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\markboth{\hfil Some remarks on the Melnikov function \hfil EJDE--2002/13}
{EJDE--2002/13\hfil Flaviano Battelli \&  Michal Fe\v ckan \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2002}(2002), No. 13, pp. 1--29. \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} \\
 %
  Some remarks on the Melnikov function 
 %
\thanks{ {\em Mathematics Subject Classifications:} 34C23, 34C37.
\hfil\break\indent
{\em Key words:} Melnikov function, residues, Fourier coefficients.
\hfil\break\indent
\copyright 2002 Southwest Texas State University. 
\hfil\break\indent
Submitted November 27, 2001. Published February 7, 2002.
\hfil\break\indent
F. Battelli was partially supported by G.N.A.M.P.A. - INdAM (Italy).
\hfil\break\indent
M. Fe\v ckan was partially supported by G.N.A.M.P.A. - INdAM (Italy) 
and by  \hfil\break\indent
Grant GA-MS 1/6179/00
} }
\date{}
%
\author{Flaviano Battelli \&  Michal Fe\v ckan}
\maketitle

\begin{abstract} 
  We study the Melnikov function  associated with a periodic 
  perturbation of a differential   equation having a homoclinic orbit.
  Our main interest is the  characterization of perturbations that
  give rise to vanishing  or non-vanishing of the Melnikov function.
  For this purpose we show that, in some cases, the Fourier coefficients
  of the Melkinov function can be evaluated by means of the calculus of
  residues. We apply this result, among other things, to the construction
  of  a second-order equation whose Melnikov function vanishes identically
  for any $C^{1}$,  $2\pi$-periodic perturbation. Then we study the second
  order Melnikov function of the perturbed equation, and prove it is
  non-vanishing for a large class of perturbations.
\end{abstract}


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\newtheorem{corollary}[theorem]{Corollary}
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\section{Introduction}

Melnikov's method has shown to be an easy and effective method
to detect chaotic dynamics in differential equations. The
starting point is an autonomous system $\dot x = f(x)$, where
$x$ belongs to an open subset $\Omega\subset\mathbb{R}^n$, having
a hyperbolic fixed point $x_0$ and a {\it non-degenerate}
homoclinic orbit $\phi (t)$, that is a non constant solution
$\phi (t)$ such that $\ds\lim_{t\to\pm\infty}\phi(t) = x_0$
and $\dot\phi (t)$ spans the space of bounded solutions of the
variational system
\begin{equation}
 \dot x = f'(\phi(t)) x. \label{eq:1}
\end{equation}
Actually some extensions to the case where the variational
system (\ref{eq:1}) has a higher dimensional space of bounded
solutions have also been given in the literature, see for example
\cite{BL,F,G}, however we are not interested in such
generalizations here. Then, associated to a given time periodic
sufficiently smooth perturbation $\varepsilon h(t,x,\varepsilon)$, with $\varepsilon$
sufficiently small, there is the so called Melnikov function:
$$
M(\alpha) := \int_{-\infty}^{+\infty}
\psi^{*}(t)h(t+\alpha,\phi(t),0) {\rm d} t
$$
with $\psi (t)$ being the unique (up to a multiplicative
constant) bounded solution of the variational system
$$
\dot x = -f'(\phi (t))^{*} x.
$$
Note that $M(\alpha)$ is a periodic function having the same
period as $h(t,x,\varepsilon)$. The basic result states that $M(\alpha)$
gives a kind of $O(\varepsilon)$-measure of the distance between the
stable and unstable manifolds of the (unique) hyperbolic periodic
solution $x_0(t,\varepsilon)$ of the perturbed system
\begin{equation}
 \dot x = f(x) + \varepsilon h(t+\alpha,x,\varepsilon) \label{eq:2}
\end{equation}
which is at a $O(\varepsilon)$-distance from $x_0$ (see \cite{H}).
Thus if $M(\alpha)$ has a simple zero at some points, the implicit
function theorem implies that these two manifolds intersect
transversally along a solution $\phi(t,\varepsilon)$ of (\ref{eq:2})
which is homoclinic to $x_0(t,\varepsilon)$. This transversality implies,
by the classical Smale horseshoe construction (see \cite{S}), that
a suitable iterate of the Poincar\'e map of the perturbed system
exhibits chaotic behavior (for a more analytical proof of this fact
see \cite{P}).

In this paper, we will mainly consider the case where
$h(t,x,0)=q(t)$ is a $T$-periodic perturbation independent of $x$,
although in the next Section some results are derived for the more
general case. We also assume that $q(t)$ is $C^{1}$. Our first
remark is that the Melnikov function is a bounded linear map from
the space of $T$-periodic functions into itself, as it can be
easily checked using the fact that $|\psi (t)| \le Ce^{-\sigma
|t|}$, for some positive real number $\sigma$. Moreover the
average of $M(\alpha)$ is:
$$ \bar{M} = \int_{-\infty}^{+\infty}
\psi^{*}(t) {\rm d} t \cdot \bar{q}.
$$
Now, in many interesting cases,
for example when one deals with a second order conservative
equation on $\mathbb{R}$, one has
$$ \int_{-\infty}^{+\infty} \psi^{*}(t)
{\rm d} t = 0 $$
so that $\bar{M} =0$. In this case the Melnikov
function can either be zero or there are $\alpha_1$ and
$\alpha_2$ such that $M(\alpha_1) < 0 < M(\alpha_2)$. This
means that the Brouwer degree of $M(\alpha)$ in the interval whose
end points are $\alpha_1$ and $\alpha_2$ is different from
zero. This, in turns, implies a kind of chaotic behavior of some
iterate of the Poincar\'e map (see \cite{BF}). This seems to be a
good reason to study the kernel of the Melnikov map:
$$
q(t)\mapsto \int_{-\infty}^{+\infty} \psi^{*}(t) q(t+\alpha) {\rm d} t.
$$
This is the purpose of this paper whose content we now briefly
explain. In Section 2 we give a method to evaluate the Melnikov
function when $\phi(t) = \Phi(e^{t})$ for some rational function
$\Phi(u)$, $u\in\mathbb{C}$. The method states that the Melnikov function
can be evaluated by means of the calculus of residues. Even if the
use of the calculus of residues for the study of the splitting of
separatrices seems to be a quite standard tool \cite{Ge}, our 
Theorem \ref{thm1} does not seem to follow directly from previous results.
Section 3 is devoted to the study of $M(\alpha)$ for a second order
equation on $\mathbb{R}$ with a $T$-periodic perturbation $\varepsilon q(t)$. Finally,
in Section 4 the results of Section 3 are used to construct some
second order equations whose Melnikov map has an infinite dimensional
kernel, possibly vanishing on the whole space of ($C^{1}$)
$2\pi$-periodic perturbations. For this class of equation we also
study the second order Melnikov function $M_2(\alpha)$, that is
the coefficient of $\varepsilon^{2}/2$ in the Taylor expansion of the
bifurcation function. We prove that for a large class of perturbations,
$M_2(\alpha)$ is not identically zero and changes sign provided
a certain simmetry condition is satisfied. This fact is important
because when the first Melnikov function is identically zero, it
is $M_2(\alpha)$ that determines the chaotic behaviour of the
system.

\section{Melnikov Function and Calculus of Residues}

Given the system $\dot x=f(x)+\varepsilon h(t,x,\varepsilon)$, $x\in\Omega
\subset\mathbb{R}^n$, $|\varepsilon|<2\varepsilon_0$, $t\in\mathbb{R}$, such that $\dot x=
f(x)$ has a non--degenerate homoclinic orbit $\phi(t)$ whose
closure is contained in $\Omega$ and $h(t,x,\varepsilon)$ is a
$C^{1}$-function, bounded together with its derivatives on
$\mathbb{R}\times\Omega\times [-\varepsilon_0,\varepsilon_0]$, the Melnikov
function of the system is given by
$$
M(\alpha)=\int_{-\infty}^{+\infty}\psi^*(t)h(t+\alpha,\phi(t
),0){\rm d} t
$$
where $\psi (t)$ is the unique, up to a multiplicative
constant, bounded solution of the variational system
$\dot y=-f^\prime(\phi (t))^* y(t)$. When $h(t,x,\varepsilon)=
h(t+T,x,\varepsilon)$, $T>0$, the existence of a simple zero of
$M(\alpha)$ implies the existence of a transversal
homoclinic  orbit for the Poincar\'e (period $T$) map of
the system $\dot x=f(x)+\varepsilon h(t,x,\varepsilon)$ with the induced
chaotic behavior.

Here we assume that
\begin{description}
 \item{(a)} $\phi(t)=\Phi(e^t)$, where $\Phi(u)$ is a
 rational function on $\mathbb{C}$ such that  $\Phi(u)\to 0$,
 and $\Phi(1/u) \to 0$ as $u\to 0$;
 \item{(b)} $\psi (t) =e^t \Psi(e^t)$, where
   $\Psi(u)u\to 0$ as $|u|\to +\infty$, and $\Psi(u)$
 is a rational function on $\mathbb{C}$.
\end{description}
From $h(t,x,\varepsilon)=h(t+T,x,\varepsilon)$ we deduce that $M(\alpha)$
is $T$-periodic. Set $\omega=\frac{2\pi}{T}$ and $M_0(\alpha) =
M(\alpha)\chi_{[-T/2,T/2]}(\alpha)$, $h_0(t,x)=h(t,x,0)
\chi_{[-T/2,T/2]}$ and, for any $n\in\mathbb{Z}$, consider:
\begin{eqnarray*}
 \hat M_0(n) & = & \ds{1\over T}\int_{-T/2}^{T/2}
 M_0(\alpha)e^{-in\omega\alpha}{\rm d}\alpha  \\
  & = & \ds{1\over T}\int_{-T/2}^{T/2}
  \int_{-\infty}^{+\infty}\psi^*(t)h(t+\alpha,\phi(t),0)
  e^{-in\omega\alpha}{\rm d} t{\rm d}\alpha \\
  & = & \ds\int_{-\infty}^{+\infty}\psi^*(t) {1\over T}
  \int_{-T/2}^{T/2} h(t+\alpha,\phi(t),0)e^{-in\omega\alpha}
  {\rm d}\alpha{\rm d} t\\
  & = & \ds\int_{-\infty}^{+\infty}e^t \Psi^*(e^t)\hat
  h(n,\Phi(e^t)) e^{in\omega t}{\rm d} t
\end{eqnarray*}
where
\begin{equation}
 \hat h(n,x):={1\over T}\int_{-\infty}^{+\infty}
 h_0(t,x)e^{-in\omega t}{\rm d} t = {1\over T}\int_{-T/2}^{T/2}
 h(t,x,0)e^{-in\omega t}{\rm d} t \label{eq:3}
\end{equation}
is the $n$-th Fourier coefficient of $h_0(t,x)$. We assume
that
\begin{description}
 \item{(c)} For any $n\in\mathbb{Z}$ the function $\hat h(n,\Phi(x))$
 extends to a meromorphic function $\hat h(n,\Phi(u))$ on $\mathbb{C}$
 having the same poles as $\Phi(u)$.
\end{description}
Thus
$$
F(n,u):=\Psi^*(u)\hat h(n,\Phi(u))
$$
is meromorphic in $\mathbb{C}$, for any fixed $n\in\mathbb{Z}$, and its
poles are either those of $\Psi(u)$ or those of $\Phi(u)$. Let
us make some comments about the function $F(n,u)$. As $\Psi(u)$
and $\Phi(u)$ take real values when $u\in\mathbb{R}_{+}$, Schwartz
reflection principle gives:
\begin{equation}
 \overline{\Psi(\bar u)}=\Psi(u),\quad\hbox{and}\quad
 \overline{\Phi(\bar u)}=\Phi(u).
 \label{eq:4}
\end{equation}
Second, we notice that, being $\overline{\hat h(n,x)}=
\hat h(-n,x)$ for any $x\in\Omega\subset\mathbb{R}^n$ we obtain
$$
\overline{\hat h(n,\Phi(\bar u))}=\hat h(-n,\Phi(u))
$$
because of the uniqueness of the analytical extension and
hence
\begin{equation}
 \overline{F(n,\bar u)}=F(-n,u). \label{eq:5}
\end{equation}
From (\ref{eq:4}) it follows that $\Psi(u)$ and $\Phi(u)$
have complex conjugate poles and hence the same holds for
$F(n,u)$. Let $w_j=u_j\pm iv_j$, $j=1,\ldots , r$ be the
poles of $F(n,u)$ (independent of $n\in\mathbb{Z}$). Note that the
$w_j$ do not belong to an angular sector around the positive
real half--line, otherwise $\Psi(e^t)$ will have
singularities on the real line. Thus the poles of
$F(n,e^z)$ are $z=\Log w_j:=\log |w_j|+i\Arg w_j$ where
$\Arg (w)\in (\beta,2\pi-\beta )$ for some $\beta >0$ and
$\Log w_j$ is the logarithm principal value. We remark
that $\Arg (\bar w_j)=2\pi - \Arg (w_j)$. Finally, for any
$u\in\mathbb{C}\setminus \{0\}$ such that $0 \le \Arg (u) < 2\pi$,
we set
$$
u^{i\omega n} := e^{in\omega Log u}.
$$
Then we integrate the meromorphic function $e^z F(n,e^z)
e^{in\omega z}$ on the boundary of the rectangle
$\{ -\rho \le \Re z \le \rho, 0 \le \Im z \le 2\pi \}$.
For $\rho$ sufficiently large Cauchy theorem implies:
$$
\begin{array}{l}
  \ds 2\pi i \sum_j \mathop{\rm Res}(e^z F(n,e^z)e^{in\omega z},\Log w_j)
  = \int_{-\rho}^{\rho}e^t F(n,e^t)e^{in\omega t}{\rm d} t \\
  \ds - \int_{-\rho}^{\rho} e^t F(n,e^t)e^{in\omega t}e^{-2\pi n\omega}
  {\rm d} t + \int_0^{2\pi}e^\rho e^{iy} F(n,e^\rho e^{iy})
  e^{in\omega\rho}e^{-n\omega y}i{\rm d} y \\
  \ds - \int_0^{2\pi}e^{-\rho}e^{iy} F(n,e^{-\rho}e^{iy})
  e^{-in\omega\rho}e^{-n\omega y}i{\rm d} y .
 \end{array}
$$
Now, the last two integrals in the right tend to zero as
$\rho\to+\infty$ uniformly with respect to $n$. Hence, for any
$n \neq 0$ we get
 \begin{eqnarray*}
  \ds \int_{-\infty}^{+\infty}e^t F(n,e^t)e^{in\omega t}{\rm d} t & =
  & \ds {2\pi i\over 1-e^{-2\pi n\omega}} \sum_j
  \mathop{\rm Res}\left( e^z F(n,e^z)e^{in\omega z},\Log w_j \right ) \\
  & = & \ds 2\pi i\sum_j \mathop{\rm Res}\left( { F(n,u)u^{in\omega}\over
  1-e^{-2\pi n\omega}}, w_j \right ).
 \end{eqnarray*}
Thus we have proved the following.

