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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 65, pp. 1--11.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/65\hfil Optimization of the principal eigenvalue]
{Optimization of the principal eigenvalue of the one-dimensional
Schr\"odinger operator}

\author[B. Emamizadeh, R. I. Fernandes\hfil EJDE-2008/65\hfilneg]
{Behrouz Emamizadeh, Ryan I. Fernandes}  % in alphabetical order

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

\address{Ryan I. Fernandes \newline
Department of Mathematics, The Petroleum Institute\\
P. O. Box 2533, Abu Dhabi, UAE} 
\email{rfernandes@pi.ac.ae }


\thanks{Submitted January 23, 2008. Published April 28, 2008.}
\subjclass[2000]{34B05, 34L15}
\keywords{Schr\"odinger equation; principal eigenvalue;
minimization; \hfill\break\indent
maximization; rearrangements of functions; fixed points}

\begin{abstract}
 In this paper we consider two optimization problems related
 to the principal eigenvalue of the one dimensional
 Schr\"odinger operator. These optimization problems
 are formulated relative to the rearrangement of a fixed function.
 We show that both problems have unique solutions, and each of
 these solutions is a fixed point of an appropriate function.
\end{abstract}

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

\section{Introduction}

In this paper we investigate two optimization problems related to
the following one-dimensional Schr\"{o}dinger eigenvalue problem:
\begin{equation}
\begin{gathered} -u''+\alpha f(x)u=\lambda u \quad \mbox{in $(-1,1)$} \\
u(-1)=u(1)=0,
\end{gathered} \label{schrodinger}
\end{equation}
where $\alpha$ is a non-negative constant, $f\in L^{\infty}(-1,1)$ is a
non-negative potential, and $\lambda$ denotes an eigenvalue.

We fix $f_0\in L^\infty (-1,1)$, and consider the rearrangement
class $\mathcal{R}$ generated by $f_0$, see the next section for
definition. By $\lambda_1(\alpha,f)$ we denote the principal
eigenvalue of \eqref{schrodinger} for a given $\alpha$ and $f\in
L^\infty (-1,1)$. We are interested in the following optimization
problems:
\begin{equation}
\inf_{f\in \mathcal{R}}\lambda_1(\alpha,f)\quad \mbox{and}\quad
\sup_{f\in \mathcal{R}}\lambda_1(\alpha,f) \label{optimization}
\end{equation}
Note that \eqref{schrodinger} is a scaled version of the one
dimensional steady state Schr\"{o}dinger eigenvalue equation
governing a particle of mass $m$, moving in a potential $V(x)$:
\begin{equation}
\begin{gathered}
-(\overline{h}/{2m})\;u''+V(x)\;u=\Lambda\;u \quad\mbox{in $(-1,1)$} \\
u(-1)=u(1)=0,
\end{gathered} \label{original}
\end{equation}
where \={h} stands for the Planck's constant. It is well known that for
(\ref{original}) there exists a countable collection of solutions
$u_n$ (quantum states), and real numbers $\Lambda_n$ (energies).
Therefore a physical interpretation of the problems
\eqref{optimization} is: we seek potentials that minimize
(maximize) the principal energy corresponding to the state
equations \eqref{schrodinger} relative to $\mathcal{R}$.

There is another physical motivation for considering the problems
\eqref{optimization} which we describe as follows. If we assume
$f_0=\chi_{D_0}$, the characteristic function of a given set
$D_0\subset (-1,1)$, then it turns out that $\mathcal{R}$ can be
identified with the set
\[
S=\{D\subset (-1,1):\;|D|=|D_0|\},
\]
where $|\cdot |$ denotes the one dimensional Lebesgue measure.
Therefore, the problems \eqref{optimization} can be reformulated
as
\begin{equation}
\inf_{D\subset (-1,1), D\in
S}\lambda_1(\alpha,\chi_D)\quad \mbox{and}\quad
\sup_{D\subset (-1,1), D\in S}\lambda_1(\alpha,\chi_D).
\label{optimizationn}
\end{equation}
The minimization problem in \eqref{optimizationn} is addressed in
\cite{c1}, where the problem is posed in any space dimension. There,
amongst other results, it is shown that the optimal solution
$\hat D$ provides an answer to the following physical problem,
specialized to one dimension:

\subsection*{Problem}  Build a string out of two given materials (of
varying densities) in such a way that the string has a prescribed
mass and so that the basic frequency of the resulting string (with
fixed ends) is as small as possible.
\smallskip

To avoid redundancy in \eqref{optimization} we impose a geometric
condition on $f_0$; namely, we assume that $f_0$ has no flat
sections on its graph where it is positive, see the next section
for precise formulation of this condition. Clearly, taking into
account this geometric stipulation rules out the identification
between $\mathcal{R}$ and $S$, hence the problems
\eqref{optimization} and \eqref{optimizationn} are no longer
equivalent.

