\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 254, pp. 1--18.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/254\hfil Multiple positive solutions]
{Multiple positive solutions to a fourth-order boundary-value problem}

\author[A. Cabada, R. Precup, L. Saavedra, S. A. Tersian \hfil EJDE-2016/254\hfilneg]
{Alberto Cabada, Radu Precup, Lorena Saavedra, Stepan A. Tersian}

\address{Alberto Cabada \newline
Instituto de Matem\'aticas, Facultade de Matem\'aticas,
Universidade de Santiago de Compostela, Santiago de Compostela, Galicia,
Spain}
\email{alberto.cabada@usc.es}

\address{Radu Precup \newline
Department of Mathematics, Babe\c{s}-Bolyai University, 400084
Cluj-Napoca, Romania}
\email{r.precup@math.ubbcluj.ro}

\address{Lorena Saavedra \newline
Instituto de Matem\'aticas, Facultade de Matem\'aticas,
Universidade de Santiago de Compostela, Santiago de Compostela, Galicia,
Spain}
\email{lorena.saavedra@usc.es}

\address{Stepan A. Tersian \newline
Department of Mathematics, University of Ruse, 7017 Ruse, Bulgaria.\newline
Institute of Mathematics and Informatics,
Bulgarian Academy of Sciences, 1113 - Sofia, Bulgaria}
\email{sterzian@uni-ruse.bg}

\thanks{Submitted May 16, 2016. Published September 21, 2016.}
\subjclass[2010]{34B18, 47J30}
\keywords{Fourth-order differential equation; boundary-value problem; 
\hfill\break\indent positive solution; critical point; fixed point}

\begin{abstract}
 We study the existence, localization and multiplicity of positive solutions
 for a nonlinear fourth-order two-point boundary value problem.
 The approach is based on critical point theorems in conical shells,
 Krasnosel'ski\u{\i}'s compression-expansion theorem, and unilateral
 Harnack type inequalities.
\end{abstract}

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

\section{Introduction}

The fourth-order boundary-value problems appear in the elasticity theory
describing stationary states of the deflection of an elastic beam. In the
last decade a lot of studies have been devoted to the existence of positive
solutions for such problems, applying the Leray-Schauder continuation
method, topological degree theory, fixed point theorems in cones, critical
point theory and lower and upper solution methods (see, for example,
\cite{CaT, CG, CFM, GKKY, ES, LZL, LZ, Ma, YCY, WIF}).

In this article, we study the existence and multiplicity of positive
solutions for nonlinear fourth-order two-point boundary value problem with
cantilever boundary conditions. More exactly, we consider the fourth-order
boundary value problem
\begin{equation}
\begin{gathered}
u^{(4)}(t)-\,f(t,u(t))=0, \quad  0<t<1, \\
u(0)=u'(0)=u''(1)=u'''(1)=0, 
\end{gathered}  \label{p}
\end{equation}
where the function $f\colon [ 0,1]\times \mathbb{R}\to \mathbb{R}$
is continuous, $f(t,\mathbb{R}_{+})\subset \mathbb{R}_{+}$ for all
$t\in[ 0,1]$, and the solution is sought in $C^{4}[ 0,1]$.

Our approach is based on critical point theorems for functionals in conical
shells (see \cite{Pre, Pre2}) and Krasnosel'ski\u{\i}'s
compression-expansion theorem. As one can see along the paper, the arguments
developed here can be applied to other boundary value problems associated to
fourth and sixth order differential equations. Because the estimates are
connected with specific boundary conditions, we concentrate only on the
model problem \eqref{p}.

The paper is organized as follows. In Section 2 we state the critical point
theorems in conical shells and Krasnosel'ski\u{\i}'s compression-expansion
theorem. We also present the fixed point formulation and the variational
formulation of the problem. In Section 3, the main existence and
multiplicity results
Theorems 3.2, 3.3, 3.5 and 3.6 are stated and proved.
Their proofs are based on the mentioned above theorems and on the
inequalities proved in Lemma 3.1 and Lemma 3.4.
Finally in Section 4, an example is presented.

\section{Preliminaries}

\subsection{Critical point theorems in conical shells}

In this subsection we introduce the results given in \cite{Pre} which we are
going to apply to the fourth order problem \eqref{p}.

For any real Hilbert space $H$ with inner product $(\cdot ,\cdot )_H$ and
norm $|\cdot |_H$, we let $H'$ be its dual space. Denoting by
$\langle \cdot ,\cdot \rangle $ the duality between $H$ and
$H'$, i.e. $\langle u^{\ast },u\rangle =u^{\ast }(u)$ for
$u^{\ast }\in H'$ and $u\in H$, according to the Riesz
representation theorem, we can consider the canonical isomorphism
$L_H:H\to H'$, given by
\begin{equation}
(u,v)_H=\langle L_H\,u,v\rangle \quad \text{for all  }u,\;
v\in H\,,  \label{bf}
\end{equation}
and its inverse $J_H:H'\to H$ for which
\begin{equation}
(J_Hu,v)_H=\langle u,v\rangle \quad \text{for }u\in
H'\,,\; v\in H\,.  \label{p2}
\end{equation}
Using this isomorphism we may identify $H$ with $H'$, letting
$L_Hu\equiv u$, $J_Hu\equiv u$, and so
$L_H=J_H=I_H\ (\text{ identity map of }H)$.

In what follows we consider two real Hilbert spaces, $X$ with inner product
and norm $(\cdot ,\cdot )_{X}$, $|\cdot |_{X}$, $\ $and $Y$ with inner
product and norm $(\cdot ,\cdot)_Y$, $| \cdot|_Y;$ we assume that $X$
is continuously embedded into $Y$ and that $Y$ is identified with $Y'$.
Then, from $X\subset Y$, one has $Y'\subset X'$, and therefore
\begin{equation}
X\subset Y\equiv Y'\subset X'.  \label{lll}
\end{equation}
Note that for every $u\in X$, the notation $J_{X}u$ will be used to denote
the element $J_{X}L_Yu$. Also, if $u,v\in Y$, then according to \eqref{bf}
and the identification $L_Yu=u$,
\begin{equation*}
\langle u,v\rangle =(u,v)_Y.
\end{equation*}
This is the reason for using the symbol $\langle \cdot ,\cdot
\rangle $ instead of $(\cdot ,\cdot)_Y$. In what
follows, for simplicity, the inner product and norm will be denoted by
$(\cdot ,\cdot) $ and $| \cdot | $ for $X$,
and by \thinspace $\langle \cdot ,\cdot \rangle $ and $\|
\cdot \| $ for $Y$. Also, we shall use the notations $L$ and $J$
instead of $L_{X}$ and $J_{X}$.

Let $K$ be a cone in $X$, i.e. a convex closed nonempty set $K$,
$K\neq \{ 0\}$, with $\lambda u\in K$ for every $u\in K$ and
$\lambda \geq 0$, and $K\cap (-K) =\{ 0\} $. For any
two positive numbers $R_0$ and $R_1$, we denote by $K_{R_0R_1}$ the
\textit{conical shell}
\begin{equation*}
K_{R_0R_1}:=\{ u\in K:\| u\| \geq R_0
\text{ and } | u| \leq R_1\} .
\end{equation*}
Such a set may be empty (even if $R_0<R_1)$ and may be disconnected. Let
$\phi \in K\setminus \{ 0\} $ be a fixed element with
$|\phi | =1$. If $R_0<\| \phi \| R_1$, then
$\mu \phi \in K_{R_0R_1}$ for every
$\mu \in [ R_0/\| \phi \| ,R_1] $, and $\mu \phi $ is an interior point of
$K_{R_0R_1}$, in the sense that $\| \mu \phi \| >R_0$
and $| \mu \phi | <R_1$, for $\mu \in (R_0/\| \phi \| ,R_1) $.
In particular, any two elements of $K_{R_0R_1}$ of the form $\mu \phi $,
 with $\mu \in [R_0/\| \phi \| ,R_1] $, belong to the same
connected component of $K_{R_0R_1}$.

