\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 149, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2009/149\hfil
Symmetry in rearrangement optimization problems]
{Symmetry in rearrangement optimization problems}
\author[B. Emamizadeh, J. V. Prajapat\hfil EJDE-2009/149\hfilneg]
{Behrouz Emamizadeh, Jyotshana V. Prajapat} % in alphabetical order
\address{Behrouz Emamizadeh \newline
Department of Mathematics, The Petroleum Institute, Abu Dhabi, UAE}
\email{bemamizadeh@pi.ac.ae}
\address{Jyotshana V. Prajapat \newline
Department of Mathematics, The Petroleum Institute, Abu Dhabi, UAE}
\email{jprajapat@pi.ac.ae}
\thanks{Submitted October 28, 2009. Published November 25, 2009.}
\subjclass[2000]{49K20, 35P15, 35J10, 74K15}
\keywords{Minimization and maximization problems; rearrangements;
\hfill\break\indent principal eigenvalue; optimal solutions; symmetry}
\begin{abstract}
This article concerns two rearrangement optimization problems.
The first problem is motivated by a physical experiment in
which an elastic membrane is sought, built out of several materials,
fixed at the boundary, such that its frequency is minimal.
We capture some features of the optimal solutions, and prove
a symmetry property. The second optimization problem is motivated
by the physical situation in which an ideal fluid flows over a
seamount, and this causes vortex formation above the seamount.
In this problem we address existence and symmetry.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\section{Introduction}
In this article, we consider two rearrangement optimization
problems which are physically relevant. A rearrangement
optimization problem is referred to an optimization problem where
the admissible set is a rearrangement class, see section 2 for
precise definition. In both problems our focus will be on (radial)
symmetry. More precisely, we will show that when the physical
domain is a ball then the optimal solutions will be radial as
well.
The first problem is concerned with the following non-linear
eigenvalue problem:
\begin{equation}
\begin{gathered}
-\Delta_pu+V(x)|u|^{p-2}u =\lambda g(x)|u|^{p-2}u,\quad\text{in }\Omega \\
u=0\quad \text{on }\partial\Omega.
\end{gathered}\label{eigenvalue}
\end{equation}
Here $V$ and $g$ are given functions, and $\lambda$ is an eigenvalue.
There are some technical conditions on $V$ and $g$, but we prefer
to delay stating them until section 3. However, we mention that
in case $p=2$ and $g=1$, \eqref{eigenvalue} is the steady state
case of an elastic membrane, fixed around the boundary, made out
of various materials (this justifies placing $V$ in the differential
equation). The constant $\lambda$ denotes the frequency of the membrane.
There are infinitely many eigenvalues, but we are only interested
in the first one, often referred to as the principal eigenvalue,
and denote it by $\lambda (g,V)$ to emphasize its dependance on
$g$ and $V$. The following variational formulation for $\lambda(g,V)$
is well known:
\begin{equation}
\lambda(g,V)=\inf\big\{\int_\Omega(|\nabla u|^p+V(x)|u|^p)dx:
u\in W^{1,p}_0(\Omega),\;\int_\Omega g(x)|u|^pdx=1\big\}.
\label{eigenvaluev}
\end{equation}
Given $g_0$ and $V_0$, we are interested in the following rearrangement
optimization problem
\begin{equation} \label{P}
\inf_{g\in\mathcal{R}(g_0),\;V\in\mathcal{R}(V_0)}\lambda(g,V),
\end{equation}
where $\mathcal{R}(g_0)$ and $\mathcal{R}(V_0)$ are rearrangement
classes generated by $g_0$ and $V_0$, respectively.
This problem has already been considered in \cite{d1},
where amongst other results the authors show \eqref{P} is solvable,
see also \cite{d2,d3}. However, here we focus on some features
of optimal solutions, and prove a symmetry result.
