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\markboth{\hfil Existence of periodic solutions \hfil EJDE--1998/31}
{EJDE--1998/31\hfil Petr Girg \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 1998}(1998), No.~31, pp. 1--10. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  147.26.103.110 or 129.120.3.113 (login: ftp)}
 \vspace{\bigskipamount} \\
  Existence of periodic solutions for a semilinear ordinary differential
  equation 
\thanks{ {\em 1991 Mathematics Subject Classifications:} 34B15, 34C15, 34C25, 34C99.
\hfil\break\indent
{\em Key words and phrases:} Ordinary differential equation, periodic solutions.
\hfil\break\indent
\copyright 1998 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Submitted August 20, 1998. Published November 20, 1998.} }
\date{}
\author{Petr Girg}

\maketitle

\begin{abstract} 
Dancer [3] found a necessary and sufficient condition for the existence
of periodic solutions to the equation
$$ \ddot x +g_1(\dot x) + g_0(x) = f(t)\,.$$
His condition is based on a functional that depends on the solution to
the above equation with $g_0=0$. However, that solution is not always
explicitly known  which makes the condition unverifiable in practical 
situations. As an alternative, we find computable bounds for the 
functional that provide a sufficient condition and 
a necessary condition for the existence of solutions.
\end{abstract}

\newcommand{\esssup}{\mathop{\rm ess~sup}}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}

\section{Introduction}

In this paper, we study the existence and the non-existence of solutions
of the semilinear boundary-value problem
 \begin{eqnarray}
   &\ddot x(t) + g_1 (\dot x(t)) + g_0 (x(t)) = f(t)\,,&\label{fund_eq}\\
  &x(0) = x(T),\ \dot x(0) = \dot x(T)\,.& \label{per_con}
\end{eqnarray}
Although a necessary and sufficient condition is already known \cite{Danc},
it can not be verified in practical situations because the condition is given
by a related nonlinear boundary-value problem.
In this article we  give, on the one hand, a sufficient condition, and on the
other hand a necessary condition, which  can be
verified for any  continuous function $f$. 
In the first part of this article, we present a
survey of known results and their physical interpretation.
And in the second part, we present our main result, which
is stated as Theorem~\ref{it_the}.

Overall, we will suppose that $g_0$, $g_1$, $f$ are continuous real-valued
functions, and $f$ is $T$--periodic.
For a given $k\geq 0$, let 
\begin{eqnarray*}                                           
C^{k}_T &=& \big\{ u : u\mbox{ is $k$-times continuously differentiable on 
 $[0,T]$, with} \\
&&\quad u(0)=u(T), u'(0)=u'(T), \dots, u^{(k)}(0)=u^{(k)}(T) \big\}\,.
\end{eqnarray*}
In these spaces the maximum norm will be denoted by $\|\cdot\|_{C^k_T}$,
and $C^{0}_T$ will be denoted by $C_T$. The subspace consisting of
functions with mean value zero will be denoted by
$$
\tilde C_T^k=\big\{u\in C_T^k : \int_0^Tu(t)\,dt =0\big\}\,. 
$$
For functions with domain $[0,T]$, with distributional derivatives, 
we define:
\begin{eqnarray*}
&L^{p} = \big\{u : \int_{0}^T |u(t)|^{p} dt < \infty \big\},\ 1\leq p < 
+\infty\,;&\\
& L^\infty =\big\{ u :  \esssup_{t\in [0,T]} |u(t)|<\infty \big\}, &\\
& W^{1,2}_T = \big\{u\in L^2 : u'\in L^2,\; u(0)=u(T)\big\} \,,&\\
& W^{2,\infty}_T = \big\{ u\in L^{\infty} :  u''\in L^{\infty},
\; u(0)=u(T),\; u'(0)=u'(T) \big\}\,. &
\end{eqnarray*} 
For a subset $X$ on the space of integrable functions on $[0,T]$,
we define  
$\tilde X = \big\{u\in X : \int_0^T u(t)dt = 0\big\}$. 
For integrable functions, we use the decomposition   
$$ f = \tilde f + \bar f\,, \quad\mbox{with } 
\bar f = \frac{1}{T} \int_0^T f(t)\,dt\,.
$$

We will assume that $f$, the right-hand side of (\ref{fund_eq}),
belongs to $C_T$, and the solution $x$ belongs to $C^2_T$.
Although the results in \cite{Danc} assume that $f$ is in a certain
Lebesgue space and that $x$ is in a certain Sobolev space, it is not
hard to get analogous results for $f$ in $C_T$ and $x$ in $C^2_T$.
So when we cite results from \cite{Danc}, we do a conversion to our
function spaces (except in section 2).

