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\markboth{\hfil Landesman-Lazer condition \hfil EJDE--2001/04}
{EJDE--2001/04\hfil Petr Tomiczek \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2001}(2001), No.04, pp. 1--11. \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)}
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 %
  A generalization of the \\ Landesman-Lazer condition
 %
\thanks{ {\em Mathematics Subject Classifications:} 35J70, 35P30, 47H15.
\hfil\break\indent
{\em Key words:} resonance, eigenvalue, Landesman-Lazer condition.
\hfil\break\indent
\copyright 2001 Southwest Texas State University. \hfil\break\indent
Submitted September 8, 2000. Published January 3, 2001. \hfil\break\indent
Partially supported by the Grant Agency of Czech
Republic, No. 201/00/0376} }
\date{}
%
\author{ Petr Tomiczek }
\maketitle

\begin{abstract}
In this paper we prove the existence of solutions to the
semi-linear problem
$$\displaylines{
u''(x)+m^2 u(x)+g(x,u(x))=f(x)\cr
u(0)=u(\pi)=0
}$$
at resonance.  We assume a Landesman-Lazer type condition and use
a variational method based on the Saddle Point Theorem.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\renewcommand{\theequation}{\thesection.\arabic{equation}}
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\section{Introduction}

Let us consider the nonlinear boundary-value problem
\begin{eqnarray} \label{1.1}
&u''(x)+m^2 u(x)+g(u(x))=f(x)\,,\quad
x\in (0,\pi)\,,&\\
&u(0)=u(\pi)=0\,,& \nonumber
\end{eqnarray}
at resonance. Here $m\in \mathbb{N}$, $f\in {C}(0,\pi)$
and $g:\mathbb{R}\to \mathbb{R}$ is
a continuous function such that
\begin{equation}       %r1.2
\lim_{s\to \infty}g(s)=g_+ \quad \mbox{and} \quad
 \lim_{s\to -\infty}g(s)=g_-
\end{equation}
exist and are finite numbers. Fu\v{c}\'{\i}k~\cite[Th.6.4]{lit3}
proved that (\ref{1.1}) has at least one solution
provided that
\begin{eqnarray} \label{1.3}
\lefteqn{ \int_0^{\pi} \bigl[g_-(\sin mx)^+  -
g_+(\sin mx)^-\bigr]\,dx } \\
&<&\int_0^{\pi}f(x)\sin mx \,dx
 < \int_0^{\pi} \bigl[ g_+(\sin mx)^+ -
 g_-(\sin mx)^-\bigr] \,dx\,. \nonumber
\end{eqnarray}
Intensive study of problem (\ref{1.1}) started with
the paper \cite{lit4} by Landesman and Lazer in 1970.
Their result \cite{lit4} has been generalized in various directions.
For a survey of results and exhaustive list of
the bibliography up to 1979, we refer the reader to
Fu\v{c}\'{\i}k \cite{lit3}. A list of works published since 1980,
can be found in Dr\'{a}bek \cite{lit1}.
Chun-Lei Tang \cite{lit8} defined the
function $F(s)={{2G(s)}\over {s}}-g(s)$ and the constants
$F_+=\liminf_{s\to +\infty}{F(s)}$,
$F_-=\limsup_{s\to -\infty}{F(s)}$ to prove that
for $m=1$, Problem (\ref{1.1}) is solvable under the condition
\begin{eqnarray} \label{1.4}
\int_0^{\pi}\bigl[ F_-(\sin x)^+ -
F_+(\sin x)^-\bigr] \,dx &<&
\int_0^{\pi}f(x)\sin x \,dx  \\
& <& \int_0^{\pi} [F_+(\sin x)^+
- F_-(\sin x)^-]\,dx\,. \nonumber
\end{eqnarray}
In this paper, we generalize the Landesman-Lazer type
conditions (\ref{1.3}) and (\ref{1.4}) to
prove solvability of (\ref{1.1}) when the nonlinearity $g$
that satisfies: \begin{itemize}
\item $g(s)$ is a continuous and odd function
\item For $s\ge \delta >0$,
$g(s)=\epsilon+K\sin s$ with $K>\epsilon>0$ (see Theorem 2).
\end{itemize}
Note that for this linearity, there is no function $f$ satisfying
conditions (\ref{1.3}) or (\ref{1.4}).
We will use a variational method based on the modification of the
Saddle  Point Theorem introduced by Rabinowitz~\cite{lit6}.


\section{Preliminaries}

 It is known that the spectrum of the linear problem
\begin{eqnarray}
&u''(x)+ \lambda u(x)=0\,,\quad x\in (0,\pi)\,,&\\
&x(0)=x(\pi)=0& \nonumber
\end{eqnarray}
is the set $C=\{ \lambda : \lambda =m^2,\ m\in N\}$.