We would like to mention that recently Bonder \cite{f1} investigated a
similar problem where the state equation is a nonlinear
Sch\"{o}dinger eigenvalue problem. However, in that case the
admissible set considered is a bounded, closed and convex subset
of an appropriate Lebesgue integrable function space. The problems
\eqref{optimization} cannot be considered in the context of \cite{f1},
since the set of rearrangements $\mathcal{R}$ lacks both convexity
and closedness in $L^\infty (-1,1)$.

Let us end this section by mentioning that we have also considered
\eqref{optimization} in higher dimensions and the results will be
reported in a follow up paper. We particularly want to present the
work in one dimensional setting to ensure the presentation is
elementary and essentially self contained.

\section{Preliminaries and statement of the main result}

In this section we recall the main properties of the principal
eigenvalue and eigenfunction for \eqref{schrodinger}. We also
state the definition of a rearrangement class, and some well known
rearrangement inequalities.

For problem \eqref{schrodinger}, $\lambda\in\mathbb{R}$ is called an
eigenvalue provided there exists a non-zero $u\in H^1_0(-1,1)$
that satisfies the integral equation
\[
\int_{-1}^1u'\phi'\,dx+\alpha\int_{-1}^1fu\phi\,dx
=\lambda\int_{-1}^1u\phi\,dx,\quad \forall\;\phi\in
C^\infty_0(-1,1)
\]
The function $u$ is called an eigenfunction
corresponding to $\lambda$.

The principal (first) eigenvalue of \eqref{schrodinger} denoted
$\lambda_1(\alpha,f)$, to emphasis dependence on $\alpha$ and $f$,
is variationally formulated as follows:
\begin{equation}\label{principal}
\lambda_1(\alpha,f)=\inf_{u\in
H^1_0(-1,1)\setminus\{0\}}\frac{\int_{-1}^1{u'}^2\,dx
+\alpha\int_{-1}^1fu^2\,dx}{\int_{-1}^1u^2\,dx}.
\end{equation}
While it is well known that $\lambda_1(\alpha,f)$ is positive
and simple, we provide a short proof of the simplicity of the
principal eigenvalue based on ideas introduced in \cite{b1}. Suppose
$u_1$ and $u_2$ are eigenfunctions corresponding to
$\lambda_1(\alpha,f)\equiv\lambda$. Thus both of these functions
are positive, and we have
\begin{equation} \label{eigen}
\lambda=\frac{\int_{-1}^1{u_i'}^2\,dx
+\alpha\int_{-1}^1fu_i^2\,dx}{\int_{-1}^1u_i^2\,dx},\quad i=1,2.
\end{equation}
Clearly $v\equiv (u^2_1+u_2^2)^{1/2}\in H^1_0(-1,1)$. Thus
$v'=(u^2_1+u_2^2)^{-1/2} (u_1,u_2)\cdot
(u_1',u_2')$, where the dot denotes the usual dot
product in $\mathbb{R}^2$. Note that from the Cauchy-Schwarz inequality
we get the strict inequality
$|(u_1,u_2)\cdot(u_1',u_2')|^2<(u^2_1+u^2_2)({u_1'}^2+{u_2'}^2)$,
unless $(u_1,u_2)$ and $(u_1',u_2')$ are colinear.
Therefore by easy calculations, and (\ref{eigen}), we
find
\[
\frac{\int_{-1}^1{v'}^2\,dx+\alpha
\int_{-1}^1fv^2\,dx}{\int_{-1}^1v^2\,dx}<\lambda;
\]
which contradicts \eqref{principal}. Hence $(u_1,u_2)$ and
$(u_1',u_2')$ must be colinear, so there exists a
nonzero constant $k$ such that
$(u_1,u_2)=k(u_1',u_2')$. This, in turn, implies
$(u_1' u_2-u_2' u_1)/{u_2^2}=0$. Thus
$d/dx(u_1/u_2)=0$, hence $u_1=ku_2$. This proves that the
principal eigenvalue is simple.

For a non-negative measurable $f$ defined on the interval
$(-1,1)$, the distribution function
$\xi_f:[0,\infty)\to [0,\infty)$ is defined as
\[
\xi_f (\gamma)=|\{x\in (-1,1):\;f(x)\geq\gamma\}|.
\]
We say the non-negative measurable functions
$f,g:(-1,1)\to\mathbb{R}$
are rearrangements of each other if and only if
\[
\xi_f(\gamma)=\xi_g(\gamma),\quad \forall\gamma\in [0,\infty).
\]
If $f\in L^\infty(-1,1)$, and $g$ is a
rearrangement of $f$, then $g\in L^\infty(-1,1)$ and $\|f\|_\infty
=\|g\|_\infty$.

