\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 90, pp. 1--5.\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-2008/90\hfil A Neumann boundary-value problem]
{A Neumann boundary-value problem on an unbounded interval}

\author[P. Amster, A. Deboli\hfil EJDE-2008/90\hfilneg]
{Pablo Amster, Alberto Deboli}  % in alphabetical order

\address{Pablo Amster\newline 
Departamento de Matem\'atica,
Universidad de Buenos Aires,
Ciudad Universitaria, Pabell\'on I
(1428) Buenos Aires, Argentina. \newline
Consejo Nacional de Investigaciones Cient\'\i ficas y T\'ecnicas (CONICET)}
\email{pamster@dm.uba.ar}

\address{Alberto Deboli \newline  
Departamento de Matem\'atica,
Universidad de Buenos Aires,
Ciudad Universitaria, Pabell\'on I
(1428) Buenos Aires, Argentina }
\email{adeboli@yahoo.com.ar}

\thanks{Submitted April 11, 2008. Published June 21, 2008.}
\subjclass[2000]{34B40, 34B15}
\keywords{Boundary-value problem on the half line;
 Neumann conditions; \hfill\break\indent upper and lower solutions;
 diagonal argument}

\begin{abstract}
 We study a Neumann boundary-value problem on the half line
 for a second order equation, in which the nonlinearity
 depends on the (unknown) Dirichlet boundary data of the
 solution. An existence result is obtained by an adapted version
 of the method of upper and lower solutions, together with a
 diagonal argument.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{example}[theorem]{Example}

\section{Introduction}

We study the Neumann boundary-value problem on the half line:
\begin{equation}\label{problem}
\begin{gathered}
y''(x) = f(x,y(x),y(0),y(\infty)) \quad x\in (0,+\infty),\\
y'(0)=v_0,\quad y'(\infty)=0,
\end{gathered}
\end{equation}
where
$$
y(\infty):=\lim_{x\to +\infty} y(x),
\quad  y'(\infty):=\lim_{x\to +\infty} y'(x)
$$
 and
$f:[0,+\infty)\times \mathbb{R}^3\to \mathbb{R}$ is continuous.

Note that the nonlinearity $f$
depends on the (unknown) Dirichlet boundary
data of the solution.
For a bounded interval, this kind of problem has been
considered for example in \cite{T}, where a problem on a two-ion
electro-diffusion model is discussed.

It is worth to observe that problem \eqref{problem} is resonant,
since the kernel of the associated linear operator $\mathcal L
y:=y''$ is non-trivial. For the case of a bounded interval, the
Neumann problem has been widely studied in the literature; in this
situation, the operator $\mathcal L$ is Fredholm of index $0$, and
the problem can be solved by the use of coincidence degree
\cite{Ma}.

If the interval is unbounded then $\mathcal L$ is not a
Fredholm operator anymore, and
an alternative method of proof is needed (see e.g.
\cite{Andres,Const, Furi,Przera,Rabier,Regan, S}).

In contrast with the above mentioned works, we deal with an extra
difficulty, which arises on the fact that the nonlinear term in
problem \eqref{problem} depends also on the (unknown) Dirichlet
boundary values of the solution. In particular, it is implicitly
assumed that the limit value $y(\infty)$ exists, and it is finite;
in this sense, the problem is over-determined. We shall prove the
existence of solutions in presence of an ordered couple of a lower
and an upper solution that converge at infinity to the same limit
$L$. More precisely, we shall assume the existence of smooth
functions $\alpha,\beta:[0,\infty)\to \mathbb{R}$ such that 
$\alpha\le \beta$, and
$\alpha'(0)\ge v_0\ge \beta'(0)$, $\alpha(\infty)=\beta(\infty)=L$ and
$$
\alpha''(x)\ge f(x,\alpha(x),u,L),\quad \beta''(x)\le f(x,\beta(x),u,L)
$$
for each $x\in [0,\infty)$ and $u\in [\alpha(0),\beta(0)]$.