\paragraph{Notation:} We shall use the classical spaces
${C}(0,\pi)$, $L^p(0,\pi)$ of continuous and
measurable real-valued functions whose $p$-th power
of the absolute value is Lebesgue integrable,
respectively. $H$ is the Sobolev space of
absolutely continu\-ous functions
$u:(0,\pi)\to \mathbb{R}$ such that $u'\in L^2(0,\pi)$
and $u(0)=u(\pi)=0$. We denote by the symbols
$\| \cdot \|$, and $\| \cdot \|_2$
the norm in $H$, and in $L^2(0,\pi)$, respectively.
Let $H^-$ be the subspace of $H$ spanned by all
eigenfunctions corresponding to
the eigenvalues $1,4,\ldots ,m^2$ and let $H^+$ be the subspace of
$H$ spanned
by all eigenfunctions corresponding to  the eigenvalues greater or equal
 to $(m+1)^2$.
 Then $H=H^-\oplus H^+$, $\dim (H^-)<\infty $
and $\dim (H^+)=\infty$.

Let $I:H\to \mathbb{R}$ be a functional such that
$I\in {C}^1(H,\mathbb{R})$
(continuously differentiable). We say that $u$
is a critical point of $I$, if
$$I'(u) v = 0 \quad \mbox{for all}\quad v\in H\,.$$
We say that $I$ satisfies Palais-Smale condition (PS) if every sequence
$(u_n)$ for which $I(u_n)$ is bounded in $H$ and $I'(u_n)\to 0$
(as $n\to \infty)$ possesses a convergent subsequence.

Now  we can  formulate a variation of the Saddle Point Theorem,
due to Lupo and Micheletti~\cite{lit5}.

\begin{theorem} \label{thm1}
Let $I\in {C}^1(H,\mathbb{R})$ and
\begin{description}
\item{(a)} $\inf_{\| u \| \to \infty}I(u)=-\infty$
for $u\in H^-$
\item{(b)} $\lim_{\| u \| \to \infty}I(u)=+\infty$
for $u\in H^+$ and  I is bounded on bounded sets in
$H^+$
\item{(c)} $I$ satisfies Palais-Smale Condition (PS).
\end{description}
Then functional $I$ has a critical point in H.
\end{theorem}

In the original Saddle Point Theorem \cite{lit6},
instead of a) and b) above, the author assumes the following
two conditions
\begin{enumerate}
\item[$(\tilde a)$] There exists a bounded neighborhood $D$ of $0$
in $H^-$ and a constant
$\alpha $ such that $I/\partial D\le \alpha$,
\item[$(\tilde b)$] There is a constant $\beta > \alpha $ such that
$I/H^+ \ge \beta$.
\end{enumerate}
It is obvious, that conditions ($\tilde a$), ($\tilde b$)
follow from conditions (a), (b).

We investigate the boundary-value problem
\begin{eqnarray} \label{2.2}
&u''(x)+m^2u(x)+g(x,u(x))=f(x)\,,\quad x\in (0,\pi)\,,&\\
&u(0)=u(\pi)=0,&\nonumber
\end{eqnarray}
where $f\in L^1(0,\pi)$, $m\in \mathbb{N}$ and
$g:(0,\pi) \times \mathbb{R} \to \mathbb{R}$
is Caratheodory type function, i.e. $g(\cdot ,s)$ is measurable for all
$s\in \mathbb{R}$ and $g(x,\cdot)$ is continuous for a.e.
$x\in (0,\pi) $.

By a solution of (\ref{2.2}) we mean a function
$u\in {C}^1(0,\pi)$ such that $u'$ is absolutely
continuous, $u$
satisfies the boundary conditions and the equations (\ref{2.2}) holds a.e. in
$(0,\pi)$.

We study (\ref{2.2}) by using of varitional
methods. More precisely, we look for critical points of the functional
$J:H\to \mathbb{R}$, which is defined by
\begin{equation}
\label{2.3}
J(u)={1\over 2}\int_0^{\pi }\bigl[(u')^2-m^2u^2
\bigr]\,dx -
\int_0^{\pi }\bigl[ G(x,u)-fu\bigr] \,dx\,,
\end{equation}
where
$$G(x,s)=\int_0^{s}g(x,t)\,dt\,.$$
Every critical point $u\in H$ of the functional $J$ satisfies
$$\int_0^{\pi}\bigl[ u'v'-m^2uv \bigr] \,dx -
\int_0^{\pi }\bigl[g(x,u)v-fv\bigr] \,dx=0\quad
\mbox{for all}\quad v\in H\,.$$
Then $u$ is also a weak solution of (\ref{2.2}) and vice versa.
The usual regularity argument for ODE yields immediately (see Fu\v{c}\'{\i}k \cite{lit3})
that any weak solution of (\ref{2.2}) is also the solution in the sense
mentioned above.