The second problem considered in this paper is motivated by
fluids flowing over seamounts. More precisely, it is well accepted
that a two dimensional ideal fluid flowing over a seamount (e.g.
hills located at the bottom of an ocean) gives rise to vortex
formation located above seamounts, see \cite{n1}. Existence of such
flows turns out to be equivalent to existence of maximizers for
certain functionals, representing some kind of energy, which can
be formulated in terms of vorticity function and the height
function. Mathematically, here is the problem we are interested
in: Let us denote by $u_f\in W^{1,2}_0(\Omega)$ the unique
solution of the Poisson differential equation
\begin{gather*}
-\Delta u=f\quad \text{in }\Omega \\
u=0\quad\text{on }\partial\Omega.
\end{gather*}
Consider the energy functional:
\[
J(f,h)=\frac{1}{2}\int_\Omega fu_f\,dx+\int_\Omega hu_f\,dx.
\]
We are interested in the following rearrangement optimization
problem:
\begin{equation} \label{PP}
\sup_{f\in\mathcal{R}(f_0),\;h\in\mathcal{R}(h_0)}J(f,h).
\end{equation}
Here $f$ represents the vorticity function and $h$ the hight function
(seamount). Details related to \eqref{PP} are given in section 4,
where we discuss existence of optimal solutions for \eqref{PP}, and
address the question of symmetry. The reader can refer to
\cite{c1,c2}
for other examples of rearrangement optimization problems.
\section{Preliminaries}
In this section we review rearrangement theory with an eye on the
optimization problems \eqref{P} and \eqref{PP}. So we only mention
results that are going to assist us with the two optimization problems
in question. The reader can refer to \cite{b1,b2} for an extensive account
of rearrangement theory. Henceforth, we assume $\Omega$ is a smooth
bounded domain in $\mathbb{R}^N$, unless stated otherwise.
\noindent\textbf{Definition.}
Two functions $f:(X,\Sigma_1,\mu_1)\to \mathbb{R}$,
$g:(Y,\Sigma_2,\mu_2)\to \mathbb{R}$ are said to be rearrangements
of each other if:
\[
\mu_1(\{x\in X: f(x)\geq\alpha\})
=\mu_2(\{x\in Y: g(x)\geq\alpha\}),\quad \forall\alpha\in\mathbb{R}.
\]
In case $\mu_i$ stands for the Lebesgue measure in $\mathbb{R}^N$,
we replace it with $|\cdot|$. When $f$ and $g$ are rearrangements
of each other we write $f\sim g$. For a fixed
$f_0:(X,\Sigma,\mu)\to \mathbb{R}$, the class of rearrangements
generated by $f_0$, denoted $\mathcal{R}(f_0)$, is defined as follows:
\[
\mathcal{R}(f_0)=\{f: f\sim f_0\}.
\]
In case $\Omega$ is a ball in $\mathbb{R}^N$, say centered at the
origin, and $f:\Omega\to \mathbb{R}$ is a Lebesgue measurable function
then $f^*:\Omega\to \mathbb{R}$ and $f_*:\Omega\to \mathbb{R}$
denote the Schwarz decreasing and increasing rearrangements of $f$.
That is, $f^*\sim f$, $f_*\sim f$, and $f^*$ is a radial function
which is decreasing as a function of $r:=\|x\|$, whereas $f_*$
is a radial function which is increasing as a function of $r$.
We will use two well known rearrangement inequalities.
\begin{lemma}[\cite{h1}]\label{lem1}
Suppose $\Omega$ is a ball in $\mathbb{R}^N$.
Then
\begin{equation}
\int_\Omega f^*g_*\,dx\leq\int_\Omega fg\,dx\leq\int_\Omega
f^*g^*\,dx, \label{hardy}
\end{equation}
where $f$ and $g$ are non-negative functions.
\end{lemma}
\begin{lemma}[\cite{b4}] \label{lem2}
Suppose $\Omega$ is a ball in $\mathbb{R}^N$. For $0\leq u\in
W^{1,p}_0(\Omega)$, $u^*\in W^{1,p}_0(\Omega)$, and
\begin{equation}
\int_\Omega |\nabla u|^p\,dx\geq\int_\Omega |\nabla u^*|^p\,dx.