\begin{theorem} \label{thm1}
 Under the conditions {\rm (1)-(3)} the Fourier
coefficients of the Melnikov function $M(\alpha)$ are given by:
\begin{equation}
 \hat M_0(n)=2\pi i\sum_j \mathop{\rm Res}\left( { F(n,u)u^{in\omega}\over
 1-e^{-2\pi n\omega}}, w_j \right )
 \label{eq:6}
\end{equation}
for $n\neq 0$, while
$$
\hat M_0(0) = {1\over T}\int_{-T/2}^{T/2}M_0(\alpha)
{\rm d}\alpha = \int_{-\infty}^{+\infty} \psi^{*}(t) \hat h_0
(0,\phi(t)) {\rm d} t
$$
where $\hat h_0(n,x)$ has been defined in {\rm (\ref{eq:3})}.
\end{theorem}


Using (\ref{eq:5}) we obtain:
$$
\begin{array}{rl}
 \overline{\hat M_0}(n) & = \ds\overline
 {\int_{-\infty}^{+\infty}e^t F(n,e^t)e^{in\omega t}{\rm d} t} =
 \int_{-\infty}^{+\infty}e^t \overline {F(n,e^t)} e^{-in\omega t}
 {\rm d} t \\
 & = \ds\int_{-\infty}^{+\infty}e^t F(-n,e^t)e^{-in\omega t}
 {\rm d} t = \hat M_0(-n).
\end{array}
$$
Finally note that, since $h(t,x)$ is $T$-periodic in $t$
and $C^{1}$ we have
$$
\hat M_0(n) = 0 \; \hbox{\rm for any $n\in\mathbb{Z}$ if and only
if }\; M_0(\alpha) \equiv 0.
$$
We conclude this Section giving a first example of application of the
above result. Consider the Duffing-like equation:
\begin{equation}
    \ddot x + x\left (\frac{x}{k}-1\right ) = \varepsilon [q_1(t)x +
    q_2(t)\dot x]\label{eq:6.1}
\end{equation}
where $k>0$ and $q_1(t)$, $q_2(t)$ are $2\pi$-periodic, $C^{1}$
functions. Setting $x_1=x$ and $x_2=\dot x$ we obtain the equivalent
system
$$
\begin{array}{l}
    \dot x_1 = x_2  \\
    \ds\dot x_1 = x_1\left (1-\frac{x}{k}\right ) +
    \varepsilon [q_1(t)x_1 + q_2(t)x_2].
\end{array}
$$
Let $\ds\Phi_0(x) = \frac{6kx}{(x+1)^{2}}$. Then the homoclinic
solution of the unperturbed system is given by $\phi(t) = \Phi(e^{t})$
where
$$
\Phi(x) =\pmatrix{    \Phi_0(x)  \cr
   x\Phi_0'(x)}
$$
Moreover:
$$
\Psi(x_1,x_2) =\pmatrix{
    -x_1\Phi''_0(x_1) - \Phi_0(x_1)  \cr
    \Phi'_0(x_1)}
$$
and
$$
h(t,x_1,x_2,\varepsilon) = \pmatrix{ 0  \cr
    q_1(t)x_1+q_2(t)x_2 }
$$
Thus:
$$
F(n,u) = q_{n}^{1}\Phi_0(u)\Phi'_0(u) +q_{n}^{2}u\Phi'_0(u)^2
$$
where $q^{1}_{n}$ and $q^{2}_{n}$ are the Fourier coefficients of
$q_1(t)$ and $q_2(t)$ respectively. From Theorem \ref{thm1} we obtain
then:
$$
\hat M_0(n) = \delta^{1}_{n}q^{1}_{n} + \delta^{2}_{n}q^{2}_{n}
$$
where
$$
\delta_{n}^{1} = \frac{2\pi i}{1-e^{-2\pi n}} \mathop{\rm Res}\Phi(u)\Phi'(u)
u^{in},-1)
$$
and
$$
\delta_{n}^{2} = \frac{2\pi i}{1-e^{-2\pi n}}
\mathop{\rm Res}(\Phi'(u))^{2}u^{in+1},-1)
$$
(note that $-1$ is the unique pole of $\Phi_0(u)$). Hence, using
the fact that $\Arg (z)\in (0,2\pi)$, we obtain the following
expressions of $\delta_{n}^{1}$, $\delta_{n}^{2}$ for $n\neq 0$:
$$
\delta_{n}^{1} = -3in^2k^2(n^{2}+1)\frac{\pi}{\sinh n\pi}
$$ and
$$ \delta_{n}^{2} = -\frac{6}{5}nk^2(n^{2}+1)(n^{2}-1)\frac{\pi}
{\sinh n\pi}. $$
Thus $\hat M_0(0)=q_0^2\int \limits_{-\infty }^{+\infty}\dot
p(t)^2 dt$ and $\hat M_0(n)=0$, for $n\neq 0$, is equivalent to
\begin{equation}
    \frac{q_{n}^{1}}{q_{n}^{2}} = -\frac{\delta_{n}^{2}}{\delta_{n}^{1}}
    = \alpha_{n} := \frac{2i(n^{2}-1)}{5n}. \label{eq:6.2}
\end{equation}
Note that, obviously, $\alpha_{-n}=\bar\alpha_{n}$ and then taking, for
any integer $n\neq 0$, $q_{n}^{1}=\alpha_{n}q^{2}_{n}$, $q^{2}_0\in\mathbb{R}$,
we get the following


\begin{corollary} \label{coro1}
Given any $2\pi$-periodic function $q_2(t)
\in H^3(\mathbb{R})$ with zero mean value, there exists a unique $2\pi$-periodic
function with zero mean value, $q_1(t)\in H^2(\mathbb{R})\subset C^1(\mathbb{R})$ such
that the Melnikov function of the equation $(\ref{eq:6.1})$ vanishes
identically on $\mathbb{R}$. Actually, for $n\neq 0$, the Fourier
coefficients of $q_1(t)$ and $q_2(t)$ satisfy the relation
$(\ref{eq:6.2})$. The map $q_2(t) \mapsto q_1(t)$ is linear,
bounded and its kernel is the space $span \{\cos t, \sin t \}$.
That is, if $q_2(t)\in span \{ 1, \cos t, \sin t \}$, the
Melnikov map of equation $(\ref{eq:6.1})$ does not vanish
identically for any non constant perturbation $\varepsilon q_1(t)x$.
\end{corollary}


