\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 143, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/143\hfil Twin positive solutions]
{Twin positive solutions for fourth-order two-point boundary-value
problems with sign changing nonlinearities}

\author[Y. Tian, W. Ge\hfil EJDE-2004/143\hfilneg]
{Yu Tian, Weigao Ge} % in alphabetical order

\address{Yu Tian\hfill\break
Department of Applied Mathematics, 
Beijing Institute of Technology, Beijing 100081, China}
\email{tianyu2992@163.com}

\address{Weigao Ge \hfill\break
 Department of Applied Mathematics,
 Beijing Institute of Technology, Beijing 100081, China}
\email{gew@bit.edu.cn}

\date{}
\thanks{Submitted August 28, 2004. Published December 3, 2004.}
\thanks{Supported by grant 10371006 from the National Natural
Sciences Foundation of China}
\subjclass[2000]{34B10, 34B15}
\keywords{Fourth-order two-point boundary-value problem; fixed
point theorem; \hfill\break\indent  double cones; positive solution}

\begin{abstract}
 A new fixed point theorem on double cones is applied to obtain
 the existence of at least two positive solutions to
 \begin{gather*}
 (\Phi_p(y''(t))''-a(t)f(t,y(t),y''(t))=0,\quad 0<t<1,\\
 y(0)=y(1)=0=y''(0)=y''(1),
 \end{gather*}
 where $f:[0,1]\times[0,\infty)\times(-\infty,0]\to R,
 a\in L^1([0,1],(0,\infty)) $. We also give some examples to
 illustrate our results.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

We study the existence of multiple positive solutions for the
fourth-order two-point boundary-value problem
\begin{gather}
(\Phi_p(y''(t))''-a(t)f(t,y(t),y''(t))=0,\quad 0<t<1, \label{e1.1}\\
 y(0)=y(1)=0=y''(0)=y''(1), \label{e1.2}
 \end{gather}
where the nonlinear term $f$ is allowed to change sign,
$a\in L^1([0,1],(0,\infty)), \Phi_p x=|x|^{p-2}x, 1/p +1/q =1, p>1$.
When $p=2$ Problem \eqref{e1.1}-\eqref{e1.2} describes the deformations
of an elastic beam. The boundary conditions are given according to
the control at the ends of the beam.

 A great deal of research has been devoted to the existence
of solutions for the fourth-order two point boundary value
problem. We refer the reader to \cite{a1,d1,g2,g3,m1,y1} and their references.
Aftabizadeh \cite{a1}, \cite{y1}, del Pino and Manasevich \cite{d1}, Gupta
\cite{g2,g3}, Ma and Wang \cite{m1}, Liu \cite{l1}
 have studied the existence problem of
positive solutions of the following fourth-order two-point boundary-value problem
\begin{gather*}
y^{(4)}(t)-f(t,y(t),y''(t))=0,\quad 0<t<1,\\
 y(0)=y(1)=0=y''(0)=y''(1).
\end{gather*}
All the above works were done under assumption that the nonlinear
term $f$ is nonnegative.

In this paper, we will impose growth conditions on $f$ which
ensure the existence of at least two positive solutions for
\eqref{e1.1}-\eqref{e1.2}. The key tool in our approach is the following fixed
point theorem on double cones.

 For a cone $K$ in a Banach space $X$ with norm $\|\cdot\|$
and a constant $r>0$, let
$$
K_{r}=\{x\in K:\|x\|<r\}, \quad \partial K_{r}=\{x\in K:\|x\|=r\}.
$$
Suppose $\alpha: K\to R^+$ is
a continuously increasing functional, i.e., $\alpha$ is continuous
and $\alpha(\lambda x)\le\alpha(x)$ for $\lambda\in(0,1)$. Let
 $$
 K(b)=\{x\in K:\alpha(x)<b\}, \quad
 \partial K(b)=\{x\in K:\alpha(x)=b\}
$$
and $K_{a}(b)=\{x\in K:a<\|x\|,
\alpha(x)<b \}$. The origin in $X$ is denoted by $\theta$.

