\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 49, 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  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/49\hfil Nonlinear elastic membranes]
{Nonlinear elastic membranes involving the p-Laplacian operator}
\author[F. Cuccu, B. Emamizadeh, G. Porru\hfil EJDE-2006/49\hfilneg]
{Fabrizio Cuccu, Behrouz Emamizadeh, Giovanni Porru}  % in alphabetical order

\address[F. Cuccu]{Mathematics Department\\
Universit\'a di Cagliari\\
Via Ospedale 72, 09124 Cagliari, Italy}
\email{fcuccu@unica.it}

\address[B. Emamizadeh]{Department of Mathematics\\
The Petroleum Institute\\
P. O. Box 2533, Abu Dhabi, UAE}
\email{bemamizadeh@pi.ac.ae}

\address[G. Porru]{Mathematics Department\\
Universit\'a di Cagliari\\
Via Ospedale 72, 09124 Cagliari, Italy}
\email{porru@unica.it}

\date{}
\thanks{Submitted February 2, 2006. Published April 14, 2006.}
\subjclass[2000]{35J20, 35B38, 49Q10}
\keywords{$p$-Laplace; rearrangements; maximum principle;
      existence; \hfill\break\indent uniqueness; domain derivative}

\begin{abstract}
 This paper concerns an optimization problem related to the Poisson
 equation for the $p$-Laplace operator, subject to homogeneous
 Dirichlet boundary conditions. Physically the Poisson equation
 models, for example, the deformation of a nonlinear elastic
 membrane which is fixed along the boundary, under load. A
 particular situation where the load is represented by a
 characteristic function is investigated.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

Let $\Omega$ be a bounded smooth domain in
$\mathbb{R}^N$.  This paper is concerned with an optimization
problem related to the  Poisson boundary-value problem
\begin{equation}
\begin{gathered}
-\Delta_pu=f,\quad \mbox{in }\Omega,\\
u=0,\quad \mbox{on }\partial\Omega.
\end{gathered} \label{poisson}
 \end{equation}
Here $p\in (1,\infty)$, and $\Delta_p$ stands
for the usual $p$-Laplacian; that is, $\Delta_p u=\nabla\cdot
(|\nabla u|^{p-2}\nabla u)$. Let $f_0\in L^q(\Omega)$, with
$q=p/(p-1)$, and let $\mathcal{R}$ be the class of rearrangements of
$f_0$. We are interested in finding
\begin{equation}
\sup_{f\in\mathcal{R}}\int_\Omega f\, u_f\, dx \label{maximization}
\end{equation}
where $u_f$ is the (unique) solution of (\ref{poisson}).

The $p$-Laplace operator arises in various physical contexts: non
Newtonian fluids, reaction diffusion problems, non linear
elasticity, electrochemical machining, elastic-plastic torsional
creep, etc., see \cite{AM}, \cite{HS}, and references therein. For
a theoretical develop of the theory of the p-Laplacian we refer to
the monograph \cite{HKM}. The case of $p=2$ is the most important
and easier to discuss: it corresponds to a first approximation,
the linear case. For non ideal materials, it is often appropriate
to involve a power of the gradient $|\nabla u|$ to describe the
law governing the model. For example, problem (\ref{poisson})
models a nonlinear elastic membrane under load $f$. The solution
$u_f$ stands for the deformation of the membrane from the rest
position. Therefore, the functional $\int_{\Omega}f\, u_f\, dx$
measures the average deformation, with respect to the measure $f\,
dx,$ of the membrane. Thus, any solution to (\ref{maximization})
determines an {\em optimal} load chosen from the class $\mathcal{R}$.

Our interest in (\ref{maximization}) spans questions such as
existence, uniqueness (in case $\Omega$ is a ball), and
qualitative properties of maximizers. In case of $p=2$, the
problem is well understood, see \cite{B1}, \cite{B2}, \cite{BM},
\cite{CJP}, \cite{NE}. In this case the functional $\int_\Omega
f\, u_f\, dx $ is weakly sequentially continuous and strictly
convex, say on $L^2(\Omega)$, so the classical results of R.
Burton  are available to be applied to prove existence and some
qualitative properties of the maximizers. However, in the case
$p\neq 2$, we will use a method which does not need the convexity
of the functional. The existence in a similar situation has been
discussed in \cite{CP}.


\section{Preliminaries}

In this section we collect some well known results. Let us begin
with the definition of a weak solution of (\ref{poisson}).

