\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 27, pp. 1--9.\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-2012/27\hfil Non-trivial solutions]
{Non-trivial solutions for two-point boundary-value problems of
fourth-order Sturm-Liouville type equations}

\author[S. Heidarkhani \hfil EJDE-2012/27\hfilneg]
{Shapour Heidarkhani} 

\address{Shapour Heidarkhani \newline
Department of Mathematics, Faculty of Sciences\\
Razi University,  67149 Kermanshah, Iran \newline
School of Mathematics, Institute for Research in Fundamental Sciences (IPM)\\
P.O. Box: 19395-5746, Tehran, Iran}
\email{sh.heidarkhani@yahoo.com, s.heidarkhani@razi.ac.ir}

\thanks{Submitted September 12, 2011. Published February 15, 2012.}
\subjclass[2000]{35J35, 47J10, 58E05}
\keywords{Fourth-order Sturm-Liouville type problem; multiplicity results;
\hfill\break\indent critical point theory}

\begin{abstract}
 Using  critical point theory due to Bonanno \cite{B}, we prove the
 existence of at least one non-trivial solution for a class of
 two-point boundary-value problems for fourth-order Sturm-Liouville type
 equations.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{example}[theorem]{example}

\section{Introduction}

In this note, we prove the existence of at least one
non-trivial solution for the two-point boundary-value problem of
fourth-order Sturm-Liouville type:
\begin{gather}\label{e1}
(p_i(x) u_i''(x))'' - (q_i(x) u_i'(x))' + r_i(x) u_i(x)
 =\lambda F_{u_i}(x,u_1,\dots,u_n) \quad x \in (0,1), \notag\\
u_{i}(0)=u_{i}(1)=u_{i}''(0)=u_{i}''(1)=0
\end{gather}
for $1\leq i\leq n$, where $n\geq 1$ is an integer, $p_i$, $q_i$,
$r_i \in L^{\infty}([0,1])$ with $p_i^-:= \operatorname{ess\,inf}_{x \in
[0,1]} p_i(x)> 0$ 
for $1\leq i\leq n$, $\lambda$ is a positive parameter, 
$F:[0,1]\times \mathbb{R}^n\to \mathbb{R}$ is a
function such that $F(.,t_1,\dots,t_n)$ is measurable in $[0,1]$
for all $(t_1,\dots,t_n)\in \mathbb{R}^n$, $F(x,.,\dots,.)$ is
$C^1$ in $\mathbb{R}^n$ for every $x\in [0,1]$ and for every
$\varrho>0$,
$$
\sup_{|(t_1,\dots,t_n)|\leq \varrho} \sum_{i=1}^n
|F_{t_i}(x,t_1,\dots,t_n)|\in L^1([0,1]),
$$ 
and $F_{u_{i}}$ denotes the partial derivative of $F$ with respect to $u_{i}$ for
$1\leq i\leq n$.

 Due to importance of fourth-order
two-point boundary-value problems in describing a large class of
elastic deflection,  many authors have studied the existence and multiplicity 
of solutions for such a problem; we refer the reader
 to \cite{AHO,Ba,BD1,BD2,BD3,BDO,C,L,WHL} and references therein.

 In \cite{BD1}, the authors, employing a three critical point
 theorem due to Bonanno and Marano \cite[Theorem 2.6]{BM},
 determined an exact open interval of the parameter $\lambda$ for which  system \eqref{e1} 
in the case $n=1$,  admits at least three distinct weak solutions.

 The aim of this article is to prove the existence of at least one non-trivial weak 
solution for  \eqref{e1} for appropriate  values of the parameter $\lambda$ 
belonging to a precise real interval, which extend the results in \cite{BDO}.
Our motivation comes from the recent paper \cite{BD2}.
For  basic notation and definitions, and also for a thorough account on the subject, 
we refer the reader to   \cite{BMO,GR,KRV}.

\section{Preliminaries and basic notation}

 First we  recall for the reader's convenience
\cite[Theorem 2.5]{R} as given in \cite[Theorem 5.1]{B} (see also
\cite[Proposition 2.1]{B} for related  results) 
which is our main tool to transfer the question of existence of
at least one weak solution of \eqref{e1}
to the existence of a critical point of the Euler functional:

