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\markboth{\hfil A second order ODE \hfil EJDE--2001/75}
{EJDE--2001/75\hfil Pablo Amster \& Mar\'\i a Cristina Mariani \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2001}(2001), No. 75, pp. 1--9. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
 \vspace{\bigskipamount} \\
 %
  A second order ODE with a nonlinear final condition
 %
\thanks{ {\em Mathematics Subject Classifications:} 34B15, 34C37.
\hfil\break\indent
{\em Key words:} Nonlinear boundary-value problems, fixed point methods.
\hfil\break\indent
\copyright 2001 Southwest Texas State University. \hfil\break\indent
Submitted: October 15, 2000. Published December 10, 2001.} }
\date{}
%
\author{Pablo Amster \& Mar\'\i a Cristina Mariani}
\maketitle

\begin{abstract} 
  We study a semilinear second-order ordinary differential
  equation with initial condition $u(0)=u_0$. We prove the 
  existence of solutions satisfying a nonlinear final condition 
  $u(T)=h'(u(T))$,  under a certain growth condition. 
  Also we state conditions ensuring that any solution with 
  Cauchy data $u(0) = u_0$, $u'(0)=v_0$ is defined  on the 
  whole interval $[0,T]$.
\end{abstract}

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\newtheorem{lemma}[theorem]{Lemma}
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\section{Introduction}

We study the differential equation
\begin{equation}
u''(t)+r(t) u'(t) + g(t,u(t)) = f(t) \label{*}
\end{equation}
with initial condition $u(0) = u_0$.

In the first section, we state the basic assumptions and results
concerning the Dirichlet problem associated with (\ref{*}).
In the second section, we define a fixed point setting for
solving a problem with final value $u(T)$ depending on
the velocity at time $T$.
We prove that if $g$ satisfies a growth
condition that holds for example when $g$ is {\sl sublinear},
then there exist a class of functions $h$ such that  (\ref{*})
admits at least one solution $u$ with $u(0)=u_0$, $u(T)=h(u'(T))$.
A physical example of this equation is the forced
pendulum equation, for which existence results under Dirichlet and
periodic conditions are known, see \cite{C,H,M1} and their references.
For nonexistence results, see e.g. \cite{A,OST}.
Finally, in the third section we prove
the existence of a continuous
real function $\psi = \psi_{u_0}$ such that
a solution of (\ref{*}) with initial value $u_0$
is defined over
$[0,T]$ if and only if the equation $\psi(s)=u'(0)$
is solvable.
Furthermore, if $g$ is locally Lipschitz on $u$
the union over $u_0$ of the sets
$\{ u_0\} \times \mathop{\rm Range}(\psi_{u_0})$
is a simply connected open subset of $\mathbb{R}^2$.

\section {Basic assumptions and unique solvability of the
 Dirichlet problem}

Let $S:H^2(0,T)\to L^2(0,T)$ be the semi-linear operator
$Su= u''+ru' + g(t,u)$.
We assume throughout this paper
that $g$ is continuous and satisfies the condition
\begin{equation}
\frac {g(t,u) - g(t,v)}{u-v}\le c < \big(\frac\pi T\big)^2
\quad \text {for all } t\in [0,T], u,v \in \mathbb{R}, u\ne v \label{G1}
\end{equation}
Moreover, we shall assume that the friction coefficient
$r\in H^1(0,T)$ is non-decreasing.

Concerning the Dirichlet problem for (\ref{*}),
we recall the following results whose proofs can be found in
\cite{AM}. For related results and a
general overview of this problem, we refer the reader to \cite{D,M2}.

\begin{lemma} \label{lm1}
Let $u, v\in H^2(0,T)$ with $u-v \in H_0^1(0,T)$.
Then
$$\| Su-Sv\|_2 \ge \big((\frac \pi T)^2-c\big)\| u-v\|_2
$$
and
$$
\| Su-Sv\|_2 \ge \frac {(\pi/T)^2-c}{\pi/T} \| u'-v'\|_2
$$
\end{lemma}

\begin{theorem} \label{thm2}
The Dirichlet problem
\begin{gather*}
 Su=f(t) \quad\text {in } (0,T) \cr
 u(0)=u_0, \quad u(T)=u_T
\end{gather*}
is uniquely solvable in $H^2(0,T)$
for any $f\in L^2(0,T)$, $u_0, u_T \in \mathbb{R}$.
\end{theorem}

\begin{theorem} \label{thm3}
Let $f\in L^2(0,T)$ and $\mathcal{S} = S^{-1}(f)$ with the topology
induced by the $H^2$-norm. Then the
trace function, $\mathop{\rm Tr}: \mathcal{S}\to \mathbb{R}^2$,
given by $\mathop{\rm Tr}(u)= (u(0),u(T))$ is an homeomorphism.
\end{theorem}