In particular if $Y=X$, the spaces $X$ and $X'$ are identified, so
$L=J=I_{X}$. Also, in this case, the conical shell $K_{R_0R_1}$ is
nonempty and simply connected for every $R_0,R_1$ with $0<R_0<R_1$.

Let $E$ be a $C^{1}$ functional defined on $X$. We say that $E$ satisfies
the \textit{modified Palais-Smale-Schechter condition} (MPSS) in
$K_{R_0R_1}$, if any sequence $(u_k) $ of elements of
$K_{R_0R_1}$ for which the sequence $(E(u_k)) $
converges and one of the following conditions holds:
\begin{itemize}
\item[(i)]  $E'(u_k) \to 0$;

\item[(ii)]  $\| u_k\| =R_0$,
$(JE'( u_k) ,Ju_k) \geq 0$  and
$JE'(u_k)-\frac{(JE'(u_k) ,Ju_k) }{|Ju_k| ^2}Ju_k\to 0$;

\item[(iii)]  $| u_k| =R_1$, $(JE'(u_k) ,u_k) \leq 0$ and
$JE'(u_k) -\frac{(JE'(u_k) ,u_k) }{R_1^2}u_k\to 0$,
\end{itemize}
has a convergent subsequence.

We say that $E$ satisfies the \textit{compression boundary condition} in $
K_{R_0R_1}$ if
\begin{gather}
JE'(u) -\lambda Ju\neq 0\quad \text{for }u\in
K_{R_0R_1},\; \| u\| =R_0,\; \lambda >0;  \label{h1} \\
JE'(u) +\lambda u\neq 0\; \text{for }u\in
K_{R_0R_1},\; | u| =R_1,\; \lambda >0.  \label{h2}
\end{gather}
We say that $E$ has a \textit{mountain pass geometry} in $K_{R_0R_1}$ if
there exist $u_0$ and $u_1$ in the same connected component of
$K_{R_0R_1}, $ and $r>0$ such that $| u_0|<r<| u_1| $ and
\begin{equation*}
\max \{ E(u_0) ,E(u_1) \} <\inf
\{ E(u) : u\in K_{R_0R_1},\; | u|=r\} .
\end{equation*}
In this case we consider the set
\begin{equation}  \label{e-gamma}
\Gamma =\{ \gamma \in C([0,1];K_{R_0R_1}):\gamma (0)=u_0,\;
\gamma (1)=u_1\}
\end{equation}
and the number
\begin{equation}  \label{e-c}
c=\inf_{\gamma \in \Gamma } \max_{t\in [ 0,1]}E(\gamma (t))\,.
\end{equation}
Finally, we say that $E$ is \textit{bounded from below} in $K_{R_0R_1}$
if
\begin{equation}  \label{e-m}
m:=\inf_{u\in K_{R_0R_1}}E(u) >-\infty .
\end{equation}

We assume that the following conditions are satisfied:
\begin{equation}
(I-JE') (K) \subset K\quad (I \text{ is the identity map on }X);  \label{c1}
\end{equation}
and there exists a constant $\nu_0>0$ such that
\begin{gather}
(JE'(u) ,Ju) \leq \nu_0\quad \text{for all $u\in K$ with }\| u\| =R_0;  \label{b1}\\
(JE'(u) ,u) \geq -\nu_0\quad  \text{for all $u\in K$ with }| u| =R_1.  \label{b2}
\end{gather}
The following theorems of localization of critical points in a conical shell
appear as slight particularizations of the main results from \cite{Pre, Pre2}.

\begin{theorem}\label{t1}
 Assume that $E$ is bounded from below in $K_{R_0R_1}$ and
that there is a $\rho >0$ with
\begin{equation}
E(u) \geq m+\rho  \label{r11}
\end{equation}
($m$ given in \eqref{e-m}) for all $u\in K_{R_0R_1}$ which
simultaneously satisfy $| u| =R_1$, $\|u\| =R_0$. 
In addition assume that $E$ satisfies the (MPSS)
condition and the compression boundary condition in $K_{R_0R_1}$. Then
there exists $u\in K_{R_0R_1}$ such that
\begin{equation*}
E'(u) =0\quad \text{and}\quad E(u) =m.
\end{equation*}
\end{theorem}

\begin{theorem}\label{t2}
Assume that $E$ has the mountain pass geometry in $K_{R_0R_1}$
and that there is a $\rho >0$ with
\begin{equation}
| E(u) -c| \geq \rho  \label{r12}
\end{equation}
($c$ given in \eqref{e-c}) for all $u\in K_{R_0R_1}$ which
simultaneously satisfy $| u| =R_1$, $\|u\| =R_0$.
 In addition assume that $E$ satisfies the (MPSS)
condition and the compression boundary condition in $K_{R_0R_1}$. Then
there exists $u\in K_{R_0R_1}$ such that
\begin{equation*}
E'(u) =0\quad\text{and}\quad E(u) =c.
\end{equation*}
\end{theorem}

\begin{remark}\label{r} \rm
If the assumptions of both Theorems \ref{t1} and \ref{t2} are
satisfied, since $m<c$, then $E$ has two distinct critical points in
 $K_{R_0R_1}$.
\end{remark}

\subsection{Krasnosel'ski\u{\i}'s compression-expansion theorem}

Problem \eqref{p} can \break also be investigated by means of fixed point
techniques. In this article, we are mainly concerned with the variational
approach based on critical point theory. However, it deserves to comment
about the applicability of fixed point methods and the surplus of
information given by the variational approach.

Thus we shall report on the applicability of Krasnosel'ski\u{\i}'s
compression-ex\-pansion theorem (see \cite{GL, KZ}), which guarantees the
existence of a fixed point of a compact operator in a conical shell of a
Banach space.

\begin{theorem}[Krasnosel'ski\u{\i}] \label{tk}
Let $(X,| \cdot |) $ be a Banach
space and $K\subset X$ a cone. Let $R_0,R_1$ be two numbers with $
0<R_0<R_1$, $K_{R_0R_1}=\{u\in K:\ R_0\leq | u|
\leq R_1\}$, and let $N:K_{R_0R_1}\to K$ be a compact
operator. Let $<$ be the strict ordering induced in $X$ by the cone $K$,
i.e. $u<v$ if and only if $v-u\in K\setminus \{ 0\} $. Assume
that one of the following conditions is satisfied:
\begin{itemize}
\item[(a)] compression: \emph{(i)} $N(u) \nless u$ for all $u\in
K$ with $| u| =R_0$, and \emph{(ii)} $N(u)
\ngtr u$ for all $u\in K$ with $| u| =R_1;$

\item[(b)] expansion: \emph{(i)} $N(u) \ngtr u$ for all $u\in K$
with $| u| =R_0$, and \emph{(ii)} $N(u)
\nless u$ for all $u\in K$ with $| u| =R_1$.
\end{itemize}
Then $N$ has at least one fixed point in $K_{R_0R_1}$.
\end{theorem}

\subsection{Fixed point formulation of the problem}

For each $v\in L^2(0,1) $, the problem
\begin{equation} \label{e-4or-v}
\begin{gathered}
u^{(4)}(t)=v(t), \quad 0<t<1, \\
u(0)=u'(0)=u''(1)=u'''(1)=0 
\end{gathered}
\end{equation}
has in $H^{4}(0,1) $ a unique solution $u$ denoted by $Sv$,
namely
\begin{equation}
(Sv) (t) =\int_0^{1}G(t,s)\,v(s)\,ds\,,\quad  t\in [ 0,1] ,  \label{sol}
\end{equation}
where $G(t,s)$ is the corresponding Green's function. Obviously, 
$Sv\in C^{4} [ 0,1] $ if $v\in C[ 0,1] $. One can easily obtain the
expression of $G(t,s) $ using the Mathematica package developed
in \cite{cacima}, namely
\begin{equation}
G(t,s)=\begin{cases}
\frac{s^2}{6}(3\,t-s)\,, & 0\leq s\leq t\leq 1, \\
\frac{t^2}{6}(3\,s-t)\,, & 0\leq t<s\leq 1\,.
\end{cases}  \label{e-green}
\end{equation}
Then  problem \eqref{p} is equivalent to the integral equation
\begin{equation}
u(t)=\int_0^{1}G(t,s)\,f(s,u(s))\,ds\,,\quad  u\in C[ 0,1] .
\label{fpp}
\end{equation}
Obviously, \eqref{fpp} represents a fixed point equation associated to the
compact operator
\begin{equation}
N(u) (t) =\int_0^{1}G(t,s)\,f(s,u(s))\,ds,\quad t\in [ 0,1] .  \label{op}
\end{equation}
It is clear that $N(u) =Sf(\cdot ,u(\cdot)) $.