When $g_0=0$, for each $\tilde f \in \tilde C_T$ there exists a value
$s(\tilde f)$ such that 
\begin{equation}
\label{s_eq}
  \ddot x(t) + g_1(\dot x(t)) = \tilde f(t) + s(\tilde f)
\end{equation}
has a periodic solution, 
\cite[Theorem~1]{Danc}. Equivalently, the range of the operator
$H_1:C^2_{T} \rightarrow C_{T}$, $H_1(x)=\ddot x + g_1\circ \dot x$,
can be written as 
\begin{equation} \label{rank}
{\cal R}_1 = \{\tilde f + s(\tilde f) : \tilde f\in \tilde C_T\}\,.
\end{equation}  

Under the assumption that $g_0$ is bounded, continuous, and satisfies
$$
  g_0(-\infty):=\lim_{\xi \rightarrow -\infty}g_0(\xi)
  < \lim_{\xi \rightarrow + \infty}g_0(\xi)=: g_0(+\infty)\,,
$$ 
Dancer  \cite[Theorem~2]{Danc} showed that a function $f \in C_T$ 
belongs to ${\cal R}$, the range of $H:C^2_{T} \rightarrow C_{T}$,
$H(x)=\ddot x + g_1\circ \dot x+ g_0\circ x$, if 
\begin{equation} \label{danc_con}
g_0(-\infty) < \bar f - s(\tilde f) < g_0(+\infty)\,.
\end{equation} 
Thus, (\ref{danc_con}) is a sufficient condition for (\ref{fund_eq})
to posses a periodic solution. However, if we also have
$$
g_0(-\infty)<g_0(\xi)<g_0(+\infty)\quad \forall \xi\in{\mathbb R}\,,
$$ 
then (\ref{danc_con}) is also a necessary condition, \cite[Theorem~4]{Danc}.
Since we do not know the functional $s(\tilde f)$ explicitly,
we can not verify condition (\ref{danc_con}) in practical situations.

The aim of our work is to find estimates for the functional
$s(\tilde f)$.
In particular we find functionals $a:\tilde C_T \rightarrow {\mathbb R}$, and
$A:\tilde C_T \rightarrow {\mathbb R}$, such that $a(\tilde f) \leq s(\tilde f)
\leq A(\tilde f)$ for all $\tilde f \in C_T$ (see Theorem 2).
Using these bounds we define the sets:
\begin{eqnarray*} 
&{\cal A}_1 = \left \{f : f\in C_T \mbox{ and } a(\tilde f) \leq \bar f \leq
A(\tilde f)\right \}\,, &\\
&{\cal A} = \left \{f: f\in C_T \mbox{ and }
g_0(-\infty) + a(\tilde f) < \bar f <
g_0(+\infty) + A(\tilde f)\right \}\,, &\\
& {\cal B} = \left \{f: f\in C_T \mbox{ and }
g_0(-\infty) + A(\tilde f) < \bar f <
g_0(+\infty) + a(\tilde f)\right \}\,.& 
\end{eqnarray*} 
\paragraph{Main result} With the above definitions, ${\cal R}=H(C^2_T)$, and
${\cal R}_1 = H_1(C^2_T)$, our main result is stated as 
$${\cal R}_1\subset{\cal A}_1,\quad\mbox{and}\quad
{\cal B}\subset {\cal R}\subset{\cal A}\,.$$ 
This means that $f$ being in  ${\cal A}$ is a necessary condition, and
that $f$ being in  ${\cal B}$ is a sufficient condition for the existence
of solutions to (\ref{fund_eq}).

\section{Related results}

In this section, we present some known results, and give a physical 
interpretation for particular cases of equation (\ref{fund_eq}). 
We want to emphasize the fact that although the conditions  come
from abstract methods of functional analysis, they have
physical interpretations (For various physical examples see e.g.
\cite{Nayfeh} or \cite{Stoker}).