\section{Nonlinearities at the endpoint}

In this section we study the problem
\begin{equation}\begin{gathered}
u''+ru'+g(t,u) = f \quad \text { in } (0,T) \\
u(0)=u_0,\quad u(T)=h(u'(T))
\end{gathered} \label{e1}
\end{equation}
for $f\in L^2(0,T)$ and $h$ continuous.
First we transform the problem in a one-dimensional fixed point problem:
Indeed, for $s\in \mathbb{R}$, we define
$u_s$ as the unique solution of the problem
\begin{gather*}
u''+ru'+g(t,u) = f \quad \text {in } (0,T) \\
u(0)=u_0,\quad u(T)=h(s)
\end{gather*}
Hence, when
$\varphi_s(t)= \frac {h(s)-u_0}T t + u_0$, we have
$$
u_s(t)- \varphi_s(t) = \int_0^T (f -ru_s'-g(\theta, u_s'))
G(t,\theta)d\theta$$
where $G$ is the Green function associated with  the second order
differential operator. Namely,
$$ G(t,\theta) =
\begin{cases}
\frac{t(\theta-T)}T & \text {if } \theta \ge t \\
\frac{\theta(t-T)}T &\text {if } \theta \le t
\end{cases}
$$
By simple computation we obtain
$$ u_s'(T) = \frac {h(s)-u_0}T +
\int_0^T (f -ru_s'-g(\theta, u_s))\frac\theta T d\theta
$$
and from Theorem \ref{thm2} we have

\begin{theorem} \label{thm4}
Let $\xi:\mathbb{R} \to \mathbb{R}$ with
$$ \xi (s) = \frac {h(s)-u_0}T + \int_0^T
(f -ru_s'-g(\theta, u_s))\frac \theta T d\theta\,.
$$
Then $\xi$ is a continuous fixed point
operator for (\ref{e1}), i.e.
$u$ is a solution of (\ref{e1}) if and only if
$u=u_s$ for some $s \in \mathbb{R}$ such that
$\xi(s) = s$.
\end{theorem}

\paragraph{Proof}
Continuity of $\xi$ follows immediately from
the continuity of $\mathop{\rm Tr}^{-1}: \mathbb{R}^2  \to S^{-1}(f) $.
Moreover, if $\xi(s)=s$,
then $u_s(T)=h(u_s'(T))$, proving that $u_s$ is a solution of
(\ref{e1}). Conversely, if $u$ is a solution of (\ref{e1}),
then $u=u_s$ for $s=u'(T)$. \quad\hfill$\Box$


We establish an existence result for (\ref{e1}) assuming that
the graph of $h$ crosses the constant $u_0$.

\begin{theorem} \label{thm5}
Assume that (\ref{G1}) holds and that $h - u_0$ has nonconstant
sign on $\mathbb{R}$. Then (\ref{e1}) admits a solution for
$T$ small enough.
\end{theorem}