Then, we shall proceed in the following way: firstly, we prove the
existence of solutions of a mixed boundary-value problem for any
bounded interval $[0,N]$; then we apply a diagonal argument in
order to obtain a subsequence of $\{ u_N\}$ that converges to a
solution of the equation. Finally, under an extra assumption on
$f$ we shall prove that the Neumann boundary condition at infinity
is satisfied.

Our main theorem reads as follows:

\begin{theorem}\label{main}
Let $(\alpha,\beta)$ be an ordered couple of a lower and an upper 
solution as before.
Furthermore, assume that
$$
|f(x,y,u,L)|\le \varphi(x)
$$
for $x\ge x_0$, $\alpha(0)\le u\le \beta(0)$, $\alpha(x)\le y\le \beta(x)$ and
some $\varphi\in L^1(x_0,+\infty)$. Then  \eqref{problem} admits
at least one solution $y$, with $\alpha\le y\le \beta$.
\end{theorem}

\section{Mixed conditions on a bounded interval. Upper and lower solutions}

In this section, we consider the  problem with mixed conditions
\begin{equation} \label{problem-bounded}
\begin{gathered}
y''(x) = f(x,y(x),y(0),L) \quad x\in (0,T),\\
y'(0)=v_0,\quad y(T)=y_T,
\end{gathered}
\end{equation}
for some arbitrary constants $y_T,L\in \mathbb{R}$ and $T>0$.

We shall prove the existence of solutions of
\eqref{problem-bounded} in presence of an ordered couple of a
lower  and an upper solution. More precisely, we shall assume the
existence of some smooth functions $\alpha\le \beta$ satisfying
\begin{equation}\label{ul1}
\alpha''(x)\ge f(x,\alpha(x),u,L),\quad \beta''(x)\le f(x,\beta(x),u,L)
\end{equation}
for $0\le x\le T$ and $\alpha(0)\le u\le \beta(0)$, and
\begin{equation}\label{ul2}
\alpha'(0)\ge v_0\ge\beta'(0), \quad \alpha(T)\le y_T\le\beta(T).
\end{equation}

The following theorem is a slight variation of the standard
results in the theory of upper and lower solutions; for the sake
of completeness we give a simple proof.

\begin{theorem}\label{bounded}
Let $\alpha\le \beta$ satisfy \eqref{ul1} and \eqref{ul2}. Then
\eqref{problem-bounded} admits at least one solution $y$, with
$\alpha\le y\le \beta$.
\end{theorem}

\begin{proof} Let $\lambda >0$ be a fixed constant and consider,
for each $w\in C([0,T])$, the modified linear problem
\begin{equation} \label{modif.linear problem}
\begin{gathered}
y''(x)-\lambda y(x)=f(x,P_x(w(x)),P_0(w(0)),L)-\lambda P_x(w(x)) \\
y'(0)=v_0,\quad y(T)=y_T
\end{gathered}
\end{equation}
where $P:[0,T]\times \mathbb{R}\to \mathbb{R}$ is the
truncation function
\begin{equation}\label{truncation}
P_x(w):=P(x,w)=
\begin{cases}
\alpha &\text{if } w < \alpha(x) \\
w &\text{if } \alpha(x)\leq w\leq \beta(x) \\
\beta &\text{if } w > \beta(x)
\end{cases}
\end{equation}

