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\markboth{\hfil Neumann and periodic boundary-value problems \hfil EJDE--2000/63}
{EJDE--2000/63\hfil Petr Girg \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol.~{\bf 2000}(2000), No.~63, pp.~1--28. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
 \vspace{\bigskipamount} \\
%
  Neumann and periodic boundary-value problems for quasilinear 
  ordinary differential equations with a nonlinearity in the derivative
%  
\thanks{ {\em Mathematics Subject Classifications:} 34B15, 47H14.
\hfil\break\indent
{\em Key words:} p-Laplacian, Leray-Schauder degree, Landesmann-Lazer condition.
\hfil\break\indent
\copyright 2000 Southwest Texas State University. \hfil\break\indent
Submitted July 18, 2000. Published October 16, 2000. \hfil\break\indent
Supported by grants GA\v{C}R \# 201/97/0395, and M\v{S}MT\v{C}R \# VS 97156} }
\date{}
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\author{Petr Girg}
\maketitle

\begin{abstract} 
 We present sufficient conditions for the existence of  solutions 
 to Neumann and periodic boundary-value problems for some class of
 quasilinear ordinary differential equations. We also show that 
 this condition is necessary for certain nonlinearities. 
 Our results involve the p-Laplacian, the mean-curvature operator 
 and nonlinearities blowing up.  
\end{abstract}


\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}{Definition}[section]
\newtheorem{example}{Example}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}{Remark}[section]

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\section{Introduction}

The semilinear boundary-value problems
\begin{eqnarray}
\label{eq1d}
&u''(t)+g(u'(t))+h(u(t))=f(t) \quad t\in (0,T)\,, &\\  
&u(0)=u(T)\,,\quad u'(0)=u'(T)\,,&
\label{per_con}
%\nonumber
\end{eqnarray}
and
\begin{eqnarray}
\label{eq1n}
&u''(t)+g(u'(t))=f(t) \quad t\in (0,T)\,,& \\  
\label{Neu_con}
&u'(0)=0\,,\quad u'(T)=0\,,& 
%\nonumber
\end{eqnarray}
have been extensively studied by many authors (see e.g. \cite{CD,Danc,Maw}).
In this paper we extend their results to the quasilinear boundary-value
problems:
\begin{equation}
\label{fund_eq}
\bigl(\varphi(u'(t))\bigr)'+g(u'(t))+h(u(t)) = f(t)\quad t\in (0,T)\,,
\end{equation}
subject to (\ref{per_con}) or to (\ref{Neu_con}).

Overall, we will assume the following:
  \begin{description}
   \item[{\rm(P)}] $\varphi$ is an increasing homeomorphism of $I_1$ onto $I_2,$
   where $I_1, I_2 \subset \mathbb R$ are open intervals containing zero and
   $\varphi(0)=0,$
   \item[{\rm(G)}] $g$ is a continuous real function,
   \item[{\rm(H)}] $h$ is a continuous, bounded real function having limits
     in $\pm\infty$:
$$h(-\infty):=\lim_{\xi\rightarrow -\infty}h(\xi)<\lim_{\xi\rightarrow +\infty}h(\xi)=:h(+\infty)\,.$$
\end{description}

\noindent
We also need to impose some of the following assumptions to prove several
particular results. 
\begin{description}
   \item[{\rm(P')}] the inverse of $\varphi$ (denoted
   by $\varphi_{-1}$) is continuously differentiable,
   \item[{\rm(P'')}] $\varphi'_{-1}(0)>0$,
   \item[{\rm(PH)}] $\varphi$~is odd and there exist $c, \delta >0$ and $p > 1$
   such that for all $z\in
   (-\delta,\delta)\cap\mathop{\rm Dom}\varphi:\,c|z|^{p-1}\leq|\varphi(z)|$,
   \item[{\rm(G')}] $g$ is a continuously differentiable real function.      
\end{description}

\noindent
The results presented involve blow-up-type nonlinearities such as
$\varphi(z)=\tan(z)$, 
bounded nonlinearities of the type $\varphi(z)=\arctan(z)$,
the $p$-Laplacian when $\varphi(z) =|z|^{p-2}z, 1<p<\infty$,
and include of course the classical results
(\cite{CD, Danc, Maw}) where
$\varphi$  is considered to be 
an identity mapping on $\mathbb R$. As for the Neumann
conditions, we obtain new results even in the semilinear case with
`$\varphi=\mbox{Identity}$' on $\mathbb R$.
To extend known results for (\ref{fund_eq})--(\ref{per_con}) or
(\ref{fund_eq})--(\ref{Neu_con}), 
we mainly combine and modify methods from \cite{CD, Danc, Maw}.

Since boundary-value problems (BVPs) of the type (\ref{fund_eq})--(\ref{per_con})
and (\ref{fund_eq})--(\ref{Neu_con}) appear in a wide variety of
applications, our extension of known results is interesting not only from the theoretical
point of view but also from a practical one. 
For instance the $p$-Laplacian with $p\not = 2$ arises in 
the study of nonlinear diffusion. Blow-up-type nonlinearity 
$\varphi(z)=z/\sqrt{1-z^2}$ comes from the modeling of mechanical
oscillations with relativistic correction. As an application of bounded
nonlinearity let us mention the mean curvature (or capillary surface) operator 
$\varphi(z)=z/\sqrt{1+z^2}$,
which occurs in problems of mathematical physics. 

Related topics are the subjects of studies by many
authors. Among others let us mention the recent paper \cite{ManMaw} in
which the authors study the existence of periodic solutions
of systems of quasilinear ordinary differential equations (ODEs) 
at resonance. Let us also mention papers \cite{HMZ1,HMZ2}; even thought
the results of those papers are of different nature, some of the techniques used are similar.
The resonance problems for
the p-Laplacian in partial derivatives are treated in \cite{ArcOrs} and \cite{BoDrKu}.

The paper is organized as follows. We begin by stating our main
results concerning the existence of the solution (Theorems 1-4) and 
discuss their applicability to various problems arising from 
physics in Section 2. 
Proofs of Theorems 1--3 can be found in Section 3. A proof of
Theorem 4 is presented in Section 4. Finally, we
present some results concerning the uniqueness of the solution in Section 5.    

Before proceeding to the results we shall establish the notation that is
used throughout the paper:

For measurable functions defined on $[0,T]$
we define:
\begin{eqnarray*}
L^{p} &:=& \Big\{u : \|u\|_{L^p}:=
\left(\int_{0}^T |u(t)|^{p} dt \right)^{\frac{1}{p}}< \infty \Big\},\ 1\leq p <
+\infty\,, \\
L^\infty &:=&\Big\{ u :  \|u\|_{\infty}:= \inf\limits_{
\begin{array}{c}
{\cal N}\subset (0,T) \\
\mathop{\rm meas}{\cal N}=0 
\end{array}
}\sup\limits_{t\in (0,T)\setminus {\cal N}} |u(t)| <\infty \Big\}\,.
\end{eqnarray*}

For functions with domain $[0,T]$, with distributional derivatives, 
we define:
\begin{eqnarray*}
& W^{1,p} := \big\{u\in L^p :\, u'\in L^p \big\} \,,&\\
& W^{1,p}_T := \big\{u\in W^{1,p} :\, u(0)=u(T)\big\} \,.&
\end{eqnarray*} 

For a given $k\geq 0$, let 
\begin{eqnarray*}
C^{k}_T &:=& \big\{ u : u\in C^k[0,T],u(0)=u(T), u'(0)=u'(T), \dots, \\
 && u^{(k-1)}(0)=u^{(k-1)}(T) \big\}\,,
\end{eqnarray*}
where $C^k[0,T]$ is the standard space of $k$-times continuously
differentiable functions defined on $[0,T]$, endowed with the norm
$\|u\|_{C^k}:=\sum_{i=0}^k \|u^{(i)}\|_{\infty}$
($C[0,T]:= C^0[0,T]$, $C_T:= C^0_T$). By $C^{\infty}_0(0,T)$ we denote
the set of functions $u$ defined on $(0,T)$, possessing derivatives of any
order in $(0,T)$ and such that the closure of $\{t\in (0,T)\,:\,\,\,
u(t)\not=0\}$ is a subset $(0,T)$.


To formulate our results, we use the decomposition   
\begin{equation}
\label{red_f}
f = \widetilde f + \overline  f\,, \quad\mbox{with } 
\overline  f = \frac{1}{T} \int_0^T f(t)\,dt
\end{equation}
and the following function spaces 
\begin{eqnarray*}
& \widetilde  C[0,T] := \left\{ u\in C[0,T]: \int_0^T u(t)dt = 0
\right\}\,, &
\end{eqnarray*}
$\widetilde  C_T := C_T\cap \widetilde  C[0,T]$, $\widetilde C^1[0,T]:=
C^1[0,T]\cap \widetilde C[0,T]$, $\widetilde C^1_T := C^1_T\cap\widetilde C[0,T]$.


Since we deal with the quasilinear equation, 
we shall work with a more general concept of solutions of ODE than that of
the classical solutions.
\begin{definition} \rm
\label{def_qsol}
A function $u\in C^1[0,T]$ is called a solution of  
(\ref{fund_eq})--(\ref{per_con}) or (\ref{fund_eq})--(\ref{Neu_con})
if $\varphi(u')\in C^1[0,T]$, $u$ satisfies 
(\ref{fund_eq}) pointwise in $(0,T)$ and  $u(0)=u(T)$, $u'(0)=u'(T)$ or  
$u'(0)=0$, $u'(T)=0$, respectively.
\end{definition}

\section{Main results}

In the following we will consider the range
of $\varphi$ to be an interval $(a,b)$, where $a$ and $b$ could be
$-\infty$ and $+\infty$, respectively.

Let us consider (\ref{fund_eq})--(\ref{per_con}) and (\ref{fund_eq})--(\ref{Neu_con}) 
with $h\equiv 0$.
Substituting $w=\varphi(u')$ in (\ref{fund_eq}) we obtain the semilinear
first order equation:
\begin{equation}
\label{f_eq}
 w'(t) + g\bigl(\varphi_{-1}\bigl(w(t)\bigr)\bigr) = f(t)\,.
\end{equation}
Now it remains to find how this substitution affects the boundary
conditions. 

Taking into account
$$
u(T)-u(0)=\int_0^T u'(t)dt=\int_0^T \varphi_{-1}\bigl(w(t)\bigr)dt
$$
and the fact that $\varphi$ is an increasing homeomorphism,
the periodic conditions (\ref{per_con}) are transformed as follows:
\begin{equation}
\label{f_con}
\int_0^T \varphi_{-1}\bigl (w(t)\bigr )dt = 0\,,
\quad w(0)=w(T)\,.
\end{equation}
Analogously, an equivalent form of the Neumann conditions (\ref{Neu_con})
reads
\begin{equation}
\label{N_con}
w(0) = 0\,,\quad w(T) = 0\,.
\end{equation}

\begin{theorem}
\label{Th1}
Let the assumptions {\rm(P)}, {\rm(G)} be satisfied and $h\equiv 0$.
Then, for any $\widetilde f\in \widetilde C_T:$ $\|\widetilde
f\|_{L^2}<\sqrt{\frac{3}{T}}\min\{-a,b\}$, there exists precisely one
$s(\widetilde f)\in\mathbb R$ such that (\ref{f_eq})--(\ref{f_con}) 
has a solution if and only if 
$$\overline  f = s(\widetilde f)\,.$$
In this case the periodic boundary-value problem for (\ref{fund_eq})
has a family of solutions $u_c(t)=u(t)+c$,
where
$u(t) = \int_0^t \varphi_{-1}\bigl(w(\widetilde f, \tau)\bigr)d\tau, $
$c\in\mathbb R$ is arbitrary and $w(\widetilde f, \cdot)$ is the solution of 
(\ref{f_eq})--(\ref{f_con}). Moreover, the mapping $s:\widetilde C_T
\rightarrow \mathbb R,$ $\widetilde f \mapsto s(\widetilde f)$ is continuous and
the absolute value of $s(\widetilde f)$ satisfy
\begin{equation} 
\label{odh_s}
|s(\widetilde f)|\leq\max\left\{\left|g(\xi)\right|\,:\,\varphi_{-1}\left(-\sqrt{\frac{T}{3}}
\|\widetilde f\|_{L^2}\right)\leq\xi\leq\varphi_{-1}\left(\sqrt{\frac{T}{3}}
\|\widetilde f\|_{L^2}\right)\right\}\,.
\end{equation} 
\end{theorem}


\begin{remark}
\label{rem_01}
{\rm Let us observe that
if $\varphi$ is an homeomorphism of interval $(\alpha,\beta)$,
$\alpha<0<\beta$ onto $\mathbb R$,
then (\ref{odh_s}) yields an uniform bound for $s(\widetilde f)$:
$$|s(\widetilde f)|\leq \max_{\xi\in [\alpha,\beta]}|g(\xi)|\,.$$}
\end{remark}


\begin{theorem}
\label{Th2}
Suppose that {\rm(P)},{\rm(P')},{\rm (G)} are satisfied and $h\equiv 0$. Then
the assertion of Theorem \ref{Th1} is valid and $u\in C^2_T$.
If also {\rm (G')} holds true then the solution $w(\widetilde f,\cdot)$ of
(\ref{f_eq})--(\ref{f_con}) is unique. The mappings $w:
\widetilde C_T \rightarrow C^1_T, \widetilde f\mapsto w(\widetilde f,\cdot)$
and $s:\widetilde C_T \rightarrow \mathbb R, \widetilde f \mapsto s(\widetilde f)$ are
continuously differentiable at any $\widetilde f\in\widetilde C[0,T]:$ 
$\widetilde f\not\equiv 0.$
If {\rm (P'')} is satisfied then $w$ and $s$
are continuously differentiable also at $\widetilde f\equiv 0.$
\end{theorem}




