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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 45, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/45\hfil Three solutions for a boundary-value problem]
{Three solutions for a fourth-order boundary-value problem}

\author[G. A. Afrouzi, S. Shokooh \hfil EJDE-2015/45\hfilneg]
{Ghasem A. Afrouzi, Saeid Shokooh}

\address{Ghasem A. Afrouzi \newline
Department of Mathematics, Faculty of Mathematical Sciences,
University of Mazandaran, Babolsar, Iran}
\email{afrouzi@umz.ac.ir}

\address{Saeid Shokooh \newline
Department of Mathematics, Faculty of Mathematical
Sciences, University of Mazandaran, Babolsar, Iran}
\email{saeid.shokooh@stu.umz.ac.ir}

\thanks{Submitted November 14, 2014. Published February 17, 2015.}
\subjclass[2000]{34B15, 34B18, 58E05}
\keywords{Dirichlet boundary condition; variational method; critical point}

\begin{abstract}
 Using two three-critical points theorems, we prove the existence of at
 least three weak solutions for one-dimensional fourth-order equations.
 Some particular cases and two concrete examples are then presented.
\end{abstract}

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

\section{Introduction}

In this note, we consider the  fourth-order boundary-value problem
\begin{equation}\label{e1.1}
\begin{gathered}
u''''h(x,u')-u''=[\lambda f(x,u)+g(u)]h(x,u'),\quad  \text{in }(0,1),\\
u(0)=u(1)=0=u''(0)=u''(1),
\end{gathered}
\end{equation}
where $\lambda$ is a positive parameter,
$f:[0,1]\times\mathbb{R}\to \mathbb{R}$ is an $L^1$-Carath\'{e}odory function,
$g:\mathbb{R}\to \mathbb{R}$ is a Lipschitz continuous function with the
Lipschitz constant $L>0$, i.e.,
$$
|g(t_1)-g(t_2)|\leq L|t_1-t_2|
$$
for every $t_1,t_2\in \mathbb{R}$, with $g(0)=0$, and
$h:[0,1]\times\mathbb{R}\to [0,+\infty)$
is a bounded and continuous function with
$m:=\inf_{(x,t)\in [0,1]\times \mathbb{R}}h(x,t)>0$.

Due to the importance of fourth-order two-point boundary value problems in
describing a large class of elastic deflection, many researchers have studied the
existence and multiplicity of solutions for such a problem, we refer the reader to
\cite{AfHaRa,AfHeiRe,AfMirRa,chai,sara} and references therein.
For example, authors in \cite{AfHeiRe}, using Ricceri's Variational Principle
 \cite[Theorem 1]{Ricceri3},
established the existence three weak solutions for the problem
\begin{gather*}
u''''+\alpha u''+\beta u=\lambda f(x,u)+\mu g(x,u),\quad  \text{in }(0,1),\\
u(0)=u(1)=0=u''(0)=u''(1),
\end{gather*}
where $\alpha$, $\beta$ are real constants,
$f, g : [0, 1] \times \mathbb{R}\to \mathbb{R}$
are Carath\'eodory functions and $\lambda, \mu > 0$.

In this article, employing two three-critical points theorems which we
recall in the next section (Theorems \ref{teo:bon} and \ref{teo:can}),
 we establish the existence three weak solutions for \eqref{e1.1}.
 A special case of Theorem \ref{thm1} is the following theorem.

\begin{theorem} \label{the0}
Let $f:\mathbb{R} \to \mathbb{R} $ be a non-negative continuous
function such that
\begin{gather*}
2^{12}\int_0^{2}f(x)\,dx<\int_0^{3\sqrt3}f(x)\,dx, \\
\limsup_{|\xi|\to +\infty} \frac{\int_0^{\xi}f(x)\,dx}{\xi^2}\le 0.
\end{gather*}
Then, for each
$$
\lambda\in\Big]\frac{2^{13}(\pi^4+\pi^2+1)}{\pi^4\int_0^{3\sqrt3}f(x)\,dx},
\frac{2(\pi^4+\pi^2+1)}{\pi^4\int_0^{2}f(x)\,dx}\Big[,
$$
the problem
\begin{gather*}
u''''-u''+u=f(u),\quad  \text{in }(0,1),\\
u(0)=u(1)=0=u''(0)=u''(1)
\end{gather*}
admits at least three weak solutions.
\end{theorem}

The following result is a consequence of Theorem \ref{thm2}.