\subsection{Variational formulation of the problem}

Next we  describe the variational structure of  problem \eqref{p}
 (see \cite{TC, YCY}).

Let $X$ be the Hilbert space
\begin{equation*}
X:=\{u\in H^2(0,1): u(0)=u'(0)=0\}
\end{equation*}
with inner product and norm
\begin{gather}
(u,v):=\int_0^{1}u''(t)v''(t)dt, \nonumber\\
|u|:=\Big(\int_0^{1}(u''(t))^2dt\Big) ^{1/2}.\label{en}
\end{gather}

To problem \eqref{p} we associate the functional $E:X\to
\mathbb{R}$ defined by
\begin{equation*}
E(u):=\frac{1}{2}| u| ^2-\,\int_0^{1}F(t,u(t)) dt,
\end{equation*}
where
\begin{equation*}
F(t,u)=\int_0^{u}f(t,s)ds\,.
\end{equation*}
The functional $E$ is $C^{1}$ and for any $u,v\in X$, and
\begin{align*}
\langle E'(u),v\rangle 
&= \int_0^{1}(u''(t)\,v''(t)-f(t,u(t))\,v(t)) \,dt \\
&= (u,v) -(f(\cdot ,u(\cdot)),v)_{L^2}.
\end{align*}
Integrating by parts twice in $(N(u) ,v) $ we obtain
that $(N(u) ,v) =(f(\cdot ,u(\cdot)) ,v)_{L^2}$. Then
\begin{equation*}
\langle E'(u),v\rangle =(u-N(u),v) .
\end{equation*}
Therefore, if $X$ is identified with $X'$, which corresponds to the
choice $Y=X$ in \eqref{lll}, then
\begin{equation*}
E'(u) =u-N(u) .
\end{equation*}
If we chose $Y=Y'=L^2(0,1) $, then since
 $f(\cdot ,u(\cdot)) \in L^2(0,1) $, based on \eqref{p2}, 
$(f(\cdot ,u(\cdot)) ,v)_{L^2}=\langle f(\cdot ,u(\cdot)),v\rangle $.
 Also $(u,v) =\langle Lu,v\rangle $.
Hence
\begin{equation*}
\langle E'(u),v\rangle =\langle Lu-f(\cdot
,u(\cdot)) ,v\rangle ,
\end{equation*}
and so
\begin{equation*}
E'(u) =Lu-f(\cdot ,u(\cdot)) ,
\end{equation*}
or equivalently
\begin{equation}
JE'(u) =u-Jf(\cdot ,u(\cdot)).  \label{fr}
\end{equation}

Notice that for each $v\in L^2(0,1) $, one has
\begin{equation}
Jv=Sv,  \label{fff}
\end{equation}
that is, $u:=Jv$ is the solution in $\ H^{4}(0,1) )$ of 
problem \eqref{e-4or-v}. Indeed, based on \eqref{p2},
\begin{equation}
(Jv,w) =\langle v,w\rangle ,\quad  w\in X.  \label{ff}
\end{equation}
For $w\in C_c^{\infty }(0,1) $, one has
$(u,w)=\langle u^{(4)},w\rangle $, where $u^{(4) }$ is in
the distributional sense. Hence 
$\langle u^{(4)},w\rangle =\langle v,w\rangle $ for every 
$w\in C_c^{\infty }(0,1) $, which shows that $u^{(4)}=v$.
 Since $v\in L^2(0,1) $, we have $u\in H^{4}(0,1) $. 
It remains to check that $u''(1)=u'''(1) =0$. For this, we come back to 
\eqref{ff}, which after two successive integration by parts in the left hand
side becomes
\begin{equation*}
u''(1) w'(1) -u'''(1) w(1) +\langle u^{(4) },w\rangle =\langle v,w\rangle .
\end{equation*}
Consequently, $u''(1) w'(1)-u'''(1) w(1) =0$. Since this
holds for every $w\in X$, we must have $u''(1)=u'''(1) =0$, as desired.

\section{Main results}

\subsection{Localization in a shell defined by the energetic norm}

First we shall deal with the localization of positive solutions $u$ of 
problem \eqref{p} in a shell defined by a single norm, more exactly
\begin{equation*}
R_0\leq | u| \leq R_1,
\end{equation*}
where $| \cdot | $ is the energetic norm given by \eqref{en}. 
Therefore, in this subsection we choose $Y=X$ and consequently we
identify $X$ to $X'$. For this situation, the following unilateral
Harnack inequality is crucial.

\begin{lemma}\label{LH}
If $u\in C^{4}[ 0,1] $ satisfies $u(0)=u'(0)=u''(1)=u'''(1)=0$ and 
$u^{(4) }$ is nonnegative and nondecreasing in $[ 0,1] $, then 
$u $ is convex and
\begin{equation}
u(t) \geq M_0(t) | u| \quad \text{for all }t\in [ 0,1] ,  \label{hi}
\end{equation}
where  $M_0(t) =\sqrt{2}(1-t) t^{3}/6$.
\end{lemma}

\begin{proof}
From $u^{(4) }\geq 0$ it follows that $u''$ is
convex. This together with $u''(1) =(u'') '(1) =0$ gives that 
$u''$ is nonnegative and nonincreasing. 
Next, from $u''\geq 0$ one has that $u$ is convex, and since 
$u(0)=u'(0) =0$, $u$ must be nondecreasing and nonnegative.

On the other hand, since $u^{(4) }\geq 0$ we have that 
$u'''$ is nondecreasing and since $u'''(1) =0$, $u'''\leq 0$.
 Then $u'$ is concave; it is also nondecreasing due to $u''\geq 0$, 
and since $u'(0) =0$, we have $u'\geq 0$. Now from $u''\geq 0$, $u'\geq 0$ and 
$u(0) =0$, we see that $u$ is nonnegative, nondecreasing and convex.

Finally note that from $u^{(4)}$ nondecreasing, we have that 
$u'''$ is convex, and since $u'''(1) =0$, the graph of $u'''$ is under the
line connecting the points $(0,u'''(0)) $ and $(1,0) $, i.e.
\begin{equation}
u'''(t) \leq (1-t) u'''(0) ,\quad t\in [ 0,1] .  \label{ci}
\end{equation}

Because the function $u''$ is nonincreasing and
the function $u'''$ is nondecreasing we have:
\begin{align*}
u(t) &=\int_0^{t}\int_0^{s}u''(\tau) d\tau ds
\geq \int_0^{t}\int_0^{s}u''(s) d\tau ds \\
&=\int_0^{t}su''(s) ds 
=\frac{t^2}{2}u''(t) -\int_0^{t}\frac{s^2 }{2}u'''(s) ds \\
& \geq -\int_0^{t}\frac{s^2}{2}u'''(s) ds 
\geq -\int_0^{t}\frac{s^2}{2}u'''(t)ds\\
&=-\frac{t^{3}}{6}u'''(t) .
\end{align*}
This inequality combined with \eqref{ci} gives
\begin{equation}
u(t) \geq -\frac{(1-t) t^{3}}{6}u'''(0) .  \label{e0}
\end{equation}
Next we deal with the energetic norm wishing to connect it to $u'''(0) $. One has
\begin{equation} \label{e1}
\begin{split}
| u| ^2 &=\int_0^{1}u''(t) ^2dt
 = u''u'|_0^{1}-\int_0^{1}u'''(t) u'(t) dt  \\
&=-\int_0^{1}u'''(t) u'(t) dt\leq -u'''(0) u'(1) .
\end{split}
\end{equation}
Also
\begin{equation}\label{e2}\begin{split}
u'(1) &=\int_0^{1}u''(t) dt
 =-\int_0^{1}\int_{t}^{1}u'''(s) \,ds\,dt \\
&\leq -\int_0^{1}\int_{t}^{1}u'''(t) \,ds\,dt  
=-\int_0^{1}(1-t) u'''(t) dt \\
&\leq -\int_0^{1}(1-t) u'''(0) dt=-\frac{1}{2}u'''(0) .
\end{split}
\end{equation}
From \eqref{e1} and \eqref{e2} we deduce 
$\ | u| ^2\leq u'''(0) ^2/2$, or
\begin{equation*}
-u'''(0) \geq \sqrt{2}|u| .
\end{equation*}
This inequality and \eqref{e0} prove \eqref{hi}.
\end{proof}

