\documentclass[twoside]{article}
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\markboth {Nonlinearities in a second order  ODE}
{ Pablo Amster }
\begin{document}
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\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
USA-Chile Workshop on Nonlinear Analysis, \newline
 Electron. J. Diff. Eqns., Conf. 06, 2001, pp. 13--21\newline 
http://ejde.math.swt.edu or http://ejde.math.unt.edu 
\newline ftp  ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)}
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Nonlinearities in a second order  ODE
% 
\thanks{ {\em Mathematics Subject Classifications:}  34G20.
 \hfil\break\indent 
{\em Key words:} Second order ODE, Dirichlet and Periodic Problems. 
 \hfil\break\indent
\copyright 2001 Southwest Texas State University. 
\hfil\break\indent Published Janaury 8, 2001.  } } 

\date{}
\author{ Pablo Amster}
\maketitle

\begin{abstract} 
In this paper we study the semilinear second order ordinary differential
equation  
$$u''+r(t)u' + g(t,u) = f(t)\,. $$
Under a growth condition on $g$, we prove the existence and 
uniqueness for the Dirichlet problem 
and establish conditions for the existence of periodic solutions.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
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\newtheorem{cor}[theorem]{Corollary}

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\section{Introduction} 

The two-point boundary-value problem for a semilinear second order ODE
$$u''+ru' + g(t,u) =0, \quad
u(0)=u_0, \quad u(T)=u_T$$ 
has been studied by many authors. In his pioneering work,  Picard \cite{P}  
proved the existence of a solution by an application of the well known 
method of successive approximations under a Lipschitz condition on $g$ 
and a smallness condition on $T$. Sharper results were 
obtained by Hamel \cite{H} in the special case of a forced pendulum equation 
(see also \cite{M1}, \cite{M2}). The existence of periodic solutions 
for this equation was first considered by Duffing \cite{Du} 
in 1918. In the absence of friction (i.e. $r=0$), 
variational methods have been applied by Lichtenstein \cite{L},
who considered the functional 
$$I(u)= \int_0^T \frac {(u')^2}2 - G(t,u(t)) dt,$$
where $G(t,x)=\int_0^xg(t,s)ds$.
Finally,  we want to mention the topological approach introduced 
in 1905 by Severini \cite{S} who used a shooting method. He also presented
and gave a survey of results obtained using Leray-Schauder techniques 
and degree theory. For further results, see \cite{M3}.

In this work, we prove the existence and uniqueness of a solution
to the Dirichlet problem under a growth condition on $g$.
Then, we apply this result for finding periodic solutions.

Let $S:H^2(0,T)\to L^2(0,T)$ be the semilinear operator given by
$$Su= u''+ru' + g(t,u)\,.$$ 
Assume that the function $g$ satisfies the growth condition 
\begin{equation}
 \label{g1}
\frac {g(t,u)-g(t,v)}{u-v}\le \frac{c_p}{p(t)} \quad 
\mbox{for } t\in [0,T] \mbox{ and } u,v \in \mathbb{R} \quad (u\neq v)\,,
\end{equation}
where $p\in C^1([0,T])$ is strictly positive, 
$r_0 := pr - p' \in H^1(0,T)$ is non-decreasing,
and $c_p < \lambda_p$ with $\lambda_p$  the
first eigenvalue of the problem
$$-(pu')'=\lambda_p u, \quad u(0)=u(T)=0\,.$$

To state a general existence and uniqueness result for the Dirichlet 
problem associated to our equation,
we need the following apriori bounds.

\begin{lemma}
Assume that $g$ satisfies \eqref{g1} and let 
$u, v\in H^2(0,T)$ with $\mathop{\rm Tr}(u)=\mathop{\rm Tr}(v)$. Then
$$\| p(Su-Sv)\|_2 \ge (\lambda_p-c_p)\| u-v\|_2$$
and
$$\| p(Su-Sv)\|_2 \ge \frac {\lambda_p-c_p}{\sqrt{\lambda_p}}
\big( \int_0^T p(u'-v')^2\big)^{1/2}$$
\end {lemma}

