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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 125, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/125\hfil Multiple solutions]
{Existence of multiple solutions for a mixed boundary-value problem}

\author[S. Heidarkhani, G. A. Afrouzi, A. Hadjian \hfil EJDE-2013/125\hfilneg]
{Shapour Heidarkhani, Ghasem Alizadeh Afrouzi, Armin Hadjian}  % in alphabetical order

\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{s.heidarkhani@razi.ac.ir}

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

\address{Armin Hadjian \newline
Department of Mathematics, Faculty of Mathematical Sciences,
University of Mazandaran, Babolsar, Iran}
\email{a.hadjian@umz.ac.ir}

\thanks{Submitted October 31, 2012. Published May 21, 2013.}
\subjclass[2000]{34B15, 35B38, 58E05}
\keywords{Mixed boundary value problem; critical points; variational
methods}

\begin{abstract}
 Using three critical points theorems, we
 prove the existence of at least three solutions for a
 second-order mixed boundary-value problem.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction}

In this article, we show the existence of at least three weak solutions 
for the  mixed boundary-value problem
\begin{equation}\label{e1.1}
\begin{gathered}
-(p u')'+q u=\lambda f(x,u)+g(u)\quad \text{in } (0,1),\\
u(0)=0,\quad u'(1)=0,
\end{gathered}
\end{equation}
where $p,q\in L^\infty([0,1])$ are such that
$$
p_0:=\operatorname{ess\,inf}_{x\in [0,1]}p(x)>0,\quad 
q_0:=\operatorname{ess\,inf}_{x\in [0,1]}q(x)\geq 0,
$$
$\lambda$ is a positive parameter,
$f:[0,1]\times\mathbb{R}\to\mathbb{R}$ is an
$L^1$-Carath\'{e}odory function and
$g:\mathbb{R}\to\mathbb{R}$ is a Lipschitz continuous
function with  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 $g(0)=0$.

Motivated by the fact that such  problems are used to
describe a large class of physical phenomena, many authors looked
for existence and multiplicity of solutions for second-order
ordinary differential nonlinear equations, with mixed conditions
at the ends. For an overview on this subject, we cite the
papers \cite{AveBuccTor, AveSa, AveGiovTorn, BonaTor, DHM, Sa}.
For instance, in \cite{BonaTor}, Bonanno and Tornatore, using
Ricceri's Variational Principle \cite{Ricceri1}, established the
existence of infinitely many weak solutions for the 
mixed boundary-value problem
\begin{gather*}
-(p u')'+q u=\lambda f(x,u)\quad \text{in } (a,b),\\
u(a)=u'(b)=0,
\end{gather*}
where $p,q\in L^\infty([a,b])$ such that
$$
p_0:=\operatorname{ess\,inf}_{x\in [a,b]}p(x)>0,\quad
q_0:=\operatorname{ess\,inf}_{x\in [a,b]}q(x)\geq 0,
$$
$f:[a,b]\times\mathbb{R}\to\mathbb{R}$ is a Carath\'{e}odory
function and $\lambda$ is a positive real parameter.

We also refer the reader to  \cite{HeiMot} which, by
means of an abstract critical point result of Ricceri
\cite{Ricceri2}, shows the existence of at least three solutions for the
 two-point boundary-value problem
\begin{gather*}
u''+(\lambda f(t,u)+g(u))h(t,u')=\mu p(t,u)h(t,u')\quad \text{ in } (a,b),\\
u(a)=u(b)=0,
\end{gather*}
where $\lambda$ and $\mu$ are positive parameters,
$f:[a,b]\times\mathbb{R}\to\mathbb{R}$ is continuous,
$g:\mathbb{R}\to\mathbb{R}$ is Lipschitz continuous with
$g(0)=0$, $h:[a,b]\times\mathbb{R}\to\mathbb{R}$ is bounded,
continuous, with $m:=\inf h>0$, and
$p:[a,b]\times\mathbb{R}\to\mathbb{R}$ is
$L^1$-Carath\'{e}odory function.


The goal of the present paper is to establish some new criteria
for  \eqref{e1.1} to have at least three weak solutions
(Theorems \ref{the3.1}-\ref{the3.3}). Our analysis is mainly based
on three recent critical point theorems that are contained in
Theorems \ref{the2.1}-\ref{the2.3} below. In fact, employing
rather different three critical points theorems, under different
assumptions on the nonlinear term $f$, we obtain the 
exact collections of $\lambda$ for whihc \eqref{e1.1}
admits at least three weak solutions in the space 
$\{u\in W^{1,2}([0,1]) : u(0)=0\}$.


A special case of our main results is the following theorem.

\begin{theorem}\label{the1.1}
Let $p,q\in L^\infty([a,b])$ such that
$$
p_0:=\operatorname{ess\,inf}_{x\in [a,b]}p(x)>0,\quad
q_0:=\operatorname{ess\,inf}_{x\in [a,b]}q(x)\geq 0,
$$
$g:\mathbb{R}\to\mathbb{R}$ be a Lipschitz continuous
function with the Lipschitz constant $L>0$ and $g(0)=0$ such that
$L<p_0$. Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function
and put $F(t)=\int_0^tf(\xi)d\xi$ for each $t\in \mathbb{R}$. Assume
that $F(d)>0$ for some $d>0$ and $F(\xi)\geq 0$ in $[0,d]$ and
$$
\liminf_{\xi\to 0}\frac{F(\xi)}{\xi^{2}}=0,\quad
\limsup_{|\xi|\to +\infty}\frac{F(\xi)}{\xi^{2}}=0.
$$
Then, there is $\lambda^*>0$ such that for each $\lambda>\lambda^*$
the problem
\begin{gather*}
-(p u')'+q u=\lambda f(u)+g(u)\quad \text{in } (0,1),\\
u(0)=0,\quad u'(1)=0,
\end{gather*}
admits at least three weak solutions.
\end{theorem}

\section{Preliminaries}

First we here recall for the reader's convenience our main tools to
prove the results; in the first one and the second one the
coercivity of the functional $\Phi-\lambda\Psi$ is required, while
in the third one a suitable sign hypothesis is assumed. The first
result has been obtained in \cite{Bonanno}, the second one in
\cite{BM} and the third one in \cite{AveBona2}. We recall the third
as given in \cite{BonaCan}.