We will suppose that $g$ satisfies the growth restriction
\begin{equation}
\label{2.4}
|g(x,s)|\le c|s|+p(x)\quad\mbox{for a.e. } x\in (0,\pi)
\quad\mbox{and  for all } s\in \mathbb{R}\,,
\end{equation}
with $p\in L^1(0,\pi)$ and $c>0$. Moreover,
\begin{equation}\label{2.5}
\lim_{|s|\to \infty }{{g(x,s)}\over {s}}=0 \quad
\mbox{uniformly for a.e. }
 x\in (0,\pi)\,.
\end{equation}
We define
$$G_+(x)=\liminf_{s\to +\infty}{{G(x,s)}\over s}\,,\quad
G_-(x)=\limsup_{s\to -\infty}{{G(x,s)}\over s}\,.$$
Assume that the following potential
Landesman-Lazer type condition holds:
\begin{eqnarray} \label{2.6}
\lefteqn{\int_0^{\pi}\bigl[ G_-(x)(\sin mx)^+  -
G_+(x)(\sin mx)^-\bigr] \,dx } \\
&<&\int_0^{\pi}f(x)\sin mx \,dx
< \int_0^{\pi}\bigl[ G_+(x)(\sin mx)^+ -
 G_-(x)(\sin mx)^-\bigr] \,dx\,. \nonumber
\end{eqnarray}
Set
$$\displaylines{
F(x,s)=\left\{\begin{array}{ll}
\displaystyle{{2G(x,s)}\over {s}}-g(x,s)& s\ne 0\,,\\
g(x,0)& s=0\,,
\end{array}\right.\cr
g_+(x)=\liminf_{s\to +\infty} {g(x,s)}\,,\quad
g_-(x)=\limsup_{s\to -\infty} {g(x,s)},\cr
F_+(x)=\liminf_{s\to +\infty}{F(x,s)}\,,\quad
F_-(x)=\limsup_{s\to -\infty}{F(x,s)}\,.
}$$
We generalize  conditions (\ref{1.3}) and
(\ref{1.4}) by assuming that $f$ and $g$ satisfy one of the following
set of inequalities: Either
\begin{eqnarray} \label{2.7}
\lefteqn{\int_0^{\pi}\bigl[ g_-(x)(\sin mx)^+  -
g_+(x)(\sin mx)^-\bigr] \,dx }\\
&<&\int_0^{\pi}f(x)\sin mx \,dx
< \int_0^{\pi}\bigl[ g_+(x)(\sin mx)^+ -
 g_-(x)(\sin mx)^-\bigr] \,dx\,, \nonumber
\end{eqnarray}
or
\begin{eqnarray} \label{2.8}
\lefteqn{\int_0^{\pi}\bigl[ F_-(x)(\sin mx)^+  -
F_+(x)(\sin mx)^-\bigr] \,dx }\\
&<& \int_0^{\pi}f(x)\sin mx \,dx
< \int_0^{\pi}\bigl[ F_+(x)(\sin mx)^+ -
 F_-(x)(\sin mx)^-\bigr] \,dx\,. \nonumber
\end{eqnarray}
We shall prove that for any $g$,
\begin{equation} \label{2.9}
g_+(x)\le G_+(x)\,,\quad F_+(x)\le
G_+(x)\,,\quad G_-(x)\le g_-(x)\,,\quad G_-(x)\le F_-(x)\,.
\end{equation}
Therefore, the potential Landesman-Lazer type condition
(\ref{2.6}) is more general than the conditions (\ref{2.7})
and (\ref{2.8}). \medskip

Let us prove the first two inequalities in (\ref{2.9}). The proof of
the other two is similar.
It follows from the definition of the function $g_+(x)$ that
$\forall\varepsilon>0$, $\exists R>0$ such that
$\forall s>R$, $ g(x,s)\ge g_+(x)-\varepsilon$ for $x\in(0,\pi)$.
Then for $s > R$ and $x\in(0,\pi)$ we have
$$\frac{G(x,s)}{s}\ge\frac 1s \int_0^R g(x,t)\,dt +
(g_+(x)-\varepsilon)\frac{s-R}{s}.$$
Hence, the inequality $g_+(x)\le G_+(x)$ follows.

We use the argument from C.-L.~Tang \cite{lit8}
and prove that $F_+(x)\le G_+(x)$.
It follows from (\ref{2.8}) that $F_+(x)>-\infty$ for a.e.
$x\in (0,\pi)$.
For these $x$ and arbitrary $\varepsilon > 0$, we set
$$
F_{\varepsilon}(x)=
\left\{
\begin{array}{ll}
F_+(x)-\varepsilon &\mbox{if } F_+(x)\in\mathbb{R}\,,\\[3pt]
1/\varepsilon &\mbox{if } F_+(x)=\infty\,.
\end{array}
\right.
$$
Then for any $\varepsilon>0$ there exists $K(x)>0$
such that
$$
F(x,s)\ge F_{\varepsilon}(x)\quad \mbox{for all}\quad
 s\ge K(x)\,.
$$
Since
$${{\partial}\over {\partial \tau}}\biggl(-{{G(x,\tau)}
\over {\tau^2}} \biggr)= -{{g(x,\tau)\cdot \tau^2-
2G(x,\tau)\cdot \tau}\over {\tau^4}}=\frac{F(x,\tau)}{\tau^2}
\ge\frac{F_{\varepsilon}(x)}{\tau^2}$$
for any $s>t>K(x)$, we have
$$\int_t^s{{\partial}\over {\partial \tau}}\biggl(
-{{G(x,\tau)}\over {\tau^2}} \biggr)\,d\tau
\ge\int_t^s\frac{F_{\varepsilon}(x)}{\tau^2}\,d\tau\,;$$
i.e.
$$
{{G(x,t)}\over {t^2}}-{{G(x,s)}\over {s^2}}
\ge F_{\varepsilon}(x)\bigl(t^{-1}-s^{-1}\bigr)\,.
$$
Assumption (\ref{2.5}) implies that
$G(x,s)/s^2 \to 0$. Since $\varepsilon>0$
was arbitrary, passing to the limit as $s\to+\infty$
in the last inequality, we obtain
$G(x,t)/t^2 \ge F_{\varepsilon}(x)/t$
and so $G_+(x)\ge F_+(x)$.

