\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 164, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/164\hfil Infinitely many solutions]
{Infinitely many solutions for a fourth-order boundary-value problem}

\author[S. M. Khalkhali, S. Heidarkhani, A. Razani \hfil EJDE-2012/164\hfilneg]
{Seyyed Mohsen Khalkhali, Shapour Heidarkhani, Abdolrahman Razani} 

\address{Seyyed Mohsen Khalkhali \newline
 Department of Mathematics, Science and Research branch,
 Islamic Azad University, Tehran, Iran}
\email{sm.khalkhali@srbiau.ac.ir}

\address{Shapour Heidarkhani \newline
 Department of Mathematics, Faculty of Sciences,
 Razi University, 67149 Kermanshah, Iran}
\email{s.heidarkhani@razi.ac.ir}

\address{Abdolrahman Razani \newline
Department of Mathematics, Science and Research Branch,
Islamic Azad University, Tehran, Iran \newline
School of Mathematics, Institute for Research in Fundamental
Sciences (IPM), P. O. Box 19395-5746, Tehran, Iran}
\email{razani@ikiu.ac.ir}

\thanks{Submitted June 28, 2012 Published September 22, 2012.}
\thanks{A. Razani was  supported by grant  91470122 from IPM}
\subjclass[2000]{34B15}
\keywords{Fourth-order equation; infinitely many solutions; critical point theory;
\hfill\break\indent variational methods}

\begin{abstract}
 In this article we consider the existence of infinitely many
 solutions to the fourth-order boundary-value problem
 \begin{gather*}
  u^{iv}+\alpha u''+\beta(x) u=\lambda f(x,u)+h(u),\quad x\in]0,1[\\
  u(0)=u(1)=0,\\
  u''(0)=u''(1)=0\,.
 \end{gather*}
 Our approach is based  on variational methods and critical point theory.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

The deformations of an elastic beam in equilibrium, whose two ends are
simply supported, can be described by the nonlinear fourth-order boundary-value
problem \
\begin{gather*}
  u^{iv}=g(x,u,u',u''),\quad x\in]0,1[\\
  u(0)=u(1)=0,\\
  u''(0)=u''(1)=0,
\end{gather*}
where $g:[0,1]\times\mathbb{R}\times\mathbb{R}\times\mathbb{R}\to\mathbb{R}$
is continuous \cite{G1,G2}. The importance of existence and multiplicity
of solutions of this problem for physicists puts it and its variants
at the center of attention of many works in mathematics.
The fourth-order boundary-value problem
\begin{gather*}
  u^{iv}+\alpha u''+\beta u=\lambda f(x,u),\quad x\in]0,1[\\
  u(0)=u(1)=0,\\
  u''(0)=u''(1)=0
\end{gather*}
where $\alpha, \beta$ are some real constants, is the subject of many recent
researches by different approaches (See \cite{LL1, LL2, TC, BD1, BD3, GST}).
In \cite{LL1, TC} the authors by means of
a version of Mountain-Pass Theorem of Rabinowitz \cite[Theorem 9.12]{Ra}
obtain their results and in \cite{LL2} by decomposition of operators
shown by Chen, and in \cite{BD3} by means of a Variational theorems
of Ricceri and Bonanno, and in \cite{GST} by means of Morse Theory.

In this work, by employing Ricceri's Variational Principle
\cite[Theorem 2.5]{R1} and applying the similar methods used in
 \cite{BD3}, albeit with different calculations that it
 seems practically has significant difference with respect to \cite{BD3},
we ensure the existence of infinitely many
solutions for
\begin{equation}
\begin{gathered} %\tag{$B_\lambda$}
\label{p}
  u^{iv}+\alpha u''+\beta(x) u=\lambda f(x,u)+h(u),\quad x\in]0,1[\\
  u(0)=u(1)=0,\\
  u''(0)=u''(1)=0,
\end{gathered}
\end{equation}
where $\alpha$ is a real constant, $\beta(x)$ is a continuous function
 on $[0,1]$ and $\lambda$ is a positive parameter,
$f:[0,1]\times\mathbb{R}\to \mathbb{R}$ is an
$L^2$-Carath\'eodory function and
$h:\mathbb{R}\to \mathbb{R}$ be a Lipschitz continuous function
 with the Lipschitz constant $L\geqslant 0$; i.e.,
\begin{equation}\label{eq.8}
|h(t_1)-h(t_2)|\leqslant L|t_1-t_2|
\end{equation}
for all $t_1, t_2\in\mathbb{R}$, satisfying $h(0)=0$.

 To be precise,  using Ricceri's Variational
Principle \cite{R1} (see Theorem \ref{t.1}), under some appropriate
hypotheses on the behavior of the potential of $f$, under some
conditions on the potentials of $h$, at infinity, we establish the
existence of a precise interval of parameters $\Lambda$ such that,
for each $\lambda\in\Lambda$, the problem \eqref{p} admits a
sequence of weak solutions which are unbounded in the Sobolev
space $W^{2,2}([0,1])\cap W_0^{1,2}([0,1])$; see Theorem
\ref{t.2}. Further, replacing the conditions at infinity of the
potentials of $f$ and $h$, by a similar one at zero, the same
results hold and, in addition, the sequence of weak solutions
uniformly converges to zero; see Theorem \ref{t.3}.

 Existence of infinitely many
solutions for boundary value problems using Ricceri's Variational
Principle \cite{R1} and its variants has been
widely investigated (see \cite{BM2,MM}). We refer the reader to the papers
\cite{D,FJ,GHK,HH,K,R2,BD3}, and \cite{BM3}-\cite{C}.  
We refer the reader also to\cite{AHO,B,BD1,BD2,Ch,L,SR} and their references, 
in which fourth-order boundary value problems have been studied.