\begin{theorem} \label{the01}
Let $f:\mathbb{R} \to \mathbb{R} $ be a non-negative continuous
function such that
\begin{gather*}
2^{11}\int_0^{2}f(x)\,dx<\int_0^3f(x)\,dx, \\
\int_0^{2^{10}}f(x)\,dx<2^{7}\int_0^3f(x)\,dx,
\end{gather*}
Then, for each
$$
\lambda\in\Big]\frac{2^{13}(\pi^4+\pi^2+1)}{\pi^4\int_0^3f(x)\,dx},
\frac{(\pi^4+\pi^2+1)}{\pi^4}\min\Big\{\frac{2}{\int_0^{2}f(x)\,dx},
\frac{2^{20}}{\int_0^{1024}f(x)\,dx}\Big\}\Big[,
$$
the problem
\begin{gather*}
u''''-u''-u=f(u),\quad  \text{in }(0,1),\\
u(0)=u(1)=0=u''(0)=u''(1)
\end{gather*}
admits at least three weak solutions.
\end{theorem}

\section{Preliminaries}

We now state two critical point theorems established by Bonanno and
coauthors \cite{BONCAND,BONMAR}
which are the main tools for the proofs of our results.
The first result has been obtained in \cite{BONMAR} and it is a more precise
version of Theorem 3.2 of \cite{BONCAND}. The second one has been established
in \cite{BONCAND}.
In the first one the coercivity of the functional $\Phi-\lambda\Psi$
is required, in the second
one a suitable sign hypothesis is assumed.

\begin{theorem}[{\cite[Theorem 2.6]{BONMAR}}] \label{teo:bon}
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 $X^*$, $\Psi:X \to \mathbb{R}$ be a sequentially
weakly upper semicontinuous, continuously G\^ateaux differentiable functional
whose G\^ateaux derivative is compact, such that
$$
\Phi(0)= \Psi(0)=0 \,.
$$
Assume that there exist $r >0$ and $\bar{x}\in X$, with $r<\Phi(\bar{x})$
such that
\begin{itemize}
\item[(i)] $\sup_{\Phi(x)\leq r}\Psi(x)<r \Psi(\bar{x})/ \Phi(\bar{x})$,

\item [(ii)] for each $\lambda$ in
\[
\Lambda_r:=\Bigl] \frac{\Phi(\bar{x})}{\Psi(\bar{x})},
\frac{r}{\sup_{\Phi(x)\leq r}\Psi(x)}\Bigr[\,,
\]
the functional
$\Phi-\lambda \Psi$ is coercive.
\end{itemize}
Then, for each $\lambda \in \Lambda_r$ the functional $\Phi-\lambda \Psi$
has at least three distinct critical points in $X$.
\end{theorem}

\begin{theorem}[{\cite[Theorem 3.2]{BONCAND}}] \label{teo:can}
Let $X$ be a reflexive real Banach space; $\Phi:X \to \mathbb{R}$
be a convex, coercive and continuously G\^ateaux differentiable functional
whose G\^ateaux derivative admits a continuous inverse on $X^*$,
$\Psi:X \to \mathbb{R}$ be a continuously G\^ateaux differentiable
functional whose G\^ateaux derivative is compact, such that
$$
\inf_{X}\Phi=\Phi(0)= \Psi(0)=0 \,.
$$
Assume that there exist two positive constants  $r_1,r_2 >0$ and
$\bar{x}\in X$, with $2r_1<\Phi(\bar{x})<r_2/2$, such that
\begin{itemize}
\item[(j)] $\sup_{\Phi(x)\leq r_1}\Psi(x)/ r_1<(2/3)\Psi(\bar{x})/\Phi(\bar{x})$,

\item[(jj)] $\sup_{\Phi(x)\leq r_2}\Psi(x)/r_2 <(1/3)\Psi(\bar{x})/\Phi(\bar{x})$,

\item[(jjj)] for each $\lambda$ in
\[
\Lambda_{r_1,r_2}^*
:=\Bigl] \frac{3}{2}\frac{\Phi(\bar{x})}{\Psi(\bar{x})},
\min\big\{\frac{r_1}{\sup_{\Phi(x)\leq r_1}\Psi(x)},
\frac{r_2}{2\sup_{\Phi(x)\leq r_2}\Psi(x)}\big\}\Bigr[
\]
and for every
$x_1,x_2 \in X$, which are local minima for the functional
$\Phi-\lambda \Psi$, and such that
$\Psi(x_1)\geq 0$ and $\Psi(x_2)\geq 0$, one has
$\inf_{t \in [0,1]}\Psi(tx_1+(1-t)x_2) \geq 0$.
\end{itemize}
Then, for each $\lambda \in \Lambda_{r_1,r_2}^*$ the functional
$\Phi-\lambda \Psi$ has at least three distinct critical points which lie
in $\Phi^{-1}(]-\infty, r_2[)$.
\end{theorem}