If the function $g_0$ is bounded and $g_1(\xi) = \lambda \xi$ for 
some $\lambda\in {\mathbb R}$, then
(\ref{fund_eq}) becomes the ``classical'' Landesman-Lazer equation.
   \begin{equation}
     \label{L_L}
     \ddot x(t) + \lambda \dot x(t) + g_0 (x(t)) = f(t)\,.
   \end{equation}
A short review of applicable results for this equation with boundary 
conditions (\ref{per_con}) is as follows: \begin{itemize}

\item A sufficient condition for (\ref{L_L}) to have a $T$--periodic
solution is the so called {\it Landesman--Lazer condition} \cite{Lan_Laz},
\begin{equation} \label{con_L_L}
g_0(-\infty) < \bar f < g_0(+\infty)\,.
\end{equation}

\item Condition (\ref{con_L_L}) is also necessary when  
$g_0(-\infty) < g_0(\xi) < g_0(+\infty)$ for all $\xi \in {\mathbb R}$.

\item The range of the operator
$\ddot x + \lambda \dot x + g_0 \circ x : C^2_T \rightarrow C_T$ is a set
of functions in $C_T$, enclosed by two parallel hyper-planes.
\end{itemize}
From a physical point of view, when $g_0(-\infty) < 0< g_0(+\infty)$,
this boundary-value problem is a model for vibrations with linear damping and nonlinear
restoring force.
When $\lambda$ is equal to zero, we have a conservative oscillator.
Condition (\ref{con_L_L}) can be interpreted as representing
the restoring force being able to overcome the mean value 
of the external forcing term $f$.

For $g_0=0$, a brief summary of results is as follows:
\begin{itemize}
 \item For $g_1$ continuous, Dancer \cite{Danc} proved
 that for all $\tilde f \in \tilde L^{\infty}$
 there exists exactly one $s(\tilde f)\in{\mathbb R}$ such that
 (\ref{s_eq}) has  solution $x$ in $W^{2,\infty}_T$,
 in the sense of distributions. Furthermore,
 the functional $s: \tilde L^{\infty}\rightarrow{\mathbb R}$ is continuous.
 \item For $g_1$ continuous, Mawhin \cite{Maw} showed that for
 all $\tilde f \in \tilde L^{1}$
 there exists $s(\tilde f)\in{\mathbb R}$ such that (\ref{s_eq})
 has a strong solution.
 \item For $g_1$ continuously differentiable,
 Ca\~nada, Dr\'abek \cite{CD} proved  that
 for all $\tilde f\in \tilde C_T$ there exists exactly one
 $s(\tilde f)\in{\mathbb R}$ such that (\ref{s_eq}) has a classical solution.
 Furthermore, the functional $s: \tilde C_T\rightarrow{\mathbb R}$ is
 continuously differentiable, and the range of $H_1$ can be written
 as in (\ref{rank}). 

\item The functional $s$  gives the necessary and sufficient condition
for the solvability of the boundary-value problem, namely
$$ \bar f = s(\tilde f)\,.
$$ 
But $s(\tilde f)$ is given in terms of the solution,  which we do not know 
a priori. Thus,  we can not formulate the condition explicitly
as is done in the Landesman--Lazer result.
\end{itemize}

From a physical point of view, (\ref{s_eq}) describes the periodically
forced vibration of a mass on a damper. 
The damping term makes the system unbalanced and $s(\tilde f)$
represents a constant force which tends to compensate for the damping term.
In this example, we consider the dissipative case:
$\dot xg_1(\dot x) > 0$, or $\dot xg_1(\dot x) < 0$, which represents a 
self--excitation (positive damping). \medskip

For general functions $g_1$ and $g_0$, with $g_0$ bounded as in the
Landesman--Lazer case, Dancer \cite{Danc} proved that the range $H(W^{2,\infty}_T)$ is 
enclosed by two manifolds parallel to the range  $H_1(W^{2,\infty}_T)$.
A sufficient condition for the solvability of Problem 
(\ref{fund_eq})--(\ref{per_con}) is given by (\ref{danc_con}).
Note that if $g_0(-\infty)<0<g_0(+\infty)$, then from (\ref{danc_con}) it follows that
the range ${\cal R}_1$ of the operator $H_1$
is a subset of the range ${\cal R}$ of the operator $H$.
In this case  
(\ref{fund_eq}) is a model for vibrations  with nonlinear  damping
and nonlinear restoring force.