Recall that a function $f:[0,1]\times\mathbb{R}\to \mathbb{R}$
is said to be an \emph{$L^2$-Carath\'eodory function}, if
\begin{itemize}
  \item[(C1)] the function $x\to f(x,t)$ is measurable for every $t\in\mathbb{R}$;
  \item[(C2)] the function $t\to f(x,t)$ is continuous for almost every $x\in [0,1]$;
  \item[(C3)] for every $\rho>0$ there exists a function $\ell_\rho\in L^2([0,1])$ such that
  \[
\sup_{|t|\leqslant\rho}|f(x,t)|\leqslant\ell_\rho(x)\quad\text{for a.e.}\ x\in[0,1].
\]
\end{itemize}
 A special case of our main result is the following theorem.

\begin{theorem}\label{1.1}
Let $f:[0,1]\times\mathbb{R}\to \mathbb{R}$
be a non-negative continuous function and denote by
$F(x,\xi)$ its antiderivative with respect to its second argument at
 any $x\in [0,1]$ such that $F(x,0)=0$. Assume that
$\ell_\xi\in L^2([0,1])$ satisfies  condition {\rm (C3)} for every
$\xi>0$. Suppose $\pi^4>|\alpha|\pi^2+\|\beta\|_\infty+L$ and
 $$
\liminf_{\xi\to +\infty}\frac{\|\ell_\xi\|_2}{\xi}=0\quad \text{and}\quad
\limsup_{\xi\to +\infty}\frac{\int_{a}^{b}F(x,\xi)dx}{\xi^{2}}=+\infty
$$
for some
$[a,b]\subset ]0,1[$ then, the problem
\begin{gather*}
  u^{iv}+\alpha u''+\beta(x) u=f(x,u)+h(u),\quad x\in]0,1[\\
  u(0)=u(1)=0,\\
  u''(0)=u''(1)=0
\end{gather*}
 admits a sequence of pairwise distinct classical solutions.
\end{theorem}

Our main tool to investigate the existence of infinitely many
solutions for the problem \eqref{p} is a smooth version of
\cite[Theorem 2.1]{BM2} which is a more precise version of Ricceri's
Variational Principle \cite[Theorem 2.5]{R1},
which we now recall.

\begin{theorem}\label{t.1}
Let $X$ be a reflexive real Banach space and $\Phi ,\Psi :X\to \mathbb{R}$
be two G\^ateaux differentiable functionals
such that $\Phi $ is sequentially weakly lower semicontinuous, strongly continuous,
and coercive and $\Psi$
is sequentially weakly upper semicontinuous. For every $r>\inf_X\Phi$ put
\[
\varphi (r):=\inf_{u\in\Phi^{-1}(]-\infty,r[)}
\frac{\sup_{v\in\Phi^{-1}(]-\infty,r])}\Psi(v)
-\Psi(u)}{r-\Phi(u)}\]
and
\[
\gamma:=\liminf_{r\to +\infty}\varphi (r),\quad
\delta:=\liminf_{r\to (\inf_X\Phi)^+}\varphi(r).
\]
Then
\begin{itemize}
  \item[(a)] for every $r>\inf_X\Phi$ and every $\lambda\in ]0, \frac{1}{\varphi(r)}[$
 the restriction of the functional $I_\lambda=\Phi-\lambda\Psi$ to
 $\Phi^{-1}(]-\infty,r[)$ admits a global minimum, which is a critical
point (local minimum) of $I_\lambda$ in $X$;

  \item[(b)] if $\gamma<+\infty$ then for every $\lambda\in]0,\frac{1}{\gamma}[$
 either $I_\lambda$ has a global minimum or there is a sequence $\{u_n\}$ of
critical points (local minimum) of $I_\lambda$ such that
  \[
\lim_{n\to +\infty}\Phi(u_n)=+\infty;
\]

  \item[(c)] if $\delta<+\infty$ then for every $\lambda\in]0,\frac{1}{\delta}[$
either there is a global minimum of $\Phi$ which is a local minimum of $I_\lambda$
or there is a sequence of pairwise distinct critical points (local minimum)
of $I_\lambda$ which weakly converges to a global minimum of $\Phi$.
\end{itemize}
\end{theorem}

\section{Preliminaries and basic lemmas}

 Hereafter, let $X=W^{2,2}([0,1])\cap W_0^{1,2}([0,1])$ with its
usual norm inherited from $W^{2,2}([0,1])$ and $\|\cdot\|_2$ denotes the
usual norm of $L^2([0,1])$; i.e.,
\[
\|u\|_2=\Big(\int_0^1|u(x)|^2dx\Big)^{1/2}.
\]
Since $\beta(x)$ in \eqref{p}, by assumption, is continuous on $[0,1]$,
there exist $\beta_1,\beta_2\in\mathbb{R}$ such that
\[
\beta_1\leqslant\beta(x)\leqslant\beta_2
\]
for every $x\in [0,1]$. Therefore,
\begin{align*}
\int_0^1|u''(x)|^2-\alpha|u'(x)|^2+\beta_1|u(x)|^2dx
&\leqslant\int_0^1|u''(x)|^2-\alpha|u'(x)|^2+\beta(x)|u(x)|^2dx\\
&\leqslant\int_0^1|u''(x)|^2-\alpha|u'(x)|^2+\beta_2|u(x)|^2dx.
\end{align*}
 We need the following Poincar\'e type inequality.

\begin{lemma}[{\cite[Lemma 2.3]{PTV}}] \label{l.4}
For every $u\in X$
\begin{equation}\label{eq.12}
\|u\|_2\leqslant\frac{1}{\pi^2}\|u''\|_2.
\end{equation}
\end{lemma}

 From which we have as a consequence
\begin{equation}\label{eq.13}
\|u'\|_2\leqslant\frac{1}{\pi}\|u''\|_2.
\end{equation}
Now put
\begin{itemize}
\item $\sigma_1:=1-\frac{\alpha}{\pi^2}+\frac{\beta_1}{\pi^4}$,
  $\sigma_2:=1$ when $\beta_2\leqslant 0$ and $\alpha\geqslant 0$;
\item $\sigma_1:=1+\frac{\beta_1}{\pi^4}$, $\sigma_2:=1-\frac{\alpha}{\pi^2}$
  when $\beta_2\leqslant 0$ and $\alpha< 0$;
\item $\sigma_1:=1-\frac{\alpha}{\pi^2}$ and $\sigma_2:=1+\frac{\beta_2}{\pi^4}$
 when $\beta_1\geqslant 0$ and $\alpha\geqslant 0$;
\item $\sigma_1:=1$ and $\sigma_2:=1-\frac{\alpha}{\pi^2}+\frac{\beta_2}{\pi^4}$
 when $\beta_1\geqslant 0$ and $\alpha< 0$;
\item $\sigma_1:=1-\frac{\alpha}{\pi^2}+\frac{\beta_1}{\pi^4}$ and
 $\sigma_2:=1+\frac{\beta_2}{\pi^4}$ when $\beta_1<0<\beta_2$
 and $\alpha\geqslant 0$;
\item $\sigma_1:=1+\frac{\beta_1}{\pi^4}$ and
 $\sigma_2:=1-\frac{\alpha}{\pi^2}+\frac{\beta_2}{\pi^4}$ when
 $\beta_1<0<\beta_2$ and $\alpha< 0$.
\end{itemize}
In each of these cases, if $\sigma_1>0$ and
\begin{equation}\label{eq.19}
\theta_i:=\sqrt{\sigma_i}\quad (i=1,2)
\end{equation}
then by \eqref{eq.12} and \eqref{eq.13}
\begin{equation}\label{eq.14}
\theta_1\|u''\|_2\leqslant\|u\|\leqslant\theta_2\|u''\|_2
\end{equation}
where
\[
\|u\|=\Big(\int_0^1\big(|u''(x)|^2-\alpha|u'(x)|^2+\beta(x)|u(x)|^2\big)dx\Big)^{1/2}
\]
and so, $\|\cdot\|$ defines a norm on $X$ equivalent to usual norm of $X$
inherited from $W^{2,2}([0,1])$.

