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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 60, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/60\hfil Optimization problems]
{Optimization problems involving Poisson's equation in $\mathbb{R}^3$}

\author[F. Bahrami, H. Fazli\hfil EJDE-2011/60\hfilneg]
{Fariba Bahrami, Hossain Fazli}  % in alphabetical order

\address{Fariba Bahrami \newline
Department of Applied Mathematics,
University of Tabriz, Tabriz, Iran}
\email{fbahram@tabrizu.ac.ir}

\address{Hossain Fazli \newline
Department of Applied Mathematics, University of Tabriz,
Tabriz, Iran}
\email{fazli64@gmail.com}

\thanks{Submitted February 16, 2011. Published May 10, 2011.}
\subjclass[2000]{49J20, 49K20}
\keywords{Rearrangements class; variational problem;
Poisson's equation; \hfill\break\indent
energy functional; minimization}

\begin{abstract}
 In this article, we prove the existence of minimizers for
 integrals associated with a second-order elliptic problem.
 For this three-dimensional optimization problem, the admissible
 set is a rearrangement class of a given function.
\end{abstract}

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

\section{Introduction}

We consider the  Poisson's equation
\begin{equation} \label{poson}
\begin{gathered}
-\Delta u=f-2h \quad \text{in }  \mathbb{R}^3\\
\lim_{|x|\to +\infty} u(x)=0, \quad f\in L^p_b(\mathbb{R}^3),
\end{gathered}
\end{equation}
where $L^p_b(\mathbb{R}^3)=\{f\in L^p(\mathbb{R}^3): f
\text{ has  bounded support}\}$ and $p>3$.
Here $h$ is a given non-negative function in
$h\in L^\infty(\mathbb{R}^3)$ with bounded support.  For the sake of
convenience in the discussions, we have $2h$ instead of $h$, but it
can be replaced by $h$.  By standard results on elliptic equations,
problem \eqref{poson} has a unique solution $u\in
W^{2,p}_{\rm loc}(\mathbb{R}^3)$; see \cite{gilbarg}.
Let $u_{f}$ be the solution of \eqref{poson},
 we define energy functional
 corresponding to \eqref{poson} as
 \begin{equation}
 \Psi_\lambda(f)=\frac{1}{2}\int_{\mathbb{R}^3}f u_{f}
+\lambda \int_{\mathbb{R}^3}gf,
 \end{equation}
for $f \in L^p_b(\mathbb{R}^3)$ where  $g\in C^2(\mathbb{R}^3)$,
$\lim_{|x|\to+\infty}g=+\infty$ and $\Delta g>c$ for some
$c>0$  and $\lambda\geq 0$. In this paper we  minimize the
functional $\Psi_\lambda$ on rearrangement class of a fixed
function. We separate the investigation of the  particular case
$\lambda=0 $, since the discussion in the case $\lambda>0$ does not
carry over the case $\lambda=0$. The same optimization problems have
been investigated in bounded domains for the Laplacian  operator in
\cite{Burton1, cuccu, ema},  for the p-Laplacian operator
in \cite{fabriz, marras}, for semilinear operators in  \cite{zivari}.
For the current problem we face two mathematical difficulties:
firstly the awkward nature of rearrangements class,
and secondly a loss of compactness which is
caused by the unboundness of the domain $\mathbb{R}^3$.
To overcome these difficulties we first investigate the problem
in a bounded domain. Then using Burton's theory on rearrangements
class, we show that a solution valid in a sufficiently large bounded
domain is in fact valid in the whole space.


\section{Notation, definitions and statement of the main result}

Henceforth we assume $p\in(3,\infty)$ and $p'$ is the conjugate
exponent of $p$; that is, $\frac{1}{p}+\frac{1}{p'}=1$. Points in
$\mathbb{R}^3$ are denoted by $x = (x_1, x_2,x_3)$,
$y=(y_1,y_2,y_3)$, and so on. By $B_r (x)$ we denote the ball
centered at $x\in \mathbb{R}^3$ with radius $r$; if the center is the
origin,  we write $B_r$. Measure will refer to Lebesgue measure on
$\mathbb{R}^3$, and if  $A\subseteq \mathbb{R}^3 $ is measurable
then  $|A|$ will denote the measure of $A$.
If $A \subset \mathbb{R}^3$ is a measurable set,
then we say $x\in A$ is a density point of $A$ whenever
$$
|B_\varepsilon(x) \cap A|>0,
$$
for all positive $\varepsilon$.


