsilon^{2},-\epsilon^{2}),$ (A.6)

where $\eta_{k}=\frac{D\phi(X_{k})}{\lvert D\phi(X_{k})\rvert}$ if
$D\phi(X_{k})\neq 0$, and is an arbitrary unit vector if $D\phi(X_{k})=0$. We
will determine $b_{k}$ later. Let $x_{k}$ and $t_{k}$ denote the spatial and
time component of $X_{k}$ respectively hereafter. Let $k_{\epsilon}$ be the
maximal $k$ satisfying

$X_{j}\in N_{\delta/2}((\hat{x},\hat{t}))\ {\rm for\ any}\ j=0,1,\ldots,k-1.$

Indeed such $k_{\epsilon}$ exists because of the definition of the sequence
$\\{X_{k}\\}$. We prove by induction that
$\Psi_{+}(x_{k})>u^{\epsilon}(X_{k})$ and

$u^{\epsilon}(X_{0})-u^{\epsilon}(X_{k})\geq\epsilon^{2}\sum_{m=0}^{k-1}f(x_{m})$
(A.7)

for all $k<k_{\epsilon}$. These inequalities hold for $k=0$. We assume that
these hold for some $k<k_{\epsilon}$. Then the Dynamic Programming Principle
(A.2) and $\Psi_{+}(x_{k})>u^{\epsilon}(X_{k})$ imply

$u^{\epsilon}(X_{k})=\max\\{\Psi_{-}(x_{k}),\sup_{v,w\in S^{1}}\min_{b=\pm
1}u^{\epsilon}(x_{k}+\sqrt{2}\epsilon bv+\nu
w\epsilon^{2},t_{k}-\epsilon^{2})+\epsilon^{2}f(x_{k})\\}.$

Thus we have

$u^{\epsilon}(X_{k})\geq\min_{b=\pm 1}u^{\epsilon}(x_{k}+\sqrt{2}\epsilon
b\eta_{0}^{\perp}+\nu\eta_{0}\epsilon^{2},t_{k}-\epsilon^{2})+\epsilon^{2}f(x_{k}).$
(A.8)

We determine $b_{k}$ in $(\ref{seq1})$ as a minimizer $b$ in (A.8). We then
get

$u^{\epsilon}(X_{k})-u^{\epsilon}(X_{k+1})\geq\epsilon^{2}f(x_{k}).$

Adding (A.7), we have

$u^{\epsilon}(X_{0})-u^{\epsilon}(X_{k+1})\geq\epsilon^{2}\sum_{m=0}^{k}f(x_{m})$

and consequently

$u^{\epsilon}(X_{0})+M(k+1)\epsilon^{2}\geq u^{\epsilon}(X_{k+1}).$ (A.9)

If $k+1<k_{\epsilon}$, we see from the definition of $k_{\epsilon}$ that
$X_{k+1}\in N_{\delta/2}((\hat{x},\hat{t}))$ and thus

$\lvert x_{k+1}-x_{0}\rvert+\lvert t_{k+1}-t_{0}\rvert\leq\lvert
x_{k+1}-\hat{x}\rvert+\lvert x_{0}-\hat{x}\rvert+\lvert
t_{k+1}-\hat{t}\rvert+\lvert t_{0}-\hat{t}\rvert\leq\delta.$ (A.10)

From the Lipschitz continuity of $\Psi_{+}$, we have

$\lvert\Psi_{+}(x_{k+1})-\Psi_{+}(x_{0})\rvert\leq L\lvert
x_{k+1}-x_{0}\rvert\leq L(\delta-\lvert
t_{k+1}-t_{0}\rvert)=L(\delta-(k+1)\epsilon^{2}).$ (A.11)

Combining (A.9) and (A.11), we obtain

$\displaystyle\Psi_{+}(x_{k+1})-u^{\epsilon}(X_{k+1})\geq\Psi_{+}(x_{0})-L(\delta-(k+1)\epsilon^{2})-u^{\epsilon}(X_{0})-M(k+1)\epsilon^{2}$
$\displaystyle\geq\Psi_{+}(x_{0})-\max\\{L,M\\}(\delta-(k+1)\epsilon^{2})-u^{\epsilon}(X_{0})-\max\\{L,M\\}(k+1)\epsilon^{2}$
$\displaystyle\geq\Psi_{+}(x_{0})-u^{\epsilon}(X_{0})-\max\\{L,M\\}\delta$
$\displaystyle>\frac{a}{2}-\frac{a}{2}=0$

and conclude the induction.

Next we take the continuous path that affinely interpolates among
$\\{X_{k}\\}$, i.e., $X(s)=X_{k}+\left(s\epsilon^{-2}-k\right)(X_{k+1}-X_{k})$
for $k\epsilon^{2}\leq s\leq(k+1)\epsilon^{2}$, and we write
$X(s)=(x(s),t_{\epsilon}^{0}-s)$. Using Taylor’s theorem for $\phi(X(t))$ at
$t=k\epsilon^{2}$, we get

$\phi(X_{k+1})-\phi(X_{k})=\epsilon^{2}\\{-\partial_{t}\phi(X_{k})+\nu\lvert
D\phi(X_{k})\rvert+\langle
D^{2}\phi(X_{k})\eta_{k}^{\perp},\eta_{k}^{\perp}\rangle\\}+\Psi_{k}(\epsilon),$
(A.12)

where $\Psi_{k}(\epsilon)=o(\epsilon^{2}).$ Moreover, from the assumption
(A.4), we have

$\phi(X_{k+1})-\phi(X_{k})\geq\epsilon^{2}(\theta_{0}-f_{\ast}(x_{k}))+\Psi_{k}(\epsilon).$

This inequality is also obtained in the case $D\phi(X_{k})=0$, using (A.12)
and (A.5). Summing up both sides, we have

$\phi(X_{k})-\phi(X_{0})\geq
k\epsilon^{2}\theta_{0}-\epsilon^{2}\sum_{m=0}^{k-1}f_{\ast}(x_{m})+\sum_{m=0}^{k-1}\Psi_{m}(\epsilon).$
(A.13)

Provided $k<k_{\epsilon}$, we have

$\lvert\Psi_{k}(\epsilon)\rvert\leq C\epsilon^{3},$

where $C$ depends on $\phi$ in $\delta$ neighborhood around
$(\hat{x},\hat{t})$, and does not depend on $k$. This estimation is derived
from the Taylor expansion (A.12). Hence (A.13) becomes

$\phi(X_{k})-\phi(X_{0})\geq
k\epsilon^{2}\theta_{0}-\epsilon^{2}\sum_{m=0}^{k-1}f_{\ast}(x_{m})+kC\epsilon^{3},~{}~{}~{}~{}~{}k\leq
k_{\epsilon}.$