\begin{theorem}[\cite{g1}] \label{thm1.1}
Let $X$ be a real Banach space with norm $\|\cdot\|$ and $K, K'\subset X$
two solid cones with $K'\subset K$. Suppose $T: K\to K$ and
$T': K'\to K'$ are two completely continuous operators and
$\alpha:K'\to R^+$ a continuously increasing functional
satisfying $\alpha(x)\le\|x\|\le M\alpha(x)$ for all $x$ in $K'$,
where $M\ge 1$ is a constants $b>a>0$ such that
\begin{itemize}
\item[(C1)] $\|Tx\|<a$, for $x\in \partial K_a$
\item[(C2)] $\|T'x\|<a$, for $x\in \partial K_{a}'$ and
$\alpha(T'x)>b$ for $x\in \partial K^{'}(b)$
\item[(C3)] $Tx=T'x$, for $x\in K'_{a}(b)\cap\{u:T'u=u\}$
\end{itemize}
then $T$ has at least two fixed points $y_1$ and $y_2$ in  $K$
such that
$$
0\le \|y_1\|<a<\|y_2\|,\quad  \alpha(y_2)<b.
$$
\end{theorem}


\section{Existence of positive solutions}

\begin{lemma}\label{lem2.1}
Suppose $g(\cdot)\in C[0,1]$, then
\begin{gather}
(\Phi_p(y''(t)))''-g(t)=0,\quad 0<t<1, \label{e2.1}\\
 y(0)=y(1)=0=y''(0)=y''(1), \label{e2.2}
\end{gather}
has a unique solution
\begin{equation}
y(t)=\int_{0}^{1}G(t,s)\Phi_q
\big(\int_{0}^{1}G(s,\tau)g(\tau)d\tau\big)ds,
\label{e2.3}
\end{equation}
 where
$$
G(t,s)= \begin{cases}
t(1-s),& 0\le t\le s \le 1;\\
 s(1-t),& 0\le s\le t \le 1.
\end{cases}
$$
\end{lemma}

\begin{proof} Let $\Phi_p y''(t)=u(t)$, then  \eqref{e2.1}-\eqref{e2.2}
 becomes
\begin{gather*}
u''-g(t)=0,\quad 0<t<1;\\
 u(0)=u(1)=0.
\end{gather*}
 It is clear that the above boundary-value problem has a unique solution,
$$
u(t)=\int_{0}^{1}G(t,s)g(s)ds,
$$
where $G(t,s)= \begin{cases}
t(1-s),& 0\le t\le s \le 1;\\
 s(1-t),& 0\le s\le t \le 1.
\end{cases}$.
 Then
 $\Phi_p y''(t)=u(t)$, i.e., $y''(t)=(\Phi_q u)(t)$.
By the boundary  condition we know that
 $$
y(t)=\int_{0}^{1}G(t,s)(\Phi_q u)(s)ds=\int_{0}^{1}G(t,s)\Phi_q
\big(\int_{0}^{1}G(s,r)g(r)dr\big)ds.
$$
 The proof is completed.
\end{proof}

In this paper, we assume the following conditions:
\begin{itemize}
\item[(H1)]  $f:[0,1]\times[0,\infty)\times(-\infty,0]\to R$ is
continuous, $a\in L^1([0,1],(0,\infty))$