\begin{theorem}[{\cite[Theorem 3.1]{Bonanno}}]\label{the2.1}
Let $X$ be a separable and reflexive real Banach space,
$\Phi:X\to\mathbb{R}$ a nonnegative continuously G\^{a}teaux
differentiable and sequentially weakly lower semicontinuous
functional whose G\^{a}teaux derivative admits a continuous inverse
on $X^{*}$, $\Psi:X\to\mathbb{R}$ a continuously G\^{a}teaux
differentiable functional whose G\^{a}teaux derivative is compact.
Assume that there exists $x_0\in X$ such that
$\Phi(x_0)=\Psi(x_0)=0$ and that
$$
\lim_{\|x\|\to+\infty}(\Phi(x)-\lambda\Psi(x))=+\infty\quad
\text{for all }\lambda\in[0,+\infty[.
$$
Further, assume that there are $r>0$, $x_1\in X$ such that
$r<\Phi(x_1)$ and
$$
\sup_{x\in\overline{\Phi^{-1}(]-\infty,r[)}^{w}}\Psi(x)
<\frac{r}{r+\Phi(x_1)}\Psi(x_1);
$$
here $\overline{\Phi^{-1}(]-\infty,r[)}^{w}$ denotes the closure
of $\Phi^{-1}(]-\infty,r[)$ in the weak topology. Then, for each
$$
\lambda\in \Lambda_1:=\Big]
\frac{\Phi(x_1)}{\Psi(x_1)-\sup_{x\in\overline{\Phi^{-1}(]-\infty,r[)}^{w}}\Psi(x)},
\frac{r}{\sup_{x\in\overline{\Phi^{-1}(]-\infty,r[)}^{w}}\Psi(x)}\Big[,
$$
the equation
\begin{equation}\label{e2.1}
\Phi'(u)-\lambda \Psi'(u)=0
\end{equation}
has at least three solutions in $X$ and, moreover, for each $h>1$,
there exist an open interval
$$
\Lambda_2\subseteq\Big[0,\frac{hr}{r\frac{\Psi(x_1)}{\Phi(x_1)}
-\sup_{x\in\overline{\Phi^{-1}(-\infty,r[)}^{w}}\Psi(x)}\Big]
$$
and a positive real number $\sigma$ such that, for each
$\lambda\in\Lambda_2$,  equation \eqref{e2.1} has at least three
solutions in $X$ whose norms are less than $\sigma$.
\end{theorem}


\begin{theorem}\cite[Theorem 3.6]{BM}\label{the2.2}
Let $X$ be a reflexive real Banach space, let $ \Phi:X \to \mathbb{R}$ be a sequentially weakly lower semicontinuous, coercive and
continuously G\^{a}teaux differentiable whose G\^{a}teaux derivative
admits a continuous inverse on $X^*$, and let $\Psi:X\to \mathbb{R}$ be a sequentially weakly upper semicontinuous and continuously
G\^{a}teaux differentiable functional whose G\^{a}teaux derivative
is compact. Assume that there exist $r\in \mathbb{R}$ and $u_1\in
X$ with $0<r<\Phi(u_1)$, such that
\begin{itemize}
\item[(A1)] $\sup_{u\in\Phi^{-1}(]-\infty,r])}\Psi(u)<r\frac{\Psi(u_1)}{\Phi(u_1)}$;

\item[(A2)] for each $\lambda\in
\Lambda_{r}:=]\frac{\Phi(u_1)}{\Psi(u_1)},
\frac{r}{\sup_{u\in\Phi^{-1}(]-\infty,r])}\Psi(u)}[$ 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[Corollary 3.1]{BonaCan}}]\label{the2.3}
Let $X$ be a reflexive real Banach space, $\Phi:X \to \mathbb{R}$
be a convex, coercive and continuously G\^{a}teaux differentiable
functional whose derivative admits a continuous inverse on
$X^\ast$, $\Psi:X \to \mathbb{R}$ be a continuously G\^{a}teaux
differentiable functional whose derivative is compact, such that
\begin{enumerate}
\item $\inf_{X}\Phi=\Phi(0)=\Psi(0)=0$;
\item for each $\lambda>0$ and for every $u_1,\ u_2$ which are local
minima for the functional $\Phi-\lambda\Psi$ and such that
$\Psi(u_1)\geq 0$ and $\Psi(u_2)\geq 0$, one has
$$
\inf_{s\in[0,1]}\Psi(su_1+(1-s)u_2)\geq 0.
$$
\end{enumerate}
Assume that there are two positive constants $r_1,r_2$ and
$\overline{v}\in X$, with $2r_1<\Phi(\overline{v})<\frac{r_2}{2}$,
such that
\begin{itemize}
\item[(B1)] $\frac{\sup_{u\in\Phi^{-1}(]-\infty,r_1[)}\Psi(u)}{r_1}<
\frac{2\Psi(\overline{v})}{3\Phi(\overline{v})}$;
\item[(B2)] $\frac{\sup_{u\in\Phi^{-1}(]-\infty,r_2[)}\Psi(u)}{r_2}<
\frac{1}{3}\frac{\Psi(\overline{v})}{\Phi(\overline{v})}$.
\end{itemize}
Then, for each
$\lambda$ in
\[
\Big]\frac{3}{2}\frac{\Phi(\overline{v})}{\Psi(\overline{v})},
\min\{ \frac{r_1}{\sup_{u\in\Phi^{-1}(]-\infty,r_1[)}\Psi(u)},
\frac{\frac{r_2}{2}}{\sup_{u\in\Phi^{-1}(]-\infty,r_2[)}\Psi(u)}\}\Big[\,,
\]
the functional $\Phi-\lambda \Psi$ has at least three distinct
critical points which lie in $\Phi^{-1}(]-\infty,r_2[)$.
\end{theorem}

Let $f:[0,1]\times\mathbb{R}\to\mathbb{R}$ be an
$L^1$-Carath\'{e}odory function and
$g:\mathbb{R}\to\mathbb{R}$ be 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 $g(0)=0$.

Put
$$
F(x,t):=\int_0^t f(x,\xi)d\xi,\quad
G(t):=-\int_0^t g(\xi)d\xi
$$
for all $x\in [0,1]$ and $t\in\mathbb{R}$. Denote
$$
X:=\big\{u\in W^{1,2}([0,1]) : u(0)=0\big\};
$$
the usual norm in $X$ is defined by
$$
\|u\|_X:=\Big(\int_0^1 (u(x))^2dx+\int_0^1
(u'(x))^2dx\Big)^{1/2}.
$$
For every $u,v\in X$, we define
\begin{equation}\label{e2.2}
(u,v):=\int_0^1 p(x)u'(x)v'(x)dx+\int_0^1 q(x)u(x)v(x)dx.
\end{equation}
Clearly, \eqref{e2.2} defines an inner product on $X$ whose
corresponding norm is
$$
\|u\|:=\Big(\int_0^1 p(x)(u'(x))^2dx+\int_0^1
q(x)(u(x))^2dx\Big)^{1/2}.
$$
Then, it is easy to see that the norm $\|\cdot\|$ on $X$ is
equivalent to $\|\cdot\|_X$. In the following, we will use
$\|\cdot\|$ instead of $\|\cdot\|_X$. Note that $X$ is a
separable and reflexive real Banach space.

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

By standard regularity results, if $f$ is a continuous function,
$p\in C^1([0,1])$ and $q\in C^0([0,1])$, then weak solutions of
the problem \eqref{e1.1} belong to $C^2([0,1])$, thus they are
classical solutions.