\begin{theorem}
\label{Th3}
Let the assumptions {\rm(P)}, {\rm(G)} be satisfied and $h\equiv 0$.
Then, for any  
$\widetilde f\in \widetilde C[0,T]$: $\|\widetilde
f\|_{L^2}<\sqrt{\frac{3}{T}}\min\{-a,b\}$, there exists
precisely one $s(\widetilde f)\in\mathbb R$
such that (\ref{f_eq})--(\ref{N_con}) has a solution
if and only if $\overline  f = s(\widetilde f)$.
In this case the Neumann boundary-value problem
for (\ref{fund_eq}) has a family of solutions $u_c(t)=u(t)+c,$ where
$u(t)=\int_0^t\varphi_{-1}\bigl(w(\widetilde f,\tau)\bigr)d\tau,$
$c\in \mathbb R$ is
arbitrary and $w(\widetilde f,\cdot)$ is the solution of (\ref{f_eq})--(\ref{N_con}).
The mapping $s:\widetilde C[0,T]\rightarrow \mathbb R, \widetilde f\mapsto s(\widetilde f)$
is continuous and the
absolute value of $s(\widetilde f)$ is estimated by (\ref{odh_s}).
If {\rm(P')} is satisfied  then $u\in C^2_T$. If also {\rm (G')} holds
true then   
mappings $w: \widetilde C_T\rightarrow C^1_T,\,\widetilde f\mapsto
w(\widetilde f, \cdot)$ and $s$ are continuously differentiable at
any point $\widetilde f\in \widetilde C[0,T]$.
\end{theorem}




\begin{remark}
\label{rem_prv}
{\rm
Let us see that we do not need the assumption {\rm(P'')} in Theorem \ref{Th3} 
in order to get that $w:\widetilde f\mapsto w(\widetilde f,\cdot)$ and
$s:\widetilde f\mapsto s(\widetilde f)$ are continuously differentiable at
$\widetilde f\equiv 0$, as we do in Theorem \ref{Th2}. The reason is the
following one:
The boundary conditions (\ref{N_con}) of the first order semilinear
problem corresponding to the Neumann BVP are
linear. Thus, they provide more regularity
than the nonlinear ones (\ref{f_con}) corresponding to the periodic BVP.} 
\end{remark}



Since the results considering (\ref{fund_eq})--(\ref{per_con}) and  
(\ref{fund_eq})--(\ref{Neu_con}), respectively, have the same formal
structure, we need to formulate them for the BVP
(\ref{fund_eq})--(\ref{per_con}) only. However, the statement, corresponding to
the BVP (\ref{fund_eq})--(\ref{Neu_con}), can be obtained by replacing the
expressions standing in front of brackets with the bracketed ones in
Theorem \ref{Th4}. We keep this notation in Section 4, when formulating
Lemmas 4, 6, 7 used in the proof of Theorem 4.
Note, that in the following theorem we suppose that $\varphi$ is an odd mapping, so that $b=-a$.

\begin{theorem}
\label{Th4}
 Assume {\rm (P)}, {\rm(PH)}, {\rm (G)}, {\rm(H)} and
\begin{equation}
\label{star} 
\sqrt{\frac{3}{T}}b-\sqrt{T}\sup_{\xi\in\mathbb R}|h(\xi)|>0\,.
\end{equation}
Then for any $\widetilde f\in\widetilde C_T$ 
( $\widetilde f\in\widetilde C[0,T]$ ) 
satisfying 
\begin{equation}\label{M_con}
\|\widetilde f\|_{L^2}<\sqrt{\frac{3}{T}}b-\sqrt{T}\sup_{\xi\in\mathbb R}|h(\xi)|
\end{equation} 
the BVP (\ref{fund_eq})--(\ref{per_con})  
( (\ref{fund_eq})--(\ref{Neu_con}) ) has a solution if
\begin{equation}
\label{D_like}
s(\widetilde f) + h(-\infty) < \overline  f< s(\widetilde f) + h(+\infty)\,,
\end{equation} 
where $s(\widetilde f)$ is given by Theorem \ref{Th1} ( Theorem \ref{Th3} ).

Suppose, moreover, that  
$$h(-\infty)<h(\xi)<h(+\infty)\, \quad \mbox{for all }\xi\in\mathbb R\,.$$
Then, if (\ref{D_like}) is false and
$\widetilde f\in\widetilde C_T$ ( $\widetilde f\in\widetilde C[0,T]$ )
satisfies (\ref{M_con}), the BVP (\ref{fund_eq})--(\ref{per_con}) 
( (\ref{fund_eq})--(\ref{Neu_con}) ) does not admit any solution.
\end{theorem}

\begin{remark}
{\rm
As one can expect, an analogous result to Theorem \ref{Th4} holds true also
for $h$ satisfying 
$$\quad\quad\quad
h(+\infty)<h(-\infty)\,.\quad\quad$$ 
In this case, for any $\widetilde f\in\widetilde C_T$ ( $\widetilde f\in\widetilde C[0,T]$ )
satisfying (\ref{M_con}), the sufficient condition on $\overline f$ reads as follows:
\begin{equation}
\label{fduka}
s(\widetilde f) + h(+\infty) <\overline  f< s(\widetilde f) + h(-\infty)\,.
\end{equation}
If moreover $h$ satisfies 
$$\quad h(+\infty)<h(\xi)<h(-\infty)\quad\mbox{for all }\xi\in\mathbb R$$
then (\ref{fduka}) is also necessary.

This is in full agreement with the semilinear case
for the periodic BVP studied in \cite{Danc}}.
\end{remark}

\begin{remark}
\label{rem_02}
{\rm
Notice, that if the range of $\varphi$ is $\mathbb R$ then $b=+\infty$ and
the conditions 
(\ref{star}) and
(\ref{M_con}) are satisfied identically. On the other hand, if $b<+\infty$
then one can apply Theorem \ref{Th4} provided that
$T<\frac{\sqrt{3} b}{\sup_{\xi\in\mathbb R}|h(\xi)|}$. Observe that the bigger $T$ is, the weaker 
condition (\ref{M_con}) is.
}
\end{remark}

\begin{remark}
\label{rem_00}
{\rm
Since we treat the problems (\ref{fund_eq})--(\ref{per_con}) and 
(\ref{fund_eq})--(\ref{Neu_con}) using transformed problems 
(\ref{f_eq})--(\ref{f_con}) and (\ref{f_eq})--(\ref{N_con}), respectively, 
we need suitable additional condition,
which ensure that solution $w$ of (\ref{f_eq})--(\ref{f_con}) or 
(\ref{f_eq})--(\ref{N_con}), respectively, satisfies
$w(t)\in \mbox{Im}\,\varphi$ for all $t\in [0,T]$.
One of such a possible condition is 
$\|\widetilde f\|_{L^2}<\sqrt{\frac{3}{T}}\min\{-a,b\}$ (which is a consequence
of (\ref{ODH1})) provided $h\equiv 0$. 

However, we would like to point out that the restrictive condition on the
`greatness' of $\widetilde f$ is given not only by the specific limitation of
the employed method, but that it also arises directly from the nature of the problem.
Let us consider (\ref{fund_eq})--(\ref{per_con}) or
(\ref{fund_eq})--(\ref{Neu_con}), respectively. 
Taking into account the boundary conditions (\ref{per_con}) or
(\ref{Neu_con}), respectively,
there exists $t_0\in [0,T]: u'(t_0)=0$.
Hence integrating (\ref{fund_eq}) we find: 
$$ \varphi(u'(t))=\int_{t_0}^t\left(\widetilde f(\tau) + \overline f -
  g\left(u'(\tau)\right) - h\left(u(\tau)\right)\right)d\tau\,, 
$$ 
which requires 
\begin{equation}
\label{intnec}
a<\int_{t_0}^t \left(\widetilde f(\tau) + \overline f -
  g\left(u'(\tau)\right) - h\left(u(\tau)\right)\right)d\tau<b\,.
\end{equation}
Integrating (\ref{fund_eq}) from $0$ to $T$ we also find
$$\overline f = \frac{1}{T}\int_0^T \left(g(u'(t))+h(u(t))\right)\,dt\,,$$
which implies 
\begin{equation}
\label{ljfds}
\inf\limits_{\xi\in\mathbb R}g(\xi)+\inf\limits_{\xi\in\mathbb R}h(\xi)\leq
\overline f\leq \sup\limits_{\xi\in\mathbb R}g(\xi)+\sup\limits_{\xi\in\mathbb R}h(\xi)\,.
\end{equation}

Suppose, in addition, that $g$ is bounded. 
If $\overline f$ does not satisfy (\ref{ljfds}) or one of the following two
inequalities holds:
\begin{equation}
\label{disaster1}
\sup_{t_0\in [0,T]}\inf_{t\in [0,T]}\left\{\int_{t_0}^t\left(\widetilde f(\tau)+\overline
    f
-\inf_{\xi\in\mathbb R} g(\xi)-\inf_{\xi\in\mathbb R} h(\xi)\right)d\tau\right\} < a\,,
\end{equation}
\begin{equation}
\label{disaster2}
\inf_{t_0\in [0,T]}\sup_{t\in [0,T]}\left\{\int_{t_0}^t\left(\widetilde f(\tau)+\overline
    f
-\max_{\xi\in\mathbb R} g(\xi)- \max_{\xi\in\mathbb R} h(\xi)\right)d\tau\right\} > b\,,
\end{equation}
then the BVP (\ref{fund_eq})--(\ref{per_con}) or
(\ref{fund_eq})--(\ref{Neu_con}), respectively, does not admit any solution
with $f=\widetilde f+\overline f$. Compare also with results in \cite{KaSe}. 
}
\end{remark}


\begin{remark}
\label{rem1}
{\rm
It is worth noting that if we consider the periodic  BVP
for quasilinear ODE:
$$(\varphi(u'))' + \lambda u' + h(u) = f,\quad t\in (0,T)\,,$$
where $\lambda\in\mathbb R$, 
then the condition (\ref{D_like}) from Theorem \ref{Th4} is reduced to
\begin{equation}
\label{LanLaz}
h(-\infty)<\overline  f<h(+\infty)\,.
\end{equation}  
It means that we obtain the Landesman-Lazer condition as a particular result.

Here we present a proof. 
If $g(z)\equiv \lambda z$, where $\lambda\in\mathbb R$, then 
integrating (\ref{f_eq}) from $0$ to
$T$ and using boundary conditions (\ref{f_con}) we find  
$0=\int_0^Tf(t)dt=(T\overline  f)$.  Since (\ref{f_eq})--(\ref{f_con}) 
has a solution if and only if $\overline  f=s(\widetilde f)$ (see Theorem \ref{Th1}),
we have that $s(\widetilde f)\equiv 0$. 
Consequently, if $g(z)\equiv \lambda z$ then the condition (\ref{D_like})
is reduced to (\ref{LanLaz}).

Note, that in order to get $s(\widetilde f)\equiv 0$ for the Neumann
boundary-value problem, we have to impose $g(z)\equiv 0$.}  
\end{remark}
\smallskip

\begin{example}
{\rm
Consider $\varphi(z)=m_0 z/\sqrt{1-\frac{z^2}{c^2}}$, $m_0>0$ and
$g,h \in C(\mathbb R,\mathbb R)$, $h$ has finite
limits $h(-\infty)<h(+\infty)$. 
Then it follows from Theorem \ref{Th4} that the periodic BVP for
$$
\left(\frac{m_0u'(t)}{\sqrt{1-\frac{(u')^2(t)}{c^2}}}\right)' + g\bigl(
u'(t)\bigr) + h\bigl(u(t)\bigr)= \widetilde f(t) + \overline  f
$$
has a solution for every couple $\widetilde f,\overline  f$ satisfying condition (\ref{D_like}):
$$h(-\infty)+s(\widetilde f)<\overline  f<h(+\infty)+s(\widetilde f)\,,$$
where $s:\widetilde C_T \rightarrow \mathbb R$ comes from Theorem \ref{Th1}.

Let us remark that the conditions (\ref{star}) and (\ref{M_con}) are
satisfied for any $\widetilde f\in\widetilde C_T$. 
This follows from the fact that the range of $\varphi$ is $\mathbb R$
(i.e. $b=+\infty$).