 In the remainder, we suppose $\theta_1$ defined by \eqref{eq.19} satisfies
$\theta_1>0$ and therefore \eqref{eq.14} holds.
The following result is useful for proving our main result.

\begin{proposition}\label{l.1}
For every $u\in X$.
\[
\|u\|_\infty\leqslant\frac{1}{2\pi\theta_1}\|u\|
\]
\end{proposition}

\begin{proof}
Similar to the proof of \cite[Proposition 2.1]{BD1}, considering \eqref{eq.13}
and \eqref{eq.14} and using well-known
inequality $\|u\|_\infty\leqslant\frac{1}{2}\|u'\|_2$ yields the conclusion.
\end{proof}

 A function $u:[0,1]\to\mathbb{R}$ is said a
\emph{generalized solution to the problem} \eqref{p}, if $u\in
C^3([0,1])$, $u'''\in AC([0,1])$, $u(0)=u(1)=0,\ u''(0)=u''(1)=0$
and $u^{iv}+\alpha u''+\beta u=\lambda f(x,u)+h(u)$. If $f$ is
continuous in $[0,1]\times \mathbb{R}$, then each generalized
solution of the problem \eqref{p} is a classical one. Standard
methods (see \cite[Proposition 2.2]{BD1}) show that a weak
solution to \eqref{p} is a generalized one when $f$ is an
$L^2$-Carath\'eodory function.

We define
\begin{equation}\label{eq.10}
 F(x,\xi)=\int_0^{\xi}f(x,t)dt\quad\text{and}\quad H(\xi)=\int_0^{\xi}h(x)dx
\end{equation}
 for every $x\in [0,1]$ and $\xi\in\mathbb{R}$.

\begin{lemma}\label{l.2}
Suppose $h:\mathbb{R}\to\mathbb{R}$ satisfies \eqref{eq.8} and $H(\xi)$
defined by \eqref{eq.10} for every $\xi\in\mathbb{R}$.
Then the functional $\Theta:X\to\mathbb{R}$ defined by
\begin{equation}\label{eq.11}
  \Theta(u):=\int_0^1H(u(x))dx
\end{equation}
is a G\^{a}teaux differentiable sequentially weakly continuous functional on $X$
with compact derivative
\[\Theta'(u)[v]=\int_0^1h(u(x))v(x)dx\] for every $v\in X$.
\end{lemma}
\begin{proof}
If $u_n\rightharpoonup u$ in $X$ then compactness of embedding
$X\hookrightarrow C([0,1])$ implies $u_n\to u$ in $C([0,1])$ i.e.
$u_n\to u$ uniformly on $[0,1]$ (see Proposition 2.2.4 of
\cite{Dr}). Hence, for some constant $M>0$ and any
$n\in\mathbb{N}$ we have $\|u_n\|_\infty\leqslant M$, and so
\begin{equation*}
  |H(u_n(x))-H(u(x))|dx\leqslant L\Big|\int_{u(x)}^{u_n(x)}|t|dt\Big|
  \leqslant\frac{L}{2}(|u_n(x)|^2+|u(x)|^2)
  \leqslant\frac{L}{2}(M^2+\|u\|_\infty^2)
\end{equation*}
for every $n\in\mathbb{N}$ and $x\in [0,1]$. Furthermore, $H(u_n(x))\to H(u(x))$
at any $x\in [0,1]$ and therefore,  the Lebesgue Convergence Theorem yields
\[
\Theta(u_n)=\int_0^1H(u_n(x))dx\to\int_0^1H(u(x))dx=\Theta(u).
\]
For proving G\^{a}teaux differentiability of $\Theta$ suppose
$u,v\in X$ and $t\neq 0$ then
\begin{align*}
&\Big|\frac{\Theta(u+tv)-\Theta(u)}{t}-\int_0^1h(u(x))v(x)dx\Big|\\
&\leqslant  \int_0^1\Big|\frac{H(u+tv)-H(u)}{t}-h(u(x))v(x)\Big|dx\\
 &=\int_0^1|h\big(u(x)+t\zeta(x)v(x)\big)-h(u(x))||v(x)|dx\\
 &\leqslant L\|v\|_\infty^2|t|
\end{align*}
in which $0<\zeta(x)<1$ for every $x\in [0,1]$. Therefore, $\Theta:X\to\mathbb{R}$ is a G\^{a}teaux differentiable at every $u\in X$ with derivative
\[
\Theta'(u)[v]=\int_0^1h(u(x))v(x)dx
\]
for every $v\in X$. Also, since
\begin{equation*}
 \big(\Theta'(u)-\Theta'(v)\big)[w]\leqslant L\int_0^1|u(x)-v(x)||w(x)|dx\leqslant
 \frac{L}{2\pi\theta_1}\|u-v\|_\infty\|w\|
\end{equation*}
for every three elements $u, v \text{ and }w$ of $X$, then
\[
\|\Theta'(u)-\Theta'(v)\|_{X^*}\leqslant\frac{L}{2\pi\theta_1}\|u-v\|_\infty
\]
which implies compactness of $\Theta':X\to X^*$.
\end{proof}

\begin{lemma}\label{l.3}
Let $f:[0,1]\times\mathbb{R}\to\mathbb{R}$ be an $L^2$-Carath\'eodory
function and $F(x,\xi)$ defined by \eqref{eq.10}.
Then $\Psi:X\to\mathbb{R}$ defined by
\begin{equation*}
 \Psi(u):=\int_0^1F\big(x,u(x)\big)dx
\end{equation*}
is a G\^{a}teaux differentiable sequentially weakly continuous functional on $X$.
\end{lemma}