Adding this relation to (A.7), we have

$\displaystyle u^{\epsilon}(X_{0})-\phi(X_{0})$ $\displaystyle\geq
u^{\epsilon}(X_{k})-\phi(X_{k})+\epsilon^{2}\sum_{m=0}^{k-1}(f(x_{m})-f_{\ast}(x_{m}))+k\epsilon^{2}\theta_{0}+kC\epsilon^{3}$
$\displaystyle\geq
u^{\epsilon}(X_{k})-\phi(X_{k})+k\epsilon^{2}\theta_{0}+kC\epsilon^{3}.$

For sufficiently small $\epsilon$, we get

$-\frac{k}{2}\epsilon^{2}\theta_{0}\geq
u^{\epsilon}(X_{k})-u^{\epsilon}(X_{0})-\phi(X_{k})+\phi(X_{0}),~{}~{}~{}~{}~{}k\leq
k_{\epsilon}.$ (A.14)

By the definition of $k_{\epsilon}$, we see that
$Y_{\epsilon}\in\overline{N_{3\delta/4}((\hat{x},\hat{t}))\setminus
N_{\delta/2}((\hat{x},\hat{t}))}$, where we substitute $Y_{\epsilon}$ for
$X_{k_{\epsilon}}$. So there is a subsequence $\\{Y_{\epsilon_{n}}\\}_{n}$
such that

$\lim_{n\to\infty}Y_{\epsilon_{n}}=(x^{\prime},t^{\prime}),$

where $(x^{\prime},t^{\prime})\in B_{\delta}((\hat{x},\hat{t}))$ and
$(x^{\prime},t^{\prime})\neq(\hat{x},\hat{t})$. From (A.14), we have

$u^{\epsilon_{n}}(x_{\epsilon_{n}}^{0},t_{\epsilon_{n}}^{0})-\phi(x_{\epsilon_{n}}^{0},t_{\epsilon_{n}}^{0})\geq
u^{\epsilon_{n}}(Y_{\epsilon_{n}})-\phi(Y_{\epsilon_{n}}).$

Letting $n$ go to $\infty$, we obtain

$\underline{u}(\hat{x},\hat{t})-\phi(\hat{x},\hat{t})\geq\underline{u}(x^{\prime},t^{\prime})-\phi(x^{\prime},t^{\prime}).$

This is a contradiction since $\underline{u}-\phi$ attains its unique minimum
at $(\hat{x},\hat{t})$. ∎

The following lemma can be found in [14, Lemma 2.3].

######  Lemma A.6.

Let $\phi$ be a $C^{3}$ function on a compact subset $K$ of $\mathbb{R}^{2}$.
Let $x\in K$ and $\epsilon\in[0,\infty)$. If $D\phi(x)\neq 0$, there exists a
constant $C_{1}$ (depending only on the $C^{2}$ norm of $\phi$) with the
following two properties for all unit vectors $v\in\mathbb{R}^{2}$.

  1. 1.

If $\sqrt{2}\lvert\langle D\phi(x),v\rangle\rvert\geq C_{1}\epsilon$,

$\sqrt{2}\epsilon\lvert\langle D\phi,v\rangle\rvert+\epsilon^{2}\langle
D^{2}\phi v,v\rangle\geq\epsilon^{2}\left\langle
D^{2}\phi\frac{D^{\perp}\phi}{\lvert D\phi\rvert},\frac{D^{\perp}\phi}{\lvert
D\phi\rvert}\right\rangle$

at $x$.

  2. 2.

If $\sqrt{2}\lvert\langle D\phi(x),v\rangle\rvert\leq C_{1}\epsilon$, there
exists a constant $C_{2}$ (depending only on the $C^{2}$ norm of $\phi$) such
that

$\sqrt{2}\epsilon\lvert\langle D\phi,v\rangle\rvert+\epsilon^{2}\langle
D^{2}\phi v,v\rangle\geq\epsilon^{2}\left\langle
D^{2}\phi\frac{D^{\perp}\phi}{\lvert D\phi\rvert},\frac{D^{\perp}\phi}{\lvert
D\phi\rvert}\right\rangle-\frac{C_{2}\epsilon^{3}}{\lvert D\phi\rvert}$

at $x$.

######  Proposition A.7.

The function $\overline{u}$ is a viscosity subsolution of (2.1) in
$\mathbb{R}^{2}\times(0,\infty)$.

###### Proof.

As in Proposition A.5, we obtain
$\Psi_{-}(x)\leq\overline{u}(x,t)\leq\Psi_{+}(x)$ and argue by contradiction
to prove the viscosity inequality. We prepare the following conditions:

$\partial_{t}\phi(x,t)-\nu\lvert D\phi(x,t)\rvert-\left\langle
D^{2}\phi(x,t)\frac{D^{\perp}\phi(x,t)}{\lvert
D\phi(x,t)\rvert},\frac{D^{\perp}\phi(x,t)}{\lvert
D\phi(x,t)\rvert}\right\rangle-f^{\ast}(x)\geq\theta_{0}>0,$ (A.15)
$\partial_{t}\phi(x,t)-\nu\lvert
D\phi(x,t)\rvert-\sup_{\lvert\zeta\rvert=1}\left\langle
D^{2}\phi(x,t)\zeta,\zeta\right\rangle-f^{\ast}(x)\geq\theta_{0}>0.$ (A.16)

We assume that there exist a smooth function $\phi$ and $(\hat{x},\hat{t})$
such that $(\hat{x},\hat{t})$ is a strict local maximum of
$\overline{u}-\phi$, $\overline{u}>\Psi_{-}$ at $(\hat{x},\hat{t})$ and, for
some $\theta_{0}>0$, (A.15) is satisfied at $(\hat{x},\hat{t})$ provided
$D\phi(\hat{x},\hat{t})\neq 0$, and (A.16) is satisfied at $(\hat{x},\hat{t})$
provided $D\phi(\hat{x},\hat{t})=0$. If $D\phi(\hat{x},\hat{t})\neq 0$, we
take a $\delta$ neighborhood of $(\hat{x},\hat{t})$ where $\overline{u}-\phi$
attains its unique maximum at $(\hat{x},\hat{t})$, $\lvert
D\phi\rvert>\theta_{1}$ for some $\theta_{1}>0$, and (A.15) holds, retaking
smaller $\theta_{0}>0$ if necessary. If $D\phi(\hat{x},\hat{t})=0$, we take a
$\delta$ neighborhood of $(\hat{x},\hat{t})$ where $\overline{u}-\phi$ attains
its unique maximum at $(\hat{x},\hat{t})$ and (A.16) holds, retaking smaller
$\theta_{0}>0$ if necessary. We take $\delta>0$ small enough to satisfy
$\delta\leq\frac{b}{3\max\\{L,M\\}}$, where
$b:=\overline{u}(\hat{x},\hat{t})-\Psi_{-}(\hat{x})$. From the definition of
$\overline{u}$, there are some sequences $\\{\epsilon_{n}\\}$,
$\\{x_{\epsilon_{n}}^{0}\\}$, and $\\{t_{\epsilon_{n}}^{0}\\}$ satisfying