It is worth noting that this problem arises from
the relativistic dynamics and describes damped oscillations.} 
\end{example}

\begin{example}
{\rm
With respect to
Theorem \ref{Th1} for all $\widetilde f\in C_T$ there exists unique $\overline  f$ such that
the periodic boundary-value problem for
$$
u''(t)+k\,\bigl(u'(t)\bigr)^2 =\widetilde f(t) + \overline  f
$$
has a solution.
This boundary-value problem describes a steady-state process of gas
burning in jets of rockets (cf. \cite{WFA}).}
\end{example}

\begin{example}
{\rm
Consider the following periodic boundary-value problem: 
$$\displaylines{
\left(u'(t)-(u'(t))^3\right)'+ u'(t) + \arctan(u(t))=\widetilde f(t) 
+ \overline  f \quad\mbox{in}\quad (0,T)\,, \cr
u(0)= u(\pi),\quad u'(0)=u'(\pi)\,.
}$$

Let $\sqrt{\frac{3}{T}}\frac{2\sqrt{3}}{9}-\frac{\pi}{2}\sqrt{T}>0$.
Then,  for all $\widetilde f\in \widetilde C_T\,:\quad \|\widetilde f\|_{L^2} <
\sqrt{\frac{3}{T}}\frac{2\sqrt{3}}{9}-\frac{\pi}{2}\sqrt{T}$, this problem
has a solution if and only if $\overline f\in \mathbb R$ satisfy 
\begin{equation}
\label{ex_eq}
-\frac{\pi}{2}<\overline  f<\frac{\pi}{2}\,.
\end{equation}

To prove this result, we can use Theorem \ref{Th4}.
Indeed, $\varphi$ is odd homeomorphism of
$(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ 
onto $(-\frac{2\sqrt{3}}{9},\frac{2\sqrt{3}}{9})$, 
so that $b=-a=\frac{2\sqrt{3}}{9}$. 
Moreover, 
$$\lim_{z\pm\infty}\arctan\,z = \pm \frac{\pi}{2}\quad \mbox{and}\quad 
-\frac{\pi}{2}<\arctan\, z < \frac{\pi}{2},\\ \mbox{for all } z\in\mathbb R\,.$$ 
Hence the condition (\ref{ex_eq}) follows from (\ref{D_like}), where $s(\widetilde
f)\equiv 0$, because $g(z)\equiv z$ (see Remark \ref{rem1}). 

This problem describes forced oscillations of voltage $u$ in an 
electrical circuit with {\sl ferro-resonance} (see e.g. 
\cite{KotKub}). Such nonlinear circuits are used in 
radiotechnics.} 
\end{example} 


\begin{table} 
\begin{center}
{\small \tabcolsep=1pt
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|} \hline &
\multicolumn{4}{|c|}{Periodic, $g(s)=s^3$} & 
\multicolumn{3}{|c|}{Neumann, $g(s)=s^3$} & 
\multicolumn{2}{|c|}{both, $h(s)=\arctan s$} \\ \cline{2-10} 
$\varphi(z)$ & \multicolumn{1}{|c|}{Thm 1} & 
\multicolumn{3}{|c|}{Theorem 2} & \multicolumn{3}{|c|}{Theorem 3} 
& \multicolumn{2}{|c|}{Theorem 4} \\ \hline 
 & (P)+(G) & +(P') & +(G') & +(P'') & (P)+(G) & +(P') & +(G') &
\multicolumn{2}{|c|}{(P)+(G)+(PH)}\\ \hline  
$\begin{array}{c} |z|^{p-2}z, \\ 1<p<2 \end{array}$ & {\sf A} &
\multicolumn{2}{|c|}{\sf A} & {\sf NA} & \multicolumn{3}{|c|}{\sf A} &\multicolumn{2}{|c|}{\sf A} 
\\ \hline
$z $ & \multicolumn{4}{|c|}{$\begin{array}{c} 
\sf A: \quad \mbox{\cite[Thm 1]{Danc}},\\ \mbox{\cite[Thm 3.4]{CD},  
\cite[Thm 4]{Maw}} \end{array}$} & 
    \multicolumn{3}{|c|}{$\begin{array}{c} \sf A: 
    \mbox{\cite[Thm 3.3]{CD},}\\ \mbox{\cite[Thm 1]{Maw}}\end{array}$}
& \multicolumn{2}{|c|}{$\begin{array}{c} \sf A \\ \mbox{\cite[Thm 2]{Danc}}^{\#}\end{array}$}  
\\ \hline
$\begin{array}{c} |z|^{p-2}z, \\ p>2 \end{array}$ & {\sf A} &
\multicolumn{3}{|c|}{\sf NA} & {\sf A} & \multicolumn{2}{|c|}{\sf NA} & \multicolumn{2}{|c|}{\sf A}
\\ \hline
$\frac{z}{\sqrt{1-z^2}}$ &  {\sf A} &  \multicolumn{3}{|c|}{\sf A} &
\multicolumn{3}{|c|}{\sf A} &  \multicolumn{2}{|c|}{\sf A}
\\ \hline
$\frac{z}{\sqrt{1+z^2}}$ &  {\sf A$^{*}$} &
    \multicolumn{3}{|c|}{\sf A$^{*}$} &
\multicolumn{3}{|c|}{\sf A$^{*}$} & $\begin{array}{c} T<\frac{2\sqrt{3}}{\pi} \\ {\sf A}^{\dag}
\end{array}$ & $\begin{array}{c} T>\frac{2\sqrt{3}}{\pi} \\ {\sf NA} \end{array}$
\\ \hline
\multicolumn{1}{|c|}{$\exp\left(\frac{1}{|z|}\right)z$} & {\sf A} & \multicolumn{2}{|c|}{\sf A} & {\sf NA} & \multicolumn{3}{|c|}{\sf A}
& \multicolumn{2}{|c|}{\sf NA} \\ \hline
\end{tabular} }
\end{center}
\caption{\quad  Legend: $^\#$ With periodic conditions  only,\hfill\break
$^{*}$ Provided that $\|\widetilde f\|_{L^2} < \sqrt{\frac{3}{T}}$,
\quad $^{\dag}$ Assuming that 
$\|\widetilde f\|_{L^2}<\sqrt{\frac{3}{T}}-\frac{\pi}{2}\sqrt{T}$.}
\label{tbl1} \end{table}


Table \ref{tbl1} illustrates the applicability of Theorems 
\ref{Th1}--\ref{Th4} for some particular nonlinearities $\varphi$. 
The symbol {\bf\sf A} (applicable) indicates that the 
corresponding assumption is satisfied and thus using an 
appropriate Theorem one obtains the desired conclusion; on the 
other hand {\bf\sf NA} (not applicable) indicates that the 
corresponding assumption is not satisfied and that the selected 
Theorem is not applicable in that case). We also relate our 
results to the results already known. 

\newpage

\section{Proofs of Theorems \ref{Th1} -- \ref{Th3}}

We start this section with the following lemma:

\begin{lemma}
\label{lemma1}
Let
$f\in C[0,T]$,
$q:J_1\rightarrow J_2$ be continuous,
where $J_1,J_2$ are nonempty intervals and $J_1$ contains zero in its interior.
Let $w\in C^1[0,T]$, $w(0)=w(T)$ and $w$ satisfies
\begin{equation}
\label{qeqa}
\quad w'(t)+q(w(t))=f(t),\quad t\in(0,T)
\end{equation}
then
\begin{equation}
\label{ODH2}
|w(t)-w(s)|\leq |t-s|^{\frac{1}{2}}\|\widetilde f\|_{L^2}\,.
\end{equation}
Moreover, if there exists $t_0\in [0,T]$ such that $w(t_0)=0$
then 
\begin{equation}
\label{ODH1}
\|w\|_{C} \leq \sqrt{\frac{T}{3}}\|\widetilde f\|_{L^2}\,.
\end{equation}
\end{lemma}

\begin{remark}
{\rm 
The previous lemma will be used to estimate solutions of (\ref{f_eq})--(\ref{f_con})
or (\ref{f_eq})--(\ref{N_con}), respectively. Indeed, the second of
(\ref{f_con}) yields $w(0)=w(T)$ directly. On the other hand, the first of
(\ref{f_con}), the continuity of $w$ and the fact, that $\varphi$ satisfies
{\rm (P)} imply that there exists $t_0\in [0,T]$ such that $w(t_0)=0$. In
the case of (\ref{N_con}) we have $w(0)=w(T)=0$. 
}
\end{remark}

\paragraph{Proof} 
Using the Cauchy--Schwarz inequality it is easy to show that
\begin{equation}
\label{wswth}
|w(t)-w(s)|=\left|\int_s^t w'(\tau)d\tau\right|\leq |t-s|^{\frac{1}{2}}\|w'\|_{L^2}\,.  
\end{equation}
Multiplying both sides of equation (\ref{f_eq}) by $w'$, integrating from $0$
to $T$, splitting $f$ as in (\ref{red_f}) and using $w(0)=w(T)$,
we find that $\| w'\|^2_{L^2}=\left(\widetilde f,  w'\right)_{L^2}$ (where
$\left(\widetilde f,  w'\right)_{L^2}:=\int_0^T\widetilde f w'$ is the scalar
product in $L^2$).
The Cauchy-Schwarz inequality yields
$\|w'\|_{L^2}\leq\|\widetilde f\|_{L^2}$,
which together with (\ref{wswth}) establishes (\ref{ODH2}).

Now we are going to estimate $\|w\|_{C}.$
Decompose the function
$w$ as $w = \widetilde w + \overline  w$, where $\int_0^T\widetilde w(t)dt=0$ and
$\overline w\in \mathbb R$.
Since $\widetilde w\in C^1[0,T] \subset W^{1,2}[0,T]$ and $w(0)=w(T)$,  
the Sobolev inequality 
\cite[Proposition 1.3]{MawW} yields  $\|\widetilde  w\|_{C} \leq \sqrt{\frac{T}{12}}\| 
w'\|_{L^2}\leq\sqrt{\frac{T}{12}}\|\widetilde f\|_{L^2}.$
Since we assume that there
exists a $t_0\in [0,T]$ such that $w(t_0)=0$, it shall be 
$|\overline w|\leq \|\widetilde w\|_{C}$. 
Hence $\|w\|_{C}\leq 2\sqrt{\frac{T}{12}}\|\widetilde
f\|_{L^2}=\sqrt{\frac{T}{3}}\|\widetilde f\|_{L^2},$
which is the desired inequality (\ref{ODH1}).
\hfill$\diamondsuit$\medskip


Now we proceed to proofs of Theorems \ref{Th1}--\ref{Th3}. 

\paragraph{Proof of Theorem \ref{Th1}}
We divide the proof into six steps:

\noindent{\bf Step 1} {\it Let $q$ be a continuously differentiable and bounded
real function. We prove that for all $\widetilde  f\in \widetilde  C_T$ there
exists  $\overline  f\in \mathbb R$ such that the equation
\begin{equation}
\label{a_w}
w'(t) + q\bigl(w(t)\bigr) = \widetilde  f(t) + \overline  f
\end{equation}
has a solution $w\in C^1[0,T]$ satisfying (\ref{f_con}).}

Let $\widetilde  f\in \widetilde  C_T$ be given.
Since $q$ is continuously differentiable and bounded function, the solution ( denoted by
$w(t,\overline  f,\alpha)$ )
to the initial-value problem
\begin{eqnarray} \label{In_a}
&w'(t) + q\bigl(w(t)\bigr) = \widetilde  f(t) + \overline  f,&\\ 
&w(0)  =  \alpha& \nonumber
\end{eqnarray}
exists on $(0,T)$, it is unique and is continuously differentiable with
respect to parameters $\alpha$ and $\overline  f$ (see e.g \cite{CodLew}). 
Let $M=\sup_{y\in\mathbb R}|q(y)|$. Taking $\overline  f >
M+\varepsilon$ and integrating (\ref{a_w}) from 0 to T, we find
 $w(T,\overline  f, \alpha) > \alpha$. On the
other hand, if $\overline  f < -M-\varepsilon$ then we get $w(T,\overline  f,
\alpha)<\alpha.$
Since $w$ depends continuously on $\overline  f$, for all $\alpha\in\mathbb R$, there
exists $\overline  f_{\alpha}\in\mathbb R$ such that $w(T,\overline  f_{\alpha}, \alpha)=\alpha.$
Moreover, the partial derivative with respect to the parameter $\overline f$
, $w_{\overline  f}(t,\overline  f, \alpha)$, is a solution to the 
linear initial-value problem
\begin{eqnarray} \label{inz}
&z'(t)  =  1-q'\bigl(w(t,\overline  f, \alpha)\bigr )z(t)\,, &\\ 
&z(0) =  0\,.& \nonumber
\end{eqnarray}
The explicit formula of the solution of (\ref{inz}) yields
$w_{\overline  f}(T,\overline  f,\alpha) > 0$, which means that $w(T,\cdot, \alpha)$ is
increasing and $\overline  f_{\alpha}$ is unique. Thus, we can define the
mapping $\psi_1:\mathbb R\rightarrow [-M,M]$ by $\alpha\mapsto\overline
f_{\alpha}$. As $w_{\overline  f}(T,\overline  f,\alpha) > 0$, by the
abstract implicit function theorem (see \cite[Theorem 2.2.3]{Am_Pro}), we
find that $\psi_1:\mathbb R\rightarrow [-M,M]$ is
continuous.