\section{Bounds for $s(\tilde f)$}

Estimates for $s(\tilde f)$ are derived from the study of equation 
(\ref{s_eq}). Putting $w=\dot x$, problem (\ref{s_eq}) subject to
 (\ref{per_con}) becomes
\begin{eqnarray} 
&\dot w(t) + g_1(w(t)) = \tilde f(t)+s(\tilde f)\,,& \label{fst_eq}\\
&w(0) = w(T)\,,\quad \int_0^{T} w(\tau)\,d\tau = 0\,.& \label{w_cond}
\end{eqnarray} 

\begin{theorem} \label{Can_Dra}
Let $g_1$ be a continuously differentiable function satisfying
$|g_1(\xi)| \leq K$ for all $\xi\in{\mathbb R}$.
Then for each $\tilde f\in\tilde C_T$ there exists precisely
one  $s(\tilde f)$ such that (\ref{s_eq}) has a periodic solution.
In this case problem (\ref{s_eq}) has a family of solutions
$ x_c(t) = x(t) + c$,
where $c\in{\mathbb R}$ is arbitrary and 
$$ x(t) = \int_0^t w_{s(\tilde f)}(\tau)\,d\tau\,, $$
with $w_{s(\tilde f)}$ the unique solution of (\ref{fst_eq}) subject to
(\ref{w_cond}). Moreover,
the map $s: \tilde f \mapsto s(\tilde f)$ from
$\tilde C_T$ to ${\mathbb R}$ is continuously
differentiable and $-K\leq s(\tilde f)\leq K$. 
\end{theorem}

The proof of the above theorem can be found in \cite{CD}.
Existence results considering a continuous function $g_1$ are studied in
\cite{Danc}. The analogous a priori bound for $\|w\|_{C_T}$ as in the following
Lemma is also given in \cite{Danc}.

\begin{lemma} \label{lm_0}
Let $g_1$ be a continuous function, and $w$ be the solution of (\ref{fst_eq})
subject to (\ref{w_cond}). Then 
$$
\|w\|_{C_T} \leq \|\tilde f\|_2\sqrt{\frac{T}{12}}\,.
$$ 
\end{lemma}

\paragraph{Proof}
Multiplying both sides of equation (\ref{fst_eq}) by $\dot w$ and
integrating  from $0$ to $T$,  using the boundary condition $w(0)=w(T)$,
we see that $\| \dot w\|^2_2=\langle\tilde f, \dot w\rangle_2$.
The Cauchy-Schwartz inequality yields that 
$\|\dot w\|_2^2\leq\|\tilde f\|_2\|\dot w\|_2$, so that
$\|\dot w\|_2\leq\|\tilde f\|_2$.
Since $w \in C^1_T \subset W^{1,2}_T$ and $\int_0^Tw(t)dt=0$,
we can use a Sobolev inequality, \cite[Prop. 1.3]{MawW},  to obtain
$$
\|w\|_\infty \leq \|\dot w\|_2\sqrt{\frac{T}{12}} \leq
\|\tilde f \|_2\sqrt{\frac{T}{12}}\,.
$$ 
Since $w$ is continuous, $\| w\|_{C_T} = \|w\|_\infty$
which is the desired inequality. \hfill$\diamondsuit$\medskip

As a consequence of the above lemma, Theorem~\ref{Can_Dra} can be applied 
for a function $g_1$ that is not necessarily bounded.
This is so because the argument of $g_1$ lies on a bounded interval.

An estimate for $s(\tilde f)$ is obtained as follows:
Integrate each term in (\ref{fst_eq}) from $0$ to $T$, use the boundary conditions 
(\ref{w_cond}) to eliminate terms with $\dot w, cw, \tilde f$, and divide 
by  $T$, to obtain
$$ 
\frac{1}{T}\int_0^T g_1(w(t))\,dt = s(\tilde f)\,.
$$  
As in \cite[Theorem 1]{Danc}, the minimum and the maximum values of $g_1$ 
provide  bounds for $s(\tilde f)$. To obtain the basic estimate we use the
fact that $\int_0^T w=0$. First subtract $cw$ in the integrand above, and 
then compute the infimum and the supremum over $c\in {\mathbb R}$:
\begin{equation} \label{bas_est}
\sup_{c\in{\mathbb R}} \min_{|\xi| \leq b} (g_1(\xi)-c\xi)
\leq s(\tilde f) \leq
\inf_{c\in{\mathbb R}} \max_{|\xi| \leq b} (g_1(\xi)-c\xi)\,,
\end{equation} 
where $\|w\|_{C_T}\leq b$ (for instance we can set
$b=\|\tilde f\|_2\sqrt{T/12}$ due to Lemma \ref{lm_0}).