\item[(H2)] $a(t)f(t,0,0)\not\equiv 0, f(t,0,0)\ge 0$ for $t\in[0,1]$.
\end{itemize}
Let $X=\{x\in C^2[0,1]: x(0)=x(1)=0=x''(0)=x''(1) \}$. Then $X$
is a Banach space with the norm
$\|x\|=\sup_{t\in[0,1]}|x''(t)|$. Define
 $$
K=\{x\in X: x \mbox{ is nonnegative and concave on }[0,1] \},
$$
and $K'=\{x\in X: x$ is nonnegative and concave on $[0,1]$,
$\alpha(x)\ge \delta^{q-1}\|x\|$, $\delta \in(0,1/2) \}$,
where
$\alpha(x)=\min_{t\in [\delta,1-\delta]}\{-x''(t)\}$.
Obviously $K, K'\subset X$ are two cones with $K'\subset K$. From
the definition $K'$, we know that
$\alpha(x)\le \|x\|\le\frac{1}{\delta^{q-1}}\alpha(x)$, $\delta\in(0,1/2)$. Denote
$$
(Tx)(t)=\Big[\int_{0}^{1}G(t,s)\Phi_q\big(\int_{0}^{1}G(s,\tau)
a(\tau)f(\tau,x(\tau),x''(\tau))d\tau\big) ds\Big]^{+},
$$
where $B^+=\max\{B,0\}$.
$$
(Ax)(t)=\int_{0}^{1}G(t,s)\Phi_q
\big(\int_{0}^{1}G(s,\tau)a(\tau)f(\tau,x(\tau),x''(\tau))d\tau\big) ds.
$$
For $x\in X$, define $\theta:X\to K$ by
$(\theta x)(t)=\max\{x(t),0\}$, then $T=\theta\circ A$.
For $x\in K'$, let
$$
(T'x)(t)=\int_{0}^{1}G(t,s)\Phi_q\big(\int_{0}^{1}G(s,\tau)a(\tau)f^{+}
(\tau,x(\tau),x''(\tau))d\tau\big)ds,
$$
where $f^+(t,x(t),x''(t))=\max\{f(t,x(t),x''(t)),0\}$.

\begin{lemma}\label{lem2.2}
For $x\in X$, we have  $\|x\|_{\infty}\le\|x''\|_{\infty}$ and
$\|x'\|_{\infty}\le\|x''\|_{\infty}$ where
$\|x\|_{\infty}=\sup_{t\in[0,1]}|x(t)|$.
\end{lemma}

\begin{proof} From
$$
x(t)=\int_{0}^{1}G(t,s)\Phi_q\big(\int_{0}^{1}G(s, r)g(r)dr\big)ds
$$
and
$$
x''(t)=-\Phi_q\big(\int_{0}^{1}G(t, r)g(r)dr\big)\,,
$$
we have
\begin{align*}
|x(t)|
&=\big|\int_{0}^{1}G(t,s)\Phi_q\big(\int_{0}^{1}G(s,r)g(r)dr\big)ds\big|\\
&=\big|\int_{0}^{1}G(t,s)|x''(s)|ds\big|\\
&\le\|x''\|_{\infty}\int_{0}^{1}G(t,s)ds\\
&=\|x''\|_{\infty}\big[\int_{0}^{t}s(1-t)ds+\int_{t}^{1}t(1-s)ds\big]\\
&=\|x''\|_{\infty}\big(\frac{t^2 (1-t)}{2}+\frac{t(1-t)^2}{2}\big).
\end{align*}
So
$\|x\|_{\infty}=\sup_{t\in[0,1]}|x(t)|\le\frac{1}{8}\|x''\|_{\infty}
<\|x''\|_{\infty}$.
At the same time, from $x'(t)=\int_{0}^{t}x''(s)ds$, we have
$\|x'\|_{\infty}=\sup_{t\in[0,1]}|x'(t)|\le \|x''\|_{\infty}$.
 The proof is complete.
 \end{proof}

Note that $X$ is a Banach space
with the norm $\|x\|=\sup_{t\in[0,1]}|x''(t)|$.