  For a given non-empty set $X$, and two functionals
$\Phi,\Psi:X\to\mathbb{R}$, we define the following two functions:
\begin{gather*}
\beta(r_1,r_2)=\inf_{v\in \Phi^{-1}(]r_1,r_2[)}
\frac{\sup_{u\in \Phi^{-1}(]r_1,r_2[)}\Psi(u)-\Psi(v)}{r_2-\Phi(v)},
\\
\rho(r_1,r_2)=\sup_{v\in \Phi^{-1}(]r_1,r_2[)}
\frac{\Psi(v)-\sup_{u\in \Phi^{-1}(]-\infty,r_1[)}\Psi(u)}{\Phi(v)-r_1}
\end{gather*}
for all $r_1,r_2\in\mathbb{R}$, $r_1<r_2$. 

\begin{theorem}[{\cite[Theorem 5.1]{B}}] \label{thm1} 
Let $X$ be a reflexive real Banach space,
$\Phi:X\to\mathbb{R}$ be a sequentially weakly lower semicontinuous,
coercive and continuously G\^ateaux differentiable functional whose
G\^ateaux derivative admits a continuous inverse on $X^*$ and
$\Psi:X\to\mathbb{R}$ be a continuously G\^ateaux differentiable
functional whose G\^ateaux derivative is compact. Put
$I_\lambda=\Phi-\lambda\Psi$ and assume that there are
$r_1,r_2\in\mathbb{R}$, $r_1<r_2$, such that
$$\beta(r_1,r_2)<\rho(r_1,r_2).$$
Then, for each
$\lambda\in]\frac{1}{\rho(r_1,r_2)},\frac{1}{\beta(r_1,r_2)}[$ there
is $u_{0,\lambda}\in\Phi^{-1}(]r_1,r_2[)$ such that
$I_{\lambda}(u_{0,\lambda})\leq I_\lambda(u)$ $\forall u\in
\Phi^{-1}(]r_1,r_2[)$ and $I'_\lambda(u_{0,\lambda})=0$.
\end{theorem}

 Let us introduce notation that  will be used later.
 Assume that
\begin{equation}\label{e2}
    \min \big\{ \frac{q_i^-}{\pi^2}, \frac{r_i^-}{\pi^4}, \frac{q_i^-}{\pi^2} 
+ \frac{r_i^-}{\pi^4} \big\} > - p_i^-,
\end{equation}
where 
$$
p_i^- := \operatorname{ess\,inf}_{x \in [0,1]} p_{i}(x) > 0,\quad
q_i^- := \operatorname{ess\,inf}_{x \in [0,1]} q_i(x), \quad
r_i^- := \operatorname{ess\,inf}_{x \in [0,1]} r_i(x),
$$
 for $1\leq i\leq n$.
Moreover, set
\[
\sigma_i := \min \big\{ \frac{q_i^-}{\pi^2}, \frac{r_i^-}{\pi^4},
\frac{q_i^-}{\pi^2} + \frac{r_i^-}{\pi^4},0 \big\},
\quad
\delta_i := \sqrt{ p_i^- + \sigma_i},
\]
for $1\leq i\leq n$.
Let $Y:= H^{2}([0,1])\cap H_0^{1}([0,1])$ be the Sobolev space
endowed with the usual norm. We recall the
 following Poincar\'{e} type inequalities (see, for instance, \cite[Lemma 2.3]{PTV}):
\begin{gather}\label{e3}
   \|u_i'\|_{L^2([0,1])}^2 \leq \frac{1}{\pi^2} \|u_i''\|_{L^2([0,1])}^2,\\
\label{e4}
 \|u_i\|_{L^2([0,1])}^2 \leq \frac{1}{\pi^4} \|u_i''\|_{L^2([0,1])}^2
\end{gather}
for all $u_i \in Y$ for $1\leq i\leq n$. Therefore, taking into
account \eqref{e2}-\eqref{e4}, the norm
$$
\|u_i\|= \Big( \int_0^1 ( p_i(x) |u_i''(x)|^2 + q_i(x)
|u_i'(x)|^2 + r_i(x) |u_i(x)|^2) dx \Big)^{1/2}
$$
for $1\leq i\leq n$ is equivalent to the usual norm and, in particular, one has
\begin{equation}\label{e5}
 \|u_i''\|_{L^2([0,1])} \leq \frac{1}{\delta_i} \|u_i\|
\end{equation}
for $1\leq i\leq n$.
We need the following proposition in the proof of Theorem \ref{thm2}.

\begin{proposition}[{\cite[Proposition 2.1]{BD1}}] \label{prop1}
Let $u_i \in Y$ for $1\leq i\leq n$. Then
$$
\|u_i\|_{\infty} \leq \frac{1}{2 \pi \delta_i} \|u_i\|
$$
for
$1\leq i\leq n$.
\end{proposition}

 Put
$$
k_i:= \Big( \|p_i\|_\infty + \frac{1}{\pi^2} \|q_i\|_\infty +
\frac{1}{\pi^4}\|r_i\|_\infty \Big)^{1/2}
$$
for $1\leq i\leq n$.
It is easy to see that $k_i > 0$ and $ \delta_i < k_i$ for $1\leq
i\leq n$.
Set $\underline{\delta}:=\min\{\delta_{i};\ 1\leq i\leq n\}$ and
 $\overline{k}:=\max\{k_{i};\ 1\leq i\leq n\}$.
Here and in the sequel, $X:=Y\times\dots\times Y$.