\paragraph{Remarks} If the function $g_1$ is a polynomial of degree 1, then
$a(\tilde f) = A(\tilde f) = g_1(0)$, and this is the exact value of
$s(\tilde f)$. 
On the other hand if $g_1(w)=w^2$, then $a(\tilde f) < A(\tilde f)$
and the direct estimate in Dancer \cite[Theorem 1]{Danc} is the same as
the basic estimate.

Finding the infimum and the supremum over all real numbers is not amenable
for computations; hence, we need to find a finite set of suitable values for
$c$. For example, the upper bound can be interpreted as an error in a
minimax polynomial approximation. In which case, we are looking for a
polynomial $q(\xi)=c\xi+d$ such that
$\|g_1-q\|_\infty$ is as small as possible.  With 
interpolation nodes $\{-b,0,b\}$, we obtain $c=(g_1(b)-g_1(-b))/(2b)$, and
we avoid calculating the supremum over $\mathbb R$.

Notice that the upper bound minus the lower bound in (\ref{bas_est}) is
an increasing function of $b$, the bound for $\|\dot x\|_\infty$.
Therefore, our strategy is to  decrement $b$, which is done
by using the following two lemmas. 

\begin{lemma} \label{lm_2}
Let $k$ and $K$ be positive constants, and $w\in \tilde C_T$ be absolutely 
continuous  with $- k \leq \dot w(\xi) \leq K$  a.e. on $[0,T]$.
Then 
$$ \|w\|_{C_T} \leq \frac{T k K}{2(k + K)}\,. $$ 
\end{lemma}

\paragraph{Proof}
On the contrary, suppose that $\|w\|_{C_T}>\frac{TkK}{2(k+K)}$. 
Without lost of generality, we may assume that the maximum norm is attained
at a point $t_0=\frac{kT}{2(k+K)}$, $0\leq t_0\leq T/2$.
If necessary multiply $w$ by $-1$, interchange the roles of $k$ and $K$,
 and shift $w$ suitably in time. Then
$$w(t_0)=\|w\|_{C_T} > \frac{TkK}{2(k+K)}=Kt_0\,.
$$
Our strategy is to prove the following two inequalities for 
$t\in[0,\frac{T}{2}]$:
\begin{eqnarray} 
&w(t)>\min\big\{Kt, k(\frac{T}{2}-t) \big\}\,,&\label{le1}\\
&w(t+\frac{T}{2})>-\min\big\{kt,K(\frac{T}{2}-t) \big\}\,.&\label{le2}
\end{eqnarray} 
Which lead us to the contradiction  that $\int_0^T w=0$ and  
$$
\int_0^T w(t)\,dt=\int_0^{T/2}w(t)\,dt+\int_0^{T/2}
w(t+\frac{T}{2})\,dt > 0\,.
$$ 
For (\ref{le1}), we consider the two cases:
If $0<t\leq t_0$, then
$$w(t)=w(t_0)+\int_t^{t_0}\bigl(-\dot w(\tau)\bigr)\,d\tau > Kt_0+
(t_0-t)(-K)=Kt\,.$$
and if  $t_0<t\leq \frac{T}{2}$, then
$$w(t)=w(t_0)+\int_{t_0}^t\dot w(\tau)\,d\tau > Kt_0 + (t-t_0)(-k)=
k(\frac{T}{2}-t)\,. $$

For (\ref{le2}), we put $u(t)=w(t+\frac{T}{2})$ and notice that
$u(0)=w(\frac{T}{2})>0$ and $u(\frac{T}{2})=w(T)=w(0)>0$. For
$t$ in $[0,\frac{T}{2}]$ we have
$$w\left(t+\mbox{$\frac{T}{2}$}\right)=u(t)=
u(0)+\int_0^t\dot u(\tau)\,d\tau > -kt$$
and
$$w\left(t+\mbox{$\frac{T}{2}$}\right)=u(t)=u\left(\mbox{$\frac{T}{2}$}\right)
+\int_t^{T/2}\bigl(-\dot u(\tau)\bigr)\,d\tau>-K(\frac{T}{2}-t)\,.$$
Hence 
$$
w(t+\mbox{$\frac{T}{2}$})>\max\big\{-kt, -K(\frac{T}{2}-t)\big\}=
-\min\big\{kt, K(\frac{T}{2}-t)\big\}.
$$
Which concludes the present proof. \hfill $\diamondsuit$