Consider the cone
\begin{equation*}
K:=\{ u\in X:u\text{ convex and } u(t) \geq M_0(
t) | u| \text{ on }[ 0,1] \} .
\end{equation*}
Note that $K\neq \{ 0\} $, since according to Lemma \ref{LH}, for
any nonzero nonnegative nondecreasing function $v\in C[ 0,1] $,
the function $u=Sv$ given by \eqref{sol} belongs to $K$ and is different
from zero. Also, since any convex function with $u(0) =u'(0) =0$ 
is nondecreasing, all the elements of $K$ are
nondecreasing functions. Consequently, if $f$ is nondecreasing on
 $[ 0,1] \times \mathbb{R}_{+}$ in each of its variables, then the composite
function $f(\cdot ,u(\cdot)) $ is nonnegative and
nondecreasing in $[ 0,1] $ for every $u\in K$ and so Lemma \ref
{LH} can be applied to any solution $u\in K$ of  \eqref{p}
guaranteeing the invariance condition $N(K) \subset K$, and
consequently, the condition \eqref{c1}.

Denote
\begin{equation*}
M_1(t) :=\frac{2}{3}t^{3/2},\quad  t\in [ 0,1] .
\end{equation*}
Our assumptions on $f$ are as follows:
\begin{itemize}
\item[(H1)] $f$ is nondecreasing on $[ 0,1] \times \mathbb{R}_{+}$
in each of its variables;

\item[(H2)] there exist $R_0,R_1$ with $0<R_0<R_1$ such that
\begin{itemize}
\item[(a)] $\int_0^{1}M_0(t) f(t,M_0(t)R_0) dt\geq R_0$,

\item[(b)] $\int_0^{1}M_1(t) f(t,M_1(t) R_1) dt\leq R_1$.
\end{itemize}

\item[(h3)] there exist $u_0,u_1\in K_{R_0R_1}=\{u\in K:R_0\leq
| u| \leq R_1\}$ and $r>0$ such that $|
u_0| <r<| u_1| $ and
\begin{equation*}
\max \{ E(u_0) , E(u_1) \} 
<\inf \{ E(u) : u\in K, | u| =r\} .
\end{equation*}
\end{itemize}

\begin{theorem}\label{t3}
Assume that {\rm (H1), (H2)} are satisfied. 
Let $\Gamma $, $m$ and $c$ be defined as in \eqref{e-gamma}, \eqref{e-c} and 
\eqref{e-m} respectively. Then the fourth-order problem \eqref{p}
has at least one positive solution $u_{m}$ in $K_{R_0R_1}$ such that
\begin{equation*}
E(u_{m}) =m.
\end{equation*}
If in addition {\rm (H3)} holds, then a second positive solution $u_c$
exists in $K_{R_0R_1}$ with
\begin{equation*}
E(u_c) =c.
\end{equation*}
\end{theorem}

\begin{proof}
We apply Theorems \ref{t1} and \ref{t2}. Recall that here we identify $X$ to
$X'$ and thus $J=I$, the identity map on $X$.

First note that the (MPSS) condition holds in $K_{R_0R_1}$ due to the
compactness of the operator $N=I-E'$. Also the boundedness of 
$(E'(u) ,u) $ on the boundaries of $K_{R_0R_1}$, i.e. \eqref{b1} and 
\eqref{b2}, is guaranteed since $E'$ maps bounded sets into bounded sets.

To check \eqref{c1}, let $u$ be any element of $K$. Hence $u$ is nonnegative
and nondecreasing on $[ 0,1] $. Then, from (h1) we also have that
$f(t,u(t))$ is nonnegative and nondecreasing in $[0,1] $. 
Now, Lemma \ref{LH} implies that $N(u) \in K$. But $N(u)=(I-E') (u) $.
 Thus \eqref{c1} holds.

Next, let us note that for any $u\in K_{R_0R_1}$, we have
\begin{equation}\label{r14}
\begin{split}
u(t) &=\int_0^{t}\int_0^{s}u''(\tau
) d\tau ds\leq \int_0^{t}\sqrt{s}(\int_0^{s}u''(\tau) ^2d\tau) ^{1/2}ds   \\
&\leq | u| \int_0^{t}\sqrt{s}ds=\frac{2}{3}t^{3/2}| u| =M_1(t) | u| .
\end{split}
\end{equation}
Then
\begin{equation*}
E(u) =\frac{1}{2}| u|
^2-\int_0^{1}F(t,u(t)) dt\geq \frac{1}{2}
R_0^2-F(1,\frac{2}{3}R_1) .
\end{equation*}
Hence $E$ is bounded from below on $K_{R_0R_1}$.

Furthermore, we check the boundary conditions \eqref{h1}. Assume that $
E'(u) -\lambda u=0$ for some $u\in K$ with $|
u| =R_0$ and $\lambda >0$. Then $\ u$ solves the problem
\begin{gather*}
u^{(4)}(t)-\,f(t,u(t))-\lambda u(t) =0, \quad 0<t<1, \\
u(0)=u'(0)=u''(1)=u'''(1)=0 
\end{gather*}
and
\begin{align*}
R_0^2 &=| u| ^2=\int_0^{1}[ f(t,u(t)) +\lambda u(t) ] u(t) dt \\
&\geq \int_0^{1}[ f(t,M_0(t) R_0)
+\lambda M_0(t) R_0] M_0(t) R_0dt \\
&> R_0 \int_0^{1}f(t,M_0(t) R_0)
M_0(t)\,dt,
\end{align*}
which contradicts  assumption (H2) (a). Hence $E'(u)
-\lambda u\neq 0$ for all $u\in K$ with $| u| =R_0$ and
$\lambda >0$.

Assume now that $E'(u) +\lambda u=0$ for some $u\in K$
with $| u| =R_1$ and $\lambda >0$. Then $u$ solves the
problem
\begin{gather*}
(1+\lambda) u^{(4)}(t)-\,f(t,u(t))=0, \quad 0<t<1, \\
u(0)=u'(0)=u''(1)=u'''(1)=0. 
\end{gather*}
Then
\begin{equation*}
R_1^2=| u| ^2=\frac{1}{1+\lambda }
\int_0^{1}f(t,u(t)) u(t) dt.
\end{equation*}
Using \eqref{r14} we deduce
\begin{equation*}
R_1^2<\int_0^{1}M_1(t) R_1f(t,M_1(t) R_1) dt,
\end{equation*}
which contradicts (H2) (b). Hence $E'(u) +\lambda u\neq 0$ for all
 $u\in K$ with $| u| =R_1$ and $\lambda >0$.