Let us introduce some notation which will be used later.
Define
\begin{gather*}
H_0^1([0,1]):=\big{\{}u\in L^2([0,1]) : u'\in L^2([0,1]),\;u(0)=u(1)=0\big\},\\
H^2([0,1]):=\big{\{}u\in L^2([0,1]) : u',u''\in
L^2([0,1])\big\}.
\end{gather*}
Take $X= H^2([0,1])\cap H_0^1([0,1])$ endowed with
the usual norm
$$
\|u\|:=\Big(\int_0^1|u''(x)|^2\,dx\Big)^{1/2}.
$$
We recall the following Poincar\'{e} type inequalities
(see, for instance, \cite[Lemma 2.3]{PTV}):
\begin{gather}\label{4}
   \|u'\|_{L^2([0,1])}^2 \leq \frac{1}{\pi^2} \|u\|^2,\\  \label{5}
 \|u\|_{L^2([0,1])}^2 \leq \frac{1}{\pi^4} \|u\|^2
\end{gather}
for all $u \in X$.
For the norm in $C^1([0,1])$,
$$
\|u\|_\infty:=\max\Big\{\max_{x\in[0,1]}|u(x)|,\max_{x\in[0,1]}|u'(x)|\Big\},
$$
we have the following relation.
\begin{proposition}\label{p1}
 Let $u \in X$. Then
\begin{equation}\label{e2.2}
\|u\|_{\infty} \leq \frac{1}{2 \pi } \|u\|.
\end{equation}
\end{proposition}

\begin{proof}
Taking \eqref{4} into account, the conclusion follows from the
well-known inequality
$\|u\|_{\infty} \leq \frac{1}{2} \|u'\|_{L^2([0,1])}$.
\end{proof}

For an excellent overview of the most significant mathematical methods
employed in this paper  we refer to \cite{ciarlet,Rad}.

 Let $g:\mathbb{R}\to \mathbb{R}$ is a Lipschitz continuous function with the
Lipschitz constant $L>0$, i.e.,
$$
|g(t_1)-g(t_2)|\leq L|t_1-t_2|
$$
for every $t_1,t_2\in \mathbb{R}$, and $g(0)=0$,
$h:[0,1]\times\mathbb{R}\to [0,+\infty)$
is a bounded and continuous function with
$m:=\inf_{(x,t)\in [0,1]\times \mathbb{R}}h(x,t)>0$,
and $f:[0,1]\times\mathbb{R}\to\mathbb{R}$ be an
$L^1$-Carath\'eodory function.

We recall that $f:[0,1]\times\mathbb{R}\to\mathbb{R}$ is an
 $L^1$-Carath\'eodory function if
\begin{itemize}
\item[(a)] the mapping $x\mapsto f(x,\xi)$ is measurable for every
 $\xi\in\mathbb{R}$;

\item[(b)] the mapping $\xi\mapsto f(x,\xi)$ is continuous for almost every
$x\in [0,1]$;

\item[(c)] for every $\rho>0$ there exists a function $l_\rho\in L^1([0,1])$
such that
$$
\sup_{|\xi|\leq \rho}|f(x,\xi)|\leq l_{\rho}(x)
$$
for almost every $x\in [0,1]$.
\end{itemize}
Corresponding to $f,g$ and $h$ we introduce the functions
$F:[0,1]\times\mathbb{R}\to\mathbb{R}$, $G:\mathbb{R}\to \mathbb{R}$ and
 $H:[0,1]\times \mathbb{R}\to [0,+\infty)$, respectively, as follows
\begin{gather*}
F(x,t):=\int_0^t f(x,\xi)\,d\xi,\quad
G(t):=-\int_0^t g(\xi)\,d\xi, \\
H(x,t):=\int_0^t\Big(\int_0^\tau\frac{1}{h(x,\delta)}d\delta\Big)d\tau
\end{gather*}
for all $x\in [0,1]$ and $t\in\mathbb{R}$.

In the following, we let $M:=\sup_{(x,t)\in [0,1]\times \mathbb{R}}h(x,t)$
and suppose that the Lipschitz constant $L$ of the function $g$ satisfies
 $0<L<\pi^4$.

We say that a function $u\in X$ is a \textit{weak solution} of \eqref{e1.1} if
\begin{align*}
&\int_0^1u''(x)v''(x)\,dx
+\int_0^1\Big(\int_0^{u'(x)}\frac{1}{h(x,\tau)}d\tau\Big) v'(x)\,dx\\
&-\lambda\int_0^1 f(x,u(x))v(x)\,dx
 -\int_0^1g(u(x))v(x)\,dx=0
\end{align*}
holds for all $v\in X$.