For a measurable function  $f:\mathbb{R}^3 \to \mathbb{R}^+$,
the strong support or simply the support of
 $f$  is denoted $\operatorname{supp}( f)$  and is defined by
$$
\operatorname{supp}( f) = \{x :f(x)>0\}.
$$
 For a measurable function $f : \mathbb{R}^3\to \mathbb{R}^+$
we define
$$
\|f\|_{-\infty}=\operatorname{ess\,inf}(f)=\sup\{M\geq 0 : f(x)\geq M,
\text{ for  almost all }x\}.
$$
 When $f$ and $g$ are non-negative
measurable functions that vanish outside sets of finite measure in
$\mathbb{R}^3$, we say $f$ is a rearrangement of $g$ whenever
$$
|\{x \in\mathbb{R}^3 :f(x)\geq\alpha\}|
= |\{x \in\mathbb{R}^3 :g(x)\geq\alpha\}|,
$$
for every positive $\alpha$.

 For any  real integrable and non-negative function $f$ vanishing
outside a bounded set $\Omega\subset\mathbb{R}^3$ of  measure $m$,
we can define a  decreasing rearrangement $f^\Delta$
which is a decreasing function on the interval $(0,m)$ satisfying
$$
|\{s \in(0,m) : f^\Delta(s)\geq\alpha\}| =|\{x \in\Omega  :
f(x)\geq\alpha\}|,
$$
for every positive $\alpha$. Also there exists a Schwarz
rearrangement $f^*$ for $f$, that is a rearrangemet of $f$ as a
radial decreasing function on a ball.

Let us fix $f_0 \in L^p(\mathbb{R}^3)\cap L^\infty(\mathbb{R}^3)$
to be a measurable and non-negative function vanishing outside
a set of measure
$4\pi a^3/3$, for some positive $a\in \mathbb{R}$. The set of all
rearrangements on $\mathbb{R}^3$ of $f_0$ with bounded support is
denoted by $\mathcal{R}$. The subset of $\mathcal{R}$ containing
functions vanishing outside the ball  $B_r$, where $r\geq a$,
is denoted by $\mathcal{R}(r)$; henceforth we assume $r\geq a$
in order that $\mathcal{R}(r)$ is non-empty.
The weak closure in $L^p(B_r )$ of $\mathcal{R}(r)$ is
denoted by $\overline{\mathcal{R}(r)^w}$.

 Now we are ready to introduce our minimizing problems
$P_\lambda$ as follows:
\begin{equation}
\label{variation} \min_{f\in\mathcal{R}}\ \Psi_\lambda (f).
\end{equation}

The set of solutions of $P_\lambda$ is denoted by $S_\lambda$.
Similarly, for $r\geq a$ we define $P_\lambda(r)$ as follows:
\begin{equation}
\label{bvariation} \min_{f\in\mathcal{R}(r)}\ \Psi_\lambda (f),
\end{equation}
 and the set of
solutions is denoted by $S_\lambda(r)$. Our main results are the
following:

\begin{theorem} \label{thm1}
There exists $\lambda_0>0$ such that for every $\lambda>\lambda_0$,
the optimization  problem $P_\lambda$ has a solution. Moreover, if
${f_{\lambda}}\in S_{\lambda}$ and $u_{ f_{\lambda}}$ be the
solution of \eqref{poson} corresponding with energy minimizer,  then
there exists a decreasing function $\varphi_\lambda$ such that
\begin{equation}\label{partial differential}
f_{\lambda}=\varphi_\lambda  \circ  ( u_{ f_{\lambda}}+\eta+\lambda g),
\end{equation}
almost everywhere in $\mathbb{R}^3$ where $\eta$ will be presented
later.
\end{theorem}

\begin{theorem}\label{thm2}
Let  $f_0$ and  $h$ be as introduced above. Let
$|\operatorname{supp}(f_0)|=4\pi a^3/3$  and
$|\operatorname{supp}(h)|=4\pi b^3/3$
for some $a, b$ positive real numbers.  We assume
\begin{equation}\label{necess-cond-supp}
b>\sqrt{3}a,\quad \| f_0\|_{\infty} < \| h\|_{-\infty}.
\end{equation}
Then the optimization problem $P_0$ has a solution.
\end{theorem}

\section{Preliminary results}

In this section we  state and/or prove some lemmas which are
essential in our analysis. We begin with a result proved by
Burton in \cite{Burton2}.