\paragraph{Proof.}
A simple computation shows that 
$$\| p(Su-Sv)\|_2\| u-v\|_2\ge 
\int_0^T p(u' -v')^2 - \int_0^T r_0(u-v) (u' -v')
- c_p \| u-v\|_2^2$$
and because $-\int_0^T r_0(u-v) (u' -v') = \frac 12 \int_0^T r_0'(u-v)^2 \ge 0$, 
the result follows since 
$\| u-v\|_2^2 \le \frac 1{\lambda_p}\int_0^T p(u'-v')^2$. 
\hfill$\diamondsuit$\smallskip

\noindent{\bf Remarks}
i) For simplicity and by the previous lemma, we may 
denote by $k_1$ the best constant such that $\| u - v\|_{1,2} 
\le k_1 \| p(Su - Sv)\|_{2}$ 
for $u, v\in H^2(0,T)$ with $\mathop{\rm Tr}(u)=\mathop{\rm Tr}(v)$.
\\
ii) In particular, if $r\in H^1(0,T)$ is non-decreasing, 
the result holds for $p\equiv 1$ and 
$c_1 < \lambda_1 = \left( \frac {\pi}T\right)^2$.

\begin{theorem}
Let $g$ satisfy \eqref{g1}. Then the Dirichlet 
problem
\begin{equation} \gathered
Su=f(t) \quad\mbox{in } (0,T) \\
u(0)=u_0, \quad u(T)=u_T 
\endgathered  \label{e1}
\end{equation}
is uniquely solvable in $H^2(0,T)$ for any $f\in L^2(0,T)$ and 
arbitrary boundary data.
\end{theorem}

\paragraph{Proof.}
Without loss of generality, we may suppose that $p\equiv 1$. 
For $0\le  \sigma \le 1$ we consider the operator $S_\sigma$ given by 
$S_\sigma u:= u''+ ru' +\sigma g(t,u)$. We remark that if $k_\sigma$ is the 
constant of lemma 1.1 for $S_\sigma$, then $k_\sigma \le k_1$.

 From the theory of linear operators, for fixed $\overline u\in H^1(0,T)$ 
we may define $u=K\overline u$ as the unique solution of the problem
\begin{gather*}
S_{0}u=f(t)- g(t,\overline u) 
\quad \mbox{ in } (0,T) \\
u(0)=u_0, \quad u(T)=u_T\,. 
\end{gather*}
Continuity of $K:H^1(0,T)\to H^1(0,T)$ follows 
immediately from the inequality
$$\|K\overline u - K\overline v\|_{1,2}\le k_1
\| S_{0}(K\overline u) - S_{0}(K\overline v)\|_2 =
k_1\| g(\cdot,\overline u) - g(\cdot,\overline v)\|_2$$
and the fact that $\| g(\cdot,\overline u) - g(\cdot,\overline v)\|_2 \to 0$ 
for $\overline u \to \overline v$ in $H^1(0,T)\hookrightarrow C([0,T])$. 
Moreover, if $\varphi(t)= \frac{u_T-u_0}T t + u_0$ we have that
$$\|K\overline u -\varphi \|_{1,2} \le 
k_1\|f - g(\cdot ,\overline u) - S_0\varphi\|_2 \le C$$
for some constant $C= C(R)$. 
Moreover, as
$$\| (K\overline u)''\|_2 = 
\| f - g(\cdot ,\overline u) - r(K\overline u)'\|_2$$
it follows that $K(B_R)$ is $H^2$-bounded. Thus, 
by the compactness of the imbedding
$H^2(0,T)\hookrightarrow H^1(0,T)$ 
we conclude that $K$ is compact. 

Let us assume that $u = \sigma Ku$ for some $\sigma \in (0,1]$. 
Then $u'' + ru' + \sigma g(t,u)=\sigma f$, and
$$\| u-\sigma\varphi\|_{1,2} \le k_1
\| S_\sigma u - S_\sigma(\sigma\varphi)\|_2 =
k_1\| \sigma f -  S_\sigma(\sigma\varphi)\|_2$$
This proves that the set $\{ u: u = \sigma Ku \}$ 
is uniformly bounded, and by Leray-Schauder theorem $K$ has a fixed
point. Uniqueness of the solution follows from lemma~1.1.
\hfill$\diamondsuit$\smallskip