We fix a non-negative function $f_0\in L^\infty (-1,1)$ satisfying
the following geometric condition:
\begin{itemize}
\item[(H)] $f_0$ has negligible flat sections over its strong
support; that is,
\[
|f_0^{-1}(\{C\})\cap\{f_0>0\}|=0,\quad \forall\;C>0.
\]
We define the rearrangement class $\mathcal{R}$, generated by $f_0$,
as follows
\[
\mathcal{R}=\{g:(-1,1)\to
[0,\infty):\;f_0\;\mbox{and}\;g\;\mbox{are rearrangements of each
other}\}.
\]
\end{itemize}
For a non-negative function $h$ defined on $(-1,1)$, the
decreasing rearrangement of $h$, denoted $h^\Delta$, is defined on
$(-1,1)$ as follows:
\[
h^\Delta (s)=\sup\{\gamma:\;\xi_h(\gamma)\geq s\},
\]
and the symmetric decreasing and increasing rearrangements of $h$,
denoted $h^*,\;h_*$, respectively, are given by:
\[
h^*(x)=h^\Delta (2|x|),\quad
h_*(x)=h^\Delta(2-2|x|).
\]
The following inequality is standard:
 \begin{equation}
\int_{-1}^1f_*(x)g^*(x)dx\leq\int_{-1}^1f(x)g(x)dx
\leq\int_{-1}^1f^*(x)g^*(x)dx.\label{inequality1}
\end{equation}
It is also known that if $u\in H^1_0(-1,1)$ is a non-negative
function, then $u^*\in H^1_0(-1,1)$ and
\begin{equation}
\int_{-1}^1{{u^*}'}^2dx\leq\int_{-1}^1{u'}^2dx,
\label{inequality2}
\end{equation}
see for example \cite{h1}. Now we state the main
results of this paper

\begin{theorem}[Minimization] \label{thm1}
 For every $\alpha\geq 0$, $f_*$ is the unique
solution of
\[
\inf_{f\in\mathcal{R}}\lambda_1(\alpha,f),
\]
that is, $\lambda_1(\alpha,f_*)=\inf_{f\in\mathcal{R}}\lambda_1(\alpha,f)$.
\end{theorem}

\begin{theorem}[Maximization] \label{thm2}
 There is $\hat\alpha>0$ such that
for $\alpha\leq\hat\alpha$, $f^*$ is the unique solution of
\[
\sup_{f\in\mathcal{R}}\lambda_1(\alpha,f),
 \]
that is, $\lambda_1(\alpha,f^*)=\sup_{f\in\mathcal{R}}\lambda_1(\alpha,f)$.
\end{theorem}

\section{Proofs of Theorems}
In this section we give proofs of our main results.

\begin{proof}[Proof of Theorem \ref{thm1}]
 Let us first show that $f_*$ is a
solution. For any $\alpha\geq 0$ and $f\in \mathcal{R}$, let $u$
denote the unique eigenfunction corresponding to
$\lambda_1(\alpha,f)\equiv\lambda (f)$, satisfying the condition
\begin{equation}
\int_{-1}^1u^2(x)\,dx=1.\label{normal}
\end{equation}
Then from \eqref{principal}, (\ref{inequality1}) and (\ref{inequality2}) we
deduce
\begin{equation}
\lambda(f)=\int_{-1}^1{u'}^2\,dx
+\alpha\int_{-1}^1fu^2\,dx\geq\int_{-1}^1{{u^*}'}^2\,dx
+\alpha\int_{-1}^1f_* {u^*}^2\,dx\geq\lambda (f_*). \label{inequality3}
\end{equation}
Since $f_*=(f_0)_*\equiv F_0$, we infer $\lambda(f)\geq \lambda(F_0)$.
This completes the existence part of the theorem.

To prove uniqueness we assume $g$ is also a minimizer. Denoting
the normalized eigenfunction corresponding to $\lambda
(g)\equiv\lambda_1(\alpha, g)$ by $u$, as in (\ref{inequality3}) ,
we obtain
\begin{equation}
\begin{aligned}
\lambda (g)&=\int_{-1}^1{u'}^2\,dx+\alpha\int_{-1}^1gu^2\,dx\\
&\geq \int_{-1}^1{{u^*}'}^2\,dx+\alpha\int_{-1}^1g u^2\,dx \\
&\geq \int_{-1}^1{{u^*}'}^2\,dx+\alpha\int_{-1}^1g_*
{u^*}^2\,dx  \\
&\geq \lambda (g_*)\geq\lambda(g).
\end{aligned}\label{ryan1}
\end{equation}
Thus all inequalities in (\ref{ryan1}) in fact equalities. In
particular we deduce
\begin{itemize}
\item[(a)] $\int_{-1}^1{{u^*}'}^2\,dx=\int_{-1}^1{u'}^2\,dx$