\label{polya}
\end{equation}
If the equality holds in \eqref{polya}, and the set
$\{x\in\Omega: \nabla u(x)=0,\;0__0,\;\|V\|_\infty<\frac{A}{C_p(A+\|g\|_\infty)}\},
\]
where $A$ is a positive constant, and $C_p$ is the constant in the
Poincar\`{e} inequality:
\[
\int_\Omega |u|^p\,dx\leq
C_p\int_\Omega |\nabla u|^p\,dx,\quad u\in W^{1,p}_0(\Omega).
\]
Next, we fix $(g_0,V_0)\in \mathcal{S}$, and let $\mathcal{R}(g_0)$
and $\mathcal{R}(V_0)$ to denote the rearrangement classes
generated by $g_0$ and $V_0$, respectively. Note that from the
definition of rearrangements, it readily follows that
$\mathcal{R}(g_0)\times \mathcal{R}(V_0)\subset \mathcal{S}$.
The question of existence of (optimal) solutions of \eqref{P} has
already been addressed in \cite{d1,d2,d3}; where amongst other results
the following result is proved.
\begin{lemma}\label{lem5}
(a) Problem \eqref{P} is solvable; that is, there exists
$(\hat g,\hat V)\in \mathcal{R}(g_0)\times \mathcal{R}(V_0)$ such that
\[
\lambda(\hat g,\hat V)=\inf_{g\in\mathcal{R}(g_0),\;V\in\mathcal{R}
(V_0)}\lambda(g,V).
\]
(b) If $(\hat g,\hat V)$ is an optimal solution of \eqref{P}, then
\begin{gather} \label{euler1}
\hat g=\phi(\hat u),\quad\text{a.e. in }\Omega, \\
\hat V=\psi(\hat u),\;\;\text{a.e. in}\;\Omega, \label{euler2}
\end{gather}
where $\phi$ and $\psi$ are increasing and decreasing functions
unknown a priori. Here $\hat u$ stands for the unique eigenfunction
corresponding to $\lambda(\hat g,\hat V)$.
\end{lemma}
Let us point out some consequences of \eqref{euler1} and \eqref{euler2}
before addressing the question of symmetry.
\begin{lemma} \label{lem6}
Suppose $(\hat g,\hat V)$ is an optimal solution of \eqref{P}. Then
\begin{itemize}
\item[(a)] The function $\hat u$ attains its smallest values on
the support of $\hat V:=\{x\in\Omega: \hat V>0\}$.
In fact, for some $t>0$,
\begin{equation}
\{x\in\Omega: \hat V>0\}=\{x\in\Omega: \hat u0\}=\hat V^{-1}(0,\infty)=\hat
u^{-1}(\psi^{-1}(0,\infty)).
\]
Since $\psi$ is decreasing,
$\psi^{-1}(0,\infty)$ must be an interval of the form
$(-\infty,t)$ or $(-\infty,t]$, for some $t\in\mathbb{R}$. Clearly the
assertion is proved once we show that the set
$E:=\{x\in\Omega: \hat u=t\}$ has zero measure, since it is
obvious that $t$ can not be a non-positive constant. Let us assume
the contrary and derive a contradiction. Specializing the
differential equation \eqref{eigenvalue} to the set $E$, we get:
\[
\hat V(x)\hat u^{p-1}=\lambda(\hat g,\hat V)\hat g(x)\hat
u^{p-1},\quad \text{in }E.
\]
Since $\hat u$ is positive in
$\Omega$, in turn, we get $\hat V(x)=\hat \lambda \hat g(x)$,
where we have replaced $\lambda(\hat g,\hat V)$ by $\hat \lambda$,
for simplicity. We show that this last equation can not hold. To
see this observe that:
\begin{equation}
\hat\lambda=\int_\Omega(|\nabla\hat
u|^p+\hat V(x)\hat u^p)\,dx\geq \int_\Omega|\nabla\hat
u|^p\,dx.\label{inequality2}
\end{equation}
On the other hand, since $\hat u$
is a normalized function, we have
\[
1=\int_\Omega\hat g\hat u^p\,dx\leq\|g_0\|_\infty,\quad
\int_\Omega\hat u^p\,dx\leq C_p\|g_0\|_\infty,\quad
\int_\Omega|\nabla\hat u|^p\,dx,
\]
hence
\begin{equation} \label{inequality3}
\int_\Omega|\nabla\hat u|^p\,dx\geq\frac{1}{C_p\|g_0\|_\infty}.