\begin{lemma}\label{lem2.3}
$T'(K')\subset K'$.
\end{lemma}

\begin{proof} For any $x\in K'$, it is clear that $(T'x)(t)$ is
nonnegative from the definition of $T'$. From
$(T'x)''(t)=-\Phi_q\big(\int_{0}^{1}G(t,s)f^+(s,x(s),x''(s))ds\big)$,
we know $(T'x)''(t)\le 0$. So $T'x$ is concave on $[0,1]$. Then
\begin{align*}
-(T'x)''(t)
&=\Phi_q\big(\int_{0}^{1}G(t,s)f^+(s,x(s),x''(s))ds\big)\\
&\le \Phi_q\big(\int_{0}^{1}G(s,s)f^+(s,x(s),x''(s))ds\big),
\end{align*}
which implies
\[
\|-(T'x)''\|_{\infty}
\le \Phi_q\big(\int_{0}^{1}G(s,s)f^+(s,x(s),x''(s))ds\big),
\]
and
\begin{align*}
\alpha(T'x)&=\min_{t\in[\delta,1-\delta]}[-(T'x)''(t)]\\
&=\min_{t\in[\delta,1-\delta]}\Phi_q
\big(\int_{0}^{1}G(t,s)f^+(s,x(s),x''(s))ds\big)\\
&= \min_{t\in[\delta,1-\delta]}\Phi_q
\Big(\int_{t}^{1}t(1-s)f^+(s,x(s),x''(s))ds\\
&\quad+ \int_{0}^{t}s(1-t)f^+(s,x(s),x''(s))ds\Big)\\
&\ge\min_{t\in[\delta,1-\delta]}\Phi_q\Big(\int_{t}^{1}\delta
s(1-s)f^+(s,x(s),x''(s))ds\\
&\quad +\int_{0}^{t}\delta s(1-s)f^+(s,x(s),x''(s))ds\Big)\\
&=\Phi_q\Big(\delta\int_{0}^{1}G(s,s)f^+(s,x(s),x''(s))ds\Big)\\
&\ge\delta^{q-1}\|T'x\|.
\end{align*}
The proof is complete. \end{proof}

For convenience, we denote
\begin{gather*}
Q=\max_{t\in[0,1]}\big\{\Phi_q\big(\int_{0}^{1}G(t,s)a(s)ds\big)\big\},\\
m=\min_{t\in[\delta,1-\delta]}\big\{\Phi_q
\big(\int_{\delta}^{1-\delta}G(t,s)a(s)ds\big)\big\},\\
m_i=\min_{t\in[\delta_i,1-\delta_i]}
\big\{\Phi_q\big(\int_{\delta_i}^{1-\delta_i}G(t,s)a(s)ds\big)\big\}.
\end{gather*}
It is clear that $0<m<Q<\infty$.

 From the continuity of $f, a\in L^{1}([0,1],(0,\infty))$. It is easy  to
see $A:K\to X$ and $T':K'\to K'$ are completely
continuous. So $T:K\to K$ is completely continuous.


\begin{theorem} \label{thm2.4}
Suppose (H1) and (H2) are satisfied and there exist $a, b, d$ such that
$0<d<\delta^{q-1}a<a<\delta^{q-1} b<b$. Assume that $f$ satisfies the following
conditions:
\begin{itemize}
\item[(H3)] For $(t,u,v)\in[0,1]\times[0,b]\times[-b,-d]$, $f(t,u,v)\ge 0$
\item[(H4)] For $(t,u,v)\in[0,1]\times[0,a]\times[-a,0]$,
  $f(t,u,v)<\Phi_p(\frac{a}{Q})$.
\item[(H5)]  For $(t,u,v)\in[0,1]\times[0,b]\times[-b,-\delta^{q-1} b],
f(t,u,v)\ge \Phi_p(\frac{b}{m})$.
\end{itemize}
 Then \eqref{e1.1}-\eqref{e1.2} has at least two positive solutions
$y_1, y_2$ such that
\begin{equation}
0<\|y_1\|<a<\|y_2\|,\quad \alpha(y_1)<\delta^{q-1}b,\quad
\|y_1\|_{\infty}<a, \quad \|y_2\|_{\infty}<b \label{e2.4}
\end{equation}
\end{theorem}