\begin{lemma}\label{rearangement}
For $r\geq a$ and $q\geq1$,  we have
\begin{itemize}
\item[(i)]  $\| f\|_q=\| f_0\|_q$,   for  $f\in\mathcal{R}(r)$;
\item[(ii)]  $\overline{\mathcal{R}(r)^w}$ is weakly sequentially
compact in $L^q(B_r)$;
\item[(iii)] $\overline{\mathcal{R}(r)^w}=\{f\in L^1(B_r):
  \int_{0}^sf^\Delta(t)dt\leq\int_0^sf_0^\Delta(t)dt,\;
 0<s\leq 4\pi r^3/3,\; \int_{B_r}f=\int_{B_r}f_0\}$.
\end{itemize}
\end{lemma}


\begin{lemma}\label{bound}
Let  $\lambda\geq 0$ and $f\in L^p_b(\mathbb{R}^3)$. Then
\begin{itemize}
\item[(i)]
 for the energy functional $\Psi_\lambda$ we have
\begin{equation}\label{energy}
\Psi_{\lambda}(f) = \frac{1}{2}\int_{\mathbb{R}^3}f
Kf-\int_{\mathbb{R}^3}\eta f+\lambda\int_{\mathbb{R}^3}g f,
\end{equation}
where
$$
Kf(x)= \frac{1}{4\pi}\int_{\mathbb{R}^3}
\frac{1}{|x-y|}f(y)dy,
$$
 and $\eta= Kh$.

\item[(ii)] for $f\in L^p_{b}(\mathbb{R}^3)$
\begin{equation}\label{boundedness of K}
 |Kf(x)|\leq C\| f\|_{p},\quad  \forall x\in\mathbb{R}^3,
\end{equation}
where $C$ depends only on $p$ and $|\operatorname{supp}(f)|$.
\end{itemize}
\end{lemma}

\begin{proof}
Using the fundamental solution of  $-\Delta$ on $\mathbb{R}^3$ and
asymptotic behavior of the solutions  in \eqref{poson}, we derive
the unique solution of the problem \eqref{poson}, $u_{f}=Kf-2Kh$,
this yields \eqref{energy}.
 We note that $u_f(x)=O(\frac{1}{|x|})$ as $|x|\to +\infty$.
 Indeed, for large $|x|$, we have $|x-y|>|x|/2$ for
$y\in \operatorname{supp}(f)\cup \operatorname{supp}(h)$.
Thus,  $K(f-2h)$    is dominated by $2\|f-2h\|_1/|x| $.

To prove (ii), let $f$ be as in the lemma, we have
$$
|Kf(x)|\leq\frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{|f(y)|}{|x-y|}dy
\leq\frac{1}{4\pi}\int_{B_{r^*}(x)}\frac{{f^*}(y)}{|x-y|}dy,
$$
where ${f^*}$ is the Schwarz rearrangement of $f$ with respect to
$x$ and
\[
r^*=\big(\frac{3|\operatorname{supp}(f)|}{4\pi}\big)
^{1/3}.
\]
The inequality is a consequence of Hardy-Littlewood inequality
\cite{Hardy}. Now by H\"older's inequality, we obtain
the assertion  where
\begin{equation}\label{fix c}
C=\frac{1}{4\pi}\Big(\int_{B_{r^*}(x)}\frac{1}{|x-y|^{p'}}dy\Big)
^{1/p'}=\frac{(3|\operatorname{supp}(f)|)^{\frac{1}{p'}-\frac{1}{3}}}
{(4\pi)^{2/3}(3-p')^{1/p'}},
\end{equation}
and $p'$ is the conjugate exponent of $p$.
\end{proof}

\begin{lemma}\label{Laplasian}
Let $K$ be as the above  lemma.
\begin{itemize}
\item[(i)] If   $U$ is a bounded open subset in $\mathbb{R}^3$,
$K:L^p(U)\to L^{p'}(U)$ is a linear compact operator.