For example if $q_2(t)=\cos 2t$ then $q_1(t)=-\frac{3}{5}\sin 2t$

\section{The case of the second order equation}

In this Section we consider the Melnikov function for the second
order equation
$$
 \ddot x = f(x) + \varepsilon q(t)
$$
where $x$ belongs to an open interval $I\subset\mathbb{R}$, $f\in C^1(I,\mathbb{R})$,
$q(t)$ is a $C^{1}$, $T$-periodic function, and $\ddot x = f(x)$ is
assumed to have the hyperbolic fixed point $x=0\in I$ and an associated
homoclinic orbit $p(t)\in I$. In this case we have
$$
\phi^{*}(t) = (p(t)\quad \dot p(t)), \quad
\psi^{*}(t) = (-\ddot p(t)\quad \dot p(t)), \quad
h^{*}(t,x) = (0\quad q(t))
$$
and hence $\psi^{*}(t)h(t,x) = \dot p(t) q(t)$. As in the
previous section we assume $p(t)=\Phi_0(e^t)$. Then
$\dot p(t) = e^t\Phi_0'(e^t)$, and
$$
\Phi (u) = \pmatrix{ \Phi_0(u)  \cr  u\Phi_0'(u)}
 \quad \Psi(u) = \pmatrix{-u\Phi_0''(u) -\Phi_0'(u)  \cr
 \Phi_0'(u)}.
$$
Note that we have $F(n,u) = R^{*}(u)\hat h(n,\Phi(u))=
\Phi_0'(u)\hat q_{n}$ where $\hat q_{n}$ is the $n$-th Fourier
coefficient of the periodic function $q(t)$. Thus in order that
the analysis of the previous Section is valid we see that we only
need that assumption (2) is satisfied, that is $\Phi_0'(u)$ is
rational (we do not need that $\Phi_0(u)$ satisfies this
condition), and $\lim_{u\to\infty}\Phi_0'(u)u = 0$. Anyway, for
simplicity, we also assume that $\Phi_0(u)$ is rational with the
same poles as $\Phi_0'(u)$. Next:
$$
\hat M_0(0) = \int_{-\infty}^{+\infty} \dot p(t){\rm d} t \cdot
\hat q_0 = 0 $$
because $p(t)$ is homoclinic, and, from
(\ref{eq:6}) we obtain for $n\in\mathbb{Z}$, $n\neq 0$:
$$
 \hat M_0(n) = 2\pi i\sum_{w_j} \mathop{\rm Res}\left(
{\Phi_0'(u) \hat q_n u^{in\omega}\over 1-e^{-2\pi n\omega}}, w_j \right )
=\delta_{n}\hat q_n
$$
where $\hat q_n$ is the $n$-th Fourier coefficient of the
periodic function
$q(t)$, $w_j$ are the poles of $\Phi_0(u)$ and
\begin{equation}
 \delta_{n} = {2\pi i\over 1-e^{-2\pi n\omega}}\sum_{w_j} \mathop{\rm Res}(
 \Phi_0'(u)u^{in\omega}, w_j). \label{eq:7}
\end{equation}
Now, let $\gamma_j$ be a circle around $w_j$ such that no
other pole $w_{i}$, $i\neq j$, is inside $\gamma_j$. We have,
integrating by parts:
\begin{equation}
 2\pi i \mathop{\rm Res}( \Phi_0'(u)u^{in\omega}, w_j) =
 \int_{\gamma_j}\Phi_0'(u) u^{in\omega}{\rm d} u =
 -in\omega\int_{\gamma_j}\Phi_0(u)u^{in\omega-1}{\rm d} u \label{eq:8}
\end{equation}
and then
\begin{equation}
 \delta_{n} = {2\pi n\omega\over 1-e^{-2\pi n\omega}}\sum_{w_j}
 \mathop{\rm Res}(\Phi_0(u)u^{in\omega-1}, w_j). \label{eq:9}
\end{equation}
Next, assume that $w_j$ is a pole of $\Phi_0 (u)$ of
multiplicity $k$. We have:
\begin{equation}
 \begin{array}{rl}
  & \mathop{\rm Res}(\Phi_0 (u)u^{in\omega-1},w_j) =
  \ds{1\over (k-1)!}
  {d^{k-1} \over du^{k-1}} \left [ (u-w_j)^k\Phi_0
  (u)u^{in\omega-1} \right ]_{u=w_j}  \\
  & \ds = {1\over (k-1)!} \sum_{m=0}^{k-1} 
  { k-1\choose  m } 
  \Big \{ {d^{k-m-1}\over
  du^{k-m-1}} \left [ (u-w_j)^k \Phi_0 (u) \right ]
  \cdot  \\
  &\quad (in\omega-1)\ldots (in\omega-m)u^{in\omega-m-1} \Big \}_{u=w_j}
  \\
  & \ds = \sum_{m=0}^{k-1} {1\over m!} \mathop{\rm Res}\left(
  (u-w_j)^m \Phi_0 (u), w_j\right ) i^m
  {(n\omega+i)\ldots (n\omega+mi)\over w_j^{m+1}}e^{in\omega\Log w_j}.
 \end{array}\label{eq:10}
\end{equation}
Now, $\mathop{\rm Res}\left( (u-w_j)^m\Phi_0 (u), w_j\right ) =0$
for $m\ge k$ because $(u-w_j)^m\Phi_0 (u)$ is holomorphic
in a neighborhood of $w_j$ when $m\ge k$. Thus, denoting by
$r$ the maximum of the multiplicities of the poles $w_j$,
we can extend the above sum up to $r-1$ obtaining:
\begin{equation}
 \begin{array}{rl}
  \ds{\delta_{n}(1-e^{-2\pi n\omega})\over 2\pi n\omega} &\ds =
  \sum_{j=1}^N \sum_{m=0}^{r-1} {1\over m!} \mathop{\rm Res}
  ((u-w_j)^m \Phi_0 (u), w_j ) i^m  \\
  & \ds\quad (n\omega+i)(n\omega+2i)\ldots(n\omega+mi){e^{-n\omega \Arg w_j}
  \over w_j^{m+1}} e^{in\omega\log |w_j|}
 \end{array}\label{eq:11}
\end{equation}
$w_1,\ldots,w_{N}$ being the poles of $\Phi_0(u)$. Let
$\beta_0=\min\{\Arg (w_j) : j=1\ldots N\}\in(0,\pi]$ and $r_0$ be the
greatest multiplicity of the poles of $\Phi_0(u)$ that
belong to the half line $\Arg (u)=\beta_0$. Multiplying
both sides of equation (\ref{eq:11}) by $e^{n\omega\beta_0}$
we see that ${1\over 2\pi n\omega}e^{n\omega\beta_0}\delta_{n}
(1-e^{-2\pi n\omega})$ is asymptotic, as $n\to+\infty$, to
$$
\sum_{\Arg (w_j)=\beta_0} \mathop{\rm Res}( \Phi_0 (u)u^{in\omega-1},
w_j)e^{n\omega\beta_0}.
$$
Now, again from equation (\ref{eq:10}) we see that for any
pole $w_j$ such that $\Arg (w_j)=\beta_0$, and
multiplicity less than $r_0$ the quantity $n^{1-r_0}
\mathop{\rm Res}\Phi_0(u)u^{in\omega-1},w_j)e^{n\omega\beta_0}$
tends to zero as $n\to\infty$. Thus the leading term in
$\ds\sum_{w_j} \mathop{\rm Res}(\Phi_0(u)u^{in\omega-1}, w_j)$ is:
$$
\sum_{{\Arg (w_j)=\beta_0\atop mult(w_j)=r_0}}\mathop{\rm Res}(\Phi_0(u)u^{in\omega-1},w_j)
$$
$mult(w_j)$ being the multiplicity of $w_j$. As a
consequence, using also $\delta_{-n}=\overline{\delta_{n}}$,
we obtain the following:


\begin{proposition} \label{prop1}
Let $\beta_0=\min\{ \Arg (w_j) :
j=1\ldots N\} \in(0,\pi]$ and $r_0=\max\{ mult (w_j) :
\Arg (w_j) = \beta_0 \}$. Then, if
\begin{equation}
 \liminf_{n\to\infty}{e^{n\omega\beta_0}\over n^{r_0-1}}\Big |
 \sum_{{\Arg (w_j) = \beta_0\atop mult(w_j)=r_0}} 
 \mathop{\rm Res}
 ( \Phi_0 (u)u^{in\omega-1}, w_j) \Big | \neq 0, \label{eq:12}
\end{equation}
there exists $\bar n$ such that for any $n\in\mathbb{Z}$ such that
$|n|\ge \bar n$ we have $\delta_{n}\neq 0$. As a consequence
the space of periodic functions $q(t)$ such that the
associated Melnikov function is identically zero is
finite-dimensional.
\end{proposition}


Condition (\ref{eq:12}) can be simplified a bit looking at
equation (\ref{eq:10}). In fact setting $r_0$ in the
place of $k$ in that equation and multiplying by
$e^{\beta_0n}n^{1-r_0}$ we see that only the term with
$m=r_0-1$ survives. Thus we obtain the following

\begin{proposition} \label{prop2} 
Let $\beta_0$, $r_0$ be as in Proposition \ref{prop1}
 Then, if
\begin{equation}
 \liminf_{n\to\infty}\Big | \sum_{{\Arg (w_j)=\beta_0\atop
 mult(w_j)=r_0}} {1\over w_j^{r_0}}Res
 ( (u-w_j)^{r_0-1}\Phi_0 (u), w_j)
 e^{in\omega\log|w_j|} \Big | \neq 0, \label{eq:13}
\end{equation}
there exists $\bar n$ such that for any $n\in\mathbb{Z}$ such that
$|n|\ge \bar n$
we have $\delta_{n}\neq 0$. As a consequence the space of
periodic functions $q(t)$ such that the associated Melnikov
function is identically zero is finite-dimensional.
\end{proposition}

\paragraph{Proof.} As we have already observed, for any pole
$w_j$ such that $\Arg (w_j)=\beta$ and $mult(w_j)=r$
the quantity:
$$
\begin{array}{l}
  \ds\left ({e^{n\omega\beta}\over n^{r-1}}\right )Res
  (\Phi_0(u)u^{in\omega-1},
  w_j) - {i^{r-1}\over (r-1)!} \prod_{k=1}^{r-1} \left (
  \omega+{ki\over n} \right )    \\
  \ds\times {1\over w_j^{r}}\mathop{\rm Res}( (u-w_j)^{r-1}
  \Phi_0(u), w_j) e^{in\omega\log|w_j|}
\end{array}
$$
tends to zero as $n$ tends to infinity, and then the result
follows from:
$$
\omega^{r-1}\le \left | \left ( \omega +{i\over n}\right )\cdot\ldots
\cdot\left ( \omega +{(r-1)i\over n} \right ) \right | \le \left (
\omega + {r-1\over n} \right )^{r-1}.
$$
The proof is finished.

\paragraph{Remark.} If $\Phi_0(u)$ has only one pole on the line
$\Arg (u) = \beta_0$ with maximum multiplicity $r_0$,
condition (\ref{eq:13}) of Proposition \ref{prop2} is certainly satisfied.
In fact in this case the left hand side of (\ref{eq:13}) reads:
$$
{1\over w_j^{r_0}}\lim_{u\to
w_j}(u-w_j)^{r_0}\Phi_0(u)
$$
and cannot be zero because $w_j$ is a pole of multiplicity
$r_0$ of $\Phi_0 (u)$. \smallskip


Equation (\ref{eq:11}) has an interesting consequence when
$\Phi(u)$ has only the simple poles $w$ and $\bar w$ (we do not exclude
that $w=\bar w$). In fact in this case we have the following result:

\begin{theorem} \label{thm2}
Assume that $\Phi_0(u)$ satisfies the
assumption of the previous section and, moreover, that it has
only the simple poles $w$ and $\bar w$ (including the case that
$\Phi_0(u)$ has only one simple pole $w=\bar w$).
Then $\delta_{n}\neq 0$ for any $n\in\mathbb{Z}$, $n\neq 0$. Thus, for any
$2\pi$-periodic, nonconstant function, the associated Melnikov function
is not identically zero.
\end{theorem}

\paragraph{Proof.}
Let us consider, first the case where $\Phi_0(u)$ has only
the simple pole $w=\bar w < 0$. We have
$$
\begin{array}{rl}
 \ds\frac{1-e^{-2\pi n\omega}}{2\pi n\omega}\delta_{n} &\ds  =
 \frac{1}{w}\mathop{\rm Res}(\Phi_0(u),w)e^{-n\pi\omega}e^{in\omega\log |w|}   \\
 & \ds  = e^{-n\pi\omega}e^{in\omega\log |w|} \lim_{z\to 1}(z-1)
 \Phi_0(zw) \neq 0
\end{array}
$$
because $w$ is a simple pole of $\Phi_0(u)$. Now consider the case
where $w\neq \bar w$ and assume, without loss of generality, that
$0<\Arg (w)<\pi$. We have:
$$
\frac{1}{w} \mathop{\rm Res}(\Phi_0(u),w) = \frac{1}{w}\lim_{u\to
w}\Phi_0(u)(u-w) = \lim_{z\to 1}\Phi_0(zw)(z-1)
$$
and
$$
\frac{1}{\bar w} \mathop{\rm Res}(\Phi_0(u),\bar w) = \frac{1}{\bar
w}\lim_{u\to \bar w} \Phi_0(u)(u-\bar w) =
\lim_{z\to 1}\Phi_0(z\bar w)(z-1).
$$
Since the above two limits exist we can evaluate them
changing $z$ with $x\in\mathbb{R}$. We get:
$$
\frac{1}{\bar w} \mathop{\rm Res}(\Phi_0(u),\bar w) = \lim_{x\to
1}\Phi_0 (\overline{xw})(x-1) = \lim_{x\to 1}
\overline{\Phi_0(xw)(x-1)} =
\overline{\frac{1}{w} \mathop{\rm Res}(\Phi_0(u),w)}.
$$
Setting $\lambda = w^{-1}\mathop{\rm Res}(\Phi_0(u),w)$ we obtain,
from the above equation, and (\ref{eq:11}):
$$
\frac{1-e^{-2\pi n\omega}}{2\pi n\omega}\delta_{n} = \lambda w^{in\omega}
+\bar\lambda\bar{w}^{in\omega}.
$$
Thus, when $n\neq 0$, $\delta_{n}=0$ if and only if
\begin{equation}
 \frac{\lambda}{\bar\lambda} = -\left (
 \frac{\bar w}{w}\right )^{in\omega}. \label{eq:14}
\end{equation}
Now
$$
\left (\frac{\bar w}{w}\right )^{in\omega} = e^{-n(\Arg\bar w -
\Arg w)\omega} = e^{-2n(\pi-\Arg w)\omega} = \alpha^{2}_{n} > 0.
$$
Thus $\lambda = -\alpha^{2}_{n}\bar\lambda$ and then
$\bar\lambda = -\alpha^{2}_{n}\lambda$. So $\lambda =
\alpha^{4}_{n}\lambda$ and hence $\alpha^{2}_{n}=1$,
because $\lambda\neq 0$. But this means that $w$ is
real and negative and this contradicts $w\neq \bar w$.
The proof is finished.