\end{equation}
Thus, from (\ref{inequality2}) and (\ref{inequality3}), we infer
\[
\hat\lambda\geq\frac{1}{C_p\|g_0\|_\infty}.
\]
This, in turn, implies $\hat \lambda\hat g-\hat V>0$, thanks to
the facts that $\hat g\geq A$, and
$\|V\|_\infty<\frac{A}{C_p(A+\|g\|_\infty)}$.
This completes the proof of part (a).
(b) Since $\hat u=0$ on $\partial\Omega$, and
$\hat u\in C(\overline{\Omega})$, it follows from part (a) that
$\{x\in\Omega: \hat V(x)>0\}$ contains a tubular domain around
$\partial\Omega$. To show that $\{x\in\Omega: \hat V(x)>0\}$
is connected, it suffices to show that the boundary of every
component of the support of $\hat V$ must intersect $\partial\Omega$.
To derive a contradiction, we assume the contrary. Let us assume
$E$ is a component of the support of $\hat V$ such that
$\partial E\cap\partial\Omega$ is empty. From part (a), we observe
that $\partial E\subseteq \{x\in\Omega: \hat u=t\}$, for some
positive $t$. We set $w(x):=\hat u(x)-t$, so
$-\Delta_pw=-\Delta_p\hat u=(\hat\lambda\hat g-\hat V){\hat u}^{p-1}>0$,
in $E$. In addition, $w(x)=0$, for $x\in\partial E$.
Therefore, by the strong maximum principle, we derive $w(x)>0$, in $E$.
Hence, $\hat u>t$ in $E$, which is a contradiction to the assertion
in part (a).
(c) We only show $\phi=\hat g^\Delta(\mu_{\hat u})$, since
$\psi=\hat g_\Delta(\mu_{\hat u})$ is proved similarly. As in the proof
of part (a), one can show that the graph of $\hat u$ has no flat
sections; that is, sets of the form $\{x\in\Omega: \hat u=\beta\}$
have zero measure. Thus, $\hat u^\Delta$ is strictly decreasing;
hence, the inverse of $\hat u^\Delta$ exists and coincides
with $\mu_{\hat u}$. On the other hand, from \eqref{euler1},
we infer $\hat g^\Delta={\phi(\hat u)}^\Delta$.
But, ${\phi(\hat u)}^\Delta=\phi(\hat u^\Delta)$, since
$\phi(\hat u)\sim\phi(\hat u^\Delta)$. Thus, we have
$\hat g^\Delta=\phi(\hat u^\Delta)$, hence
$\hat g^\Delta(\mu_{\hat u})=\phi(\hat u^\Delta\circ\mu_{\hat u})=\phi$.
\end{proof}
Now we state the main result of this section.
\begin{theorem} \label{thm1}
Suppose $\Omega$ is a ball centered at the origin. Suppose that
$(\hat g,\hat V)$ is an optimal solution of \eqref{P}. Then
\begin{equation}
\hat g={g_0}^*,\quad \hat V=(V_0)_*,\label{optimal}
\end{equation}
modulo sets of measure zero.
\end{theorem}
\begin{proof}
Again we write $\hat\lambda$ in place of
$\lambda(\hat g,\hat V)$. Recall
$\int_\Omega \hat g\hat u^p\,dx=1$, hence
$1\leq\int_\Omega {\hat g}^*(\hat u^*)^p\,dx=:\gamma$.