\paragraph{Proof}
First we give a slightly different formulation of the equality
$\xi(s)=s$.
Integrating by parts, we see that
$$\int_0^Tr(\theta)u_s'(\theta) \theta d\theta=
r(T)Th(s) -
\int_0^T[r(\theta) + \theta r'(\theta)] u_s(\theta) d\theta
 $$
and then
$$\xi (s) = (\frac 1T - r(T)) h(s) +
\frac 1T \left[\int_0^T \theta f(\theta)d\theta - u_0\right]
+\frac 1T \int_0^T (r+\theta r')u_s -\theta g(\theta, u_s)d\theta$$
Hence, $s$ is a fixed point of $\xi$ if and only if
\begin{equation}
sT= (1-r(T)T)h(s) - u_0 + \int_0^T (r+\theta r')u_s
 -\theta g(\theta, u_s)d\theta + \int_0^T \theta f(\theta)d\theta
\label{e2}
\end{equation}
>From Lemma \ref{lm1},
$$\| u_s-\varphi_s\|_2\le \frac {T^2}{\pi^2 - cT^2}
\| Su_s - S\varphi_s\|_2 = \frac {T^2}{\pi^2 - cT^2}
\| f - r\varphi_s' - g(\cdot ,\varphi_s)\|_2
$$
and
$$\| u_s-\varphi_s\|_\infty
\le \frac {\pi T^{3/2}}{\pi^2 - cT^2}
\| f - r\varphi_s' - g(\cdot ,\varphi_s)\|_2
$$
Moreover,
$$\|\varphi_s\|_2 =
\sqrt{\frac T3
(h(s)^2+h(s)u_0+u_0^2)} := c(s) \sqrt{T}$$
and as
$$\|\varphi_s\|_\infty = \max \{ | u_0|, | h(s)|\}, \quad \quad
\varphi_s' =\frac {h(s)-u_0}T
$$
then letting $T\to 0$
for fixed $s$ we have that
$\| u_s\|_2 \to 0$ and $\| u_s\|_\infty$ is bounded. Hence,
we conclude that
the right-hand side of (\ref{e2}) converges to $h(s) - u_0$.

Setting
$s_\pm \in \mathbb{R}$ such
that $h(s_+) < u_0 < h(s_-)$,
it follows, for small $T$, that
$$T\xi (s_+) \le
h(s_+) - u_0 + B(s_+)
$$
and
$$T\xi (s_-) \ge
h(s_-) - u_0 + B(s_-)
$$
for some $B$ such that $B(s_\pm)\to 0$. Hence
it suffices to
take $T$ such that
$$h(s_+) - u_0+ B(s_+) \le Ts_+,
\quad
h(s_-) - u_0+ B(s_-) \ge Ts_-$$
\hfill$\Box$

For the next existence result, we
assume that $g$ grows at most linearly, i.e.
\begin{equation}
|g(t,x)| \le \alpha |x| + \beta \label{G2}
\end{equation}
for some positive constants $\alpha$, $\beta$.
We remark that (\ref{G1}) and (\ref{G2}) are independent: for example,
$g(x) = -x^3$ satisfies (\ref{G1}) but not (\ref{G2}).
Conversely, $g(x) = \sin (Kx)$ does not satisfy (\ref{G1}) for $K \ge
\left(\frac {\pi}{T}\right)^2$.
For simplicity we define the constants
$$c_T=
\sqrt{\frac T3}
+ \frac {T^2}{\pi^2 - cT^2}
\Big(\alpha \sqrt{\frac T3} +  \frac {\| r\|_2}T\Big), \quad
M = \Big(\| r+\theta r'\|_2 + \sqrt{\frac {T^3}3} \alpha\Big) c_T$$
and the functions
$$C_\pm(s) =
\Big( (1-r(T)T) \mathop{\rm sgn} \big(\frac {h(s)}s\big)
\pm M\Big) \Big| \frac {h(s)}s\Big|\,.
$$
                       
\begin{theorem} \label{thm6}
Assume that (\ref{G1}) and (\ref{G2}) hold.
Then (\ref{e1}) admits at least one solution $u\in H^2(0,T)$ in each of
the following cases

\noindent Case A: $M < |1-r(T)T|$, with
\begin{equation}
 T < \limsup_{s\to +\infty} C_-(s)
\quad \text {or} \quad
 T > \liminf_{s\to -\infty} C_+(s) \label{A1}
\end{equation}
and
\begin{equation}
\quad T < \limsup_{s\to -\infty} C_-(s)
\quad \text {or} \quad
 T > \liminf_{s\to +\infty} C_+(s)\label{A2}
\end{equation}

\noindent Case B:
$M > |1-r(T)T|$, with
$T > \liminf_{s\to \pm\infty} C_+(s)$

\noindent Case C: $M = |1-r(T)T|$,
and there exist sequences $s^-_j \to -\infty$,
$s^+_j \to +\infty$ such
that $T > C_+(s^\pm_j)$ for every $j$,
each one of them satisfying one of the following conditions:
\begin{equation}
 \mathop{\rm sgn} \big(\frac {h(s_j)}{s_j}\big)
= \mathop{\rm sgn}(1-r(T)T)
\quad\text{for every $j$} \label{C1}
\end{equation}
or
\begin{equation}
\lim_{j\to \infty} \frac {h(s_j)}{s_j^2} = 0 \label{C2}
\end{equation}
\end{theorem}