It is well known that $(X,\|\cdot\|)$ is compactly embedded in
$(C^0([0,1]),\|\cdot\|_\infty)$ and
\begin{equation}\label{e2.3}
\|u\|_\infty \leq \frac{1}{\sqrt{p_0}}\|u\|
\end{equation}
for all $u\in X$ (see, e.g., \cite{Talenti}).

Also, we use the following notation:
$$
\|p\|_\infty:=\operatorname{ess\,sup}_{x\in [0,1]}p(x),\quad
\|q\|_\infty:=\operatorname{ess\,sup}_{x\in [0,1]}q(x).
$$

Suppose that the Lipschitz constant $L>0$ of the function $g$
satisfies $L<p_0$. Finally, put
$$
k:=\frac{3p_0}{6\|p\|_\infty+2\|q\|_\infty}, \quad
\tau:=\frac{p_0-L}{p_0+L}.
$$
For other basic notations and definitions, we refer the reader to
\cite{Hei, Zeidler}.

\section{Main results}

Our main results are the following theorems.

\begin{theorem}\label{the3.1}
Assume that there exist a function $w\in X$, a positive function
$a\in L^1$ and two positive constants $r$ and $\gamma$ with
$\gamma<2$ such that
\begin{itemize}
\item[(A1)] $\|w\|^2>\frac{2p_0\,r}{p_0-L};$
\item[(A2)] $\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx
 <r\frac{\int_0^1 F(x,w(x))dx}{r+\frac{p_0+L}{2p_0}\|w\|^2};$ 
\item[(A3)] $F(x,t)\leq a(x)(1+|t|^\gamma)$ for almost every $x\in [0,1]$ and
for all $t\in \mathbb{R}$.
\end{itemize}
Then, for each $\lambda$ in
\[
\Lambda_1:=\Big]\frac{\frac{p_0+L}{2p_0}\|w\|^2}{\int_0^1
F(x,w(x))dx-\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}} F(x,t)dx},
\frac{r}{\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}\Big[,
\]
problem \eqref{e1.1} admits at least three weak solutions in
$X$ and, moreover, for each $h>1$, there exist an open interval
$$
\Lambda_2\subseteq\Big[0,\frac{hr}{\frac{2p_0\,r}{(p_0+L)\|w\|^2}\int_0^1
F(x,w(x))dx-\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}\Big]
$$
and a positive real number $\sigma$ such that, for each
$\lambda\in\Lambda_2$, the problem \eqref{e1.1} admits at least
three weak solutions in $X$ whose norms are less than $\sigma$.
\end{theorem}

\begin{theorem}\label{the3.2}
Assume that there exist a function $w\in X$ and a positive constant
$r$ such that
\begin{itemize}
\item[(B1)] $\|w\|^2>\frac{2p_0\,r}{p_0-L}$;
\item[(B2)] $\frac{\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}{r}
<\frac{2p_0}{p_0+L}\frac{\int_0^1 F(x,w(x))dx}{\|w\|^2}$;
\item[(B3)] $ \frac{2}{p_0-L}\limsup_{|t|\to +\infty
}\frac{F(x,t)}{t^2}<\frac{\int_0^1\sup_{|t|
\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}{r}$.
\end{itemize}
Then, for each
$$
\lambda\in\Big]\frac{p_0+L}{2p_0}\frac{\|w\|^2}{\int_0^1
F(x,w(x))dx},\
\frac{r}{\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}
F(x,t)dx}\Big[,
$$
problem \eqref{e1.1} admits at least three weak solutions.
\end{theorem}

\begin{theorem}\label{the3.3}
Suppose that $f:[0,1]\times \mathbb{R}\to \mathbb{R}$ satisfies the
condition $f(x,t)\geq 0$ for all $x\in [0,1]$ and $t\in\mathbb{R}$. Assume
that there exist a function $w\in X$ and two positive constants
$r_1$ and $r_2$ with
$\frac{4p_0\,r_1}{p_0-L}<\|w\|^2<\frac{p_0\,r_2}{p_0+L}$ such that
\begin{itemize}
\item[(C1)] 
\[
\frac{\int_0^1\sup_{|t|\leq\sqrt{\frac{2r_1}{p_0-L}}}
F(x,t) dx }{r_1}<\frac{4p_0}{3(p_0+L)}\frac{\int_0^1 F(x,w(x))dx}
{\|w\|^2};
\]
\item[(C2)] 
\[
\frac{\int_0^1\sup_{|t|\leq\sqrt{\frac{2r_2}{p_0-L}}}
F(x,t) dx }{r_2}<\frac{2p_0}{3(p_0+L)}\frac{\int_0^1 F(x,w(x))dx}
{\|w\|^2}.
\]
\end{itemize}
Then, for each
$$
\lambda\in\Big]\frac{3(p_0+L)}{4p_0}\frac{\|w\|^2}{\int_{0}^1F(x,w(x))dx},\
\Theta_1\Big[,
$$
where
$$
\Theta_1:=\min\Big\{\frac{r_1}{\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}
F(x,t)dx},\frac{\frac{r_2}{2}}{\int_0^1
\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}\Big\},
$$
problem \eqref{e1.1} admits at least three nonnegative weak
solutions $v^1,v^2,v^3$ such that
$$
|v^{j}(x)|<\sqrt{\frac{2r_2}{p_0-L}}
$$
for each $x\in [0,1]$ and $j=1,2,3$.
\end{theorem}

Let us give particular consequences of Theorems
\ref{the3.1}-\ref{the3.3} for a fixed test function $w$.


\begin{corollary}\label{cor3.4}
Assume that there exist a positive function $a\in L^1$ and three
positive constants $c$, $d$ and $\gamma$ with $c<\sqrt{2}d$ and
$\gamma<2$ such that Assumption {\rm (A3)} in Theorem \ref{the3.1}
holds. Furthermore, suppose that
\begin{itemize}
\item[(A4)] $F(x,t)\geq 0$ for all $(x,t)\in
[0,\frac{1}{2}]\times[0,d]$;