\item[(b)] $\int_{-1}^1gu^2\,dx=\int_{-1}^1g_*{u^*}^2\,dx$
\end{itemize}
At this stage we show that the graph of $u$ has no significant
flat sections. Let us consider the differential equation satisfied
by $u$
\begin{equation}
\label{ryan2} -u''+\alpha gu=\lambda (g)u.
\end{equation}
Restricting (\ref{ryan2}) to the set where $g$ is zero yields
$-u''=\lambda (g)u$, hence if $u=c$, $c$ is a
constant, then $c\lambda (g)=0$, which is not possible, since
$\lambda (g)>0$, and $c$ is positive by the positivity of the
eigenfunctions. On the other hand, if we restrict (\ref{ryan2}) to
the set where $g$ is positive, then, assuming $u=c$, we find
$\alpha cg=\lambda (g)c$, so $g=\lambda (g)/\alpha$, which is
again impossible since $g$ satisfies the geometric condition
(H).

Since $u$ has no significant flat sections on its graph, the set
on which $u'$ vanishes must have measure zero. This result
in conjunction with (a) above and  \cite[Theorem 1.1]{f2} imply $u=u^*$
almost everywhere in $(-1,1)$, see also \cite{k1}. So from (b) above we
deduce
\[
\int_{-1}^1g{u^*}^2\,dx=\int_{-1}^1g_*{u^*}^2\,dx.
 \]
Now recalling (\ref{inequality1}), we deduce that $g$ and
$g_*=F_0$ are both minimizers of the linear functional
$\mathcal{L}:L^\infty (-1,1)\to\infty$, defined by:
\[
\mathcal{L}(h)=\int_{-1}^1h\;{u^*}^2\,dx,
\]
relative to $h\in \mathcal{R}$.
Since the graph of $u$ has no significant flat sections we can use
\cite[Lemma 2.9]{b3}, \cite[Theorem 3]{b2} and \cite[Lemma 2.2]{c2}
which guarantees
that $\mathcal{L}$ has a unique minimizer relative to $\mathcal{R}$.
This obviously implies $g=F_0$, as desired.
\end{proof}


\begin{remark} \label{rmk1}
From Burton's theory one can also establish the
following functional relation
\[
F_0=\phi_\alpha(u),
\]
which holds pointwise almost
everywhere in $(-1,1)$, where $\phi_\alpha$ is a decreasing
function in terms of $u^\Delta$ and $g_\Delta$. This relation
shows that the largest parts of $F_0$ are concentrated where $u$
is the smallest. Hence the largest parts of $u$ must be
concentrated near zero.
\end{remark}

We now proceed to prove Theorem \ref{thm2}.
Henceforth $u_{\alpha,f}$
denotes the normalized eigenfunction, see (\ref{normal}),
corresponding to $\lambda_1(\alpha,f)$. We need the following
lemmas.

\begin{lemma} \label{lem1}
If $f\in L^\infty (-1,1)$ is a non-negative even function, then
$u_{\alpha,f}$ is even.
\end{lemma}

\begin{proof}
 For simplicity we set $u\equiv u_{\alpha,f}$, and
introduce $w(x)=u(-x)$. Note that $w$ satisfies the normality
condition (\ref{normal}). Moreover, observe that
\[
\int_{-1}^1{u'}^2\,dx=\int_{-1}^1{w'}^2\,dx\quad \mbox{and}\quad
\int_{-1}^1fu^2\,dx=\int_{-1}^1fw^2\,dx,
\] where in the second equation we have used $f(x)=f(-x)$. Thus
we derive
 \[
\lambda (f)=\lambda_1(\alpha,f)
 =\int_{-1}^1{w'}^2\,dx+\alpha\int_{-1}^1fw^2\,dx.
\]
Hence $w$ is also an eigenfunction corresponding to
$\lambda_1(\alpha,f)$, so by uniqueness we obtain $u=w$, as
desired.
\end{proof}


\begin{lemma} \label{lem2}
Suppose $f\in L^\infty (-1,1)$ is a non-negative
function such that $f=f^*$. There exists $\hat\alpha>0$ such that,
if $\alpha <\hat\alpha$, then $u_{\alpha,f}=u_{\alpha,f}^*$.
\end{lemma}