\paragraph{Remarks:}
i) The left-hand-side in condition \ref{A1} (resp. \ref{A2}) implies
$$\limsup_{s\to +\infty} \frac {h(s)}{s }\mathop{\rm sgn}(1-r(T)T)
> \frac T{|1-r(T)T| - M}
\quad \text { (resp. $s\to -\infty$) }$$
ii) The following assumptions are sufficient
for the right-hand-side in condition \ref{A1}
(resp. \ref{A2}) to be satisfied.
$$ %\text {2')}
\liminf_{s\to -\infty}
\Big| \frac {h(s)}s\Big| < \frac T{M + |1-r(T)T|}
\quad \text { (resp. $s\to +\infty$)}$$
or
$$ %\text {2'')}
\mathop{\rm sgn}\big(\frac {h(s_j)}{s_j}\big)
= -\mathop{\rm sgn}(1-r(T)T)
$$
for a sequence $s_j \to -\infty$ (resp. $s_j \to +\infty$).
\smallskip

\noindent iii) Conditions in case B are not fulfilled when
$$|h(s)|\ge a|s| + b, \quad \text{with}\quad a \ge
\frac T{M-|1-r(T)T|}$$

In the same way,
conditions in case C imply
$$\liminf_{|s|\to \infty} \Big| \frac {h(s)}s\Big|  < \frac T{2M} $$

\paragraph{Proof of Theorem \ref{thm6}}
As in the previous theorem,
$$\| u_s\|_2 \le \sqrt{T} c(s)
+ \frac {T^2}{\pi^2 - cT^2} \big(\alpha \sqrt{T}
c(s)
+ |h(s)-u_0| \frac {\| r\|_2}T +\| f\|_2 + \beta\big):=A(s)$$
and then
$$\| u_s\|_2 \le c_T|h(s)| + \gamma |h(s)|^{1/2}+ \delta$$
for some constants $\gamma, \delta \in \mathbb{R}$.
Moreover,
$$\Big|\int_0^T (r+\theta r')u_s
-\theta g(\theta, u_s)d\theta\Big| \le
\Big(\| r+\theta r'\|_2 + \sqrt{\frac {T^3}3} \alpha\Big) c_T|h(s)|+ R(s)$$
with $R(s) \le
C_1|h(s)|^{1/2} + C_2$ for some constants $C_1,C_2$.
We remark that $\frac {R(s)}s \to 0$ for $|s|\to \infty$ if
$h$ is {\sl subquadratic}
(i.e. $\frac {h(s)}{s^2}\to 0$ for $|s|\to \infty$).
Hence,
\begin{align*}
[(1-r(T)T) &-M \mathop{\rm sgn}(h(s))] h(s) - R(s)\\
 \le& T\xi(s) \\
\le&
[(1-r(T)T) + M \mathop{\rm sgn}(h(s))] h(s) + R(s)
\end{align*}
and it suffices to find
$s_\pm$ satisfying:
\begin{gather}
s_- T\le [(1-r(T)T) -M \mathop{\rm sgn}(h(s_-))] h(s_-) - R(s_-)\label{e-}
\\
s_+ T\ge [(1-r(T)T) + M \mathop{\rm sgn}(h(s_+))] h(s_+) + R(s_+)
\label{e+}
\end{gather}
Assuming that $s_- > 0$ then (\ref{e-}) is equivalent to
$$T\le \Big[ \mathop{\rm sgn} \big(\frac{h(s_-)}{s_-}\big)
(1-r(T)T) -M\Big] \Big|\frac{h(s_-)}{s_-}\Big| - \frac {R(s_-)}{s_-}$$
Hence, if
$M < |1-r(T)T|$ then left-hand-side of (\ref{A1}) is a sufficient
 condition for (\ref{e-}): indeed, if
$T < k  \Big|\frac{h(s_j)}{s_j}\Big|$ for $s_j \to +\infty$ and some
$k>0$, then
$$
k\Big|\frac{h(s_j)}{s_j}\Big| -
\frac {R(s_j)}{s_j} = \Big|\frac{h(s_j)}{s_j}\Big| \Big( k -
\frac {R(s_j)}{|h(s_j)|}\Big)
$$
As $|h(s_j)| \to \infty$, we have that
$R(s_j)/|h(s_j)| \to 0$ and the result follows.