As a simple consequence, we have an existence result for the general 
Dirichlet problem 
\begin{equation} \gathered
Su=f(t,u,u') \quad\mbox{ in } (0,T) \\
u(0)=u_0, \quad u(T)=u_T 
\endgathered \label{e2}
\end{equation}

\begin{cor}
Let $f$ be continuous and $g$ satisfy \eqref{g1}.  
Assume that the growing condition
\begin{equation}
\label{f}
|f(t,u,x)| \le c|(u,x)| + d 
\end{equation}
holds for some constant 
$c < \frac 1{k_1\|p\|_\infty}$. 
Then \eqref{e2} is solvable in $H^2(0,T)$. 
\end {cor}

\paragraph{Proof.}
By \eqref{f} and the previous theorem, 
the operator $K:H^1(0,T) \to H^1(0,T)$ given by $K \overline u =u$, 
with $u$ the unique solution of
\begin{gather*}
Su=f(t,\overline u,\overline u') \quad\mbox{ in } (0,T) \\
u(0)=u_0, \quad u(T)=u_T 
\end{gather*}
is well defined and compact. Moreover, as
$$\|K \overline u -\varphi\|_{1,2}\le k_1 
\|p(S(K \overline u) - S\varphi)\|_2 \le
k_1 \|p\|_\infty
\left( \|S\varphi)\|_2 + c\|\overline u\|_{1,2} + d\right)$$
then $K(B_R)\subset B_R$ for $R$ large and
the result follows from Schauder Theorem.

\section{ Solutions to the periodic problem } 

In this section we'll apply the previous results to the periodic
problem
\begin{equation} \gathered 
Su=f(t) \quad\mbox{ in } (0,T) \\
u(0)= u(T), \quad u'(0)= u'(T)
\endgathered \label{per}
\end{equation}
It is well known that the forced pendulum 
equation $u'' + b\sin (u) = f$ admits periodic solutions 
for constant $b$ if $f$ is
periodic and {\sl orthogonal to constants}. 
We'll show in the general case that in the presence of friction 
this orthogonality condition can be reinterpreted in terms of a certain 
$p_1>0$. More precisely, we'll 
show that in some cases -including the generalized 
pendulum equation- \eqref{per} is not solvable 
for any $f$ such that $\big< p_1,f\big>$ is large enough.

\begin{lemma}
For any $c \in \mathbb{R}$ there exists a unique 
$p_c$ such that $p_c (0) = p_c (T)=c$ and
$p_c' - r p_c$ is constant. 
Furthermore, $p_c = c p_1$, and 
$p_1$ is strictly positive. 
\end{lemma}

\paragraph{Proof.} From the equation 
$p_c' - r p_c = k_c$ we obtain that
$$p_c = \left( c + k_c \int_0^t e^{-\int_0^s r}ds\right) e^{\int_0^t r}$$
and from the condition $p_c (0) = p_c (T)=c$ we conclude that
$$k_c = c \frac {1- e^{\int_0^T r}}{\int_0^T e^{\int_s^T r}ds}= ck_1$$
Thus, $p_c = cp_1$. 
Moreover, if $k_1 \ge 0$ it's immediate that 
$p_1 >0$, and if $k_1 < 0$, assuming that 
$p_1$ vanishes there exists 
$t_0 \in (0,T)$ such that $p_1 (t_0) = 0 \le p_1'(t_0)$. Then 
$ k_1 = p_1'(t_0) \ge 0$, a contradiction.
\hfill$\diamondsuit$\smallskip

Using the preceding lemma we'll see 
that periodic solutions of $Su=f$ satisfy an orthogonality condition. 
Indeed, from $$u'' + ru' +g(t,u)= f$$
we obtain 
$$(p_1u')' -k_1u' +p_1g(t,u)= p_1f\,.$$
By the equality $ p_1u'\Big|_0^T = u\Big|_0^T = 0$ we have
$$ \int_0^T p_1g(t,u)= \int_0^T p_1f\,.$$

\begin{cor}
With the previous notation, let us assume that  $g(t,u) \le g_{max}$ 
for any $t\in [0,T]$, $u\in \mathbb{R}$ and some constant $ g_{max}\in \mathbb{R}$
(respectively, $g(t,u) \ge g_{min}$ for any $t\in [0,T]$, $u\in \mathbb{R}$ 
and some constant $ g_{min}\in \mathbb{R}$). 
Then \eqref{per} is not solvable for any $f\in L^2(0,T)$ such that
$\big< p_1,f\big> > g_{max}\| p_1\|_1$ 
(resp. $\big< p_1,f \big> < g_{min}\| p_1\|_1$).
\end{cor}
\rm