\item[(ii)]For $f\in L^p(U)$, $Kf\in W^{2,p}(U)$ and
$-\Delta Kf=f$, almost everywhere in $U$.
\end{itemize}
\end{lemma}

\begin{proof}
Since $W^{1,2}(U)$ is compactly embedded into  $L^{p'}(U)$ for
$p>3$, in order to show the compactness of $K$  it is sufficient
 to prove the bondedness of $K$ as a map from $L^p(U)$ into
$W^{1,2}(U)$. To do this, let $f\in L^p(U)$ we have
\begin{equation}
|\nabla Kf(x)|
\leq \frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{|f(y)|}{
|x-y|^2}dy, \quad  x\in U.
\end{equation}
Similar to the proof of the lemma  above and the fact that
$p'<\frac{3}{2}$, we deduce
\[
 \|\nabla Kf\|_{2}\leq C\|f\|_{p},
\]
where $C$ depends on $|U|$ and $p$. This completes the proof.
For a proof of part (ii) see \cite{Em2000}.
\end{proof}

The following lemma is a simple variation of \cite[Lemma 2.15]{Burton2}.

\begin{lemma}\label{2.15}
Let $r\geq a$ and  $\upsilon\in L^{p'}(B_r)$. Denote by
$L_{\alpha}(\upsilon)$ the level set of $\upsilon$ at height
$\alpha$; that is,
$$
L_{\alpha}(\upsilon)=\{x\in B_r : \upsilon(x)=\alpha\}.
$$
Let ${ T} : L^p(B_r)\to \mathbb{R}$ be the linear functional
defined by
$$
{T}(f)=\int_{B_r}f \upsilon.
$$
If $\hat{f}$ is a minimizer of $T$ relative to
$\overline{\mathcal{R}(r)^w}$ and if
$$
\left|L_\alpha(\upsilon)\cap \operatorname{supp}(\hat{f})\right|=0,
$$
for every $\alpha\in\mathbb{R}$, then $\hat{f}\in\mathcal{R}(r)$ and
$$
\hat{f}=\varphi o \upsilon,
$$
almost everywhere in $B_r$, for some decreasing function $\varphi$.
\end{lemma}

\section{Investigation in the case: $\lambda>0$}

In this section we consider the case in which  $\lambda>0$. First we
are concerned with the existence of minimizers for the energy
functional in a bounded domain, then we will demonstrate the
problem in the unbounded domain.

\subsection{Bounded domains}
We begin with the following lemma.

\begin{lemma}\label{burton-lemma}
\begin{itemize}
\item[(i)] The energy functional $\Psi_{\lambda}$  attains
its minimum relative to $\overline{\mathcal{R}(r)^w}$
 for $r\geq a$.

\item[(ii)] If $f_{r,\lambda}$ is any minimizer for
$\Psi_{\lambda}$ relative to
$\overline{\mathcal{R}(r)^w}$, then $f_{r,\lambda}$
is a solution of the following variational problem
\begin{equation}\label{minimum}
\inf_{f\in\overline{\mathcal{R}(r)^w}}\int_{\mathbb{R}^3}
f(u_{f_{r,\lambda}}+\eta+\lambda g),
\end{equation}
where $u_{f_{r,\lambda}}$ is the solution of \eqref{poson}
corresponding to $f_{r,\lambda}$ and $\eta=Kh$.
\end{itemize}
\end{lemma}

\begin{proof}
From Lemma \ref{bound}, the optimization problem \eqref{bvariation}
is equivalent to
\[
\inf_{f\in \mathcal{R}(r)}\big\{\frac{1}{2}\int_{\mathbb{R}^3}f
Kf-\int_{\mathbb{R}^3}\eta
f+\lambda\int_{\mathbb{R}^3}g f\big\}.
\]
By Lemma \ref{Laplasian}, $K$ is compact and symmetric,
 then $\Psi_{\lambda}$ is a weakly sequentially continuous and
G\^ateaux differentiable functional. From Lemma
\ref{rearangement}, $\overline{\mathcal{R}(r)^w}$ is weakly
sequentially compact, hence $\Psi_{\lambda}$ attains its minimum on
it. If $f_{r,\lambda}$ is a minimizer of $\Psi_{\lambda}$ on
$\overline{\mathcal{R}(r)^w}$, since the G\^ateaux
differential of $\Psi_{\lambda}$ at $f_{r,\lambda}$ is
$Kf_{r,\lambda}-\eta+{\lambda}g$, then by \cite[Theorem 3.3]{Burton2},
$f_{r,\lambda}$ is a solution of the variational
problem \eqref{minimum}.
\end{proof}