\begin{lemma}\label{lm_3}
Let $w$ be a solution to Problem (\ref{fst_eq})-(\ref{w_cond}), with 
$\|w\|_{C_T}\leq b$ and $\tilde f\not\equiv 0$.
Then 
$$ -k\leq \dot w(t) \leq K\,,$$
where $k$ and $K$ are the positive constants: 
$-k= \min_{t\in [0,T]} \tilde f(t)+m$ and
$K=\max_{t\in [0,T]} \tilde f(t) +M$,  where 
\begin{eqnarray*}
&m = \sup_{c\in{\mathbb R}}\min_{|\xi|\leq b}(g_1(\xi)-c\xi)-\max_{|\xi|\leq b}
g_1(\xi)\,,&\\
&M = \inf_{c\in{\mathbb R}}\max_{|\xi|\leq b}(g_1(\xi)-c\xi)-\min_{|\xi|\leq b}
g_1(\xi)\,.& 
\end{eqnarray*}
\end{lemma}

\paragraph{Proof}
From (\ref{fst_eq}) we obtain
$$
 \min_{t\in [ 0,T ]}\left(\tilde f(t) + s(\tilde f) - g_1(w(t))\right)
  \leq \dot w(t) \leq
\min_{t\in [ 0,T ]}\left(\tilde f(t) + s(\tilde f) - g_1(w(t))\right)\,.
$$
Using the estimates  for $s(\tilde f)$ in (\ref{bas_est}), we  obtain 
the desired  inequality. Notice that because $g_1$ is continuous and the extrema is
computed on a bounded interval, then
\begin{eqnarray*}
&-\infty<\min_{|\xi|\leq b}(g_1(\xi)-0\cdot\xi)-\max_{|\xi|\leq b} g_1(\xi) 
\leq m \,,&\\
& M \leq 
\max_{|\xi|\leq b}(g_1(\xi)-0\cdot\xi)-\min_{|\xi|\leq b}g_1(\xi)<\infty\,.&
\end{eqnarray*}
It is left only to check that $k$ and $K$ are
positive. This follows from the fact that $-k<\dot w(t)<K$ on [0,T], 
$\int_0^T\dot w(\tau)d\tau=w(T)-w(0)=0$ and $w$ is not a constant function if $\tilde f\not\equiv 0$.
\hfill$\diamondsuit$

\paragraph{Iterated estimates}
As an initial value put $b_0>0$, such that $\|w\|_{C_T}\leq b_0$
(for instance: $b_0=\|\tilde f\|_2\sqrt{T/12}$ due to Lemma \ref{lm_0}).
Then for $n=0,1,2,\dots$,
let $k_n, K_n$ be the constants obtained in Lemma~\ref{lm_3} with $b=b_n$, 
and let $$b_{n+1}= \frac{Tk_nK_n}{2(k_n+K_n)}\,.$$

\begin{lemma} Let $b_n, k_n, K_n$ be defined as above.
If $b_1\leq b_0$, then $b_{n+1}\leq b_n$ for all $n\geq 1$.
\end{lemma}

\paragraph{Proof} We proceed by induction. First notice that 
$b_1\leq b_0$ is one of the hypotheses. 
Now assume that $b_n\leq b_{n-1}$. Then in the statement of Lemma~\ref{lm_3}
we see that
$$ 0\geq m_n\geq m_{n-1}\quad\mbox{and}\quad 0\leq M_n \leq M_{n-1}\,.$$
Thus, $k_n\leq k_{n-1}$ and $K_n\leq K_{n-1}$. 
Since $\frac{TkK}{2(k+K)}$ is a decreasing function of $k$, and of $K$, 
we have $b_{n+1}\leq b_n$. \hfill$\diamondsuit$\medskip

From the above lemma, iterations can be repeated indefinitely.
However, in practice the process should stop when the decrement in
$b_n$ is less than a predetermined value. 
Now, we define the lower and upper bounds for $s(\tilde f)$.