Now we'll give some existence results 
for \eqref{per}, assuming that $g$ satisfies \eqref{g1}. 
Our method is based in the existence and uniqueness 
result given by Theorem 1.2: indeed, for fixed $s\in \mathbb{R}$ 
we may define $u_s$ as the unique solution of the
problem
\begin{gather*}
Su=f(t) \quad\mbox{ in } (0,T) \\
u(0)= u(T) = s 
\end{gather*}

\begin{lemma}
The mapping  $s\to u_s$ is continuous for the $H^1$-norm.
\end{lemma}

\paragraph{Proof.} For 
$s\to s_0$ and $w_s=u_s-u_{s_0}$ 
we have
\begin{eqnarray*}
0 &=& \int_0^T p(Su_s-Su_{s_0})w_s\\
&\le& 
pw_s'w_s\Big|_0^T-\int_0^T p(w_s')^2 
+ \frac {r_0w_s^2}2 \Big|_0^T- \int_0^T r_0' \frac {w_s^2}2
+ c_p \int_0^T w_s^2
\end{eqnarray*}
Because $\int_0^T r_0' \frac {w_s^2}2 \ge 0$, we conclude that 
$$0\le (1- \frac {c_p}{\lambda_p})  \int_0^T p(w_s')^2 \le 
pw_s'w_s\Big|_0^T + r_0 \frac {w_s^2}2 \Big|_0^T\,.$$
Since $w_s(0) = w_s(T) = s-s_0 \to 0$
it suffices to prove that 
$\|w_s \|_{1,\infty}$ is bounded. 
As 
$\| u_s - s\|_{1,2}\le k_1\|p(f-g(\cdot , s))\|_2$, 
we deduce that $w_s$ is
$H^1$-bounded. 
Moreover, from the equality 
$u_s'' = f - ru_s'-g(t,u_s)$ we obtain
that $\|w_s\|_{2,2}$ is bounded, and from the imbedding 
$H^2(0,T) \hookrightarrow C^1([0,T])$ 
the proof is complete.
\hfill$\diamondsuit$\smallskip

 From the previous remarks, the 
solvability of \eqref{per} is equivalent to the solvability 
of the equation $\psi(s) = \int_0^T p_1f$, where $\psi:\mathbb{R}\to \mathbb{R}$ is 
given by $\psi (s)=\int_0^T p_1g(t,u_s)$. 
Continuity of $\psi$ follows immediately 
from the previous lemma, and hence \eqref{per} will admit 
a solution if and only if there exist $s^\pm$ such that
$$\psi(s^+)\ge \big< p_1,f\big> \ge \psi(s^-)$$

\paragraph{Remark.} 
Writing $u_s(t) -s = \int_0^t p^{-1/2} p^{1/2}u_s'$ we obtain that
$$\| u_s -s\|_\infty \le \delta_p \|p(f-g(\cdot ,s))\|_2 $$
for $\delta_p = \left( \int_0^T\frac 1p \right)^{1/2} 
\frac {\sqrt{\lambda_p}}{\lambda_p-c_p}$.

Thus, if we consider the condition
\begin{equation}
\label {g2}
\|p g(\cdot ,s)\|_2 \le c|s| + d \quad \mbox{ with } c\delta_p < 1
\end{equation}
then
$ u_s(t) \in {\mathcal J}_s^\varepsilon$ for any 
$t\in [0,T]$, where 
${\mathcal J}_s^\varepsilon$ is the interval centered in $s$ with radius 
$\delta_p(c|s| + d) + \varepsilon$, $\varepsilon = \delta_p \|pf\|_2$. 
As a simple consequence we have the following

\begin{theorem}
Let $g$ satisfy \eqref{g1}-\eqref{g2}, and 
assume that there exist $s^\pm$ such that
$$g|_{[0,T]\times {\mathcal J}_{s^+}^\varepsilon} \ge 
\frac {\int_0^T p_1f}{\| p_1\|_1} \ge
g|_{[0,T]\times {\mathcal J}_{s^-}^\varepsilon}$$
for $\varepsilon = \delta_p \|pf\|_2$. 
Then \eqref{per} admits a solution 
$u_s$ for some $s$ between $s^-$ and $s^+$. 