The conclusions follow from Theorem \ref{t1} and Theorem \ref{t2}.
\end{proof}

\begin{example} \label{examp3.1} \rm 
We give an example of a function $f(t,u) =f(u)$ which satisfies
the conditions (H1) and (H2) of Theorem 3.2. Note that 
\begin{gather*}
0\leq \frac{\sqrt{2}}{6}(1-t)t^{3}<0.03\quad \text{if }0\leq t\leq 1, \\
\int_0^{1}(\frac{\sqrt{2}}{6}(1-t)t^{3}) ^2dt=\frac{1}{4536},
\end{gather*}
and $4600\times 3/100=138$. Define
\begin{equation*}
f(u) =\begin{cases}
0, & u\leq 0, \\
4600 u, & 0\leq u\leq 0.03, \\
138, & u\geq 0.03.
\end{cases}
\end{equation*}
Taking $R_0=1$ and $R_1=37$, by
\begin{equation*}
\int_0^{1}138\,\frac{2}{3}t^{3/2}dt=\frac{184}{5}=36.8,
\end{equation*}
we obtain that the conditions (H1) and (H2) are satisfied.
\end{example}

For the autonomous case $f(t,u) =f(u) $, where $f$
is nonnegative and nondecreasing on $\mathbb{R}_{+}$, we may replace the
conditions of (H2) by a couple of simpler inequalities as shows the next
result.

\begin{theorem}\label{t4}
Assume that $f:\mathbb{R}_{+}\to \mathbb{R}_{+}$ is
continuous nondecreasing and that for some numbers $a\in (0,1) $,
$R_0$ and $R_1$ with $0<R_0<R_1$, one has
\begin{equation}
\frac{f(M_0(a) R_0) }{M_0(a) R_0}
\geq \frac{1}{(1-a) M_0(a) ^2},\quad 
\frac{f(\frac{2}{3}R_1) }{R_1}\leq \frac{15}{4}.  \label{f2}
\end{equation}
Then \eqref{p} has at least one positive solution $u_{m}$ in 
$K_{R_0R_1}$ with $E(u_{m}) =m$. If in addition {\rm (H3)}
holds, then a second positive solution $u_c$ exists in $K_{R_0R_1}$
with $E(u_c) =c$.
\end{theorem}

\begin{proof}
Since $M_1(t) \leq 2/3$ for every $t\in [ 0,1] $,
we have
\begin{equation*}
\int_0^{1}M_1(t) f(M_1(t) R_1)
dt\leq f\big(\frac{2}{3}R_1\big) \int_0^{1}\frac{2}{3}t^{3/2}dt
=\frac{4}{15}f\big(\frac{2}{3}R_1\big) .
\end{equation*}
Then the inequality
\begin{equation*}
\frac{4}{15}f\big(\frac{2}{3}R_1\big) \leq R_1,
\end{equation*}
or equivalently the second inequality in \eqref{f2} is a sufficient
condition for (H2)(b) to hold. As concerns the first inequality in 
\eqref{f2}, let us remark that if $E'(u) -\lambda u=0$ for some 
$u\in K$ with $| u| =R_0$ and $\lambda >0$, then
\begin{equation}\label{f3}
\begin{split}
R_0^2 &=| u| ^2
=\int_0^{1}[ f(u(t)) +\lambda u(t) ] u(t) dt \\
&\geq \int_{a}^{1}[ f(u(t)) +\lambda u(
t) ] u(t) dt.  
\end{split}
\end{equation}
The function $u$ being nondecreasing, one has 
$u(t) \geq u(a) $ for all $t\in [ a,1] $. Also, since $u \in K$, 
$u(a) \geq M_0(a) | u| $. Then from \eqref{f3},
\begin{align*}
R_0^2 
&\geq (1-a) [ f(M_0(a) R_0) +\lambda M_0(a) R_0] M_0(a) R_0 \\
&> (1-a) f(M_0(a) R_0) M_0(a) R_0.
\end{align*}
Hence
\begin{equation*}
R_0>(1-a) M_0(a) f(M_0(a)R_0) ,
\end{equation*}
i.e.~the opposite of the first inequality in \eqref{f2}.
\end{proof}

Clearly the inequalities \eqref{f2} express the oscillation of the function 
$f(t) /t$ up and down the values $1/(1-a) M_0(a) ^2$ and $45/8$.

\begin{remark}[Existence asymptotic conditions] \rm
The existence of two numbers $R_0$, $R_1$ satisfying \eqref{f2} is
guaranteed by the asymptotic conditions
\begin{equation}
\limsup_{\tau \to 0}\frac{f(\tau) }{\tau }>\frac{
3^{11}}{32}\quad \text{and}\quad
\liminf_{\tau \to \infty }\frac{f(\tau) }{\tau }<\frac{45}{8}.  \label{ac1}
\end{equation}
\end{remark}

\begin{remark}[Multiplicity] \rm
Theorems \ref{t3} and \ref{t4} can be used to obtain multiple positive
solutions. Indeed, if their assumptions are fulfilled for two pairs 
$(R_0,R_1) $, $(\overline{R}_0,\overline{R}_1) $,
then we obtain four solutions, provided that the sets $K_{R_0R_1}$ and 
$K_{\overline{R}_0\overline{R}_1}$ are disjoint. This happens if 
$0<R_0<R_1<\overline{R}_0<\overline{R}_1$. We can even obtain
sequences of positive solutions; for instance, in connection with Theorem
\ref{t4}, if
\begin{equation}
\limsup_{\tau \to 0}\frac{f(\tau) }{\tau }>\frac{
3^{11}}{32}\quad \text{and}\quad
\liminf_{\tau \to 0}\frac{f(\tau) }{\tau }<\frac{45}{8},  \label{ac2}
\end{equation}
then there exists a sequence $(u_k) $ of positive solutions
with $u_k\to 0$ as $k\to \infty $. Also, if
\begin{equation}
\limsup_{\tau \to \infty }\frac{f(\tau) }{\tau }>
\frac{3^{11}}{32}\quad \text{and}\quad 
\liminf_{\tau \to \infty }
\frac{f(\tau) }{\tau }<\frac{45}{8},  \label{ac3}
\end{equation}
then there exists a sequence $(u_k) $ of positive solutions
with $| u_k| \to \infty $ as $k\to \infty $.

Note that $3^{11}/32\simeq 5536$ is the minimal value of 
$1/(1-a) M_0(a) ^2$ for $a\in (0,1) $, which
is reached at $a=2/3$. A better estimation than \eqref{hi} would allow the
replacement of this value by a smaller number.
\end{remark}

For examples of functions satisfying asymptotic conditions of type \eqref{ac1}, 
\eqref{ac2} or \eqref{ac3}, we refer to the recent paper \cite{h}.

\begin{remark}[Fixed point approach] \label{rmk3.3} \rm 
Under the assumptions of Theorem \ref{t3}, the existence of a solution in 
$K_{R_0R_1}$ can also be obtained via Krasnosel'ski\u{\i}'s theorem.
Indeed,  problem \eqref{p} is equivalent to the fixed point problem 
\eqref{fpp} in $X$ for the compact operator $N:K_{R_0R_1}\to K$ given
by \eqref{op}.

Let us check the condition (a)(i) from Theorem \ref{tk}. Assume the
contrary, i.e. $N(u) <u$ for some $u\in K$ with $|u| =R_0$. 
Then $N(u) =u-v$ for some $v\in K\setminus\{ 0\} $. This means that
 $(u-v) ^{(4) }=f(t,u) $ in the sense of distributions. Now multiply by $u$
and integrate to obtain
\begin{equation*}
| u| ^2-\int_0^{1}u''(t)v''(t) dt=\int_0^{1}f(t,u(t)) u(t) dt.
\end{equation*}
Since $v,u-v\in K$, one has $v''\geq 0$ and $u''-v''\geq 0$ in $[ 0,1] $. Hence
\begin{equation*}
\int_0^{1}u''(t) v''(t) dt\geq \int_0^{1}v''(t) ^2dt=| v| ^2>0.
\end{equation*}
Then
\begin{equation*}
R_0^2=| u| ^2>\int_0^{1}f(t,u(t)) u(t) dt.
\end{equation*}
Next we use \eqref{hi} to derive a contradiction to (H2) (a).

The condition (a)(ii) from Theorem \ref{tk} can be proved similarly.

Notice that under assumptions (H2)(a) and (b), a solution exists in 
$K_{R_1R_0}$ in case that $R_1<R_0$. However this is not guaranteed
by the variational approach.

We may conclude that, compared to the fixed point approach, the variational
method gives an additional information about the solution, namely of being a
minimum for the energy functional. Moreover, a second solution of mountain
pass type can be guaranteed by the variational approach.
\end{remark}

The above approach was essentially based on the monotonicity assumption on 
$f$, which was required by the Harnack type inequality \eqref{hi}. Thus a
natural question is if such an inequality can be established for functions
 $u $ satisfying the boundary conditions and $u^{(4) }\geq 0$,
without the assumption that $u^{(4) }$ is nondecreasing. In the
absence of the answer to this question, an alternative approach is possible
in a shell defined by two norms as shown in the next section.