$\epsilon_{n}\searrow
0,~{}~{}(x_{\epsilon_{n}}^{0},t_{\epsilon_{n}}^{0})\to(\hat{x},\hat{t}),~{}~{}u^{\epsilon_{n}}(x_{\epsilon_{n}}^{0},t_{\epsilon_{n}}^{0})\to\overline{u}(\hat{x},\hat{t}).$

We may substitute $\epsilon$ for $\epsilon_{n}$ hereafter. We take $\epsilon$
small enough to satisfy $(x_{\epsilon}^{0},t_{\epsilon}^{0})\in
N_{\delta/2}((\hat{x},\hat{t}))$ and
$u^{\epsilon}(x_{\epsilon}^{0},t_{\epsilon}^{0})-\Psi_{-}(x_{\epsilon}^{0})\geq
b/2$.

We construct the sequence $\\{X_{k}\\}$ and the functions
$\Psi_{k}:S^{1}\times S^{1}\to\mathbb{R}$ inductively as follows. We first let

$X_{0}:=(x_{\epsilon}^{0},t_{\epsilon}^{0}),$

and

$\Psi_{0}(v,w):=\min_{b=\pm 1}u^{\epsilon}(x_{\epsilon}^{0}+\sqrt{2}\epsilon
bv+\nu\epsilon^{2}w,t_{\epsilon}^{0}-\epsilon^{2})+\epsilon^{2}f(x_{\epsilon}^{0}).$

Then let $(v_{0},w_{0})$ satisfying

$\Psi_{0}(v_{0},w_{0})\geq\sup_{(v,w)\in S^{1}\times
S^{1}}\Psi_{0}(v,w)-\epsilon^{3},$

and we determine

$X_{1}=X_{0}+(\sqrt{2}\epsilon
b_{0}v_{0}+\nu\epsilon^{2}w_{0},-\epsilon^{2}),$

where we will decide $b_{0}$ later. For any $k\in\mathbb{N}$, we similarly
define

$\Psi_{k}(v,w):=\min_{b=\pm 1}u^{\epsilon}(x_{k}+\sqrt{2}\epsilon
bv+\nu\epsilon^{2}w,t_{k}-\epsilon^{2})+\epsilon^{2}f(x_{k}),$

and

$X_{k+1}:=X_{k}+(\sqrt{2}\epsilon
b_{k}v_{k}+\nu\epsilon^{2}w_{k},-\epsilon^{2}),$ (A.17)

where $(v_{k},w_{k})\in S^{1}\times S^{1}$ satisfies

$\Psi_{k}(v_{k},w_{k})\geq\sup_{(v,w)\in S^{1}\times
S^{1}}\Psi_{k}(v,w)-\epsilon^{3}.$

Define $k_{\epsilon}$ as in the proof of Proposition A.5. We prove by
induction that $u^{\epsilon}(X_{k})>\Psi_{-}(x_{k})$ and

$u^{\epsilon}(X_{0})-u^{\epsilon}(X_{k})\leq\epsilon^{2}\sum_{m=0}^{k-1}[f(x_{m})]+k\epsilon^{3}$
(A.18)

for all $k<k_{\epsilon}$. These inequalities hold for $k=0$. We assume that
these hold for some $k<k_{\epsilon}$. Then the Dynamic Programming Principle
(A.2) and $u^{\epsilon}(X_{k})>\Psi_{-}(x_{k})$ imply

$u^{\epsilon}(X_{k})=\min\\{\Psi_{+}(x_{k}),\sup_{v,w\in S^{1}}\min_{b=\pm
1}u^{\epsilon}(x_{k}+\sqrt{2}\epsilon bv+\nu
w\epsilon^{2},t_{k}-\epsilon^{2})+\epsilon^{2}f(x_{k})\\}.$

Thus we have

$\displaystyle u^{\epsilon}(X_{k})$ $\displaystyle\leq\sup_{v,w\in
S^{1}}\Psi_{k}(v,w)$ $\displaystyle\leq\Psi_{k}(v_{k},w_{k})+\epsilon^{3}$
$\displaystyle=u^{\epsilon}(x_{k}+\sqrt{2}\epsilon b_{k}v_{k}+\nu
w_{k}\epsilon^{2},t_{k}-\epsilon^{2})+\epsilon^{2}f(x_{k})+\epsilon^{3}$

and hence

$u^{\epsilon}(X_{k})-u^{\epsilon}(X_{k+1})\leq\epsilon^{2}f(x_{k})+\epsilon^{3},$
(A.19)

which means (A.18) holds for $k+1$. From the Lipschitz continuity of
$\Psi_{-}$ and (A.10), we have

$-\Psi_{-}(x_{k+1})\geq-\Psi_{-}(x_{0})-L(\delta-(k+1)\epsilon^{2})$ (A.20)

provided $k+1<k_{\epsilon}$. Combining (A.19) and (A.20), we obtain

$\displaystyle u^{\epsilon}(X_{k+1})-\Psi_{-}(x_{k+1})$ $\displaystyle\geq
u^{\epsilon}(X_{0})-\Psi_{-}(x_{0})-M(k+1)\epsilon^{2}-L(\delta-(k+1)\epsilon^{2})-(k+1)\epsilon^{3}$
$\displaystyle\geq
u^{\epsilon}(X_{0})-\Psi_{-}(x_{0})-\max\\{M,L\\}\delta-(k+1)\epsilon^{3}.$

We notice that $(k+1)\epsilon^{3}\leq L\delta\epsilon$ from (A.11). Therefore
$u^{\epsilon}(X_{k+1})>\Psi_{-}(x_{k+1})$ holds for sufficiently small
$\epsilon$ and we conclude the induction.