\begin{lemma}\label{level set1}
Let $r\geq a$ and $f_{r,\lambda}$ be a minimizer of $\Psi_{\lambda}$
relative to $\overline{\mathcal{R}(r)^w}$. Let
$\psi_{r,\lambda}=Kf_{r,\lambda}-\eta+\lambda g$
 and denote by $L_\alpha(\psi_{r,\lambda})$ the level set of $\psi_{r,\lambda}$  at height $\alpha$.
Then there exists $\lambda_0>0$ such that for every
$\lambda>\lambda_0$,
$$
\left|L_{\alpha}(\psi_{r,\lambda})\cap \operatorname{supp}
(f_{r,\lambda})\right|=0,\quad \forall\alpha\in\mathbb{R}.
$$
\end{lemma}

 \begin{proof}
Let $r\geq a$. From Lemma \ref{burton-lemma}, for every
$\lambda>0$, the minimizer  $f_{r,\lambda}$ of $\Psi_\lambda$ on
$\overline{\mathcal{R}(r)^w}$ exists. Suppose there exists
$\hat{\alpha}\in\mathbb{R}$  such that
$\left|L_{\hat{\alpha}}(\psi_{r,\lambda})\cap
\operatorname{supp}(f_{r,\lambda})\right|>0$.
Let $S_{\hat{\alpha}}=
L_{\hat{\alpha}}(\psi_{r,\lambda})\cap
\operatorname{supp}(f_{r,\lambda})$. Since
$\psi_{r,\lambda}=u_{f_{r,\lambda}}+\eta+\lambda g$,
 using \cite[Theorem 7.7]{gilbarg},
 lemma \ref{Laplasian} and  equation \eqref{poson},  we have
\begin{equation}\label{1}
-\Delta\psi_{r,\lambda}=f_{r,\lambda}-h-\lambda\Delta g=0, \quad
\text{ a.e. in }\ S_{\hat{\alpha}}.
\end{equation}
 On the other hand, by Lemma \ref{rearangement},
\begin{equation}\label{2}
\int_{0}^sf_{r,\lambda}^\Delta(t)dt\leq\int_0^sf_0^\Delta(t)dt,
\quad  s>0.
\end{equation}
Then we deduce
\begin{equation}\label{3}
\| f_{r,\lambda}^\Delta\|_{\infty} \leq \| f_0^\Delta\|_{\infty}.
\end{equation}
Since $f_{r,\lambda}^\Delta$  is a rearrangement of
$f_{r,\lambda}\in \overline{\mathcal{R}(r)^w}$ and $f_0^\Delta$ is a
rearrangement of $f_0$, from  equation \eqref{3}, we conclude
\begin{equation}\label{4}
\| f_{r,\lambda}\|_{\infty} \leq \| f_0\|_{\infty}\,.
\end{equation}
 If we assume that  $\lambda_0=\|
f_0\|_\infty / \| \Delta g \|_{-\infty}$, then
for every $\lambda>\lambda_0$, we have
\begin{equation}\label{4'}
\| f_0\|_{\infty} < \| h+\lambda \Delta g \|_{-\infty}
\end{equation}
Finally, from \eqref{4} and  \eqref{4'}  for every $\lambda>
\lambda_0$, we deduce
\begin{equation}\label{5}
\| f_{r,\lambda}\|_{\infty} < \| h+\lambda \Delta g\|_{-\infty}
\end{equation}
which is a contradiction to \eqref{1}. This completes the proof.
 \end{proof}

\begin{lemma}\label{increasing function}
Let $\lambda_0$ be as in the lemma above. Then for every
$\lambda>\lambda_0$, the  variational problem $P_\lambda(r)$ has a
solution for $r\geq a$. If $f_{r,\lambda}$ is any solution of
$P_\lambda(r)$, then
\begin{equation}\label{e6}
f_{r,\lambda}=\varphi_\lambda\circ (u_{f_{r,\lambda}}+\eta+\lambda
g),
\end{equation}
almost everywhere in $B_r$, for a decreasing  unknown function
$\varphi_\lambda$.
\end{lemma}

 \begin{proof}
Let $r\geq a$. From Lemma \ref{burton-lemma}, there exists
$f_{r,\lambda}\in\overline{\mathcal{R}(r)^w}$ such that
$f_{r,\lambda}$ is a minimizer of $\Psi_\lambda$ relative to
$\overline{\mathcal{R}(r)^w}$ and a solution of \eqref{minimum}.
By Lemma \ref{level set1}, for every $\lambda>\lambda_0$,
the level sets of
$\psi_{r,\lambda}=Kf_{r,\lambda}-\eta+\lambda g$ on
$\operatorname{supp}(f_{r,\lambda}) $  have zero measure.
We can use Lemma
\ref{2.15} to deduce equation \eqref{e6}.
 \end{proof}