\begin{proof}
First we show that $T$ has a fixed point $y_1\in K$ with
$\|y_1\|\le a$. In fact, for any $y\in \partial K_a$, we have
$\|y\|=a$. So $0\le y(t)\le a, -a\le y''(t)<0, t\in[0,1]$. Let
$I=\{t\in[0,1]:f(t, y(t), y''(t))\ge 0\}$.
\begin{align*}
\|Ty\|&=\max_{t\in[0,1]}|(Ty)''(t)|\\
&=\max_{t\in[0,1]}\max\big\{\Phi_q\big(\int_{0}^{1}G(t,s)a(s)f(s,y(s),y''(s))ds
\big), 0\big\}\\
&\le \max_{t\in[0,1]}\Phi_q\big(\int_{I}G(t,s)a(s)f(s,y(s),y''(s))ds\big)\\
&\le \Phi_q\Big(\max_{t\in[0,1],0\le u\le a,-a\le v\le
0}f(t,u,v)\max_{t\in[0,1]}\big\{\int_{I}G(t,s)a(s)ds\big\}\Big)\\
&< \frac{a}{Q}\max_{t\in[0,1]}\big\{\Phi_q\big(\int_{0}^{1}G(t,s)a(s)ds
\big)\big\}\\
&= a.
\end{align*}
The existence of $y_1$ is proved by using condition (C1) of Theorem \ref{thm1.1}
and $0\le y_1 \le a$, $-a\le y''_{1}\le 0$. Obviously, $y_1$ is a
solution of \eqref{e1.1}-\eqref{e1.2}. Suppose this is not true, then there
is $t_0\in(0, 1)$ such that  $y_1(t_0)\neq (Ay_1)(t_0)$. It must
be $(Ay_1)(t_0)<0=y_1(t_0)$. Let $(t_1, t_2)$ be the maximum
interval such that $(Ay_1)(t_0)<0$ for $t\in (t_1, t_2)$. We claim
$[t_1, t_2]\neq [0,1]$ because of $a(t)f(t, 0, 0)\not\equiv 0$ for
$t\in[0, 1]$.

