\documentclass[twoside]{article}
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\markboth{\hfil Two-sided Mullins-Sekerka flow \hfil}%
{\hfil U. F. Mayer \hfil}
\begin{document}
\setcounter{page}{171}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc Differential Equations and Computational Simulations III}\newline
J. Graef, R. Shivaji, B. Soni, \& J. Zhu (Editors)\newline
Electronic Journal of Differential Equations, Conference~01, 1997, pp. 171--179. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp 147.26.103.110 or 129.120.3.113 (login: ftp)}
\vspace{\bigskipamount} \\
Two-sided Mullins-Sekerka flow\\ does not preserve convexity
\thanks{{\em 1991 Mathematics Subject Classifications:}\/ 35R35,
35J05, 35B50, 53A07. \hfil\break\indent
{\em Key words and phrases:}\/ Mullins-Sekerka flow, Hele-Shaw flow,
Cahn-Hilliard equation, \hfil\break\indent
free boundary problem, convexity, curvature.
\hfil\break\indent
\copyright 1998 Southwest Texas State University and University of
North Texas. \hfil\break\indent
Published November 12, 1998.} }
\date{}
\author{ Uwe F. Mayer}
\maketitle
\begin{abstract}
The (two-sided) Mullins-Sekerka model is a nonlocal evolution model
for closed hypersurfaces, which was originally proposed as a model for
phase transitions of materials of negligible specific heat. Under this
evolution the propagating interfaces maintain the enclosed volume
while the area of the interfaces decreases. We will show by means of an
example that the Mullins-Sekerka flow does not preserve convexity in
two space dimensions, where we consider both the Mullins-Sekerka model
on a bounded domain, and the Mullins-Sekerka model defined on the
whole plane.
\end{abstract}
\newcommand\dist{\mathop{\rm dist}}
\newcommand\eqref[1]{(\ref{#1})}
\newtheorem{thm}{Theorem}
\newtheorem{prop}{Proposition}
\newtheorem{lem}{Lemma}
\section*{Introduction}
The (two-sided) Mullins-Sekerka flow is a nonlocal generalization of
the mean curvature flow arising from physics~\cite{MulSek63}, and was
originally proposed as an isotropic model for solidification and
liquidation of materials of negligible specific heat. It is also a
sharp-interface model describing phase transition in quenched binary
alloys. More precisely, it has been shown by Alikakos, Bates, and Chen
that the Mullins-Sekerka model arises as a singular limit for the
movement of level sets of solutions to the Cahn-Hilliard
equation~\cite{AliBatChe94}, the Cahn-Hilliard equation being a
fourth-order nonlinear partial differential equation modeling the
process of phase separation and coarsening in a melted
alloy~\cite{CahHil59}. This relationship between the Mullins-Sekerka
model and the Cahn-Hilliard equation was first formally derived by
Pego~\cite{Peg89}, and has likewise been established by
Stoth~\cite{Sto96} for the radially symmetric case, but for somewhat
more general boundary conditions. In the literature the
Mullins-Sekerka model has sometimes been called the Hele-Shaw model
(with surface tension), and it is also known as the quasi-stationary
Stefan problem. The problem studied in this paper is the same as the
one considered in
\cite{AliBatChe94,Che93,Che96,CheHonYi96,EscSim97a,EscSim98}.
There are two independent proofs for existence of smooth solutions for
the two-sided version of the Mullins-Sekerka model in a bounded
domain, due to Chen, Hong, and Yi~\cite{CheHonYi96}, and due to Escher
and Simonett~\cite{EscSim97a}; the existence of weak solutions had
been established earlier by Chen for the two-dimensional
case~\cite{Che93}. However, the author is unaware of any explicit
result about the existence of smooth solutions of the Mullins-Sekerka
model for unbounded domains.