 \item[(A5)] $\int_0^1\sup_{t\in [-c,c]}F(x,t)dx
<(k\tau c^2) \frac{\int_{1/2}^{1}F(x,d)dx}{k\tau c^2+d^2}$.
\end{itemize}
Then, for each $\lambda$ in
$$
\Lambda'_1:=\Big]
\frac{\frac{p_0+L}{2k}d^2}{\int_{1/2}^{1}F(x,d)dx
 -\int_0^1\sup_{t\in[-c,c]}F(x,t)dx},
\frac{(p_0-L)c^2}{2\int_0^1\sup_{t\in [-c,c]}F(x,t)dx}\Big[,
$$
problem \eqref{e1.1} admits at least three weak solutions in
$X$ and, moreover, for each $h>1$, there exist an open interval
$$
\Lambda'_2\subseteq\Big[0,\frac{(p_0-L)h c^2/2}{\frac{2k\tau
c^2}{d^2}\int_{1/2}^{1}F(x,d)dx-\int_0^1\sup_{t\in
[-c,c]}F(x,t)dx}\Big]
$$
and a positive real number $\sigma$ such that, for each
$\lambda\in\Lambda'_2$,  problem \eqref{e1.1} admits at least
three weak solutions in $X$ whose norms are less than $\sigma$.
\end{corollary}

\begin{proof}
We claim that all the assumptions of Theorem \ref{the3.1} are
fulfilled with $w$ given by
\begin{equation}\label{e3.1}
w(x):=\begin{cases}
2 d^2 x, & x\in [0,1/2[,\\
d, & x\in [1/2,1].
\end{cases}
\end{equation}
and $r:=(p_0-L)c^2/2$. It is easy to verify that $w\in X$ and, in
particular, one has
$$
2p_0 d^2\leq\|w\|^2\leq\frac{p_0 d^2}{k}.
$$
Hence, taking into account that $c<\sqrt{2}d$, we have
$$
\|w\|^2>\frac{2p_0\,r}{p_0-L}.
$$
Thus, (A1) holds. Since $0\leq w(x)\leq d$ for each
$x\in[0,1]$, the condition (A4) ensures that
$$
\int_{0}^{1/2}F(x,w(x))dx\geq 0,
$$
so from (A5),
\begin{align*}
\int_0^1\sup_{t\in [-c,c]}F(x,t)dx &<(k\tau c^2)
\frac{\int_{1/2}^{1}F(x,d)dx}{k\tau c^2+d^2}\\
&=\frac{(p_0-L)kc^2}{(p_0-L)kc^2+(p_0+L)d^2}\int_{1/2}^{1}F(x,d)dx\\
&=\frac{(p_0-L)c^2}{2}\frac{\int_{1/2}^{1}F(x,d)dx}{\frac{(p_0-L)c^2}{2}+\frac{(p_0+L)d^2}{2k}}\\
&\leq r\frac{\int_{ 0}^1F(x,w(x))dx}{r+\frac{p_0+L}{2p_0}\|w\|^2},
\end{align*}
and thus (A2) holds. Next notice that
\begin{align*}
&\frac{\frac{p_0+L}{2p_0}\|w\|^2}{\int_0^1
F(x,w(x))dx-\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}\\
&\leq\frac{\frac{p_0+L}{2k}d^2}{\int_{1/2}^{1}F(x,d)dx-\int_{0}^1\sup_{t\in[-c,c]}F(x,t)dx}
\end{align*}
and
$$
\frac{r}{\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}
=\frac{(p_0-L)c^2}{2\int_{0}^1 \sup_{t\in[-c,c]}F(x,t)dx}.
$$
In addition note
\begin{align*}
&\frac{\frac{p_0+L}{2k}d^2}{\int_{1/2}^{1}F(x,d)dx-\int_{0}^1\sup_{t\in[-c,c]}F(x,t)dx}\\
&<\frac{\frac{p_0+L}{2k}d^2}{\Big{(}\frac{\frac{(p_0-L)c^2}{2}+
\frac{(p_0+L)d^2}{2k}}{\frac{(p_0-L)c^2}{2}}-1\Big{)}
\int_0^1\sup_{t\in[-c,c]}F(x,t)dx}\\
&=\frac{(p_0-L)c^2}{2\int_{0}^1 \sup_{t\in[-c,c]}F(x,t)dx}.
\end{align*}
Finally note that
\begin{align*}
&\frac{hr}{\frac{2p_0\,r}{(p_0+L)\|w\|^2}\int_0^1
F(x,w(x))dx-\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}\\
&\leq\frac{(p_0-L)h c^2/2}{\frac{2k\tau
c^2}{d^2}\int_{1/2}^{1}F(x,d)dx-\int_0^1\sup_{t\in [-c,c]}F(x,t)dx},
\end{align*}
 and taking into account that $\Lambda'_1\subseteq \Lambda_1$
and $\Lambda_2\subseteq \Lambda'_2$, we have the desired conclusion
directly from Theorem \ref{the3.1}.
\end{proof}

\begin{corollary}\label{cor3.5}
Assume that there exist two positive constants $c$ and $d$ with
$c<d$ such that the assumption {\rm (A4)} in Corollary \ref{cor3.4}
holds. Furthermore, suppose that
\begin{itemize}
\item[(B4)] $\int_0^1\sup_{t\in [-c,c]}F(x,t)dx<\frac{k\tau
c^2}{d^2}\int_{1/2}^{1}F(x,d)dx$;
 \item[(B5)] $\limsup_{|t|\to+\infty
}\frac{F(x,t)}{t^2}<\frac{\int_0^1\sup_{t\in
[-c,c]}F(x,t)dx}{c^2}$.
\end{itemize}
Then, for each
$$
\lambda\in\Big]\frac{p_0+L}{2k}\frac{d^2}{\int_{1/2}^{1}F(x,d)dx},
\frac{(p_0-L)c^2}{2\int_0^1\sup_{t\in [-c,c]}F(x,t)dx}\Big[,
$$
problem \eqref{e1.1} admits at least three weak solutions.
\end{corollary}

\begin{proof}
All the assumptions of Theorem \ref{the3.2} are fulfilled by
choosing $w$ as given in \eqref{e3.1} and $r:=(p_0-L)c^2/2$, and
bearing in mind that
$$
2p_0 d^2\leq\|w\|^2\leq\frac{p_0 d^2}{k}.
$$
and recalling
$$
\int_{0}^{1/2}F(x,w(x))dx\geq 0.
$$
Hence, by applying Theorem \ref{the3.2} we have the conclusion.
\end{proof}

\begin{proof}[Proof of Theorem \ref{the1.1}]
Fix $\lambda>\lambda^*:= \frac{(p_0+L)d^2}{kF(d)}$ for some $d>0$.
Since
$$
\liminf_{\xi\to 0}\frac{F(\xi)}{\xi^{2}}=0,
$$
there is
$\{c_m\}_{m\in \mathbb{N}}\subseteq ]0,+\infty[$ such that
$\lim_{m\to +\infty} c_{m}=0$ and
$$
\lim_{m\to +\infty}\frac{\sup_{|\xi| \leq c_{m}}F(\xi)}{c_m}=0.
$$
In fact, one has
$$
\displaystyle \lim _{m\to +\infty}\frac{\sup_{|\xi| \leq
c_{m}}F(\xi)}{c_m}=\lim_{m\to
+\infty}\frac{F(\xi_{c_m})}{\xi_{c_m}^{2}}.
\frac{\xi_{c_m}^{2}}{c_m}=0,
$$
where $F(\xi_{c_m})=\sup_{|\xi|\leq c_m}F(\xi)$. Hence, there is
$\overline{c}>0$ such that
$$
\frac{\sup_{|\xi| \leq\overline{c}}F(\xi)}{\overline{c}^{2}}
<\min\Big\{\frac{k\tau F(d)}{2d^{2}};\
\frac{p_0-L}{2\lambda}\Big\}
$$
and $\overline{c}<d$. From
Corollary \ref{cor3.5} we have the desired conclusion.
\end{proof}