\begin{proof}
If $u_n\rightharpoonup u$ in $X$, in Lemma \ref{l.2} was proved
that $u_n\to u$ uniformly on $[0,1]$ and there exists $M>0$ such
that $\|u_n\|_\infty\leqslant M$ for any $n\in\mathbb{N}$. Since
$F(x,\xi)$ is differentiable with respect to $\xi$ for a.e. $x\in
[0,1]$ so $F\big(x,u_n(x)\big)\to F\big(x,u_n(x)\big)$ a.e. on
$[0,1]$. Moreover, by the assumption (C3) on $f(x,t)$
\[F\big(x,u_n(x)\big)\leqslant M\ell_M(x)\]
and by the Lebesgue Convergence Theorem
\begin{equation*}
 \Psi(u_n)=\int_0^1F\big(x,u_n(x)\big)dx\to \int_0^1F\big(x,u(x)\big)dx=\Psi(u).
\end{equation*}
Therefore $\Psi$ is a sequentially weakly continuous functional on
$X$.
For proving the G\^{a}teaux differentiability of $\Psi$, let
$u,v\in X$ with $\|u\|<2\pi\theta_1 M$ and $\|v\|<2\pi\theta_1 M$ for
some $M>0$. Then for $t\neq 0$ by the Mean Value Theorem
\begin{align*}
&\Big|\frac{\Psi(u+tv)-\Psi(u)}{t}-\int_0^1 f(x,u(x))v(x)dx\Big|\\
&\leqslant\int_0^1 \big|f(x,u(x)+t\zeta(x)v(x))-f(x,u(x))\big||v(x)|dx\\
&\leqslant \|v\|_\infty \int_0^1 \big|f(x,u(x)+t\zeta(x)v(x))-f(x,u(x))\big|dx
\end{align*}
where $0<\zeta(x)<1$ for every $x\in [0,1]$ for which $F(x,\xi)$ is
 differentiable with respect to $\xi$. Since the assumption \textbf{C}$_2$ on $f(x,t)$ implies
\[
\lim_{t\to 0} f(x,u(x)+t\zeta(x)v(x))=f(x,u(x))\quad\text{for a.e. }x\in[0,1]
\]
and by Proposition \ref{l.1} we have $\|v\|_\infty \leqslant M$
and $\|u\|_\infty \leqslant M$, then by the assumption
(C3) on $f(x,t)$ we have
\[
|f(x,u(x)+t\zeta(x)v(x))-f(x,u(x))|\leqslant\ell_{2M}(x)+\ell_M(x)
\]
for any $|t|<1$. Therefore the Lebesgue Convergence Theorem implies
\[
\lim_{t\to 0}\frac{\Psi(u+tv)-\Psi(u)}{t}=\int_0^1 f(x,u(x))v(x)dx.
\]
Since for every $v\in X$, some constant $M>0$ can be found so that
both of inequalities $\|u\|<2\pi\theta_1 M$ and $\|v\|<2\pi\theta_1 M$
hold, thus $\Psi$ is G\^{a}teaux differentiable at every $u\in X$.
\end{proof}

\section{Main Results}

Our approach closely depends on the test function
$v_0\in X$ defined by
\begin{equation*}
v_0(x)= \begin{cases}
    \frac{2ax-x^2}{a^2} &\text{if }  x\in [0,a[,\\
 1 &\text{if }  x\in [a,b],\\
 \frac{2bx-x^2-2b+1}{(1-b)^2} &\text{if }  x\in ]b,1].
  \end{cases}
\end{equation*}
Let
  \begin{equation*}
    K(a,b):=\frac{4\pi^2\theta_1^2}{\|v_0\|^2}
  \end{equation*}
for every $0<a\leqslant b<1$, we get a positive continuous function
\begin{equation}\label{eq.20}
k(\epsilon):=\min\big\{K(a,b): a,b\in [\epsilon, 1-\epsilon],\ a\leqslant b\,\big\}
\end{equation}
which is defined for every $0<\epsilon<1/2$.

\begin{theorem}\label{t.2}
  Suppose that $L<\pi^4\theta_1^2$. Let $f:[0,1]\times\mathbb{R}\to\mathbb{R}$
be an $L^2$-Carath\'e-dory function  and $F(x,\xi)$ defined by \eqref{eq.10}.
Assume that $\ell_\xi\in L^2([0,1])$ satisfies {\rm(C3)} condition on $f(x,t)$ for every
$\xi>0$. Furthermore, suppose that there exist an interval
$[a,b]\subset [\epsilon,1-\epsilon]$ for some $0<\epsilon<\frac{1}{2}$
for which $k(\epsilon)$    defined by \eqref{eq.20} and two positive constants
$T$ and $p$ and a function $q\in L^2([0,1])$ such that
  \begin{itemize}
    \item[(i)] $f(x,t)\geqslant q(x)-p|t|$ for every
 $(x,t)\in \big([0,a]\cup [b,1]\big)\times\{t\in\mathbb{R}\,|\,t\geqslant T\}$;

    \item[(ii)] ${\liminf_{\xi\to \infty}
\frac{\|\ell_\xi\|_2}{(\pi^4\theta_1^2-L)\xi}
  <\frac{\pi k(\epsilon)}{2(\pi^4\theta_1^2+L)}
          \limsup_{\xi\to\infty}\frac{\int_a^bF(x,\xi)\,dx}{\xi^2}}$.
\end{itemize}
Then, for every
    \begin{equation*}
\lambda\in\Lambda:=\Big]\frac{2(\pi^4\theta_1^2+L)}{\pi^2k(\epsilon)}
\frac{1}{{\limsup_{\xi\to\infty}
    \frac{\int_a^bF(x,\xi)\,dx}{\xi^2}}},\
    \limsup_{\xi\to \infty}
 \frac{(\pi^4\theta_1^2-L)\xi}{\pi\|\ell_\xi\|_2}\Big[
  \end{equation*}
 problem \eqref{p} has an unbounded sequence of generalized
solutions in $X$.
  \end{theorem}