\begin{proof}
Note that $u\equiv u_{\alpha,f}$ is an even function
by Lemma \ref{lem1}. Hence to complete the proof of the lemma it suffices
to show $u'$ is non-positive on $(0,1)$. We first show that
$\lambda_1(\cdot,f)$ is right continuous at zero; that is,
$\lim_{\alpha\to 0^+}\lambda_1(\alpha,f)=\lambda$, where
$\lambda$ is the principal eigenvalue of \eqref{schrodinger} with
$f=0$ (or $\alpha=0$). From \eqref{principal}, for any $\alpha >0$,
 we have
\[
\lambda_1(\alpha,f)=\int_{-1}^1{u'}^2\,dx+\alpha\int_{-1}^1fu^2\,dx
\geq\int_{-1}^1{u'}^2\,dx\geq\lambda.
\]
 Therefore, $\liminf_{\alpha\to 0^+}\lambda_1({\alpha,f)\geq\lambda}$.
Also, for $\epsilon >0$, there exists $v\in H^1_0(-1,1)$
normalized such that
\[
\lambda +\epsilon\geq\int_{-1}^1{v'}^2\,dx
\geq\lambda_1(\alpha,f)-\alpha\int_{-1}^1fv^2\,dx,\quad
\forall\;\alpha >0.
\]
Therefore we obtain
$\lambda+\epsilon\geq\limsup_{\alpha\to 0^+}\lambda_1(\alpha,f)$.
Since $\epsilon$ is arbitrary we get
$\lambda\geq\limsup_{\alpha\to 0^+}\lambda_1(\alpha,f)$,
as desired.

Next, we fix a non-negative $\phi\in H^1_0(-1,1)$. Multiplying the
differential equation:
 \[
-u''+\alpha fu=\lambda_1(\alpha,f)u,
\]
by $\phi$ and integrating over $(-1,1)$ yields
\begin{equation}
\int_{-1}^1u'\phi'\,dx=\int_{-1}^1(\lambda_1(\alpha,f)-\alpha
f)u\phi\,dx. \label{ali2}
\end{equation}
Now, since $f=f^*$,
\begin{equation} \lambda_1(\alpha,f)-\alpha
f\geq\lambda_1(\alpha,f)-\alpha f(0),\label{ali1}
\end{equation}
and since $\lambda_1(\cdot,f)$ is right-continuous at zero, the function
$q(\alpha)\equiv \lambda_1(\alpha,f)-\alpha f(0)$ is also
right-continuous at zero. But since $q(0)>0$ we infer the
existence of $\hat\alpha>0$ such that $q$ is positive on
$[0,\hat\alpha]$. With $\alpha\in [0,\hat\alpha]$, from
(\ref{ali1}) and (\ref{ali2}) we deduce
\[
\int_{-1}^1u'\phi'\,dx=\int_{-1}^1(\lambda_1(\alpha,f)-\alpha
f)u\phi\,dx\geq q(\alpha)\int_{-1}^1u\phi\,dx\geq 0.
\]
Hence
$u''\leq 0$ in the sense of distributions. Note that
since $(\lambda_1(\alpha,f)-\alpha f)u\in L^\infty(-1,1)$, $u\in
W^{2,p}(-1,1)$ for every $p\in [1,\infty)$. This implies that in
fact $u''\leq 0$ holds almost everywhere in $(-1,1)$.
Therefore the following inequality holds:
\[
\int_{0}^xu''\,dt\leq 0,
\]
for $x\in (0,1)$. Since $u'$ is absolutely continuous on $[-1,1]$,
we can apply the Fundamental Theorem of Calculus to deduce
\[
u' (x)-u' (0)\leq 0,\quad x\in (0,1)
 \]
Recall that $u$ is an even function, thus clearly we have
\[
(u(x)-u(-x))'=0,\quad x\in (-1,1).
\]
Thus, in particular, we find $u' (0)=0$. Hence we obtain $u'
(x)\leq 0$ for almost every $x$ in $(0,1)$. In fact, since
$u'$ is continuous we have $u' (x)\leq 0$ for every
$x$ in $(0,1)$. This completes the proof of the lemma.
\end{proof}

 From Lemmas \ref{lem1} and  \ref{lem2}, the following result is
immediate.

\begin{lemma} \label{lem3}
If $\alpha\leq\hat\alpha$ and $f\in L^\infty (-1,1)$, then
$u_{\alpha, f^*}$ satisfies
\[
u_{\alpha,f^*}=(u_{\alpha,f^*})^*,
\]
almost everywhere in $(-1,1)$.
\end{lemma}