\noindent {\bf Definition.}
A function $u\equiv u_f\in W_0^{1,p}(\Omega)$ is
a weak solution of (\ref{poisson}) provided
\[
 \int_\Omega
|\nabla u|^{p-2}\nabla u\cdot\nabla v\,dx=\int_\Omega
fv\,dx,\;\quad\forall \;v\in W_0^{1,p}(\Omega).
\]
It is a standard
result that (\ref{poisson}) has a unique weak solution $u_f$, for
which the following equations hold
\begin{equation}
\int_\Omega fu_f\,dx=\int_\Omega |\nabla u_f|^p\,dx
=\frac{1}{p-1}\sup_{u\in
W_0^{1,p}(\Omega)}\int_\Omega \bigl(pf\, u- |\nabla u|^p\bigr)dx.
\label{identities}
\end{equation}

\noindent{\bf Definition.} Suppose
$f:(X,\Sigma,\mu)\to\mathbb{R}^+$ and $g:(X ',\Sigma
',\mu ')\to\mathbb{R}^+$ are measurable functions. We
say $f$ and $g$ are rearrangements of each other if and only if
\[
\mu (\{x\in X|\;f(x)\geq\alpha\})=\mu ' (\{x\in X
'|\;g(x)\geq\alpha\}),\quad\forall \alpha\geq 0.
\]
Henceforth we fix $f_0\in L_+^q(\Omega)$, with $q=p/(p-1)$. The
set of all rearrangements of $f_0$ is denoted by $\mathcal{R}$. Thus, for
any $f\in\mathcal{R}$, we have
\[
 \mathcal{L}_N(\{x\in\Omega|\;f(x)\geq\alpha\})=\mathcal{
L}_N(\{x\in\Omega|\;f_0(x)\geq\alpha\}),\quad\forall \alpha\geq 0,
\]
where $\mathcal{L}_N$ denotes the $N$-dimensional Lebesgue
measure. For $f:\Omega\to\mathbb{R}^+$, $f^\Delta$ and $f^*$
denote the decreasing and Schwarz rearrangements of $f$,
respectively. Recall that $f^\Delta$ is defined on $(0,\mathcal{
L}_N(\Omega))$, and $f^*$ is defined on $B$, the ball centered at
the origin with volume equal to $\mathcal{L}_N(\Omega)$.

\begin{lemma}  \label{lem1}
Let $q=p/(p-1)$, $f\in L^q_+(\Omega)$,
$g\in L^p_+(\Omega)$. Suppose that every level set of $g$ $($that
is, sets of the form $g^{-1}(\{\alpha\}))$, has measure zero. Then
there exists an increasing function $\phi$ such that $\phi\circ g$
is a rearrangement of $f$.
\end{lemma}

\begin{lemma} \label{lem2}
Suppose $\zeta\in L^p_+(\Omega)$, and $f\in L^q_+(\Omega)$. Suppose there
exists an increasing function $\phi$ such that $\phi\circ \zeta\in
\mathcal{R}(f)$, the set of rearrangements of $f$. Then $\phi\circ
\zeta$ is the unique maximizer of the linear functional
$\int_\Omega h\, \zeta\, dx$, relative to $h\in\overline {\mathcal{R}}(f)$,
where $\overline{\mathcal{R}}(f)$ denotes the weak closure of
$\mathcal{R}(f)$
in $L^q(\Omega)$.
\end{lemma}

\begin{lemma} \label{lem3}
Suppose $g\in L^p_+(\Omega)$. Then there exists $\hat f\in \mathcal{
R}(f_0)$ which maximizes the linear functional $\int_\Omega
hg\,dx$, relative to $h\in \overline{\mathcal{R}}(f_0)$; that is,
\[
\int_\Omega \hat fg\,dx=\sup_{h\in \overline{\mathcal{R}}(f_0)}\int_\Omega
hg\,dx.
\]
\end{lemma}

 For the proof of Lemma \ref{lem1}, see \cite[Lemma 2.4]{BM}, and for
Lemmas \ref{lem2} and \ref{lem3}, see \cite[Lemma 2.4]{B2}.

Next we recall a well known rearrangement inequality. If $u\in
W^{1,p}_0(\Omega)$ is non-negative and if $u^*$ denotes the
Schwarz rearrangement of $u$, then $u^*\in W^{1,p}_0(\Omega)$ and
the inequality
\begin{equation}
\int_B|\nabla u^*|^p\,dx\leq \int_\Omega |\nabla u|^p\,dx \label{polya}
\end{equation}
holds. The case of equality in
(\ref{polya}) has been considered in \cite{BZ}. The following
result can be deduced from Lemma 3.2, Theorem 1.1, and Lemma
2.3(v), in \cite{BZ}.

