\documentclass[reqno]{amsart}
\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 71, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE--2003/71\hfil Geometric properties]
{Geometric properties of solutions to maximization
problems}
\author[F. Cuccu, K. Jha \& G. Porru\hfil EJDE--2003/71\hfilneg]
{Fabrizio Cuccu, Kanhaiya Jha, \& Giovanni Porru}
\address{Fabrizio Cuccu \newline
Mathematics Department, Universit\`a di Cagliari\\
via Ospedale 72 \\
09124 Cagliari, Italy}
\email{fcuccu@unica.it}
\address{Kanhaiya Jha \newline
Kathmandu University, Mathematics Department\\
Kathmandu, Nepal}
\email{jhakn@ku.edu.np}
\address{Giovanni Porru \newline
Mathemaiics Department, Universit\`a di Cagliari\\
via Ospedale 72 \\
09124 Cagliari, Italy}
\email{porru@unica.it}
\date{}
\thanks{Submitted December 2, 2002. Published June 24, 2003.}
\thanks{Partially supported by ``Contributo regionale Legge N. 43"}
\subjclass[2000]{35J25, 49K20, 49J20}
\keywords{Functionals, maximization, symmetry}
\begin{abstract}
We investigate the geometric configuration of the maxima
of some functionals associated with solutions to Dirichlet problems
for special elliptic equations. We also discuss the symmetry breaking
and symmetry preservation of the solutions in some particular cases.
\end{abstract}
\maketitle
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\section{Introduction}
Let $\Omega\subset {\mathbb R}^N$ be a bounded domain with a smooth
boundary $\partial \Omega$ and let $D\subset\Omega$ be Lebesgue measurable.
Consider the Dirichlet problem
\[ \begin{gathered}
-\Delta u(x)=\chi_D(x)\quad \hbox{in }\Omega,\\
u(x)=0\quad \hbox{on } \partial\Omega,
\end{gathered} \eqno(1.1)
\]
where $\chi_D(x)=1$ if $x\in D$ and $\chi_D(x)=0$ if $x\in \Omega\setminus D$.
Since $\chi_D(x)$ is not continuous, (1.1) is understood
in the weak sense. By standard results on elliptic equations, problem (1.1)
has a unique solution $u\in H^2(\Omega)\cap C^1(\Omega)\cap
C^0(\overline\Omega)$ [14]. Of course, the solution does not
change if $D$ is replaced by a new set which differs from $D$ by a
subset of measure zero.
In a previous paper [8], we have introduced and discussed the maximization
problem
$$
\max_{|G|=\alpha, |D|=\beta}\int_\Omega \chi_Gu_Ddx, \eqno(1.2)$$
where $G\subset \Omega$, $0<\alpha\le |\Omega|$, $0<\beta<|\Omega|$ and
$u_D$ is the solution to problem (1.1). The sets $D$ and $G$ are
defined apart subsets of measure zero. In [8]
we have found a result of existence for general $\Omega$, $\alpha$, $\beta$,
and a result of uniqueness in case $\Omega$ is a ball or, for general $\Omega$,
in case $\alpha=|\Omega|$. We also have proved that when
$\alpha=\beta$ problem (1.2) reduces to the maximization of the
energy integral
$$\max_{|D|=\beta}\int_\Omega |\nabla u_D|^2dx, \eqno(1.3)
$$
extensively investigated in [1,5,6,7,9,11].
In section 2 of the present paper, we shall prove that if $\beta\le\alpha$ and
$(D,G)$ is a solution to problem (1.2) then $D\subset G$. As a consequence,
for $\beta=\alpha$ we must have $D=G$. This special case has been
investigated in [8] by using a different argument. The interest of our result
relies in that the solution of problem (1.2) is not unique in general.
In section 3 we shall consider a special example to show that if
$\Omega$ is symmetric, a solution $(D,G)$ to problem (1.2) may not
be symmetric. This phenomenon, known as symmetry breaking, was
already observed in [6,9,11] for problem (1.3). Of course, in this
situation we have multiple solutions.