Let $v=\gamma^{-1/p}{\hat u}^*$, so
$\int_\Omega \hat g^*v^p\,dx=1$. We also have
\begin{equation}
\begin{aligned}
\hat\lambda
&=\int_\Omega |\nabla\hat u|^p\,dx+\int_\Omega \hat V(x)\hat u^p\,dx\\
&\geq \int_\Omega |\nabla\hat u^*|^p\,dx+\int_\Omega \hat V(x)\hat u^p\,dx \\
&\geq \int_\Omega |\nabla\hat u^*|^p\,dx+\int_\Omega \hat V_*(x)({\hat u}^*)^p\,dx \\
&= \gamma \Big(\int_\Omega|\nabla v|^p\,dx+\int_\Omega
\hat V_*v^p\,dx\Big) \\
&\geq \int_\Omega|\nabla v|^p\,dx+\int_\Omega \hat V_*v^p\,dx\\
&\geq\lambda(\hat g^*,\hat V_*)\geq\hat\lambda,
\end{aligned} \label{inequality8}
\end{equation}
where in the first inequality we
have used Lemma \ref{lem2}, and in the second one we have used
Lemma \ref{lem1}.
Therefore all inequalities in (\ref{inequality8}) are in fact
equalities. In particular, we infer
\begin{gather*}
\int_\Omega |\nabla \hat u|^p\,dx=\int_\Omega|\nabla \hat u^*|^p\,dx,\\
\int_\Omega \hat V\hat u^p\,dx=\int_\Omega \hat V_*(\hat
u^*)^p\,dx,\\
\int_\Omega \hat g\hat u^p\,dx=\int_\Omega {\hat g}^*(\hat u^*)^p\,dx.
\end{gather*}
To complete the proof, by \cite[Lemma 2.9 and Lemma 2.4 (ii)]{b2},
it suffices to show that $\hat u=\hat u^*$.
From $\int_\Omega |\nabla \hat u|^p\,dx=\int_\Omega|\nabla \hat
u^*|^p\,dx$, in conjunction with Lemma \ref{lem2}, we need to show the set
$Q=\{x\in\Omega: \nabla\hat u=0,\;0<\hat u(x)0$, in $N$, thanks to the fact
that $(\hat g,\hat V)\in \mathcal{S}$. Also, $\partial N\subset
\{x\in\Omega: \hat u(x)=\hat u(x_0)\}$. So by the strong maximum
principle we see that $w>0$, in the interior of $N$. Since
$w(x_0)=0$, we can apply the Hopf boundary point lemma to conclude
that $\frac{\partial w}{\partial\nu}(x_0)<0$, where $\nu$ stands
for the outward unit normal to $\partial N$ at $x_0$. This, in
turn, implies that $\frac{\partial \hat u}{\partial\nu}(x_0)<0$,
hence $\nabla \hat u(x_0)\neq 0$. So $Q$ is empty, as
desired.
\end{proof}
\section{Problem \eqref{PP}}
In this section we study problem \eqref{PP}, where we address both
questions of existence and symmetry. Henceforth we assume
$1\leq p<\infty$, and $p>\frac{2N}{N+2}$, where $N\geq 2$ is the space
dimension. We fix two non-negative functions $f_0$ and $h_0$ in
$L^p(\Omega)$, and as before let $\mathcal{R}(f_0)$ and
$\mathcal{R}(h_0)$ denote rearrangement classes generated by $f_0$
and $h_0$, respectively. The energy functional
$J:L^p(\Omega)\times L^p(\Omega)\to \mathbb{R}$ is defined by:
\[
J(f,h)=\frac{1}{2}\int_\Omega fu_f\,dx+\int_\Omega hu_f\,dx.
\]
Note that $J$ is finite. Indeed, for $f,g\in L^p(\Omega)$,
\[
|J(f,h)|\leq \frac{1}{2} \|f\|_p\|u_f\|_q+\|h\|_p\|u_f\|_q,
\]
where $q$ is the conjugate of $p$; that is, $q=\frac{p}{p-1}$. An
application of the embedding $W^{1,2}_0(\Omega)\to
L^q(\Omega)$, since $p>\frac{2N}{N+2}$, implies
\begin{equation}
|J(f,h)|\leq\frac{1}{2}C\|f\|_p^2+C\|h\|_p\|f\|_p, \label{finite}
\end{equation}
where $C$ is a positive constant, thus $J$ is finite. Note
that from (\ref{finite}), we readily infer that $J$ is bounded on
$\mathcal{R}(f_0)\times \mathcal{R}(h_0)$. Now we prove problem
\eqref{PP} is solvable.