As for the geometry of solutions, only little is known. It has been
shown that smooth initial configurations that are close to being
spherical (in a $C^{2+\alpha}$ sense) will converge exponentially fast
to a sphere~\cite{EscSim98}; an analogous result had been obtained
before for weak solutions in two space dimensions~\cite{Che93}. In the
Mullins-Sekerka model, the normal velocity of a moving interface
depends on the jump of the normal derivative of a function across the
interface, the function being the harmonic extension of the mean
curvature of the propagating interface. One can ask whether the
Mullins-Sekerka flow shares properties with the mean curvature flow,
since both flows are in some way driven by surface tension. Not all
results can be expected to generalize, due to the nonlocal character
of the Mullins-Sekerka problem, in particular not those that rest on a
local argument for the mean curvature flow. It is known that the mean
curvature flow preserves convexity \cite{GagHam86,Hui84}. It is
therefore a natural question to ask whether this is also true for the
Mullins-Sekerka flow. Under the assumption of short-term existence of
sufficiently smooth solutions this question was answered negatively
for the one-sided Mullins-Sekerka model in a previous paper by the
author~\cite{May93}. The current paper is an extension of the previous
one to the two-sided model. The loss of convexity for the two-sided
Mullins-Sekerka model in all of ${\mathbb R}^2$ has also been numerically
evidenced by Bates, Chen, and Deng~\cite{BatCheDen95}, for an improved
algorithm see the paper by Zhou, Chen, and Hou~\cite{ZhuCheHou96}.
\section*{The two-sided Mullins-Sekerka model}
We look at a curve $\Gamma_0$ contained in a fixed domain
$\Omega\subset{\mathbb R}^2$, and we consider the free boundary problem
governed by the evolution law given by
\begin{equation}\label{MS2}
\begin{array}{rlll}
\Delta u \!& = \!& 0 & \mbox{on } \Omega\setminus\Gamma_t\,, \\
\frac{\partial u}{\partial n} \!& = \!& 0 & \mbox{on } \partial\Omega\,,\\
u \!& = \!&\kappa & \mbox{on }\Gamma_t\,, \\
V \!& = \!&\left[\frac{\partial u}{\partial n}\right]&\mbox{on }\Gamma_t\,.
\end{array}
\end{equation}
The above assumes $\Omega$ to be bounded, in case $\Omega={\mathbb R}^2$ one
needs to replace the second line in this system with
\[
|u| = O(1) \quad \mbox{as } |x|\rightarrow\infty\,.
\]
Also in the equations above, $n$\/ is the outer unit normal to~$\Gamma_t$
and to $\partial\Omega$, while $V$ and $\kappa$\/ are the normal
velocity and the curvature of~$\Gamma_t$, respectively. The signs are
chosen in such a way that a circle has positive curvature and a
shrinking curve has negative velocity. The expression on the
right-hand side of the equation for $V$ denotes the jump of the normal
derivative of $u$ across~$\Gamma_t$, that is $\left[\frac{\partial
u}{\partial n}\right] = \frac{\partial u^+}{\partial n} -
\frac{\partial u^-}{\partial n}$, where the superscripts $+$ and $-$
indicate the regions outside and inside of~$\Gamma_t$. We define
$n^+=n$ and $n^-=-n$, which are the inner unit normals to the outside
region $\Omega^+$ and the inside region $\Omega^-$ of~$\Gamma_t$, and
then we rewrite the equation for the normal velocity as
\[
V = \frac{\partial u^+}{\partial n^+}
+ \frac{\partial u^-}{\partial n^-}\,.
\]
The principal idea is to consider an initial curve $\Gamma_0$ for
which one can make qualitative statements about the initial velocity,
and then to use continuity to forecast the shape of the evolving
curves. We look at a shape given by a straight tube with two circular
end caps. We will show that both $\frac{\partial u^+}{\partial n^+}$
and $\frac{\partial u^-}{\partial n^-}$ are positive quantities, but
that they are smaller in the center of the straight part of the
figure, whence the normal velocity will be smaller in the center, and
the figure will become nonconvex. Notice that by the strong maximum
principle one has $\frac{\partial u^+}{\partial n^+}>0$ and
$\frac{\partial u^-}{\partial n^-}>0$ at any point at which the
curvature attains its minimum, provided the curve is not a
circle. Thus any piece of a convex curve at which the curve loses its
convexity must necessarily move outwards, and convexity can therefore
only be lost when some parts of the curve move out slower than others,
just as for the example given herein. However, one may construct
examples that lose their convexity at one place only, as compared to
the example herein, which simultaneously loses convexity at both
straight line pieces.