If $t_2<1$. $y_1(t)=0, t\in[t_1,t_2]$. $(Ay_1)(t_2)=0,
(Ay_1)(t)<0$, for $t\in(t_1,t_2)$. Then $(Ay_1)'(t_2)\ge 0$. For
$t\in(t_1,t_2)$, we have
\begin{align*}
(Ay_1)''(t)&=-\Phi_q(\int_{0}^{1}G(t,s)a(s)f(s,y(s),y''(s))ds)\\
&=-\Phi_q(\int_{0}^{1}G(t,s)a(s)f(s,0,0)ds)<0.
\end{align*}
So $(Ay_1)'(t)$ is decreasing for $t\in(t_1,t_2)$. Noticing
$(Ay_1)'(t_2)\ge 0, $ so $t_1=0$
and $(Ay_1)'(t)> 0, t\in[0,t_2), (Ay_1)(0)<0$, which contradicts (1.2).
If $t_1>0$. So $y_1(t)=0, (Ay_1)(t_1)=0, (Ay_1)(t)<0$ for
$t\in(t_1,t_2). (Ay_1)'(t_1)\le 0$.
\begin{align*}
(Ay_1)''(t)&=-\Phi_q(\int_{0}^{1}G(t,s)a(s)f(s,y(s),y''(s))ds)\\
&=-\Phi_q(\int_{0}^{1}G(t,s)a(s)f(s,0,0)ds)\le 0.
\end{align*}
So $(Ay_1)'(t)$ is decreasing for $t\in(t_1,t_2)$. So $t_2=1,
(Ay_1)(1)<0$, which contradicts boundary condition (1.2).
So $y_1$ is a solution of \eqref{e1.1}-\eqref{e1.2}.
We now show that (C2) of Theorem \ref{thm1.1} is satisfied. For
$x\in\partial K^{'}_{a}$, i.e., $\|x\|=a$, then $0<x(t)<a$,
$-a<x''(t)<0$ for $t\in[0,1]$.
\begin{align*}
\|T'y\|&=\max_{t\in[0,1]}|(T'y)''(t)|\\
&=\max_{t\in[0,1]}\big\{\Phi_q\big(\int_{0}^{1}G(t,s)a(s)f^+(s,y(s),y''(s))ds
\big)\big\}\\
&\leq\Phi_q\Big[\max \big\{f^+(t,u,v): t\in[0,1],0\le u\le
a,-a\le v\le 0\big\}\\
&\quad\times \max_{t\in[0,1]}\big\{\Phi_q\big(\int_{0}^{1}G(t,s)a(s)ds\big)\big\}
\Big]\\
&<\frac{a}{Q}\int_{0}^{1}G(t,s)a(s)ds \\
&=a.
\end{align*}
For $y\in\partial K'(\delta^{q-1} b)$, we have
$\alpha(y)=\delta^{q-1} b$. So $ \delta^{q-1} b\le \|y\|\le b$,
i.e., $-b\le y''(t)\le -\delta^{q-1} b$ for $t\in[\delta,
1-\delta]$, at the same time $\|y\|_{\infty}\le b$. Then
\begin{align*}
\alpha(T'y)&=-\min_{t\in[\delta,1-\delta]}(T'y)''(t)\\
&=\min_{t\in[\delta,1-\delta]}\Phi_q
\big(\int_{0}^{1}G(t,s)a(s)f^+(s,y(s),y''(s))ds\big)\\
&\ge\Phi_q\big(\min_{t\in[\delta,1-\delta]}\int_{\delta}^{1-\delta}G(t,s)a(s)f^+
(s,y(s),y''(s))ds\big)\\
&> \Phi_q\Big(\min\big\{f(t,u,v):t\in[0,1],u\in[0,b],v\in[-b,-\delta^{q-1} b]\big\} \\
&\quad \times \min_{t\in[\delta,1-\delta]}
\int_{\delta}^{1-\delta}G(t,s)a(s)ds\Big)\\
&= b>\delta^{q-1} b.
\end{align*}
Finally we show that $(C_3)$ of
Theorem \ref{thm1.1} is also satisfied. Let $x\in K^{'}_{a}(\delta^{q-1}
b)\cap\{u: T'u=u\}$, then $\|x\|<\frac{1}{\delta^{q-1}}\alpha(x)$.
From $\alpha(x)\le\|x\|\le \frac{1}{\delta^{q-1}}\alpha(x)$, we
have
$$
\min_{t\in[\delta, 1-\delta]}\{-x''(t)\}=\alpha(x)\ge
\delta^{q-1}\|x\|>\delta^{q-1} a>d\,.
$$
So $-x''\in[d, b]$. From
(H3), we have $f(t,u,v)=f^+(t,u,v)$, which implies $Ty=T'y$.
Therefore, there exist two positive solutions $y_1,y_2$ satisfying
\eqref{e2.2}.
 \end{proof}

\noindent{\bf Remark.} When $p=2, a(t)\equiv 1, f(t,u,v)>0, \delta=1/4$,
Theorem \ref{thm2.4} reduces to \cite[Theorem 3.1]{l1}.
 But our result shows at least two positive solutions,
whereas there is at least one positive solution in B. Liu
\cite[Theorem 3.1]{l1}.

\begin{theorem} \label{thm2.5}
Suppose (H1),(H2) hold. Also assume that
 \begin{itemize}
\item[(H6)] $\delta_i\in(0,1/2)$, $i=1,2,\dots, n$,
$0<\int_{\delta_i}^{1-\delta_i}a(s)ds<\infty$