\subsection{Localization in a shell defined by two norms}

In the previous section, a unilateral Harnack inequality was established for
functions $u$ satisfying the two point boundary conditions and with 
$u^{(4) }$ nonnegative and nondecreasing in $[ 0,1] $,
in terms of the energetic norm. If we renounce to the monotonicity of 
$u^{(4) }$, then we have the following lemma in terms of the max
norm $\| \cdot \|_{\infty }$, and finally in the $L^2$-norm $\| \cdot \| $. 
This allows us to find positive
solutions of  \eqref{p}, first in a conical shell defined by the
norm $\| \cdot \|_{\infty }$ using Krasnosel'ski\u{\i}'s
fixed point theorem, and next in a conical shell defined by two norms, the
energetic norm $| \cdot | $ of $X$ and the norm $\| \cdot \| $ of
 $Y=L^2(0,1) $, using variational methods.

\begin{lemma} \label{LH2}
If $u\in C^{4}[ 0,1] $ satisfies $u(0)=u'(0)=u''(1)=u'''(1)=0$ and 
$u^{(4) }\geq 0$ in $[ 0,1] $, then
\begin{equation}
u(t)\geq M(t) \| u\|_{\infty }\quad \text{for all  }
t\in [ 0,1]\,,  \label{r15}
\end{equation}
where  $M(t) :=(3-t)t^2/3$.
\end{lemma}

\begin{proof}
We use the following estimations of Green's function:
\begin{equation}
\frac{t^2}{6}(3-t)s^2\,\leq G(t,s)\leq \frac{s^2}{2}\quad 
\text{for all }(t,s)\in [ 0,1]\times [ 0,1]\,.  \label{ge}
\end{equation}
To prove them, we first assume that $0\leq t<s\leq 1$. Then
\begin{align*}
\frac{t^2}{6}(3-t)s^2
&\leq \frac{t^2}{6}\{ (3-t)s^2+(1-s) [ s(1-t) +(s-t) +s] \} \\
&= \frac{t^2}{6}(3s-t) 
=G(t,s) \\
&\leq \frac{s^2}{6}(3-0) =\frac{s^2}{2}.
\end{align*}
Similarly, for $0\leq s\leq t\leq 1$, one has
\begin{align*}
\frac{t^2}{6}(3-t)\,s^2 
&\leq \frac{t^2}{6}(3-s)\,s^2 \\
&\leq \frac{s^2}{6}\{ (3-s) t^2+(1-t)[ t(1-s) +(t-s) +t] \} \\
&=\frac{s^2}{6}(3t-s) =G(t,s) \\
&\leq \frac{s^2}{6}(3-0) =\frac{s^2}{2}.
\end{align*}
Hence \eqref{ge} is proved.

Let $u\in C^{4}[ 0,1] $ satisfy $u(0)=u'(0)=u''(1)=u'''(1)=0$ and $u^{(4) }\geq 0$
in $[ 0,1] $. Then
\begin{equation}
\begin{aligned}
u(t) 
&=\int_0^{1}G(t,s)u^{(4) }(s) \,ds\\
&\geq \int_0^{1}\frac{(3-t)\,t^2\,s^2}{6}\,u^{(4) }(s) \,ds   \\
&=\frac{(3-t)\,t^2}{3}\int_0^{1}\,\frac{s^2}{2}u^{(4)
}(s) \,ds   \\
&\geq \frac{(3-t)\,t^2}{3}\int_0^{1}\,\{ \max_{t\in [
0,1]}G(t,s)\} \,u^{(4) }(s) \,ds \\
&\geq \frac{(3-t)\,t^2}{3}\max_{t\in [ 0,1]}\{
\int_0^{1}G(t,s)u^{(4) }(s) \,ds\} \\
& =\frac{(3-t)\,t^2}{3}\| u\|_{\infty }.
\end{aligned}
\end{equation}
Hence \eqref{r15} is proved.
\end{proof}

Notice that, since $\| u\| \leq \| u\|_{\infty }$, where $\| \cdot \| $ 
is the $L^2(0,1) $-norm, inequality \eqref{r15} also gives
\begin{equation*}
u(t)\geq M(t)\| u\| \quad \text{for all } t\in [ 0,1].
\end{equation*}

Using Lemma \ref{LH2}, the existence of a positive solution can be
immediately obtained via Krasnosel'ski\u{\i}'s compression-expansion theorem.

\begin{theorem}
Assume that there exist positive numbers $\alpha ,\beta $,
 $\alpha \neq \beta $ such that
\begin{equation}
\alpha \leq (S\underline{f_{\alpha }}) (1) \quad
\text{and}\quad \beta \geq (S\overline{f_{\beta }}) (1) ,  \label{r0}
\end{equation}
where $S$ is the solution operator defined by \eqref{sol} and
\begin{align*}
\underline{f_{\alpha }}(t) =\min \{ f(t,u) :
 M(t) \alpha \leq u\leq \alpha \} , \\
\overline{f_{\beta }}(t) =\max \{ f(t,u) :
M(t) \beta \leq u\leq \beta \} .
\end{align*}
Then  problem \eqref{p} has at least one positive solution $u$
such that
\begin{equation*}
R_0\leq \| u\|_{\infty }\leq R_1,
\end{equation*}
where $R_0=\min \{ \alpha ,\beta \} $ and $R_1=\max \{\alpha ,\beta \} $.
\end{theorem}

\begin{proof}
As shown in Section 2.3,  problem \eqref{p} is equivalent to the fixed
point problem $N(u) =u$ in $C[ 0,1] $.

In the space $C[ 0,1] $ we consider the cone
\begin{equation*}
K=\{ u\in C[ 0,1] : \text{ }u(0) =0,\;
 u(t) \geq M(t) \| u\|_{\infty }\text{ for all }t\in [ 0,1] \} .
\end{equation*}
From Lemma \ref{LH2} and the properties of $f$, it follows that 
$N(K) \subset K$. Also $N$ is a compact operator.

Now we show that the required boundary conditions from Krasnosel'ski\u{\i}'s
theorem are satisfied. Assume by contradiction that $N(u) <u$
for some $u\in K$ with $\| u\|_{\infty }=\alpha $. Then 
$N(u) =u-v$ for some $v\in K\setminus \{ 0\} $. Hence
\begin{equation}
u(t) -v(t) =N(u) (t) . \label{r1}
\end{equation}
We have
\begin{equation*}
M(t) \alpha \leq u(t) \leq \alpha \quad \text{for all }t\in [ 0,1] .
\end{equation*}
Hence
\begin{equation*}
f(t,u(t)) \geq \underline{f_{\alpha }}(t) .
\end{equation*}
Since Green's function is positive, the solution operator $S$ preserves the
ordering, so that $N(u) (t) \geq (S\underline{f_{\alpha }}) (t) $.  
Returning to \eqref{r1}, we deduce that
\begin{equation}
u(t) -v(t) \geq (S\underline{f_{\alpha }}) (t) .  \label{r2}
\end{equation}
Since $v(t) \geq M(t) \| v\|_{\infty }>0$, for $t>0$, \eqref{r2} yields
\begin{equation*}
\alpha =\| u\|_{\infty }\geq u(1) >u(1) -v(1) \geq (S\underline{f_{\alpha }})(1) ,
\end{equation*}
a contradiction to our first assumption from \eqref{r0}.

Next assume that $N(u) >u$ for some $u\in K$ with $\|u\|_{\infty }=\beta $. 
Then $N(u) =u+v$ for some $v\in K\setminus \{ 0\} $ and, since 
$G(t,s)\leq G(1,s)$ for all $t,s\in [ 0,1]$, we have
\begin{equation}
u(t) +v(t) =N(u) (t) 
\leq (S\overline{f_{\beta }}) (t) \leq (S\overline{ f_{\beta }}) (1) .  \label{r3}
\end{equation}
Let $t_0$ be such that $u(t_0) =\| u\|_{\infty }=\beta >0$. Since $u(0) =0$,
 one has $t_0>0$ and so $v(t_0) \geq M(t_0) \| v\|_{\infty }>0$. 
Then, for $t=t_0$, \eqref{r3} gives
\begin{equation*}
\beta <(S\overline{f_{\beta }}) (1) ,
\end{equation*}
which contradicts our second assumption from \eqref{r0}. Thus Theorem 
\ref{tk} applies.