We now give a closer look at the case where $\Phi_0(u)$
has two poles of multiplicity $r_0$ on the half-line
$\Arg (u)=\beta_0$. Since
$$
{1\over w_j^{r_0}}\mathop{\rm Res}( (u-w_j)^{r_0-1}\Phi_0 (u),
w_j) =\lim_{z\to 1} (z-1)^{r_0}\Phi_0 (w_jz),
$$
we see that we have to study the equation:
\begin{equation}
 \lambda_1|w_1|^{in\omega} + \lambda_2|w_2|^{in\omega} = 0
\label{eq:15}
\end{equation}
where $\lambda_j = \lim_{z\to 1} (z-1)^{r_0}\Phi_0
(w_jz)$.
Now, equation (\ref{eq:15}) has a solution $n\in\mathbb{N}$ if and
only if
$$
\left | {w_1\over w_2} \right |^{in\omega} =
-{\lambda_2\over\lambda_1}.
$$
Thus we have the following cases:
\begin{description}
 \item{(i)} $|\lambda_2|\neq|\lambda_1|$. In this case
 $\liminf_{n\to\infty}\left | \lambda_1 |w_1|^{in\omega} +
 \lambda_2 |w_2|^{in\omega} \right | \neq 0$;
 \item{(ii)} $|\lambda_2|=|\lambda_1|$ and
 $\log(|\frac{w_1}{w_2}|)$ is a rational multiple of
 $T=\frac{2\pi}{\omega}$. In this case equation (\ref{eq:15}) either
 has a (least) solution $n_0\in\mathbb{N}$, and hence it has infinite
 solutions of the type: $n=n_0+kq$, $k\in\mathbb{Z}$, and $q\in\mathbb{Z}$
 such that $\frac{q}{T}\log(|{w_1\over w_2}|)\in\mathbb{Z}$,
 or $\liminf_{n\to\infty}\left | \lambda_1 |w_1|^{in\omega} +
 \lambda_2|w_2|^{in\omega} \right | \neq 0$;
 \item{(iii)} $|\lambda_2|=|\lambda_1|$ and
 $\log(|{w_1\over w_2}|)$ is not a rational multiple
 of $T$. In this case $\liminf_{n\to\infty}\left | \lambda_1
 |w_1|^{in\omega} + \lambda_2|w_2|^{in\omega} \right | = 0$.
\end{description}
As a consequence, Proposition \ref{prop2} applies if either
$|\lambda_2|\neq|\lambda_1|$, or $\log(|{w_1\over
w_2}|)$ is a rational multiple of $T$ and
$-{\lambda_2\over\lambda_1}$ is not one of the (finite)
values of $\big| {w_1\over w_2} \big|^{in\omega}$.

\paragraph{Remarks.}
(i) In this section we have assumed that
$\Phi_0(u)$ and $\Phi'_0(u)$ are both rational functions on $\mathbb{C}$ with
the same poles. However it may well happen that the poles of $\Phi'_0(u)$
correspond to essential singularities of $\Phi_0(u)$.
Nonetheless, the argument of this Section hold even in this case,
we simply do not have to integrate by parts as in (\ref{eq:8}) and
use (\ref{eq:7}) instead of (\ref{eq:9}). For example equation
(\ref{eq:11}) reads: $$
\begin{array}{rl}
  \ds{\delta_{n}(1-e^{-2\pi n\omega})\over 2\pi i} &\!\!\!\ds =
  \sum_{j=1}^N \sum_{m=0}^{r-1} {1\over m!} Res
  ((u-w_j)^m \Phi_0' (u), w_j ) i^m e^{in\omega\log |w_j|} \\
  &\!\!\!\ds\quad n\omega (n\omega + i)(n\omega + 2i)\ldots(n\omega + (m-1)i)
  {e^{-n\omega \Arg w_j}\over w_j^{m}}
\end{array}
$$
with $r$ being the multiplicity of the pole $w_j$ of $\Phi_0'(u)$.
Thus Proposition \ref{prop1} and \ref{prop2} hold with the following 
changes: \\
$r_0$ is the maximum of the multiplicities of the poles of
$\Phi_0'(u)$ and in equations (\ref{eq:12}), (\ref{eq:13}),
$\Phi_0(u)$ and $u^{in\omega-1}$ have to be changed with
$\Phi_0'(u)$, and $u^{in\omega}$ respectively. Moreover, 
Theorem \ref{thm2}
holds as is (with $\Phi_0'(u)$ instead of $\Phi_0(u)$ of
course). The proof goes almost in the same way, apart that
equation (\ref{eq:14}) has to be written as:
$$
 \frac{\lambda w}{\overline{\lambda w}} = -\left (
 \frac{\bar w}{w}\right )^{in\omega} = -\alpha_{n}^{2}\le 0.
$$
The rest of the proof is the same with $\lambda w$ instead of
$\lambda$.
\newline\noindent
Note that the function $\Phi_0(u)={\rm tan}^{-1}\left (
\frac{3u} {2(u^{2}+1)}\right )$ is an example of such a
situation. In fact $2i{\rm tan}^{-1}(u) =
\log\left ( \frac{1+iu}{1-iu}\right )$ has, at the points
$u=\pm i$, essential singularities and is defined, for
example, outside the set $\{ z=iy\; | \; |y|\ge 1\}$. In
the next Section we will give a method to construct a second
order differential equation satisfied by $p(t):=
\Phi_0(e^t)$. Following this method we see that $p(t)$
satisfies:
$$
\ddot p = \frac{1}{9}\frac{9-41{\rm tan}^{2}p}{({\rm
tan}^{2}p+1)^{2}}{\rm tan}p.
$$
(ii) From the previous Section we know that $\delta_{n}$ is
also given by:
$$
\delta_{n} = \int_{-\infty}^{+\infty} \dot p(t) e^{in\omega t} dt.
$$
Now, the function:
$$
\delta (\xi) = \int_{-\infty}^{+\infty} \dot p(t) e^{-i\omega\xi t} dt
$$
tends to zero as $|\xi|\to\infty$ and the same holds for $i\xi\delta(\xi)$
and $(i\xi)^{2}\delta(\xi)$ because $p(t)$, $\dot p(t)$ tend to zero
exponentially fast as $|t|\to\infty$ (and hence belong to $L^{2}(\mathbb{R})$),
and $p(t)$ satisfies the equation $\ddot p=f(p)$. In fact we have, for
example, integrating by parts:
$$
i\omega\xi\delta(\xi) = \int_{-\infty}^{+\infty} \ddot p(t) e^{-i\omega\xi t}
dt = \int_{-\infty}^{+\infty} f(p(t)) e^{-i\omega\xi t} dt
$$
and
$$
(i\omega\xi)^{2}\delta(\xi) = \int_{-\infty}^{+\infty} f'(p(t))\dot p(t)
e^{-i\omega\xi t} dt.
$$
As a consequence $\xi\delta(\xi)\to 0$, $\xi^{2}\delta(\xi)\to 0$ as
$|\xi|\to\infty$ and then $\delta_{n}\in L^1(\mathbb{Z})\cap L^{2}(\mathbb{Z})$.
Thus the series:
$$
\sum_{n\in\mathbb{Z}} \delta_{n}e^{in\omega t}
$$
is totally convergent to a continuous, $T$-periodic function
$\Delta(t)$ whose $n$-th Fourier coefficient is precisely
$\delta_{n}$. Note that, being $\delta_{-n}=\bar\delta_{n}$
we have:
$$
\Delta(t) = \delta_0 + 2\sum_{n=0}^{+\infty} ({\rm Re}\delta_{n})
\cos nt - ({\rm Im}\delta_{n})\sin nt.
$$
Now, let $\phi_1(t)$ and $\phi_2(t)$ be two $T$-periodic
functions on $\mathbb{R}$. For $t\in\mathbb{R}$, we set:
$$
\phi_1*\phi_2(t) = \frac{1}{T}\int_{-T/2}^{T/2} \phi_1(t-s)
\phi_2(s) ds.
$$
Then $\phi_1*\phi_2(t)$ is $T$-periodic and its $n$-th Fourier
coefficient is:
$$
\begin{array}{l}
 \displaystyle{1\over T^{2}} \int_{-T/2}^{T/2}
 \int_{-T/2}^{T/2} \phi_1(t-s)\phi_2(s) ds e^{-in\omega t} dt  \\
 \displaystyle = {1\over T^{2}} \int_{-T/2}^{T/2} \Big\{
 \int_{-T/2}^{T/2} \phi_1(\tau)e^{-in\omega\tau} d\tau \Big \}
 \phi_2(s) e^{-in\omega s} ds = \phi_1^{(n)}\phi_2^{(n)}
\end{array}
$$
$\phi_j^{(n)}$ being the $n$-th Fourier coefficient of $\phi_j(t)$.
As a consequence $\delta_{n}\hat q_{n}$ is the Fourier coefficient of
both $M(\alpha)$ and $\Delta*q(\alpha)$ that is
$$
M(\alpha) = \Delta*q(\alpha) = {1\over T}\int_{-T/2}^{T/2}
\Delta(\alpha-s)q(s) ds = {1\over T}\int_{-T/2}^{T/2}
\Delta(s)q(\alpha-s) ds.
$$
Finally, we note that the function $\Delta (\alpha)$ can be expressed by
means of $\dot p(t)$ as follows. We have:
$$
\begin{array}{rl}
 M(\alpha ) &  \ds = \int_{-\infty}^{+\infty} \dot p(t-\alpha)
 q(t) dt = \sum_{k\in\mathbb{Z}} \int_{(2k-1)T/2}^{(2k+1)T/2} \dot p(s-\alpha)
 q(s) ds  \\
 &  \ds = \int_{-T/2}^{T/2}\sum_{k\in\mathbb{Z}} \dot p(s+kT-\alpha)
 q(s) ds.
\end{array}
$$
Now, the function $\sum_{k\in\mathbb{Z}} \dot p(kT-t)$ is $T$-periodic
and continuous (actually analytic, since so is $\dot p(t)$). Thus:
$$
\Delta(t) = T\sum_{k\in\mathbb{Z}} \dot p(kT-t).
$$