\begin{corollary}\label{cor3.6}
Suppose that $f:[0,1]\times\mathbb{R}\to\mathbb{R}$ satisfies the
condition $f(x,t)\geq 0$ for all $x\in [0,1]$ and $t\in\mathbb{R}$. Assume
that there exist three positive constants $c_1,c_2$ and $d$ with
$c_1<d$ and $\sqrt{\frac{2}{k\tau}}d<c_2$ such that
\begin{itemize}
\item[(C3)] $\int_0^1\sup_{t\in
[-c_1,c_1]}F(x,t)dx<\frac{2k\tau
c_1^2}{3d^2}\int_{1/2}^{1}F(x,d)dx$;
\item[(C4)] $\int_0^1\sup_{t\in [-c_2,c_2]}F(x,t)dx<\frac{\tau c_2^2}{3d^2}\int_{1/2}^{1}F(x,d)dx$.
\end{itemize}
Then, for each
$$
\lambda\in\Big]\frac{3(p_0+L)}{4k}\frac{d^2}{\int_{1/2}^{1}F(x,d)dx},\,\Theta_2\Big[,
$$
where
$$
\Theta_2:=\min\Big\{\frac{(p_0-L)c_1^2}{2\int_0^1\sup_{t\in
[-c_1,c_1]}F(x,t)dx},\frac{(p_0-L)c_2^2}{4\int_0^1\sup_{t\in
[-c_2,c_2]}F(x,t)dx}\Big\},
$$
problem \eqref{e1.1} admits at least three nonnegative weak
solutions $v^1,\,v^2,\,v^3$ such that $|v^{j}(x)|<c_2$ for each
$x\in [0,1]$ and $j=1,2,3$.
\end{corollary}

\begin{proof}
Following the same way as in the proof of Corollary \ref{cor3.5}, we
achieve the stated assertion by applying Theorem  \ref{the3.3} with
$w$ as given in \eqref{e3.1}, $r_1:=(p_0-L)c_1^2/2$ and
$r_2:=(p_0-L)c_2^2/2$.
\end{proof}

We point out that, applying Theorems \ref{the3.1}-\ref{the3.3}, we
have the relevant results of Corollaries \ref{cor3.4}-\ref{cor3.6}
for the following mixed boundary value problem with a complete
equation
\begin{equation}\label{e3.3}
\begin{gathered}
-(\bar{p}u')'+\bar{r}u'+\bar{q}u=\lambda f(x,u)+g(u)\quad \text{in } (a,b),\\
u(0)=0,\quad u'(1)=0,
\end{gathered}
\end{equation}
where $f:[0,1]\times\mathbb{R}\to\mathbb{R}$ is a continuous
function, $g:\mathbb{R}\to\mathbb{R}$ is a Lipschitz
continuous function with the Lipschitz constant $L>0$ and $g(0)=0$,
$\bar{p}\in C^1([0,1]),\,\bar{q},\bar{r}\in C^0([0,1])$ and
$\lambda$ is a positive parameter. Moreover, $\bar{p}$ is
nonnegative and $R$ is a primitive of $\bar{r}/\bar{p}$.

 If fact, since the solutions of problem \eqref{e3.3} are solutions 
of the problem
\begin{gather*} 
-(e^{-R}\bar{p} u')'+e^{-R}\bar{q}u=\Big{(}\lambda f(x,u)
 +g(u)\Big{)}e^{-R}\quad \text{in } (0,1),\\
u(0)=0,\quad u'(1)=0,
\end{gather*}
assuming the Lipschitz constant $L>0$ of the function $g$
satisfies 
\[
L<\min_{x\in [0,1]}e^{-R(x)}\bar{p}(x),
\]
 and setting
$$
k':=\frac{3\min_{x\in
[0,1]}e^{-R(x)}\bar{p}(x)}{6\|e^{-R}\bar{p}\|_\infty+2\|e^{-R}\bar{q}\|_\infty},
\quad
\tau':=\frac{\min_{x\in [0,1]}e^{-R(x)}\bar{p}(x)-L}{\min_{x\in
[0,1]}e^{-R(x)}\bar{p}(x)+L},
$$
under the assumptions of Corollary \ref{cor3.4} but with 
(A5) replaced by the assumption
$$
\int_0^1\sup_{t\in [-c,c]}e^{-R(x)}F(x,t)dx<(k'\tau' c^2)
\frac{\int_{1/2}^{1}e^{-R(x)}F(x,d)dx}{k'\tau' c^2+d^2},
$$ 
by the same reasoning as in the proof of Corollary \ref{cor3.4}, using
Theorem \ref{the3.1}, for each $\lambda$ in
\begin{align*}
\Lambda''_1:=\Big]&\frac{\frac{\min_{x\in
[0,1]}e^{-R(x)}\bar{p}(x)+L}{2k'}d^2}{\int_{1/2}^{1}e^{-R(x)}F(x,d)dx-\int_0^1\sup_{t\in[-c,c]}e^{-R(x)}F(x,t)dx},\\
&\frac{(\min_{x\in
[0,1]}e^{-R(x)}\bar{p}(x)-L)c^2}{2\int_0^1\sup_{t\in
[-c,c]}e^{-R(x)}F(x,t)dx}\Big[,
\end{align*}
problem \eqref{e3.3} admits at least three classical solutions
in $X$;  moreover, for each $h>1$, there exist an open interval
$$
\Lambda''_2\subseteq\Big[0,\frac{(\min_{x\in
[0,1]}e^{-R(x)}\bar{p}(x)-L)h c^2/2}{\frac{2k'\tau'
c^2}{d^2}\int_{1/2}^{1}e^{-R(x)}F(x,d)dx-\int_0^1\sup_{t\in
[-c,c]}e^{-R(x)}F(x,t)dx}\Big]
$$
and a positive real number $\sigma$ such that, for each
$\lambda\in\Lambda''_2$, problem \eqref{e3.3} admits at least
three classical solutions in $X$ whose norms are less than
$\sigma$. Moreover, under the assumptions of Corollary
\ref{cor3.5}, but replacing Assumptions (B4) and (B5) by the
assumptions 
$$
\int_0^1\sup_{t\in [-c,c]}e^{-R(x)}F(x,t)dx
<\frac{k'\tau' c^2}{d^2}\int_{1/2}^{1}e^{-R(x)}F(x,d)dx
$$ 
and
$$
\limsup_{|t|\to+\infty}\frac{e^{-R(x)}F(x,t)}{t^2}
<\frac{\int_0^1\sup_{t\in
[-c,c]}e^{-R(x)}F(x,t)dx}{c^2},
$$
respectively, by the same
reasoning as given in the proof of Corollary \ref{cor3.5}, using
Theorem \ref{the3.2}, for each $\lambda$ in
$$
\Big]\frac{\min_{x\in
[0,1]}e^{-R(x)}\bar{p}(x)+L}{2k'}
\frac{d^2}{\int_{1/2}^{1}e^{-R(x)}F(x,d)dx},\frac{(\min_{x\in
[0,1]}e^{-R(x)}\bar{p}(x)-L)c^2}{2\int_0^1\sup_{t\in
[-c,c]}e^{-R(x)}F(x,t)dx}\Big[,
$$
problem \eqref{e3.3} admits at least three classical
solutions. Also, under the assumptions of Corollary \ref{cor3.6},
but replacing the condition $\sqrt{\frac{2}{k\tau}}d<c_2$,
Assumptions (C3) and (C4) by the condition
$\sqrt{\frac{2}{k'\tau'}}d<c_2$, the assumptions
$$
\int_0^1\sup_{t\in [-c_1,c_1]}e^{-R(x)}F(x,t)dx<\frac{2k'\tau'
c_1^2}{3d^2}\int_{1/2}^{1}e^{-R(x)}F(x,d)dx
$$ 
and
$$
\int_0^1\sup_{t\in [-c_2,c_2]}e^{-R(x)}F(x,t)dx <\frac{\tau'
c_2^2}{3d^2}\int_{1/2}^{1}e^{-R(x)}F(x,d)dx,
$$
respectively, by the same reasoning as in the proof of Corollary \ref{cor3.6}, 
using Theorem \ref{the3.3}, for each
$$
\lambda\in\Big]\frac{3(\min_{x\in
[0,1]}e^{-R(x)}\bar{p}(x)+L)}{4k'}\frac{d^2}{\int_{1/2}^{1}e^{-R(x)}F(x,d)dx},\,\Theta_3\Big[,
$$
where
$$
\Theta_3:=\min\Big\{\frac{(\min_{x\in
[0,1]}e^{-R(x)}\bar{p}(x)-L)c_1^2}{2\int_0^1\sup_{t\in
[-c_1,c_1]}e^{-R(x)}F(x,t)dx},\frac{(\min_{x\in
[0,1]}e^{-R(x)}\bar{p}(x)-L)c_2^2}{4\int_0^1\sup_{t\in
[-c_2,c_2]}e^{-R(x)}F(x,t)dx}\Big\},
$$
problem \eqref{e3.3} admits at least three nonnegative
classical solutions $v^1,v^2,v^3$ such that $|v^{j}(x)|<c_2$
for each $x\in [0,1]$ and $j=1,2,3$.