\section{Main results}
Put
$$
A:=\frac{\pi^4-L}{2\pi^4},\quad B:=\frac{\pi^2+m(\pi^4+L)}{2m\pi^4},
$$
and suppose that $B\leq 4A\pi^2$. We formulate our main results as follows.

\begin{theorem} \label{thm1}
Assume that there exist two positive constants $c,d$, satisfying
$c<32d/(3\sqrt{3}\pi)$, such that
\begin{itemize}
\item [(A1)]  $F(x,t)\geq 0$ for all
 $(x, t)  \in \left([0,3/8]\cup [5/8,1]\right) \times [0,d]$;

\item [(A2)]
\[
 \frac{\int_0^1 \max_{|t|\leq c}F(x,t)\,dx}{c^2}
<\frac{27}{4096}\frac{\int_{3/8}^{5/8}F(x,d)\,dx}{d^2};
\]

\item [(A3)]
\[
\limsup_{|\xi|\to +\infty} \frac{\sup_{x \in [0,1]}F(x,\xi)}{\xi^2}
\le \frac{\pi^4A}{B}
    \frac{\int_{0}^{1}\max_{|t|\leq c}F(x,t)\,dx}{c^2}.
\]
\end{itemize}
Then, for every $\lambda$ in
\[
\Lambda:=\Big]\frac{4096Bd^2}{27\int_{3/8}^{5/8}F(x,d)\,dx},
 \frac{Bc^2}{\int_0^1\max_{|t|\leq c}F(x,t)\,dx}\Big[ ,
\]
problem \eqref{e1.1} has at least three distinct weak solutions.
\end{theorem}

\begin{proof}
Fix $\lambda$ as in the conclusion. Our aim is to apply Theorem \ref{teo:bon}
to our problem. To this end, for every $u\in X$, we introduce the functionals
 $\Phi,\Psi:X\to \mathbb{R}$
by setting
\begin{gather*}
\Phi(u):=\frac{1}{2}\|u\|^2+\int_0^1H(x,u'(x))\,dx+\int_0^1G(u(x))\,dx, \\
\Psi(u):=\int_0^1 F(x,u(x))\,dx,
\end{gather*}
and putting
$$
I_{\lambda}(u):=\Phi(u)-\lambda\Psi(u)\quad \forall\ u\in X.
$$
Note that the weak solutions of \eqref{e1.1} are exactly the critical points of
$I_{\lambda}$. The functionals $\Phi,\Psi$ satisfy the regularity assumptions of
Theorem \ref{teo:bon}. Indeed, by standard arguments, we have that
$\Phi$ is G\^{a}teaux differentiable
and sequentially weakly lower semicontinuous and its G\^{a}teaux derivative
is the functional $\Phi'(u)\in X^*$, given by
\[
\Phi'(u)(v)=\int_0^1u''(x)v''(x)\,dx
+\int_0^1\Big(\int_0^{u'(x)}\frac{1}{h(x,\tau)}d\tau\Big) v'(x)\,dx
-\int_0^1g(u(x))v(x)\,dx
\]
for any $v\in X$. Furthermore, the differential $\Phi':X\to  X^*$ is
a Lipschitzian operator. Indeed, taking \eqref{4} and \eqref{5} into account,
for any $u,v\in X$, there holds
\begin{align*}
\|\Phi'(u)-\Phi'(v)\|_{X^*}
&=\sup_{\|w\|\leq 1}|(\Phi'(u)-\Phi'(v),w)|\\
&\leq\ sup_{\|w\|\leq 1}\int_0^1|u''(x)-v''(x)||w''(x)|\,dx\\
&\quad +\sup_{\|w\|\leq 1}\int_0^1\big|\int_{u'(x)}^{v'(x)}
\frac{1}{h(x,\tau)}d\tau\big| \,|w'(x)|\,dx\\
&\quad +\sup_{\|w\|\leq 1}\int_0^1|g(u(x))-g(v(x))||w(x)|\,dx\\
& \leq \|u-v\|+\frac{1}{m}\sup_{\|w\|\leq 1}\|u'-v'\|_{L^2(0,1)}\|w'\|_{L^2(0,1)}\\
&\quad +
L\sup_{\|w\|\leq 1}\|u-v\|_{L^2(0,1)}\|w\|_{L^2(0,1)}\\
&\leq \big(1+\frac{1}{m\pi^2}+\frac{L}{\pi^4}\big)\|u-v\|
=2B\|u-v\|.
\end{align*}
Recalling that $g$ is Lipschitz continuous and $h$ is bounded away from zero,
the claim is true. In particular, we derive that $\Phi$ is continuously
differentiable.
Also, for any $u,v\in X$, we have
\begin{align*}
(\Phi'(u)-\Phi'(v),u-v)
&=\|u-v\|^2+\int_0^1\Big(\int_{u'(x)}^{v'(x)}
 \frac{1}{h(x,\tau)}d\tau\Big)(u'(x)-v'(x))\,dx\\
&\quad -\int_0^1(g(u(x))-g(v(x)))(u(x)-v(x))\,dx\\
&\geq \|u-v\|^2+\frac{1}{M}\|u'-v'\|^2_{L^2(0,1)}-L\|u-v\|^2_{L^2(0,1)}\\
&\geq \|u-v\|^2-\frac{L}{\pi^4}\|u-v\|^2=2A\|u-v\|^2.
\end{align*}
By the assumption $L<\pi^4$, it turns out that $\Phi'$ is a strongly
 monotone operator.
So, by applying Minty-Browder theorem \cite[Theorem 26.A]{ZII1},
$\Phi':X\to  X^*$ admits a Lipschitz continuous inverse.
On the other hand, the fact that $X$ is compactly embedded into $C^0([0,1])$ implies
that the functional $\Psi$ is well defined, continuously G\^{a}teaux differentiable
and with compact derivative, whose G\^{a}teaux derivative is given by
$$
\Psi'(u)(v)=\int_0^1f(x,u(x))v(x)\,dx
$$
for any $v\in X$.