\section{Some examples}

In this Section we apply the result of the previous Section to
construct a second order equation in $\mathbb{R}$ whose Melnikov function
vanishes identically on an infinite number of (independent)
$2\pi$-periodic functions, or on any $2\pi$-periodic function
$q(t)$ (actually, we will give an example of case (ii) of the
previous Section). To do this we will first prove a result
allowing us to construct second order equations satisfied by
prescribed homoclinic solutions. For completeness we will also
give an example showing that this procedure can also produce non
rational differential equations. To start with we make some
remarks on the properties of the function $p(t)$ and the
associated $\Phi_0(x)$. Since $p(t)\to 0$ as $|t|\to\infty$,
there exists $t_0$ such that $\dot p(t_0)=0$. Without loss of
generality we can assume that $t_0=0$. Thus $p(t)= p(-t)$
because both satisfy the Cauchy problem: $$
\begin{array}{c}
 \ddot x = f(x)  \\
 x(0) = p(0) \quad \dot x(0) =\dot p(0)\, .
\end{array}
$$
Possibly changing $f(x)$ with $-f(-x)$, we can also assume
that $p_0:=p(0)>0$. Thus $t\dot p(t)<0$ for any $t\neq 0$.
In fact if $\dot p(\tau)=0$ for some $\tau>0$ then $p(t)$ would be
$2\tau$-periodic contradicting the fact that $p(t)\to 0$ as
$|t|\to\infty$. Now, let $\Phi_0(x)$ be as in the previous
section. Since we want that the equality $\Phi_0(x)=
p(\log x)$ holds for any $x>0$, we see that we have to
assume that:
\begin{equation}
 \Phi_0(x)=\Phi_0(1/x)\label{eq:16}
\end{equation}
for any $x>0$ and then
$\Phi_0(u)=\Phi_0(1/u)$ because of uniqueness of the
analytical extension. Thus, besides the pole $w_j$,
$\Phi_0(u)$ has also the pole $w_j^{-1}$ whose argument
is $2\pi - \Arg (w_j)$ (here we assume that
$0<\Arg (w_j)\le\pi$). Then, the function $\Phi_0(x)$ is
increasing in $[0,1]$ and decreasing in $[1,\infty)$,
moreover $\Phi_0(1)=p_0$. Thus there exist two functions
$x_{\pm}(p)$ defined on $(0,p_0]$ such that
\begin{description}
 \item{(i)} $x_{+}(p)$ is decreasing on $(0,p_0]$ and
$x_{+}(p_0)=1$,
 \item{(ii)} $x_{-}(p)$ is increasing on $[0,p_0]$ and
$x_{-}(p_0)=1$,
 $x_{-}(0)=0$
\end{description}
and satisfy:
\begin{equation}
 \Phi_0(x_{\pm}(p)) = p.
 \label{eq:17}
\end{equation}
Note that, because of (\ref{eq:16}), we obtain:
$$
x_{+}(p) = {1\over x_{-}(p)}
$$
for any $p\in(0,p_0]$, moreover, being $\Phi_0'(1)=0$ we
get: $\ds\lim_{p\to p_0} x_{\pm}'(p_0) = \mp\infty.$
Now, $p(t)$ satisfies the equation:
$$
\ddot p(t) = F(e^{t})
$$
where $F(x)=x^{2}\Phi_0''(x)+x\Phi_0'(x)$ is a rational function
defined on a neighborhood of $x\ge 0$. Thus the point is to see
whether $F(e^t)=f(p(t))$ for some $C^{1}$-function $f(p)$.
We note the following:
$$
F(1/x) = {1\over x^{2}}\Phi_0''(1/x) +{1\over
x}\Phi_0'(1/x)
$$
and, using (\ref{eq:16}), we get:
$$
-{1\over x^{2}}\Phi_0'(1/x) = \Phi_0'(x), \quad
{2\over x^{3}}\Phi_0'(1/x)+{1\over x^{4}}\Phi_0''(1/x) =
\Phi_0''(x).
$$
Thus it is easy to see that $F(x) = F(1/x)$ and then
$F(x_{-}(p))=F(x_{+}(p))$. Note that we also get $x^{2}F'(x)
= -F'(1/x)$. This last equation implies $F'(1)=0$ (note that
a similar conclusion holds for $\Phi_0'(1)$). We set
$$
f(p) := F(x_{-}(p))\; (=F(x_{+}(p))), \quad p\in[0,p_0].
$$
Note that we choose $x_{-}(p)$ so that $f(p)$ is continuous
up to $p=0$; moreover from (\ref{eq:17}) we see that either
$x_{-}(p(t))=e^t$ or $x_{+}(p(t))=e^t$, but then, in any case
$f(p(t)) = F(e^{t}) = \ddot p(t)$. Thus we want to show that $f(p)$
can be extended in a $C^{1}$ way in a neighborhood of $[0,p_0]$.
To this end we simply have to show that the limits:
$\lim_{p\to p_0}{d\over dp}F(x_{-}(p))$ and
$\lim_{p\to 0}{d\over dp}F(x_{-}(p))$ exist in $\mathbb{R}$. We have:
$$
\lim_{p\to p_0}{d\over dp}F(x_{-}(p)) = \lim_{p\to p_0}
{F'(x_{-}(p))
\over \Phi_0'(x_{-}(p))} = \lim_{x\to 1}{F'(x) \over
\Phi_0'(x)} =
\lim_{x\to 1}{F''(x) \over \Phi_0''(x)} = {F''(1) \over
\Phi_0''(1)}
\in\mathbb{R} .
$$
Note that we have to assume that $\Phi_0''(1)\neq 0$
because otherwise $f(p_0)=0$ and then $p(t)\equiv p_0$
will be another solution of the Cauchy problem $\ddot p =
f(p)$, $p(0)=p_0$, $\dot p(0) = 0$. Hence $f(p)$ cannot
even be Lipschitz continuous in any neighborhood of $p_0$.
Next we prove that the limit
$$
\lim_{p\to 0}{d\over dp}F(x_{-}(p))=\lim_{x\to 0}{F'(x)
\over \Phi_0'(x)}
$$
exists in $\mathbb{R}$. To this end we observe that, $\Phi_0(x)$
being analytic, $x=0$ has to be a zero of finite multiplicity,
say $k\ge 1$, of $\Phi_0(x)$, that is $\Phi_0(x)=x^kG(x)$,
with $G(0)\neq 0$. Then $F(x) = k^{2}\Phi_0(x) + O(x^{k+1})$
and
$$
\lim_{x\to 0}{F'(x) \over \Phi_0'(x)} = \lim_{x\to0}
{k^{2} \Phi_0'(x) + O(x^{k}) \over \Phi_0'(x)} = k^{2}.
$$
Let us briefly recall what we have seen so far. Let $\Phi_0(x)$
be a non negative rational function on $[0,+\infty)$ such that
$\Phi_0(x)=\Phi_0(1/x)$ and $\Phi_0(x)$ is strictly
increasing from $0$ to $\Phi_0(1)$ on $[0,1]$ (and of course
strictly decreasing from $\Phi_0(1)$ to $0$ on $[1,\infty)$)
and $\Phi_0''(1)\neq 0$. Then $p(t):=\Phi_0(e^t)$ is a
homoclinic solution of a $C^{1}$-equation $\ddot p = f(p)$.
We now want to show that this equation can have further
smoothness properties, for example that can be $C^{2}$. To do
this we show that the limits $\ds\lim_{p\to p_0}
{d^{2}\over dp^{2}}F(x_{-}(p))$ and $\ds\lim_{p\to
0}{d^{2}\over dp^{2}}F(x_{-}(p))$ exist in $\mathbb{R}$. As for the
first we will see that the result holds without any further
assumption on $\Phi_0(x)$, but the same does not hold
in general for the second. We have:
$$
{d^{2}\over dp^{2}}F(x_{-}(p)) =
{F''(x_{-}(p))\Phi_0'(x_{-}(p))-
F'(x_{-}(p)) \Phi_0''(x_{-}(p)) \over
\Phi_0'(x_{-}(p))^{3}}
$$
and hence we are led to evaluate the two limits:
\begin{equation}
 \lim_{x\to 1}{F''(x)\Phi_0'(x)-F'(x)\Phi_0''(x) \over
 \Phi_0'(x)^{3}}\label{eq:18}
\end{equation}
and
\begin{equation}
 \lim_{x\to 0}{F''(x)\Phi_0'(x)-F'(x)\Phi_0''(x) \over
 \Phi_0'(x)^{3}}.\label{eq:19}
\end{equation}
Let us consider, first, the limit in (\ref{eq:18}). Since
$F'(1)=\Phi_0'(1)=0$
we apply L'Hopital rule and get:
$$
\begin{array}{l}\ds
 \lim_{x\to 1}{F''(x)\Phi_0'(x)-F'(x)\Phi_0''(x) \over
 \Phi_0'(x)^{3}} = {1\over 3\Phi_0''(1)}\lim_{x\to 1}
 {F'''(x)\Phi_0'(x)-F'(x)\Phi_0'''(x) \over
 \Phi_0'(x)^{2}} \\  \\
 \ds = {1\over 6\Phi_0''(1)^{2}}\lim_{x\to 1}{F^{(iv)}(x)
 \Phi_0'(x) + F'''(x)\Phi_0''(x) - F''(x)\Phi_0'''(x)
 - F'(x)\Phi_0^{(iv)}(x) \over \Phi_0'(x)}
\end{array}
$$
provided the last limit exists. Now, from $\Phi_0(1/x)=
\Phi_0(x)$ we get:
$$
-\Phi_0'''(1/x) = x^{6}\Phi_0'''(x) +
6x^{5}\Phi_0''(x) +
6x^{4}\Phi_0'(x)
$$
and then $\Phi_0'''(1) = -3\Phi_0''(1)$. Similarly
$F'''(1) = -3F''(1)$. Thus we can apply again L'Hopital
rule and obtain:
$$
\lim_{x\to 1}{F''(x)\Phi_0'(x)-F'(x)\Phi_0''(x) \over
\Phi_0'(x)^{3}} = {F^{(iv)}(1)\Phi_0''(1) -
F''(1)\Phi_0^{(iv)}(1)
\over 3\Phi_0''(1)^{3}}\in \mathbb{R}.
$$
Now we consider $\lim_{p\to 0} f''(p)$ that is the limit in
(\ref{eq:19}). Recall that we set $\Phi_0(x)=
x^kG(x)$, with $G(0)\neq 0$ and note that also $F(x)$ has
$x=0$ as a zero of multiplicity $k$. Thus the numerator of
(\ref{eq:19}) has $x=0$ as a zero of multiplicity (at least)
$2k-3$ while the denominator has $x=0$ as a zero of
multiplicity $3(k-1)$. Now a simple computation shows that
$x=0$ is actually a zero of the numerator of multiplicity at
least $2(k-1)$, but in general this is the maximum we can
expect. In fact one has:
\begin{equation}
 F''(x)\Phi_0'(x)-F'(x)\Phi_0''(x) =
 k(k+1)(2k+1)x^{2(k-1)}G(x)G'(x)
 + O(x^{2k-1}).\label{eq:20}
\end{equation}
Of course this is not enough to prove that $f(p)$ is $C^{2}$
up to $p=0$, unless $k=1$. So we assume
that $k\in\mathbb{N}$, $k>1$, and the following holds:
$$
\Phi_0(x) = x^{k}G_0(x^{k})
$$
where $G_0(0)\neq 0$. In this case in fact the left hand side
of equation (\ref{eq:20}) vanishes at $x=0$ (since $G'(0)=0$) and
we actually have:
$$
F''(x)\Phi_0'(x)-F'(x)\Phi_0''(x) =
6k^{5}G_0(x^k)G_0'(x^k)x^{3(k-1)} + O(x^{4k-3})
$$
and then
$$
\lim_{x\to 1}{F''(x)\Phi_0'(x)-F'(x)\Phi_0''(x) \over
\Phi_0'(x)^{3}} = {6k^{2}G_0'(0)\over G_0(0)^{2}}
\in\mathbb{R}.
$$
Let us rewrite what we have done as a theorem:

\begin{theorem} \label{thm3}
Let $\Phi_0(u)=u^kG(u)$, $k\ge 1$, be a
rational function such that $G(0)\ne 0$ and the following hold:
\begin{description}
 \item {(i)} $\Phi_0(u)=\Phi_0(1/u)$ (that is $G(1/u)=
 u^{2k}G(u)$),
 \item {(ii)} $\Phi_0(x)>0$ when $x$ is real and $x>0$,
 \item {(iii)} $\Phi_0'(x)=0$ on $x>0$ is equivalent to $x=1$,
 \item{(iv)} $\Phi_0''(1)\neq 0$.
\end{description}
Then $\lim_{u\to\infty}u\Phi_0'(u)=0$ and there exists a
$C^{1}$--function $f(p)$ in a neighborhood of $[0,\Phi_0(1)]$
such that $p(t)=\Phi_0 (e^{t})$ is the solution of the equation
$\ddot p = f(p)$. Moreover, if $G(u)=G_0(u^k)$ for some rational
function $G_0(u)$, $G_0(0)\ne 0$, the function $f(p)$ is $C^{2}$
in a neighborhood of $[0,\Phi_0(1)]$.
\end{theorem}

\paragraph{Proof}. We only have to prove that
$\lim_{u\to\infty}u\Phi_0'(u)=0$. To this end we note that
$G(1/u)=u^{2k}G(u)$ implies $G'(1/u)=-2ku^{2k+1}G(u) -
u^{2k+2}G'(u)$ and then
$$
\begin{array}{rl}
 \ds\lim_{u\to\infty} u\Phi_0'(u) & \ds =
 \lim_{u\to 0} {\Phi_0'(1/u)\over u} = \lim_{u\to 0}
 {k G(1/u)\over u^k} + {G'(1/u)\over u^{k+1}}   \\
 & \ds = \lim_{u\to 0} -u^{k}\{ uG'(u) + k G(u)\} = 0.
\end{array}
$$
Finally note that condition $\Phi_0''(1)\neq 0$ can be
also stated in terms of $G(u)$ since condition (i) implies
$G'(1)=-kG(1)$ and then $\Phi_0''(1) = G''(1) - k(k+1) G(1)$.
The proof is finished.