In particular, if there exist $s^\pm$ such that
$$g|_{[0,T]\times \mathcal J_{s^+}^\varepsilon} \ge 0 \ge
g|_{[0,T]\times \mathcal J_{s^-}^\varepsilon}$$
then \eqref{per} admits a solution 
$u_s$ for some $s$ between $s^-$ and $s^+$
for any $f\perp p_1$  
such that
$ \delta_p \|pf\|_2\le \varepsilon$.

\end{theorem}

\paragraph{Proof.}

As $u_s^\pm([0,T])\subset \mathcal J_{s^\pm}^\varepsilon$, we obtain: 
$$\int_0^T p_1g(t,u_{s^+}) \ge \int_0^T p_1f \ge 
\int_0^T p_1g(t,u_{s^-})$$
and the result holds. 
\hfill$\diamondsuit$\smallskip

Using the fact that $|s| - \delta_p(c|s| + d)\to + \infty$
we deduce the following existence results:

\begin{cor}

Let $g$ satisfy 
\eqref{g1}-\eqref{g2}, and 
assume, for some $M >0$ that
$$g(t,x) sg (x) \ge 0 \mbox{ for } \quad |x| \ge M$$
or 
$$g(t,x) sg (x) \le 0 \mbox{ for } \quad |x| \ge M$$
Then \eqref{per} is solvable for any $f\perp p_1$. 

\end {cor}

\begin{cor}

Let $g$ satisfy 
\eqref{g1}-\eqref{g2}, and 
assume that 
$$\lim_{|x| \to +\infty}g(t,x)sg (x) = +\infty  \quad 
\mbox{or } \quad \lim_{|x| \to +\infty}g(t,x)sg (x) = -\infty$$
uniformly on $t$. Then \eqref{per} is solvable for any $f$. 
\end {cor}

\paragraph{Proof.} Under the first assumption, 
there exists $M$ such that 
$$g(t,x)sg(x) \ge \frac 
{|\int_0^T p_1f|}{\|p_1\|_1}\quad \mbox{ for }|x|\ge M $$
Hence, taking $s >0$ such that $s - \delta_p (cs + d + \| pf\|_2)\ge M$ 
we have
$$\int_0^T p_1 g(t,u_s) \ge |\int_0^T p_1 f| \ge \int_0^T p_1 f\,.$$
In the same way, for $s<0$ with
$s + \delta_p (-cs + d + \| pf\|_2)\le -M$ we obtain 
$\int_0^T p_1 g(t,u_s) \le \int_0^T p_1 f$ and the proof is complete. 
The case $g(t,x)sg (x)\to -\infty$ is analogous. 
\hfill$\diamondsuit$\smallskip

\paragraph{Remark.} 
In the previous corollaries (2.5)-(2.6), we also have that
all the solutions belong to a compact arc of $H^1(0,T)$, 
namely
$\{ u_s: -S \le s \le S \}$
with
$$S = \frac {M +\delta_p(d+\| pf\|_2|)}{1-\delta_p c}\,.$$
\medskip 
 
We may also apply theorem (2.4)
to the forced pendulum equation with friction 
\begin{equation} \label{p}
u'' + ru' + b\sin (u) = f\,.
\end{equation} 
We first remark that in this case condition 
\eqref{g1} reads
\begin{equation}
\label{b}
|b(t)| \le \frac{c_p}{p(t)} \quad \mbox{for any } t\in [0,T]
\end{equation}
for some $p>0$ with $pr-p'$ nondecreasing and $c_p < \lambda_p$.