\begin{theorem} \label{thm2}
Problem \eqref{PP} is solvable; that is, there exists
$(\overline{f},\overline{h})\in \mathcal{R}(f_0)\times \mathcal{R}(h_0)$ such that
\begin{equation}
J(\overline{f},\overline{h})=\sup_{f\in\mathcal{R}(f_0),\;h\in
\mathcal{R}(h_0)}J(f,h)=:I.
\label{existence}
\end{equation}
\end{theorem}
\begin{proof}
Let $(f_n,h_n)\in \mathcal{R}(f_0)\times
\mathcal{R}(h_0)$ be a maximizing sequence. Since
$\|f_n\|_p=\|f_0\|_p$ and $\|h_n\|_p=\|h_0\|_p$, we infer
existence of $\hat f\in \overline{\mathcal{R}}(f_0)$ and $\hat h\in
\overline{\mathcal{R}}(h_0)$ such that
\[
f_n\rightharpoonup \hat f,\quad
h_n\rightharpoonup\hat h,
\]
where ``$\rightharpoonup$" stands for weak convergence in $L^p(\Omega)$.
Hence we obtain $I=J(\hat f,\hat h)$, since $J$ is weakly
sequentially continuous in $L^p(\Omega)\times L^p(\Omega)$. This,
in particular, implies that $J(\hat f,\hat h)\geq J(\hat f,h)$,
for every $h\in \mathcal{R}(h_0)$. Thus,
\[
\frac{1}{2}\int_\Omega \hat fu_{\hat f}\,dx+\int_\Omega \hat
hu_{\hat f}\,dx\geq \int_\Omega \hat fu_{\hat f}\,dx+\int_\Omega
hu_{\hat f}\,dx,\quad \forall h\in \mathcal{R}(h_0).
\]
So, $\int_\Omega \hat hu_{\hat f}\,dx\geq\int_\Omega hu_{\hat f}\,dx$,
for every $h\in\mathcal{R}(h_0)$. That is to say, $\hat h$
maximizes the linear functional
$l(h):=\int_\Omega hu_{\hat f}\,dx$, relative to
$h\in\mathcal{R}(h_0)$. Hence an application
of Lemma \ref{lem3} implies existence of $\overline{h}\in \mathcal{R}(h_0)$ such
that $l(\overline{h})\geq l(h)$, relative to $h\in\mathcal{R}(h_0)$. By
continuity of $l: L^p(\Omega)\to \mathbb{R}$, we obtain
$l(\overline{h})\geq l(h)$, for every $h\in \overline{\mathcal{R}}(h_0)$, the weak
closure of $\mathcal{R}(h_0)$ in $L^p(\Omega)$. In particular, we
deduce $l(\overline{h})\geq l(\hat h)$, hence we must have
$l(\overline{h})=l(\hat h)$. Whence
\[
J(\hat f,\hat h)=\frac{1}{2}\int_\Omega
\hat fu_{\hat f}\,dx+\int_\Omega \hat hu_{\hat f}\,dx=\int_\Omega
\hat fu_{\hat f}\,dx+\int_\Omega \overline{h}u_{\hat f}\,dx.
\]
Next, we introduce $\Phi : L^p(\Omega)\to \mathbb{R}$ by
\[
\Phi(f)=\frac{1}{2}\int_\Omega fu_f\,dx+\int_\Omega \overline{h}u_f\,dx.
\]
It is easy to check that $\Phi$ is strictly convex, weakly
sequentially continuous in $L^p(\Omega)$. Hence, by Lemma \ref{lem4},
$\Phi$ has a maximizer, say $\overline{f}$, relative to
$\mathcal{R}(f_0)$. By weak continuity of $\Phi$ we deduce $\Phi
(\overline{f})\geq\Phi (\hat f)$. Thus,
\[
J(\overline{f},\overline{h})=\Phi (\overline{f})\geq\Phi (\hat f)
= \frac{1}{2}\int_\Omega \hat fu_{\hat f}\,dx
+\int_\Omega \overline{h}u_{\hat f}\,dx
= J(\hat f,\hat h)\geq J(\overline{f},\overline{h}).