Next we take the continuous path $X(s)$ and use the Taylor’s theorem in the
same way as in Proposition A.5. Then we have

$\displaystyle\phi(X_{k+1})-\phi(X_{k})$ $\displaystyle=\sqrt{2}\epsilon
b_{k}\langle
D\phi(X_{k}),v_{k}\rangle+\epsilon^{2}\\{-\partial_{t}\phi(X_{k})+\nu\langle
D\phi(X_{k}),w_{k}\rangle+\langle
D^{2}\phi(X_{k})v_{k},v_{k}\rangle\\}+\Phi_{k}(\epsilon),$

where $\Phi_{k}(\epsilon)=o(\epsilon^{2}).$ By taking $b_{k}$ in (A.17)
properly, we get

$\begin{split}&\phi(X_{k+1})-\phi(X_{k})\\\
&\leq-\sqrt{2}\epsilon\lvert\langle
D\phi(X_{k}),v_{k}\rangle\rvert+\epsilon^{2}\\{-\partial_{t}\phi(X_{k})+\nu\lvert
D\phi(X_{k})\rvert+\langle
D^{2}\phi(X_{k})v_{k},v_{k}\rangle\\}+\Phi_{k}(\epsilon).\end{split}$ (A.21)

We first consider the case $D\phi(\hat{x},\hat{t})\neq 0$. If
$k<k_{\epsilon}$, we just consider $\phi$ in $N_{\delta}((\hat{x},\hat{t}))$.
We now use the assumption (A.15) and Lemma A.6 replacing $\phi$ by $-\phi$ to
get

$\displaystyle\phi(X_{k+1})-\phi(X_{k})$
$\displaystyle\leq\epsilon^{2}\left\\{-\partial_{t}\phi(X_{k})+\nu\lvert
D\phi(X_{k})\rvert+\left\langle
D^{2}\phi(X_{k})\frac{D^{\perp}\phi(X_{k})}{\lvert
D\phi(X_{k})\rvert},\frac{D^{\perp}\phi(X_{k})}{\lvert
D\phi(X_{k})\rvert}\right\rangle\right\\}$
$\displaystyle+\frac{C_{2}\epsilon^{3}}{\lvert
D\phi(X_{k})\rvert}+\Phi_{k}(\epsilon)$
$\displaystyle\leq\epsilon^{2}\left(-\frac{\theta_{0}}{2}-f^{\ast}(x_{k})\right),$
(A.22)

for sufficiently small $\epsilon$. This inequality is also obtained in the
case $D\phi(\hat{x},\hat{t})=0$, using (A.21) and the assumption (A.16). From
(A.19) and (A.22), we have

$u^{\epsilon}(X_{0})-u^{\epsilon}(X_{k})-\phi(X_{0})+\phi(X_{k})\leq-\frac{k}{4}\epsilon^{2}\theta_{0},~{}~{}~{}~{}~{}k\leq
k_{\epsilon}.$ (A.23)

By the same argument as in Proposition A.5, we obtain

$\overline{u}(\hat{x},\hat{t})-\phi(\hat{x},\hat{t})\leq\overline{u}(x^{\prime},t^{\prime})-\phi(x^{\prime},t^{\prime}),$

where $(x^{\prime},t^{\prime})\neq(\hat{x},\hat{t})$. This is a contradiction
since $\overline{u}-\phi$ attains its unique maximum at $(\hat{x},\hat{t})$. ∎

The proof of Proposition A.3 is completed by checking that $\overline{u}$ and
$\underline{u}$ satisfy the initial condition. To prove the last proposition,
we need additional property of the solution to (4.3) and strategies of the
game.

######  Lemma A.8.

Let $\delta\in(0,\nu^{-1}]$. For sufficiently small $\epsilon>0$, we have

$\frac{r_{2}-\delta}{\delta^{-1}-\frac{\nu}{2}}\leq
t_{\epsilon}(r_{1},r_{2}),$

for any $r_{1}$ and $r_{2}$ satisfying $0\leq r_{1}\leq\delta\leq
r_{2}\leq\nu^{-1}$.

###### Proof.

Let $\delta\leq r\leq\nu^{-1}$. We take $\epsilon>0$ small enough to satisfy
$T_{\epsilon^{2}}(r)\geq r$. Concretely we assume $\nu\epsilon^{2}\leq\delta$.
Then the inequality (4.4) implies

$T_{\epsilon^{2}}(r)-r\leq\left(\delta^{-1}-\nu\right)\epsilon^{2}+\frac{\nu^{2}\epsilon^{4}}{2\delta}.$

Hence we obtain

$t_{\epsilon}(r_{1},r_{2})\geq
t_{\epsilon}(\delta,r_{2})\geq\frac{r_{2}-\delta}{\left(\delta^{-1}-\nu\right)\epsilon^{2}+\frac{\nu^{2}\epsilon^{4}}{2\delta}}\epsilon^{2}\geq\frac{r_{2}-\delta}{\delta^{-1}-\nu+\frac{\nu}{2}}=\frac{r_{2}-\delta}{\delta^{-1}-\frac{\nu}{2}}.$

∎

######  Definition A.9 (Reversed concentric strategy).

Let $\nu>0$, $\epsilon>0$ and $z\in\mathbb{R}^{2}$. Let $x\in\mathbb{R}^{2}$
be the current position of the game. Let $(v,w)\in S^{1}\times S^{1}$ be a
choice by Paul in the same round. A choice $b\in\\{\pm 1\\}$ by Carol is
called a reversed $z$ concentric strategy if

$\langle bv,x+\nu\epsilon^{2}w-z\rangle\leq 0.$

If Carol takes reversed $z$ concentric strategy through the game, we get
$\lvert x_{n}-z\rvert\leq P_{n}$, where $P_{n}$ satisfies

$P_{n+1}=\sqrt{(P_{n}+\nu\epsilon^{2})^{2}+2\epsilon^{2}}$ (A.24)

with $P_{0}=\lvert x_{0}-z\rvert$. We define $t_{\epsilon}(a,b)$ in the same
way as (4.6), replacing the operator $T_{h}$ as follows:

$T_{h}(R):=\sqrt{(R+\nu h)^{2}+2h}.$

######  Lemma A.10.

Let $\delta>0$. For sufficiently small $\epsilon>0$, we have

$\frac{r_{2}-\delta}{\delta^{-1}+\nu+1}\leq t_{\epsilon}(r_{1},r_{2})$

for any $r_{1}$ and $r_{2}$ satisfying $0\leq r_{1}\leq\delta\leq r_{2}$.

###### Proof.

The proof is similar to that of Lemma A.8, so is omitted. ∎

######  Proposition A.11.

Let $u_{0}$ be a continuous function. Then
$\overline{u}(x,0)=\underline{u}(x,0)=u_{0}(x)$ for all $x\in\mathbb{R}^{2}$.