\paragraph{Example.}
We suppose that the nonlinearities $g_i(x,s)=g_i(s)$
($i=1,2,3,4$)
are continuous, odd and for all $s\ge \epsilon >0$:
$g_1(s)=1$, $g_2(s)=|\sin s|$, $g_3(s)={4\over {\pi }}-|\sin s|$,
$g_4(s)=1+\sin s$.
We set
\begin{eqnarray*}
&M_6=\{f\in L^1(0,\pi)\mbox{ such that $f$ satisfies
(\ref{2.6})}\},&\\
&M_7=\{f\in L^1(0,\pi)\mbox{ such that $f$ satisfies
(\ref{2.7})}\},\\
&M_{8}=\{f\in L^1(0,\pi)\mbox{ such that $f$ satisfies
(\ref{2.8})}\}.
\end{eqnarray*}
Then for $g_1$, one has $M_6=M_7=M_8\neq\emptyset$,
$g_+=F_+=G_+=1$; \\
for $g_2$, one has $M_7=\emptyset$,
$\emptyset \neq M_8 \subset \subset M_6$,
$g_+=0<F_+=\frac{4}{\pi}-1<G_+=\frac{2}{\pi}$;\\
for $g_3$, one has $M_8=\emptyset$,
$\emptyset\neq M_7 \subset \subset M_6$,
$F_+=0<g_+=\frac{4}{\pi}-1<G_+=\frac{2}{\pi}$;\\
for $g_4$, one has
$M_7= M_8=\emptyset$, $\emptyset \neq M_6$,
$g_+=F_+=0<G_+=1$.

\section{Main result}

\begin{theorem} \label{thm2}
Under the assumptions (\ref{2.4}), (\ref{2.5}), and (\ref{2.6}),
Problem (\ref {2.2}) has at least one solution in H.
\end{theorem}

\paragraph{Proof}
Firstly, we note that by  (\ref{2.5}),
\begin{equation} \label{3.4}
\lim_{\| u \|\to \infty}\int_0^{\pi}
{{G(x,u)-f u}\over {{\| u \| }^2}} \,dx\,= 0\,.
\end{equation}
We shall prove that the functional $J$ defined by (\ref{2.3})
satisfies the assumptions in Theorem~\ref{thm1} (Saddle Point Theorem).