\subsection{Unbounded domain}
We proved that the variational problem $P_\lambda(r)$ has a
solution for $\lambda>\lambda_0$ and $r\geq a$. Now   we will show
that if $r$ is chosen large enough, it ceases to have any influence
whatever on the variational problem, $P_\lambda(r)$. To do this, we
now perform some calculations to deduce the following result.

\begin{lemma}\label{exist-unb-domain}
Let $\lambda>\lambda_0$. Then, there exists $r_0>a$  such that for
$r\geq r_0$ and $f_{r,\lambda}\in S_\lambda(r)$  we have
\[
\operatorname{supp}(f_{r,\lambda})\subset B_{r_0}.
\]
\end{lemma}

 \begin{proof}
To prove this lemma, it is sufficient to show that the
 support of $f_{r,\lambda}$ does not have any dense point on the
 boundary of $B_r$ when $r$ is chosen large enough.
Let $r_h>a$ be the smallest positive number for which
$\operatorname{supp}(h)\subset B_{r_h}$.  We consider $r>r_h+1$ and
$f_{r,\lambda}\in S_\lambda(r)$. From Lemma \ref{increasing function}
we have
\begin{equation}\label{7}
f_{r,\lambda}=\varphi_\lambda\circ (u_{f_{r,\lambda}}+\eta+\lambda
g),
\end{equation}
almost everywhere in $B_r$, for a decreasing  unknown function
$\varphi_\lambda$
 where $u_{f_{r,\lambda}}$ is the  solution of \eqref{poson}
 corresponding with $f_{r,\lambda}$.
To seek a contradiction suppose the assertion is false. Then there
exists $x_0\in den(\operatorname{supp}(f_{r,\lambda} ))$(set of dense points of
support) such that $|x_0|=r$. Let $A=\operatorname{supp}(f_{r,\lambda})\cap
B_1(x_0)$, then $|A|>0$. For $x\in A$
\begin{equation}  \label{8}
 Kf_{r,\lambda}(x)
=\frac{1}{4\pi} \int_{B_r}\frac{1}{|x-y|}f_{r,\lambda}(y)dy\\
\geq \frac{1}{4\pi}\frac{\| f_0\|_1}{2r}
\end{equation}
and
\begin{equation} \label{9}
 \eta(x)=\frac{1}{4\pi} \int_{B_{r_h}}\frac{1}{|x-y|}h(y)dy
\leq \frac{1}{4\pi}\frac{\| h\|_1}{r-r_h-1}
\end{equation}
From \eqref{8},  \eqref{9} and relation $u_{f_{r,\lambda}}=
Kf_{r,\lambda}-2\eta$, we obtain
 \begin{equation}\label{lower bound}
u_{f_{r,\lambda}}(x)+\eta(x)+\lambda g(x)\geq
\frac{1}{4\pi}\Big(\frac{1}{2r}\|
f_0\|_1-\frac{1}{r-r_h-1}\|
h\|_1\Big)+\lambda g(x).
\end{equation}
Since $|\operatorname{supp}(f_{r,\lambda} )|=4\pi a^3/3$ and
$r_h>a$, there exists $D\subset B_{r_h}$  such that
$D\cap \operatorname{supp}(f_{r,\lambda})$ is
empty and $|D|>0$.  For $z\in D$ from Lemma \ref{bound} we have
\begin{equation} \label{11}
 Kf_{r,\lambda}(z)=\frac{1}{4\pi} \int_{B_r}\frac{1}{|z-y|}
f_{r,\lambda}(y)dy
\leq C\| f_0 \|_p,
\end{equation}
where $C$ depends on $p$ and $|\operatorname{supp}(f_{r,\lambda})|$.
Also
\begin{equation}
\eta(z)=\frac{1}{4\pi} \int_{B_{r_h}}\frac{1}{|z-y|}h(y)dy
\geq\frac{1}{4\pi}\frac{1}{2r_h} \| h\|_1.
\label{12}
\end{equation}
Then, from \eqref{11} and \eqref{12} we derive
\begin{equation}
u_{f_{r,\lambda}}(z)+\eta(z)+\lambda g(z)\leq \lambda g(z)-C_1.
\label{12'}
\end{equation}
Now,   since $|z|\leq r_h$ for $z\in D$ and \eqref{12'},  we deduce
\begin{equation}\label{upper bound}
u_{f_{r,\lambda}}(z)+\eta(z)+\lambda g(z)\leq C_2,  \quad  z\in D,
\end{equation}
where $C_2$ is a constant independent of $r$. If we make $r$ large
we derive from \eqref{lower bound} and \eqref{upper bound}
$$
(u_{f_{r,\lambda}}(x)+\eta(x)+\lambda g(x))
-(u_{f_{r,\lambda}}(z)+\eta(z)+\lambda g(z))>0,
$$
for $x\in A$ and $z\in D$. Since $|A|>0$, $|D|>0$, this is a
contradiction to \eqref{7}.
 \end{proof}