\section*{The inside of the curve}
First we consider only the region $\Omega^-$ inside of the curve. The
statement made above about $\frac{\partial u^-}{\partial n^-}$ follows
from repeated applications of the maximum principle, the precise
argument was presented in a previous paper~\cite{May93}. We will
repeat here the flow of the argument for the sake of completeness, and
we write $u$ for $u^-$ in this section. Let us place the figure into
a standard Cartesian coordinate system, the straight parts parallel to
the $x$-axis, and the figure being centered about the origin. We
restrict our attention to the right half of the figure. As $u$ has a
maximum on the circular part this implies $\frac{\partial u}{\partial
n}>0$, and hence $u_x>0$. On the straight part we have $u_x\equiv 0$,
as $u$\/ is identically zero there. We also have $u_x\equiv 0$\/ on
the $y$-axis by the symmetry of $u$. Of course, one has to take care
of the non-smoothness of the curve where the circular arc connects to
the straight parts. This is done in the usual fashion by inserting a
small $C^\infty$ transition piece, and it can be shown that even on
this transition piece one can maintain the condition $u_x\geq0$, for
the technical details see~\cite{May93}. We conclude $u_x>0$\/ in the
interior of the right half of $\Omega^-$ by the maximum principle.
As $u_x\equiv 0$\/ on the upper straight part, we must have
$\frac{\partial}{\partial n}u_x =u_{xy}<0$\/ on the right half of it
by another application of the maximum principle. Yet another
application of the maximum principle for the function $u$ itself tells
us that $\frac{\partial u}{\partial n}=u_y<0$\/ on the upper straight
line. Therefore on the right half of the upper line the quantity
$\frac{\partial u}{\partial n^-}=|u_y|$ decreases towards the
center. By symmetry the same effect happens on the left half of the
curve.
\section*{The outside of the curve for the case $\Omega={\mathbb R}^2$}
We start out with this case because it illustrates the main idea, and
it is technically easier than the case of a bounded domain. We will
place the curve $\Gamma_0$ again into a Cartesian coordinate system,
but shifted down from the position chosen in the previous section, so
that the upper straight part falls onto the $x$-axis, see
Figure~\ref{figrectangle}. This way we can use Poisson's formula to
represent the harmonic function $u=u^+$ in the upper half plane if we
know its restriction to the $x$-axis,
\[
u(x,y)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{y}{(x-\xi)^{2}
+y^{2}}\,u(\xi,0)\,d\xi\,.
\]
As the upper straight part of the curve is on the $x$-axis we know
that $u\equiv0$ on this part of the $x$-axis, say on the interval
$[-2\varepsilon,2\varepsilon]\times\{0\}$. Using the symmetry of
$u$ we obtain
\[
u(x,y)=\frac{1}{\pi}\int_{2\varepsilon}^{\infty}
\left(\frac{y}{(x-\xi)^{2}+y^{2}}+\frac{y}{(x+\xi)^{2}+y^{2}}\right)
u(\xi,0)\,d\xi\,.
\]
Also, since $\Gamma_0$ is convex, we know that $u$ is nonnegative
everywhere on $\Gamma_0$, hence positive everywhere outside of $\Gamma_0$,
and so in particular on the rest of the $x$-axis. Using methods from
elementary calculus it is easily seen that the kernel
\[
k(x)=\frac{y}{(x-\xi)^{2}+y^{2}}+\frac{y}{(x+\xi)^{2}+y^{2}}
\]
is increasing for $02\varepsilon$ and $00$ on the interior of
${\cal R}$, and, by the strong maximum principle, $u_{xy}>0$ on the
interval $(0,\varepsilon)\times\{0\}$, because $u_x$ vanishes on this
part of the boundary of the rectangle~${\cal R}$. For use in the
next section we note that for any sufficiently small $\delta>0$ there
is a constant $c>0$ such that
\begin{equation}\label{lowerbound}
\frac{\partial^{2}u}{\partial x\partial
y}(x,0)>c\quad\mbox{for } \delta\leq x\leq\varepsilon-\delta\,.