For Assumption (a), we argue by contradiction. Suppose that
$$\inf_{\| u \| \to \infty}J(u)=-\infty \quad\mbox{for}
\quad u\in H^-$$
is not true.
Then there is a sequence $(u_n)\subset H^-$ such that
$\| u_n \| \to \infty$ and a constant $c_-$ satisfying
\begin{equation} %3.1
\label{3.1}
\liminf_{n\to \infty}J(u_n)\ge c_-.
\end{equation}
 From the definition of $J$ and from (\ref{3.1}) it follows that
\begin{equation}   %3.2
\label{3.2}
\liminf_{n\to \infty}\biggl[ {1\over 2}\int_0^{\pi}
{{(u'_n)^2-m^2 u^2_n}\over {{\| u_n \| }^2}}\,dx
-\int_0^{\pi}
{{G(x,u_n)-f u_n}\over {{\| u_n \| }^2}} \,dx \biggr] \ge 0.
\end{equation}
For $u\in H^-$ we have
\begin{equation}   %3.3
\label{3.3}
\int_0^{\pi}[(u')^2-m^2 u^2]\,dx= {\| u \|}^2- m^2
{\| u \|}^2_2\le 0
\end{equation}
and the equality in (\ref{3.3}) holds only for $u=k\sin mx$,
$k\in \mathbb{R}$.
Set $v_n=u_n/\| u_n \|$.
Since $\dim H^-<\infty $ there is $v_0\in H^-$ such
that $v_n\to v_0$ strongly in $H^-$
(also strongly in $L^2(0,\pi))$.
Then (\ref{3.4}), (\ref{3.2}), and (\ref{3.3}) yield
$$v_0=k\sin mx\,,$$
where $k={1\over m}\sqrt{{2\over {\pi}}}$ or
$k=-{1\over m}\sqrt{{2\over {\pi}}}$, ($\| v_0\|=1$).
Let $k={1\over m}\sqrt{{2\over {\pi}}}$.
We divide (\ref{3.1}) by ${\| u_n \|}$ to obtain
\begin{equation}    %3.5
\label{3.5}
\liminf_{n\to \infty}\biggl[ {1\over 2}\int_0^{\pi}
{{(u'_n)^2-m^2 u^2_n}\over {{\| u_n \| }}}\,dx
-\int_0^{\pi}
{{G(x,u_n)-f u_n}\over {{\| u_n \| }}}\,dx \biggr] \ge 0\,.
\end{equation}
Because $u_n\in H^-$ the first integral in (\ref{3.5}) is less than
or equal to zero and we have
\begin{equation}
\label{3.6}
\liminf_{n\to \infty}\biggl[ \int_0^{\pi}
-{{G(x,u_n)}\over {u_n }}v_n \,dx \biggr]
+k\int_0^{\pi}f \sin mx \,dx \ge 0\,.
\end{equation}
We know that $v_n\to k\sin mx$, $k>0$ in $H^-$.
Because of the compact imbedding
$H^-\subset {C}(0,\pi)$,
we have
$v_n\to k\sin mx$ in ${C}(0,\pi)$
and we get
$$
\lim_{n\to \infty}u_n(x) = \left\{
\begin{array}{l}
+\infty\quad\mbox{for $x\in(0,\pi)$ such that
 $\sin mx>0$}\,,\\
-\infty\quad\mbox{for $x\in(0,\pi)$ such that
 $\sin mx<0$}\,.
\end{array}\right.
$$
We note that from (\ref{2.6}) it follows that there exist
constants  $R$, $r$ and functions
$A_+(x), A_-(x)\in L^1(0,\pi)$ such that
$A_+(x)\le G_+(x)$, $G_-(x)\le A_-(x)$
for a.e. $x\in(0,\pi)$ and for all
$s\ge R$, $s\le r$, respectively.
We obtain from Fatou's lemma and (\ref{3.6})
\begin{equation}
\label{3.7}
\int_0^{\pi} f(x)\sin mx \,dx \ge \int_0^{\pi}
\bigl[ G_+(x)(\sin mx)^+ -G_-(x)(\sin mx)^-\bigr]
\,dx\,,
\end{equation}
a contradiction to (\ref{2.6}). We proceed for the case
$k=-{1\over m} \sqrt{2\over {\pi}}$. Then Assumption a) of
Theorem 1 is verified. \smallskip

For Assumption (b), we prove that
$$\lim_{\| u \| \to
\infty}J(u)=\infty\quad\mbox{for all } u\in H^+$$
and that $J$ is bounded on bounded sets.
Because of the compact imbedding of H into ${C}(0,\pi)$
$(\| u \|_{{C}(0,\pi)}\leq c_1\| u \|)$, and of H
into L$^2(0,\pi)$ $(\| u \|_2\leq c_2 \| u \|)$,
and the assumption (\ref{2.4}) one has
\begin{eqnarray} \label{3.8}
\int_0^{\pi}\Bigl[G(x,u(x))-f(x)u(x) \Bigr]
\,dx &\leq& \int_0^{\pi}\Bigl[c\frac{u^2(x)}{2}
+ p(x)|u(x)| - f(x)u(x)\Bigr]\,dx \nonumber \\
 &\leq& c_2\frac{c}{2}\| u \|^2+(\| p \|_1+\| f \|_1)
c_1\| u \|\,.
\end{eqnarray}
Hence $J$ is bounded on bounded subsets of H.

Since $u\in H^+$, we have
\begin{equation}   %3.9
\label{3.9}
\| u \|^2\ge(m+1)^2\| u
\|_2^2\,.
\end{equation}
The definition of $J$, (\ref{3.4}), and (\ref{3.9}) yield
\begin{equation}   %3.10
\label{3.10}
\lim_{\| u \|\to \infty} \frac{J(u)}{\| u \|^2}
\ge \lim_{\| u \|\to \infty}\frac{1}{2}(2m+1)
\frac{\| u\|_2^2}{\| u \|^2}\,.
\end{equation}
If $\| u\|_2^2/ \| u \|^2 \to 0$
then it follows from the definition of $J$ and (\ref{3.4}) that
\begin{equation}   %3.11
\label{3.11}
\lim_{\| u \|\to \infty}
\frac{J(u)}{\| u \|^2}
= \frac{1}{2}\,.
\end{equation}
Then (\ref{3.10}) and (\ref{3.11}) imply
$\lim_{\| u \|\to \infty} J(u)= \infty$; therefore,
Assumption b) of Theorem 1 is satisfied. \smallskip