###### Proof.

Let $x\in\mathbb{R}^{2}$. For the initial position $y\in\mathbb{R}^{2}$, the
terminal time $s>0$ and the step size $\epsilon>0$, we define
$V^{-}(y,s,\epsilon)$ as the minimum total cost when Paul takes a $x$
concentric strategy through the game. Similarly we define
$V^{+}(y,s,\epsilon)$ as the supremum total cost when Carol takes reversed $x$
concentric strategy through the game. It is clear by the property of the value
functions that

$V^{-}(y,s,\epsilon)\leq u^{\epsilon}(y,s)\leq V^{+}(y,s,\epsilon).$

It is sufficient to show

$\varliminf_{\begin{subarray}{c}(y,s)\to(x,0)\\\ \epsilon\searrow
0\end{subarray}}V^{-}(y,s,\epsilon)\geq u_{0}(x)$ (A.25)

and

$\varlimsup_{\begin{subarray}{c}(y,s)\to(x,0)\\\ \epsilon\searrow
0\end{subarray}}V^{+}(y,s,\epsilon)\leq u_{0}(x).$ (A.26)

We first analyze $V^{-}$. We denote by $V^{-}_{quit}(y,s,\epsilon)$ (resp.
$V^{-}_{end}(y,s,\epsilon)$) the minimum total cost when Paul takes $x$
concentric strategy through the game and Carol quits (resp. does not quit) the
game on the way. Then we write

$V^{-}(y,s,\epsilon)=\min\\{V^{-}_{end}(y,s,\epsilon),V^{-}_{quit}(y,s,\epsilon)\\}.$

Furthermore we analyze $V^{-}_{end}$. We denote by $V^{-}_{run}(y,s,\epsilon)$
(resp. $V^{-}_{ter}(y,s,\epsilon)$) the minimum running cost (resp. terminal
cost) in the same situation as $V^{-}_{end}(y,s,\epsilon)$. Obviously we have

$V^{-}_{end}(y,s,\epsilon)\geq
V^{-}_{run}(y,s,\epsilon)+V^{-}_{ter}(y,s,\epsilon).$

Since Paul takes a $x$ concentric strategy, he can stay in $B(x,2\nu^{-1})$ by
Lemma 4.2. So the running cost is at most $M:=\sup_{z\in B(x,2\nu^{-1})}\lvert
f(z)\rvert$, and at least $-M$ per round. Hence we have

$\displaystyle\lvert V^{-}_{run}(y,s,\epsilon)\rvert$
$\displaystyle\leq\epsilon^{2}NM=\epsilon^{2}\left\lceil
s\epsilon^{-2}\right\rceil
M\leq\epsilon^{2}\left(s\epsilon^{-2}+1\right)M=(s+\epsilon^{2})M.$ (A.27)

We denote by $Ter(y,s,\epsilon)$ a terminal point $x_{N}$ in the situation of
$V^{-}_{end}(y,s,\epsilon)$. Since $u_{0}$ is continuous, what we have to
prove about the terminal cost is

$\lim_{\begin{subarray}{c}(y,s)\to(x,0)\\\ \epsilon\searrow
0\end{subarray}}Ter(y,s,\epsilon)=x$

for any choices of Carol. Let
$\\{(y_{n},s_{n},\epsilon_{n})\\}\subset\mathbb{R}^{2}\times(0,\infty)\times(0,\infty)$
be any sequence satisfying

$\epsilon_{n}\searrow 0,~{}~{}y_{n}\rightarrow x,~{}~{}s_{n}\rightarrow 0.$

Let $\delta>0$. Then we shall show that $\lvert
Ter(y_{n},s_{n},\epsilon_{n})-x\rvert<\delta$ for sufficiently large $n$.
Indeed, from Lemma A.8, there exists $\tilde{\epsilon}>0$ such that

$\frac{\delta/3}{3\delta^{-1}-\nu+1}\leq t_{\epsilon}(r,2\delta/3)$ (A.28)

for all $r\in[0,\delta/3)$ and all $\epsilon\in(0,\tilde{\epsilon})$. We take
$n$ large enough so that

$\lvert y_{n}-x\rvert<\delta/3,~{}\lvert
s_{n}\rvert<\frac{\delta/3}{3\delta^{-1}-\nu+1},~{}\epsilon_{n}<\tilde{\epsilon}.$

Hence, together with (A.27), we obtain

$\lim_{\begin{subarray}{c}(y,s)\to(x,0)\\\ \epsilon\searrow
0\end{subarray}}V^{-}_{run}(y,s,\epsilon)+V^{-}_{ter}(y,s,\epsilon)=u_{0}(x).$
(A.29)

If Carol quits the game on the way, the game positions $\\{x_{n}\\}$ are in
$B(x,\lvert Ter(y,s,\epsilon)-x\rvert)$. Thus we have

$V^{-}_{quit}(y,s,\epsilon)\geq\inf\\{\Psi_{+}(z)\mid z\in B(x,\lvert
Ter(y,s,\epsilon)-x\rvert)\\}$

and hence

$\displaystyle\varliminf_{\begin{subarray}{c}(y,s)\to(x,0)\\\ \epsilon\searrow
0\end{subarray}}V^{-}_{quit}(y,s,\epsilon)$
$\displaystyle\geq\lim_{\begin{subarray}{c}(y,s)\to(x,0)\\\ \epsilon\searrow
0\end{subarray}}\inf\\{\Psi_{+}(z)\mid z\in B(x,\lvert
Ter(y,s,\epsilon)-x\rvert)\\}$ $\displaystyle=\Psi_{+}(x)\geq u_{0}(x).$

Together with (A.29), we obtain (A.25).

We next estimate $V^{+}$. We denote by $V^{+}_{quit}(y,s,\epsilon)$ (resp.
$V^{+}_{end}(y,s,\epsilon)$) the supremum total cost when Carol takes a
reversed $x$ concentric strategy through the game and Paul quits (resp. does
not quit) the game on the way. Then we write

$V^{+}(y,s,\epsilon)=\max\\{V^{+}_{end}(y,s,\epsilon),V^{+}_{quit}(y,s,\epsilon)\\}.$

We further denote by $V^{+}_{run}(y,s,\epsilon)$ (resp.
$V^{+}_{ter}(y,s,\epsilon)$) the supremum running cost (resp. terminal cost)
in the same situation as $V^{+}_{end}(y,s,\epsilon)$. Obviously we have

$V^{+}_{end}(y,s,\epsilon)\leq
V^{+}_{run}(y,s,\epsilon)+V^{+}_{ter}(y,s,\epsilon).$

From Lemma A.10, Paul is forced to stay in a compact set for sufficiently
small $s$. Thus as in (A.27), we have

$\lvert
V^{+}_{run}(y,s,\epsilon)\rvert\leq\epsilon^{2}NM\leq(s+\epsilon^{2})M.$

The values $V^{+}_{ter}$ and $V^{+}_{quit}$ are also estimated in the same way
as $V^{-}_{ter}$ and $V^{-}_{quit}$ respectively. The only difference is

$\frac{\delta/3}{3\delta^{-1}+\nu+1}\leq t_{\epsilon}(r,2\delta/3)$

instead of (A.28). Thus (A.26) is also obtained. ∎

##  Appendix B Set theory

The following are supplementary propositions related to general topology and
convex sets.