\begin{theorem} \label{thm1}
Let $u\in W^{1,p}_0(\Omega)$ be non-negative, and suppose equality
holds in $(\ref{polya})$. Then $u^{-1}(\alpha,\infty)$ is a
translate of ${u^*}^{-1}(\alpha,\infty)$, for every $\alpha\in
[0,M]$, where $M$ is the essential supremum of $u$ over $\Omega$,
modulo sets of measure zero. Moreover, if
\begin{equation}
\mathcal{L}_N(\{x\in\Omega|\;\nabla u(x)=0,\;\;0<u(x)<M\})=0,
\label{measure}
\end{equation}
 then $u(x)=u(x-x_0)$, for some $x_0\in
\mathbb{R}^N$; that is, $u$ is a translation of $u^*$.
\end{theorem}

\section{Main Results}
We begin with the following result.

\begin{theorem} \label{thm2}
The maximization problem $(\ref{maximization})$ is solvable; that
is, there exists $\hat f\in \mathcal{R}(f_0)$ such that
\[
\int_\Omega \hat f\hat u\,dx=\sup_{f\in\mathcal{R}(f_0)}\int_\Omega
fu_f\,dx,
 \]
 where $\hat u=u_{\hat f}$.
\end{theorem}

\begin{proof}
 Let
\[
 I\:=\sup_{f\in \mathcal{R}(f_0)}\int_\Omega fu_f\,dx.
\]
 We first show that $I$ is finite. Consider $f\in
\mathcal{R}(f_0)$; then from (\ref{identities}) followed by
H\"older's inequality we find
\begin{equation}
 \int_\Omega |\nabla
u_f|^p\,dx=\int_\Omega fu_f\,dx\leq \|f\|_q\;\|u_f\|_p.
\label{ali1}
\end{equation}
Since $\|f\|_q=\|f_0\|_q$, it follows from
(\ref{ali1}) and the Poincar\'e inequality that $I$ is finite.

Let $\{f_i\}$ be a maximizing sequence and let $u_i=u_{f_i}$. From
(\ref{ali1}) it is clear that $\{u_i\}$ is bounded in
$W^{1,p}(\Omega)$, hence it has a subsequence (still denoted
$\{u_i\}$) that converges weakly to $u\in W^{1,p}_0(\Omega)$. We
also infer that $\{u_i\}$ converges strongly to $u$ in
$L^p(\Omega)$. On the other hand, since $\{f_i\}$ is bounded in
$L^q(\Omega)$, it must contain a subsequence (still denoted
$\{f_i\}$) converging weakly to $\eta\in L^q(\Omega)$. Note that
$\eta\in\overline{\mathcal{R}}$, the weak closure of $\mathcal{R}$ in
$L^q(\Omega)$. Thus, using the weak lower semi-continuity of the
$W^{1,p}_0(\Omega)$-norm and (\ref{identities}) we obtain
\begin{equation}
\begin{aligned}
I=\lim_{i\to\infty}\int_\Omega f_iu_i\,dx
&= \frac{1}{p-1}\lim_{i\to\infty}\int_\Omega
\bigl(pf_iu_i-|\nabla u_i|^p\bigr)\,dx  \\
&\leq  \frac{1}{p-1}\int_\Omega \bigl(p\eta\, u-|\nabla u|^p\bigr)\,dx.
\end{aligned} \label{ali2}
\end{equation}
Note that from Lemma \ref{lem3} we infer the existence of $\hat f\in \mathcal{
R}(f_0)$ that maximizes the linear functional $\int_\Omega
hu\,dx$, relative to $h\in \overline{\mathcal{R}}(f_0)$. As a consequence we
obtain
\begin{equation}
 \int_\Omega \eta\, u\,dx\leq\int_\Omega \hat f\,
u\,dx. \label{ali3}
\end{equation}
Applying (\ref{identities}), (\ref{ali2})
and (\ref{ali3}) we find
\[
I\leq\frac{1}{p-1}\int_\Omega
\bigl(p\hat f u-|\nabla u|^p\bigr)\,dx\leq
\frac{1}{p-1}\int_\Omega \bigl(p\hat f \hat u-|\nabla\hat
u|^p\bigr)\,dx=\int_\Omega \hat f\hat u\, dx \leq I.
 \]
Recall
that $\hat u=u_{\hat f}$. Thus $\hat f$ is a solution to
(\ref{maximization}), as desired.
\end{proof}

The next issue addressed is the so called Euler-Lagrange equation
for solutions of (\ref{maximization}).

\begin{theorem}\label{thm3}
Suppose $\hat f$ is a solution of $(\ref{maximization})$ with
$f_0$ non negative. Then there exists an increasing function
$\phi$ such that
\begin{equation}
 \hat f=\phi\circ\hat u \label{euler}
\end{equation}
almost everywhere in $\Omega$, where $\hat u= u_{\hat f}$.
Equation $(\ref{euler})$ is referred to as the Euler-Lagrange
equation for $\hat f$.
\end{theorem}

To prove Theorem \ref{thm3} we need some preparations. Let us begin with
the following result.