We note that if $\alpha <\beta $, then \eqref{r0} represents the 
\textit{compression} condition, while if $\alpha >\beta $, then \eqref{r0} 
expresses the \textit{expansion} condition.
\end{proof}

Next we are interested into two positive solutions for \eqref{p}. We shall
succeed this by the variational approach based on Theorems \ref{t1} and 
\ref{t2} applied to the Hilbert spaces $X=\{u\in H^2(0,1):$ $\ u(0)=u'(0)=0\}$ 
with norm $| \cdot | $ given by \eqref{en}, and
$Y=L^2(0,1) $ with the usual norm denoted by $\| \cdot \| $.

Let us consider the cone
\begin{equation*}
K=\{ u\in X:\text{ }u(t)\geq M(t) \|u\|  \text{for all }t\in [ 0,1]\}
\end{equation*}
and a fixed element $\phi \in K\setminus \{ 0\} $ with 
$| \phi | =1$. Such an element can be $Jv$, where $v$ is
any nonzero nonnegative continuous function on $[ 0,1] $, because of
Lemma \ref{LH2}. In addition, consider two numbers $R_0$, $R_1$ such
that $0<R_0<\| \phi \| R_1$, and let
\begin{equation*}
K_{R_0R_1}=\{ u\in K: \| u\| \geq R_0,\; | u| \leq R_1\} .
\end{equation*}
Denote
\begin{gather*}
\underline{g}(t) = \min \{ f(t,u) : M(t) R_0\leq u\leq c_{\infty }R_1\} , \\
\overline{g}(t)  = \max \{ f(t,u) : M(t) R_0\leq u\leq c_{\infty }R_1\} ,
\end{gather*}
where $c_{\infty }>0$ is such that $\| v\|_{\infty }\leq c_{\infty }| v| $ for all 
$v\in K$. For example we may take $c_{\infty }=2/3$, since for any $v\in K$, 
H\"{o}lder's inequality gives
\begin{equation}
v(t) =\int_0^{t}\int_0^{s}v''(\tau) d\tau ds
\leq | v| \int_0^{t}\sqrt{s}\leq \frac{2}{3}| v| .  \label{r13}
\end{equation}

Our assumptions are as follows:
\begin{itemize}
\item[(H1)] There exist $R_0,R_1$ with $0<R_0<\| \phi \| R_1$ such that
\begin{itemize}
\item[(a)] $R_0\leq \| S\underline{g}\| $,

\item[(b)] $R_1\geq c_{\infty }\| \overline{g}\|_{L^{1}(0,1) }$.
\end{itemize}
\item[(H2)] The functional $E$ has the mountain pass geometry in 
$K_{R_0R_1}$ and there exists $\rho >0$ such that
\begin{equation}
E(u) \geq c+\rho  \label{r10}
\end{equation}
for all $u\in K_{R_0R_1}$ which simultaneously satisfy 
$\|u\| =R_0$ and $| u| =R_1$.
\end{itemize}

\begin{theorem} \label{t3.6}
Under assumptions {\rm (H1), (H2)},  problem \eqref{p}
has at least two positive solutions $u_{m},u_c\in K_{R_0R_1}$ with 
$E(u_{m}) =m$ and $E(u_c) =c$, with $m$ and $c$
defined by \eqref{e-m} and \eqref{e-c} respectively.
\end{theorem}

\begin{proof}
For $u\in K_{R_0R_1}$, one has
\begin{equation*}
M(t) R_0\leq M(t) \| u\| 
\leq u(t)\leq \| u\|_{\infty }\leq c_{\infty }\,|u|\leq
c_{\infty }\,R_1.
\end{equation*}
It follows that
\begin{equation*}
F(t,u(t)) \leq \omega :=\max \{F(t,u)
: 0\leq t\leq 1,\; M(t) R_0\leq u\leq c_{\infty }R_1\},
\end{equation*}
whence, for all $u\in K_{R_0R_1}$, it is fulfilled that
\begin{equation*}
E(u) =\frac{1}{2}| u|^2-\int_0^{1}F(t,u(t)) dt\geq -\omega ,
\end{equation*}
and so $m>-\infty $.

Next, from $c>m$ we see that \eqref{r10} guarantees both \eqref{r11} and 
\eqref{r12}. It remains to check the compression boundary condition given by 
\eqref{h1}, \eqref{h2}. Assume first that \eqref{h1} does not hold. Then 
$JE'(u) -\lambda Ju=0$ for some 
$u\in K_{R_0R_1}$, $\| u\| =R_0$ and $\lambda >0$. Hence,
in view of \eqref{fr},
\begin{equation*}
u=J(f(\cdot ,u(\cdot)) +\lambda u) .
\end{equation*}
Since $u\in C[ 0,1] $, one has 
$f(\cdot ,u(\cdot)) +\lambda u\in C[ 0,1] $, and using \eqref{fff} we
have
\begin{equation*}
u=S(f(\cdot ,u(\cdot)) +\lambda u) .
\end{equation*}
For $t>0$, one has $f(t,u(t)) +\lambda u(t) >f(t,u(t)) \geq \underline{g}(t) $.
 As a result $S(f(\cdot ,u(\cdot)) +\lambda u) >S\underline{g}$ on $(0,1]$. 
Taking the $L^2$-norm, we deduce
\begin{equation*}
R_0=\| u\| >\| S\underline{g}\| ,
\end{equation*}
which contradicts (H1)(a).

Next assume that $JE'(u) +\lambda u=0$ for some 
$u\in K_{R_0R_1},\ | u| =R_1$ and $\lambda >0$. Then
\begin{equation*}
(1+\lambda) u^{(4) }=f(t,u(t)) ,
\end{equation*}
whence, arguing as in the proof of Theorem \ref{t3}, we deduce that
\begin{equation*}
(1+\lambda) R_1^2=\int_0^{1}u(t) f(t,u(t)) dt.
\end{equation*}
Consequently
\begin{equation*}
R_1<c_{\infty }\| \overline{g}\|_{L^{1}(0,1)},
\end{equation*}
which contradicts (H1)(b).
\end{proof}

\begin{remark}\label{r3.4} \rm
In the autonomous case, if $f=f(u) $, and $f$ is
nondecreasing on $\mathbb{R}_{+}$, a sufficient condition for (H1)(a) to
hold is
\begin{equation*}
R_0\leq f(M(a) R_0) \| S\chi_{[a,1]}\| ,
\end{equation*}
where $a$ is some number from $(0,1) $ and $\chi_{[ a,1] }$ is the characteristic 
function of the interval $[ a,1]$. Also in this case, (H1)(b) reduces to
\begin{equation*}
R_1\geq c_{\infty }f(c_{\infty }R_1) .
\end{equation*}
\end{remark}

\section{An example}

We present an example inspired by that in \cite{Pre}, to which we
can apply either Theorem \ref{t4} (working in a shell defined only by the
energetic norm), or Theorem \ref{t3.6} (in a shell defined by two norms).
Consider  problem \eqref{p}, where the function $f(t,u)=f(u) $ is 
\begin{equation}
f(u)=\begin{cases}
pu^{p}\,, & 0\leq u\leq 1\,, \\
pu^2\,, & 1\leq u\leq b\,, \\
p((u-b)^{p}+b^2)\,, & u\geq b,\,
\end{cases}  \label{pp}
\end{equation}
with $\ 0<p\leq 1/2\ $ and a sufficiently large number $b>2$ as chosen
below.

Note that the function $f$ is positive and nondecreasing in 
$\mathbb{R}_{+}$. Also note as a typical behavior, that $f$ is 
first \textit{sublinear} near zero (here in $[ 0,1] )$, next \textit{superlinear} 
(on a sufficiently large finite interval $[ 1,b] )$, and again 
\textit{sublinear} towards infinity (on $[b,\infty ))$.