\subsection{Proof of Theorem \ref{thm1}}
Let $r_0$ be as in Lemma \ref{exist-unb-domain}.
Assume $f_{r,\lambda}$ to be a solution of $P_\lambda(r)$
for $r\geq r_0$ and $\lambda>\lambda_0$.
 From Lemma \ref{exist-unb-domain},
$\operatorname{supp}(f_{r,\lambda})\subset B_{r_0}$ for $r>r_0$,
therefore we obtain the inclusion $S_\lambda(r_0)\subset
S_\lambda$ that it means $P_{\lambda}$ has a solution. Let
$f_{\lambda}\in S_{\lambda}$ for $\lambda>\lambda_0$. To prove
the last part of theorem,  if ${f}_\lambda\in S_\lambda$ we have
by applying Lemma \ref{increasing function}
\begin{equation}\label{14}
{f}_\lambda=\varphi_\lambda\circ (u_{f_{\lambda}}+\eta+\lambda g),
\end{equation}
almost everywhere in $B_r$ for $r>r_0$ and  a decreasing unknown
function $\varphi_\lambda$. Notice that we can suppose
$\varphi_\lambda\geq0$. Since $u_{f_{\lambda}}+\eta+\lambda g$ is a
continuous function on the compact set $B_{r_0}$, and
$\operatorname{supp}(f_{\lambda})\subset B_{r_0}$,
there exists $k\in\mathbb{R}$ such that
\begin{equation}\label{15}
u_{f_{\lambda}}+\eta+\lambda g< k\quad \text{a.e }
\operatorname{supp}(f_{\lambda}).
\end{equation}
On the other hand, by applying condition \eqref{lower bound} we have
$u_{f_{\lambda}}+\eta+\lambda g\to +\infty$, as $|x|\to\infty$. Then
we can find $r > r_0$ such that
\begin{equation}\label{16}
u_{f_{\lambda}}+\eta+\lambda g\geq k\quad\text{a.e outside } B_r.
\end{equation}
Now define
 $$
\hat{\varphi}_\lambda(t)=\begin{cases}
{\varphi}_\lambda(t) &t<k\\
0&\text{otherwise}.
\end{cases}
$$
Clearly $\hat{\varphi}_\lambda$ is a decreasing function
and ${f}_\lambda=\hat{\varphi}_\lambda\circ (u_{f_{\lambda}}
+\eta+\lambda g)$
almost everywhere on $\mathbb{R}^3$.

\section{The case $\lambda=0$}

To derive the existence result in this case we assume
some conditions. Here
we suppose  $f_0$ and $h$ satisfy all conditions mentioned in the
Theorem \ref{thm2}. Now  we deduce the following result in bounded
domain.

\begin{lemma}\label{level set}
Let $r\geq a$ and $f_r$ be a minimizer of $\Psi_0$ relative to
$\overline{\mathcal{R}(r)^w}$. Let $\psi_r=u_{f_r}+\eta$ where $u_r$ is
a  solution of \eqref{poson} corresponding  to $f_r$. Denote by
$L_\alpha(\psi_r )$ the level set of $\psi_r$  at height $\alpha$.
Then
$$
|L_{\alpha}(\psi_r)\cap \operatorname{supp}(f_r)|=0,\quad
 \forall\alpha\in\mathbb{R}.
$$
\end{lemma}