\begin{lemma} \label{lem4}
Suppose $\hat f$ and $\hat u$ are as
in Theorem \ref{thm3}. Then $\hat f$ maximizes the linear functional
$\int_\Omega h\hat u\,dx$, relative to $h\in \mathcal{R}(f_0)$.
\end{lemma}

\begin{proof}
 Since $\hat f$ is a solution of (\ref{maximization}),
the following inequality holds for every $f\in \mathcal{R}(f_0)$
\begin{equation}
\int_\Omega \hat f\hat u\,dx\geq\int_\Omega fu_f\,dx. \label{ali4}
\end{equation}
Next, applying (\ref{identities}) to the right hand side of
(\ref{ali4}) yields
\begin{equation}
\int_\Omega \hat f\hat
u\,dx\geq\frac{1}{p-1}\int_\Omega \bigl(pf\hat u-|\nabla\hat
u|^p\bigr)\, dx, \label{ali5}
\end{equation}
 for every $f\in \mathcal{R}(f_0)$.
We also have
\begin{equation}
 \int_\Omega \hat f\hat
u\,dx=\frac{1}{p-1}\int_\Omega \bigl(p\hat f\hat u-|\nabla\hat
u|^p\bigr)\, dx. \label{ali6}
\end{equation}
Combination of (\ref{ali5}) and
(\ref{ali6}) implies
\[
\int_\Omega \hat f\hat
u\,dx\geq\int_\Omega f\hat u\,dx,
 \]
 for every $f\in \mathcal{R}(f_0)$. This completes the proof.
\end{proof}

In what follows we shall write $\inf_{x\in S}f(x)$ ($\sup_{x\in
S}f(x)$) for the essential inferior (superior) of $f(x)$ in $S$.

\begin{lemma} \label{lem5}
Let $\hat f$ and $\hat u$ be as in Theorem \ref{thm3}, and let $S(\hat
f)=\{x\in\Omega|\;\hat f(x)>0\}$. Set
\[
\gamma=\inf_{x\in S(\hat f)}\hat u(x),\quad
\delta=\sup_{x\in \Omega\setminus S(\hat
f)}\hat u(x).
\]
 Then $\gamma\geq\delta$.
\end{lemma}

\begin{proof}
 To derive a contradiction assume $\gamma <\delta$.
Let us fix $\gamma <\xi_1<\xi_2<\delta$. Since $\xi_1>\gamma$,
there exists a set $A\subset S(\hat f)$, with positive measure,
such that $\hat u\leq\xi_1$ on $A$. Similarly, $\xi_2<\delta$
implies that there exists a set $B\subset \Omega\setminus S(\hat
f)$, with positive measure, such that $\hat u\geq\xi_2$ on $B$.
Without loss of generality we may assume that $\mathcal{L}_N(A)=\mathcal{
L}_N(B)>0$ (otherwise we consider suitable subsets of $A$ and $B$
having the same measures). Next, consider a measure preserving map
$T:A\to B$, see \cite{R}. Using $T$ we define a particular
rearrangement of $\hat f$, denoted $\overline f$.
\[
\overline f(x)=\begin{cases} \hat f(Tx), & x\in A \\
\hat f(T^{-1}x) & x\in B \\
\hat f(x) & \Omega\setminus (A\cup B).
\end{cases}
\]
Thus
\begin{align*}
\int_\Omega \overline f\hat u\,dx-\int_\Omega\hat f\hat u\,dx
&= \int_{A\cup B}\overline f\hat u\,dx-\int_{A\cup B}\hat f\hat
u\,dx \\
&=\int_B\overline f\hat u\,dx-\int_A\hat f\hat u\,dx \\
&\geq \xi_2\int_B\overline fdx-\xi_1\int_A\hat fdx\\
&=(\xi_2-\xi_1)\int_A\hat f\,dx>0.
\end{align*}
Therefore, $\int_\Omega\overline f\hat u\,dx>\int_\Omega\hat f\hat
u\,dx$, which contradicts the maximality of $\hat f$
(see Lemma \ref{lem4}).
\end{proof}