\section{Proofs}

\begin{proof}[Proof of Theorem \ref{the3.1}]
Our aim is to apply Theorem \ref{the2.1} to our problem. To this
end, for each $u\in X$, we let the functionals
$\Phi,\Psi:X\to\mathbb{R}$ be defined by
$$
\Phi(u):=\frac{1}{2}\|u\|^2 + \int_0^1 G(u(x))dx,\quad
\Psi(u):=\int_0^1 F(x,u(x))dx,
$$
and put
$$
I_{\lambda}(u):=\Phi(u)-\lambda\Psi(u)\quad\forall\ u\in X.
$$
 The functionals $\Phi$ and
$\Psi$ satisfy the regularity assumptions of Theorem \ref{the2.1}.
Indeed, by standard arguments, we have that $\Phi$ is G\^{a}teaux
differentiable and sequentially weakly lower semicontinuous and
its G\^{a}teaux derivative at the point $u\in X$ is the functional
$\Phi'(u)\in X^\ast$, given by
$$
\Phi'(u)(v)=\int_0^1 p(x)u'(x)v'(x)dx+\int_0^1
q(x)u(x)v(x)dx-\int_0^1 g(u(x))v(x)dx
$$
for every $v\in X$. Furthermore, the differential
$\Phi':X\to X^\ast$ is a Lipschitzian operator. Indeed, for
any $u,v\in X$, there holds
\begin{align*}
&\|\Phi'(u)-\Phi'(v)\|_{X^\ast}\\
&=\sup_{\|w\| \leq 1}|(\Phi'(u)-\Phi'(v),w)|\\
&\leq \sup_{\|w\|\leq 1} |(u-v,w)|+\sup_{\|w\|\leq
1}\int_0^1 |g(u(x))-g(v(x))|\,|w(x)|dx\\
&\leq \sup_{\|w\|\leq 1} \|u-v\|\,\|w\|
+\sup_{\|w\|\leq 1} \Big(\int_0^1
|g(u(x))-g(v(x))|^2\Big)^{1/2} \Big(\int_0^1
|w(x)|^2\Big)^{1/2}.
\end{align*}
Recalling that $g$ is Lipschitz continuous and the embedding
$X\hookrightarrow L^2([0,1])$ is compact, the claim is true. In
particular, we derive that $\Phi$ is continuously differentiable.
The inequality \eqref{e2.3} yields for any $u,v\in X$ the estimate
\begin{align*}
(\Phi'(u)-\Phi'(v),u-v) 
&= (u-v,u-v)-\int_0^1
\big{(}g(u(x))-g(v(x))\big{)}\big{(}u(x)-v(x)\big{)}dx\\
&\geq \frac{p_0-L}{p_0}\|u-v\|^2.
\end{align*}
By the assumption $L<p_0$, it turns out that $\Phi'$ is a strongly
monotone operator. So, by applying Minty-Browder theorem 
\cite[Theorem 26.A]{Zeidler}), $\Phi':X\to X^\ast$ 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 Gˆateaux
derivative at the point $u\in X$ is given by
$$
\Psi'(u)(v)=\int_0^1 f(x,u(x))v(x)dx
$$
for every $v\in X$. Note that the weak solutions of \eqref{e1.1}
are exactly the critical points of $I_{\lambda}$. Also, since $g$
is Lipschitz continuous and satisfies $g(0)=0$, we have from
\eqref{e2.3} that
\begin{equation}\label{e4.1}
\frac{p_0-L}{2 p_0}\|u\|^2\leq\Phi(u)\leq\frac{p_0+L}{2
p_0}\|u\|^2,
\end{equation}
for all $u\in X$, and so $\Phi$ is coercive.