\begin{proof}[Proof of Theorem \ref{thm2}]
 Let $\hat\alpha$ be given by Lemma \ref{lem2}, and let $\alpha\leq\hat\alpha$.
Fix $f\in \mathcal{R}$, and set
$\lambda (f)\equiv \lambda_1(\alpha,f)$, $u_f\equiv u_{\alpha,f}$
for simplicity. Then from \eqref{principal}, (\ref{inequality1})
and Lemma \ref{lem3} we have
\begin{align*}
\lambda (f)&\leq\int_{-1}^1u^{\prime 2}_{f^*}\,dx
 +\alpha\int_{-1}^1f{u_{f^*}}^2\,dx\\
&\leq\int_{-1}^1 u{\prime 2}_{f^*}\,dx+ \alpha\int_{-1}^1f^*{(u_{f^*})^*}^2\,dx
\\
&=\lambda (f^*)=\lambda (F^0),
\end{align*}
 where $F^0\equiv f^*_0$. This completes
the existence part of the theorem. To prove uniqueness, assume
$g\in \mathcal{R}$ is another maximizer besides $F^0$. Then
 \[
\lambda (g)\leq\int_{-1}^1{u^{\prime
2}_{g^*}}\,dx+\alpha\int_{-1}^1g{u^2_{g^*}}\,dx
\leq\int_{-1}^1{u^{\prime 2}_{g^*}}\,dx
+ \alpha\int_{-1}^1g^*{u^2_{g^*}}\,dx=\lambda
(g^*)\leq\lambda (g).
\]
Thus all inequalities above are in fact
equalities, hence in particular we infer
\begin{equation}
\int_{-1}^1g^*{u^2_{g^*}}\,dx=\int_{-1}^1g{u^2_{g^*}}\,dx.
\label{ali4}
\end{equation}
As in the minimization case, we show that the
graph of $u_{g^*}$ has no significant flat sections. Recall that
$u_{g^*}$ satisfies the following differential equation
\begin{equation}
-u''_{g^*}+\alpha g^*u_{g^*}=\lambda_1(\alpha,
g^*)u_{g^*}. \label{ali3}
\end{equation}
Assume that $u_{g^*}=c$ on a subset
of $E\subseteq\{g^*=0\}$, where $E$ has positive measure. Whence
restricting (\ref{ali3}) to $E$ yields $c\lambda_1(\alpha,g^*)=0$,
which is impossible since both $c$ and $\lambda_1(\alpha,g^*)$ are
positive constants. On the other hand if we assume $u_{g^*}=c$ on
a subset $K\subseteq\{g^*>0\}$, where $K$ has positive measure,
then restricting (\ref{ali3}) to $K$ implies $\alpha
cg^*=c\lambda_1(\alpha,g^*)$, from which it follows that the graph
of $g^*$ has a flat section on a set of positive measure which is
not possible because $g^*$ satisfies the condition (H). This
shows that the graph of $u_{g^*}$ has no significant flat
sections.

 From (\ref{ali4}) and (\ref{inequality1}) it follows that $g$ and
$g^*$ both maximize the linear functional $\mathcal{N}:L^\infty
(-1,1)\to\mathbb{R}$ defined by
 \[
\mathcal{N}(h)=\int_{-1}^1hu^2_{g^*}\,dx,
\]
relative to $h\in \mathcal{R}$.
However, by Burton's theory \cite{b2}, $u_{g^*}$ not having any
significant flat sections on its graph, $\mathcal{N}$ has a unique
maximizer relative to $\mathcal{R}$, hence we
must have $g=g^*=F^0$, as desired.
\end{proof}


\begin{remark}  \label{rmk2} \rm
 As in the case of the minimization, the Burton's
theory provides a functional relation as follows:
 \[
F^0=\psi_\alpha (u),
\] almost everywhere in $(-1,1)$, where
$u\equiv u_{F^0}$, and $\psi_\alpha$ is an increasing function.
\end{remark}

\section{$F_0$ and $F^0$ as fixed points}

In this section we show the solutions to the optimization problems
\eqref{optimization}; namely, $F_0$ and $F^0$ are fixed points of
appropriate operators. We begin by proving a continuity result for
the principal eigenvalue.

\begin{lemma} \label{lem4}
For every $\alpha\geq 0$, the function
$\lambda_1(\alpha, \cdot):\;L^\infty(-1,1)\to\mathbb{R}$ is
continuous.
\end{lemma}