\begin{proof}[Proof of theorem \ref{thm3}]
Notice that from \eqref{poisson} and \cite[Lemma 7.7]{GT},
it is clear that the level sets of $\hat u$, restricted to
$S(\hat f)$, have measure
zero. Therefore applying Lemma \ref{lem1}, we infer existence of an
increasing function $\tilde\phi$ such that $\tilde\phi\circ\hat u$
is a rearrangement of $\hat f$ relative to the set $S(\hat f)$.
Equivalently, $\tilde\phi\circ\hat u$, restricted to $S(\hat f)$,
is a rearrangement of $\hat f^\Delta$, restricted to the interval
$(0,s)$, where $s=\mathcal{L}_N(S(\hat f))$. Now, define
\[
\phi (t)=\begin{cases}
 \tilde\phi (t) & t\geq\gamma \\
  0 & t<\gamma,
\end{cases}
\]
where $\gamma=\inf_{S(\hat f)}\hat u(x)$. Note that, since
$\tilde\phi$ is non-negative, $\phi$ is an increasing function.
Moreover, $\phi\circ\hat u$ is a rearrangement of $\hat f^\Delta$,
on $(0,\omega)$, where $\omega=\mathcal{L}_N(\Omega)$. Thus
$\phi\circ\hat u\in \mathcal{R}(f_0)$, hence we can apply Lemma \ref{lem2} to
deduce that $\phi\circ\hat u$ is the unique maximizer of the
linear functional $\int_\Omega h\hat u\,dx$, relative to $h\in
\mathcal{R}(f_0)$. This obviously implies $\hat f=\phi\circ\hat u$,
almost everywhere in $\Omega$.
\end{proof}

\noindent{\bf Remark.}
The function $\phi$ in above can be
extended to all of $\mathbb{R}$. Thus from (\ref{euler}) we infer $S(\hat
f)=\hat u^{-1}\bigl(\phi ^{-1}(0,\infty)\bigr)$. Since $\phi$ is
increasing the set $\phi ^{-1}(0,\infty)$ is either the interval
$(\gamma,\infty)$ or $[\gamma,\infty)$. In both cases, since the
level sets of $\hat u$ on $S(\hat f)$ have measure zero, we can
write $S(\hat f)=\{\hat u>\gamma\}$. If we assume $f_0\in
L^\infty(\Omega)$ then we have the continuity of the solution
$\hat u$ (see \cite{T}). In this situation the boundary of $S(\hat
f)$, denoted $\partial S(\hat f)$, satisfies
\begin{equation}
\partial S(\hat f)\subset\{\hat u=\gamma\}, \label{boundary}
\end{equation}
 thanks to the continuity of $\hat u$. Note that
$\mathcal{L}_N(S(f_0))=\mathcal{L}_N(S(\hat f))$.
Therefore, if $\mathcal{L}_N(S(f_0))<\mathcal{L}_N(\Omega)$,
it follows that $\gamma$ is strictly positive.

An example of interest is the following.
\smallskip

\noindent{\bf Example.}
Suppose $f_0=\chi_{E_0}$, where $\chi_{E_0}$ is the characteristic
function of the measurable set $E_0\subset\Omega$, and let
$\mathcal{L}_N(E_0)<\mathcal{L}_N(\Omega)$. Denoting a solution of
(\ref{maximization}) by $\hat f$, it is clear that
$\hat f=\chi_{\hat E}$, for some measurable set $\hat E\subset\Omega$,
having the same measure as $E_0$. From the last Remark we infer
that $\hat u=u_{\hat f}$ is constant on $\partial\hat E$. Also,
$\partial\hat E$ does not intersect $\partial\Omega$. So
physically speaking, in order to maximize the average deformation
of the nonlinear elastic membrane under uniform loads (given by
the appropriate rearrangement class) it is best to place the load
away from the boundary (independently of the geometry of the
membrane). We will return to this example in the last section.

We now address the question of uniqueness in a ball.

\begin{theorem} \label{thm4}
Suppose $\Omega$ is a ball centered at the origin. Then the
maximization problem $(\ref{maximization})$ with $f_0$ non
negative and essentially bounded has a unique solution, namely,
$f_0^*$.
\end{theorem}

For the rest of this section $\Omega$ is always a ball. We need
the following result.

\begin{lemma} \label{lem6}
If $f\geq 0$, then
\begin{equation}
\frac{1}{p}\int_\Omega|\nabla
u_f^*|^p\,dx+\frac{1}{q}\int_\Omega f^*u_{f^*}\, dx\geq\int_\Omega
f^*u_f^*\,dx. \label{ali7}
 \end{equation}
\end{lemma}