\]
Therefore, $J(\overline{f},\overline{h})=I$. Hence
$(\overline{f},\overline{h})\in \mathcal{R}(f_0)\times \mathcal{R}(h_0)$ is
an optimal solution of the problem \eqref{PP}, as desired.
\end{proof}
\begin{corollary} \label{coro4.1}
Let $(\overline{f},\overline{h})$ be an optimal solution of \eqref{PP},
and assume $|\{x\in: f_0(x)>0\}|<|\Omega|$.
Then $\partial\{x\in\Omega: \overline{f}(x)>0\}$, boundary of the support
of $\overline{f}$, does not intersect $\partial\Omega$.
\end{corollary}
\begin{proof}
Since $(\overline{f},\overline{h})$ is a solution of \eqref{PP}, we deduce,
in particular, that $J(\overline{f},\overline{h})\geq J(f,\overline{h})$, for every
$f\in \mathcal{R}(f_0)$. That is, $\overline{f}$ maximizes $J(f,\overline{h})$
relative to $f\in \mathcal{R}(f_0)$. The functional $J(\cdot, \overline{h})$
is strictly convex, and weakly sequentially continuous in $L^p(\Omega)$.
For fixed $g\in L^p(\Omega)$, and $t>0$, it is straightforward to obtain
\begin{align*}
J(\overline{f}+tg,\overline{h})
&= J(\overline{f},\overline{h})+t\int_\Omega \overline{f}u_g\,dx+\frac{1}{2} t^2
\int_\Omega gu_g\,dx+t\int_\Omega \overline{h} u_g\,dx \\
&= J(\overline{f},\overline{h})+t\int_\Omega (\overline{f}+\overline{h})u_g\,dx+\frac{1}{2}t^2\int_\Omega gu_g\,dx.
\end{align*}
This, in turn, implies, $\partial_1J(\overline{f},\overline{h})$, the subdifferential
of $J$ at $\hat f$, for fixed $\overline{h}$, can be identified with
$u_{\overline{f}+\overline{h}}$; note that this is a consequence of the
following symmetry:
\[
\int_\Omega (\overline{f}+\overline{h})u_g\,dx=\int_\Omega gu_{\overline{f}+\overline{h}}\,dx.
\]
Now we can apply
Lemma \ref{lem4} to deduce that $\overline{f}=\phi(u_{\overline{f}+\overline{h}})$, almost
everywhere in $\Omega$, for some increasing function $\phi$.
Hence, the largest values of $u_{\overline{f}+\overline{h}}$ are obtained on
$\{x\in\Omega: \overline{f}(x)>0\}$. On the other hand, we know that
$u_{\overline{f}+\overline{h}}$ vanishes on $\partial\Omega$, hence
$\partial\{x\in\Omega: \overline{f}(x)>0\}$ must avoid $\partial\Omega$,
as desired.
\end{proof}
We need the following lemma before stating our symmetry result.
\begin{lemma} \label{lem7}
Let $\Omega$ be a ball centered at the origin, and
$f\in L^p(\Omega)$. Suppose $u_f^*(0)=u_{f^*}(0)$.
Then $u_f^*(x)=u_{f^*}(x)$ in $\Omega$.
\end{lemma}
\begin{proof}
Let us denote the distribution function of $u_f$ by
$\mu (t)$; that is,
\[
\mu(t)=|\{x\in\Omega: u_f(x)\geq t\}|,
\quad 0\leq t\leq M:=\sup_{\overline{\Omega}}u_f.