 We say that $u=(u_1,\dots,u_n)$ is a weak solution to the  \eqref{e1}
if $u=(u_1,\dots,u_n)\in X$ and
\begin{align*}
&\sum_{i=1}^n\int_0^1 \left( p_i(x) u_i''(x)
v_i''(x)+q_i(x)u_i'(x)v_i'(x)+r_i(x)u_i(x)v_i(x)\right)dx\\
&-\lambda\sum_{i=1}^n\int_0^1
F_{u_i}(x,u_1,\dots,u_n)v_i(x)dx=0
\end{align*}
for every $v=(v_1,\dots,v_n)\in X$.
For $\gamma>0$ we denote the set
  \begin{equation}\label{e6}
K(\gamma)=\big\{(t_1,\dots,t_n)\in \mathbb{R}^n:\ \sum_{i=1}^n|t_i|\leq \gamma\big\}.
\end{equation}

  \section{Results}

For a given non-negative constant $\nu$ and a
positive constant $\tau$ with
$2(\frac{\underline{\delta}\pi\nu}{n})^{2}\neq
\frac{4096}{54}n(\overline{k}\tau)^2$, put
   $$
a_\tau(\nu):=\frac{\int_0^{1}\sup_{(t_1,\dots,t_n)\in
K(\nu)}F(x,t_1,\dots,t_n)dx-\int_{\frac{3}{8}}^{\frac{5}{8}}F(x,\tau,\dots,\tau)dx}{2(\frac{\underline{\delta}\pi\nu}{n})^{2}-
\frac{4096}{54}n(\overline{k}\tau)^2}
$$ 
where
$K(\nu)=\big\{(t_1,\dots,t_n)\in \mathbb{R}^n:
\sum_{i=1}^n|t_i|\leq \nu\big\}$ (see \eqref{e6}).

 We formulate our main result as follows:

\begin{theorem} \label{thm2}
Assume that there exist a non-negative constant $\nu_1$ and two positive constants
 $\nu_2$ and $\tau$ with
$\pi\nu_1<n\sqrt{\frac{4096}{108}n}\tau$ and
$n\sqrt{\frac{4096}{108}n}\overline{k}\tau<\underline{\delta}\pi\nu_2$
such that
\begin{itemize}
\item[(A1)]  $F(x,t_1,\dots,t_n)\geq 0$  for each
  $(x,t_1,\dots,t_n)\in([ 0, 3/8 ] \cup [5/8 ,1])\times [0,\tau]^n$;

\item[(A2)] $a_\tau(\nu_2)<a_\tau(\nu_1)$.
\end{itemize}
Then, for each
$\lambda\in]\frac{1}{a_\tau(\nu_1)},\ \frac{1}{a_\tau(\nu_2)}
[$, system \eqref{e1} admits at least one non-trivial weak
solution $u_0=(u_{01},\dots,u_{0n})\in X$ such that
$$
4(\frac{\underline{\delta}\pi\nu_1}{n})^{2}<\sum_{i=1}^n\|u_{0i}\|^{2}
<4(\frac{\underline{\delta}\pi\nu_2}{n})^{2}.
$$
\end{theorem}