\begin{proof} From the variational characterization of $u_{f^*}$
the following inequality is clear
\begin{align*}
\frac{1}{p}\int_\Omega|\nabla u_f^*|^p\,dx +
\frac{1}{q}\int_\Omega f^*u_{f^*}\, dx
& \geq
\frac{1}{p}\int_\Omega |\nabla u_{f^*}|^p\,dx-\int_\Omega
f^*u_{f^*}\,dx \\
&\quad + \frac{1}{q}\int_\Omega f^*
u_{f^*}\,dx+\int_\Omega f^*u_f^*\,dx. \label{ali8}
\end{align*}
Since the first three terms on the right hand side of the above
inequality drop out thanks to (\ref{identities}), we obtain
(\ref{ali7}).
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm4}]
 Suppose $f$ is a
solution of (\ref{maximization}). Then, from Lemma \ref{lem6} we have
\[
\frac{1}{p}\int_\Omega|\nabla
u_f^*|^p\,dx+\frac{1}{q}\int_\Omega f^*u_{f^*}\, dx\geq\int_\Omega
f^*u_f^*\,dx.
\]
 Applying the Hardy-Littlewood inequality
$\int_\Omega f^*u_f^*\,dx\geq\int_\Omega fu_f\,dx$ to the right
hand side of the above inequality we find
\begin{equation}
\frac{1}{p}\int_\Omega|\nabla u_f^*|^p\,dx+\frac{1}{q}\int_\Omega
f^*u_{f^*}\,dx\geq\int_\Omega fu_f\,dx=\frac{1}{p}\int_\Omega
fu_f\,dx+\frac{1}{q}\int_\Omega fu_f\,dx. \label{beh1}
\end{equation}
Since
$f$ is a solution of (\ref{maximization}) we have $\int_\Omega
fu_{f}\,dx\geq\int_\Omega f^*u_{f^*}\,dx$. Moreover, an
application of (\ref{identities}) yields $\int_\Omega
fu_{f}\,dx=\int_\Omega |\nabla u_f|^p\,dx.$ Therefore,
(\ref{beh1}) implies
\begin{equation}
\int_\Omega |\nabla u_f^*|^p\,dx\geq
\int_\Omega |\nabla u_f|^p\,dx. \label{beh2}
\end{equation}
Hence, from (\ref{polya}), we obtain equality in (\ref{beh2}).

Next we show that $u_f=u_f^*$. According to Theorem \ref{thm1}, we only
need to show that (\ref{measure}) holds. Let us consider
$x\in\Omega$ such that $0<u(x)<\max_\Omega u(x)$, and set
$S=\{z\in\Omega:\;u(z)\geq u(x)\}$, which is a closed ball by
Theorem \ref{thm1}. If we define $v(z)=u(z)-u(x)$, we have $-\Delta
_pv(z)=-\Delta _pu(z)\geq 0$, since $f$ is non-negative. Since $v$
vanishes on $\partial S$, by the strong maximum principle
\cite[Theorem 5]{V} we have $v>0$ in $S$. Therefore, $u(z)>u(x)$ for all
$z\in S$. Hence $x$ must be a boundary point of $S$. So, by the
Hopf boundary lemma \cite[Theorem 5]{V} we derive $\frac{\partial
u}{\partial\nu}(x)=\frac{\partial v}{\partial\nu}(x)\neq 0$, where
$\nu$ stands for the outward unit normal to $\partial S$ at $x$.
Thus (\ref{measure}) holds, as desired. Finally, from
(\ref{euler}), we deduce that $f$ coincides with its Schwarz
rearrangement, so $f=f^*=f_0^*$. This completes the proof of the
theorem.
\end{proof}

\section{Domain derivative}

This section is devoted to the example mentioned earlier. We have
seen that if $\hat f=\chi _{\hat D}$ is a solution of
(\ref{maximization}), then $\hat u=u_{\hat f}$ is constant on the
free boundary $\partial\hat D$. A natural question arises: Does
the same result hold if $\chi _{\hat D}$ is {\em any} critical
point of the functional $\int_\Omega fu_f$, relative to the class
of rearrangements of $\chi_{\hat D}$? We give an affirmative
answer to this question under some restrictions on $\hat D$. In
order to put things in the right context we need to introduce the
notion of domain derivative \cite{S1}, \cite{S2}, that is
specialized to our situation.

Let $D$ be an open smooth subset of $\Omega$ with
$\mathop{\rm dist}(D,\partial\Omega)>0$. Let $V$ be a regular (smooth)
vector field with support in $\Omega$. Define $D^t=(Id +tV)(D)$,
with small $t\in R^+$ such that $D^t\subset\Omega$. Here $Id$
denotes the identity map. Note that for small $t$, the operator
$Id +tV$ is a diffeomorphism. In particular, $D^t$ is an open and
smooth set. If $D\Delta D^t$ denotes the familiar symmetric
difference of $D$ and $D^t$, then
\begin{equation}
\mathcal{L}_N(D\Delta D^t)\leq
ct, \label{symmetric}
 \end{equation}
where $c$ is a positive constant
independent of $t$. As a consequence of (\ref{symmetric}), the
function $\chi_{D^t}-\chi_D$ tends to zero in $L^q(\Omega)$ (for
any $q>1$) as $t$ tends to zero. Let us define
\[
I(D)=\int_Du\,dx,
\]
where $u$ satisfies
\begin{equation}
-\Delta_pu=\chi_D\quad\mbox{in }\Omega, \quad
u=0\quad \mbox{on }\partial\Omega.
\label{poisson2}
 \end{equation}
Also
\[
 I(D^t)=\int_{D^t}u^t\,dx,
\]
where $u^t$ satisfies
\begin{equation}
-\Delta_pu^t=\chi_{D^t}\quad \mbox{in }\Omega,\quad u^t=0\quad
\mbox{on }\partial\Omega. \label{poisson3}
 \end{equation}
For the sake of the following definition we
introduce $\mathcal{V}$ to be the set of all regular vector
fields with support in $\Omega$. \smallskip