Rewrite (\ref{In_a}) as an integral equation:
$$
w(t)=\int_0^t \left(\widetilde f(\tau)-q(w(\tau))\right)\,d\tau +\overline f
  t +\alpha\,.
$$ 
Taking  $\alpha > \int_0^T|\widetilde f(\tau)|d\tau+2TM$, we see that
$w(t,\overline  f,\alpha)>0$ for all $t\in[0,T]$ and $\overline  f \in [-M,M]$.  
Consequently $\int_0^T\varphi_{-1}\bigl (w(\tau,\overline  f,\alpha)\bigr )d\tau >0.$
Conversely, if $\alpha<-\int_0^T|\widetilde f(\tau)|d\tau-2TM$ then
$w(t,\overline  f, \alpha)<0$ for all $t\in [0,T]$ and $\overline  f\in [-M,M]$. Hence
$\int_0^T\varphi_{-1}\bigl(w(\tau,\overline  f,\alpha)\bigr)d\tau<0.$
Since $\varphi_{-1}$
is a continuous real function, the mapping $\alpha \mapsto
\int_0^T\varphi_{-1}\bigl(w(t,\overline  f,\alpha)\bigr)dt$ is continuous and
there exists at least one $\alpha_{\overline  f}\in\mathbb R$ such that
$\int_0^T\varphi_{-1}\bigl(w(t,\overline  f,\alpha)\bigr)dt = 0.$
Moreover the partial derivative of $w(t,\overline  f,\alpha)$ 
with respect to initial condition $\alpha$, $w_{\alpha}(t,\overline  f,\alpha)$, 
is the solution of the linear initial-value problem 
\begin{eqnarray}\label{in_da}
&v'(t) + q'\bigl(w(t,\overline  f,\alpha)\bigr)v(t)=0\,, &\\ 
&v(0) = 1\,.&\nonumber
\end{eqnarray}
Taking into account explicit formula of the solution of (\ref{in_da}),
it is easy to verify that $w_{\alpha}(t,\overline  f, \alpha)>0$
for all $t\in [0,T]$ and $\overline  f, \alpha\in\mathbb R$. Thus, for
all $\overline  f\in\mathbb R$ there exists unique $\alpha=\alpha_{\overline  f}$ such that
$\int_0^T\varphi_{-1}\bigl(w(t,\overline  f,\alpha_{\overline f})\bigr)dt=0$. 
By the same reason as above, the mapping $\psi_2:\mathbb
R\rightarrow\mathbb R$ defined by
$\overline  f\mapsto \alpha_{\overline  f}$ is continuous.

Let us consider the continuous
mapping $\psi_1\circ\psi_2:[-M,M]\rightarrow [-M,M]$. 
Due to the Brouwer fixed point theorem there exists
at least one $\overline  f_0\in [-M,M]$ such that $\psi_1\bigl(\psi_2(\overline 
f_0)\bigr)=\overline  f_0$. Hence the function $w(t,\overline 
f_0, \psi_2(\overline  f_0))$ is
a solution of the equation (\ref{a_w})
subject to (\ref{f_con}).

\noindent {\bf Step 2} {\it We show that if {\rm (P)}, {\rm (G)} are satisfied
then, for all $\widetilde  f\in\widetilde  C_T$, there exists at least one $\overline 
f\in\mathbb R$ such that the boundary-value problem for the equation
\begin{equation}
\label{wfg}
w'(t)+g\bigl(\varphi_{-1}\bigl(w(t)\bigr)\bigr)=\widetilde  f(t) +\overline  f
\end{equation}
subject to (\ref{f_con}) has at least one solution.}

Let $w$ be the solution of (\ref{wfg})--(\ref{f_con}).
Integrating (\ref{wfg}) from $0$ to $T$ and then dividing by $T$,
we obtain
$$\frac{1}{T}\int_0^T g\bigl(\varphi_{-1}\penalty0{}(w(t))\bigr)d\,t=\overline 
f\,.$$
With respect to (\ref{ODH1}) (consider $q:=g\circ\varphi_{-1}$ in Lemma \ref{lemma1}) 
the following a priori bound on $w$ results: $\|w\|_{C}
\leq\sqrt{\frac{T}{3}}\|\widetilde  f\|_{L^2}$.
Hence we estimate $\overline  f$ as follows:
\begin{equation}
\label{ODH3}
|\overline  f|\leq\sup\left\{|g(\xi)|:
\varphi_{-1}\left(-\sqrt{\frac{T}{3}}\|\widetilde  f\|_{L^2}\right)\leq\xi\leq
\varphi_{-1}\left(\sqrt{\frac{T}{3}}\|\widetilde  f\|_{L^2}\right)\right\}.
\end{equation}

As $\|w\|_{C}\leq\sqrt{\frac{T}{3}}\|\widetilde f\|_{L^2}$,
the restriction of $g\circ\varphi_{-1}$ on
$I=\left[-\sqrt{\frac{T}{3}}\|\widetilde  f\|_{L^2}\right.,$
$\left.\sqrt{\frac{T}{3}}\|\widetilde  f\|_{L^2}\right]$ is essential
in further considerations. Note that the assumption $\|\widetilde 
f\|_{L^2}\leq\sqrt{\frac{T}{3}}\mbox{min}\{|a|,b\}$ imply $I\subset (a,b)$
$(=\mathop{\rm Dom}\varphi_{-1})$; hence $g\circ \varphi_{-1}$ is well defined
on $I$. 
Let us introduce the following sequence of 
functions $\gamma_n(x):=n\int_{I}\varrho\left (\frac{x-y}{n}\right
)g\bigl(\varphi_{-1}(y)\bigr )dy,$ where $\varrho:\mathbb R\rightarrow\mathbb R$ is the
regularization kernel given by
\begin{equation}
\varrho(x):=\left\{
\begin{array}{ll}
  c_0 e^{\frac{1}{|x|^2-1}} &\mbox{for } |x|<1\,, \\[1pt]
  0 & \mbox{for } |x|\geq 1\,,
\end{array}
\right.
\end{equation}
where $c_0$ is a constant such that $\int_{-1}^1 \varrho(x)dx = 1.$
It is easy to verify that
$\bigl\{\gamma_n\bigr\}_{n=1}^{\infty}$ converges to
$g\circ\varphi_{-1}$ uniformly on $I,$ that $\gamma_n$ is continuously
differentiable for any $n\in\mathbb N$ and that
$\|\gamma_n\|_{C(I)}\leq\|g\circ\varphi_{-1}\|_{C(I)}$ ( $C(I)$ denotes the
space of continuous real functions defined on $I$ ).

From the first step we know that there exist $w_n$ and $\overline  f_n$ such that
$$ w'_{n}(t) + \gamma_n\bigl(w_n(t)\bigr)=\widetilde f(t) + \overline  f_n $$
and
$$\int_0^T\varphi_{-1}\bigl(w_n(t)\bigr)\,dt=0,\quad w_n(0)=w_n(T).$$
Utilizing (\ref{ODH1}) and $\|\gamma\|_{C(I)}\leq
\|g\circ\varphi_{-1}\|_{C(I)}<+\infty$ we find that 
$\bigl\{\overline  f_n\bigr\}_{n=1}^{\infty}$ is a bounded
sequence. By (\ref{ODH1}), $w_n$ are equibounded and, by 
(\ref{ODH2}), $w_n$ are equicontinuous (we apply Lemma \ref{lemma1}
with $q:=\gamma_n$ for each $n\in\mathbb N$, employing the fact that resulting
inequalities do not contain $q$). Hence we can select subsequences
$w_{n_k}, \overline  f_{n_k}$ such that $w_{n_k}\rightarrow w$ in $C_T$
and $\overline 
f_{n_k}\rightarrow \overline  f$. Since 
$$ w_{n_k}(t)=w_{n_k}(0)+\int_0^t\left(\widetilde  f(\tau)+\overline  f_{n_k}-\gamma_{n_k}(w_{n_k}(\tau))\right)d\tau\,,$$
passing to the limit we obtain 
$$w(t)=w(0)+\int_0^t\left(\widetilde  f(\tau) + \overline  f 
-g\bigl(\varphi_{-1}(w(\tau))\bigr) \right)d\tau\,.$$ 
As the integrand in the right-hand-side is continuous, $w\in C^{1}_T$ and
satisfies (\ref{wfg}) in $(0,T)$. Thus Step 2 is over.

\noindent{\bf Step 3} {\it We prove that if $\overline  f_1, \overline  f_2 \in\mathbb R$ and the
equation (\ref{f_eq}) with $f = \widetilde  f+\overline  f_i, i=1,2$ has a
solution satisfying (\ref{f_con}), then $\overline  f_1 = \overline  f_2$}.
Conversely, assume that $\overline  f_1 > \overline  f_2$ and there exists $w_i\in
C^1_T, i=1,2$ such that 
$w'_i(t) +g\left(\varphi_{-1}(w_i(t))\right)=\widetilde  f(t)+\overline 
f_i,$ $i=1,2$ subject to (\ref{f_con}).
Then from (\ref{per_con}) it follows that the function
$\varphi_{-1}\bigl(w_1(t)\bigr)-\varphi_{-1}\bigl(w_2(t)\bigr)$
is a $T$-periodic function with mean value zero, so that there exist
 $t_0$ and $\delta_1>0$ such that
$\varphi_{-1}\bigl(w_1(t_0)\bigr)-\varphi_{-1}\bigl(w_2(t_0)\bigr)=0$ and
$\varphi_{-1}\bigl(w_1(t)\bigr)-\varphi_{-1}\bigl(w_2(t)\bigr)<0$
for all $t_0<t<t_0+\delta_1$ (note that as we consider the periodic problem, we
can shift the problem suitably in $t$ if it is necessary). 
Taking into account that $\varphi$ is an increasing homeomorphism,
we get $w_1(t_0)=w_2(t_0)$ and $w_1(t)-w_2(t)<0$ for all $t_0<t<t_0+\delta_1.$
Since  $w_1(t_0)=w_2(t_0)$, $\overline  f_1>\overline  f_2$ and since 
the functions $g\left(\varphi_{-1}(w_1(\cdot))\right)$ and 
$g\left(\varphi_{-1}(w_2(\cdot))\right)$ are continuous, there exists $\delta_2>0$  
such that $\left|g\left(\varphi_{-1}(w_1(t))\right)
-g\left(\varphi_{-1}(w_2(t))\right)\right|<(\overline f_1-\overline f_2)/2$
for any $t_0<t<t_0+\delta_2$.
Taking $\delta=\min\{\delta_1, \delta_2\}$, we arrive at 
$$\left(w_1-w_2\right)'(t)=\overline f_1-\overline f_2 +
g(\varphi_{-1}(w_1(t)))-g(\varphi_{-1}(w_2(t)))>0$$ 
for all $t_0<t<t_0+\delta$. However, $w_1(t)-w_2(t)<0$ for all $t_0<t<t_0+\delta$,
a contradiction.

\noindent {\bf Step 4} {\it Estimate (\ref{odh_s}).}

The estimate (\ref{odh_s}) is a direct consequence of (\ref{ODH3}).

\noindent {\bf Step 5} {\it Continuity of $s$.}

The proof of the continuity of $s$ is analogous to that one in
\cite{Danc} (page 256, step 5) and so is omitted.

\noindent{\bf Step 6} {\it Completion of the proof of Theorem \ref{Th1}.}

Taking into account the relation between the solution $u$ of the problem
(\ref{fund_eq})--(\ref{per_con}) with $h\equiv 0$ and the solution $w$ of the problem
(\ref{f_eq})--(\ref{f_con}): $u'=\varphi_{-1}(w)$,
the assertion of Theorem \ref{Th1} follows. \hfill$\diamondsuit$

\paragraph{Proof of Theorem \ref{Th2}}

Since $u'=\varphi_{-1}(w)$, using {\rm(P')} we get $u\in C^2_T$.

Assumptions {\rm (P')} and {\rm (G')} imply that
$g\circ\varphi_{-1}:(a,b)\rightarrow \mathbb R$ is
a continuously differentiable real function. Let $q$ be a continuously
differentiable and bounded real function satisfying
$q(y)=g\bigl(\varphi_{-1}(y)\bigr)$ for all
$|y|\leq\sqrt{\frac{T}{3}}\|\widetilde f\|_{L^2}$.


Taking into account a priori bound (\ref{ODH1}) and our
definition of $q$, any solution of (\ref{f_eq})--(\ref{f_con}) satisfies (\ref{a_w})--(\ref{f_con})
and vice versa. Thus we are reduced to prove the uniqueness of the solution
to (\ref{a_w})--(\ref{f_con}).  Due to Theorem \ref{Th1}, for each $\widetilde f\in\widetilde C_T: \|\widetilde f\|_{C_T}<\sqrt{\frac{3}{T}}\min\{|a|,b\}$,
there exists precisely one $s(\widetilde f)\in\mathbb R$ such that (\ref{f_eq})--(\ref{f_con})
possesses solution; $s(\widetilde f)$ is also the unique value corresponding to
$\widetilde f$ such that (\ref{a_w})--(\ref{f_con}) has solution. 
Since $q$ satisfies assumptions of the Step 1 of the proof
of Theorem \ref{Th1}, 
the initial value $w(0)=\alpha$ is uniquely determined by $\alpha
=\phi_2(s(\widetilde f))$. Now, as $q$ is a continuously differentiable
function, the solution of the
initial value problem (\ref{In_a}), with 
$\overline f=s(\widetilde f), \alpha=\phi_2(\overline f)$, is unique (see
\cite{CodLew}) and so is that of (\ref{a_w})--(\ref{f_con}) with $\overline
f = s(\widetilde f)$.

What remains to prove is that $w$ and $s$ are continuously differentiable
with respect to $\widetilde f$. 
Following the idea of the proof of Theorem 3.4 in \cite{CD}
we define $G:C_T\times\mathbb R\times\mathbb R\times\widetilde  C_T\rightarrow
C_T\times\mathbb R\times\mathbb R$ by the formula
\begin{equation}
G(w,\overline  f,\alpha,\widetilde  f) :=
\left (
\begin{array}{c}
w(t)-\overline  f t - \int_0^t\widetilde  f(\tau)d\tau + \int_0^t q\bigl
(w(\tau)\bigr )d\tau - \alpha \\[2pt]
\int_0^T \varphi_{-1}\bigl(w(\tau)\bigr)d\tau \\[2pt]
w(T)-\alpha
\end{array}
\right ).
\end{equation}

As in \cite{CD} one can check that the operator equation
\begin{equation}
\label{op_eq}
G(w,\overline  f,\alpha,\widetilde  f) = 0
\end{equation}
is equivalent to the boundary-value problem (\ref{a_w})--(\ref{f_con}).
As a consequence of this fact we obtain that for all $\widetilde  f_0\in\widetilde  C_T$
there exists a unique triple
$(w_0,\overline  f_0,\alpha_0)\in C_T^1\times\mathbb R\times\mathbb R$ such that $G(w_0,\overline 
f_0,\alpha_0,\widetilde  f_0)=0$.