\begin{proof}
 Let us fix $\alpha\geq 0$, and assume $f_n\to
f$ in $L^\infty(-1,1)$. We want to show $\lambda_1(\alpha,
f_n)\to\lambda_1(\alpha,f)$, as $n\to\infty$. For
simplicity we use the notation
$\lambda (f_n)\equiv\lambda_1(\alpha,f_n)$ and
$\lambda (f)\equiv\lambda_1(\alpha,f)$; also $u_n\equiv u_{\alpha,f_n}$ and
$u\equiv u_{\alpha,f}$.
 From \eqref{principal}, we get
\[
\lambda (f_n)\leq\int_{-1}^1{u'}^2\,dx+\alpha\int_{-1}^1f_nu^2\,dx,
\quad n\in\mathbb{N}
\]
Therefore
\begin{equation}
\limsup_{n\to\infty}\lambda (f_n)\leq\int_{-1}^1{u'}^2\,dx
+\alpha\int_{-1}^1fu^2\,dx=\lambda
(f).\label{ali5}
\end{equation}
On the other hand for every $n\in\mathbb{N}$ we have
\[
\lambda (f_n)=\int_{-1}^1{u^{\prime2}_n}\,dx+\alpha\int_{-1}^1f_nu_n^2\,dx.
\]
Thus, since $\{\lambda (f_n)\}$ and $\{f_n\}$ are bounded, it follows that
$\{u_n\}$ is bounded in $H^1_0(-1,1)$. Hence there exist a
subsequence of $\{u_n\}$, still denoted $\{u_n\}$, and a function
$\hat u\in H^1_0(-1,1)$ such that
\[
u_n\rightharpoonup\hat u,\quad \mbox{in }H^1_0(-1,1),\quad
u_n\to\hat u, \mbox{ uniformly on $(-1,1)$}.
\]
Whence
\begin{equation}\label{ali6}
\begin{aligned}
\liminf_{n\to\infty}\lambda (f_n)
&\geq \liminf_{n\to\infty}\Big(\int_{-1}^1{u^{\prime
2}_n}\,dx+\alpha\int_{-1}^1f_nu_n^2\,dx\Big)  \\
& \geq \int_{-1}^1({\hat u'})^2\,dx+\alpha\int_{-1}^1f{\hat
u}^2\,dx\geq\lambda (f).
 \end{aligned}
\end{equation}
 Clearly, (\ref{ali5}), (\ref{ali6}) and the fact that the
 sequence $\{f_n\}$ is arbitrary prove the assertion of the lemma.
\end{proof}

\begin{lemma} \label{lem5}
Let $\alpha\geq 0$, and $f_n\to f$ in $L^\infty (-1,1)$.
Then $u_{\alpha,f_n}\to u_{\alpha,f}$ uniformly on
$(-1,1)$.
\end{lemma}

\begin{proof}
 From Lemma \ref{lem4}, $\lambda_n\to\lambda$ as
$n\to\infty$. Hence $\{\lambda_n\}$ is bounded; which
implies that $\{u_n\}$ is bounded in $H^1_0(-1,1)$. Thus there
exist a subsequence, still denoted $\{u_n\}$, and a function
$\hat u\in H^1_0(-1,1)$ such that
\begin{equation}
u_n\rightharpoonup\hat u,\quad \mbox{in }H^1_0(-1,1),\quad
u_n\to\hat u,\quad \mbox{uniformly on }(-1,1). \label{ali7}
\end{equation}
We show that $\hat u=u$; this proves the assertion of the lemma,
since $u$
is the only accumulation point of $\{u_n\}$. From
\eqref{principal}, (\ref{ali7}) and the fact that the $H^1_0$-norm
is weakly sequentially lower semi-continuous we deduce
\begin{equation}
\begin{aligned}
\lambda&\leq\int_{-1}^1{{\hat u}^{\prime 2}}\,dx
+\alpha\int_{-1}^1f{\hat u}^2\,dx \\
&\leq \liminf_{n\to\infty}
\Big(\int_{-1}^1{u'_n}^2\,dx+\alpha\int_{-1}^1f_nu_n^2\,dx\Big) \\
&= \liminf_{n\to \infty}\lambda_n=\lambda.
\end{aligned}
\end{equation}
Thus we obtain
\[
\lambda=\int_{-1}^1{{\hat u}^{\prime 2}}\,dx
 +\alpha\int_{-1}^1f{\hat u}^2\,dx,
\]
which shows that $\hat u$ is also an eigenfunction corresponding to
$\lambda$.
However, since $\hat u$ satisfies (\ref{normal}), it follows from
uniqueness that $u=\hat u$, as desired.
\end{proof}

To state the next result we need some more notation. The principal
eigenvalue of \eqref{schrodinger} is a map
$\lambda_1:\mathbb{R}\times L^\infty (-1,1)\to\mathbb{R}$.
The Fr\'{e}chet derivative of $\lambda_1$ with respect to the
second variable, if it exists, denoted $\partial_2\lambda_1(\alpha,\cdot)$.
For a function $f\in L^\infty (-1,1)$, $\partial_2\lambda_1(\alpha,f)$
would be a linear functional from $L^\infty(-1,1)$ into the reals. By
$\partial_2\lambda_1(\alpha,f)[h]$ we denote the Fr\'{e}chet
derivative at $f$ applied to $h\in L^\infty(-1,1)$.