\begin{proof}
 Put
 \begin{equation*}
\Phi(u)=\frac{1}{2}\|u\|^2-\int_0^1H\big(u(x)\big)dx=\frac{1}{2}\|u\|^2-\Theta(u) ,
\quad \Psi(u)=\int_0^1F\big(x,u(x)\big)dx
 \end{equation*}
 for every $u\in X$. Since \eqref{eq.8} holds for every $t_1,t_2\in\mathbb{R}$ and
$h(0)=0$, we have $|h(t)|\leqslant L|t|$ for every
$t\in\mathbb{R}$, and so using \eqref{eq.12}, \eqref{eq.14} and Lemma \ref{l.2}
we obtain
\begin{equation}
  \Phi(u)\geqslant\frac{1}{2}\|u\|^2-\Big|\int_0^1H\big(u(x)\big)dx\Big|
  \geqslant\frac{1}{2}\|u\|^2-\frac{L}{2}\int_0^1|u(x)|^2dx
  \geqslant\Big(\frac{1}{2}-\frac{L}{2\pi^4\theta_1^2}\Big)\|u\|^2 \label{eq.4},
\end{equation}
and similarly
\begin{equation}
  \Phi (u)\leqslant\frac{1}{2}\|u\|^2+\Big|\int_0^1H\big(u(x)\big)dx\Big|\leqslant
  \Big(\frac{1}{2}+\frac{L}{2\pi^4\theta_1^2}\Big)\|u\|^2. \label{eq.5}
\end{equation}
Also, since $\Phi+\Theta$ is a continuous functional on $X$ and $\Theta$,
by Lemma \ref{l.2}, is a G\^{a}teaux
differentiable weakly continuous and therefore continuous functional on $X$
 then $\Phi$ is a continuous functional on $X$
and by a routine argument can be
proved that $\Phi$ is a G\^{a}teaux differentiable functional with the differential
\begin{equation*}
  \Phi'(u)[v]=\int_0^1[u''(x)v''(x)-\alpha u'(x)v'(x)+\beta(x) u(x)v(x)]dx
-\int_0^1h(u(x))v(x)dx
\end{equation*}
and it is sequentially weakly lower semicontinuous since $\Theta$
is sequentially weakly continuous, and if $u_n\rightharpoonup u$
in $X$ then
\begin{equation*}
\liminf_{n\to\infty}\Phi(u_n)
=\liminf_{n\to\infty}\frac{1}{2}\|u_n\|^2-\lim_{n\to\infty}\Theta(u_n)
\geqslant\frac{1}{2}\|u\|^2-\Theta(u)=\Phi(u).
\end{equation*}
 It is easy to see that the critical points of the functional
$I_\lambda=\Phi-\lambda\Psi$ and the weak solutions
 (and therefore generalized solutions) of the problem \eqref{p} are
the same and by Theorem \ref{t.1} we prove our
result.

Assume that $\{\xi_n\}_{n=1}^\infty$ is a sequence of
positive numbers such that $\xi_n\to \infty$ and
\[
\lim_{n\to \infty}\frac{\|\ell_{\xi_n}\|_2}{(\pi^4\theta_1^2-L)\xi_n}=
\liminf_{\xi\to
\infty}\frac{\|\ell_\xi\|_2}{(\pi^4\theta_1^2-L)\xi}
\]
and
let $r_n=\frac{2(\pi^4\theta_1^2-L)}{\pi^2}\xi_n^2$  then by \eqref{eq.4}
for any $v\in X$ such that $\Phi(v)<r_n$ we have
\begin{equation}\label{eq.2}
\|v\|\leqslant \pi^2\theta_1\sqrt{\frac{2\Phi(v)}{\pi^4\theta_1^2-L}}
< \pi^2\theta_1\sqrt{\frac{2r_n}{\pi^4\theta_1^2-L}}=2\pi\theta_1\xi_n
\end{equation}
and by Proposition \ref{l.1},
\begin{equation}\label{eq.15}
  \|v\|_\infty<\xi_n.
\end{equation}
On the other hand, by condition (C3) on $f(x,t)$ and \eqref{eq.15}
\[
|F(x,v(x))|\leqslant\Big|\int_0^{v(x)}\ell_{\xi_n}(x)dt\Big|
=|v(x)|\ell_{\xi_n}(x)
\]
and so by the H\"{o}lder inequality and Lemma \ref{l.4} and \eqref{eq.14}
\begin{equation}\label{eq.16}
|\Psi(v)|\leqslant\int_0^1|v(x)||\ell_{\xi_n}(x)|dx\leqslant
\frac{1}{\pi^2\theta_1}\|v\|\|\ell_{\xi_n}\|_2.
\end{equation}
Therefore, since $L<\pi^4\theta_1^2$, by \eqref{eq.4}
\begin{equation}\label{eq.17}
\sup_{v\in\Phi^{-1}(]-\infty,r_n[)}\Psi(v)
=\sup_{v\in\Phi^{-1}([0,r_n[)}\Psi(v)
\leqslant\frac{2\xi_n}{\pi}\|\ell_{\xi_n}\|_2
\end{equation}
and then by \eqref{eq.16} and \eqref{eq.17}
\begin{align*}
\varphi(r_n)
&\leqslant\inf_{u\in\Phi^{-1}([0,r_n[)}
\frac{\sup_{v\in\Phi^{-1}([0,r_n[)}\Psi(v)-\Psi(u)}{r_n-\Phi(u)}\\
&\leqslant\inf_{u\in\Phi^{-1}([0,r_n[)}
\frac{\|\ell_{\xi_n}\|_2}{\pi^2\theta_1}\frac{2\pi\theta_1\xi_n+\|u\|}{r_n-\Phi(u)}\\
&\leqslant\frac{\pi\|\ell_{\xi_n}\|_2}{(\pi^4\theta_1^2-L)\xi_n}
\end{align*}
and hence
  \begin{equation}
\gamma\leqslant\liminf_{\xi \to \infty}
    \frac{\pi\|\ell_{\xi}\|_2}{(\pi^4\theta_1^2-L)\xi}<+\infty.\label{eq.1}
  \end{equation}
Then \eqref{eq.1} in conjunction with the assumption (ii) imply
  \[
\Lambda\subset ]0,\frac{1}{\gamma}[
\]
  and by \eqref{eq.4} the functional $\Phi$ is coercive,
since $L<\pi^4\theta_1^2$. Therefore part \textit{b)} of Theorem \ref{t.1} implies
  either the functional $I_\lambda=\Phi-\lambda\Psi$ has a global minimum
or there exists a sequence $\{u_n\}$ of
  weak solutions of problem \eqref{p} such that $\lim_{n\to \infty}\|u_n\|=\infty$
for every $\lambda\in\Lambda$.