Now applying the abstract implicit function theorem we conclude the
proof. To do this we shall prove:
{\it The assumptions of the implicit function theorem (see
\cite[Theorem 2.2.3]{Am_Pro}) are satisfied at any point 
$(w_0,\overline  f_0,\alpha_0,\widetilde 
f_0)\in C^1_T\times\mathbb R\times\mathbb R\times\widetilde  C_T$,
 $\widetilde  f_0\not\equiv 0$
(and if {\rm (P'')} also for $\widetilde  f_0\equiv 0),$
at which the equation (\ref{op_eq}) holds.}

One can see that the operator $G$ and the partial Fr\'echet derivatives 
$G_{(w,\overline f,\alpha)}$ and $G_{\widetilde  f}$ 
(the first partial Fr\'echet derivatives of $G$ with respect to 
$(w,\overline f,\alpha)$ and $\overline f$, respectively) are continuous
in the neighbourhood of $(w_0,\overline  f_0,\alpha_0,\widetilde  f_0)$. By a
direct calculation, we get that the partial
Fr\'echet derivative $G_{(w,\overline  f, \alpha)}(w_0,\overline  f_0, \alpha_0,
\widetilde  f_0):C^1_T\times\mathbb R\times\mathbb R\rightarrow\ C^1_T\times\mathbb R\times\mathbb R,$ is
given by the formula:
\begin{equation}
\label{op_der}
(\omega, \sigma, \kappa) \mapsto
\left (
\begin{array}{c}
\omega(t)+\int_0^t q'\bigl(w_0(\tau)\bigr)\omega(\tau)d\tau-\sigma
t-\kappa \\[2pt]
\int_0^T\varphi_{-1}\bigl(w_0(\tau)\bigr)\omega(\tau)d\tau \\[2pt]
\omega(T)-\kappa
\end{array}
\right ).
\end{equation}

Let $(\widetilde \phi, \beta, r)\in \widetilde  C_T\times\mathbb R\times\mathbb R$. Then the operator
equation
$$G_{(w,\overline  f,\alpha)}\left(w_0,\overline  f_0, \alpha_0, \widetilde
f_0\right)(\omega,\sigma,\kappa)= \left(\int_0^t\widetilde \phi(\tau)\,d\tau\,,\,\,\beta\,,\,\, r\right)^{T}$$
is equivalent to the initial-value problem
\begin{eqnarray} \label{df_eq}
&\omega'(t)+q'\bigl(w_0(t)\bigr)\omega(t)=\sigma + \widetilde\phi(t)\,,&\\
&\omega(0) = \kappa\,,&\nonumber
\end{eqnarray}
subjected to the additional conditions
\begin{equation}
\label{df_con}
\int_0^T\varphi_{-1}'\bigl(w_0(\tau)\bigr)\omega(\tau)d\tau
=\beta\,,\quad \omega(T)=\kappa+r.
\end{equation}

Let $A(t):=e^{\int_0^t q'\left(w_0(\tau)\right)d\tau}>0$.
The solution of the linear initial value problem (\ref{df_eq})
has the explicit form:
\begin{equation}
\label{exp_sol}
\omega(t)=\kappa\frac{1}{A(t)} + \frac{1}{A(t)}
\int_0^t\widetilde\phi(s)A(s)ds +
\sigma \frac{1}{A(t)}\int_0^t A(s)ds\,.
\end{equation}
Substituting (\ref{exp_sol}) into (\ref{df_con}) we obtain system of two
linear equations for $\sigma$ and $\kappa$. This system can be uniquely 
solved if corresponding determinant $\mbox{D}$ is different from zero.
 By a straightforward calculation we obtain:
\begin{eqnarray}\label{deter}
\mbox{D}&=& \frac{1}{A(T)}\Big [\int_0^T A(s)ds
\int_0^T\frac{\varphi_{-1}'\left(w_0(t)\right)}{A(t)}dt  \\ 
&&- \int_0^T\frac{\varphi_{-1}'(w_0(t))}{A(t)}\int_0^tA(s)\,ds\,dt\Big]+
\int_0^T\frac{\varphi_{-1}'(w_0(t))}{A(t)}\int_0^tA(s)\,ds\,dt 
\nonumber\\
&=& \frac{1}{A(T)}\int_0^T
A(s)\int_0^s\frac{\varphi'_{-1}(w_0(t))}{A(t)}\,dt\,ds +
\int_0^T\frac{\varphi'_{-1}\bigl(w_0(t)\bigr)}{A(t)}\int_0^t 
A(s)\,ds\,dt\,. \nonumber
\end{eqnarray}
Since $\varphi_{-1}:I_2\rightarrow I_1$ is an increasing homeomorphism, it follows that
$\varphi_{-1}'(z)>0$ a.e. in $I_2$. By a contradiction, one can show that
if $\widetilde  f\not\equiv 0$ then $w_0$ satisfying
(\ref{f_eq})--(\ref{f_con}) 
is not a constant function. Thus
$\varphi_{-1}'\left(w_0(t)\right)>0$ on some subset of $[0,T]$ of
positive measure,  which taking into account (\ref{deter}) implies
$\mbox{D}>0$. 

Now let us consider $\widetilde f\equiv 0$.
We have to impose $\varphi'(0)>0$ to conclude $D>0$.
The reason consists in the fact that 
$w_0\equiv 0$ is the unique solution of (\ref{f_eq})--(\ref{f_con}) with
$\widetilde f \equiv 0$.


We proved that  $G_{(w,\overline  f,\alpha)}(w_0,\overline
f_0,\alpha_0):C^1_T\times\mathbb R\times\mathbb R\rightarrow C^1_T\times\mathbb R\times\mathbb R$
is bijective linear mapping. Then the inverse of
$G_{(w,\overline  f,\alpha)}(w_0,\overline f_0,\alpha_0)$ is continuous 
due to the Banach open mapping theorem (see \cite{Zeid}).
Thus $G_{(w,\overline  f,\alpha)}(w_0,\overline f_0,\alpha_0)$ is an
isomorphism of $C^1_T\times\mathbb R\times\mathbb R$ onto itself
and the assumptions of the implicit function theorem \cite[Theorem 2.2.3]{Am_Pro} are satisfied.
This completes the proof of Theorem \ref{Th2}.
\hfill$\diamondsuit$


\paragraph{Proof of Theorem \ref{Th3}}
At first we perform the proof under the assumptions {\rm (P)}, {\rm (P')} and
{\rm (G')}. Considering (\ref{ODH1}) we can define
continuously differentiable and bounded function $q$ satisfying
$q(y)=g\bigl(\varphi_{-1}(y)\bigr)$ for all
$|y|\leq\sqrt{\frac{T}{3}}\|\widetilde  f\|_{L^2}.$

Since the related first order problem (\ref{f_eq})--(\ref{N_con})
is the same as that one considered in \cite{CD}, the existence of the solution and
the differentiability of $w(\widetilde  f)$ and $s(\widetilde  f)$ follows from
\cite[Theorem 3.4]{CD}.

The assumption 
{\rm(P')} can be omitted and the assumption {\rm(G')} can be replaced by {\rm(G)}
using the same argument as in the Step 2 of the proof of Theorem \ref{Th1}.
From \cite[Theorem 3]{Maw} we obtain the uniqueness of $\overline  f$ corresponding to a fixed $\widetilde  f$.
The continuity of $s(\widetilde  f)$ can be proved in the same manner as in  
Step 5 of the proof of Theorem \ref{Th1}. 
A priori bound (\ref{odh_s}) is a consequence of (\ref{ODH1}) (cf. Step 4
in the proof of Theorem \ref{Th1}).
\hfill$\diamondsuit$

\section{Proof of Theorem \ref{Th4}}

To prove Theorem \ref{Th4}, we need the following comparison results:

\begin{lemma}\label{lem2}
Let $v'(t)+g\biggl(\varphi_{-1}\bigl(v(t)\bigr)\biggr)\leq\alpha+
\widetilde f(t)$ on $[0,T]$ and $v$ satisfies (\ref{f_con}). Then $\alpha\geq
s(\widetilde f)$ (where $s(\widetilde f)$ comes from Theorem \ref{Th1}).
\end{lemma}

\paragraph{Proof}
Assume conversely that $\alpha<s(\widetilde  f)$.
Let $w$ satisfies (\ref{f_eq})
and (\ref{f_con}).
Since 
$$\varphi_{-1}(v(0))-\varphi_{-1}(w(0)) = \varphi_{-1}(v(T))-\varphi_{-1}(w(T))$$ and 
$$\int_0^T \bigl(\varphi_{-1} \bigl(v(t)\bigr)-
\varphi_{-1}\bigl(w(t)\bigr)\bigr) dt=0\,,$$ there exists a $t_0\in [0,T)$
and $\delta>0$ such that $\varphi_{-1}\bigl(v(t_0)\bigr)-
\varphi_{-1}\bigl(w(t_0)\bigr)=0$ and $\varphi_{-1}\bigl(v(t)\bigr)-
\varphi_{-1}\bigl(w(t)\bigr) > 0$ for all $t_0<t<t_0+\delta$ (note that we consider
periodic problem so if necessary we can shift suitably the problem in $t$).
Taking into account that $\varphi$ is an increasing homeomorphism, 
we get $v(t_0)=w(t_0)$ and $v(t)>w(t)$ for all $t_0<t<t_0+\delta$.
On the other hand, there exists $\delta'>0$ such that
$$( v'- w')(t)\leq\alpha-s(\widetilde  f)-g\biggl(\varphi_{-1}\bigl(v(t)\bigr)\biggr)
+g\biggl(\varphi_{-1}\bigl(w(t)\bigr)\biggr)<0$$
for all  $t: |t-t_0|<\delta'$. 
Since $v(t_0)=w(t_0)$, we obtain $v(t)-w(t)<0$
for all $t_0<t<t_0+\delta'$, which is a contradiction.
\hfill$\diamondsuit$\medskip

\begin{lemma}\label{lem3}
Let $ v'(t)+g\biggl(\varphi_{-1}\bigl(v(t)\bigr)\biggr)\leq\alpha+
\widetilde f(t)$ on $[0,T]$ and $v$ satisfies (\ref{N_con}), then $\alpha\geq
s(\widetilde f)$ (here $s(\widetilde f)$ is taken from Theorem \ref{Th3}).
\end{lemma}

\paragraph{Proof} 
Suppose the contrary, i.e.  $\alpha<s(\widetilde  f)$.
Let $w$ satisfies (\ref{f_eq})--(\ref{N_con}), then 
$$ v'(0)\leq-g\biggl(\varphi_{-1}\bigl(v(0)\bigr)\biggr)+\widetilde  f(0)+\alpha<
-g\biggl(\varphi_{-1}\bigl(w(0)\bigr)\biggr)+\widetilde  f(0)+s(\widetilde  f)=
w'(0).
$$
From here and (\ref{N_con}) we get that there exists $\varepsilon>0$,
such that $v(t)<w(t)$ for all $t\in (0,\varepsilon)$.

If $v(t)<w(t)$ for all $t\in (0,T]$ then $v(T)<w(T)$ and either $v$ or $w$
can not satisfy (\ref{N_con}) at $T$. Thus $\varepsilon \leq T$ and there exists
$t_0\in (0,T]$ such that
$$v(t)<w(t) \mbox{ for all } t\in (0,t_0) \mbox{ and } v(t_0)=w(t_0)\,.$$
For $t\in (0,t_0)$ we have
$$\frac{v(t)-v(t_0)}{t-t_0}>\frac{w(t)-w(t_0)}{t-t_0}\,,$$
i.e. $ v'(t_0)\geq w'(t_0)$. On the other hand, we have
\begin{eqnarray*}
 w'(t_0)&=&-g\bigl(\varphi_{-1}\bigl(w(t_0)\bigr)\bigr)+\widetilde  
f(t_0)+s(\widetilde  f) \\
&=&-g\bigl(\varphi_{-1}\bigl(v(t_0)\bigr)\bigr)+\widetilde  f(t_0)
+s(\widetilde  f)\\
&>&-g\bigl(\varphi_{-1}\bigl(v(t_0)\bigr)\bigr)+\widetilde  f(t_0)
+\alpha\geq v'(t_0)\,,
\end{eqnarray*}
which is a contradiction.\hfill$\diamondsuit$

\begin{remark}
{\rm
It is possible to show that the assertions of  Lemmas \ref{lem2} and \ref{lem3} hold true also with inverted inequalities. This is in agreement
with the semilinear periodic problem studied in \cite{Danc}.
These inequalities are used to prove the `dual' version of Theorem \ref{Th4} with $h(+\infty)<h(-\infty)$.}
\end{remark}


\begin{lemma}\label{lem4}
Let $\varphi$ be an increasing homeomorphism of $\mathbb R$ onto itself and there exist
$c,C>0$ and $p>1$: $c|z|^{p-1}\leq|\varphi(z)|
\leq C(|z|^{p-1}+1)$ for all $z\in\mathbb R$. Then, for any $y\in C_T$ 
( $y\in C[0,T]$ ), there exists exactly one $u\in C^1[0,T]$ with
$\varphi(u')\in C^1[0,T]$ and satisfying
\begin{equation}
\label{p_eq}
\bigl(\varphi(u'(t))\bigr)'-\varphi(u(t))=y(t)\quad t\in (0,T)
\end{equation}
subject to (\ref{per_con}) ( (\ref{Neu_con}) ).
\end{lemma}

\paragraph{Proof}
For the sake of brevity we present the proof only for the
periodic conditions. In the case of the Neumann conditions, 
the proof is analogous.