\begin{proof} 
To apply Theorem \ref{thm1} to our problem, arguing as in \cite{BDO,BP}, 
we introduce the functionals
$\Phi,  \Psi:X \to \mathbb{R}$ for each $u=(u_1,\dots,u_n) \in X$,
as follows
\begin{gather*}
\Phi(u)=\sum_{i=1}^n\frac{\|u_{i}\|^{2}}{2},\\
\Psi(u)=\int_0^{1}F(x,u_1(x),\dots,u_n(x))dx.
\end{gather*}
It is well known that $\Phi$ and $\Psi$ are well defined and
continuously differentiable functionals whose derivatives at the
point $u=(u_1,\dots,u_n)\in X$ are the functionals
$\Phi'(u),\Psi'(u)\in X^{*}$, given by
\begin{gather*}
\Phi'(u)(v)=\sum_{i=1}^n\int_0^1 \left( p_i(x) u_i''(x)
v_i''(x)+q_i(x)u_i'(x)v_i'(x)+r_i(x)u_i(x)v_i(x)\right)dx, \\
\Psi'(u)(v)=\int_0^{1}\sum_{i=1}^nF_{u_i}(x,u_1(x),\dots,u_n(x))v_{i}(x)dx
\end{gather*}
for every $v=(v_1,\dots,v_n)\in X$, respectively. Moreover,
$\Phi$ is sequentially weakly lower semicontinuous, $\Phi'$ admits
a continuous inverse on $X^{*}$ as well as $\Psi$ is sequentially
weakly upper semicontinuous. Furthermore, $\Psi':X \to X^{*}$ is a
compact operator. Indeed, it is enough to show that $\Psi'$ is
strongly continuous on $X$. For this, for fixed $(u_1,\dots,u_n)\in
X $ let $(u_{1m},\dots,u_{nm})\to(u_1,\dots,u_n)$ weakly in $X$ as
$m\to +\infty$, then we have $(u_{1m},\dots,u_{nm})$ converges
uniformly to $(u_1,\dots,u_n)$ on $[0,1]$ as $m\to +\infty$(see
\cite{Z}). Since $F(x,.,\dots,.)$ is $C^1$ in $\mathbb{R}^n$ for
every $x\in [0,1]$, the derivatives of $F$ are continuous in
$\mathbb{R}^n$ for every $x\in [0,1]$, so for $1\leq i\leq n$,
$F_{u_i}(x,u_{1m},\dots,u_{nm})\to F_{u_i}(x,u_1,\dots,u_n)$ strongly
as $m\to +\infty$ which follows $\Psi'(u_{1m},\dots,u_{nm})\to
\Psi'(u_1,\dots,u_n)$ strongly as $m\to +\infty$. Thus we proved
that $\Psi'$ is strongly continuous on $X$, which implies that
$\Psi'$ is a compact operator by Proposition 26.2 of \cite{Z}. Set
$w(x)= (w_1(x),\dots,w_n(x))$ such that for $1 \leq i\leq n$,
$$
w_{i}(x)= \begin{cases}
-\frac{64 \tau}{9} ( x^2 - \frac{3}{4}x ) & x\in [0, \frac{3}{8}[,\\
\tau & x\in [\frac{3}{8}, \frac{5}{8}],\\
-\frac{64 \tau}{9} ( x^2 - \frac{5}{4}x + \frac{1}{4} ) &
x\in ] \frac{5}{8}, 1],
\end{cases}
$$ 
$r_1 = 2(\frac{\underline{\delta }\pi \nu_1}{n})^2$ and 
$r_2 = 2(\frac{\underline{\delta} \pi \nu_2}{n})^2$.
 It is easy to verify that $w=(w_1,\dots,w_n)\in X$, and in particular, one has
$$
\frac{4096}{27} \delta_{i}^2 \tau^2 \leq \|w_{i}\|^2 \leq
\frac{4096}{27} k_{i}^2 \tau^2
$$
for $1 \leq i\leq n$. So, from the definition of $\Phi$, we have
$$
\frac{4096}{54}n(\underline{\delta}\tau)^2
\leq\sum_{i=1}^n\frac{4096}{27} \delta_{i}^2 \tau^2\leq
\Phi(w)\leq\sum_{i=1}^n\frac{4096}{27} k_{i}^2
\tau^2\leq\frac{4096}{54}n(\overline{k}\tau)^2.
$$ 
From the conditions $\pi\nu_1<n\sqrt{\frac{4096}{108}n}\tau$ and
$n\sqrt{\frac{4096}{108}n}\overline{k}\tau<\underline{\delta}\pi\nu_2$,
we obtain
$$
r_1<\Phi(w)<r_2.
$$
Moreover, from Proposition \ref{prop1} one has 
$$
\sup_{x\in [0,1]}\sum_{i=1}^n|u_{i}(x)|^{2}\leq \frac{1}{(2 \pi
\underline{\delta})^2} \sum_{i=1}^n\|u_{i}\|^{2}
$$
for each
$u=(u_1,\dots,u_n)\in X$, so
 from the definition of $\Phi$, we observe that
\begin{align*}
\Phi^{-1}(]-\infty,r_2[)
&= \{ (u_1,\dots,u_n)\in X; \Phi(u_1,\dots,u_n)< r_2\}\\
&= \big\{
(u_1,\dots,u_n)\in X;\sum_{i=1}^n\|u_{i}\|^{2}< 2r_2\big\}\\
&\subseteq \big\{ (u_1,\dots,u_n)\in X;
\sum_{i=1}^n|u_{i}(x)|^{2}<\frac{\nu_2^{2}}{n^{2}}\ \
\textrm{for all}\ x\in [0,1]\big\}\\
&\subseteq \big\{(u_1,\dots,u_n)\in X; \sum_{i=1}^n|u_{i}(x)|<\nu_2\ \
\textrm{for all}\ x\in [0,1]\big\},
\end{align*}
from which it follows
\begin{align*}
\sup_{(u_1,\dots,u_n)\in\Phi^{-1}(]-\infty,r_2[)}\Psi(u)
&= \sup_{(u_1,\dots,u_n)\in\Phi^{-1}(]-\infty,r_2[)}\int_0^{1}F(x,u_1(x),\dots,u_n(x))dx\\
&\leq  \int_0^{1}
 \sup_{(t_1,\dots,t_n)\in K(\nu_2)}F(x,t_1,\dots,t_n)dx.
\end{align*}
 Since for $1\leq i\leq n$, $0\leq w_i(x)\leq \tau$ for each $x\in
[0,1]$, the condition (A1) ensures that
$$
\int_0^{\frac{3}{8}}F(x,w_1(x),\dots,w_n(x))dx
+\int_{\frac{5}{8}}^{1}F(x,w_1(x),\dots,w_n(x))dx \geq 0.
$$
So, one has
\begin{align*}
\beta(r_1,r_2)&\leq \frac{\sup_{u\in \Phi^{-1}(]-\infty,r_2[)}\Psi(u)-\Psi(w)}{r_2-\Phi(w)}\\
&\leq \frac{\int_0^{1}\sup_{(t_1,\dots,t_n)\in
K(\nu_2)}F(x,t_1,\dots,t_n)dx-\Psi(w)}{r_2-\Phi(w)}\\
&\leq a_\tau(\nu_2).
\end{align*}
On the other hand, by similar reasoning as before, one has
\begin{align*}
\rho(r_1,r_2)
&\geq  \frac{\Psi(w)-\sup_{u\in \Phi^{-1}(]-\infty,r_1[)}\Psi(u)}{\Phi(w)-r_1}\\
&\geq  \frac{\Psi(w)-\int_0^{1}\sup_{(t_1,\dots,t_n)\in
K(\nu_1)}F(x,t_1,\dots,t_n)dx}{\Phi(w)-r_1}\\
&\geq  a_\tau(\nu_1).
\end{align*}
Hence, from Assumption (A2), one has $\beta(r_1,r_2)<\rho(r_1,r_2)$. 
Therefore, from Theorem \ref{thm1}, taking into account that the weak solutions of the
system \eqref{e1} are exactly the solutions of the equation
$\Phi'(u)-\lambda \Psi'(u)=0$, we have the conclusion.
\end{proof}