  Now we prove unboundedness of $I_\lambda$ from below under
condition (ii) and
thus the existence of infinitely many solutions of problem \eqref{p} is proved.
If $\lambda\in\Lambda$, then
\[
\frac{2(\pi^4\theta_1^2+L)}{\pi^2k(\epsilon)}<\lambda
\limsup_{\xi\to\infty}\frac{\int_a^bF(x,\xi)dx}{\xi^2}
\]
  and there exist a constant $c$ and a sequence of reals $\{\eta_n\}$ so that,
 $\eta_n\geqslant n$ and
  \[
\lim_{n\to\infty}\frac{\int_a^bF(x,\eta_n)dx}{\eta_n^2}=
  \limsup_{\xi\to\infty}\frac{\int_a^bF(x,\xi)dx}{\xi^2}
\]
  and in addition
\begin{equation}
    \frac{2(\pi^4\theta_1^2+L)}{\pi^2k(\epsilon)}<c
<\lambda\lim_{n\to\infty}\frac{\int_a^bF(x,\eta_n)dx}{\eta_n^2}.\label{eq.3}
  \end{equation}
    Let $\{v_n\}$ be a sequence in $X$ which is
defined by
  \begin{equation*}
   v_n(x)=v_0(x)\eta_n=
  \begin{cases}
 \eta_n \frac{2ax-x^2}{a^2} &\text{if }  x\in [0,a[,\\
 \eta_n &\text{if }  x\in [a,b],\\
 \eta_n\frac{2bx-x^2-2b+1}{(1-b)^2} &\text{if }  x\in ]b,1],
  \end{cases}
  \end{equation*}
then from \eqref{eq.20} and \eqref{eq.5} we observe that
  \begin{equation}\label{eq.18}
  \Phi(v_n)\leqslant\frac{\pi^4\theta_1^2+L}{2\pi^4\theta_1^2}\|v_n\|^2
  \leqslant\frac{2(\pi^4\theta_1^2+L)}{\pi^2k(\epsilon)}\eta_n^2.
  \end{equation}
On the other hand, by assumption (i),
\[
F(x,u(x))\geqslant -|q(x)||u(x)|-\frac{p|u(x)|^2}{2}-|u(x)|\ell_T(x)
\quad (x\in [0,a[\,\cup\,]b,1])
\]
and then the H\"{o}lder inequality, Lemma \ref{l.4} and
\eqref{eq.14} imply
\begin{equation}\label{eq.7}
\int_0^aF(x,u(x))dx+\int_b^1F(x,u(x))dx\geqslant
-\frac{2\|\ell_T\|_2+2\|q\|_2+p}{2\pi^2\theta_1}\|u\|.
\end{equation}
So by \eqref{eq.3}, \eqref{eq.18} and \eqref{eq.7}, there exists
 $N\in\mathbb{N}$ such that for any
  $n\geqslant N$
  \begin{align*}
   I_\lambda(v_n)
&=\Phi(v_n)-\lambda\Psi(v_n)\leqslant
\frac{2(\pi^4\theta_1^2+L)}{\pi^2k(\epsilon)}\eta_n^2-
          \lambda\int_0^1F(x,v_n(x))dx\\
&\leqslant\frac{2(\pi^4\theta_1^2+L)}{\pi^2k(\epsilon)}\eta_n^2-
\lambda\int_a^bF(x,\eta_n)dx+\frac{p+2\|q\|_2+2\|\ell_T\|_2}{2\pi^2\theta_1}\|v_n\|\\
   &\leqslant\Big(\frac{2(\pi^4\theta_1^2+L)}{\pi^2k(\epsilon)}-c\Big)\eta_n^2
+\frac{p+2\|q\|_2+2\|\ell_T\|_2}{\pi\sqrt{k(\epsilon)}}\eta_n.
  \end{align*}
Since $\lim_{n\to\infty}\eta_n=\infty$ then \eqref{eq.3} implies the functional
$I_\lambda$  is unbounded from below and the proof is completed.
\end{proof}

\begin{remark}\rm
  If in Theorem \ref{t.2} instead of \textit{i)} we assume $F(x,t)\geqslant 0$
for every $(x,t)\in [0,a[\,\cup\,]b,1]\times\mathbb{R}$
  then the assumption \textit{ii)} can be replaced by a more general one like
\begin{itemize}
\item[(ii')] There exist two sequences $\{a_n\}$ and
$\{b_n\}$ of real numbers such that $|a_n|\leqslant
b_n\sqrt{k(\epsilon)\frac{\pi^4\theta_1^2-L}{\pi^4\theta_1^2+L}}$
for every $n\in\mathbb{N}$ and ${\lim_{n\to\infty}a_n}=+\infty$ and
\[
\liminf_{n\to +\infty}\frac{\|\ell_{b_n}\|_2(b_n{\sqrt{k(\epsilon)}}+a_n)}
{\frac{\pi^4\theta_1^2-L}{\pi^4\theta_1^2+L}b_n^2k(\epsilon)-
a_n^2}<\frac{\pi {\sqrt{k(\epsilon)}}}{2}
\limsup_{\xi\to +\infty}\frac{\int_a^bF(x,\xi)dx}{\xi^2}.
\]
\end{itemize}
Obviously, from (ii') we
obtain (ii), by choosing $a_n=0$ for all $n\in\mathbb{N}$.
Moreover, if we assume \textit{ii')} instead of (ii) and set
$r_n=\frac{2(\pi^4\theta_1^2-L)b_n^2}{\pi^2}$ for all $n\in \mathbb{N}$,
by the same arguing as inside in Theorem \ref{t.2},
 we have
\begin{align*}
\varphi(r_n)&=
\inf\Big\{\frac{\sup_{v\in\Phi^{-1}(]-\infty,r_n[)}\Psi(v)-\Psi(u)}{r_n-\Phi(u)}
: u\in\Phi^{-1}(]-\infty,r_n[)\Big\}\\
&\leqslant\frac{\sup_{v\in\Phi^{-1}([0,r_n[)}\Psi(v)-\Psi(v_n)}{r_n-
\Phi(v_n)}\\
&\leqslant\frac{\|\ell_{b_n}\|_2}{\pi^2\theta_1}\frac{2\pi\theta_1b_n+\|v_n\|}
{\frac{2(\pi^4\theta_1^2-L)}{\pi^2}b_n^2-\frac{\pi^4\theta_1^2+L}{2\pi^4\theta_1^2}\|v_n\|^2}\\
&\leqslant\frac{\pi\|\ell_{b_n}\|_2}{\pi^4\theta_1^2+L}\frac{b_n+\frac{a_n}{\sqrt{k(\epsilon)}}}
{\frac{\pi^4\theta_1^2-L}{\pi^4\theta_1^2+L}b_n^2-
\frac{a_n^2}{k(\epsilon)}}
\end{align*}
where
 $$
v_n(x)=  \begin{cases}
    a_n \frac{2ax-x^2}{a^2} &\text{if }  x\in [0,a[,\\
 a_n &\text{if } x\in [a,b],\\
 a_n\frac{2bx-x^2-2b+1}{(1-b)^2} &\text{if } x\in ]b,1],
  \end{cases}
$$
Therefore $\gamma<\infty$. Similarly,
the second part of the proof of Theorem \ref{t.2} can be improved
so that the conclusion of the theorem can be obtained for the
interval
\[
\Lambda'=\Big]
\frac{2(\pi^4\theta_1^2+L)}
{\pi^2k(\epsilon){\limsup_{\xi\to +\infty}\frac{\int_a^bF(x,\xi)dx}{\xi^2}}},
\frac{\pi^4\theta_1^2+L}{\pi{\sqrt{k(\epsilon)}}}
\limsup_{n\to +\infty}\frac{\frac{\pi^4\theta_1^2-L}{\pi^4\theta_1^2+L}b_n^2k(\epsilon)-
a_n^2}{\|\ell_{b_n}\|_2\big(b_n{\sqrt{k(\epsilon)}}+a_n\big)}
\Big[
\]
instead of $\Lambda$.
\end{remark}