\]
It is well known that the function
\[
\xi (t):=\frac{1}{N^2C_N^{2/N}}(-\mu'
(t))\mu(t)^{-2+2/N}\int_0^{\mu(t)}f^\Delta (s)\,ds,
\]
where $C_N$ is the measure of the the $N$-dimensional unit ball,
satisfies $\xi (t)\geq 1$ and
\begin{equation}
\int_0^{u_f^*(x)}\xi(t)\,dt=u_{f^*}(x), \label{talenti}
\end{equation}
see \cite{t1}. We claim $\xi(t)\equiv 1$. To prove the claim we assume the
contrary and derive a contradiction. To this end, we assume
$\xi(t)>1$ on a set of positive measure. Then, from
(\ref{talenti}), we obtain
\[
u_{f^*}(0)=\int_0^{u_f^*(0)}\xi(t)\,dt>u_f^*(0).
\]
This is a
contradiction to our hypothesis that $u_{f^*}(0)=u_f^*(0)$. Thus
we must have $\xi(t)\equiv 1$. This, in conjunction with
(\ref{talenti}), implies
\[
u_{f^*}(x)=\int_0^{u_f^*(x)}\xi(t)\,dt=
\int_0^{u_f^*(x)}dt=u_f^*(x).
\]
This completes the proof of the lemma.
\end{proof}
Our symmetry result is the following, compare with \cite{b3}.
\begin{theorem} \label{thm3}
Let $\Omega$ be a ball centered at the origin. Let $(f,h)$ be an
optimal solution of \eqref{PP}. Then
\begin{equation}
f=f_0^*,\quad h=h_0^*. \label{symmetry}
\end{equation}
In particular, we deduce that \eqref{PP} has a
unique solution, $(f_0^*,h_0^*)$.
\end{theorem}
\begin{proof} From \cite{t1}, we know $u_f^*\leq u_{f^*}$. Hence, we have
\begin{equation} \label{inequality6}
\begin{aligned}
J(f,h)
&=\int_\Omega fu_f\,dx+\int_\Omega hu_f\,dx\\
&\leq \int_\Omega f^*u_f^*\,dx+\int_\Omega hu_f\,dx \\
&\leq \int_\Omega f^*u_{f^*}\,dx +\int_\Omega hu_f\,dx \\
&\leq \int_\Omega f^*u_{f^*}\,dx+\int_\Omega h^*u_f^* \\
&\leq \int_\Omega f^*u_{f^*}\,dx+\int_\Omega h^*u_{f^*}\,dx \\
&= J(f^*,h^*)\leq J(f,h)
\end{aligned}
\end{equation}
where in the first and third inequalities we have used Lemma \ref{lem1},
whereas the last inequality follows from the optimality of $(f,h)$.
Hence, all inequalities in (\ref{inequality6}) are in fact equalities.
This, in turn, implies
\begin{gather}
\int_\Omega f^*u_f^*\,dx=\int_\Omega f^*u_{f^*}\,dx, \label{ea}\\
\int_\Omega h^*u_f^*\,dx=\int_\Omega h^*u_{f^*}. \label{eb}
\end{gather}
From \eqref{ea}, we derive $\int_\Omega f^*(u_{f^*}-u_f^*)\,dx=0$.
So, we must have $u_{f^*}(x)=u_f^*(x)$, for every $x$ in the support
of $f^*$, thanks to the fact that $u_f^*\leq u_{f^*}$. Hence,
in particular, $u_{f^*}(0)=u_f^*(0)$.
Now we can apply Lemma \ref{lem7} to
deduce $u_f^*=u_{f^*}$, in $\Omega$. Note that from
(\ref{inequality6}), we get $\int_\Omega fu_f\,dx=\int_\Omega
f^*u_f^*\,dx$. This coupled with $u_f^*=u_{f^*}$ yield
$\int_\Omega fu_f\,dx=\int_\Omega f^*u_{f^*}\,dx$. This clearly
implies $\int_\Omega |\nabla u_f|^2\,dx=\int_\Omega |\nabla
u_{f^*}|^2\,dx=\int_\Omega |\nabla u_f^*|^2\,dx$. Similarly to the
proof of Theorem \ref{thm1}, one can show that the set
$\{x\in\Omega: \nabla u_f(x)=0,\;0__