In section 4 we prove that if $\Omega$ is Steiner symmetric and if
$(D,G)$ is a solution to problem (1.2) then both $D$ and $G$ are
Steiner symmetric. This fact was already observed in [8] by using
a result described in [2]. Here we use a different approach which
may have independent interest.
Several open problems remain. One is the uniqueness of the solution to (1.2)
for larger classes of domains $\Omega$ and general $\alpha$ and $\beta$. We
think that the convexity of $\Omega$ should be sufficient to have uniqueness.
Another problem is the investigation of the shape of the optimal
pair $(D,G)$ in case uniqueness holds. We believe that $D$ and $G$ are
convex when $\Omega$ is convex.
A physical model of problem (1.2) is described in [8].
Many others models leading to equation (1.1) and its
generalizations are discussed in [10].
\section{Geometric properties}
Problem (1.2) with $\alpha=|\Omega|$ reduces to
$$
\max_{|D|=\beta}\int_\Omega u_Ddx=\max_{|D|=\beta}\int_Dw\,
dx,$$
where $w=w(x)$ is the solution to the Saint-Venant problem
\begin{gather*}
-\Delta w(x)=1\quad\hbox{in } \Omega,\\
w(x)=0\quad \hbox{on }\partial\Omega.
\end{gather*}
The maximizing domain $D$ is unique in this case and can be expressed as
$D=\{x\in \Omega:w(x)>t\}$ for a particular $t$ (see [8]).
Therefore, from now on we consider $\alpha<|\Omega|$. We state now
our main result of this section.
\begin{theorem} Let $|\Omega|>\alpha\ge \beta>0$
and let $(D,G)$ be a solution to problem $(1.2)$. Then $D\subset
G$. Moreover, there are positive numbers $t\le \tau$ and positive functions
$u_D(x)\le u_G(x)$ such that $D=\{x\in \Omega:u_G(x)>\tau\}$ and
$G=\{x\in \Omega:u_D(x)>t\}$ up to sets of measure zero.
\end{theorem}
\begin{proof} Let $(D,G)$ be a solution to problem $(1.2)$ and
let $u_D$ and $u_G$ satisfy
\[ \begin{gathered}
-\Delta u_D(x)=\chi_D(x)\quad \hbox{in } \Omega,\\
u_D(x)=0\quad\hbox{on } \partial\Omega,
\end{gathered} \eqno(2.1)
\]
\[ \begin{gathered}
-\Delta u_G(x)=\chi_G(x)\quad \hbox{in } \Omega,\\
u_G(x)=0\quad\hbox{on } \partial\Omega.
\end{gathered} \eqno(2.2)
\]
By [8] we know that
$$\{u_D(x)>t\}\subset G\subset \{u_D(x)\ge t\}, \eqno(2.3)$$
$$\{u_G(x)>\tau\}\subset D\subset \{u_G(x)\ge \tau\} \eqno(2.4)$$
for some non negative $t$, $\tau$. Here and in the sequel, we
denote by $\{u(x)>t\}$ the set $\{x\in \Omega:u(x)>t\}$.
Since $|D|>0$ (and $\Omega$ is connected) we have $u_D(x)>0$ in $\Omega$.
If we had $t=0$ then, by (2.3), we would have $G=\Omega$. But this contradicts
the hypothesis $\alpha<|\Omega|$. Similarly, one shows that $\tau>0$.
Let us prove that
$$
\int_\Omega |\nabla(u_G-u_D)|^2dx\le (\tau-t)(\alpha-\beta). \eqno(2.5)$$
Indeed, subtracting equation (2.1) from equation (2.2), multiplying by $(u_G-u_D)$
and integrating we find
$$
\int_\Omega |\nabla(u_G-u_D)|^2dx=\int_\Omega
(u_G-u_D)(\chi_G-\chi_D)dx$$
$$=\int_{G\setminus D}(u_G-u_D)dx+\int_{D\setminus
G}(u_D-u_G)dx.$$
Using (2.3) and (2.4) we find
$$\int_{G\setminus D}(u_G-u_D)dx\le\int_{G\setminus D}(\tau-t)dx=(\tau-t)|G\setminus D|$$
and
$$\int_{D\setminus G}(u_D-u_G)dx\le\int_{D\setminus
G}(t-\tau)dx=(t-\tau)|D\setminus G|.$$
Since $|G\setminus D|=|G|-|D\cap G|$ and $|D\setminus G|=|D|-|D\cap G|$,
inequality (2.5) follows. Recalling that $\alpha\ge\beta$, (2.5) implies that $t\le\tau$.