In the same way, if we assume that $s_- < 0$, then (\ref{e-})
is equivalent to
$$T\ge \Big[ \mathop{\rm sgn} \Big(\frac{h(s_-)}{s_-}\Big)
(1-r(T)T) +M\Big] \Big|\frac{h(s_-)}{s_-}\Big| - \frac {R(s_-)}{s_-}$$
and right-hand-side of (\ref{A1}) is sufficient, as
well as conditions in cases B and C.
The same conclusions can be obtained for (\ref{e+}), which completes
the proof. \hfill$\Box$

\paragraph{Example}
We consider the forced pendulum equation
\begin{equation}
u''(t) + \sin u = f(t) \label{eP}
\end{equation}
for which it is clear that (\ref{G2}) holds, and
(\ref{G1}) holds when $T <\pi$. In this case $c_T= \sqrt{\frac T3}$,
$M=0$, and $C_-(s) = C_+(s) =\frac {h(s)}s $.
If we assume, further, that
$$\lim_{s\to \pm \infty}\frac {h(s)}s = L_\pm$$
then (\ref{e1}) is solvable, unless
$$
L_- \le T \le L_+ \quad \text {or} \quad L_+ \le T \le L_-
$$
In particular, (\ref{e1}) is solvable when $h$ is sublinear or superlinear
(and obviously when $h$ is linear, $h(s) = as + b$, for $T\neq a$).

It is well known that (\ref{eP}) admits
$T$-periodic solutions when
$f$ is $T$-periodic and $\int_0^T f=0$.
Furthermore, in 
\cite{C} it has been proved that
for any $2\pi$-periodic $f_0 \in L^2(0,2\pi)$
such that $\int_0^{2\pi} f_0 =0$
there exist two numbers
$d(f_0) \le 0 \le D(f_0)$ such that
(P) admits $2\pi$-periodic solutions for
$f(t) = f_0(t) + f_1$
if and only if
$$d(f_0) \le f_1\le D(f_0)$$


\paragraph{Remark}
Assuming (\ref{G1}) and (\ref{G2}) we may define the functions
$\xi^\pm:\mathbb{R}\to \mathbb{R}$ as
\begin{align*}
\xi^\pm(s)=& \frac 1T \Big( (1-r(T)T)h(s)\pm
\Big[ \| r+\theta r'\|_2 A(s) +
\sqrt{\frac {T^3}3} (\alpha A(s) + \beta) \Big] \\
&+ \int_0^T\theta f(\theta)d\theta - u_0\Big)
\end{align*}
with $A(s)$ as in the previous proof.
Then a sufficient condition for the solvability of (\ref{e1}) is the
existence of $s_\pm \in \mathbb{R}$ such that
$s_- \le \xi^-(s_-)$ and $\xi^+(s_+) \le s_+$. Indeed,
from the previous computations
we have
$$|\int_0^T (r+\theta r')u_s -\theta g(\theta, u_s)d\theta| \le
\| r+\theta r'\|_2 A(s) +
\sqrt{\frac {T^3}3} (\alpha A(s) + \beta)$$
Then $\xi^- \le \xi\le \xi^+$ and the result
the result follows from Theorem \ref{thm4}. \hfill$\Box$

\section{Blow-up results}

In this section we study the behavior of
the solutions of the Cauchy problem
\begin{equation}\begin{gathered}
u''+ru'+g(t,u) = f \quad \text { in } (0,T) \\
u(0)=u_0, \quad u'(0)= v_0
\end{gathered}\label{e3}
\end{equation}
As a simple remark, under condition (\ref{G1})
we see that if $g$ is locally Lipschitz on $u$, then
there exists an interval $I(u_0)$ such that
$v_0\in I(u_0)$ if and only if $u$ is defined over
$[0,T]$.
Indeed, it suffices to show that the set
$$I:= \{ v_0: \text{ the local solution of (\ref{e3})
does not blow up on $[0,T]$}\} $$
is connected.
Let $v_0, v_2 \in I$ and $v_1 \notin I$ such that
$v_0 <v_1 < v_2$. Then the corresponding solution $u_1$
intersects $u_0$ or $u_2$ in $(0,T]$, and from
the uniqueness in Theorem \ref{thm2},
a contradiction yields.