One might wonder what kind of system one obtains starting with
functions $\Phi_0(x)$ as in Theorem \ref{thm3}. 
Actually, since $\Phi_0(x)$
is a rational function one might expect that the function $F(x_{-}(p))$
is a rational function of $p$. However this is not generally true
because $x_{\pm}(p)$ are in general far from being rational. To show
this we start with the function
$$
\Phi_0(x) := \frac{x(x^2+1)}{x^4+4x^2+1}.
$$
There is no particular reason for the coefficient 4. It only has to be
different from 2, otherwise the expression of $\Phi_0(x)$ can be
simplified. It is easy to see that all the conditions of 
Theorem \ref{thm2} are satisfied. In particular we have
$$
\Phi_0(x) = \Phi_0\left ( {1\over x}\right );\quad \Phi_0(1) =
{1\over 3}; \quad \Phi_0'(1) = 0;\quad \Phi_0'(0) = 1; \quad
\Phi_0''(1) = -{1\over 9}.
$$
Moreover we obtain the following expression for $F(x) = x^{2}
\Phi_0''(x) + x\Phi_0'(x)$:
$$
\begin{array}{rl}
    F(x) & \ds\!\!\! = \frac{x(x^{2}+1)(x^{8}-16x^{6}+18x^{4}-16x^{2}+1)}
    {(x^{4}+4x^{2}+1)^{3}}  \\
    & \ds\!\!\! = \Phi_0(x)\frac{(x^{8}-16x^{6}+18x^{4}-16x^{2}+1)}
    {(x^{4}+4x^{2}+1)^{2}}.
\end{array}
$$
In order to apply the above described procedure we have to solve the
equation:
\begin{equation}
    x(x^{2}+1) = (x^{4}+4x^{2}+1) p \label{eq:21}
\end{equation}
for $x$ as a function of $p$. Since $\Phi_0(x)=\Phi_0(1/x)$ we can
solve (\ref{eq:21}) multiplying it by $x^{-2}$ and setting $z =
x+x^{-1}$. We obtain:
$$
pz^2-z+2p = 0
$$
which has the solution
\begin{equation}
    z_{\pm}(p) = \frac{1\pm\sqrt{1-8p^2}}{2p}.
    \label{eq:22}
\end{equation}
Now, $x_-(p)$ and $x_+(p) = x_-(p)^{-1}$ are both solutions of the
equation $x+x^{-1} = z_+(p)$, and not $x+x^{-1} = z_-(p)$, because,
for $p = p_0 = \Phi_0(1) = 1/3$ we have $z_{+}(p_0) =
2$, $z_{-}(p_0) = 1$ and $x_+(p_0) = x_-(p_0) = 1$.
Now, we want to construct $f(p) = F(x_{-}(p))$ where $x_{-}(p)$ is
the unique solution of $\Phi_0(x) = p$ such that $0\le
x_{-}(p)\le 1$. We have $\Phi_0(x_{-}(p)) = p$ for any $0\le p
\le {1\over 3}$, and $x_{-}(\Phi_0(x)) = x$ for any $0\le x \le
1$. So: $$ f(p) = p f_0(x_{-}(p)) $$ where $$ f_0(x) =
\frac{x^{8}-16x^{6}+18x^{4}-16x^{2}+1}{(x^{4}+4x^{2} +1)^{2}} =
\frac{x^{4}-16x^{2}+18-16x^{-2}+x^{-4}}{(x^{2}+4 +x^{-2})^{2}}. $$
Since $x_{-}(p)+x_{-}(p)^{-1} = z_{+}(p)$ we have
$$
x_{-}^{2}(p)+x_{-}(p)^{-2} = z_{+}^{2}(p)-2,
$$ and $$
x_{-}^{4}(p)+x_{-}(p)^{-4} = z_{+}^{4}(p)-4z_{+}^{2}(p)+2. $$
So $$ f_0(x_{-}(p)) =
\frac{z_{+}^{4}(p)-20z_{+}^{2}(p)+52} {(z_{+}^{2}(p)+2)^{2}}. $$
Plugging (\ref{eq:22}) in the above equation we obtain, after some
algebra: $$ f_0(x_{-}(p)) = 7-6\sqrt{1-8p^2}-48p^2. $$ Thus we
have seen that the second order equation
\begin{equation}
    \ddot x = x(7-6\sqrt{1-8x^2}-48x^2)    \label{eq:23}
\end{equation}
has the homoclinic solution $\ds p(t) = \frac{e^{t}(e^{2t}+1)}
{e^{4t}+4e^{2t}+1}$. Note that the equation (\ref{eq:23}) is defined on the
interval $(-\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}})$ that contains
$[0,\frac{1}{3}]$.


We now give an example of equations whose associated Melnikov function
vanishes on an infinite dimensional space of $C^{1}$, $2\pi$-periodic
functions. Take $a\in\mathbb{R}$, $a^{2}\neq 0,1$ and set:
$$
\Phi_0(x) = {|a^{4}-1|x^{2}\over
(x^{2}+a^{2})(a^{2}x^{2}+1)}.
$$
Note that $\Phi_0(x)>0$, for $x\neq 0$, and changing $a$
with $a^{-1}$, we obtain the same function, so we assume
$a^{2}>1$. Moreover $\Phi_0(x)$ satisfies all the assumptions
of Theorem \ref{thm3} including $\Phi_0(u)=u^kG_0(u^k)$ with $k=2$.
For example one has:
$$
\Phi_0''(1) = {8a^2(1-a^2) \over (1+a^2)^3}
$$
which is different from zero when $a^{2}\neq 0,1$. Now, the
(simple) poles of $\Phi_0(u)$ are
$$
w_1:=ia,\quad \bar w_1=-ia, \quad w_2:=ia^{-1}, \quad
\bar w_2=-ia^{-1}
$$
and we have
$$
\begin{array}{l}
 \lambda_1 = \lim_{z\to 1} (z-1)\Phi_0(w_1z) =
 -1/2  \\
 \bar\lambda_1 = \lim_{z\to 1} (z-1)\Phi_0(\bar w_1z)
 = -1/2  \\
 \lambda_2 = \lim_{z\to 1} (z-1)\Phi_0(w_2z) = 1/ 2  \\
 \bar\lambda_2 = \lim_{z\to 1} (z-1)\Phi_0(\bar w_2z)
 = 1/2.
\end{array}
$$ Thus equation (\ref{eq:11}) gives, after some algebra: $$
\delta_{n} = \frac{\pi in}{\sinh (n\frac{\pi}{2})}\sin(n\log a) $$
and we obtain the following:
\begin{description}
 \item{(a)} taking $a=e^{m\pi}$, $m\in\mathbb{N}$, we can construct a family
 of second order equation whose Melnikov function is identically
 zero, no matter what the ($2\pi$-periodic) perturbation is;
 \item{(b)} taking $a=e^{m\pi/2}$, $m\in\mathbb{N}$, we can construct a family
 of second order equation whose Melnikov function is identically zero
 on an infinite number of independent $2\pi$-periodic
 perturbations but not for all.
\end{description}
To obtain an analytical expression of such systems, we proceed as
in the previous example. The equation $\Phi_0(x) = p$ reads: $$
pa^2(x^2+\frac{1}{x^{2}})-a^4+p+pa^4+1 = 0 $$ and again can be
solved by setting $z=x+x^{-1}$. We obtain:
$$
z^2=(a^{2}-1)\frac{a^{2}+1-p(a^{2}-1)}{pa^2} $$
which has the solutions
$$ z_{\pm}(p) =
\pm\frac{\sqrt{p[a^4-1-p(a^2-1)^{2}]}}{ap} $$
It is not necessary
to solve the equations $x+x^{-1} = z_{\pm}$. We only have to
note that both $x_{-}(p)$ and $x_{+}(p) = x_-(p)^{-1}$ are
solutions of the equation $x+x^{-1} = z_{+}(p)$, and not
$x+x^{-1} = z_{-}(p)$, because $z_+(p_0) = 2$, $z_-(p_0) =
-2$ and $x_-(p_0) = x_+(p_0) = 1$. Next we compute $F(x)=x^{2}
\Phi_0''(x) + x\Phi_0'(x)$. Since $F(x) = F(1/x)$ we expect
that $F(x)$ can be expressed in terms of $z = x+x^{-1}$. An
annoying computation shows that, in fact, $F(x) = G(x+x^{-1})$,
where: $$ G(z) = 4a^{2}(a^{4}-1)
\frac{a^2z^4-(a^{4}+4a^2+1)z^2+2(a^2-1)^{2}}
{(a^2z^2+(a^2-1)^2)^{3}}. $$ Thus $f(p,a) = F(x_{-}(p)) =
G(z_{+}(p))$. After some algebra, we get:
\begin{equation}
    f(p,a) = 4p\left ( 2p^{2}-3{a^{4}+1\over a^{4}-1}p+1\right )
    = 4p[2p^{2}-3p\coth(2\log a)+1].
    \label{eq:24}
\end{equation}
Thus, in this case, $p(t)=\Phi_0(e^t)$ is the solution of an
analytic second order equation $\ddot x = f(x,a)$ such that, when
$a=e^{m\pi}$ (or $a=e^{m\pi/2}$), $m\in\mathbb{N}$, its Melnikov function
vanishes identically on any $2\pi$-periodic functions (or it is
identically zero for infinitely many independent $2\pi$-periodic
functions but not for all). The geometrical meaning of this is
that, in spite of the fact that the perturbation of the equation
is of the order $O(\varepsilon)$, the distance between the stable and
unstable manifolds of the perturbed equation, along a transverse
direction, is of the order (at least) $O(\varepsilon^{2})$. This means
that in order to study the intersection of the stable and the
unstable manifolds, we have to look at the second order Melnikov
function. For a $C^{2}$-equation like $\ddot x + f(x) = \varepsilon q(t)$
this second order Melnikov function is given by: $$ M_2(\alpha)
= \int_{-\infty}^{+\infty} \dot p(t) f''(p(t)) v_{\alpha}^{2}(t)
dt $$ where $v_{\alpha}(t)$ is any fixed bounded solution of the
equation
\begin{equation}
    \ddot x = f'(p(t))x + q(t+\alpha). \label{eq:24a}
\end{equation}
This solution exists thanks to the fact that $M(\alpha)=0$. Note that
any two of these bounded solutions differ for a multiple of $\dot p(t)$,
and hence $v_{\alpha+2\pi}(t) = v_{\alpha}(t) +\lambda\dot p(t)$, for
some $\lambda\in\mathbb{R}$. On the other hand $M_2(\alpha)$ does not
depend on the particular solution $v_{\alpha}(t)$ we choose. This fact
easily follows from the fact that $\ddot p(t)$ is a bounded solution
of the non homogeneous system
$$
\ddot x = f'(p(t))x + f''(p(t))\dot p(t)^{2}
$$
and $\dot v_{\alpha}(t)$ is a bounded solution of
$$
\ddot x = f'(p(t))x + f''(p(t))\dot p(t)v_{\alpha} + \dot q(t+\alpha).
$$
Hence:
$$
\int_{-\infty}^{+\infty}\dot p(t)f''(p(t))\dot p(t)^{2} dt = 0
$$
and
$$
\int_{-\infty}^{+\infty}\dot p(t)f''(p(t))\dot p(t)v_{\alpha}(t) dt =
-\int_{-\infty}^{+\infty}\dot p(t)\dot q(t+\alpha) = M'(\alpha) = 0.
$$
Thus $M_2(\alpha)$ is $2\pi$-periodic. This fact, however, also
follows from the more general fact that the bifurcation function itself
is $2\pi$-periodic. We now prove the following result.

\begin{theorem}\label{thm4}
 For any $m\in\mathbb{N}$ and $c\ne 0$, the second
order Melnikov function $M_2(\alpha)$ associated to the equation
\begin{equation}
    \ddot x = 4x(2x^{2}-3x\coth (2 m\pi)+1) +\varepsilon \left (\frac{c}{2} +
    q_{odd}(t) \right ) \label{eq:25}
\end{equation}
does not vanish identically on a dense subset ${\cal S}$ of the
space $C^1_{odd,2\pi }$ of all $C^{1}$-smooth, $2\pi-$periodic and
odd functions $q_{odd}(t)$. Actually, ${\cal S}$ is the complement
of a codimension one closed linear subspace of $C^1_{odd,2\pi}$. Moreover
if a positive integer $k\in\mathbb{N}$ exists such that $q_{odd}(t+\frac{\pi}{k})
= -q_{odd}(t)$, $M_2(\alpha)$ changes sign in the interval
$[0,\frac{\pi}{k}]$. 
\end{theorem}

\paragraph{Proof.} We emphasize the fact that many of the
arguments of this proof can be used even for more general
equations than (\ref{eq:25}) having a homoclinic orbit. For this
reason we will write $f(x)$ instead of $4x(2x^{2}-3x\coth (2
m\pi)+1)$, $q(t)$ instead of $\frac{c}{2}+q_{odd}(t)$ and $p(t)$
for the orbit homoclinic to the hyperbolic fixed point $x=0$, in
the first part of the proof. Note that the hyperbolicity of $x=0$
implies that $f'(0)>0$.