######  Lemma B.1.

Let $A\subset\mathbb{R}^{d}$ be an open set. Let $K\subset A$ be a compact
set. Then, for sufficiently small $\delta>0$,

$B_{\delta}(K)\subset A.$

###### Proof.

We can assume without loss of generality that $A$ is bounded. We define
$f(x):=\sup\\{\delta>0\mid B_{\delta}(x)\subset A\\}$ for $x\in A$ and check
that it is a lower semicontinuous function. Let $\epsilon>0$. For $x,y\in A$
satisfying $\lvert x-y\rvert<\epsilon$, it is clear that
$B_{f(x)-\epsilon}(y)\in A$ and then $f(y)\geq f(x)-\epsilon$.

Since $f$ is lower semicontinuous, it has a minimizer $\hat{x}$ in $K$ by the
extreme value theorem. Letting $\delta=f(\hat{x})$, we obtain the conclusion.
∎

######  Lemma B.2.

Let $A\subset\mathbb{R}^{2}$ be a connected open set. Then $Co(A)=\\{x\in
l_{a,b}\mid a,b\in A\\}$.

###### Proof.

It is clear that $Co(A)\supset\\{x\in l_{a,b}\mid a,b\in A\\}$. By
Carathéodory’s theorem, we have

$Co(A)=\left\\{\sum_{i=1}^{3}\lambda_{i}x_{i};\ x_{i}\in
A,\lambda_{i}\in[0,1],\sum_{i=1}^{3}\lambda_{i}=1\right\\}.$

Fix any element $x\in Co(A)$. Then we can write
$x=\sum_{i=1}^{3}\lambda_{i}x_{i}$ for some $x_{i}\in A$ and
$\lambda_{i}\in[0,1]$. It suffices to consider the case $x_{1},x_{2},x_{3}$
are different and $\lambda_{i}\in(0,1)$. We can assume $x=(0,0)$,
$x_{1}=(0,1)$, $x_{2}\in\\{(p,q)\in\mathbb{R}^{2}\mid p<0\\}$, and
$x_{3}\in\\{(p,q)\in\mathbb{R}^{2}\mid p>0\\}$. Since $A$ is a a connected
open set, it is also path-connected. Thus there is a continuous path
$\Gamma\subset A$ that connects $x_{2}$ and $x_{3}$. By the intermediate value
theorem, the path $\Gamma$ crosses $y$-axis. If $\Gamma$ crosses
$x_{4}\in\\{(0,q)\in\mathbb{R}^{2}\mid q<0\\}$, then $x_{1},x_{4}\in A$ and
$x\in l_{x_{1},x_{4}}$. Otherwise let $l$ be the line satisfying $x\in l$ and
$l_{x_{2},x_{3}}\parallel l$. The path $\Gamma$ crosses points $x_{4}\in
l\cap\\{(p,q)\in\mathbb{R}^{2}\mid p<0\\}$ and $x_{5}\in
l\cap\\{(p,q)\in\mathbb{R}^{2}\mid p>0\\}$. Thus we have $x_{4},x_{5}\in A$
and $x\in l_{x_{4},x_{5}}$. Therefore we conclude that $x\in l_{a,b}$ for some
$a,b\in A$. ∎

######  Proposition B.3.

If $A\subset\mathbb{R}^{d}$ is open, then $Co(A)$ is open.

###### Proof.

Fix $x\in Co(A)$. By Carathéodory’s theorem, we have
$x=\sum_{i=1}^{d+1}\lambda_{i}x_{i}$ for some $x_{i}\in A$ and
$\lambda_{i}\in[0,1]$ satisfying $\sum_{i=1}^{d+1}\lambda_{i}=1$. Since $A$ is
open, we see that $\cup_{i=1}^{d+1}B_{r_{0}}(x_{i})\subset A$ for some
$r_{0}>0$. Therefore, for any unit vector $v\in S^{1}$, we have $x+rv\in
Co(A)$ for $0\leq r<r_{0}$ since $x+rv=\sum_{i=1}^{d+1}\lambda_{i}(x_{i}+rv)$.
∎

######  Proposition B.4.

Let $O_{-}\subset\mathbb{R}^{d}$. If $O_{-}$ satisfies (3.4), then
$\overline{O_{-}}$ is strictly convex.

###### Proof.

Assume that $\overline{O_{-}}$ is not strictly convex, i.e., there exist
$x,y\in\overline{O_{-}}$ such that $\lambda
x+(1-\lambda)y\in(O_{-}^{int})^{c}$ for some $\lambda\in(0,1)$.

$1)~{}\lambda x+(1-\lambda)y=:z\in\partial O_{-}$ for some $\lambda\in(0,1)$.
By (3.4) we can take an open ball $B$ that satisfies $z\in\partial B$ and
$\overline{O_{-}}\subset\overline{B}$. Since $x\in\overline{B}$ and
$z\in\partial B$, we have $y\in(\overline{B})^{c}$, which is a contradiction.

$2)~{}\lambda x+(1-\lambda)y\in(\overline{O_{-}})^{c}$ for any
$\lambda\in(0,1)$. Let $z=\frac{x+y}{2}$. Let $\delta>0$ satisfy
$B_{\delta}(z)\subset\left(\overline{O_{-}}\right)^{c}$. Since $x\in\partial
O_{-}$, there exists $w\in O_{-}$ such that $w\in B_{2\delta}(x)$. Since
$\frac{w+y}{2}\in\left(\overline{O_{-}}\right)^{c}$, we see that $\lambda
w+(1-\lambda)y\in\partial O_{-}$ for some $\lambda\in(0,1)$ and hence deduce a
contradiction. ∎

##  Appendix C Graph theory

We present the notion of graph and some related notions in the graph theory.

######  Definition C.1.

For a non empty set $V$ and a set $E$ of unordered pairs in $V$, the pair of
the sets $(V,E)$ is called a graph. A graph $H=(V^{\prime},E^{\prime})$ is
called a subgraph of $G$ if $V^{\prime}\subset V$ and $E^{\prime}\subset E$. A
subgraph $H=(V^{\prime},E^{\prime})\subset G$ is called a path of $G$ if
$V^{\prime}$ is a finite set $\\{x_{0},x_{1},\cdots,x_{n}\\}$(duplication is
permitted.) and $E^{\prime}=\\{\langle x_{i},x_{i+1}\rangle\mid
i=0,1,\cdots,n-1\\}$, where we denote unordered pairs by $\langle,\rangle$. A
graph $G=(V,E)$ is connected if for any $v_{1},v_{2}\in V$, there is a path of
$G$ whose endpoints are $v_{1}$ and $v_{2}$.