 Now we point out the following consequence of Theorem \ref{thm2}.

 \begin{theorem} \label{thm3}
Assume that Assumption (A1) in Theorem \ref{thm2} holds.
 Suppose that there exist two positive constants $\nu$ and $\tau$ with
$n\sqrt{\frac{4096}{108}n}\overline{k}\tau<\underline{\delta}\pi\nu$
such that
\begin{itemize}
\item[(A3)] 
$$
\frac{\int_0^{1}\sup_{(t_1,\dots,t_n)\in
K(\nu)}F(x,t_1,\dots,t_n)dx}{\nu^{2}}<\frac{108}{4096n^3}(\frac{\underline{\delta}
\pi }{\overline{k}})^2\frac{\int_{
\frac{3}{8}}^{\frac{5}{8}}F(x,\tau,\dots,\tau)dx}{\tau^2};
$$

\item[(A4)] $F(x,0,\dots,0)=0$ for every $x\in[0,1]$
\end{itemize}
Then, for each
$$
\lambda\in\Big]\frac{\frac{4096}{54}n(\overline{k}\tau)^2}{\int_{
\frac{3}{8}}^{\frac{5}{8}}F(x,\tau,\dots,\tau)dx},\
\frac{2(\frac{\underline{\delta}\pi\nu}{n})^{2}}{\int_0^{1}\sup_{(t_1,\dots,t_n)\in
K(\nu)}F(x,t_1,\dots,t_n)dx} \Big[,
$$ 
system \eqref{e1} admits at least one non-trivial weak solution
$u_0=(u_{01},\dots,u_{0n})\in X$ such that
$\sum_{i=1}^n\|u_{i}\|_{\infty}<\nu$.
\end{theorem}