Since $g$ is Lipschitz continuous and satisfies $g(0)=0$, while $h$ is bounded
away from zero, the inequalities \eqref{4} and \eqref{5} yield for any
 $u\in X$ the estimate
\begin{equation}\label{arsi}
A\|u\|^2\leq \Phi(u)\leq B\|u\|^2.
\end{equation}

We will verify (i) and (ii) of Theorem \ref{teo:bon}.
Put $r=Bc^2$. Taking \eqref{e2.2} into account, for every $u \in X$
such that $\Phi(u)\leq r$, one has
$\max_{x \in [0,1]}|u(x)| \leq c$.
Consequently,
$$
\sup_{\Phi(u)\leq r}\Psi(u)\leq\int_0^1\max_{|t|\leq c}F(x,t)\,dx;
$$
that is,
$$
\frac{\sup_{\Phi(u)\leq r}\Psi(u)}{r}\leq
\frac{\int_0^1\max_{|t|\leq c}F(x,t)\,dx}{Bc^2}\,.
$$
Hence,
\begin{equation}\label{eq:1}
\frac{\sup_{\Phi(u)\leq r}\Psi(u)}{r}<
\frac{1}{\lambda}\,.
\end{equation}
Put
\[
 w(x)= \begin{cases}
-\frac{64 d}{9} ( x^2 - \frac{3}{4}x ), & x\in [0, \frac{3}{8}[,\\
d, & x\in [\frac{3}{8}, \frac{5}{8}],\\
-\frac{64d}{9} ( x^2 - \frac{5}{4}x + \frac{1}{4} ), &
x\in ] \frac{5}{8}, 1 ].
\end{cases}
\]
It is easy to verify that $w\in X$ and, in particular,
$$
\|w\|^2=\frac{4096}{27}d^2.
$$
So, taking \eqref{arsi} into account, we deduce
$$
\frac{4096}{27}Ad^2\leq\Phi(w)\leq \frac{4096}{27}Bd^2.
$$
Hence, from $c<\frac{32}{3\sqrt{3}\pi}d$ and $B\leq 4A\pi^2$,
 we obtain $r<\Phi(w)$.

Since $0\leq w(x)\leq d$ for each $x\in [0,1]$, assumption
(A1) ensures that
$$
\int_0^{3/8}F(x,w(x))\,dx+\int_{5/8}^{1}F(x,w(x))\,dx\geq 0,
$$
and so
\begin{align*}
\Psi(w)\geq\int_{3/8}^{5/8}F(x,d)\,dx.
\end{align*}
Therefore, we obtain
\begin{equation}\label{eq:2}
\frac{\Psi(w)}{\Phi(w)}
\geq \frac{27}{4096}\frac{\int_{3/8}^{5/8}F(x,d) \,dx}{Bd^2}>\frac{1}{\lambda}.
\end{equation}
Therefore, from \eqref{eq:1} and \eqref{eq:2}, condition (i) of
Theorem \ref{teo:bon} is fulfilled.

Now, to prove the coercivity of the functional $I_{\lambda}$.
By (A3), we have
\[
\limsup_{|\xi|\to +\infty} \frac{\sup_{x \in [0,1]}F(x,\xi)}{\xi^2}
<\frac{\pi^4A}{\lambda}.
\]
So, we can fix $\varepsilon >0$ satisfying
\[
\limsup_{|\xi|\to +\infty} \frac{\sup_{x \in [0,1]}F(x,\xi)}{\xi^2}
<\varepsilon<\frac{\pi^4A}{\lambda}.
\]
Then, there exists a positive constant $\theta$ such that
$$
F(x,t)\leq \varepsilon |t|^{2}+\theta \quad \forall x \in [0,1], \; \forall
t \in \mathbb{R}\,.
$$
Taking into account \eqref{5} and \eqref{arsi},
we have
\[
I_{\lambda}(u)=\Phi(u)-\lambda \Psi(u)
\geq A\|u\|^2 -\lambda \varepsilon\|u\|^2_{L^2[0,1]}-\lambda \theta\geq
\big(A-\frac{\lambda \varepsilon}{\pi^4}\big)\|u\|^2-\lambda \theta
\]
for all $u \in X$. So, the functional $I_{\lambda}$ is coercive.
Now, the conclusion of Theorem \ref{teo:bon} can be used. It follows that
for every
\[
\lambda \in \Lambda\subseteq
\Bigl] \frac{\Phi({w})}{\Psi({w})},
\frac{r}{\sup_{\Phi(u)\leq r}\Psi(u)}\Bigr[\,,
\]
the functional $I_{\lambda}$ has at least three distinct critical
points in $X$, which are the weak solutions of the problem \eqref{e1.1}.
This completes the proof.
\end{proof}