Introduce the subsets of $\Omega$
\begin{gather*}
\Omega_1=\{u_G(x)-u_D(x)>\tau-t\},\\
\Omega_2=\{u_G(x)-u_D(x)=\tau-t\},\\
\Omega_3=\{u_G(x)-u_D(x)<\tau-t\}.
\end{gather*}
Of course, $\Omega_1\cup\Omega_2\cup\Omega_3=\Omega$. By
(2.4), $u_G(x)\le \tau$ outside $D$, and by (2.3),
$u_D(x)\ge t$ in $G$.
Hence, $u_G(x)-u_D(x)\le\tau-t$ in $G\setminus D$. Therefore,
$$G\setminus D\subset\Omega_2\cup\Omega_3=\Omega\setminus\Omega_1.$$
The last inclusion yields
$$\Omega_1\subset D\cup(\Omega\setminus G). \eqno(2.6)$$
On the other side, using equations (2.1)-(2.2) we find
$$-\Delta(u_G-u_D)=\chi_G-\chi_D\le 0\quad \hbox{in }
D\cup(\Omega\setminus G). \eqno(2.7)
$$
By (2.6), inequality (2.7) holds in $\Omega_1$. Since
$u_G(x)-u_D(x)=\tau-t$ on the boundary of $\Omega_1$, by the
maximum principle, we get $u_G(x)-u_D(x)\le \tau-t$ in $\Omega_1$. Recalling the
definition of $\Omega_1$, we conclude that this set must be empty.
By (2.4), $u_G(x)\ge \tau$ in $D$, and by (2.3), $u_D(x)\le t$ outside $G$.
Hence, $u_G(x)-u_D(x)\ge\tau-t$ in $D\setminus G$. Therefore,
since $\Omega_1$ is empty,
$$D\setminus G\subset\Omega_1\cup\Omega_2=\Omega_2.$$
On $\Omega_2$, $u_G(x)-u_D(x)=\tau-t$, therefore
$\Delta(u_G-u_D)=0$ almost everywhere in $D\setminus G$. On the other
side, by using equations (2.1)-(2.2) once more, we get $\Delta(u_G-u_D)=1$
on $D\setminus G$. We conclude that the measure of $D\setminus G$
must be zero, hence, $D\subset G$ up to a set of measure zero. The
first assertion of the theorem is proved.
We have $\Delta u_G=0$ almost everywhere on the set $\{u_G(x)=\tau\}\cap D$.
On the other side, $\Delta u_G=-1$ in $G$. Since $D\subset G$, the
set $\{u_G(x)=\tau\}\cap D$ must have measure zero. Therefore, by
(2.4), $D=\{u_G(x)>\tau\}$ up to a set of measure zero.
Decompose the set $\{u_D(x)=t\}\cap G$ into
$E_1=\{u_D(x)=t\}\cap G\cap D$, $E_2=\{u_D(x)=t\}
\cap G\cap (\partial D)$, $E_3=\{u_D(x)=t\}\cap G\cap(\Omega\setminus\overline D)$.
$E_1$ has measure zero because $\Delta u_D=-1$ in $D$, therefore $u_D$ cannot
be constant on a set of positive measure. $E_2$ has
measure zero because $u_G(x)=\tau$ on $\partial D$ and $\Delta u_G=-1$
on $G$. $E_3$ has measure zero because the function $u_D(x)$ is harmonic
(and positive) in the open set $\Omega\setminus\overline D$ and $u=0$
on $\partial \Omega$. Therefore, by (2.3),
we must have $G=\{u_D(x)>t\}$ up to a set of measure zero.