\noindent{\bf Definition.}
We say that $D$ (as above) is a critical point of the functional
$I$ provided
\[
dI(D;V)=c\;dVol (D;V),
\]
for some constant $c$ and every $V\in \mathcal{V}$.
Here
\begin{gather*}
\mathop{\rm dI}(D;V):=\lim_{t\to 0^+}\frac{I(D^t)-I(D)}{t},\\
\mathop{\rm  dVol} (D;V):=\lim_{t\to 0^+}\frac{\mathcal{L}_N(D^t)
-\mathcal{L}_N(D)}{t}.
\end{gather*}
Of course, if we consider measure preserving vector fields $V$
then $dVol (D;V)=0$.
We are now ready to state the main result of this section.

\begin{theorem} \label{thm5}
Suppose $D$ is an open smooth subset of $\Omega$. Suppose
$\mathop{\rm dist}(D,\partial\Omega)>0$, and $D$ is a critical point of
$I$, relative to $\Gamma$. Then $u$, the solution of
\eqref{poisson2}, is constant on $\partial D$.
\end{theorem}

\begin{lemma} \label{lem7}
Let $u$ and $u^t$ be the
solutions of \eqref{poisson2} and \eqref{poisson3},
respectively. Then $u^t\to u$, in $W^{1,p}_0(\Omega)$, as
$t\to 0^+$.
\end{lemma}

\begin{proof} Let us recall the well known inequality (see, for
example, \cite{T})
\begin{equation} \label{simon}
C(|X|^{p-2}X-|Y|^{p-2}Y,X-Y)\geq
\begin{cases} |X-Y|^p, & p\geq 2 \\
 \frac{|X-Y|^2}{(|X|+|Y|)^{2-p}}, & p\leq 2,
\end{cases}
\end{equation}
where $X$ and $Y$ are vectors in $\mathbb{R}^n$, $|X|$ (similarly $|Y|$)
denotes the Euclidean length of $X$, $C$ is a positive constant,
and $(\cdot,\cdot)$ stands for the usual dot product in $\mathbb{R}^n$.
Let us consider two cases

\noindent {\em Case 1: $p\geq 2$}: Using
(\ref{simon}) we have
\[
 \|\nabla u^t-\nabla u\|_p^p\leq
C\int_\Omega (|\nabla u^t|^{p-2}\nabla u^t-|\nabla u|^{p-2}\nabla
u)\cdot (\nabla u^t-\nabla u)\,dx.
\]
 Using (\ref{poisson2}) and
(\ref{poisson3}) we can rewrite the above inequality as
\[
\|\nabla u^t-\nabla u\|_p^p\leq C \int_\Omega
(\chi_{D^t}-\chi_D)(u^t-u)\,dx.
\]
So by applying the H\"older's
inequality followed by the Poincar\'e inequality we obtain
 \[
\|\nabla u^t-\nabla u\|_p^{p-1}\leq \tilde C\Big(\int_\Omega
|\chi_{D^t}-\chi_D|^q\,dx\Big)^{1/q}.
 \]
>From the above inequality, the assertion of the lemma follows.