\begin{proof}
Let $r\geq a$.  Suppose there exists $\hat{\alpha}\in\mathbb{R}$  such
that $|L_{\hat{\alpha}}(\psi_r)\cap \operatorname{supp}(f_r)|>0$.
Let $A_{\hat{\alpha}}= L_{\hat{\alpha}}(\psi_r)\cap
\operatorname{supp}(f_r)$. Then
from equation \eqref{poson}, we have
\begin{equation}\label{e1}
-\Delta \psi_r=f_r-h=0, \quad \text{a.e. in } A_{\hat{\alpha}}.
\end{equation}
So $A_{\hat{\alpha}}\subset \operatorname{supp}(h)$.
On the other hand, by Lemma \ref{rearangement}, we have
\begin{equation*}
\int_{0}^sf_r^\Delta(t)dt\leq\int_0^sf_0^\Delta(t)dt, \quad
s>0.
\end{equation*}
Then
\begin{equation}\label{e3}
\| f_r^\Delta\|_{\infty} \leq \|f_0^\Delta\|_{\infty}.
\end{equation}
Since $f_r^\Delta$ is a rearrangement of $f_r$ and
$f_0^\Delta$ is a rearrangement of $f_0$, from \eqref{e3} we obtain
\begin{equation}\label{e4}
\| f_r\|_{\infty} \leq \|f_0\|_{\infty}.
\end{equation}
Finally, from \eqref{e4} and condition
\eqref{necess-cond-supp}, we deduce
\begin{equation}\label{e5}
\| f_r\|_{\infty} < \| h\|_{-\infty}.
\end{equation}
which is a contradiction to \eqref{e1}.
 \end{proof}

 \subsection{Proof of Theorem \ref{thm2}}

 Since the level sets of $\psi_r$ have zero measure, similar to the
 proof of Lemma \ref{increasing function} we can claim that
there  exists  minimizer $f_r$ for $P_r$ such that
\begin{equation}\label{6}
f_r=\varphi o (u_{f_r}+\eta),
\end{equation}
almost everywhere in $B_r$,  for a decreasing unknown function
$\varphi$. To prove the existence in unbounded domain, it is enough
to show that the support of $f_r$  does not have any dense point at
the boundary of $B_r$ when $r$ is chosen large enough. Let $r_h>a$
be the smallest positive number for which $\operatorname{supp}(h)\subset B_{r_h}$.
Since $b>\sqrt{3}a$,   then  similar to presented trend in the proof
of Lemma \ref{exist-unb-domain} there exits
$A\subset \operatorname{supp}(f_r)$ with positive measure
and $D\subset
\operatorname{supp}(h)\cap(\operatorname{supp}(f_r))^c$
such that $|D|>0$ and $|z-y|<b$ for almost
every $z,y\in D$. Then,   for $r>r_h+1$ we have
\begin{gather}\label{10}
u_{f_r}(x)+\eta(x)\geq\frac{1}{4\pi}\Big(\frac{1}{2r}\|
f_0\|_1-\frac{1}{r-r_h-1}\| h\|_1\Big), \quad
\text{a.e. in } A,
\\
u_{f_r}(z)+\eta(z)\leq \frac{a^2}{2}\|
f_0\|_\infty-\frac{1}{8\pi b} \|
h\|_{-\infty}|\operatorname{supp}(h)|,\quad\text{a.e. in } D.
\end{gather}
 Utilizing  conditions mentioned in  \eqref{necess-cond-supp},
there exists $C<0$ such that
\begin{equation}\label{13}
u_{f_r}(z)+\eta(z)\leq C \quad\text{a.e. in } D.
\end{equation}
If we make $r$ large enough we derive from \eqref{10}
and \eqref{13},
$$
\big(u_{f_r}(x)+\eta(x)\big)-\big(u_{f_r}(z)+\eta(z)\big)>0,
$$
for $x\in A$ and $z\in D$. Since $|A|>0$ and  $|D|>0$, this is a
contradiction to \eqref{6}. Let $r_0$ be such that $r>r_0$, support
of $f_r$  does not touch the boundary of $B_r$ where $f_r$ is a
solution of $P(r)$ for $r\geq r_0$. Then, $\operatorname{supp}(f_r)$
does not have any density point on the boundary of $B_r$ for $r>r_0$.
This means that $\operatorname{supp}(f_r)$ has a positive distance
from the boundary of $B_r$.
Hence $\operatorname{supp}(f_r)\subset B(r_0)$.
Therefore we obtain the inclusion $S(r_0)\subset S$.
It yields  that $P$ has a solution.

\subsection*{Acknowledgement}
The first author wants to thank Dr. Behrouz
Emamizadeh for his useful suggestions.

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\end{document}