Furthermore from (A3) for any fixed $\lambda\in
[0,+\infty[$, using \eqref{e4.1}, taking \eqref{e2.3} into
account, we have
\begin{align*}
\Phi(u)-\lambda \Psi(u)
&=\frac{1}{2}\|u\|^2+\int_0^1 G(u(x))dx-\lambda\int^1_{0}F(x,u(x))dx\\
&\geq \frac{p_0-L}{2p_0}\|u\|^2-\lambda\int_{0}^{1}(a(x)(1+|u(x)|^\gamma)dx\\
&\geq \frac{p_0-L}{2p_0}\|u\|^2-\lambda\|a\|_{L^1([0,1])}
(1+\frac{1}{p_0^{\gamma/2}}\|u\|^{\gamma}),
\end{align*}
and so
$$
\lim_{\|u\|\to+\infty}(\Phi(u)-\lambda\Psi(u))=+\infty.
$$
Also according to (A1) we achieve $\Phi(w)>r$. From the
definition of $\Phi$ and by using \eqref{e4.1} we have
\begin{align*}
\Phi^{-1}(]-\infty,r[)
&=\Big\{u\in X : \Phi(u)<r\Big\}\\
&\subseteq \Big\{u\in X : \|u\|<\sqrt{\frac{2p_0r}{p_0-L}}\Big\}\\
&\subseteq \Big\{u\in X : |u(x)|<\sqrt{\frac{2r}{p_0-L}} \quad
\text{for all } x\in [0,1]\Big\}.
\end{align*}
So, we obtain
$$
\sup_{u\in\overline{\Phi^{-1}(]-\infty,r[)}^{w}}\Psi(u) \leq
\int_{0}^1 \sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx .
$$
Therefore, from (A2) and \eqref{e4.1}, we have
\begin{align*}
\sup_{u\in\overline{\Phi^{-1}(]-\infty,r[)}^{w}}\Psi(u) &\leq
\int_{0}^1 \sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx\\
&< \frac{r}{r+\frac{p_0+L}{2p_0}\|w\|^2}\int_{0}^1F(x,w(x))dx \\
&< \frac{r}{r+\Phi(w)}\Psi(w).
\end{align*}
Now, we can apply Theorem \ref{the2.1}. Note for each $x\in [0,1]$,
\begin{align*}
\frac{\Phi(w)}{\Psi(w)-\sup_{u\in\overline{\Phi^{-1}(]-\infty,r[)}^{w}}\Psi(u)}
\leq \frac{\frac{p_0+L}{2p_0}\|w\|^2}{\int_0^1 F(x,w(x))dx-\int_0^1
\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}
\end{align*}
and
$$
\frac{r}{\sup_{u\in\overline{\Phi^{-1}(]-\infty,r[)}^{w}}\Psi(u)}\geq
\frac{r}{\int_0^1 \sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}.
$$
Note also that (A2) implies
\begin{align*}
&\frac{\frac{p_0+L}{2p_0}\|w\|^2}{\int_{0}^1F(x,w(x))dx-\int_0^1
\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}\\
&<\frac{\frac{p_0+L}{2p_0}\|w\|^2}{(\frac{r+\frac{p_0+L}{2p_0}\|w\|^2}{r}
\,-\,1)\int_0^1 \sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}
\\
&=\frac{r}{\int_0^1 \sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}.
\end{align*}
Also,
\begin{align*}
&\frac{hr}{r\frac{\Psi(w)}{\Phi(w)}
 -\sup_{u\in\overline{\Phi^{-1}(-\infty,r[)}^{w}}\Psi(u)}\\
&\leq\frac{hr}{\frac{2p_0r}{(p_0+L)\|w\|^2}\int_0^1F(x,w(x))dx-
\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}=\rho.
\end{align*}
From (A2) it follows that
\begin{align*}
&\frac{2p_0r}{(p_0+L)\|w\|^2}\int_0^1 F(x,w(x))dx-
\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx\\
&>\Big(\frac{2p_0r}{(p_0+L)\|w\|^2}-\frac{r}{r+\frac{p_0+L}{2p_0}\|w\|^2}\Big)\int_0^1 F(x,w(x))dx\\
&\geq\Big(\frac{2p_0r}{(p_0+L)\|w\|^2}-\frac{2p_0r}{(p_0+L)\|w\|^2}
\Big)\int_0^1 F(x,w(x))dx=0,
\end{align*}
since $\int_0^1 F(x,w(x))dx\geq 0$ (note $F(x,0)=0$ so
$\int_{0}^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx\geq 0$
and now apply (A2)). Now with $x_0=0$ and $x_1=w$ from
Theorem \ref{the2.1} (note $\Psi(0)=0$) it follows that, for each
$\lambda\in\Lambda_1$, the problem \eqref{e1.1} admits at least
three weak solutions and there exist an open interval
$\Lambda_2\subseteq[0,\rho]$ and a real positive number
 $\sigma$ such that, for each $\lambda\in\Lambda_2$, the problem
\eqref{e1.1} admits at least three weak solutions  whose norms in
$X$ are less than $\sigma$. Thus, the conclusion is achieved.
\end{proof}


\begin{proof}[Proof of Theorem \ref{the3.2}]
 To apply Theorem \ref{the2.2} to our problem, we take the functionals
$\Phi,\Psi:X\to\mathbb{R}$ as given in the proof of Theorem
\ref{the3.1}. Let us prove that the functionals $\Phi$ and $\Psi$
satisfy the conditions required in Theorem \ref{the2.2}. The
regularity assumptions on $\Phi$ and $\Psi$, as requested in
Theorem \ref{the2.2} hold. According to (B1) we deduce
$\Phi(w)>r$. From the definition of $\Phi$ we have
$$
\Phi^{-1}(]-\infty,r[)\subseteq \Big\{u\in X :
|u(x)|<\sqrt{\frac{2r}{p_0-L}}\quad \text{for all } x\in
[0,1]\Big\},
$$
and it follows that
$$
\sup_{u\in\Phi^{-1}(]-\infty,r[)}\Psi(u)\leq\int_0^1
\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx.
$$
Therefore, due to  assumption (B2), we have
\begin{align*}
\frac{\sup_{u\in\Phi^{-1}(]-\infty,r[)}\Psi(u)}{r}
&\leq \frac{\int_{0}^1 \sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}{r}\\
&<\frac{2p_0}{p_0+L}\frac{\int_0^1
F(x,w(x))dx}{\|w\|^2}\\
&\leq\frac{\Psi(w)}{\Phi(w)}.
\end{align*}
Furthermore, from (B3) there exist two constants 
$\eta, \vartheta\in \mathbb{R}$ with
$$
\eta<\frac{\int_{0}^1
\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}{r}
$$
such that
$$
\frac{2}{p_0-L}F(x,t)\leq \eta t^2+\vartheta
$$
for all $x\in [0,1]$ and all $t\in \mathbb{R}$. Fix $u\in X$. Then
\begin{equation}\label{e4.3}
F(x,u(x))\leq \frac{p_0-L}{2}(\eta |u(x)|^2+\vartheta)
\end{equation}
for all $x\in [0,1]$. Now, to prove the coercivity of the functional
$\Phi-\lambda\Psi$, first we assume that $\eta>0$. So, for any fixed
$$
\lambda\in\Big]\frac{p_0+L}{2p_0}\frac{\|w\|^2}{\int_0^1
F(x,w(x))dx},\
\frac{r}{\int_0^1\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}
F(x,t)dx}\Big[,
$$
using \eqref{e4.3}, we have
\begin{align*}
\Phi(u)-\lambda \Psi(u)
&=\frac{1}{2}\|u\|^2+\int_0^1 G(u(x))dx-\lambda\int^1_0 F(x,u(x))dx\\
&\geq \frac{p_0-L}{2p_0}\|u\|^2-\frac{\lambda(p_0-L)}{2}\Big(\eta
\int_0^1|u(x)|^2dx+\vartheta\Big)\\
&\geq \frac{p_0-L}{2p_0}\Big{(}1-\eta\frac{r}{\int_{0}^1
\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx}\Big{)}\|u\|^2
-\frac{\lambda(p_0-L)}{2}\vartheta,
\end{align*}
and thus
$$
\lim_{\|u\|\to+\infty}(\Phi(u)-\lambda\Psi(u))=+\infty.
$$
On the other hand, if $\eta\leq 0$, clearly we obtain
$\lim_{\|u\|\to+\infty}(\Phi(u)-\lambda \Psi(u))=+\infty$.
Both cases lead to the coercivity of functional $\Phi-\lambda \Psi$.