\noindent{\em Case 2: $p\leq 2$}:
Let us begin with the following observation
\begin{align*}
\|\nabla u^t-\nabla u\|^p_p
&= \int_\Omega\frac{|\nabla u^t-\nabla u|^p}{(|\nabla u^t|+|\nabla
u|)^{\frac{p(2-p)}{2}}}\;(|\nabla u^t|+|\nabla
u|)^{\frac{p(2-p)}{2}}\,dx \\
&\leq  \Big(\int_\Omega\frac{|\nabla u^t-\nabla u|^2}{(|\nabla
u^t|+|\nabla u|)^{(2-p)}}\,dx\Big)^{p/2}
\Big(\int_\Omega (|\nabla u^t|+|\nabla
u|)^p\,dx\Big)^{(2-p)/2},
\end{align*}
which follows from the H\"older inequality, since $2/p>1$.
Note that $(u^t)$ is bounded in $W^{1,p}_0(\Omega)$. Thus from the
above inequality we find
\begin{equation}
\|\nabla u^t-\nabla u\|^p_p \leq
C\left(\int_\Omega\frac{|\nabla u^t-\nabla u|^2}{(|\nabla
u^t|+|\nabla u|)^{(2-p)}} \,dx\right)^{p/2}. \label{beh4}
\end{equation}
Now applying (\ref{simon}) to the right hand side of
(\ref{beh4}), the assertion of the lemma can be confirmed using
similar arguments as in the ending part of Case 1.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm5}]
 Let us begin with the  identity
\begin{equation}
I(D^t)-I(D)=\int_{D^t}(u^t-u)\,dx+\int_{D^t}u\,dx-\int_Du\,dx.
\label{porru1}
\end{equation}
Following \cite{S1}, we define
\begin{equation}
u' (x)=\lim_{t\to 0^+}\frac{u^t(x)-u(x)}{t}. \label{porru2}
\end{equation}
Moreover, we have
\[
 \int_{D^t}u\,dx-\int_D
u\,dx=\int_D\Big[ u(x+tV)\big|\det \big(\delta_{ij}
+t\frac{\partial V_i}{\partial x_j}\big)\big|-u(x)\Big]\,dx.
\]
Since $t$ is small, we find
\begin{equation}
\begin{aligned}
\int_{D^t}u\,dx-\int_D u\,dx
&= \int_D[(u(x)+t\nabla u\cdot
V+o(t))(1+t\nabla\cdot V+o(t))-u(x)]\,dx  \\
&= t\int_D(\nabla u\cdot V+u\nabla\cdot V)\,dx+o(t)  \\
&= t\int_D\nabla\cdot (uV)\,dx+o(t)  \\
&= t\int_{\partial
D}u(V\cdot\nu)\,d\sigma +o(t),
\end{aligned}\label{porru3}
\end{equation}
where $\nu$ is the outward unit normal to $\partial D$ and
$d\sigma$ is the surface measure. Inserting (\ref{porru3}) and
(\ref{porru2}) into (\ref{porru1}) yields
\begin{equation}
\lim_{t\to 0^+}\frac{I(D^t)-I(D)}{t}
=\int_{D}u'\,dx+\int_{\partial D}u(V\cdot\nu)\,d\sigma. \label{derivative}
\end{equation}
Now multiply (\ref{poisson2}) by $u^t$, (\ref{poisson3}) by $u$,
subtract the new equations, and finally integrate over $\Omega$.
 We find
\begin{equation}
\int_\Omega\frac{|\nabla u^t|^{(p-2)}-|\nabla
u|^{(p-2)}}{t}\;\nabla u^t\cdot\nabla u\,dx
=\frac{1}{t}\Big[\int_{D^t}u\,dx-\int_Du\,dx\Big]
+\int_D\frac{u-u^t}{t}\,dx. \label{porru4}
\end{equation}
Since $u^t$ tends to $u$ in $W^{1,p}_0(\Omega)$
as $t$ tends to zero, and also
\[
 \frac{d}{dt}|\nabla u^t|^{(p-2)}=(p-2)|\nabla u|^{(p-4)}
\nabla u\cdot\nabla
u',\quad \mbox{at } t=0,
\]
taking the limit of (\ref{porru4}), when $t$ tends to zero,
we find
\begin{equation}
(p-2)\int_\Omega |\nabla u|^{(p-2)}\nabla u\cdot\nabla
u'\,dx=\int_{\partial
D}u(V\cdot\nu)\,d\sigma-\int_Du'\,dx. \label{porru5}
\end{equation}
If we multiply (\ref{poisson2}) by $u'$, and integrate we find
(recall that the support of $V$ is in $\Omega$)
\[
\int_\Omega |\nabla u|^{p-2}\nabla u\cdot\nabla
u'\,dx=\int_Du'\,dx.
\]
Inserting this equation in (\ref{porru5}) we find
\[
\int_Du'\,dx=\frac{1}{p-1}\int_{\partial D}u(V\cdot\nu)\,d\sigma.
\]
Finally inserting the latter estimate
into (\ref{derivative}) yields
\[
dI(D,V)=q\int_{\partial D}u(V\cdot\nu)\,d\sigma.
\]
Recalling the formula for the derivative of the volume, that is,
\[
\mathop{\rm  dVol}(D,V)=\int_{\partial D}(V\cdot\nu)\,d\sigma,
\]
and the fact that $D$ is a critical
point of $I$, we derive
\[
\mathop{\rm dI}(D,V)=c\mathop{\rm dVol} (D,V)
\Leftrightarrow
u(x)=\mbox{constant},\quad \mbox{on }\partial D.
\]
This obviously completes the proof of the theorem.
\end{proof}

\subsection*{Acknowledgement}
This work was initiated when B. Emamizadeh visited University of
Cagliari. He wants to thank Professors G. Porru and F. Cuccu for
their great hospitality. He also thanks the Petroleum Institute
for its support during this visit.

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