\begin{theorem} \label{it_the}
Let $b_n$ be as defined above. Put $b=\inf\{b_0,b_1,\dots\}$, and
\begin{eqnarray*}
&a(\tilde f)= \sup_{c\in{\mathbb R}}\min_{|\xi|\leq b} 
\left (g_1(\xi)-c\xi\right )\,,&\\
&A(\tilde f) = \inf_{c\in{\mathbb R}}\max_{|\xi|\leq b} 
\left(g_1(\xi)-c\xi\right )\,.&
\end{eqnarray*} 
Then the functional $s(\tilde f)$ satisfies
$a(\tilde f) \leq s(\tilde f) \leq A(\tilde f)$.   
\end{theorem}

\paragraph{Proof} Notice that by Lemma~\ref{lm_2}, $\|w\|_\infty\leq b_n$
for all $n$. Therefore, from the basic estimate (\ref{bas_est}), the statement
of this theorem follows.  Notice that
even if $A(\tilde f)$ is not the absolute infimum over $c$, the equality 
in this Theorem is still valid. The same statement applies for 
$a(\tilde f)$. \hfill$\diamondsuit$\medskip


Computational experiments show that the iteration method refines estimates 
if the ratio $-\max(\tilde f)/\min(\tilde f)$ is much larger than one, or very 
close to zero. To illustrate this case, we study the following
boundary-value problem

\paragraph{Example 1}  Consider $\dot w(t) + g_1(w(t)) = \tilde f(t)+s(\tilde f)$,
where
$$
\tilde f(t) =\left\{ \begin{array}{ll} 
          -\sin(t)/20 & \mbox{if } 0\leq t\leq \pi\\
          \sin(20t)   & \mbox{if } \pi<t\leq 21\pi/20\,. \end{array}\right.
$$
Notice that the ratio $-\max \tilde f/\min \tilde f$ is large.
The period is $T=21\pi/20$, $\|\tilde f\|_2^2=\pi/800 +\pi/40$,
and the estimate for $\|w\|_\infty$ is $b_0=\|\tilde f\|_2\sqrt{T/12}$.

To avoid computing the maximum and the minimum over $c\in{\mathbb R}$,  
we use  $c=(g_1(b)-g_1(-b))/(2b)$; see the remark after (\ref{bas_est}). 
The following table shows the estimates obtained for several 
functions $g_1$. \smallskip

\begin{tabular}{|c|c|c|c|} \hline
& $g_1(\xi)=\xi^2$ &   $g_1(\xi)=\xi^3$  & $0.1\arctan(\xi)$ \\ \hline
min-max $g$  & $0\leq s\leq$ 2.2669e-2 &
 $|s|\leq$ 3.4131e-3 &  
 $|s|\leq$ 1.4944e-2\\ \hline
basic est.  & $0\leq s\leq$ 2.2669e-2  &
 $|s|\leq$ 1.3137e-3  & 
 $|s|\leq$ 4.3011e-5\\ \hline
iterated    & $0\leq s\leq$ 8.2592e-3  & 
 $|s|\leq$ 1.9403e-4 & 
 $|s|\leq$ 1.0002e-5 \\ \hline
\end{tabular}

\paragraph{Example 2}  For $\alpha>0$, consider the equation
$$\dot w(t) + \arctan(w(t)) = \alpha\sin(t)+s(\tilde f)\,.$$
Notice that $\max\tilde f=-\min\tilde f=1$, the period is $T=2\pi$, 
$\|\tilde f\|_2=\alpha\sqrt\pi$, and the estimate for $\|w\|_\infty$ is 
$b_0=\alpha\pi/\sqrt{6}$. The following table shows the estimates obtained 
for several values of $\alpha$. \smallskip

\begin{tabular}{|c|c|c|c|} \hline
& $\alpha=0.01$ &  $\alpha=0.1$ & $\alpha=1$ \\ \hline
min-max $g$  & $|s|\leq$ 1.2824e-2 &
 $|s|\leq$ 0.12756  &  
 $|s|\leq$ 0.90856\\ \hline
basic est.  & $|s|\leq$ 2.7064e-7 &
 $|s|\leq$ 2.6716e-4 & 
 $|s|\leq$ 0.11593\\ \hline
 iterated    & $|s|\leq$ 2.7064e-7 & 
 $|s|\leq$ 2.6716e-4 & 
 $|s|\leq$ 0.11593\\ \hline
\end{tabular}

\paragraph{Remark} For all functions $g_1$ and all $\alpha\neq 0$ in
$\dot w(t) + g_1(w(t)) = \alpha\sin(t)+s(\tilde f)$
the iterated method fails to improve the basic estimate.