At first we prove that, for any given $y\in C_T$, there exists precisely
one weak solution of (\ref{p_eq})--(\ref{per_con}), where the weak solution
of (\ref{p_eq})--(\ref{per_con}) is any function $u\in W^{1,p}_T$ such that
the following identity
\begin{equation} \label{w_eq}
\int_0^T\left\{\varphi(u')v'+\varphi(u)v\right\}=-\int_0^Tyv
\end{equation}
is satisfied for each $v\in W^{1,p}_T$. Then, using an regularity argument,
we show that this function $u$ is smooth enough and satisfies
(\ref{p_eq})--(\ref{per_con}) in the sense indicated in the assertion of
the lemma.
 
{\it Existence and uniqueness of the weak solution. } 
Let us define $\psi:W_T^{1,p}\rightarrow L^{p'}\quad$ (where $p':=p/(p-1)$)
by $u\mapsto \psi(u)$ if and only if 
$\langle\psi(u),v\rangle=\int_0^T\varphi(u')v'\,,$ $u,v\in W^{1,p}_T$ for
all $v\in W^{1,p}_T$. Since $|\varphi(z)|\leq C(|z|^{p-1}+1)$ for all
$z\in\mathbb R$, it is easy to verify that $\psi$ is a continuous operator; 
this follows from the Nemitskii theorem (see \cite[Theorem 1.2.2]{Am_Pro}).
From the fact that $\varphi$ is an increasing
function we get strict monotonicity of $\psi$. Since any
monotone continuous operator is also hemicontinuous (see \cite{Zeid}), 
we get that $\psi$ is a hemicontinuous one.
The assumption
$|\varphi(z)|\geq c|z|^{p-1}$ for all $z\in\mathbb R$ implies that $\psi$ is
weakly coercive.
Now the existence and uniqueness of the weak solution of (\ref{p_eq})--(\ref{per_con}) follows from
\cite[Theorem 32.H]{Zeid}.

{\it Regularity.} Suppose that $u$ is a weak solution of
(\ref{p_eq})--(\ref{per_con}). We show that $u\in C^1_T$,
$\varphi(u')\in C^1[0,T]$ and that (\ref{p_eq}) holds pointwise in
$(0,T)$. Integrating by parts we can rewrite 
the equation (\ref{w_eq}) into the form:
\begin{eqnarray}\label{i_eq}
\int_0^T \left(\varphi(u'(t))-\int_0^t\bigl[\varphi
\bigl(u(\tau)\bigr)+y(\tau)\bigr]d\tau\right) v'(t)dt\, + && \\
\left[\int_0^{t}\bigl(\varphi\bigl(u(\tau)\bigr)+y(\tau)\bigr)d\tau\, v(t)
\right]^T_0
&=& 0\,. \nonumber
\end{eqnarray}

Let us define a function $M:[0,T]\rightarrow \mathbb R$,
$$t\mapsto\varphi(u')-\int_0^t
\bigl[\varphi\bigl(u(\tau)\bigr)+y(\tau)\bigr]d\tau.$$
It is easy to see that $M\in L^{p'}$ and from (\ref{i_eq}) we get
$$\int_0^TM(t)v'(t)dt=0\mbox{ for all } v\in C^{\infty}_0(0,T)\,.$$
Hence
$$\int^T_0\frac{\delta M}{\delta t}v=0\mbox{ for all } v\in C^{\infty}_0(0,T),$$
where $\frac{\delta M}{\delta t}$ denotes the distributional derivative of $M$.
Since $M\in L^{p'}\hookrightarrow L^1$ and $\frac{\delta M}{\delta t}=0$, we obtain
that $M(t)=k$ a.e. in $[0,T]$ and $k\in \mathbb R$. Thus
\begin{equation}
\label{veinte}
\varphi(u'(t))-\int_0^t\bigl[\varphi
\bigl(u(\tau)\bigr)+y(\tau)\bigr]d\tau - k = 0\quad \mbox{a.e. in}\quad [0,T]\,.
\end{equation}
Since $\varphi$ is an increasing homeomorphism of $\mathbb R$ onto itself,
we can rewrite the previous equation into the following form
\begin{equation}\label{phi_eq}
u'(t)-\varphi_{-1}\left(\int_0^t\bigl[\varphi
\bigl(u(\tau)\bigr)+y(\tau)\bigr]d\tau - k\right) = 0\,.
\end{equation}
Now let us define a function $F:\mathbb R\times [0,T]\rightarrow\mathbb R,$
$$(z,t)\mapsto z-\varphi_{-1}\left(\int_0^t\bigl[\varphi
\bigl(u(\tau)\bigr)+y(\tau)\bigr]d\tau - k\right)\,.$$
It is possible to show that $F$ is continuous on $\mathbb R\times [0,T]$.
Moreover, $F(\cdot, t_0)$ is an increasing function for all $t_0\in [0,T]$, 
and 
$$\lim_{z\rightarrow -\infty}F(z,t_0)=-\infty, \quad
\lim_{z\rightarrow +\infty}F(z,t_0)=+\infty\,.$$  
Hence for each $t\in [0,T]$ there
exists exactly one $z(t)\in\mathbb R$, such that
$$F(z(t),t)=0\,.$$
Since $\frac{\partial F}{\partial z}$ is continuous and $\frac{\partial
F}{\partial z} = 1 \not = 0$, we can apply the implicit function theorem
to show that $z(\cdot)\in C[0,T]$.
From (\ref{phi_eq}) we get
$$F(u'(t),t)=0\quad\mbox{a.e. in}\quad [0,T]\,.$$
Thus we arrive at
$$z(t)=u'(t)\quad\mbox{a.e. in}\quad [0,T]\,.$$
Since $u\in W_T^{1,p}$ is absolutely continuous, this identity
holds true for all $t\in [0,T]$ and thus $u\in C^1[0,T]\cap W^{1,p}_T.$

Now it remains to prove that $\varphi(u')\in C^1[0,T].$
Let us define $G:\mathbb R\times [0,T]\rightarrow\mathbb R$,
$$(z,t)\mapsto z -\int_0^t\biggl(\varphi(u(\tau))
- y(\tau)\biggr)d\tau - k\,.$$
The function $G$ is continuous on $\mathbb R\times [0,T]$. Moreover,  for all
$t_0\in [0,T]$, $G(\cdot, t_0)$ is an increasing function and
$\lim_{z\rightarrow \pm\infty}G(z,t_0)=\pm\infty$.
Hence for each $t\in[0,T]$ there exists exactly one
$z(t)$ such that
$$G\bigl(z(t),t\bigr)=0\,.$$
Since $u\in C^1[0,T]$ and $y\in C[0,T]$,
the partial derivatives $\frac{\partial G}{\partial z}$ and
$\frac{\partial G}{\partial t}$ are continuous; moreover
$\frac{\partial G}{\partial z}=1\not = 0$.
Then the implicit function theorem yields
$z(t)\in C^1[0,T]$. Taking into account (\ref{veinte}) and the fact that $u\in C^1[0,T]$,
we see that $z(t)=\varphi(u')$ for all $t\in [0,T]$; thus $\varphi(u')\in C^1[0,T]$.
Now, since $u\in C^1[0,T]\cap W^{1,p}_T$ and $\varphi(u')\in C^1[0,T]$,
integrating (\ref{w_eq}) by parts we show that $u$ satisfies
(\ref{p_eq})
in $(0,T)$ and that
$u'(0)=u'(T)$. This concludes the proof.
\hfill$\diamondsuit$\medskip

Now we can define a solution operator $K: C_T\rightarrow C^1_T$,
\begin{equation}
\label{define_K}
 y \mapsto K(y)\,,
\end{equation}
where $K(y)$ is the solution of (\ref{p_eq})--(\ref{per_con}).
Analogously we can define a solution  operator $K': C[0,T]\rightarrow C^1[0,T]$
corresponding to the Neumann problem (\ref{p_eq})--(\ref{Neu_con}).

\begin{lemma}\label{lem5}
Let $K$ and $K'$ be defined as above. Then $K$ is compact as a mapping between 
$C_T$ and $C^1_T$ and $K'$ is compact as a mapping between $C[0,T]$ and $C^1[0,T]$.
\end{lemma}

\paragraph{Proof}
We prove the compactness of $K$. The proof of the compactness of $K'$ is analogous.

Let us consider the following sequence of equations:
\begin{equation}\label{s_eq}
\left(\varphi(u'_n(t))\right)'-\varphi(u_n(t)) = y_n(t)\,,
\end{equation}
subject to (\ref{per_con}), where
$\{y_n\}_{n=1}^{\infty}\subset C[0,T]$ is bounded. We are going to show
that one can select a convergent subsequence from
$\{u_n\}_{n=1}^{\infty}\subset C^1_T$.
Multiplying the equation (\ref{s_eq}) by $u_n$, integrating from $0$ to
$T$, integrating the first term in the left-hand-side by parts and using 
the periodic conditions (\ref{per_con}), we obtain
\begin{equation}\label{s_int}
-\int_0^T\bigl\{\varphi(u'_n(t))u'_n(t)+
\varphi(u_n(t))u_n(t)\bigr\}dt=\int_0^T y_n(t)u_n(t)dt\,.
\end{equation}
Since we assume that $|\varphi(z)|\geq c|z|^{p-1}$ for all $z\in\mathbb R$,
we find that 
$$
\|u_n\|^p_{W^{1,p}_T}\leq
\frac{1}{c}\int_0^T\bigl\{\varphi(u'_n(t))
u'_n(t)+ \varphi(u_n(t))u_n(t)\bigr\}dt\,.
$$
Using this we estimate the terms on the left-hand side of (\ref{s_int}). The
right hand-side of (\ref{s_int}) is estimated by the H\"older inequality.
Therefore we find that
$$\|u_n\|^p_{W^{1,p}_T}\leq\frac{1}{c}\|y_n\|_{L^{p'}}\|u_n\|_{L^p}\leq
\frac{1}{c}\|y_n\|_{L^{p'}}\|u_n\|_{W^{1,p}_T}\,,$$
which implies
\begin{equation}\label{w_est}
\|u_n\|_{W^{1,p}_T}\leq\frac{1}{c}\left(\|y_n\|_{L^{p'}}\right)^{\frac{1}{p-1}}\,.
\end{equation}
Thus $\{u_n\}^{\infty}_{n=1}$ is bounded in $W^{1,p}_T$.
Since $W^{1,p}_T$ is compactly imbedded in $C_T$, we can select $\{u_{n_k}\}_{k=1}^{\infty}$ such that $u_{n_k}\rightarrow w$ in $C_T$.
Let $h_k(t)=y_{n_k}(t)-\varphi\bigl(u_{n_k}(t)\bigr)$.
One can see that there exists $\beta>0$ such that $\|h_k\|_{C}\leq\beta$.
Hence $\|\left(\varphi\left(u'_{n_k}\right)\right)'\|_{C}\leq\beta$.
It follows from (\ref{per_con})  that there exists $t^k_0\in [0,T]$,
such that $u'_{n_k}(t^k_0)=0$. From (\ref{s_eq}) we obtain $\varphi(u'_{n_k}(t))=\int_{t_0^k}^t
h_k(\tau)d\tau$ and consequently $\|\varphi(u_{n_k}'(t))\|_{C}\leq T\beta$.
Thus $\varphi(u'_{n_k}(t))$ is bounded in $C^1_T$ norm.
Due to the compact imbedding of $C^1_T$ into
$C_T$ we can select $\varphi(u_{n_{k_j}}')\rightarrow v$
in $C_T$. Since $u_{n_{k_j}}'=\varphi_{-1}\bigl(\varphi(u'_{n_{k_j}})\bigr)$
and $\varphi_{-1}$ is continuous,
$u'_{n_{k_j}}\rightarrow \varphi_{-1}(v)$ in $C_T.$
On the other hand, $u_{n_{k_j}}$ can be written in the form 
$$u_{n_{k_j}}(t)=u_{n_{k_j}}(0)+\int_0^t\varphi_{-1}\left(\varphi(u'_{n_{k_j}}(\tau))\right)d\tau\,. $$
Since $u_{n_{k_j}}\rightarrow w$ as $n_{k_j}\rightarrow$ in $C_T$, we find that 
$w(t)=w(0)+\int_0^t\varphi_{-1}\left(v(\tau)\right)d\tau$. 
As the integrand is continuous, differentiating 
the former equation we get $w'=\varphi_{-1}(v)$. Thus
$u_{n_{k_j}} \rightarrow w$ in $C^1_T$.
This ends the proof.
\hfill$\diamondsuit$

\begin{remark}
{\rm
The proof of the previous lemma is based on the ideas from \cite{DelP}.}
\end{remark}


Let $g,h,\widetilde f, T$ be as in Theorem \ref{Th4} and let 
$z_0:= 2\sup_{\xi\in\mathbb R}|g(\xi)| + |h(+\infty)| +
|h(-\infty)|+2\sup_{\xi\in\mathbb R}|h(\xi)|+\|\widetilde f\|_{C}$. 
We define a function $l:\mathbb R\rightarrow\mathbb R$ by 
\begin{equation} \label{def_l}
l(z):=\left\{
\begin{array}{ll}
z & \mbox{for}\ 0\leq z\leq z_0\,, \\
z_0 & \mbox{for}\ z>z_0\,,
\end{array}\right.
\end{equation}
for $z\geq 0$ and $l(z)=-l(-z)$ for $z<0$.
We also define:  
\begin{equation}\label{dfm}
\Gamma(p,c, F) := \Big|\sqrt{\frac{T}{3}}\frac{F}{c}\Big|^{p-1}
\end{equation}
for any $p>1$, $c>0$, $F\geq 0$.