As a first step we simplify the expression of $M_2(\alpha)$ in the
following way. Let $v_{\alpha}(t)$ be a bounded solution of the equation
$\ddot x =f'(p(t))x + q(t+\alpha)$, whose existence is guaranteed by the
fact
that $M(\alpha)=0$, and $u(t)$ be the unique $2\pi$-periodic solution of
the equation $\ddot x =f'(0)x + q(t)$. Then $r_{\alpha}(t) := v_{\alpha}(t)
- u(t+\alpha)$ is a bounded solution of
$$
\ddot x = f'(p(t))x + [f'(p(t))-f'(0)]u(t+\alpha).
$$
As a consequence $r_{\alpha}(t)\to 0$ exponentially together with its
first and second derivative (uniformly with respect to $\alpha$) and
$v_{\alpha}(t) = r_{\alpha}(t) + u(t+\alpha)$. Then
$$
\begin{array}{rl}
    M_2(\alpha) = & \ds \!\!\! \int_{-\infty}^{+\infty} \frac{d}{dt}
    [ f'(p(t) -f'(0) ] v_{\alpha}^{2}(t) dt  \\
    = & \ds \!\!\! -2\int_{-\infty}^{+\infty} [f'(p(t)) - f'(0)]
    v_{\alpha}(t)\dot v_{\alpha}(t) dt  \\
    = & \ds \!\!\! -2\int_{-\infty}^{+\infty} \Big ( [
    \ddot v_{\alpha}(t) - q(t+\alpha) ]\dot v_{\alpha}(t) -
    f'(0) v_{\alpha}(t)\dot v_{\alpha}(t) \Big ) dt.
\end{array}
$$
Now we observe that
$$
\begin{array}{rl}
    \ds 2\lim_{n\to +\infty}\int_{-n\pi}^{n\pi}\ddot v_{\alpha}(t)
    \dot v_{\alpha}(t) dt = & \!\!\! \ds\lim_{n\to+\infty} \Big \{
    [\dot r_{\alpha}(n\pi) + \dot u(n\pi+\alpha)]^{2}  \\
     & \ds - [\dot r_{\alpha}(-n\pi) + \dot u(-n\pi+\alpha)]^{2}
     \Big \} = 0
\end{array}
$$
because $\dot r_{\alpha}(t) \to 0$ as $|t|\to+\infty$
and $u(t)$ is $2\pi$-periodic. Similarly, using the fact that
$r_{\alpha}(t)\to 0$ as $|t|\to+\infty$, we get:
$$
\begin{array}{rl}
    \ds 2\lim_{n\to +\infty}\int_{-n\pi}^{n\pi} v_{\alpha}(t)
    \dot v_{\alpha}(t) dt = & \!\!\! \ds\lim_{n\to+\infty} \Big \{
    [r_{\alpha}(n\pi) + u(n\pi+\alpha)]^{2}  \\
     & \ds - [r_{\alpha}(-n\pi) + u(-n\pi+\alpha)]^{2}
     \Big \} = 0.
\end{array}
$$
As a consequence
\begin{equation}
    M_2(\alpha) = 2 \lim_{n\to +\infty}\int_{-n\pi}^{n\pi}
    \dot v_{\alpha}(t)q(t+\alpha) dt \label{eq:25a}
\end{equation}
Note that $\ds\lim_{n\to +\infty}\int_{-n\pi}^{n\pi}$ in equation
(\ref{eq:25a}) cannot be replaced by $\ds\int_{-\infty}^{+\infty}$
because the convergence of this integral is not guaranteed. Now,
in order to compute $v_{\alpha}(t)$, we first look for a
fundamental matrix of the homogeneous equation $\ddot x =
f'(p(t))x$. We already know that $\dot p(t)$ is a solution of the
previous equation that satisfies also $\dot p(0) = 0$, and $\ddot
p(0) \neq 0$. So we look for a solution $y(t)$ such that
$y(0)\ddot p(0) = 1$ and $\dot y(0)=0$. If $y(t)$ is such a
solution, Liouville Theorem implies that $$ X(t) = \pmatrix{y(t) &
\dot p(t) \cr \dot y(t) & \ddot p(t)} $$ satisfies ${\rm det} X(t)
= 1$ that is $\dot p(t)\dot y(t) - \ddot p(t) y(t) = -1$.
Integrating this equation we obtain: $$ y(t) = -\dot p(t)\int^t
\frac{1}{\dot p(s)^{2}}ds. $$ Note that, no matter the constant we
add to the integral, $c\dot p(t)$ vanishes for $t=0$; however the
constant is uniquely determined by the condition $\dot y(0) = 0$
(from which the equality $y(0) \ddot p(0) = 1$ follows). Let $\mu
= \sqrt{f'(0)}$. From $p(t)=P(e^{\mu t})$, we obtain:
\begin{equation}
    y(t) = Y(e^{\mu t}) \label{eq:25b}
\end{equation}
where
\begin{equation}
    Y(x) = -\frac{1}{\mu^{2}} xP'(x)\int^{x} \frac{d\sigma}{\sigma^{3}
    [P'(\sigma)]^{2}}. \label{eq:25c}
\end{equation}
Specializing (\ref{eq:25c}) to equation (\ref{eq:25}) where $P(x) =
\frac{(a^{4}-1)x}{(x+a^{2}) (a^{2}x+1)}$, with $\mu = 2$, we obtain
$Y(x) = Y_0(x) + Y_{s}(x) + Y_{b}(x)$ where
$$
\begin{array}{l}
    \ds Y_0(x) = \frac{3}{2} \frac{a^{2}(a^{8}+3a^{4}+1)}{a^{4}-1}
    \frac{x(x^{2}-1)\log x}{(x+a^{2})^{2}(a^{2}x+1)^{2}}  \\   \\
    \ds Y_{s}(x) = \frac{a^{2}}{8(a^{4}-1)}\left ( x + x^{-1}
    \right )  \\   \\
    \ds Y_{b}(x) = \frac{3}{4}\frac{a^{4}+1}{a^{4}-1} -
    \frac{a^{16}+52a^{12}+72a^{8}-4a^{4}-1}{16a^{2}(a^{4}-1)^{2}(x+a^{2})}
    +\frac{a^{12}+29a^{8}+29a^{4}+1}{16(a^{4}-1)(x+a^{2})^{2}}  \\
    \ds \phantom{v_{b}(x) =}
    -\frac{a^{16}+4a^{12}-72a^{8}-52a^{4}-1}{16a^{4}(a^{4}-1)^{2}(a^{2}x+1)}
    +\frac{a^{12}+29a^{8}+29a^{4}+1}{16a^{4}(a^{4}-1)(a^{2}x+1)^{2}}
\end{array}
$$
Note that $Y_0(x)+Y_{b}(x)$ is bounded on $[0,+\infty)$ while $Y_{s}(x)$
is
unbounded near $x=0$ and infinity. Now, the variation of constants formula
gives, for any solution of equation (\ref{eq:24a}):
$$
\begin{array}{rl}
    v_{\alpha}(t) = &\!\!\! \ds c_1y(t) + c_2\dot p(t) +
    \int_0^{t} [\dot p(t) y(s) - \dot p(s) y(t)]q(s+\alpha)
    ds   \\
    = &\!\!\! \ds \Big [ c_1 - \int_0^{t}\dot p(s)q(s+\alpha)
    ds \Big ] y(t) + \dot p(t) \Big [ c_2 +
    \int_0^{t} y(s)q(s+\alpha) ds \Big ].
\end{array}
$$ 
Then, from the boundedness of $q(t)$, the fact that $y(t)$ is
of the order $e^{\mu |t|}$ at $\pm \infty$ and $\dot p(t)$ is of
the order $e^{-\mu |t|}$ at $\pm\infty$, we see that the second
term is bounded on $\mathbb{R}$. Hence $v_{\alpha}(t)$ will be bounded on
$\mathbb{R}$ if and only if a constant $c_1$ exists such that 
$$ \Big[
c_1 - \int_0^{t}\dot p(s)q(s+\alpha) ds \Big] y(t) $$ 
is
bounded on $\mathbb{R}$, and this can happen (if and) only if 
$$ c_1 =
\int_0^{+\infty}\dot p(s)q(s+\alpha) ds = \int_{-\infty}^{0}\dot
p(s)q(s+\alpha) ds. $$ This choice of $c_1$ is made possible by
the fact that $M(\alpha)=0$ and gives: $$ v_{\alpha}(t) =
y(t)\int_{t}^{\infty}\dot p(s)q(s+\alpha) ds + \dot p(t) \left [
c_2 + \int_0^{t} y(s)q(s+\alpha) ds\right ]. $$ Note that
$v_{\alpha}(t)$ is bounded on $\mathbb{R}$ for any value of $c_2$.
However we can make it unique by adding the condition $\dot
v_{\alpha}(0) = 0$. Since $\dot y(0) = 0$ we see that this is
equivalent to choosing $c_2=0$. That is
\begin{equation}
    v_{\alpha}(t) = y(t)\int_{t}^{\infty}\dot p(s)q(s+\alpha) ds
    + \dot p(t) \int_0^{t} y(s)q(s+\alpha) ds. \label{eq:26}
\end{equation}
It is worth of mentioning that equation (\ref{eq:26}) gives a
bounded solution of equation (\ref{eq:24a}) provided $p(t)$ is a
homoclinic solution of $\ddot x = f(x)$, and $y(t)$ is defined as
in (\ref{eq:25b}) and (\ref{eq:25c}).

Now, we write $q(t) = q_{even}(t) + q_{odd}(t)$ where
$q_{even}(-t) = q_{even}(t)$ and $q_{odd}(-t) = -q_{odd}(t)$. Then
the solution $v(t)$ of the equation $\ddot x = f'(p(t))x + q(t)$
satisfies $v(t) = v_{even}(t) + v_{odd}(t)$ where $v_{even}(t)$ is
the (unique) bounded solution of $$ \ddot x = f'(p(t))x
+q_{even}(t), \quad \dot x(0) =0 $$ while $v_{odd}(t)$ is the
(unique) bounded solution of $$ \ddot x = f'(p(t))x +q_{odd}(t),
\quad \dot x(0) =0. $$ From $p(t)=p(-t)$, and the uniqueness of
the solutions we get $v_{even}(t) = v_{even}(-t)$ and $v_{odd}(t)
= -v_{odd}(-t)$ and then $$ M_2(0) = 2 \lim_{n\to
+\infty}\int_{-n\pi}^{n\pi} \dot v_{even}(t)q_{odd}(t) + \dot
v_{odd}(t)q_{even}(t) dt. $$ Now, we consider the situation where
$q_{even}(t) = \frac{c}{2}\neq 0$ is constant and different from
zero. We obtain immediately: $$ \int_{-n\pi}^{n\pi}\dot
v_{odd}(t)q_{even}(t) dt = c\; v_{odd}(n\pi). $$ Next, let
$u_{odd}(t)$ be the unique bounded solution of $\ddot x = f'(0)x
+q_{odd}(t)$. From the uniqueness we see that $u_{odd}(t)$ is
$2\pi$-periodic and odd, moreover $v_{odd}(t) - u_{odd}(t)$ is a
bounded solution of $\ddot x = f'(0)x +
[f'(p(t))-f'(0)]v_{odd}(t)$ and hence tends to zero exponentially
as $|t|\to +\infty$. As a consequence $$ \lim_{n\to
+\infty}v_{odd}(n\pi) = \lim_{n\to +\infty}u_{odd}(n\pi). $$ On
the other hand $-u_{odd}(-n\pi) = u_{odd}(n\pi) = u_{odd}(-n\pi)$
because of oddness and periodicity. As a consequence
$u_{odd}(n\pi) = 0$ and then $$ M_2(0) = 2 \lim_{n\to
+\infty}\int_{-n\pi}^{n\pi} \dot v_{even}(t)q_{odd}(t) dt = 2
\int_{-\infty}^{\infty} \dot v_{even}(t)q_{odd}(t) dt $$ the last
equality being justified by the fact that $v_{even}(t) +
\frac{c}{2f'(0)}$ tends to zero, as $|t|\to +\infty$, together
with its first derivative, being a bounded solution of $$ \ddot x
= f'(p(t))x - \frac{c}{2f'(0)} [ f'(p(t)) - f'(0)]. $$ At this
point we note that when $q_{odd}(t+ \frac{\pi}{k}) = -q_{odd}(t)$
we have $v_{\pi/k}(t) = v_{even}(t) - v_{odd}(t)$ and hence it is
easy to see that $$
\begin{array}{cc}
    M_2(\pi/k) & \ds\!\!\! = 2\lim_{n\to+\infty}\int_{n\pi}^{n\pi}
    \dot v_{\pi/k}(t)[\frac{c}{2} - q_{odd}(t)] dt \\
    & \ds\!\!\! = -2\int_{-\infty}^{\infty} \dot v_{even}(t) q_{odd}(t)
    = -M_2(0)
\end{array}
$$
and the theorem follows provided we prove that $M_2(0)\neq 0$.