The theorem is proved.
\end{proof}
\noindent\textbf{Remarks.} By (2.1), (2.2) and Theorem 2.1, the functions
$u=u_D$ and $v=u_G$ satisfy the equations
$$
-\Delta u=H(v-\tau),\quad u|_{\partial\Omega}=0, \eqno(2.8)$$
$$
-\Delta v=H(u-t),\quad v|_{\partial\Omega}=0, \eqno(2.9)
$$
where $H(s)=0$ for $s\le 0$ and $H(s)=1$ for $s> 0$.
The system (2.8)-(2.9) may have solutions different
from $u_D,\ u_G$ even when $\alpha=\beta$. Indeed, if
$\alpha=\beta$ then $u_D=u_G=u$, and $u$ satisfies
$$
-\Delta u=H(u-t),\quad u|_{\partial\Omega}=0. \eqno(2.10)
$$
If $\Omega$ is a thin annulus and $\beta$ is small enough then problem
(1.3) has a non radial solution $u=u_D$ which satisfies (2.10) [9,11].
Let $w(x)$ be the (radial) solution to the Saint-Venant problem
associated with $\Omega$. Using the method of monotone operators
(starting from $w$) we find a radial solution $v=v(x)$ to (2.10) with
$u_D(x)\le v(x)\le w(x)$. Of course, $u_D(x)\not = v(x)$ because $u_D(x)$
is non radial.
\section{Symmetry breaking}
In [6,9,11] it was shown the symmetry breaking of the solution
to problem (1.2) in case $\alpha=\beta$. Now, we examine an
example to discuss the case $\alpha\not=\beta$.
Recall that if $\Omega$ is a ball then the
maximum of (1.2) is reached when $D$ and $G$ are balls concentric
with $\Omega$ [8]. Let $B_1$ and $B_2$ be open unit balls in
$\mathbb R^2$ centered at $(-2,0)$ and $(2,0)$ respectively, and
let $\Omega=B_1\cup B_2$.
Let $D=D_1\cup D_2$ with $D_1$ a ball concentric with $B_1$ and radius
$R$, and $D_2$ a ball concentric with $B_2$ and radius $S$. Similarly, let
$G=G_1\cup G_2$ with $G_1$ a ball concentric with $B_1$ and radius $T$,
and $G_2$ a ball concentric with $B_2$ and radius $Q$. We have
$|\Omega|=2\pi$, $|D|=\pi(R^2+S^2)$ and $|G|=\pi(T^2+Q^2)$.
Assume
$$R^2+S^2=b,\quad T^2+Q^2=a,\quad b\le a\le 2.
$$
We study the problem
$$\max_{|D|=\pi b,\; |G|=\pi a}\int_G u_Ddx \eqno(3.1)
$$
with $b/2\le R^2\le \min[1,b]$ and $a/2\le T^2\le \min[1,a]$.
Since $b\le a$, by Theorem 2.1, the maximum in (3.1) is attained
when $D\subset G$. Therefore, we may suppose $R\le T$. If $u=u_D$ is
the corresponding solution to problem (1.1) we find
$$u(r)=\begin{cases}
\frac{R^2}{4}-\frac{r^2}{4}-\frac{R^2}{2}\log R & 0\le r0:\hbox{\rm d($x$)}\mu\},\quad\Sigma_h(\mu)
=\Sigma(\mu)\cap\{x\in\Omega: \hbox{\rm d($x$)}0$ in $\Omega$, we have $w(x)\le 0$ on
$\partial\Sigma(\mu)$ with $w(x)< 0$ at some point of the boundary of
each connected component. We conclude
that $w(x)<0$ on $\Sigma(\mu)$ for such values of $\mu$. The same
conclusion holds for $z(x)$. Recall that $w(x)$ and $z(x)$ depend on
$\mu$.