Problem (\ref{modif.linear problem})
has a unique solution $y\in C^2([0,T])$, given by
the integral representation
$$
y(x)=y_0(x)+\int_0^T G(x,s)\phi(s)ds,
$$
where
\begin{gather*}
y_0(x)=\frac{v_0 \sinh(\sqrt{\lambda}(x-T)) +\sqrt{\lambda} y_T
\cosh(\sqrt{\lambda}x)}{\sqrt{\lambda} \cosh(\sqrt{\lambda}T)},\\
\phi(s)=\phi(w,s):=
f(s,P_s(w(s)),P_0(w(0)),L)-\lambda P_s(w(s))
\end{gather*}
and $G$ is the Green function
$$
G(x,s)=\begin{cases}
\frac{\cosh(\sqrt{\lambda}s)\sinh(\sqrt{\lambda}(x-T))}{\sqrt{\lambda}
\cosh(\sqrt{\lambda}T)} & \text{if } 0\leq s<x \\[3pt]
\frac{\cosh(\sqrt{\lambda}x)\sinh(\sqrt{\lambda}(s-T))}{\sqrt{\lambda}\cosh(\sqrt{\lambda}T)}
& \text{if } x\leq s\leq T.
\end{cases}
$$
Then, we  define the compact operator $K: C([0,T])\to C([0,T])$
by
$$
Kw(x) =y_0(x)+\int_0^T G(x,s)\phi(w,s)ds,
$$
with $y_0, \phi$ and $G$ as above.

It is clear that $\|\phi(w,\cdot)\|_\infty\le M$
where the constant $M$ is independent of $w$; thus, there exists
$R>0$ such that
$\|Kw\|_\infty\le R$ for every $w\in C([0,T])$.
From Schauder's Theorem it follows that
$K$ has a fixed point $y\in \overline {B_R(0)}$.

 Next, we shall prove
that $\alpha(x)\leq y(x)\leq \beta(x)$ for all $x\in [0,T]$, and hence
$y$ is a solution of the original problem.
It suffices to prove that $\alpha\le y$, since the other inequality is analogous.


By contradiction, suppose there exists
$x_0\in [0,T]$ such that $y(x_0)<\alpha(x_0)$.
We may assume that $x_0$ is an absolute maximum value of
the function $\alpha-y$.

We consider the two possible cases:

\noindent\textbf{Case 1.} $x_0\in (0,T)$.
From the definition of $P$, it follows that
$P_{x_0}(y(x_0))=\alpha(x_0)$. Moreover,
$\alpha(0)\leq P_0(y(0))\leq \beta(0)$
and hence
\[
\alpha''(x_0)\geq f(x_0,\alpha(x_0),P_0(y(0)),L).
\]
On the other hand,
$$
0\geq (\alpha- y)''(x_0)
=\alpha''(x_0)- f(x_0,\alpha(x_0),P_0(y(0)),L)
+\lambda  (\alpha(x_0)-y(x_0))>0,
$$
which is a contradiction.


\noindent\textbf{Case 2.} $x_0=0$ or $x_0=T$.
As $y(T)=y_T \ge \alpha(T)$, it follows that $x_0=0$. Moreover,
$\alpha'(0)\geq y'(0)=v_0$,
and by continuity
$$
(\alpha-y)''=\alpha''- f(\cdot,\alpha,P_0(y(0)),L) +\lambda  (\alpha-y(\cdot))>0
$$
over some interval $(0,\delta)$ with $\delta$ small enough. This implies
$(\alpha-y)' > 0$ on $(0,\delta)$, and
then $\alpha-y > (\alpha - y)(0)$ on $(0,\delta)$.
This contradicts the fact that $0$ is an absolute maximum of the function
$\alpha-y$.
\end{proof}