 Now we point out a simple consequence of Theorem \ref{t.2}.

\begin{corollary}\label{c1}
Suppose that $L<\pi^4\theta_1^2$.  Let
$f:[0,1]\times\mathbb{R}\to\mathbb{R}$ be an
$L^2$-Carath\'eodory function. Assume that $\ell_\xi\in L^2([0,1])$
satisfies (C3) condition on $f(x,t)$ for every
$\xi>0$ and there exists an interval
$[a,b]\subset [\epsilon,1-\epsilon]$ for some
$0<\epsilon<\frac{1}{2}$ such
  that assumption (i) in Theorem \ref{t.2} holds. Furthermore,
  suppose that
  \begin{itemize}
    \item[(iii)] ${\liminf_{\xi\to \infty}}\frac{\|\ell_\xi\|_2}{\xi}
<\frac{(\pi^4\theta_1^2-L)}{\pi}$;
    \item[(iv)]  ${\limsup_{\xi\to\infty}}
    \frac{\int_a^bF(x,\xi)\,dx}{\xi^2}>\frac{2(\pi^4\theta_1^2+L)}{\pi^2k(\epsilon)}$,
  \end{itemize}
then the problem
  \begin{equation}
\begin{gathered}
  u^{iv}+\alpha u''+\beta(x) u=h(u)+ f(x,u),\quad x\in(0,1)\\
  u(0)=u(1)=0,\\
  u''(0)=u''(1)=0
\end{gathered}
\end{equation}
has an unbounded sequence of generalized solutions in $X$.
  \end{corollary}

Note that Theorem \ref{1.1}  is an immediate
consequence of Corollary \ref{c1}.
Now we present the following example to illustrate our results.

\begin{example}\label{e1} \rm
Let $r>0$ be a real number and
$\{t_n\},\,\{s_n\}$ be two strictly
increasing sequences of reals that recursively defined by
\[
t_1=r,\ s_1=2r
\]
and for $n\geqslant 1$ by
\begin{gather*}
t_{2n}=(2^{2n+1}-1)t_{2n-1},\quad  t_{2n+1}=(2-\frac{1}{2^{2n+1}})t_{2n},\\
s_{2n}=\frac{t_{2n}}{2^n}=(2-\frac{1}{2^{2n}})s_{2n-1},\quad
s_{2n+1}=2^{n+1} t_{2n+1}=(2^{2n+2}-1)s_{2n}.
\end{gather*}
If $f:[0,1]\times\mathbb{R}\to\mathbb{R}$  be the function defined
as
\[
f(x,t)=\begin{cases}
2g(x)t&  (x,t)\in[0,1]\times [0,t_1],\\
g(x)\Big(s_{n-1}+\frac{s_n-s_{n-1}}{t_n-t_{n-1}}(t-t_{n-1})\Big)
& (x,t)\in[0,1]\times [t_{n-1},t_n] \\
&\text{for some } n>1
\end{cases}
\]
where $g:[0,1]\to \mathbb{R}$ is a positive continuous
function with $0<m\leqslant g(x)\leqslant M$.
Then $f(x,t)$ is an $L^2$-Carath\'eodory function and since $f(x,t)$ is
strictly increasing with respect to $t$ argument at every $x\in [0,1]$, the
function
$\ell_\xi(x):=f(x,\xi)$ satisfies in (C3) condition on $f(x,t)$; i.e.,
 \[
\sup_{|t|\leqslant\xi}|f(x,t)|\leqslant\ell_\xi(x)\quad \forall x\in[0,1].
\]
 Now we have
\begin{gather*}
\liminf_{\xi\to +\infty}\frac{\|\ell_\xi\|_2}{\xi}\leqslant\lim_{n\to\infty}
\frac{M s_{2n}}{t_{2n}}=0,
\\
\begin{aligned}
\limsup_{\xi\to +\infty}\frac{\int_{a}^bF(x,\xi)dx}{\xi^2}
&\geqslant \lim_{n\to\infty}\frac{m(b-a)(t_{2n+1}-t_{2n})(s_{2n+1}
 +s_{2n})}{2t_{2n+1}^2}\\
&\geqslant\lim_{n\to\infty}\frac{m(b-a)2^{3n+2}(2^{2n+1}-1)}{(2^{2n+2}-1)^2}
=+\infty
\end{aligned}
\end{gather*}
for every $[a,b]\subset [\epsilon,1-\epsilon]$ with $a<b$ and any
$0<\epsilon<\frac{1}{2}$. Hence by Corollary \ref{c1}, for every
$\lambda\in]0,+\infty[$, the boundary value problem
\begin{gather*}
 u^{iv}+\alpha u''+\beta(x) u= \lambda f(x,u)+u^{+},\\
u(0)=u(1)=0,\\
u''(0)=u''(1)=0
\end{gather*}
 where  $\alpha<\pi^2-\frac{1}{\pi^2}$ is a real constant and $\beta(x)$
is any non-negative continuous function on $[0,1]$
 and $u^+=\max\{u,0\}$, has an unbounded sequence of generalized solutions
in $X$ (for instance, $\alpha=9$ and $\beta(x)=\sin \pi x$).
\end{example}