\begin{proof}
The conclusion follows from Theorem \ref{thm2}, by taking $\nu_1=0$ and $\nu_2=\nu$.
Indeed, owing to our assumptions, one has
\begin{align*}
a_{\tau}(\nu_{2})
&< \frac{\Big(1-\frac{\frac{4096n^3}{108}(\frac{\tau\overline{k}}{\underline{\delta}
\pi })^2}{\nu^{2}}\Big)\int_0^{1}\sup_{(t_1,\dots,t_n)\in
K(\nu)}F(x,t_1,\dots,t_n)dx}{2(\frac{\underline{\delta}\pi\nu}{n})^{2}-
\frac{4096}{54}n(\overline{k}\tau)^2}\\
&= \frac{\Big(1-\frac{\frac{4096}{54}n(\tau\overline{k})^{2}}
{2(\frac{\underline{\delta}\pi\nu}{n})^{2}}\Big)
\int_0^{1}\sup_{(t_1,\dots,t_n)\in
K(\nu)}F(x,t_1,\dots,t_n)dx}{2(\frac{\underline{\delta}\pi\nu}{n})^{2}-
\frac{4096}{54}n(\overline{k}\tau)^2}\\&= \frac{\int_0^{1}\sup_{(t_1,\dots,t_n)\in
K(\nu)}F(x,t_1,\dots,t_n)dx}{2(\frac{\underline{\delta}\pi\nu}{n})^{2}}.
\end{align*}
On the other hand, taking Assumption (A4) into account, one has
$$
\frac{\int_{
\frac{3}{8}}^{\frac{5}{8}}F(x,\tau,\dots,\tau)dx}
{\frac{4096}{54}n(\overline{k}\tau)^2}=a_{\tau}(\nu_1).
$$
Moreover, since 
$$
\sup_{x\in [0,1]}\sum_{i=1}^n|u_{i}(x)|^{2}\leq \frac{1}{(2 \pi
\underline{\delta})^2} \sum_{i=1}^n\|u_{i}\|^{2}
$$
for each 
$u=(u_1,\dots,u_n)\in X$, an easy computation ensures that
$\sum_{i=1}^n\|u_{i}\|_{\infty}<\nu$ whenever $\Phi(u)<r_2$.
 Now, owing to Assumption (A3), it is sufficient to invoke 
Theorem \ref{thm2} for concluding the proof.
\end{proof}

Now, we point out a simple version of Theorem \ref{thm2} in the case $n=1$.
 Let $p_1=p$, $q_1=q$, $r_1=r$, $\delta_1=\delta$ and $k_1=k$. 
Let $f: [0,1] \times \mathbb{R} \to \mathbb{R}$ be an $L^2$-Carath\'eodory function.
Let $F$ be the function defined by $F(x,t)=\int_0^{t}f(x,s)d s$
for each $(x,t)\in[0,1]\times \mathbb{R}$.
 For a given non-negative constant $\nu$ and a positive constant $\tau$ with
$2(\delta\pi\nu)^{2}\neq \frac{4096}{54}(k\tau)^2$, put
   $$
b_\tau(\nu):=\frac{\int_0^{1}\sup_{|t|\leq\nu}F(x,t)dx
-\int_{\frac{3}{8}}^{\frac{5}{8}}F(x,\tau)dx}{2(\delta\pi\nu)^{2}-
\frac{4096}{54}(k\tau)^2}.
$$
Then, we have the following result.

\begin{theorem} \label{thm4}
Assume that there exist a non-negative constant $\nu_1$ and two positive 
constants $\nu_2$ and $\tau$ with
$\pi\nu_1<\sqrt{\frac{4096}{108}}\tau$ and
$\sqrt{\frac{4096}{108}}k\tau<\delta\pi\nu_2$ such
that
\begin{itemize}
\item[(B1)]  $F(x,t)\geq 0$  for each  $(x,t)\in([ 0, 3/8] \cup [5/8 ,1])\times [0,\tau]$;;

\item[(B2)] $b_\tau(\nu_2)<b_\tau(\nu_1)$.
\end{itemize}
Then, for each $\lambda\in]\frac{1}{b_\tau(\nu_1)},\ \frac{1}{b_\tau(\nu_2)}[$,
 the problem
\begin{equation}\label{e7}
\begin{gathered}
(p(x) u''(x))'' - (q(x) u'(x))' + r(x) u(x) =\lambda f(x,u) \quad x \in (0,1), \\
u(0)=u(1)=u''(0)=u''(1)=0
\end{gathered}
\end{equation}
admits at least one non-trivial weak
solution $u_0\in Y$ such that
$2\delta\pi\nu_1<\|u_0\|<2\delta\pi\nu_2$.
\end{theorem}

We remark that if  $p(x)=1$, $q(x)=-A$ and $r(x)=B$ for every $x\in[0,1]$, then
Theorem \ref{thm4} gives \cite[Theorem 3.1]{BDO}.