Let $(m,M)$ be the largest interval of $\mu$ such that both
$$w(x)<0\quad \hbox{\rm and}\quad z(x)<0 $$
hold on $\Sigma(\mu)$. By contradiction, assume $m>0$.
Since $w$ and $z$ are continuous with respect to $\mu$, we have
$$w(x)\le 0\quad \hbox{\rm and}\quad z(x)\le 0 \quad \forall x\in \Sigma(m).$$
Then, $\Delta w\ge 0$ and $\Delta z\ge 0$ on $\Sigma(m)$ by (4.9).
The strong maximum principle ([12] Theorem 2.13) and the assumption $m>0$
yield
$$w(x)<0\quad \hbox{\rm and}\quad z(x)<0 \quad \forall x\in \Sigma(m).$$
The boundary point Lemma ([12] Lemma 2.12) applied to the flat boundary
$x_1=\mu$ of $\Sigma(\mu)$, $m\le \mu\le M$ yields
$$\frac{\partial w}{\partial x_1}<0\quad \hbox{\rm and}\quad \frac{\partial z}{\partial x_1}<0
\quad \forall x\in \overline{\Sigma(m)}\setminus \partial\Omega.$$
Recalling the definition of $w$ and $z$ we must have
$$\frac{\partial u}{\partial x_1}<0\quad \hbox{\rm and}\quad \frac{\partial v}{\partial x_1}<0
\quad \forall x\in \overline{\Sigma(m)}\setminus \partial\Omega.\eqno(4.10)$$
Following again the argument described in [12] (pag. 97), for
$\epsilon>0$ and $\tau>0$ small, choose a set
$$E_\epsilon=(m-\epsilon,m+\tau]\times S,\ \ S\subset
\mathbb R^{N-1}, \ \ E_\epsilon\subset \Sigma(m-\epsilon)$$
as well as a compact subset $F\subset \Sigma(m)$. Using (4.10)
one proves that $w(x)$ and $z(x)$ are strictly negative on
$E_\epsilon$ provided $\{m\}\times S$ is a compact subset of
$\{x_1=m\}\cap\Omega$.
Let $G_\epsilon=\Sigma(m-\epsilon)\setminus(E_\epsilon\cup F)$.
$S$ and $F$ can be chosen so that, for $\epsilon$ small, $w(x)<0$ on $F$
and $G_\epsilon\subset \Sigma_h(m-\epsilon)$. Using the strong maximum principle
again one gets $w(x)<0$ and $z(x)<0$ on $\Sigma(m-\epsilon)$. This contradicts
the maximality of $(m,M)$ for the negativity of $w(x)$ or $z(x)$.
We conclude that $m=0$. Hence, $u(x_1,y)\le u(-x_1,y)$ and $v(x_1,y)\le v(-x_1,y)$
on $\Sigma(0)$. Repetition of the same proof starting from the left side of
$\Omega$ leads to the inequalities $u(x_1,y)\ge u(-x_1,y)$ and $v(x_1,y)\ge v(x_1,y)$
on $\Sigma(0)$, and the theorem follows.
\end{proof}
\noindent\textbf{Remark.} The result of Theorem 4.1 can be
extended the the more general system
\begin{gather*}
-\Delta u=h(u)+f(v),\quad u|_{\partial\Omega}=0,\\
-\Delta v=k(v)+g(u),\quad v|_{\partial\Omega}=0,
\end{gather*}
where $f$ and $g$ are as before, whereas $h$ and $k$ are locally Lipschitz
continuous in $(0,\infty)$. Indeed, in this case, instead of (4.5) one finds
$$
\Delta w+c_1(x,\mu)w\ge 0 \quad \hbox{\rm and}\quad \Delta z+c_2(x,\mu)z\ge
0\quad \forall x\in \Sigma_h(\mu),
$$
where $c_1(x,\mu)$ and $c_2(x,\mu)$ are bounded uniformly with
respect to $\mu$. The maximum principles for thin sets apply in this
situation [12].
Symmetry results for systems in case of smooth
functions are discussed in [18].
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\end{document}