\paragraph{Remark}
It is well known that if the growth condition (\ref{G2}) holds,
then any solution of (\ref{e3}) 
is defined over $\mathbb{R}$ for every $u_0$.
In other words, the solutions may blow up only when
$|g|$ grows faster than linearly.

\paragraph{Example}
Let $g(t,u)= -2u^3$ and $f= 0$. Then (\ref{G1}) holds, and for
$u_0 = 0 \neq v_0$ we have that
$$u' = \mathop{\rm sgn}(v_0) \sqrt{v_0^2 + u^4}$$
Assume for example that $u$ is defined over $[0,1]$. Then,
as $| u'| > |v_0|$ for $t>0$, we have that
$|u(\frac 12)| > \frac {v_0}2$.
Moreover, $|u'| > u^2$, and hence
$$
\frac 1 {|u(\frac 12)|} - \frac 1 {|u(1)|} > \frac 12
$$
Thus,
$$ \frac 2{|v_0|} -\frac 12 > \frac 1{|u(1)|}$$
proving that $|v_0| < 4$. This shows that
$I(0) \subset (-4,4)$. \medskip


The following theorem shows that the Lipschitz condition
is not necessary in order to prove the existence of $I(u_0)$. Further,
we give an explicit expression for
$I(u_0)$ as the range of a continuous function.

\begin{theorem} \label{thm7}
Assume that (\ref{G1}) holds.
Then there exists an interval $I(u_0)$ such that the following two
conditions are equivalent:\\
i) $v_0\in I(u_0)$ \\
ii) At least one local solution of (\ref{e3}) is defined over
$[0,T]$. \\
Moreover, if
$h(s) = u_0+ sT$ and $\psi:\mathbb{R}\to \mathbb{R}$ given by
$$\psi(s) = s
+ \int_0^T (f -ru_s'-g(\theta, u_s))\frac{\theta-T}{T}d\theta,
$$
then
$I(u_0)= \mathop{\rm Range} (\psi)$.
\end{theorem}

\paragraph{Proof}
As in Section 3, we have
$$u_s(t)- \varphi_s(t)
= \int_0^T (f -ru_s'-g(\theta, u_s))G(t,\theta)d\theta$$
with
$\varphi_s(t)= st + u_0$.
By simple computation,
$u_s'(0) = \psi(s)$, and the proof is complete. \hfill$\Box$

\paragraph{Remark}
In particular, if $g$ is locally Lipschitz on $u$
then $\psi$ is injective and hence $I(u_0)$ is open.

\begin{theorem} \label{thm8}
Assume (\ref{G1}) and that $g$ is locally Lipschitz on $u$.
Then the set
$$\bigcup_{u_0\in \mathbb{R}} \{ u_0\}\times I(u_0)$$
is open and simply connected in $\mathbb{R}^2$.
\end{theorem}

\paragraph{Proof}
Let $\mathcal{S} = S^{-1}(f)$ and consider the continuous
mapping $\rho:\mathcal{S} \to \mathbb{R}^2$, $\rho(u) = (u(0),u'(0))$.
Then $v_0 \in I(u_0)$ if and only if
$(u_0,v_0)\in \mathop{\rm Range} (\rho)$.
As $g$ is locally Lipschitz, $\rho$ is injective, and
hence
$\mathop{\rm Range} (\rho) = \rho\circ \mathop{\rm Tr}^{-1}(\mathbb{R}^2)$
is open and simply connected. \hfill$\Box$

\paragraph{Acknowledgement}
The authors want to thank Professor Alfonso Castro for the careful
reading of the manuscript and his fruitful suggestions and remarks.

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\end{thebibliography}


\noindent\textsc{Pablo Amster} (e-mail: pamster@dm.uba.ar)\\
\textsc{Maria Cristina Mariani} (e-mail: mcmarian@dm.uba.ar)\\[3pt]
Departamento de Matem\'atica, \\
Facultad de Ciencias Exactas y Naturales, \\
Universidad de Buenos Aires - CONICET, \\
Pab. I, Ciudad Universitaria,
(1428) Buenos Aires, Argentina

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