For Assumption (c), we show that $J$ satisfies the Palais-Smale condition.
First, we suppose that the sequence $(u_n)$ is unbounded and
there exists a constant $c_3$ such that
\begin{equation}   \label{3.12}
\Bigl| {1\over 2}\int_0^{\pi}
\bigl[(u'_n)^2-m^2 u^2_n\bigr] \,dx
-\int_0^{\pi}
\bigl[G(x,u_n)-f u_n\bigr]\,dx \Bigr| \le c_3
\end{equation}
and
\begin{equation}   %3.13
\label{3.13}
\lim_{n\to \infty}\| J'(u_n)
\|=0\,.
\end{equation}
Let $(w_k)$ be an arbitrary sequence bounded in $H$.
It follows from (\ref{3.13}) and the Schwarz inequality that
\begin{eqnarray} \label{3.14}
\lefteqn{ \bigl| \lim_{{n\to \infty}\atop{k\to \infty}}
\int_0^{\pi}[u_n'w_k'-m^2u_nw_k] \,dx
-\int_0^{\pi}[g(x,u_n)w_k-fw_k] \,dx \bigr| }\nonumber \\
&=&|\lim_{{n\to \infty}\atop{k\to \infty}}J'(u_n) w_k | \\
&\le& \lim_{{n\to \infty}\atop{k\to \infty}}
\| J'(u_n)\|\cdot\| w_k\| =0\,. \hspace{5cm} \nonumber
\end{eqnarray}
By (\ref{2.4}) and (\ref{2.5}) we obtain
\begin{equation}  %3.15
\label{3.15}
\lim_{{n\to \infty}\atop{k\to \infty}}
\int_0^{\pi}
\Bigl[{{g(x,u_n)}\over {\| u_n \| }}w_k
- {{f}\over {\| u_n \| }}w_k\Bigr] \,dx  =
0\,.
\end{equation}
Put $v_n=u_n/ \| u_n \|$. Due to compact imbedding
$H\subset L^2(0,\pi)$
there is $v_0\in H$ such that (up to subsequence)
$v_n\rightharpoonup v_0$ weakly in H,
$v_n\to v_0$ strongly in $L^2(0,\pi)$.
One has from (\ref{3.14}), (\ref{3.15})
\begin{equation}    %3.16
\label{3.16}
\lim_{{n\to \infty}\atop{k\to \infty}}
\int_0^{\pi}\Bigl[
{{u'_n}\over {\| u_n \|}}w_k'-
m^2{{u_n}\over {\| u_n \|}}w_k\Bigr]\,dx
=0
\end{equation}
and also
\begin{equation}    %3.17
\label{3.17}
\lim_{{{n\to \infty}\atop{m\to \infty}}
\atop{k\to \infty}}
\int_0^{\pi}\bigl[
(v_n-v_m)'w_k'-m^2(v_n-v_m)w_k\bigr] \,dx=0\,.
\end{equation}
We set $k=n$ and $w_k=v_n$, $k=m$ and $w_k=v_m$ in (\ref{3.17})
and subtract these equalities we get
\begin{equation}    %3.18
\label{3.18}
\lim_{{n\to \infty}\atop{m\to \infty}}
\Bigl[\| v_n-v_m\|^2-m^2\| v_n-v_m
\|^2_2 \Bigr]=0\,.
\end{equation}
Since $v_n\to v_0$ strongly in $L^2(0,\pi)$
then $\| v_n-v_m\|_2\to 0$. Since
 (\ref{3.18}) holds then $v_n$ is a Cauchy sequence
in $H$ and $v_n\to v_0$ strongly in $H$.
Hence it follows from (\ref{3.16}) and from the usual regularity
argument for ordinary differential equations (see
Fu\v{c}\'{\i}k \cite{lit3}) that either
$v_0=\frac{1}{m}\sqrt{\frac{2}{\pi}}\sin mx$ or
$v_0=-\frac{1}{m}\sqrt{\frac{2}{\pi}}\sin mx$ ($\| v_0 \|=1$).
Suppose that $v_0=\frac{1}{m}\sqrt{\frac{2}{\pi}}\sin mx$.
Setting $w_k=\sin mx$ in (\ref{3.14}), we get
\begin{equation}
\label{3.19}
\lim_{n\to \infty}
\int_0^{\pi}\bigl[ -g(x,u_n(x))\sin mx +
f(x)\sin mx\bigr] \,dx =0\,.
\end{equation}
Let $\overline H$ be the subspace of $H$
spanned by the eigenfunctions
$\sin x$, $\sin 2x$, \dots, $\sin(m-1)x$.
Then we write $v_n= \overline{v}_n+a_n\sin mx+\tilde{v}_n$,
where $\overline{v}_n\in {\overline H}$,
$\tilde{v}_n\in H^{+}$ and $a_n\in\mathbb{R}$
 (likewise $u_n= \overline{u}_n+\| u_n \|
 a_n\sin mx+\tilde{u}_n$).
If we set $k=n$ and $w_k= -\overline{v}_n+a_n\sin mx+\tilde{v}_n$ in