\section{A diagonal argument for problem \eqref{problem}}

In this section we give a proof of Theorem \ref{main}.
From Theorem \ref{bounded}, for every $N\in \mathbb{N}$
we  consider a solution $y_N$ of
\begin{equation}\label{problem-N}
\begin{gathered}
y_N''(x) = f(x,y_N(x),y_N(0),L) \quad x\in (0,N),\\
y_N'(0)=v_0,\quad y_N(N)=\frac{\alpha(N)+\beta(N)}2,
\end{gathered}
\end{equation}
such that $\alpha|_{[0,N]}\leq y_N\leq\beta|_{[0,N]}$.
For fixed $M$,
let us observe firstly that if $N\ge M$ then
$$
\begin{array}{l}
\|y''_N|_{[0,M]}\|_\infty=\sup_{x\in [0,M]}\{|f(x,y_N(x),y_N(0),L)|\}\leq C_2
\end{array}
$$
for some positive constant $C_2$ independent of $N$.
Moreover, writing
\[
y'_N(x)=v_0+\int_0^{x} y''_N(s)\ ds
\]
it is also seen that
$\|y'_N|_{[0,M]}\|_\infty\leq |v_0|+M C_2:=C_1$.
Finally, from the equality
\[y_N(x)=y_N(M) - \int_x^{M} y'_N(s)\ ds\]
we conclude that
$\|y_N|_{[0,M]}\|_\infty\leq C_0$ for some constant $C_0$ depending only on
$M$.
Hence, from the Arzel\'a-Ascoli theorem
we deduce the existence of
a subsequence of $\{y_N\}_{N\ge M}$ that
converges on $[0,M]$ for the $C^1$-norm.

Thus, we may proceed as follows:
for $M=1$, let us choose a subsequence
(still denoted $\{y_N\}$)
that converges in $C^1([0,1])$ to some function $y^1$.
Repeating the procedure for $M=2, 3, \ldots$, we may assume
that $y_N|_{[0,M]}$ converges to some function $y^M$ in the
$C^1$-sense.

It follows from the construction that $y^{M+1}|_{[0,M]}=y^M$,
and this implies that the function
$y:[0,+\infty)\to \mathbb{R}$ given by
$y(x)=y^M(x)$ if $0\le x\leq M$ is well defined.
Moreover, $y'(0)=v_0$, and
$y''_N$ converges uniformly
in $[0,M]$ to $f(\cdot,y(\cdot),y(0),L)$.

Thus, for any test function $\xi\in C_0^\infty(0,M)$ we obtain:
\begin{align*}
\int_0^M f(x,y,y(0),L)\xi(x)\,dx
&= \lim_{N\to\infty} \int_0^M y_N''(x)\xi(x)\,dx\\
&= \lim_{N\to\infty}\int_0^My_N(x)\xi''(x)\,dx \\
&=\int_0^M y(x)\xi''(x)\,dx,
\end{align*}
whence
$$
y''(x)=f(x,y(x),y(0),L)
$$
in $[0,M]$, in the weak sense.
Moreover, it is clear that $y(\infty)=L$, and as
$y$ is twice continuously differentiable we deduce that $y$
is a classical solution of the equation.
Finally, for each $N\ge x_0$ we may establish a point $x_N\in (N,N+1)$ such that
$y'(x_N)=y(N+1)-y(N)\to 0$; thus, for $x > x_N$ we have
$$
|y'(x)-y'(x_N)| =\big|\int_{x_N}^x f(t,y(t),y(0),L)\,dt \big|
\le \int_{x_N}^x \varphi(t)\,dt\le
\int_{x_N}^{+\infty} \varphi(t)\,dt.
$$
As $\varphi$ is integrable, we conclude that $y'(x)\to 0$ as
$x\to \infty$, and so completes the proof.

\begin{example} \label{exa3.1} \rm
Consider the problem
\begin{gather*}
y''(x) = 2(x+1)y(x)^4 + g(x,y(x),y(0),y(\infty)) \quad x\in (0,+\infty),\\
y'(0)=v_0,\quad y'(\infty)=0,
\end{gather*}
where $g:[0,+\infty)\times \mathbb{R}^3\to \mathbb{R}$ is
continuous and satisfies
$$
g(x,0, u, 0) = 0,\quad
0\le g(x,y,u,0)\le \varphi(x) \quad \text{for $x\ge 0$, $0\le y\le\frac 1{x+1}$ and $u \in
[0,1]$}
$$
for some $\varphi\in L^1([0,+\infty))$. Then, if $v_0\in
[-1,0]$ the assumptions of Theorem \ref{main} are fulfilled,
taking $\alpha\equiv 0$ and $\beta(x)=\frac 1{x+1}$.
\end{example}

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\end{document}