Now, we present a consequence of Theorem \ref{thm1}.

\begin{corollary} \label{cor1}
Let $\alpha \in L^1([0,1])$ be a non-negative
and non-zero function and let $\gamma:\mathbb{R} \to \mathbb{R}$ be a continuous
function. Put $\alpha_0:=\int_{3/8}^{5/8}\alpha(x)dx$, $\|\alpha\|_1:=\int_0^1
\alpha(x)\,dx$ and $\Gamma(t)=\int_{0}^{t}\gamma(\xi)d\xi$ for all
$t \in \mathbb{R}$, and
assume that there exist two positive constants $c,d$, with
$c<\frac{32}{3\sqrt{3}\pi}d$, such that
\begin{itemize}
\item[(A1')] $\Gamma(t)\geq 0$ for all $t  \in  [0,d]$;

\item[(A2')]
\[
\frac{\max_{|t|\leq c}\Gamma(t)}{c^2}<\frac{27}{4096}\frac{\alpha_0}{\|\alpha\|_1}
\frac{ \Gamma(d)}{d^2};
\]

\item[(A3')] $ \limsup_{|\xi|\to +\infty}  \Gamma(\xi)/\xi^2\le 0$.
\end{itemize}
Then, for every
$$
\lambda \in \Big]\frac{4096}{27}\frac{B d^2}{\alpha_0\Gamma(d)} ,
 \frac{Bc^2}{\|\alpha\|_1\max_{|t|\leq c} \Gamma(t)}\Big[\,,
$$
the problem
\begin{equation}\label{earasi}
\begin{gathered}
u''''h(x,u')-u''=[\lambda \alpha(x)\gamma(u)+g(u)]h(x,u'),\quad  \text{in }(0,1),\\
u(0)=u(1)=0=u''(0)=u''(1)
\end{gathered}
\end{equation}
has at least three weak solutions.
\end{corollary}

The proof of the above corollary follows from Theorem \ref{thm1}
by choosing $f(x,t):=\alpha(x)\gamma(t)$ for all
 $(x,t) \in [0,1]\times \mathbb{R}$.

\begin{remark} \label{rmk3.2} \rm
Clearly, if $\gamma$ is non-negative then assumption (A1') is satisfied
 and (A2') becomes
 $$
\frac{ \Gamma(c)}{c^2}<\frac{27}{4096}
\frac{\alpha_0}{\|\alpha\|_1}\,\frac{ \Gamma(d)}{d^2}.
$$
\end{remark}

\begin{remark}\label{rem3.5}\rm
Theorem \ref{the0} in the introduction is an immediate consequence of
Corollary \ref{cor1}, on choosing $g(u)=-u$, $h\equiv 1$, $c=2$ and
$d=3\sqrt{3}$.
\end{remark}

The following lemma will be crucial in our arguments.

\begin{lemma}\label{teo:positive}
Assume that $f(x,t) \geq 0$ for all $(x,t) \in [0,1]\times \mathbb{R}$.
If $u$ is a weak solution of \eqref{e1.1}, then $u(x) \geq 0$
for all $x \in [0,1]$.
\end{lemma}

\begin{proof}
Arguing by contradiction, if we assume that $u$ is negative at a point of
$[0,1]$, the set
$$
\Omega:=\{x\in [0,1]:{u}(x)<0\},
$$
is non-empty and open. Let us consider
$
\bar{v}:=\min\{{u},0\},
$
one has, $\bar{v}\in X$. So, taking into account that ${u}$ is a weak solution and
by choosing $v=\bar{v}$, from our assumptions, one has
\begin{align*}
0&\geq \lambda\int_{\Omega} f(x,{u}(x)){u}(x)\,dx\\
&= \int_{\Omega}|{u}''(x)|^2\,dx+\int_{\Omega}
\Big(\int_0^{{u}'(x)}\frac{1}{h(x,\tau)}d\tau\Big) {u}'(x)\,dx
-\int_{\Omega}g({u}(x)){u}(x)\,dx\\
&\geq \frac{\pi^4-L}{\pi^4}\|{u}\|_{H^2(\Omega)\cap H_0^1(\Omega)}^2.
\end{align*}
Therefore, $\|{u}\|_{H^2(\Omega)\cap H_0^1(\Omega)}=0$ which is absurd. Hence,
the conclusion is achieved.
\end{proof}

