\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,\;00,\;\|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