To precisely indicate the path of graph introduced in the proof of Theorem 3.2
, we present the following proposition, though the assertion seems to be
obvious.

######  Proposition C.2.

Let $G=(V,E)$ be a connected graph. Let $\langle a,b\rangle,\langle
c,d\rangle\in E$. Then there is a path $P=(V^{\prime},E^{\prime})$ such that
$a,b,c,d\in V^{\prime}$ and $\langle a,b\rangle,\langle c,d\rangle\in
E^{\prime}$.

###### Proof.

If $a=c$, $(\\{b,a,d\\},\\{\langle b,a\rangle,\langle a,d\rangle\\})$ is a
required path. Hereafter we consider the case neither $a=c$, $b=c$, $a=d$ nor
$b=d$. Let $P_{0}=(V_{0},E_{0})$ be a path with endpoints $a$ and $c$.

If $b,d\notin V_{0}$, define $V_{2}:=V_{0}\cup\\{b,d\\}$ and
$E_{2}:=E_{0}\cup\\{\langle b,a\rangle,\langle c,d\rangle\\}$.

If $b\notin V_{0}$ and $d\in V_{0}$, define $V_{1}:=V_{0}\cup\\{b\\}$ and
$E_{1}:=E_{0}\cup\\{\langle b,a\rangle\\}$. Writing

$\begin{split}V_{1}&=\\{x_{0},x_{1},\cdots,x_{n}\\},\\\ E_{1}&=\\{\langle
x_{i},x_{i+1}\rangle\mid i=0,1,\cdots,n-1\\},\end{split}$ (C.1)

we see that $x_{0}=b$, $x_{1}=a$, $x_{n}=c$ and $x_{j}=d$ for some
$j\in\\{2,3,\cdots,n\\}$. Then we further define
$V_{2}:=\\{x_{0},x_{1},\cdots,x_{j},x_{n}\\}$ and $E_{2}:=\\{\langle
x_{0},x_{1}\rangle,\langle x_{1},x_{2}\rangle,\cdots\langle
x_{j-1},x_{j}\rangle,\langle x_{j},x_{n}\rangle\\}$.

We finally consider the case $b,d\in V_{0}$. When we follow the path $P_{0}$
from $a$ to $c$, there are two cases: whether we find $b$ earlier than $d$ or
not. In the former case, writing $V_{0}$ and $E_{0}$ as (C.1), we see
$x_{j}=b$, $x_{m}=d$ for some $0\leq j<m\leq n$. We then define
$V_{2}:=\\{x_{0},x_{j},x_{j+1},\cdots,x_{m},x_{n}\\}$ and $E_{2}:=\\{\langle
x_{0},x_{j}\rangle,\langle x_{j},x_{j+1}\rangle,\cdots\langle
x_{m-1},x_{m}\rangle,\langle x_{m},x_{n}\rangle\\}$. In the latter case, we
see $x_{j}=d$, $x_{m}=b$ for some $0\leq j<m\leq n$. We define
$V_{2}:=\\{x_{l},x_{0},x_{1},\cdots,x_{j},x_{n}\\}$ and $E_{2}:=\\{\langle
x_{l},x_{0}\rangle,\langle x_{0},x_{1}\rangle,\cdots\langle
x_{j-1},x_{j}\rangle,\langle x_{j},x_{n}\rangle\\}$.

In any case, $P_{2}:=(V_{2},E_{2})$ is a required path. ∎

##  Appendix D Curve theory

To complement the proof of Thorem 3.2, we will mathematically describe the
construction of a Jordan curve $\hat{C}$ that is included in a given closed
curve $C$ and includes a given point $x\in C$. We begin with a general
property of connected sets. In what follows we especially notice that two
points in an open and connected subset of $\mathbb{R}^{d}$ can be connected by
a polygonal line.

######  Definition D.1 (polygonal line connected).

A path is called a polygonal line if it consists of finite line segments. Let
$A\subset\mathbb{R}^{d}$. The set $A$ is called polygonal line connected if
for any two points $x,y\in A$, there exists a polygonal line in $A$ that
connects $x$ and $y$.

######  Proposition D.2.

Let $A\subset\mathbb{R}^{d}$ be an open set. Then the following statements are
equivalent.

  1. 1.

$A$ is connected.

  2. 2.

$A$ is path-connected.

  3. 3.

$A$ is polygonal line connected.

###### Proof.

Without loss of generality, we can assume $A\neq\emptyset$ since otherwise all
the statements are obviously true.

3$\Rightarrow$2.

This is clear because a polygonal line is a path.

2$\Rightarrow$1.

Fix $x\in A$. Since a path is a connected set, all elements $y\in A$ are in
the connected component including $x$. Therefore $A$ is connected.

1$\Rightarrow$3.

Fix $x\in A$. We define
$O:=\\{y\in A\mid\ \mbox{there exists a polygonal line in $A$ that connects
$x$ and $y$}\\}$. We first show that $O$ is an open set. Let $y\in O$. Since
$y\in A$, there is an open ball $B_{\delta}(y)$ such that
$B_{\delta}(y)\subset A$. For any $z\in B_{\delta}(y)$, we can make a
polygonal line in $A$ that connects $x$ and $z$, combining the line segment
between $y$ and $z$ with a polygonal line between $x$ and $y$. Therefore we
have $B_{\delta}(y)\subset O$, which means $O$ is an open set.

We show that $A\setminus O$ is also an open set. Let $y\in A\setminus O$. As
before, there is an open ball $B_{\delta}(y)$ such that $B_{\delta}(y)\subset
A$. If a point $z$ in $B_{\delta}(y)$ is in $O$, we can make a polygonal line
in $A$ that connects $x$ and $y$. This is a contradiction. Hence we have
$B_{\delta}(y)\subset A\setminus O$.

Since $A$ is connected, $O$ must be $\emptyset$ or $A$. Since $x\in O$, we
have $O=A$ and conclude that $A$ is polygonal line connected. ∎

We state the condition on components of the closed curve $C$. In what follows
we call a map $f|_{A}$ injective at $t\in A$ if $s\in A$ and $f(s)=f(t)$ imply
$s=t$. Also we call a map $f|_{A}$ injective in $B(\subset A)$ if $s\in A$,
$t\in B$ and $f(s)=f(t)$ imply $s=t$. We set a class $\mathscr{C}$ of curves
in $\mathbb{R}^{2}$. We make the assumption on $\mathscr{C}$:

  * (A1)

There exists a map $\mathscr{C}\ni C\mapsto\gamma_{C}\in
C([0,1];\mathbb{R}^{2})$ such that

    1. 1.

$\gamma_{C}([0,1])=C$ and the set $\\{t\in(0,1)\mid\gamma_{C}\ \mbox{is not
injective at $t$}\\}$ is at most finite.