\end{equation}
We interpret this as the statement $\displaystyle\frac{\partial}{\partial
x}\Big(\frac{\partial u}{\partial n^+}\Big)>0$, and we see that
the normal derivative grows towards the right. The analogous statement
is true for the corresponding part of the figure on the other side of
the $y$-axis.
\begin{figure}[hbt]
\begin{center}
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\end{center}
\caption{The rectangle ${\cal R}$ and the location of the curve
$\Gamma_0$. Not drawn to scale.}
\label{figrectangle}
\end{figure}
\section*{The outside of the curve for the case $\Omega$ a disk}
Just as in the previous section we will place the curve into a
Cartesian coordinate system so that the upper straight part falls onto
the $x$-axis. For simplicity we assume that $\Omega=\Omega_R$ is the
disk of radius $R$ centered at the origin, but this is not essential.
We will now consider the family of harmonic functions $u_R=u_R^+$
obtained by varying the radius $R$ of $\Omega_R$. Let $\widetilde{u}$
be the function obtained in the previous section; $\widetilde{u}$ is
the harmonic extension of the curvature of $\Gamma_0$ to the outside
of $\Gamma_0$ in ${\mathbb R}^2$. Our objective is to show that the
functions $u_R$ do not differ much from $\widetilde{u}$, at least not
for $R$ sufficiently large. To do this we will show that the family
$\{u_R, R\geq R_0\}$ is a normal family ($R_0$ is arbitrary). First we
establish the necessary equicontinuity near $\Gamma_0$, and for this
we will need the following boundary estimate from elliptic regularity
theory.
\begin{lem}
Let $W$ be a $C^{2+\alpha}$ domain in ${\mathbb R}^m$, and let
$\Gamma_0$ be a compact component of $\partial W$ with
$\dist(\Gamma_0,\partial W\setminus\Gamma_0)>0$ in case
$\Gamma_0\not=\partial W$. Let
$v\in C^{2+\alpha}(\overline{W})$ be a solution to $\Delta v=f$, $v=0$
on $\Gamma_0$, where $f\in C^{\alpha}(\overline W)$. Then there is a
$\delta>0$ such that for all $x_0\in\Gamma_0$
one has
\begin{equation}\label{lemma}
|v|_{2+\alpha;B_\delta(x_0)\cap W}\leq C(|v|_{0;W}+|f|_{\alpha;W})
\end{equation}
where the constants $\delta$ and $C$ depend only on $m$, $\alpha$,
$\Gamma_0$, and $W$. The various norms are the usual H\"older norms.
\end{lem}
The proof of this lemma is similar to the proof of Lemma 6.5
in~\cite{GilTru83}, and is omitted here.
Recall that $\Omega_R^+$ stands for the part of $\Omega_R$ outside of
$\Gamma_0$. Let $\Omega_{R_0}$ be any fixed disk containing $\Gamma_0$
and set $W=\Omega_{R_0}^+$. Let $\phi\in C^{2+\alpha}(\overline{W})$
be any function with $\phi(x,y)=\kappa(x,y)$ for $(x,y)\in\Gamma_0$.
Finally, let $u_R$ be the harmonic function associated with
$\Omega_R^+$, $R\geq R_0$. Then we apply the lemma to $v:=u_R-\phi$
and $f:=\Delta\phi$, and we conclude in particular that there is a
neighborhood of $\Gamma_0$ in $\overline{W}$ on which we have a
uniform bound on the first derivatives of $v$, and hence on the first
derivatives of $u_R$. Thus there is a neighborhood of $\Gamma_0$ on
which the family $\{u_R, R\geq R_0\}$ is equicontinuous.
