\documentclass[twoside]{article} \usepackage{amssymb} % font used for R in Real numbers \pagestyle{myheadings} \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)} \vspace{\bigskipamount} \\ % 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}} \catcode@=11 \@addtoreset{equation}{section} \catcode@=12 \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 $$%r1.2 \lim_{s\to \infty}g(s)=g_+ \quad \mbox{and} \quad \lim_{s\to -\infty}g(s)=g_-$$ 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 $$\label{2.3} J(u)={1\over 2}\int_0^{\pi }\bigl[(u')^2-m^2u^2 \bigr]\,dx - \int_0^{\pi }\bigl[ G(x,u)-fu\bigr] \,dx\,,$$ 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 $$\label{2.4} |g(x,s)|\le c|s|+p(x)\quad\mbox{for a.e. } x\in (0,\pi) \quad\mbox{and for all } s\in \mathbb{R}\,,$$ with $p\in L^1(0,\pi)$ and $c>0$. Moreover, $$\label{2.5} \lim_{|s|\to \infty }{{g(x,s)}\over {s}}=0 \quad \mbox{uniformly for a.e. } x\in (0,\pi)\,.$$ 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 $g$, $$\label{2.9} g_+(x)\le G_+(x)\,,\quad F_+(x)\le G_+(x)\,,\quad G_-(x)\le g_-(x)\,,\quad G_-(x)\le F_-(x)\,.$$ 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 $M_7=\emptyset$, $\emptyset \neq M_8 \subset \subset M_6$, $g_+=00$ 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}) $$\label{3.7} \int_0^{\pi} f(x)\sin mx \,dx \ge \int_0^{\pi} \bigl[ G_+(x)(\sin mx)^+ -G_-(x)(\sin mx)^-\bigr] \,dx\,,$$ 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 $$%3.9 \label{3.9} \| u \|^2\ge(m+1)^2\| u \|_2^2\,.$$ The definition of $J$, (\ref{3.4}), and (\ref{3.9}) yield $$%3.10 \label{3.10} \lim_{\| u \|\to \infty} \frac{J(u)}{\| u \|^2} \ge \lim_{\| u \|\to \infty}\frac{1}{2}(2m+1) \frac{\| u\|_2^2}{\| u \|^2}\,.$$ If $\| u\|_2^2/ \| u \|^2 \to 0$ then it follows from the definition of $J$ and (\ref{3.4}) that $$%3.11 \label{3.11} \lim_{\| u \|\to \infty} \frac{J(u)}{\| u \|^2} = \frac{1}{2}\,.$$ 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 $$\label{3.12} \Bigl| {1\over 2}\int_0^{\pi} \bigl[(u'_n)^2-m^2 u^2_n\bigr] \,dx -\int_0^{\pi} \bigl[G(x,u_n)-f u_n\bigr]\,dx \Bigr| \le c_3$$ and $$%3.13 \label{3.13} \lim_{n\to \infty}\| J'(u_n) \|=0\,.$$ 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 $$%3.15 \label{3.15} \lim_{{n\to \infty}\atop{k\to \infty}} \int_0^{\pi} \Bigl[{{g(x,u_n)}\over {\| u_n \| }}w_k - {{f}\over {\| u_n \| }}w_k\Bigr] \,dx = 0\,.$$ 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}) $$%3.16 \label{3.16} \lim_{{n\to \infty}\atop{k\to \infty}} \int_0^{\pi}\Bigl[ {{u'_n}\over {\| u_n \|}}w_k'- m^2{{u_n}\over {\| u_n \|}}w_k\Bigr]\,dx =0$$ and also $$%3.17 \label{3.17} \lim_{{{n\to \infty}\atop{m\to \infty}} \atop{k\to \infty}} \int_0^{\pi}\bigl[ (v_n-v_m)'w_k'-m^2(v_n-v_m)w_k\bigr] \,dx=0\,.$$ 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 $$%3.18 \label{3.18} \lim_{{n\to \infty}\atop{m\to \infty}} \Bigl[\| v_n-v_m\|^2-m^2\| v_n-v_m \|^2_2 \Bigr]=0\,.$$ 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 $$\label{3.19} \lim_{n\to \infty} \int_0^{\pi}\bigl[ -g(x,u_n(x))\sin mx + f(x)\sin mx\bigr] \,dx =0\,.$$ 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, $$\label{3.23} 0=\lim_{n\to \infty} \frac{\| u_n^{\perp}\|^2} {\| u_n \|}\,.$$ Now we divide (\ref{3.12}) by $\| u_n \|$. We get $$\label{3.24} \lim_{n\to \infty}\biggl\{\frac{1}{2} \int_0^{\pi} \Bigl[{{u'_n}\over {\| u_n \|}} u_n'- m^2{{u_n}\over {\| u_n \|}}u_n\Bigr] \,dx - \int_0^{\pi} {{G(u_n)-fu_n}\over {\| u_n \|}} \,dx\biggr\}=0\,.$$ 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 $$\label{3.26} \int_0^{\pi} f(x)\sin mx \,dx \ge \int_0^{\pi} \bigl[ G_+(x)(\sin mx)^+ -G_-(x)(\sin mx)^-\bigr] \,dx\,.$$ 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 $$%3.28 \label{3.28} \lim_{{n\to \infty}\atop{m\to \infty}} \int_0^{\pi} \bigl[g(x,u_n)-g(x,u_m)\bigr]\bigl(u_n-u_m\bigr) \,dx =0\,.$$ If we set $w_k=u_n$, $w_k=u_m$ in (\ref{3.28}) and subtract these equalities, then $$%3.29 \label{3.29} \lim_{{n\to \infty}\atop{m\to \infty}} \int_0^{\pi}[ (u'_n-u'_m)^2-m^2(u_n-u_m)^2] \,dx =0\,.$$ 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. \begin{thebibliography}{0} {\frenchspacing \bibitem{lit1}P. Dr\'{a}bek: {\em Solvability and bifurcations of Nonlinear Equations}, Pittman Research Notes, Longman, London 1992. \bibitem{lit2}D. G. de Figueiredo, W. M. Ni: {\em Pertubations of second order linear elliptic problems by nonlinearities without Landesman-Lazer conditions}, Nonlinear analysis, Theory, Methods \& Applications, Vol 3, No.5, 629--634, (1979). \bibitem{lit3}S. Fu\v{c}\'{\i}k: {\em Solvability of Nonlinear Equations and Boundary Value problems}, D.Reidel Publ. Company, Holland 1980. \bibitem{lit8}C.-L. Tang: Solvability for Two Boundary Value Problems, Journal of Mathematical Analysis and Applications 216, 368-374(1997). \bibitem{lit4} E. M. Landesman, A. C. Lazer: {\em Nonlinear perturbations of elliptic boundary value problems at resonance},J. Math. Mech. 19(1970), 609--623. \bibitem{lit5}D. Lupo, A. M. Micheletti: {\em Some remarks by variational methods on conditions of nonresonance for semilinear elliptic problems}, preprint Universit\'{a} di Pisa, 1988/11. \bibitem{lit6}P. Rabinowitz: {\em Minimax methods in critical point theory with applications to differential equations}, CBMS Reg. Conf. Ser. in Math. no 65, Amer. Math. Soc. Providence, RI., (1986). \bibitem{lit7}J. Santanilla: {\em Solvability of a nonlinear boundary value problem without Landesman-Lazer conditions}, Nonlinear analysis, Theory, Methods \& Applications, Vol 13, No. 6, 683--693, (1989). }\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}