\begin{lemma}
\label{lem6}
Let $\varphi$ be an increasing homeomorphism of $\mathbb R$ onto $\mathbb R$
and there exist $c,C>0$ and $p>1$: $c|z|^{p-1}\leq |\varphi(z)|\leq C\left(|z|^{p-1}+1\right)$. Let
{\rm (G)} be satisfied and $g$ be bounded. Then, for any
$\lambda\in [0,1]$ and $|\overline f|\leq\sup_{\xi\in\mathbb
  R}|h(\xi)|+\sup_{\xi\in\mathbb R}|g(\xi)|$, all solutions of
\begin{equation}
\label{lam_eq}
  \bigl(\varphi(u')\bigr)'+\lambda g(u')+l(u)=\lambda (\widetilde
  f+\overline f)
\end{equation}
subject to (\ref{per_con}) ( (\ref{Neu_con}) )
are a priori bounded by \newline
$\|u\|_{C^1}\leq (1+2T)\,\Gamma\left(p,c, \|\widetilde f \|_{L^2}+\sqrt{T}z_0\right)+z_0$.
\end{lemma}

\paragraph{Proof}
We rewrite the equation (\ref{lam_eq}) as
$$\bigl(\varphi(u')\bigr)'+ \lambda g(u')=\lambda (\widetilde f+\overline f)-l(u)\,.$$
Then it follows from (\ref{ODH1}) 
(consider $q:=\lambda g$ in Lemma \ref{lemma1}) that any  solution $u$  of
(\ref{lam_eq})--(\ref{per_con}) satisfy
$$
\|\varphi(u')\|_{C}\leq\sqrt{\frac{T}{3}}\left\|\lambda \widetilde f-l(u)\right\|_{L^2}\,.  
$$
Taking into account the assumption $|\varphi(z)|\geq c|z|^{p-1}$,  
the following inequality
$$c |u'|^{p-1}\leq |\varphi(u')|\leq\sqrt{\frac{T}{3}}\left\|\lambda\widetilde f-l(u)\right\|_{L^2}$$
is satisfied for any $t\in [0,T]$. This implies
$$
\|u'\|_{C}\leq\left(\frac{1}{c}\sqrt{\frac{T}{3}}\left\|\lambda \widetilde f-l(u)\right\|_{L^2}\right)^{\frac{1}{p-1}}\,.
$$
Since  $\|\lambda\widetilde f-l(u)\|_{L^2}\leq\|\widetilde
f\|_{L^2}+\sqrt{T}z_0$ (recall that $\|\lambda \widetilde
f\|_{L^2}=|\lambda|\|\widetilde f\|_{L^2}$, where
$\lambda\in [0,1]$ and $\|l(u)\|_{L^2}\leq\sqrt{T}\sup_{\xi\in\mathbb
  R}|l(\xi)|=\sqrt{T}z_0$), 
we get 
\begin{equation}
\label{iuy}
\|u'\|_{C}\leq\Gamma\left(p,c,\|\widetilde f\|_{L^2}+\sqrt{T}z_0\right)\,.
\end{equation}

Let us split the function $u$ as $u=\widetilde u+\overline u$, where $\overline u =
\frac{1}{T}\int_0^Tu(t)dt$. 
As $\left\|\widetilde u\right\|_{C}\leq\int_0^T\left|u'\right|\leq T\left\|
u'\right\|_{C}$, from (\ref{iuy}) we have 
\begin{equation}
\label{iduy}
\|\widetilde u\|_{C}\leq T\,\Gamma\left(p,c,\|\widetilde f\|_{L^2}+\sqrt{T}z_0\right)\,.
\end{equation}
Now rewrite (\ref{lam_eq}) as
$$\bigl(\varphi(u')\bigr)'=\lambda \left(\widetilde f+\overline f - g(u')\right)-l(u)$$
and suppose that $\overline u>z_0+ T\,\Gamma\left(p,c,\|\widetilde f\|_{L^2}+\sqrt{T}z_0\right)$. 
Then (\ref{iduy}) implies $u(t)>z_0$ for all  $t\in [0,T]$ and, by the
definition of $l$, we have that $l(u)=z_0$ for all $t\in [0,T]$. Hence we
can rewrite (\ref{lam_eq}) as
$$
\left(\varphi(u')\right)'=\lambda\left(\widetilde f+\overline
  f-g(u')\right)-z_0\,.
$$
Integrating from $0$ to $T$, using $g(u')<\sup_{\xi\in\mathbb R}|g(\xi)|$
and $\widetilde f<\|\widetilde f\|_{C}$ we get that
$$\varphi(u'(T))<\left(\lambda(\|\widetilde f\|_{C}+\overline f+\sup_{\xi\in\mathbb
  R}|g(\xi)|)-z_0\right)T +
  \varphi(u'(0))\,.$$
Since $\lambda\in [0,1]$ and $z_0>\sup_{\xi\in\mathbb
  R}|g(\xi)|+\|\widetilde f\|_{C}+\overline f$, we find that $\varphi(u'(T))<\varphi(u'(0))$,
which contradicts the periodic conditions. 
Analogously as above, the possibility 
$\overline u<-z_0-T\,\Gamma\left(p,c,\|\widetilde
  f\|_{L^2}+\sqrt{T}z_0\right)$ leads to a contradiction.
Hence $|\overline  u|<z_0+T\,\Gamma\left(p,c,\|\widetilde f\|_{L^2}+\sqrt{T}z_0\right)$. 
Now, taking into account 
$\|u\|_{C^1}\leq\|\widetilde u\|_{C}+|\overline u|+\|u'\|_{C}$, 
the assertion follows. \hfill$\diamondsuit$

\begin{lemma}\label{lem7}
Assume that $\varphi$ is an odd increasing homeomorphism of $\mathbb R$ onto itself,
there exist $c,C>0$ and $p>1$: $c|z|^{p-1}\leq|\varphi(z)|\leq C(|z|^{p-1}+1)$ for all
$z\in\mathbb R$, $g$ satisfies   {\rm(G)} and is bounded.
Let us define $V : C^1_T\rightarrow C^1_T$ (~$C^1[0,T]\rightarrow
C^1[0,T]$~) by 
$$u\mapsto u-K\bigl(-\varphi(u)-g(u')-l(u)+f\bigl)\,, $$
where the operator $K$ is defined by (\ref{define_K}) for the periodic (
the Neumann )
BVP.
Then there exists $R_0>0$ such that, if $V(u)=0$ then $\|u\|_{C^1}\leq R_0$.
Moreover the Leray--Schauder degree 
$\deg (V, B(0,R), 0)$ is well
defined and non-zero if $B(0,R)=\{u\in C^1_T \ \mbox{( }C^1[0,T]\mbox{ )}$: 
$\|u\|_{C^1}<R\}$ with any $R > R_0$.
\end{lemma}

\paragraph{Proof}
Let us define $U :C^1_T\times [0,1]\rightarrow C^1_T$ by 
$$(u,\lambda) \rightarrow u-K\bigl(-\varphi(u)-\lambda g(u')-l(u)+\lambda
f\bigl)\,.$$

Since, by Lemma \ref{lem5}, $K: C_T\rightarrow C^1_T$  
is compact and
$$-\varphi(u)-\lambda g(u')-l(u)+\lambda f$$
is an continuous operator from $C^1_T$ to $C_T\,$,
$U(\cdot,\lambda)$ is a compact perturbation of the identity of $C^1_T$
onto itself for all $\lambda\in [0,1]$.
Furthermore, it follows that $V=U(\cdot,1)$ from the definition of $U$.

It is easy to see that the operator equation $U(\cdot,\lambda)=0$ is
equivalent to the equation (\ref{lam_eq}):
$$
\bigl(\varphi(u')\bigr)'+\lambda g(u')+l(u)=\lambda f
$$
subject to (\ref{per_con}).
Due to Lemma \ref{lem6}, for any $\lambda\in [0,1]$, all solutions of 
(\ref{lam_eq})--(\ref{per_con}) satisfy 
$\|u\|_{C^1}\leq (1+2T)\Gamma\left(p,c, \|\widetilde f\|_{L^2}+\sqrt{T}z_0\right)+z_0=:R_0$.

Hence the degree is well defined for every ball $B(0,R)$ with radius $R>R_0$.
Since  $\varphi$ and $l$ are odd functions, $U(\cdot,0)$ is odd mapping;
consequently $$\deg (M(\cdot,0),B(0,R),0)\not=0\,.$$
Finally, the homotopy invariance property of the degree implies
that
$$\deg (U(\cdot,\lambda),B(0,R),0)\not=0\mbox{ for all }\lambda\in[0,1].$$
\hfill$\diamondsuit$\medskip

Now, we prove Theorem \ref{Th4}.
The method follows the idea of the proof of Theorem 2 in \cite{Danc}.
For the sake of brevity we will
present a proof for the periodic conditions. The argument for the
Neumann conditions is analogous.


\paragraph{Proof of Theorem \ref{Th4}} 
The proof is divided into three steps:

\noindent{\bf Step 1} {\it First, we will prove the theorem under additional
assumptions:
$\mathop{\rm Dom}\varphi = \mathbb R$, there exists $c,C>0$ and $p>1$ such that
$c|z|^{p-1}\leq |\varphi(z)|\leq C(|z|^{p-1}+1)$ for all $z\in\mathbb R$ and $g$ is bounded.}

We use the degree argument. 
Define the operator $N:C^1_T\times\mathbb R\rightarrow C^1_T$ by
$$(u,\lambda)\mapsto u-K\left(-\varphi(u)- g(u')-\lambda
h(u)-(1-\lambda)l(u) + \widetilde  f+\overline  f\right),$$
where $K:C_T\rightarrow C^1_T$ is
introduced after Lemma \ref{lem4} and $l:\mathbb R\rightarrow\mathbb R$
is defined by (\ref{def_l}).
Now we can rewrite the boundary-value problem (\ref{fund_eq})--(\ref{per_con})
into an equivalent operator equation $N(u,1)=0$.

From the definition of $N$ it follows that $N(\cdot,\lambda)$ is a compact
perturbation of the identity of $C^1_T$ onto itself for all $\lambda\in [0,1]$. We start our
homotopy argument with $\lambda=0$. Then $N(\cdot,0)=V$,
where $V$ is defined in Lemma \ref{lem7}.
Hence, by Lemma \ref{lem7}, there exists $R_0>0$ such that if $u$ 
is a solution of $N(u,0)=0$ then $\|u\|_{C^1}\leq R_0$; moreover,
$\deg (N(\cdot,0), B(0,R), 0)\not = 0$ provided that $R>R_0$.
We show later that there exists $R_0'> R_0>0$
such that, for all $\lambda\in [0,1]$,
every solution $u$ of the operator equation $N(u,\lambda)=0$ satisfies 
$\|u\|_{C^1}<R'_0$.  
Then $\deg (N(\cdot,\lambda), B(0,R'_0),0)$
is well defined for all $\lambda\in [0,1]$ and
due to the homotopy invariance property
$\deg (N(\cdot,\lambda), B(0,R'_0),0)=
\deg (N(\cdot,0), B(0,R'_0),0)\not = 0$ for all $\lambda\in [0,1]$.

Since $\|u\|_{C^1}=\|u\|_{C} + \|u'\|_{C}$,
we shall prove that both $\|u\|_{C}$ and $\|u'\|_{C}$
are estimated by some constants independent of $\lambda$.
The equation $N(u,\lambda)=0$ is equivalent to
$$ \left(\varphi(u')\right)' + g(u') = f - \lambda h(u) - (1-\lambda)l(u)$$
subject to (\ref{per_con}). 

An argument similar to that in the proof of Lemma \ref{lem6} shows that
we can find $A>0$, independent of $\lambda$, such that $\|u'\|_{C}\leq A$.
Therefore $\|\widetilde  u\|_{C}\leq AT$
and it remains to prove that there exists a constant $j>0$, independent of $\lambda$
such that $|\overline  u|\leq j$.
Let $\gamma := \frac{1}{2}\left(h(+\infty)+s(\widetilde  f)-\overline  f\right)$;
by (\ref{D_like}) $\gamma>0$. Hence we can find $m>0$ (independent of $\lambda$) such that
\begin{equation}\label{p_int}
(1-\lambda)h(y)+\lambda l(y)>h(+\infty)-\gamma
\end{equation}
for all $y\geq m$ and $0\leq \lambda\leq 1$.
Assuming that $\overline  u>m+AT$, we find
$$\left(\varphi(u')\right)' + g(u') =
\overline  f + \widetilde  f - (1-\lambda) h(u)- \lambda l(u)
<\overline  f + \widetilde  f - h(+\infty) + \gamma = s(\widetilde  f) - \gamma + \widetilde  f.$$
Taking $\alpha=s(\widetilde  f)-\gamma$ we see from Lemma \ref{lem2}
that $s(\widetilde  f) - \gamma\geq s(\widetilde  f)$,
a contrary to $\gamma>0$.
Thus $\overline  u\leq m+AT$.
Similarly, we exclude the possibility $\overline  u\geq -m-AT$.
Hence $\|u\|_{C}\leq AT+m =: j$ and
$\|u\|_{C^1}\leq A(1+T)+m$,
where $A$ and $m$ do not depend on $\lambda$.
So that the desired radius is any $R_0'> \max\{A(1+T)+m, R_0\}$.