To prove this statement, notice that $\max\tilde f=-\min\tilde f=|\alpha|$, 
the period is $T=2\pi$, $\|\tilde f\|_2=|\alpha|\sqrt\pi$, and the estimate 
for $\|w\|_\infty$ is $b_0=|\alpha|\pi/\sqrt{6}$. As in Lemma~\ref{lm_3},
$m_0$ and $M_0$ are non-negative quantities; thus, $k_0\geq |\alpha|$ and
$K_0\geq |\alpha|$. Since $b_1$ is an increasing function of $k_0$ and of 
$K_0$, it follows that
$$
b_1 \geq \frac \pi 2|\alpha| > \frac \pi {\sqrt 6}|\alpha|=b_0\,.$$
Which indicates that the iteration method is unsuccessful in this case.


\paragraph{Example 3}
Consider $\dot w(t)+g_1(w(t))=\tilde f(t)+s(\tilde f)$ with
$$g_1(\xi)=2\bigl(\arctan\bigl(10000(\xi+0.12)\bigr)+
\arctan\bigl(10000(\xi-0.12)\bigr)\bigr)$$
and $\tilde f$ defined as in Example 1.
Note that  $g_1$ varies significantly only in the neighbourhood
of several points (namely $-0.12$ and $0.12$). In such a case it is better no to
apply the iteration method directly, but apply the
iteration method with $b_0=\|\tilde f\|_2\sqrt{T/12}$ to
$$\dot v(t) + d(v(t))=\tilde f(t)+s_d(\tilde f)\,,$$
where $d(\xi)=g_1(\xi) \mbox{ for } |\xi|<\delta$ and $d(\xi)=
g_1(\delta\mbox{sgn}(\xi))$ otherwise with some $0<\delta\leq b_0$.
If $b=\inf\{b_0,b_1,\dots\}\leq \delta$ then considering $\|v\|_{\infty}\leq b$
and $d(\xi)=g_1(\xi)$ for $|\xi|\leq\delta$ we get $w=v$. Thus $\|w\|_{\infty}
\leq b$ and $s(\tilde f)=s_d(\tilde f)$. The following table shows the
estimates obtained for direct
application of iteration method and different values of $\delta$.\smallskip

\begin{tabular}{|c|c|c|c|} \hline
& direct application &  $\delta=0.11$ & $\delta=0.1$ \\ \hline
$b$ & 1.5056e-1 & 1.1946e-1$>$0.11& 9.0409e-2$<$0.1 \\ \hline
min-max $g$  & $|s|\leq$ 6.2759 &
 $|s|\leq$ 6.2759  &
 $|s|\leq$ 9.0908e-3\\ \hline
basic est.  & $|s|\leq$ 4.8202 &
 $|s|\leq$ 4.8202 &
 $|s|\leq$ 1.4010e-3\\ \hline
 iterated    & $|s|\leq$ 4.8202 &
 $|s|\leq$ 4.8202 &
 $|s|\leq$ 1.6342e-3\\ \hline
\end{tabular}

\paragraph{Remark} Note that for $\delta=0.1$ the basic estimate yields better
result than iterated although $b=$ 9.0409e-2 $<b_0=$ 1.5056e-1.
The reason is that we avoid calculating supremum or infimum over $c$ and use
$c=\bigl(g_1(b)-g_1(b)\bigr)/2b$ in formulas for $a(\tilde f)$ and $A(\tilde f)$
in Theorem \ref{it_the}.

\paragraph{Acknowledgments}
The author is grateful to Professors Pavel Dr\' abek and Herbert Leinfelder
for  their advice; to the Grant Agency of the Czech Republic for their support 
(Grant \#201/97/0395); and to the Ministry of Education of the Czech Republic 
for their support (Grant \#VS97156). My special thanks go to Professor
Julio G. Dix for the improvements on the final version of the manuscript.


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\bigskip

{\sc Petr Girg}\\
Centre of Applied Mathematics\\
University of West Bohemia, P.O. Box 314\\
306 14, Plze\v n, The Czech Republic \\
E-mail address:  pgirg@kma.zcu.cz\\
Web page http://www-cam.zcu.cz/Girg


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