\subsection{Application of Theorem \ref{t4}}

For $1\leq u\leq 2$, we have
\begin{equation*}
F(u)=\int_0^{u}f(s) ds=\frac{p}{p+1}+\frac{p}{3}
(u^{3}-1)\leq \frac{p}{p+1}+\frac{7\,p}{3}=\frac{p\,(10+7\,p)}{3\,(p+1)}
\,.
\end{equation*}
Choose $r=2$. Then, for $u\in K$ and $|u|=2$, as for \eqref{r13}, 
$|u|_{\infty }\leq 2|u|/3<2$ and so, recalling $0<p\leq 1/2$,
\begin{equation*}
E(u)=\frac{|u|^2}{2}-\,\int_0^{1}F(u(t))\,dt\geq 2-\frac{p\,(10+7\,p)}{
3\,(p+1)}\geq \frac{1}{2}\,.
\end{equation*}
We take for $u_0$ any element of $K\setminus \{ 0\} $ with 
$| u_0| =1$. Clearly $| u_0| <r=2$.
Then
\begin{equation*}
E(u_0)=\frac{|u_0|^2}{2}-\int_0^{1}F(u_0(t) )\,dt=
\frac{1}{2}-\int_0^{1}F(u_0(t) )\,dt<\frac{1}{2}\,.
\end{equation*}
Next we take $u_1=b u_0/\| u_0\|_{\infty }$. Then 
$| u_1| =b/\| u_0\|_{\infty }>2$ if
we choose $b>2\| u_0\|_{\infty }$. Also
\begin{equation*}
E(u_1)\leq \frac{b^2}{2\| u_0\|_{\infty }^2}
-\int_{(u_1>1) }F(u_1(t)) dt.
\end{equation*}
Since $\| u_1\|_{\infty }=b>2$ and $u_1(0)=0$ the
level set $(u_1>1) $ is a proper subset of $[ 0,1]$.
Also $u_1(t) \leq b$ for all $t$. Hence on the level set $(u_1>1) $ we have
\begin{equation*}
F(u_1) =\frac{p}{p+1}+\frac{p}{3}(u_1^{3}-1) > \frac{p}{3}u_1^{3}.
\end{equation*}
Then
\begin{equation*}
E(u_1)<\frac{b^2}{2\| u_0\|_{\infty }^2}
-\frac{pb^{3}}{3\| u_0\|_{\infty }^{3}}\int_{(
u_1>1) }u_0(t) ^{3}dt.
\end{equation*}
Taking into account that the level set $(u_1>1) $ enlarges as 
$b$ increases, we can see that the right side of the last inequality tends to
$-\infty $ as $b\to +\infty $ (here we underline the role of the
superlinearity of $f$ on the interval $[ 1,b] $ guaranteeing the
term $b^{3}$ in the last inequality). Thus we may choose $b$ large enough to
have
\begin{equation*}
E(u_1)<\frac{1}{2}.
\end{equation*}
Hence the assumption (h3) of Theorem \ref{t4} is satisfied. Also, due to the
sublinearity of $f$ towards zero and infinity, one has
\begin{equation}
\lim_{\tau \to 0}\frac{f(\tau )}{\tau }=+\infty \,,\quad \text{and}
\quad \lim_{\tau \to +\infty }\frac{f(\tau )}{\tau }=0\,.
\label{af}
\end{equation}
Consequently, we may find $R_0$ (small enough) and $R_1$ (large enough),
such that $u_0$ and $u_1$ belong to $K_{R_0R_1}$ \ and the
conditions \eqref{f2} hold.

Therefore, according to Theorem \ref{t4},  problem \eqref{p} with $f$
given by \eqref{pp}, $0<p\leq 1/2$ and $b$ sufficiently large has two
positive solutions.

\subsection{Application of Theorem \protect\ref{t3.6}}

The mountain pass geometry of the energy functional can be shown as above.
It remains to guarantee the conditions (H1) and \eqref{r10} from (H2).

First note that since $f$ is nondecreasing on $\mathbb{R}_{+}$, we can use
Remark \ref{r3.4}, in view of which we may find $\,R_0,\overline{R}_1$
 such that $\,u_0,u_1\in K_{R_0R_1}$ \thinspace and (H1)
holds, for every $R_1\geq \overline{R}_1$. Recall that, this time, the
conical shell $K_{R_0R_1}$ is defined by the energetic norm and the $
L^2$-norm.

Now we fix $R_0$ and we look for an $R_1\geq \overline{R}_1$ such that
\eqref{r10} holds. For the variable parameter $R_1$, we denote by 
$\Gamma _{R_1},\,c_{R_1}$ the corresponding elements $\Gamma $ and $c$ in
Theorem \ref{t3.6}. It is clear that if~ $R_1\geq \overline{R}_1$, then
$\Gamma_{\overline{R}_1}\subset \Gamma_{R_1}$ and consequently,
 $c_{\overline{R}_1}\geq c_{R_1}$. Hence to guarantee \eqref{r10} it suffices
to find $R_1\geq \overline{R}_1$ such that for a given number
 $\rho >0$,
 $E(u) \geq c_{\overline{R}_1}+\rho$ for all
 $u\in K_{R_0R_1}$  satisfying simultaneously
$\| u\| =R_0$ and  $|u| =R_1$. We shall guarantee even more, namely that
\begin{equation}
E(u) \geq c_{\overline{R}_1}+\rho \quad \text{for all $u\in K$  with }| u| =R_1.
\label{L10}
\end{equation}
Let $u\in K$ be such that $| u| =R_1$. Then
\begin{align*}
E(u)  &=\frac{1}{2}R_1^2-\int_0^{1}F(u(t)) dt \\
&=\frac{1}{2}R_1^2-\int_{(u\leq b)}F(u(t)) dt-\int_{(u\geq b) }F(
u(t)) dt \\
&\geq \frac{1}{2}R_1^2-p\Big(\frac{1}{p+1}+\frac{b^{3}}{3}\Big)  \\
&\quad -\int_{(u\geq b) }\Big(\frac{p}{p+1}+\frac{p}{3}(
b^{3}-1) +\frac{p}{p+1}(u-b) ^{p+1}+pb^2u-pb^{3}\Big) dt \\
&\geq \frac{1}{2}R_1^2-C_1| u|_{L^{p+1}(0,1) }^{p+1}-C_{2}| u|_{L^{1}(0,1)
}-C_3,
\end{align*}
where $C_1,C_{2},C_3>0$ are constants depending only on $p\ $and $b$.
Since $X$ is continuously embedded into $L^{q}(0,1) $ for every 
$q\geq 1$, we deduce that
\begin{equation*}
E(u) \geq \frac{1}{2}R_1^2-\widetilde{C}_1R_1^{p+1}-\widetilde{C}_{2}R_1-C_3.
\end{equation*}
The expression in the right side of this inequality tends to $\infty $ as 
$R_1\to \infty $. Hence we can find $R_1\geq \overline{R}_1$
such that \eqref{L10} holds.

Therefore, according to Theorem  \ref{t3.6}, problem \eqref{p} with $f$,
given by \eqref{pp}, $0<p\leq 1/2$ and $b$ sufficiently large has two
positive solutions.

\subsection*{Acknowledgements}

The first and third authors were partially supported by Ministerio de
Ciencia y Tecnolog\'{\i}a, Spain, and FEDER, Project MTM2013-43014-P.

The second author was supported by a grant of the Romanian National
Authority for Scientific Research, CNCS -- UEFISCDI, project number
PN-II-ID-PCE-2011-3-0094.

The third author is supported by a FPU scholarship, Ministerio de
Educaci\'on, Cultura y Deporte, Spain.

The fourth author is thankful to the Department of Mathematical Analysis,
University of Santiago de Compostela, Spain, where a part of the paper was
prepared, during his visit. He is partially supported
by the Fund  ``Science research'' at University of Ruse, under Project 16-FNSE-03.

The authors are thankful to the anonymous referee whose detailed remarks and
suggestions yielded to an improved presentation of the matter.

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\end{document}