Now, from equation (\ref{eq:26}) we obtain:
$$
v_{even}(t) = \frac{c}{2}\left ( \dot p(t) \int_0^{t} y(s) ds -
p(t) y(t) \right ) = \frac{c}{2} v(t)
$$
where $v(t)$ is defined by the equality. We note that $v(t)$ is the
bounded solution of $\ddot x = f'(p(t))x + 1$, with $\dot x (0) =0$
and that $v(t) + \frac{1}{f'(0)}$ tends to zero exponentially, as
$|t| \to +\infty$, together with its derivative. Moreover
$$
M_2(0) = c \int_{-\infty}^{\infty}\dot v(t)q_{odd}(t) dt =
c \int_0^{2\pi}r(t)q_{odd}(t) dt
$$
where
\begin{equation}
    r(t) = \sum_{k\in\mathbb{Z}} \dot v(t+2k\pi)\label{eq:26a}
\end{equation}
is $2\pi-$periodic and odd. From $p(t) = P(e^{\mu t})$ and $y(t) =
Y(e^{\mu t})$ we see that $v(t) = V(e^{\mu t})$ where
$$
V(x) = xP'(x)\int_1^{x}\frac{Y(\sigma)}{\sigma} d\sigma - P(x)Y(x).
$$
Note that $V(x)$ is linear in $Y(x)$. Applying the above
considerations to equation (\ref{eq:25}) (hence with $\mu=2$) we obtain
after some integrations:
$$
\begin{array}{rl}
    \ds V(x) +\frac{1}{4} & \ds \!\!\! = \frac{3a^{2}x(a^4+1)(1-x^{2})
    \log x}{4(a^2x+1)^2(x+a^2)^2}   \\
     & \!\!\! \ds + \frac{x[(a^{12}+23a^8+23a^4+1)(x^2+1)+16a^{2}
     (a^{8}+4a^4+1)x]}{16a^2(a^2x+1)^2(x+a^2)^{2}}.
\end{array}
$$ We set $$ \widetilde M_2(\alpha) := \int_{-\infty}^{+\infty} \dot
v(t) q_{odd}(t+\alpha) dt. 
$$ 
Then $\widetilde M_2(\alpha)$ is $2\pi-$periodic and 
$M_2(0) = c\widetilde M_2(0)$. Expanding $\widetilde
M_2(\alpha)$ into its Fourier series we get: 
$$ \widetilde
M_2(\alpha) = -\sum_{n\in\mathbb{Z}} in\gamma_{n}q_{n} e^{in\alpha} 
$$
$q_{n}$ being the $n-$th Fourier coefficient of $q_{odd}(t)$ and
$$ \gamma_{n} = \int_{-\infty}^{+\infty} [V(e^{2t})+\frac{1}{4}]
e^{int} dt. $$ Note that $in\gamma_{-n}/(2\pi )$ are also the Fourier
coefficients of the function $r(t)$ defined in (\ref{eq:26a}). Since
$q_{odd}(t)$ is an odd real function we easily get $q_{n} = ic_{n}$
where $c_{n}$ are real numbers such that $c_{n}=-c_{-n}$. Thus
\begin{equation}
    \widetilde M_2(\alpha) = \sum_{n\in\mathbb{Z}} n\gamma_{n}c_{n} e^{in\alpha}.
    \label{eq:27}
\end{equation}
Being $\widetilde M(\alpha)$ a real valued function, we also get:
$\bar\gamma_{n} =\gamma_{-n}$. Moreover, arguing as in Section 2 of this
paper we can evaluate the Fourier coefficients of $\widetilde
M_2(\alpha)$ by means of residues and get, for $n\neq 0$: $$
\gamma_{n} = \frac{\pi ie^{n\pi}}{\sinh (n\pi)}\Big ( \sum_{w_j}
\mathop{\rm Res} (W(u)u^{in-1},w_j) + \frac{2\pi i}{e^{2n\pi}-1}
\sum_{w_j}\mathop{\rm Res} (H(u)u^{in-1},w_j) \Big) $$ 
where
$w_j$ runs in the set $\{ \pm ia, \pm i/a\}$ and: $$
\begin{array}{rl}
    W(u) & \ds \!\!\! = W_0(u) + H(u)\log u   \\   \\
    H(u) & \ds \!\!\! = \frac{3a^{2}(a^{4}+1)u^{2}(1-u^{4})}
    {2(a^2u^{2}+1)^2(u^{2}+a^2)^2}  \\   \\
    W_0(u) & \ds \!\!\! = \frac{u^{2}((a^{12}+23a^8+23a^4+1)(u^4+1)
    +16a^{2}(a^8+4a^4+1)u^{2})}{16a^2(a^2u^{2}+1)^2(u^{2}+a^2)^{2}}
    \end{array}
$$
Note that $W(u)$ is the extension of $V(x^{2})+\frac{1}{4}$ to the
complex field. Moreover $W(u)$ is a meromorphic function on $\mathbb{C}
\setminus\{x\in\mathbb{R}  :  x\ge 0\}$ that satisfies $\lim_{u\to\infty}
W(u) = 0$ uniformly with respect to $Arg(u)\in (0,2\pi)$. An annoying
computation shows that:
$$
\begin{array}{rl}
\mathop{\rm Res} (H(u)u^{in-1},ia) = & \ds \!\!\! \frac{3i(a^{4}+1)}
    {8(a^{4}-1)}n(\cos(n\log a)+i\sin(n\log a)) e^{-n\pi/2}  \\
\mathop{\rm Res} (H(u)u^{in-1},i/a) = & \ds \!\!\! -\frac{3i(a^{4}+1)}
    {8(a^{4}-1)}n(\cos(n\log a)-i\sin(n\log a)) e^{-n\pi/2}  \\
\mathop{\rm Res} (H(u)u^{in-1},-ia) = & \ds \!\!\! \frac{3i(a^{4}+1)}
    {8(a^{4}-1)}n(\cos(n\log a)+i\sin(n\log a)) e^{-3n\pi/2}  \\
\mathop{\rm Res} (H(u)u^{in-1},-i/a) = & \ds \!\!\! -\frac{3i(a^{4}+1)}
    {8(a^{4}-1)}n(\cos(n\log a)-i\sin(n\log a)) e^{-3n\pi/2}.
\end{array}
$$
Thus:
$$
\sum_{w_j}\mathop{\rm Res} (H(u)u^{in-1},w_j) = -\frac{3(a^{4}+1)}
{4(a^{4}-1)}n(e^{-3n\pi/2}+e^{-n\pi/2})\sin (n\log a)
$$
which is zero for $a=e^{m\pi}$. A similar computation gives:
$$
\begin{array}{rl}
    \ds\sum_{w_j}\mathop{\rm Res} (W_0(u)u^{in-1},& \!\!\!\!\! w_j) =
    \ds -\frac{in(e^{n\pi}+1)}{32e^{3n\pi/2}}\frac{a^{12}+9a^{8}
    -9a^{4}-1}{a^{4}(a^{4}-1)}\cos (n\log a)  \\
    & \!\!\!\ds -\frac{3i(e^{n\pi}+1)}{4e^{3n\pi/2}}\frac{a^{4}+1}
    {a^{4}-1}\sin (n\log a)  \\
    \ds\sum_{w_j}\mathop{\rm Res} (H(u)u^{in-1}\log u,& \!\!\!\!\! w_j) =
    \ds\frac{3in(e^{n\pi}+1)\log a}{4e^{3n\pi/2}}\frac{a^{4}+1}
    {a^{4}-1}\cos (n\log a)  \\
    & \!\!\!\ds +\frac{3i(a^{4}+1)[2(e^{n\pi}+1) - n\pi (e^{n\pi}+3)]}
    {8(a^{4}-1)e^{3n\pi/2}}\sin (n\log a)
\end{array}
$$ As a consequence, setting $a=e^{m\pi}$: $$ \gamma_{n} =
\frac{(-1)^{nm}\pi n}{8\sinh (n\frac{\pi}{2})} \left [ \cosh
^{2}(2m\pi) - 6m\pi \coth (2m\pi) + 2 \right ] $$ that is $$
M_2(0) = c\cdot C_{m}\sum_{n\in \mathbb{Z} \setminus \{0\}
}\frac{(-1)^{nm}n^{2}c_{n}}{\sinh(n\frac{\pi}{2})} = 2 c \cdot
C_{m}\sum_{n>0}\frac{(-1)^{nm}n^{2} c_{n}}{\sinh (n\frac{\pi}{2})}
$$ $C_{m}$ being a positive constant. So, for any non zero real
number $c$, we have $\gamma_{n}\neq 0$, for any
$n\in\mathbb{Z}\setminus\{0\}$ and then $r(t)\neq 0$. Since $M_2(0)\ne 0$
if and only if $\int\limits_0^{2\pi} r(t)q_{odd}(t) dt \ne 0$,
the thesis of the present theorem follows. For example the space
of $2\pi-$periodic functions for which $M_2(0)\neq 0$ is
different from zero contains functions like
$\frac{c}{2}+q_{odd}(t)$, where $c\neq 0$, and $q_{odd}(t)$ is a
$2\pi-$periodic, odd function whose Fourier coefficients $ic_{n}$
satisfy $(-1)^{nm}c_{n}\ge 0$ (resp. $\le 0$) for $n>0$.  The
proof is finished.


We conclude this Section with a remark. Letting $m\to+\infty$ in equation
\begin{equation}
    \ddot x = 4x(2x^{2}-3x\coth (2 m\pi)+1) \label{eq:30}
\end{equation}
 we obtain the equation
\begin{equation}
    \ddot x = 4x(2x^{2}-3x+1) \label{eq:31}
\end{equation}
which has two {\it heteroclinic} connections to the equilibria
$x=0$ and $x=1$. Since the Melnikov function of equation
(\ref{eq:30}) is identically zero for any $2\pi$-periodic
perturbation of the equation, one might wonder whether this fact
holds for the Melnikov functions associated to the heteroclinic
orbits of equation (\ref{eq:31}). The answer to this question is
negative as it can be easily seen by direct evaluation of the
Fourier coefficients of the Melnikov function. In fact let us
consider, for example the heteroclinic solution of (\ref{eq:31})
going from $x=0$ to $x=1$: $$ p_{\infty}(t) = {e^{2t}\over
e^{2t}+1} = R(e^{t}) $$ where $R(x) ={x^{2}\over x^{2}+1}$.
Applying the procedure described in this paper we see that the
Fourier coefficients of the Melnikov function are given by
$\delta_{n}q_{n}$ where $\delta_0=1$ and, for $n\neq 0$: $$
\delta_{n} = {2n\pi\over 1-e^{-2n\pi}} [ \mathop{\rm Res}({u^{in+1}\over
u^{2}+1},i) + \mathop{\rm Res}({u^{in+1}\over u^{2}+1},-i)] = {n\pi\over
2\sinh (n{\pi\over 2})}. $$ Geometrically, this strange behaviour
depends on the fact that the homoclinic solution of (\ref{eq:30})
gets orbitally closer and closer (as $m\to\infty$) to the {\it
heteroclinic cycle} and not to any of the heteroclinic orbits. As
a matter of fact, setting $$ p_{m}(t) = {e^{2t}(e^{4m\pi}-1)\over
(e^{2t}+e^{2m\pi}) (e^{2t+2m\pi}+1)}, $$ the Melnikov function
associated to a heteroclinic solution of (\ref{eq:31}) is the
limit, for $m\to\infty$ of either: $$ \int_0^{+\infty}\dot
p_{2m}(t)q(t+\alpha) dt $$ or $$ \int^{0}_{-\infty}\dot
p_{2m}(t)q(t+\alpha) dt $$ and these are not zero in general. To
see this, consider, for example the heteroclinic solution of
(\ref{eq:31}) $p_{\infty}(t)$. We have, for $t\le 0$:
\begin{equation}
    0 \le \dot p_{\infty}(t+m\pi) - \dot p_{m}(t) =
    {1\over 2\cosh^{2}(m\pi-t)} \le 2e^{2t}. \label{eq:32}
\end{equation}
From Lebesgue's theorem we get then:
$$
\lim_{m\to+\infty}\int_{-\infty}^{0} [\dot p_{\infty}(t+m\pi) -
\dot p_{m}(t)] b(t) dt = 0
$$
for any $L^{\infty}$-function $b(t)$, and hence:
$$
\begin{array}{l}
    \ds\int_{-\infty}^{\infty} \ds \dot p_{\infty}(t) q(t+\alpha)
    dt = \lim_{m\to+\infty}\int_{-\infty}^{2m\pi} \dot p_{\infty}(t)
    q(t+\alpha) dt =   \\
    \ds\lim_{m\to+\infty}\int_{-\infty}^{0} \dot p_{\infty}(t+2m\pi)
    q(t+\alpha ) dt =
    \ds\lim_{m\to+\infty}\int_{-\infty}^{0}\dot p_{2m}(t)
    q(t+\alpha ) dt.
\end{array}
$$
A similar argument shows that
$$
\int_{-\infty}^{\infty} \ds \dot p_{\infty}(t+\pi) q(t+\alpha)dt =
\lim_{m\to+\infty}\int_{-\infty}^{0}\dot p_{2m+1}(t) q(t+\alpha ) dt.
$$

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\noindent\textsc{Flaviano Battelli} \\
Dipartimento di  Matematica ``V. Volterra'', \\
Facolt\'a di Ingegneria -- Universit\'a di Ancona,  \\
Via Brecce Bianche 1, 60131 Ancona - Italy \\
e-mail: fbat@dipmat.unian.it \smallskip

\noindent\textsc{Michal Fe\v ckan}\\
Department of Mathematical Analysis, Comenius University, \\
Mlynsk\'a dolina, 842 48 Bratislava - Slovakia \\
e-mail: Michal.Feckan@fmph.uniba.sk
\end{document}