The equicontinuity in the interior of $W$\/ follows from the
well-known interior gradient bound for harmonic functions, see for
example Theorem 2.10 in~\cite{GilTru83}. We quote it here in its
entirety because we will need it twice.
\begin{lem}\label{interiorbound}
Let $u$ be any harmonic function in $W\subset{\mathbb R}^m$ and let
$K$ be any compact subset of $W$. Then for any multi-index
$\alpha$ we have
\begin{equation}\label{interior}
\sup_{K}|D^\alpha u|
\leq(m|\alpha|/\dist(W,K))^{|\alpha|}\,
\sup_{W}|u|\,,
\end{equation}
where $D$ stands for the derivative operator.
\end{lem}
Finally, notice that the functions $u_R$ are uniformly bounded by
their maximum on $\Gamma_0$. Hence, fixing any ball $\Omega_{R_i}$
with $R_i>R_0$ we see that by the Arzela-Ascoli Theorem we get a
subsequence of $\{u_R, R\geq R_0\}$ converging uniformly to a harmonic
function on $\Omega_{R_i}^+$ as the radius $R$ converges to infinity. Now
we choose a sequence $R_i\rightarrow\infty$ and apply the usual
diagonalization process, by which we obtain a subsequence of harmonic
functions converging uniformly to a harmonic function $\bar{u}$
defined on the complete outside of $\Gamma_0$. In other words, if we
index that subsequence with $i$ again, we find $R_i$ such that the
sequence $u_{R_i}$ converges uniformly to $\bar{u}$ on any fixed
$\Omega_R^+$. Of course, the function $u_{R_i}$ is only defined on
$\Omega_R^+$ provided $R_i\geq R$, but this is irrelevant for our
argument. Now, the function $\bar{u}$ is bounded at infinity and has
the same boundary values on $\Gamma_0$ as all the functions $u_R$. By
the unique solvability of the exterior Dirichlet problem we conclude
that $\bar{u}=\widetilde{u}$, which is the harmonic extension of the
curvature of $\Gamma_0$ to all of ${\mathbb R}^2$. Note that the dimension 2
is important here, boundedness of a harmonic function is not enough to
conclude harmonicity at infinity for dimensions greater than two. In
fact, the argument shows that any sequence $u_{R_k}$ with
$R_k\rightarrow\infty$ has a subsequence converging to
$\widetilde{u}$, and so we see that the complete family $\{u_R, R\geq
R_0\}$ converges to $\bar{u}$ as $R\rightarrow\infty$. In summary, we
have shown the following.
\begin{prop}
Let $W$ be the exterior of $\Gamma_0$ in all of ${\mathbb R}^2$, and set
$\Omega_R^+=W\cap\Omega_R$. Let $\widetilde{u}$ be the harmonic
extension to $W$ of a smooth function $\kappa$ defined on $\Gamma_0$,
and let $u_R$ denote the harmonic extension of $\kappa$ to
$\Omega_R^+$, where we constitute a zero Neumann condition on
$\partial \Omega_R$. Then as $R\rightarrow\infty$ the functions $u_R$
converge uniformly to $\widetilde u$ on any compact subset of
$W\cup\Gamma_0$.