\item[(H7)] There exists constants $a_i, b_i, d_i>0$, $i=1,2,\dots, n$,
where $0<d_i<\delta^{q-1}a_i <a_i<\delta^{q-1}_{i} b_i<b_i<d_{i+1}$
such that for $i=1,2,\dots, n$, we have
\begin{gather*}
f(t,u,v)\ge 0 \quad\mbox{for } (t,u,v)\in[0,1]\times[0,b_i]\times[-b_i, -d_i],\\
f(t,u,v)<\Phi_q(\frac{a_i}{Q}) \quad\mbox{for }
(t,u,v)\in[0,1]\times[0,a_i]\times[-a_i,0],\\
f(t,u,v)\ge \Phi_q(\frac{b_i}{m_i}) \quad\mbox{for }
(t,u,v)\in[0,1]\times[0,b_i]\times[-b_i,-\delta_i^{q-1} b_i] .
\end{gather*}
\end{itemize}
 Then \eqref{e1.1}-\eqref{e1.2} has at least $n+1$ positive solutions
$y_1, y_2, \dots,y_{n+1}$ satisfying
\begin{gather*}
0\le\|y_1\|<a_1<\|y_2\|\le b_1, \quad \alpha(y_2)<\delta_{1}^{q-1} b_1, \quad
 0<\|y_1\|_{\infty}<a_1, \\
0<\|y_2\|_{\infty}<b_1 ,\quad
a_2<\|y_3\|, \quad  \alpha(y_3)<\delta_{2}^{q-1} b_2, \quad
 0<\|y_3\|_{\infty}<b_2,\\
\dots \\
a_n<\|y_{n+1}\|,\quad  \alpha(y_{n+1})<\delta_{n}^{q-1} b_n, \quad
 0<\|y_{n+1}\|_{\infty}<b_n.
\end{gather*}
\end{theorem}

 \begin{theorem} \label{thm2.6}
 Suppose (H1), (H2),  (H6) hold. Also assume
\begin{itemize}
\item[(H8)] There exists constants $a_i$, $b_i>0$, $d$,  $i=1,2,\dots, n$,
where $0<d<\delta^{q-1}a_i<a_i <\delta_i^{q-1} b_i<b_i$,
 such that for $i=1,2,\dots, n$, we have:
\begin{gather*}
f(t,u,v)\ge 0 \quad\mbox{for }
(t,u,v)\in[0,1]\times[0,b_n]\times[-b_n, -d],\\
f(t,u,v)<\frac{a_i}{Q} \quad\mbox{for }(t,u,v)\in[0,1]\times[0,a_i]\times[-a_i,0],\\
f(t,u,v)\ge \frac{b_i}{m_i}\quad\mbox{for }
(t,u,v)\in[0,1]\times[0,b_i]\times[-b_i,-\delta_{i}^{q-1} b_i],
\end{gather*}
\end{itemize}
Then \eqref{e1.1}-\eqref{e1.2} has at least $2n$ positive solutions
$y_1, y_2, \dots,y_{2n}$ satisfying
\begin{gather*}
0\le\|y_1\|<a_1<\|y_2\|, \quad \alpha(y_2)<\delta_{1}^{q-1} b_1<\alpha(y_3), \\
 0<\|y_1\|_{\infty}<a_1, \quad 0<\|y_2\|_{\infty}<b_1, \\
 \|y_3\|<a_2<\|y_4\|,    \quad  \alpha(y_4)<\delta_{2}^{q-1} b_2<\alpha(y_5), \quad
  \|y_3\|_{\infty}<a_2,  \quad  \|y_4\|_{\infty}<b_2,
\dots\\
\|y_{2n-1}\|<a_n<\|y_{2n}\|,\quad  \alpha(y_{2n})<\delta_{n}^{q-1} b_n, \quad
\|y_{2n-1}\|_{\infty}<a_n,\quad \|y_{2n}\|_{\infty}<b_n.
\end{gather*}
\end{theorem}