\begin{lemma} \label{lem6}
Let $\alpha\geq 0$, and $f$ and $h$ be functions in $L^\infty
(-1,1)$. Then
\begin{equation}\label{derivative}
\partial_2\lambda_1(\alpha,f)[h]=\alpha\int_{-1}^1hu_{\alpha,f}^2\,dx,
\end{equation}
hence $\partial_2\lambda_1(\alpha,f)$ can be identified with
$\alpha u_{\alpha,f}^2$.
\end{lemma}

\begin{proof} From \eqref{principal} we have
\begin{equation}
\begin{aligned}
\lambda_1(\alpha,f+h)
&\leq \int_{-1}^1{u'}^2_{\alpha,f}\,dx
+\alpha\int_{-1}^1(f+h)u^2_{\alpha,f}\,dx  \\
&= \lambda_1(\alpha,f)+\alpha\int_{-1}^1hu^2_{\alpha,f}\,dx
\end{aligned}
\end{equation}
Thus we derive
\begin{equation} \label{ali8}
\lambda_1(\alpha,f+h)-\lambda_1(\alpha,f)-\alpha\int_{-1}^1
hu^2_{\alpha,f}\,dx\leq 0.
\end{equation}
On the other hand, again from \eqref{principal}, we have
\begin{align*}
\lambda_1(\alpha,f)
&\leq\int_{-1}^1{u'}^2_{\alpha,f+h}\,dx
+\alpha\int_{-1}^1fu^2_{\alpha,f+h}\,dx  \\
&=\lambda_1(\alpha,f+h)-\alpha\int_{-1}^1hu^2_{\alpha,f+h}\,dx.
\end{align*}
Thus
\begin{equation}\label{ali9}
\lambda_1(\alpha,f+h)-\lambda_1(\alpha,f)
-\alpha\int_{-1}^1hu^2_{\alpha,f}\,dx
\geq\alpha\int_{-1}^1h(u^2_{\alpha,f+h}-u^2_{\alpha,f})\,dx.
\end{equation}
 From (\ref{ali8}) and (\ref{ali9}) we deduce
\begin{equation} \label{ali10}
\begin{aligned}
|\lambda_1(\alpha,f+h)-\lambda_1(\alpha,f)
-\alpha\int_{-1}^1hu^2_{\alpha,f}\,dx|
&\leq \alpha\|h\|_\infty\int_{-1}^1|u^2_{\alpha,f+h}-u^2_{\alpha,f}|\,dx \\
&\leq 2\alpha\|h\|_\infty\|u_{\alpha,f+h}-u_{\alpha,f}\|_2 \\
&\leq 4\alpha\|h\|_\infty\|u_{\alpha,f+h}-u_{\alpha,f}\|_\infty,
\end{aligned}
\end{equation}
where in the second inequality we used the H\"{o}lder inequality,
and in the third one the fact that $u_{\alpha,f+h}$ and
$u_{\alpha,f}$ both satisfy (\ref{normal}).
Note that from Lemma \ref{lem5}, we infer that
$\|u_{\alpha,f+h}-u_{\alpha,f}\|_\infty=o(1)$, as
$\|h\|_\infty\to 0$. This result coupled with
(\ref{ali10}) imply that
\[
\lambda_1(\alpha,f+h)-\lambda_1(\alpha,f)
-\alpha\int_{-1}^1hu^2_{\alpha,f}\,dx=o(\|h\|_\infty),
\]
as $\|h\|_\infty\to 0$. This, in turn, verifies
(\ref{derivative}).
\end{proof}

To state the next result we introduce the functional
$\Phi_\alpha:L^\infty(-1,1)\to\mathbb{R}$ as
\[
\Phi_\alpha (g)=\sqrt{\frac{1}{\alpha}\partial_2
\lambda_1(\alpha,g)},\quad \alpha>0.
\]
Note that $\Phi_\alpha$ is well-defined since from
Lemma \ref{lem6}, $\partial_2\lambda_1(\alpha,g)$ can be identified with the
positive function $\alpha u^2_{\alpha,g}$.

\begin{theorem} \label{thm3}
\begin{itemize}
\item[(a)] For every $\alpha>0$, we have
$F_0=\phi_\alpha\circ\Phi_\alpha (F_0)$,
almost everywhere in $(-1,1)$.

\item[(b)] For $0<\alpha\leq\hat\alpha$, we have
$F^0=\psi_\alpha\circ\Phi_\alpha (F^0)$,  almost everywhere in
$(-1,1)$.

\end{itemize}
Where the functions $\phi_\alpha$ and $\psi_\alpha$ are the
functions stated in Remarks \ref{rmk1}, and \ref{rmk2},
respectively.
\end{theorem}

The proof of the above theorem is straightforward; so we omit it.

\subsection*{Acknowledgements}
The authors would like to acknowledge the financial support
received through separate research initiation grants from The
Petroleum Institute, Abu Dhabi, UAE. They would also like to thank
the anonymous referee for reading the manuscript carefully.

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\end{document}