(\ref{3.14}), we get
\begin{eqnarray} \label{3.20}
&&\lim_{n\to \infty}  \Big\{ \int_0^{\pi}
\bigl[ u_n'(-\overline{v}_n+a_n\sin mx+\tilde{v}_n)'
-m^2u_n(-\overline{v}_n+a_n\sin
mx+\tilde{v}_n)\bigr]\,dx \nonumber \\
&&-\int_0^{\pi}
\bigl[ g(x,u_n)(-\overline{v}_n+a_n\sin mx+\tilde{v}_n)
-f(-\overline{v}_n+a_n\sin mx+\tilde{v}_n)\bigr]\,dx
\Big\} \nonumber\\
&&=0\,.
\end{eqnarray}
It follows from (\ref{3.19}) and (\ref{3.20})
that
\begin{eqnarray} \label{3.21}
\lim_{n\to \infty}\frac{1}{\| u_n \| }
 \Bigl\{\int_0^{\pi}
\bigl[ -u_n'\overline{u}_n'+m^2u_n\overline{u}_n
+u_n'\tilde{u}_n'-m^2u_n\tilde{u}_n\bigr]\,dx&& \\
-\int_0^{\pi}
\Bigl[ \frac{g(x,u_n)}{u_n}u_n(-\overline{u}_n+\tilde{u}_n)
-f(-\overline{u}_n+\tilde{u}_n)\Bigr]\,dx \Bigr\}&=&0\,.\nonumber
\end{eqnarray}
For $\overline{u}_n \in \overline H$ we have
 $\| \overline{u}_n \|^2\le(m-1)^2\|
\overline{u}_n\|_2^2$, and for $\tilde{u}_n \in H^+$
we have  $\| \tilde{u}_n \|^2\ge(m+1)^2\|
\tilde{u}_n\|_2^2$.
It follows from the orthogonality $\int_0^{\pi}
\overline{u}_n'\tilde{u}_n'\,dx=0$
that the first integral in (\ref{3.21}) satisfies
\begin{eqnarray} \label{3.21a}
\lefteqn{ \int_0^{\pi}
\bigl[ -u_n'\overline{u}_n'+m^2u_n\overline{u}_n
+u_n'\tilde{u}_n'-m^2u_n\tilde{u}_n\bigr]\,dx }\nonumber \\
&=& -\| \overline{u}_n \|^2+ m^2
\| \overline{u}_n \|_2^2 +
\| \tilde{u}_n \|^2- m^2
\| \tilde{u}_n \|_2^2\\
&\ge& -\| \overline{u}_n \|^2+ \frac{m^2}{(m-1)^2}
\| \overline{u}_n \|^2+\| \tilde{u}_n \|^2- \frac{m^2}{(m+1)^2}
\| \tilde{u}_n \|^2 \nonumber\\
&=&\frac{2m-1}{(m-1)^2}\| \overline{u}_n \|^2+
\frac{2m+1}{(m+1)^2}\| \tilde{u}_n \|^2\,.\nonumber
\end{eqnarray}
It follows from (\ref{2.5}) and (\ref{2.4}) that
$\forall \varepsilon>0$, $\exists R>0$ such that for a.e. $x\in
(0,\pi)$ and all $|s|>R$,
$$\left|\frac{g(x,s)}{s}\right|< \varepsilon \,.$$
Also for a.e.  $x\in (0,\pi)$, and all $|s|\le R$,
$$|g(x,s)|\le c R +p(x)\,.$$
It follows from the imbedding
$H\subset L^2(0,\pi)$ ($\| u \|_2\leq\| u \|$)
that the second integral in the equations (\ref{3.21})
satisfies
\begin{eqnarray} \label{3.21b}
\lefteqn{ \int_0^{\pi}
\Bigl[ \frac{g(x,u_n)}{u_n}u_n(-\overline{u}_n+\tilde{u}_n)
-f(-\overline{u}_n+\tilde{u}_n)\Bigr]\,dx} \\
&\le &\varepsilon (\| \overline{u}_n \|^2 +
\| \tilde{u}_n \|^2)+(cR+\| p \|_1 +\| f \|_1)
(\| \overline{u} \|_{{C}(0,\pi)}
+\| \tilde{u} \|_{{C}(0,\pi)})\,.\nonumber
\end{eqnarray}
It follows from (\ref{3.21}), (\ref{3.21a}), (\ref{3.21b}),
and the imbedding
$H\subset {C}(0,\pi)$ $(\| u \|_{{C}(0,\pi)}\leq c_1\| u \|)$  that there are constants $\varrho_1>0$, $\varrho_2>0$
such that
$$
0\ge \lim_{n\to \infty} \frac{1}{\| u_n \| }
\bigl[ \varrho_1\| \overline{u}_n
\|^2 + \varrho_2\| \tilde{u}_n \|^2
 -(cR+\| p \|_1+\| f \|_1)c_1(
 \| \overline{u}_n\| +  \| \tilde{u}_n \|)\bigr]\,.
$$
Let $u_n^{\perp}=\overline{u}_n+\tilde{u}_n$. Then it holds
  $\| u_n^{\perp}\|^2 = \|\overline{u}_n\|^2+ \|\tilde{u}_n\|^2$
and $\|\overline{u}_n\|+
\|\tilde{u}_n\|\; \le \sqrt{2} \| u_n^{\perp}\|$\,.
Since there are constants $\varrho>0$ and $\beta>0$ such that