\subsection*{Example} Consider the boundary-value problem
\begin{equation}
\begin{gathered}
(\Phi_3 y''(t))''-\big[\frac{y+\pi/6}{6}\big(\sqrt{3}\cos(y''+\frac{5}{12}\pi)
\big)^{19}+\frac{t}{10}\big]=0,\quad 0<t<1,\\
 y(0)=y(1)=0=y''(0)=y''(1),
  \end{gathered} \label{e2.5}
\end{equation}
where $a(t)=t$,
$f(t,u,v)=\frac{u+\pi/6}{6}(\sqrt{3}\cos(v+\frac{5}{12}\pi))^{19}+\frac{t}{10}$,
$p=3$, $q=3/2$. Clearly $f$ is allowed to change sign on
$[0,1]\times [0, \infty)\times (-\infty, 0 )$.
$$
Q=\max_{t\in
[0,1]}\Phi_{3/2}\big(\int_{0}^{1}G(t,s)a(s)ds\big)
=\max_{t\in [0,1]}\big(\frac{t}{6}(-t^2+1)\big)^{1/2}
=(\frac{\sqrt{3}}{27})^{1/2}=3^{-\frac{5}{4}}.
$$
Note that
\begin{align*}
\int_{\delta}^{1-\delta}G(t,s)a(s)ds
&=(1-t)\int_{\delta}^{t}s^2ds+t\int_{t}^{1-\delta}(1-s)s ds\\
&=\frac{1}{6}[-t^3+t(4\delta^3-3\delta^2+1)-2\delta^3].
\end{align*}
Let $\delta=1/4$, $d=\pi/36$, $a=\pi/12$,
$b=\pi/2$. It is clear that $d<\delta^{1/2}a<a<\delta^{1/2}b$.
Then
$$
m=\min_{t\in[\delta,
1-\delta]}\Phi_{3/2}\big(\int_{\delta}^{1-\delta}G(t,s)a(s)ds\big)
>\sqrt{\frac{1}{24}}\,.
$$
For $(t, u, v)\in[0, 1]\times[0, \pi/2]\times[-\pi/2, -\pi/36]$, we have
$ f(t, u,v)=\frac{u+\pi/6}{6}(\sqrt{3}\cos(v+\frac{5}{12}\pi))^{19}+\frac{t}{10}>0$.
 So (H3)  holds.
For $(t, u, v)\in[0, 1]\times[0, \pi/12]\times[-\pi/12, 0]$,
$ f(t, u,v)=\frac{u+\pi/6}{6}\big(\sqrt{3}\cos(v+\frac{5}{12}\pi)\big)^{19}
+\frac{t}{10}<\frac{\pi}{24}\times(\frac{\sqrt{3}}{2})^{19}+\frac{1}{10}
<0.6<(\frac{\pi}{12}\times 3^{5/4})^{2}=\Phi_{3}(a/Q)$. So
(H4)  holds.
 For $(t, u, v)\in[0, 1]\times[0, \pi/2]\times[-\pi/2, -\pi/4]$,
$f(t, u, v)=\frac{u+\pi/6}{6}\big(\sqrt{3}\cos(v+\frac{5}{12}\pi)\big)^{19}
+\frac{t}{10}>(\pi\sqrt{6})^2>\Phi_{3}(b/m)$. So (H5)  holds.
Thus by Theorem \ref{thm2.4}, this boundary-value problem has at least two positive
 solutions $y_1, y_2$ such that
 $$
0<\|y_1\|<\frac{\pi}{12}<\|y_2\|,\quad
\alpha(y_1)<\frac{\pi}{16},\quad
\|y_1\|_{\infty}<\frac{\pi}{12},\quad
\|y_2\|_{\infty}<\frac{\pi}{2}.
$$

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\end{document}