 Arguing as in the proof of Theorem \ref{t.2} but using conclusion
(c) of Theorem \ref{t.1} instead of (b), the following result
holds.

  \begin{theorem}\label{t.3}
  Suppose that $L<\pi^4\theta_1^2$. Let
$f:[0,1]\times\mathbb{R}\to\mathbb{R}$ be an
$L^2$-Carath\'eo\-dory function. Assume that
$\ell_\xi\in L^2([0,1])$ satisfies in {\rm (C3)} condition on $f(x,t)$
for every $\xi>0$
and there exists an interval $[a,b]\subset [\epsilon,1-\epsilon]$ for some
  $0<\epsilon<\frac{1}{2}$ such that
    \begin{itemize}
      \item[(i)] $F(x,t)\geqslant 0$ for every
$(x,t)\in [0,a[\,\cup\,]b,1]\times\mathbb{R}$;

      \item[(ii)]
$$
{\liminf_{\xi\to 0^+}\frac{\|\ell_\xi\|_2}{(\pi^4\theta_1^2-L)\xi}
      <\frac{\pi k(\epsilon)}{2(\pi^4\theta_1^2+L)}
          \limsup_{\xi\to 0^+}\frac{\int_a^bF(x,\xi)\,dx}{\xi^2}}.
$$
    \end{itemize}
   Then, for every
    \begin{equation*}
      \lambda\in\Lambda:=\Big]\frac{2(\pi^4\theta_1^2+L)}{\pi^2k(\epsilon)}
\frac{1}{{\limsup_{\xi\to 0^+}
    \frac{\int_a^bF(x,\xi)\,dx}{\xi^2}}},\
    \limsup_{\xi\to 0^+}\frac{(\pi^4\theta_1^2-L)\xi}{\pi\|\ell_\xi\|_2}\Big[
    \end{equation*}
  Problem \eqref{p} has a sequence of non-zero generalized solutions in $X$
that  converges weakly to $0$.
  \end{theorem}

  \begin{proof}
   Since $\inf_X\Phi=\min_X\Phi=0$ as a consequence of \eqref{eq.4} and
the assumption $L<\pi^4\theta_1^2$. Exactly as in the proof of Theorem \ref{t.2}
 it can be shown that
   \[
\delta=\liminf_{r\to (\inf_X\Phi)^+}\varphi(r)\leqslant\frac{\pi}{\pi^4\theta_1^2-L}
\liminf_{\xi\to 0^+}
   \frac{\|\ell_\xi\|_2}{\xi}<+\infty
\]
   and therefore
   \[
\Lambda\subset ]0,\frac{1}{\delta}[.
\]
  If $\lambda\in\Lambda$ then
  \[
\frac{2(\pi^4\theta_1^2+L)}{\pi^2 k(\epsilon)}<\lambda\limsup_{\xi\to 0^+}
\frac{\int_a^bF(x,\xi)dx}{\xi^2}
\]
  and there exist a constant $c$ and a sequence of reals $\{\zeta_n\}$ so that,
 $\zeta_n\leqslant\frac{1}{n}$ and
  \[
\lim_{n\to\infty}\frac{\int_a^bF(x,\zeta_n)dx}{\zeta_n^2}=
  \limsup_{\xi\to 0^+}\frac{\int_a^bF(x,\xi)dx}{\xi^2}
\]
  and in addition
  \begin{equation}
    \frac{2(\pi^4\theta_1^2+L)}{\pi^2 k(\epsilon)}
<c<\lambda\lim_{n\to\infty}\frac{\int_a^bF(x,\eta_n)dx}{\eta_n^2}.\label{eq.6}
  \end{equation}
  Let $\{w_n\}$ be a sequence in $X$
defined by
  \begin{equation} \label{eq.9}
  w_n(x)=  \begin{cases}
  \zeta_n \frac{2ax-x^2}{a^2} &\text{if }  x\in [0,a[,\\	
  \zeta_n &\text{if }  x\in [a,b],\\
   \zeta_n\frac{2bx-x^2-2b+1}{(1-b)^2} &\text{if } x\in ]b,1],
  \end{cases}
  \end{equation}
then $\{w_n\}$ converges strongly to 0 in $X$ and by \eqref{eq.5}
  \begin{equation*}
  \Phi(w_n)\leqslant\frac{\pi^4\theta_1^2+L}{2\pi^4 \theta_1^2}\|w_n\|^2
\leqslant\frac{2(\pi^4\theta_1^2+L)}{\pi^2 k(\epsilon)}\zeta_n^2
  \end{equation*}
  hence, by (i) and the similar arguments as in the proof of Theorem \ref{t.2},
there exists $N\in\mathbb{N}$ such that for any
  $n\geqslant N$
  \begin{align*}
   I_\lambda(w_n)
&\leqslant \frac{2(\pi^4\theta_1^2+L)}{\pi^2 k(\epsilon)}\zeta_n^2
-\lambda\int_a^b F(x,\zeta_n)dx\\
   &\leqslant \Big(\frac{2(\pi^4\theta_1^2+L)}{\pi^2 k(\epsilon)}-c\Big)\zeta_n^2.
  \end{align*}
  Since $I_\lambda(0)=0$ therefore \eqref{eq.6} implies 0 is not a local minimum
of $I_\lambda$ and then according to (c) of Theorem \ref{t.1}
  there exists a sequence $\{u_n\}$ of local minimums of $I_\lambda$
that weakly converges to 0.
  \end{proof}

\begin{remark} \label{rmk3} \rm
 Since the embedding $X\hookrightarrow C([0,1])$ is compact, by
\cite[Proposition 2.2.4]{Dr}, every weakly convergent
  sequence in $X$ converges strongly in $C([0,1])$; i.e.,
 converges uniformly on $[0,1]$. Therefore the generalized solutions of the
  problem \eqref{p} established in Theorem \ref{t.3} converges uniformly
to zero on $[0,1]$.
  \end{remark}

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