The following result gives the existence of at least one non-trivial weak 
solution in $Y$ to the problem \eqref{e7} in the autonomous
case.
 Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. Put
 $F(t)=\int_0^{t}f(\xi)d\xi$ for all $t\in\mathbb{R}$. We have
 the following result as a direct consequence of Theorem \ref{thm4}.

\begin{theorem} \label{thm5}
Assume that there exist a non-negative constant $\nu_1$ and two positive 
constants $\nu_2$ and $\tau$ with
$\pi\nu_1<\sqrt{\frac{4096}{108}}\tau$ and
$\sqrt{\frac{4096}{108}}k\tau<\delta\pi\nu_2$ such
that
\begin{itemize}
\item[(C1)] $f(t)\geq 0$  for each  $t\in[-\nu_2,\ \max\{\nu_2,\tau\}]$;

\item[(C2)] $\frac{F(\nu_2)-\frac{1}{4}F(\tau)}{2(\delta\pi\nu_{2})^{2}-
\frac{4096}{54}(k\tau)^2}<
\frac{F(\nu_1)-\frac{1}{4}F(\tau)}{2(\delta\pi\nu_1)^{2}-
\frac{4096}{54}(k\tau)^2}$.
\end{itemize}
Then, for each $\lambda\in]\frac{2(\delta\pi\nu_1)^{2}-
\frac{4096}{54}(k\tau)^2}{F(\nu_1)-\frac{1}{4}F(\tau)},\
\frac{2(\delta\pi\nu_{2})^{2}-
\frac{4096}{54}(k\tau)^2}{F(\nu_2)-\frac{1}{4}F(\tau)} [$,
the problem
\begin{equation}\label{e8} 
\begin{gathered}
(p(x) u''(x))'' - (q(x) u'(x))' + r(x) u(x) =\lambda f(u) \quad x \in (0,1), \\
u(0)=u(1)=u''(0)=u''(1)=0
\end{gathered}
\end{equation}
admits at least one non-trivial weak solution $u_0\in Y$ such
that $\nu_1<\frac{\|u_0\|}{2\delta\pi}<\nu_2$.
\end{theorem}

\begin{proof}
Since $\delta<k$, from the conditions
$\pi\nu_1<\sqrt{\frac{4096}{108}}\tau$ and
$\sqrt{\frac{4096}{108}}k\tau<\delta\pi\nu_2$, we obtain
$2(\delta\pi\nu_1)^{2}<\frac{4096}{54}(\delta\tau)^2$ and
$\frac{4096}{54}(k\tau)^2<2(\delta\pi\nu_2)^{2}$, and so
$2(\delta\pi\nu_1)^{2}<2(\delta\pi\nu_2)^{2}$, and so
$\nu_1<\nu_2$. Therefore, Assumptions (C1) means $f(t)\geq 0$  for
each  $t\in[-\nu_1,\ \nu_1]$ and $f(t)\geq 0$  for each
$t\in[-\nu_2,\nu_2]$, which follow $\max_t\in[-\nu_1,\
\nu_1]F(t)=F(\nu_1)$ and $\max_t\in[-\nu_2,\ \nu_2]F(t)=F(\nu_2)$.
So, from Assumptions (C1) and (C2) we arrive at assumptions (B1)
and (B2), respectively. Hence, we achieve the stated assertion by
applying Theorem \ref{thm4} observing that the problem \eqref{e1}
reduces to the problem \eqref{e8}.
\end{proof}

 As an example, we point out the following special case of our main
result.

\begin{theorem} \label{thm6}
Let $g:\mathbb{R}\to\mathbb{R}$ be a nonnegative continuous function such
that $\lim_{t\to 0^{+}}\frac{g(t)}{t}=+\infty$. Then, for each
$\lambda\in]0,2(\delta\pi)^{2}\sup_{\nu>0}\frac{\nu^{2}}{\int_0^{\nu}g(\xi)d\xi}
[$, the problem
\begin{gather*}
(p(x) u''(x))'' - (q(x) u'(x))' + r(x) u(x) =\lambda g(u) \quad x \in (0,1), \\
u(0)=u(1)=u''(0)=u''(1)=0
\end{gather*}
admits at least one non-trivial weak solution in $Y$.
\end{theorem}