$$
0\ge \lim_{n\to \infty} \frac{1}{\| u_n \| }
\bigl(\varrho \| u_n^{\perp}\|^2 - \beta \| u_n^{\perp} \|\bigr)\,.
$$
Therefore,
\begin{equation} \label{3.23}
0=\lim_{n\to \infty}
\frac{\| u_n^{\perp}\|^2}
{\| u_n \|}\,.
\end{equation}
Now we divide (\ref{3.12}) by $\| u_n \|$. We get
\begin{equation}
\label{3.24}
\lim_{n\to \infty}\biggl\{\frac{1}{2}
\int_0^{\pi}
\Bigl[{{u'_n}\over {\| u_n \|}} u_n'-
m^2{{u_n}\over {\| u_n
\|}}u_n\Bigr] \,dx -
\int_0^{\pi}
{{G(u_n)-fu_n}\over {\| u_n \|}}
\,dx\biggr\}=0\,.
\end{equation}
We obtain from (\ref{3.23}) the following
equality
\begin{eqnarray} \label{3.25}
\lefteqn{\lim_{n\to \infty} \frac{1}{2}
\int_0^{\pi}\Bigl[{{u'_n}\over {\| u_n \|}} u_n'-
m^2{{u_n}\over {\| u_n \|}}u_n\Bigr] \,dx }\\
&=& \lim_{n\to \infty}\frac{1}{2} \int_0^{\pi}
\Bigl[{{({(u_n^{\perp})'})^2}\over {\| u_n \|}}
- m^2{{({u_n^{\perp}})^2}\over {\| u_n\|}}
\Bigr] \,dx = 0\,.\nonumber
\end{eqnarray}
We know that $v_n\to k\sin mx$, $k>0$ in $H$.
Due to compact imbedding $H\subset {C}(0,\pi)$
we have $v_n\to k\sin mx$ in ${C}(0,\pi)$
and we get
$$
\lim_{n\to \infty}u_n(x) = \left\{
\begin{array}{ll}
+\infty & \mbox{for $x\in(0,\pi)$ such that
 $\sin mx>0$}\,,\\[2pt]
-\infty & \mbox{for $x\in(0,\pi)$ such that
 $\sin mx<0$}\,.
\end{array}\right.
$$
Using Fatou's lemma, (\ref{3.24}), and (\ref{3.25}) we conclude
\begin{equation}
\label{3.26}
\int_0^{\pi} f(x)\sin mx \,dx \ge
\int_0^{\pi}
\bigl[ G_+(x)(\sin mx)^+ -G_-(x)(\sin mx)^-\bigr]
\,dx\,.
\end{equation}
This is a contradiction to (\ref{2.6}).
This implies that the sequence $\{\| u_n\|\}$
is bounded. Then there exists $u_0\in H$
such that $\| u_n\|\rightharpoonup u_0$ in $H$,
$\| u_n\|\to u_0$ in $L^2(0,\pi)$, ${C}(0,\pi)$
 (taking a subsequence if it is necessary).
It follows from the equality (\ref{3.14}) that
\begin{eqnarray} \label{3.27}
 \lim_{{{n\to \infty}\atop{m\to \infty}}\atop{k\to \infty}}
\Bigl\{\int_0^{\pi}
\bigl[(u_n-u_m)'w_k'-m^2(u_n-u_m)w_k\bigr] \,dx &&\\
 -\int_0^{\pi}
\bigl[g(x,u_n)-g(x,u_m)\bigr]w_k \,dx\Bigr\}
&=&0\,.\nonumber
\end{eqnarray}
The strong convergence $u_n\to u_0$ in ${C}(0,\pi)$ and the
assumption  (\ref{2.4}) imply
\begin{equation}    %3.28
\label{3.28}
\lim_{{n\to \infty}\atop{m\to \infty}}
\int_0^{\pi}
\bigl[g(x,u_n)-g(x,u_m)\bigr]\bigl(u_n-u_m\bigr) \,dx
=0\,.
\end{equation}
If we set $w_k=u_n$, $w_k=u_m$ in (\ref{3.28})
and subtract these equalities, then
\begin{equation}    %3.29
\label{3.29}
\lim_{{n\to \infty}\atop{m\to \infty}}
\int_0^{\pi}[
(u'_n-u'_m)^2-m^2(u_n-u_m)^2] \,dx
=0\,.
\end{equation}
Hence the strong convergence $u_n \to u_0$
in $L^2(0,\pi)$ and (\ref{3.29}) imply
the strong convergence $u_n \to u_0$ in $H$.
This shows that $J$ satisfies Palais-Smale condition and
the proof of Theorem 2 is complete.


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}\end{thebibliography}

\noindent {\sc Petr Tomiczek} \\
Department of Mathematics, University of West Bohemia,\\
Univerzitn\'{\i} 22, 306 14 Plze\v{n}, Czech Republic \\
e-mail: tomiczek@kma.zcu.cz

\end{document}