Our other main result is as follows.

\begin{theorem} \label{thm2}
Assume that there exist three positive constants $c_1,c_2,d$,
satisfying $\frac{3\sqrt{3}\pi}{16\sqrt{2}}c_1<d<\frac{3\sqrt{3}}{64\sqrt{2}}c_2$,
such that
\begin{itemize}
\item[(B1)] $f(x,t)\geq 0$ for all $(x, t)  \in [0,\,1] \times [0,c_2]$;

\item[(B2)]
\[
 \frac{\int_0^1  F(x,c_1)\,dx}{c_1^2}<\frac{9}{2048}
\frac{\int_{3/8}^{5/8}F(x,d)\,dx}{d^2};
\]

\item[(B3)]
\[ \frac{\int_0^1  F(x,c_2)\,dx}{c_2^2}<\frac{9}{4096}
\frac{\int_{3/8}^{5/8}F(x,d)\,dx}{d^2}.
\]
\end{itemize}
Let
\[
\Lambda':=\Big]\frac{2048}{9}\frac{Bd^2}{\int_{3/8}^{5/8}F(x,d)\,dx},
 B\,\min \Big\{\frac{c_1^2}{\int_0^1 F(x,c_1)\,dx},
\frac{c_2^2}{2\int_0^1 F(x,c_2)\,dx} \Big\}\Big[.
\]
Then, for every $\lambda \in \Lambda'$ the problem \eqref{e1.1}
has at least three weak solutions $u_i$, $i=1,2,3$, such that
$0<\|u_i\|_{\infty}\leq c_2$.
\end{theorem}

\begin{proof}
Without loss of generality, we can assume $f(x,t) \geq 0$ for all
$(x,t) \in [0,1] \times \mathbb{R}$.
Fix $\lambda$ as in the conclusion and take $X, \Phi$ and
$\Psi$ as in the proof of Theorem \ref{thm1}.
Put $w$ as in Theorem \ref{thm1}, $r_1=Bc_1^2$ and
$r_2=Bc_2^2$. Therefore, one has $2r_1<\Phi(w)<\frac{r_2}{2}$
and we have
\begin{align*}
\frac{1}{r_1}\sup_{\Phi(u)<r_1}\Psi(u)
&\leq \frac{1}{Bc_1^2}\int_0^1 F(x,c_1)\,dx<\frac{1}{\lambda}\\
 &< \frac{9}{2048}\frac{\int_{3/8}^{5/8}F(x,d)\, dx}{Bd^2}\\
&\leq \frac{2}{3} \frac{\Psi(w)}{\Phi(w)} \,,
\end{align*}
and
\begin{align*}
\frac{2}{r_2}\sup_{\Phi(u)<r_2}\Psi(u)
&\leq \frac{2}{Bc_2^2}\int_0^1 F(x,c_2)\,dx<\frac{1}{\lambda}\\
 &<\frac{9}{2048}\frac{\int_{3/8}^{5/8}F(x,d)\, dx}{Bd^2}\\
&\leq \frac{2}{3} \frac{\Psi(w)}{\Phi(w)} \,.
\end{align*}
So, conditions (j) and (jj) of Theorem \ref{teo:can} are satisfied.
Finally, let $u_1$ and $u_2$ be two local minima for $\Phi-\lambda\Psi$.
 Then, $u_1$ and $u_2$ are critical points for $\Phi-\lambda\Psi$, and so,
they are weak solutions for the problem \eqref{e1.1}.
Hence, owing to Lemma \ref{teo:positive}, we obtain $u_1(x)\geq 0$ and
$u_2(x)\geq 0$ for all $x \in [0,1]$. So, one has $\Psi(s u_1+(1-s)u_2)\geq 0$
for all $s \in [0,1]$.
 From Theorem \ref{teo:can} the functional $\Phi-\lambda\Psi$ has at least
three distinct critical points which are weak solutions of \eqref{e1.1}.
This complete the proof.
\end{proof}

Now, we present a consequence of Theorem \ref{thm2}.

\begin{corollary} \label{corsi}
Let $\alpha\in L^1([0,1])$ be such that $\alpha(x)\geq 0$ a.e. $x\in
[0,1],\; \alpha\not\equiv 0$, and let
$\gamma:\mathbb{R}\to \mathbb{R}$ be a continuous
function. Put $\alpha_0:=\int_{3/8}^{5/8}\alpha(x)dx$, $\|\alpha\|_1:=\int_0^1
\alpha(x)\,dx$ and $\Gamma(t)=\int_{0}^{t}\gamma(\xi)d\xi$ for all
$t \in \mathbb{R}$, and
assume that there exist three positive constants $c_1,c_2,d$, with
$\frac{3\sqrt3\pi}{16\sqrt2}c_1<d<\frac{3\sqrt3}{64\sqrt{2}}c_2$, such that
\begin{itemize}
\item[(B1')] $\gamma(t)\geq 0$ for all $t  \in  [0,c_2]$;