\noindent{\bf Step 2} {\it Now we remove the additional assumptions 
on $\varphi$ and $g$.}

Let us take fixed $\widetilde f\in C_T$ and define
$$
M:=\varphi_{-1}\Big(\sqrt{\frac{T}{3}}\big(\|\widetilde  f\|_{L^2}+
\sqrt{T}\sup_{\xi\in\mathbb R}|h(\xi)|\big)\Big)>0\,.
$$
Note that (\ref{M_con}) imply that $0<\varphi(M)<b\,\,\,(=-a)$. 
As a consequence $[-M,M]\subset \mathop{\rm Dom}\varphi\,\,\,(= I_1)$.
Define $\psi:\mathbb R\rightarrow\mathbb R$ by
$$\psi(y) := \left\{
\begin{array}{ll}
\varphi(y) & \mbox{if} \ |y|\leq M\,, \\[2pt]
\varphi\bigl(M\mathop{\rm sgn}(y)\bigr)+y|y|^{p-2}-M^{p-1}\mathop{\rm sgn}(y)
& \mbox{otherwise}
\end{array}
\right.
$$
and
$q:\mathbb R\rightarrow\mathbb R$ by
$$q(y) := \left\{
\begin{array}{ll}
g(y) &\mbox{if } \ |y|\leq M\,,   \\
g(M\,\mathop{\rm sgn} (y)) & \mbox{otherwise}\,.
\end{array}
\right.
$$
Observe that both $\psi$ and $q$ satisfy
conditions of Step 1. 

Thus, due to Step 1, the following problem
\begin{eqnarray}\label{odugfasy}
&\bigl(\psi(y')\bigr)'+q(y')=\widetilde f+\overline f -h(y)\,,& \\
&y(0)=y(T)\,,\quad y'(0)=y'(T)& \nonumber
\end{eqnarray}
has a solution provided that
\begin{equation}
\label{dajkf}
r(\widetilde f)+h(-\infty)<\overline f<r(\widetilde f)+h(+\infty) \,,
\end{equation}
where $r(\widetilde f)$ is the (only admissible) value of the integral
$\frac{1}{T}\int_0^T f(t)dt$ such that the associated first order problem: 
\begin{eqnarray} \label{adjkf}
&v'+q\left(\psi_{-1}(v)\right)=f\,, &\\
&\int_0^T\psi_{-1}(v(t))dt\,,\quad v(0)=v(T) &\nonumber   
\end{eqnarray}
has a solution 
(note that the existence and uniqueness of such $r(\widetilde f)$ follows from
Theorem \ref{Th1} by setting $\varphi:=\psi$ and $g:=q$ in Theorem \ref{Th1}).

Using (\ref{ODH1}) we estimate $v$ by
$\|v\|_{C}\leq\sqrt{\frac{T}{3}}\|\widetilde f\|_{L^2}<\varphi(M)$, which,
by definition of $\psi$, implies $\psi_{-1}(v)=\varphi_{-1}(v)$. 
Then the definition of $q$ and the fact 
that $M>\varphi_{-1}\left(\sqrt{\frac{T}{3}}\|\widetilde f\|_{L^2}\right)$ yield
$q(\varphi_{-1}(v))=g(\varphi_{-1}(v))$.
So that any solution to (\ref{adjkf}) satisfies also
\begin{eqnarray}\label{lj6rl}
&v'+g(\varphi_{-1}(v))=\widetilde f + s(\widetilde f)\,,& \\
&\int_0^T\varphi_{-1}(v(t))dt=0\,,\quad v(0)=v(T)&\nonumber 
\end{eqnarray}
and vice versa.
Integrating equations in (\ref{adjkf}) and (\ref{lj6rl}),respectively, from $0$ to $T$
and taking into account corresponding boundary conditions, we arrive at
$$s(\widetilde f)=\frac{1}{T}\int_0^T q\left(\psi_{-1}(v(t))\right)\,dt=\frac{1}{T}
\int_0^T q\left(\varphi_{-1}(v(t))\right)\,dt=r(\widetilde f)\,.$$
Thus (\ref{dajkf}) becomes (\ref{D_like}): 
$$s(\widetilde f)+h(-\infty)<\overline f<s(\widetilde f)+h(+\infty)\,.$$

Now let us look at the equation (\ref{odugfasy}) again.
Using (\ref{ODH1}) we estimate the norm of $\varphi(y')$ by 
$$\|\varphi(y')\|_C\leq\sqrt{\frac{T}{3}}\left(\|\widetilde
  f\|_{L^2}+\sqrt{T}\sup_{\xi\in\mathbb R}|h(\xi)|\right)\,;$$
hence $\|y'\|_C\leq M$. So that $\psi(y')=\varphi(y')$ and any
solution of (\ref{odugfasy}) satisfy also (\ref{fund_eq})--(\ref{per_con}).
Since (\ref{odugfasy}) possesses solution provided that (\ref{D_like}), so does
(\ref{fund_eq})--(\ref{per_con}). This concludes Step 2. 

\noindent{\bf Step 3} 
Assume that $h(-\infty)<h(\xi)<h(+\infty)$ for all $\xi\in\mathbb R$.
An argument similar to that of \cite[Theorem 4]{Danc} shows that 
the condition (\ref{D_like}) is necessary for the existence of a
solution to (\ref{fund_eq})--(\ref{per_con}). Details are left to the
reader. We note that one should use Lemma \ref{lem2} instead of Lemma 2 from \cite{Danc}.
( Lemma \ref{lem3} shall be used in the case of the BVP (\ref{fund_eq})--(\ref{per_con}) ).
\hfill$\diamondsuit$

\section{Some remarks about uniqueness}

In this section we would like to present one result concerning the
uniqueness of the solution to the BVPs studied in the precedent
sections.  At first we formulate and
prove two more general statements: 

\begin{proposition}
\label{prop_un}
Suppose that $F: [0,T]\times\mathbb R^2\rightarrow\mathbb R$ is continuous,
$F(0,\cdot,\cdot)=F(2\pi,\cdot,\cdot)$ and 
\begin{equation}
\label{con_inc}
\xi>\eta\quad \mbox{ implies }\quad F(\cdot, \xi, \cdot)>F(\cdot, \eta,\cdot)\,,
\end{equation}
then there exists at most one solution of the BVP:
\begin{eqnarray}\label{equ_incp}
&\left(\varphi(u')\right)'=F(t,u,u')\,,\quad t\in(0,T)& \\
&u(0)=u(2\pi),\quad u'(0)=u'(2\pi)\,. &\nonumber
\end{eqnarray}
\end{proposition}

\paragraph{Proof}
If $u_1,\,\,\, u_2$ are two distinct solutions of (\ref{equ_incp}) then
\begin{equation}
\label{substr_eq}
\left(\varphi(u_1')\right)'-\left(\varphi(u_2')\right)' = F(t,u_1,u_1')-F(t,u_2,u_2')\,.
\end{equation}
Set $z=u_1-u_2$; obviously $z\in C[0,T]$ and
hence $z$ attains its maximum value, say at $t_M$ on [0,T].
Without lost of generality, we can suppose that $u_1(t_M)>u_2(t_M)$. This
implies $\max_{t\in [0,T]}z(t)>0$. Since $z(0)=z(T)$, we may assume $t_M\in
[0,T)$. Taking into account that $z\in C^1[0,T]$ and that $z$ satisfies 
(\ref{per_con}),  it follows that $z'(t_M)=0$, 
which is equivalent to $u_1'(t_M)=u'_2(t_M)$.

Upon integrating from $t_M$ to $t$, the equation (\ref{substr_eq}) becomes
$$\varphi\left(\rule{0pt}{3mm}u_1'(t)\right)-\varphi\left(\rule{0pt}{3mm}u_2'(t)\right)
= \int_{t_M}^{t}
\left(\rule{0pt}{5mm} F\left(\rule{0pt}{3mm}\tau,u_1\left(\tau\right),
u_1'\left(\tau\right)\right)-F\left(\rule{0pt}{3mm}\tau,u_2\left(\tau\right),u_2'\left(\tau\right)
\right)\right)d\tau\,.$$
Considering $u_1(t_M)>u_2(t_M)$, $u_1'(t_M)=u_2'(t_M)$, the continuity of 
\newline
$F\left(\rule{0pt}{3mm}t,u_1(t),u_1'(t)\right)-F\left(\rule{0pt}{3mm}t,u_2(t),u_2'(t)\right)$ 
and (\ref{con_inc}), we find that there exists
$\delta>0$ such that
$$\varphi(u_1'(t))-\varphi(u_2'(t))>0 \mbox{ for all }
t\in(t_M,t_M+\delta)\,.$$
Hence $u'_1(t)>u_2'(t)$ for all $t\in(t_M, t_M+\delta)$, which implies
$z'(t)>0$ for all $t\in (t_M, t_M+\delta)$,
a contrary to $z(t_M)\geq z(t)$ for all $t\in [0,T]$. 
\hfill$\diamondsuit$

\begin{proposition}
\label{prop_un_Neu}
Suppose that $F: [0,T]\times\mathbb R^2\rightarrow\mathbb R$ is continuous and satisfies
(\ref{con_inc}).
Then the BVP
\begin{eqnarray}\label{equ_incn}
&\left(\varphi(u')\right)'=F(t,u,u')\,,\quad t\in(0,T)& \\
&u'(0)=0,\quad u'(T)=0 & \nonumber
\end{eqnarray}
admits at most one solution.
\end{proposition}

\paragraph{Proof} The only difference to the proof of Proposition
\ref{prop_un} is that we have to exclude the possibility $t_M=T$.

Suppose that $z=u_1-u_2$ attains its maximum at $T$. Integrating
(\ref{equ_incn}) we find that 
$$\varphi(u_1'(t))-\varphi(u_2'(t)) = - \int_{t}^{T}
\left(\rule{0pt}{5mm} F\left(\rule{0pt}{3mm}\tau,u_1\left(\tau\right),
u_1'\left(\tau\right)\right)-F\left(\rule{0pt}{3mm}\tau,u_2\left(\tau\right),u_2'\left(\tau\right)
\right)\right)d\tau\,.$$
Taking into account $u_1(T)>u_2(T)\,,\quad u_1'(T)=u'_2(T)=0\,,$ 
the continuity of 
$F(t,u_1(t),u_1'(t))-F(t,u_2(t),u_2'(t))$ 
and (\ref{con_inc}), we find that there exists
$\delta>0$ such that
$$\varphi(u_1'(t))-\varphi(u_2'(t))<0 \mbox{ for all }
t\in(T-\delta,T)\,.$$
Hence $u'_1(t)-u'_2(t)<0$ for all $t\in (T-\delta,T)$, which implies
$z'(t)<0$ for all $t\in (T-\delta,T)$,
a contrary to $z(T)\geq z(t)$ for all $t\in [0,T]$. 
\hfill$\diamondsuit$
\medskip

Taking a particular function $F(t, u, v) =  -h(u) - g(v) +
f$, where $h\in C(\mathbb R, \mathbb R)$, is bounded and strictly
decreasing, $g\in C(\mathbb R,
\mathbb R)$ and $f\in C[0,T]$, we
obtain uniqueness for the BVPs (\ref{fund_eq})--(\ref{per_con}) and 
(\ref{fund_eq})--(\ref{Neu_con}) studied previously:
\begin{theorem}
\label{uniquenessTH}
Suppose {\rm (P)}, {\rm (G)},  $f\in C_T$ ( $f\in C[0,T]$ ), 
$h\in C(\mathbb R,\mathbb R)$
and moreover impose that $h$ is strictly decreasing. Then
the BVP (\ref{fund_eq})--(\ref{per_con}) ( (\ref{fund_eq})--(\ref{Neu_con}) )
admits at most one solution.
\end{theorem}

We would like to point out that the monotonicity assumption, $h$ being decreasing, was
essential to prove the uniqueness.
The following simple example shows that there are no
analogous statements to the previous theorem assuming $h$ increasing, 
even not for semilinear BVP:

\begin{example}   
\label{un_cex}
{\rm
Consider BVP:
\begin{equation}
\begin{array}{cr}
\label{un_cex_eq}
u''(t)+h(u(t))=\cos 2t\,,&\quad t\in(0,T) \\
u(0)=u(2\pi),\quad u'(0)=u'(2\pi) &
\end{array}
\end{equation}
with 
$$
h(s)\left\{
\begin{array}{lc}
s \quad & \mbox{for } |s|\leq 1 \\
(2-\frac{1}{|s|}){\rm sgn}(s) \quad & \mbox{for } |s|> 1
\end{array}
\right.
$$
then $u=-\frac{1}{3}\cos(t) + c\sin(t+\phi)$, with parameters $|c|\leq\frac{2}{3}$ and
$\phi\in [0,2\pi)$, is a two dimensional subset of the set of all solutions
to (\ref{un_cex_eq}).}
\end{example}
For some particular results in this direction (the semilinear problem) we refer to
\cite{Danc}.


\paragraph{Acknowledgment.}
The author wish to thank to Prof. Dr\'abek for his advice while writing
this paper and to Prof. Man\'asevich for stimulating discussion during the
conference EDPCHILE '99. The author would also like to thank to referees for their
suggestions and for including references \cite{HMZ1}, and \cite{HMZ2}. 

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\noindent{\sc Petr Girg} \\
Department of Mathematics \\
University of West Bohemia \\
P. O. Box  314 \\
306 14  Plze\v n, Czech Republic \\
e-mail: pgirg@kma.zcu.cz

\end{document}