\begin {theorem}
With the previous notation, let us assume that 
\\
i) $b$ satisfies \eqref{b} and does not vanish in $(0,T)$.\\
ii) $\|p(f\pm b)\|_2 \le \frac c{\delta_p}$ for some $c< 
\frac{\pi}2$.\\
iii) $|\int_0^T p_1f| \le cos (c) \|p_1b\|_1$

\noindent Then there exist 
$s_1 \in [-\frac {\pi}2, \frac {\pi}2]$, 
$s_2 \in [\frac {\pi}2, \frac {3}2\pi]$
such that 
$u_{s_i}+ 2k\pi$ is a periodic solution of 
\eqref{p} for any integer $k$.
\end{theorem}

\paragraph{Proof.} From the previous computations 
for $s = \frac {\pi}2 + k\pi$ we obtain that
$$\|u_s-s\|_\infty \le \delta_p \| p(f- (-1)^k b\|_2 
\le c < \frac {\pi}2$$
As $\sin u_s = (-1)^k \cos(u_s-s)$, taking $k$ such that $(-1)^kb>0$
we conclude that 
$$\int_0^T p_1b\sin u_s = \int_0^T p_1|b|\cos (u_s -s) \ge cos (c)\| p_1b\|_1
\ge \int_0^T p_1f$$
In the same way, for 
$s = \frac {\pi}2 + (k\pm 1)\pi$
$$\int_0^T p_1b\sin u_s = 
-\int_0^T p_1|b|\cos (u_s -s) \le -cos (c)\| p_1b\|_1\le \int_0^T p_1f
$$
and the result holds.
\hfill$\diamondsuit$\smallskip

\paragraph{Remark.}
In particular, condition iii) is 
fulfilled if $f$ is orthogonal to $p_1$.

\medskip 
If we assume that $\|pf\|_2\le \frac {\pi}{2\delta_p}$ we also obtain 
existence under slightly different conditions.

\begin {theorem}
With the previous notation, let us assume that 
\\
i) $b$ satisfies \eqref{b} and does not vanish in $(0,T)$
\\
ii) $\|pf\|_2\le \frac {\pi}{2\delta_p}$, 
$ \|p(f-|b|)\|_2 < \frac c{\delta_p}$ for some $c< \frac{\pi}2$.
\\
iii) $\sin (\delta_p \|pf\|_2) \le \frac {\int_0^Tp_1f}{\|p_1b\|_1} 
\le  \cos (\delta_p \|p(f-|b|)\|_2)$.

\noindent Then,  if $b>0$ (resp. $b<0$) there exist 
$s_1 \in [0, \frac {\pi}2]$, $s_2 \in [\frac {\pi}2, \pi]$ 
(resp. $s_1 \in [-\frac {\pi}2, 0]$, $s_2 \in [\pi, \frac 32 {\pi}]$) 
such that $u_{s_i}+ 2k\pi$ is a periodic solution of 
\eqref{p} for any integer $k$.

\noindent Moreover, if we replace
ii) and iii) by  \\
ii') $\|pf\|_2\le \frac {\pi}{2\delta_p}$, 
$ \|p(f+|b|)\|_2 < \frac c{\delta_p}$ for some $c< \frac{\pi}2$.
\\
iii') $\sin (\delta_p \|pf\|_2) \le \frac {-\int_0^Tp_1f}{\|p_1b\|_1} \le  
\cos (\delta_p \|p(f+|b|)\|_2)$

\noindent then 
if $b<0$ (resp. $b>0$) there exist 
$s_1 \in [0, \frac {\pi}2]$, $s_2 \in [\frac {\pi}2, \pi]$ 
(resp. $s_1 \in [-\frac {\pi}2, 0]$, $s_2 \in [\pi, \frac 32 {\pi}]$) 
such that $u_{s_i}+ 2k\pi$ is a periodic solution of \eqref{p} 
for any integer $k$.
\end{theorem}

\paragraph{Proof.}  It follows like in the previous theorem, 
using the fact that if $s=k\pi$ then
$\|u_s-s\|_\infty \le \delta_p\| pf\|_2$, and 
$$|\int_0^T p_1b\sin u_s | \le \|p_1 b\|_1 \sin (\delta_p \|pf\|_2)\,.$$

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\bibitem{S} Severini, C., Sopra gli integrali delle equazione differenziali del 
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\noindent{\sc Pablo Amster} \\
Depto. de Matem\'atica,  CONICET \\
Facultad de Cs. Exactas y Naturales \\ 
Universidad de Buenos Aires \\
Ciudad Universitaria, Pab. I - (1428)\\
Buenos Aires, Argentina \\
e-mail: pamster@dm.uba.ar

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