So, the assumptions (A1) and (A2) in Theorem
\ref{the2.2} are satisfied. Hence, by using Theorem \ref{the2.2},
the problem \eqref{e1.1} admits at least three distinct weak
solutions in $X$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{the3.3}]
 Let $\Phi$ and $\Psi$ be as in
the proof of Theorem \ref{the3.1}. Let us apply Theorem
\ref{the2.3} to our functionals. Obviously, $\Phi$ and $\Psi$
satisfy the condition (1) of Theorem \ref{the2.3}.

 Now, we show that the functional $\Phi-\lambda\Psi$ satisfies the
assumption (2) of Theorem \ref{the2.3}. Let $u^\ast$ and
$u^{\ast\ast}$ be two local minima for $\Phi-\lambda\Psi$. Then
$u^\ast$ and $u^{\ast\ast}$ are critical points for
$\Phi-\lambda\Psi$, and so, they are weak solutions for the
problem \eqref{e1.1}, and in particular they are nonnegative.
Indeed, by the similar reasoning as given in 
\cite[Theorem 3.1]{DHM}, let $u_0$ be a weak solution of the problem
\eqref{e1.1}. Arguing by a contradiction, assume that the set
$A=\big\{x \in ]0,1] : u_0(x)<0\big\}$ is nonempty and of positive
measure. Put $\bar v(x)=\min\{0, u_0(x)\}$ for all $x \in [0,1]$.
%
Clearly, $\bar v\in X$ and, taking into account that $u_0$ is a weak 
solution and by choosing $v=\bar v $, one has
\begin{align*}
&\int_0^1 p(x)u'_0(x)\bar v'(x)dx +\int_0^1 q(x)u_0(x)\bar v(x)dx \\
&- \lambda\int_0^1 f(x,u_{0}(x))\bar v(x)dx-\int_0^1 g(u_{0}(x))\bar
v(x)dx =0\,.
\end{align*}
Thus, from our sign assumptions on the data, we have
$$
\int_A p(x)|u'_0(x)|^2dx+\int_A q(x)|u_0(x)|^2dx-\int_A
g(u_0(x))u_0(x)dx\leq 0\,.
$$
On the other hand,
\begin{align*}
&\frac{p_0-L(m(A))^2}{p_0}\|u_0\|_{W^{1,2}(A)}^2\\
&\leq\int_A p(x)|u'_0(x)|^2dx+\int_A q(x)|u_0(x)|^2dx
-\int_A g(u_0(x))u_0(x)dx,
\end{align*}
where $m(A)$ is the Lebesgue measure of the set $A$. Hence,
$u_0\equiv0$ on $A$ which is absurd. 
Then, $u^{\ast}(x)\geq
0$ and $u^{\ast\ast}(x)\geq 0$ for every $x\in [0,1]$. Thus, it
follows that $su^{\ast}+(1-s)u^{\ast\ast}\geq 0$ for all $s\in
[0,1]$, and that
$$
f(x,su^{\ast}+(1-s)u^{\ast\ast})\geq 0,
$$
and consequently, $\Psi(su^{\ast}+(1-s)u^{\ast\ast})\geq 0$,
for every $s\in [0,1]$.

 Moreover, from the condition
$\frac{4p_0\,r_1}{p_0-L}<\|w\|^2<\frac{p_0\,r_2}{p_0+L}$, we
observe $2r_1<\Phi(w)<\frac{r_2}{2}$. From the definition of
$\Phi$ we have
$$
\Phi^{-1}(]-\infty,r[)\subseteq \Big\{u\in X :
|u(x)|<\sqrt{\frac{2r}{p_0-L}}\quad \text{for all } x\in
[0,1]\Big\},
$$
and it follows that
$$
\sup_{u\in\Phi^{-1}(]-\infty,r[)}\Psi(u)\leq\int_0^1
\sup_{|t|\leq\sqrt{\frac{2r}{p_0-L}}}F(x,t)dx.
$$
Therefore, due to the assumption (C1), we infer that
\begin{align*}
\frac{\sup_{u\in\Phi^{-1}(]-\infty,r_1[)}\Psi(u)}{r_1}
&\leq \frac{\int_{0}^1 \sup_{|t|\leq\sqrt{\frac{2r_1}{p_0-L}}}F(x,t)dx}{r_1}\\
&<\frac{4p_0}{3(p_0+L)}\frac{\int_0^1
F(x,w(x))dx}{\|w\|^2}\\
&\leq\frac{2}{3}\frac{\Psi(w)}{\Phi(w)}.
\end{align*}
As above, from assumption (C2), we deduce that
\begin{align*}
\frac{\sup_{u\in\Phi^{-1}(]-\infty,r_2[)}\Psi(u)}{r_2}
&\leq \frac{\int_0^1 \sup_{|t|\leq\sqrt{\frac{2r_2}{p_0-L}}}F(x,t)dx}{r_2}\\
&<\frac{2p_0}{3(p_0+L)}\frac{\int_0^1 F(x,w(x))dx}{\|w\|^2}\\
&\leq\frac{1}{3}\frac{\Psi(w)}{\Phi(w)}.
\end{align*}
So, the assumptions (B1) and (B2) in Theorem
\ref{the2.3} are satisfied. Hence, by using Theorem \ref{the2.3},
the problem \eqref{e1.1} admits at least three distinct weak
solutions in $X$. This completes the proof.
\end{proof}

\subsection*{Acknowledgments}
The authors express their sincere gratitude to the
referees for reading this paper and specially for their
valuable suggestions leading to improvements.

The research of Shapour Heidarkhani was partially supported by grant 91470046
from IPM.

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\end{document}