    2. 2.

For any $C,D\in\mathscr{C}$, $\\{t\in(0,1)\mid\gamma_{C}(t)\notin D\\}$ is at
most a finite union of open intervals.

We now state the assumption on the closed curve $C$:

  * (A2)

For some $C_{1},C_{2},\cdots,C_{N}\in\mathscr{C}$, $C=\cup_{i=1}^{N}C_{i}$,
$\gamma_{C_{i}}(1)=\gamma_{C_{i+1}}(0)$ for $i\in\\{1,2,\cdots,N-1\\}$ and
$\gamma_{C_{N}}(1)=\gamma_{C_{1}}(0)$.

######  Remark D.3.

The set of line segments and arcs in $\mathbb{R}^{2}$ satisfies
$\mathrm{(A1)}$. Hence the closed curve in the proof of Theorem 3.2 satisfies
$\mathrm{(A2)}$.

Set $\gamma:[0,N]\to\mathbb{R}^{2}$ as

$\gamma(t):=\gamma_{C_{i}}(t-[t]),$

if $i-1\leq t\leq i$. Here we denote by $[t]$ the maximal integer that is no
more than $t$. For a point $x\in C$, there is no loss of generality to assume
$\gamma(0)=x$. The reason is the following:

######  Proposition D.4.

Let $\mathscr{C}$ satisfy $\mathrm{(A1)}$. Let $C^{\prime}\in\mathscr{C}$.
Then
$\mathscr{C}^{\prime}:=\mathscr{C}\cup\\{C^{\prime}_{1},C^{\prime}_{2}\\}$
also satisfies $\mathrm{(A1)}$, where we define

$C^{\prime}_{1}:=\gamma_{C^{\prime}}([0,a])\ \mbox{and}\
C^{\prime}_{2}:=\gamma_{C^{\prime}}([a,1])$

for $a\in(0,1)$.

###### Proof.

Let $\gamma_{C^{\prime}_{1}}(t)=\gamma_{C^{\prime}}(t/a)$ and
$\gamma_{C^{\prime}_{2}}(t)=\gamma_{C^{\prime}}\left(a+(1-a)t\right)$. The
proof is done by checking the assumption $\mathrm{(A1)}$ directly. ∎

We assume that $\gamma|_{[0,N)}$ is injective in a neighborhood of $0$. i.e.,

  * (A3)

There exists $\delta>0$ such that $\gamma|_{[0,N)}$ is injective in
$[0,\delta)$.

######  Remark D.5.

The closed curve $C\cup\hat{\Gamma}$ in the proof of Theorem 3.2 satisfies
$\mathrm{(A3)}$ because $B_{3\delta}(\hat{\Gamma})\subset L$ and $x\notin L$.

We inductively define

$t_{1}:=0,$ $s_{i}:=\sup\\{\tau\mid\gamma|_{(t_{i},N]}\ \mbox{is injective
in}\ (t_{i},\tau)\\},$ $t_{i+1}:=\sup\\{\tau\mid\gamma(s_{i})=\gamma(\tau)\\}$

for $i=1,2,\cdots$.

######  Proposition D.6.

For some $m\in\mathbb{N}$, $s_{m}=N$ or $t_{m}=N$.

###### Proof.

We first prove $t_{j}<s_{j}$ for all $j$. The assumption that
$\gamma|_{[0,N)}$ is injective in a neighborhood of $0$ implies $t_{1}<s_{1}$.
If $s_{1}<N$, then fix $j\in\\{2,3,\cdots\\}$. Let $i=[t_{j}]+1$. Set

$A_{i}:=\bigcap_{i+1\leq k\leq N}\\{t\in(i-1,i)\mid\gamma(t)\notin C_{k}\\}$

for $1\leq i\leq N-1$ and $A_{N}:=(N-1,N)$. We also set

$B_{i}:=\\{t\in(i-1,i)\mid\gamma|_{(i-1,i)}\ \mbox{is not injective at}\
t\\}.$

From the assumption $\mathrm{(A1)}$, $A_{i}$ is a finite union of open
intervals and $B_{i}$ is at most finite. By the definition of $t_{j}$ we see
that if $i-1<t_{j}<i$, then $t_{j}\in A_{i}$. Hence, if $i-1<t_{j}<i$, we have
$(t_{j},t_{j}+\delta)\subset A_{i}\cap B_{i}^{c}$ for some $\delta>0$. Also
this assertion holds for $t_{j}=i-1$. To show it, we prove by contradiction
that $\inf A_{i}=i-1$. We assume that $\inf A_{i}>i-1$. Then we have
$\inf\\{t\in(i-1,i)\mid\gamma(t)\notin C_{k}\\}>i-1$ for $i+1\leq k\leq N$.
From the continuity of $\gamma$, we obtain $\gamma(t_{j})\in C_{k}$, which is
a contradiction with the definition of $t_{j}$.

Now it turns out that there exists $l\geq 0$ such that $i-1\leq t_{j}<i$
implies $i\leq s_{j+l}$ or $i\leq t_{j+l+1}$. Indeed, if $s_{j}<i$, then
$s_{j}\in\partial A_{i}\cup B_{i}$. If $s_{j}<i$ and $s_{j}\in\partial A_{i}$,
then $i\leq t_{j+1}$. If $s_{j}<i$ and $s_{j}\in B_{i}$, then $t_{j+1}\in
B_{i}$. Thus the proof is complete. ∎

If $s_{m}=N$ or $t_{m+1}=N$, then let $T=\sum_{i=1}^{m}(s_{i}-t_{i})$. Define
$\hat{\gamma}:[0,T]\to\mathbb{R}^{2}$ as follows:

$\hat{\gamma}(t):=\gamma(t)$

if $t\leq s_{1}$ and

$\hat{\gamma}(t):=\gamma\left(t+t_{k+1}-\sum_{i=1}^{k}(s_{i}-t_{i})\right)$

if $\sum_{i=1}^{k}(s_{i}-t_{i})\leq t\leq\sum_{i=1}^{k+1}(s_{i}-t_{i})$.

######  Proposition D.7.

$\hat{C}:=\hat{\gamma}([0,T])$ is a Jordan closed curve.

###### Proof.

This assertion is obvious from the definitions of $t_{i}$ and $s_{i}$. ∎

$x$ Figure 19: An example of $C$ and $x$ that we avoid

$x$ Figure 20: An example of $C$ that includes an infinite number of loops

######  Remark D.8.

By assuming $\rm{(A3)}$, we can avoid a closed curve $C$ and a point $x$ in it
such as Figure 20. We also avoid a closed curve $C$ that includes an infinite
number of loops such as Figure 20 by the assumption $\rm{(A2)}$.

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