\end{prop}
Now consider the interval $I=[\delta,\varepsilon-\delta]\times\{0\}$
from the previous section, and set $u=\widetilde{u}-u_{R}$ in
Lemma~\ref{interiorbound}. We want to use $\Omega_{R_0}^+$ as
the set $W$ in Lemma~\ref{interiorbound}, however, the interval $I$ is
not in the interior of $\Omega_{R_0}^+$. We will therefore enlarge all
the domains $\Omega_R^+$ of the harmonic functions $u_R$ slightly by
reflecting near the interval $I$ across the $x$-axis. This extension
is possible since $\widetilde{u}\equiv u_R\equiv0$ on this part of
$\Gamma_0$. The convergence statement made in the proposition above is
of course also true for these slightly bigger regions, and hence we
can use the interior derivative estimate~\eqref{interior} from
Lemma~\ref{interiorbound} for the interval $I$. Finally, making the
radius $R$ large enough, we see that then
\[
\sup_{I}\left|
\frac{\partial^{2}(\widetilde{u}-u_{R})}{\partial x\partial y}\right|
\leq(4/\delta)^{2}\,
\sup_{\Omega_{R_0}^+}|\widetilde{u}-u_{R}|
\leq c/2\,,
\]
where $c$ is the lower bound from~\eqref{lowerbound}. Hence,
using~\eqref{lowerbound}, we get
\[
\frac{\partial^{2}{u_{R}}}{\partial x\partial
y}(x,0)>c/2\quad\mbox{for } \delta\leq x\leq\varepsilon-\delta\,.
\]
As in the previous section, we interpret this as the statement
$\displaystyle\frac{\partial}{\partial x}\Big(\frac{\partial {u_{R}}}{\partial
n^+}\Big)>0$, whence we see the normal derivative grows towards the
right.
\section*{The conclusion}
We have shown in the previous sections that the two normal
derivatives making up the normal velocity grow as we go away from the
center (where we start a distance $\delta>0$ away from the center),
and hence so does the normal velocity, which is their sum. Hence the
straight part of the curve moves out, but the center moves slower than
the ends, and thus by continuity of the evolution the curve will
become non-convex, see Figure~\ref{figvelocity}.
\begin{figure}[hbt]
\begin{center}
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\end{center}
\caption{A schematic sketch of the initial velocity of the interval
$I$ belonging to the straight part of $\Gamma_0$.}
\label{figvelocity}
\end{figure}
As mentioned in the introduction, the existence of smooth solutions
for the Mullins-Sekerka flow on a bounded domain has been established
recently~\cite{CheHonYi96,EscSim97a}. In fact, as follows from the
proofs in~\cite{EscSim97a}, the solution constitutes a semi-flow on a
space of curves (in two dimensions) parameterized over a reference
curve, and one has continuous dependence of the solution on the
initial data, measured in the $C^{2+\alpha}$ norm. The example
presented herein that leads to a loss of convexity can therefore be
slightly perturbed, and it will still evolve into a nonconvex shape. In
particular we can perturb it in such a fashion that the resulting
initial curve is strictly convex.
\vbox{
\begin{thm} There are strictly convex smooth initial configurations
that will evolve into nonconvex curves under the two-sided
Mullins-Sekerka flow on a large disk. In particular one can choose a
small smooth perturbation of a curve consisting of a straight tube
with two end caps, where the flat tube can be arbitrarily short, so
that these initial curves can be chosen arbitrarily close in the
$C^1$ norm to a circle.
\end{thm}
}
As noted above, the example herein also leads to a loss of convexity
in case of the Mullins-Sekerka model on all of ${\mathbb R}^2$. In fact, the
analytic argument for the loss of convexity was easier for the
unbounded case than for the bounded case. It is clear that if one has
continuous dependence of the solution on the initial data that one can
also go over to a strictly convex example.
\begin{thm} Assume that the two-sided Mullins-Sekerka flow on all of
${\mathbb R}^2$ allows a smooth solution provided the initial configuration
is $C^\infty$. Then there are convex smooth initial configurations
consisting of a straight tube with two end caps that will evolve into
nonconvex curves. The flat tube can be arbitrarily short. In
particular these initial curves can be chosen arbitrarily close
in the $C^1$ norm to a circle.
\end{thm}
{
\frenchspacing
\begin{thebibliography}{10}
\bibitem{AliBatChe94}
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}
\bigskip
\noindent
{\sc Uwe F. Mayer }\\
Department of Mathematics\\
Vanderbilt University\\
Nashville, TN 37240, USA\\
E-mail address: mayer\@@math.vanderbilt.edu
\end{document}