\begin{proof} 
For fixed $\lambda$ as in the conclusion, there
exists positive constant $\nu$ such that
$$
\lambda<2(\delta\pi)^{2}\frac{\nu^{2}}{\int_0^{\nu}g(\xi)d\xi}.
$$
Moreover,  $\lim_{t\to 0^{+}}\frac{g(t)}{t}=+\infty$
implies $\lim_{t\to
0^{+}}\frac{\int_0^{t}g(\xi)d\xi}{t^{2}}=+\infty$.
 Therefore, a positive constant $\tau$ satisfying
$\sqrt{\frac{4096}{108}}k\tau<\delta\pi\nu$ can be chosen such
that
$$
\frac{1}{\lambda}(\frac{4\times 4096}{54}k^2)<\frac{\int_0^{\tau}g(\xi)d\xi}{\tau^{2}}.
$$
Hence, arguing as in the proof of Theorem \ref{thm3}, the conclusion
follows from Theorem \ref{thm5} with $\nu_1=0$, $\nu_2=\nu$ and
$f(t)=g(t)$ for every $t\in \mathbb{R}$.
\end{proof}

\begin{remark}\label{rmk1} \rm
For fixed $\rho$ put 
$\lambda_\rho:=2(\delta\pi)^{2}\sup_{\nu\in]0,\rho[}
\frac{\nu^{2}}{\int_0^{\nu}g(\xi)d\xi}$.
The result of Theorem \ref{thm6} for every
$\lambda\in]0,\lambda_\rho[$ holds with $|u_0(x)|<\rho$ for all
$x\in [0,1]$ where $u_0$ is the ensured non-trivial weak solution
in $Y$ (see \cite[Remark 4.3]{BDO}).
 \end{remark}

 We close this article by presenting the following examples to
 illustrate our results.
 
\begin{example} \rm
 Consider the problem 
\begin{equation} \label{e9}
\begin{gathered}
(3e^xu'')''-((x^2-\pi^2)u')'+(x^2-\pi^4) u
=\lambda (1+e^{-u^+}(u^+)^2(3-u^+))\quad x\in(0,1),\\
u(0)=u(1)=0,\quad u''(0)=u''(1)=0
\end{gathered}
\end{equation}
where $u^+=\max\{u,0\}$. Let
$$
g(t)=1+e^{-t^+}(t^+)^2(3-t^+)
$$ 
for all $t\in\mathbb{R}$ where $t^+=\max\{t,0\}$. 
It is clear that $\lim_{t\to 0^{+}}\frac{g(t)}{t}=+\infty$.
 Note that $p^-=3$, $q^-=-\pi^2$ and
$r^-=-\pi^4$, we have $\sigma=-2$, and so $\delta=1$. Hence,
taking Remark \ref{rmk1} into account, since
$\sup_{\nu\in]0,1[}\frac{\nu^{2}}{\int_0^{\nu}g(\xi)d\xi}=\sup_{\nu\in]0,1[}
\frac{\nu^{2}}{\nu+e^{-\nu}\nu^3}=\frac{e}{1+e}$,
by applying Theorem \ref{thm6}, for every
$\lambda\in]0,\frac{2\pi^2e}{1+e}[$ the problem \eqref{e9} has at
least one non-trivial classical solution $u_0\in Y$ such that
$\|u_0\|_{\infty}<1$.
 \end{example}

 \begin{example} \rm
 Put $p(x)=1$, $q(x)=\pi^2$, $r(x)=x-\pi$ for all $x\in[0,1]$ and $g(t)=(1+t)e^t$ 
for every $t\in\mathbb{R}$. Clearly, one has
  $\sigma=(1-\frac{1}{\pi^3})^{\frac{1}{2}}$. Hence, since
$$
\sup_{\nu\in]0,1[}\frac{\nu^{2}}{\int_0^{\nu}g(\xi)d\xi}
=\sup_{\nu\in]0,1[}\frac{\nu^{2}}{\nu e^\nu}=\frac{1}{e},
$$
from Theorem \ref{thm6}, taking Remark \ref{rmk1} into account, for
every $\lambda\in]0,\frac{2\pi^2(1-\frac{1}{\pi^3})}{e}[$ the
problem 
\begin{equation}  \label{e10}
\begin{gathered}
u^{iv}-\pi^2u''+(x-\pi)u=\lambda (1+u)e^u\quad x\in(0,1),\\
u(0)=u(1)=0,\quad u''(0)=u''(1)=0
\end{gathered}
\end{equation}
 has at least one non-trivial classical solution
$u_0\in Y$ such that $\|u_0\|_{\infty}<1$.
 \end{example}

\subsection*{Acknowledgements}
The author expresses his sincere gratitude to the
referees for reading this paper very carefully and specially for
valuable suggestions concerning improvement of the manuscript.
 This research was partially supported by a grant 90470020 
from IPM.

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\end{document}