\item[(B2')]
\[
\frac{\Gamma(c_1)}{c_1^2}<\frac{9}{2048}\frac{\alpha_0}{\|\alpha\|_1}
\frac{ \Gamma(d)}{d^2};
\]

\item[(B3')]
\[
\frac{\Gamma(c_2)}{c_2^2}<\frac{9}{4096}\frac{\alpha_0}{\|\alpha\|_1}
\frac{ \Gamma(d)}{d^2}.
\]
\end{itemize}
Then, for every
$$
\lambda \in \Big]\frac{2048}{9}\frac{B d^2}{\alpha_0\Gamma(d)} ,
 B\min\Big\{\frac{c_1^2}{\|\alpha\|_1 \Gamma(c_1)},
\frac{c_2^2}{2\|\alpha\|_1 \Gamma(c_2)}\Big\}\Big[\,,
$$
the problem \eqref{earasi} has at least three weak solutions $u_i$, $i=1,2,3$,
 such that $0<\|u_i\|_{\infty}\leq c_2$.
\end{corollary}

The proof of the above corollary follows from Theorem \ref{thm2}
by choosing $f(x,t):=\alpha(x)\gamma(t)$ for all
$(x,t) \in [0,1]\times \mathbb{R}$.

\begin{remark}\label{rem3.6}\rm
Theorem \ref{the01} in the introduction is an immediate consequence of
Corollary \ref{corsi}, on choosing $g(u)=u$, $h\equiv 1$, $c_1=2$, $c_2=2^{10}$,
and $d=3$.
\end{remark}

Finally, we present the following examples to illustrate our results.

\begin{example}\label{ex3.11}\rm
Consider the following problem
\begin{equation}\label{arsa}
\begin{gathered}
u''''-u''(2+x+\cos u')+u=\lambda f(u),\quad  \text{in }(0,1),\\
u(0)=u(1)=0=u''(0)=u''(1),
\end{gathered}
\end{equation}
where $f:\mathbb{R}\to \mathbb{R}$ is defined by
$$
f(t)=\begin{cases}
 2^{-10} &\text{if }|t| \leq 1,\\
   2^{-10}\,t^4 &\text{if }1<|t| \leq 32,\\
  2^{20}\,t^{-2} &\text{if } |t| > 32.
 \end{cases}
$$
Here, $g(t)=-t$ and $h(x,t)=(2+x+\cos t)^{-1}$
for all $x\in [0,1]$ and $t\in \mathbb{R}$. It is easy
to verify that (A2') and (A3') are satisfied with
$c=1$ and $d=32$.
 From Corollary \ref{cor1}, for each parameter
$$
\lambda\in \Big]\frac{48(\pi^4+4\pi^2+1)}{\pi^4},
\frac{512(\pi^4+4\pi^2+1)}{\pi^4}\Big[,
$$
problem \eqref{arsa} admits at least three weak solutions.
\end{example}

\begin{example}\label{ex3.12}\rm
Consider the  problem
\begin{equation}\label{arsa1}
\begin{gathered}
u''''-u''(3+\sin u')-2u=\lambda f(u),\quad  \text{in }(0,1),\\
u(0)=u(1)=0=u''(0)=u''(1),
\end{gathered}
\end{equation}
where $f:\mathbb{R}\to \mathbb{R}$ is defined by
$$
f(t)=\begin{cases}
 2^{-20} &\text{if }|t| \leq 2^{-5},\\
   t^4 &\text{if }2^{-5}<|t| \leq 1,\\
  t^{-2} &\text{if } |t| > 1.
 \end{cases}
$$
Here, $g(t)=2t$ and $h(x,t)=(3+\sin t)^{-1}$
for all $x\in [0,1]$ and $t\in \mathbb{R}$. It is easy
to verify that (B2') and (B3') are satisfied with
$c_1=2^{-5}$, $d=1$ and $c_2=2^{10}$.
 From Corollary \ref{corsi}, for each parameter
$$
\lambda\in \Big]\frac{2276(\pi^4+4\pi^2+2)}{\pi^4},
\frac{2^{14}(\pi^4+4\pi^2+2)}{\pi^4}\Big[,
$$
problem \eqref{arsa1} admits at least three weak solutions $u_i$, $i=1,2,3$,
such that $0<\|u_i\|_{\infty}\leq 1024$.
\end{example}

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\end{document}