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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 154, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/154\hfil Positive solutions]
{Positive solutions for third-order Sturm-Liouville boundary-value
problems with $p$-Laplacian}

\author[C. B. Zhai,C. M. Guo \hfil EJDE-2009/154\hfilneg]
{Chengbo Zhai, Chunmei Guo}  % in alphabetical order

\address{Chengbo Zhai\newline
School of Mathematical Sciences, Shanxi University,
 Taiyuan  030006,  Shanxi, China}
\email{cbzhai@sxu.edu.cn}

\address{Chunmei Guo \newline
School of Mathematical Sciences, Shanxi University,
 Taiyuan  030006,  Shanxi, China}
\email{guocm@sxu.edu.cn}

\thanks{Submitted September 1, 2008. Published November 28, 2009.}
\subjclass[2000]{34K10}
\keywords{Positive solution; Sturm-Liouville
boundary value problem;\hfill\break\indent $p$-Laplacian operator;
concave functional; fixed point}

\begin{abstract}
 In this article, we consider the third-order Sturm-Liouville  boundary
 value problem, with $p$-Laplacian,
 \begin{gather*}
 (\phi_p(u''(t)))'+f(t,u(t))=0, \quad t\in (0,1),\\
 \alpha u(0)-\beta u'(0)=0,\quad \gamma u(1)+\delta u'(1)=0,\quad u''(0)=0,
 \end{gather*}
 where $\phi_p(s)=|s|^{p-2}s$, $p>1$.
 By means of the Leggett-Williams fixed-point theorems, we prove
 the existence of multiple positive solutions.
 As an application, we give an example that illustrates our result.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction}

In this paper, we study the existence of multiple positive solutions
for the following third-order Sturm-Liouville boundary value problem
with $p$-Laplacian
\begin{gather}
 (\phi_p(u''(t)))'+f(t,u(t))=0, \quad t\in (0,1),\label{e1.1}\\
 \alpha u(0)-\beta u'(0)=0,\quad \gamma u(1)+\delta u'(1)=0,\quad
 u''(0)=0,\label{e1.2}
\end{gather}
 where $\phi_p(s)=|s|^{p-2}s$, $p>1$, $(\phi_p)^{-1} = \phi_q$,
 $\frac 1p + \frac 1q = 1$, $\alpha, \beta, \gamma, \delta\geq 0$.

During the past decades, wide attention has been paid to the study
equations with $p$-Laplacian operator,  which arises in the
modelling of different physical and natural phenomena, non-Newtonian
mechanics \cite{d1,j1}, combustion theory \cite{r1}, population
biology \cite{o2,o3}, nonlinear flow laws \cite{g1,l4,l5}, and
system of Monge-Kantorovich partial differential equations
\cite{e1}. There exist a very large number of papers devoted to the
existence of solutions of the $p$-Laplacian operator. The
second-order problem,
$$
 (\phi_p(u'(t)))'+f(t,u(t))=0, \quad t\in (0,1),
$$
with various boundary conditions has been studied by many authors,
see \cite{h2,l3,m1,o1,w1,w2,z1,z2}
 and the references therein.
However, to the best of our knowledge, few papers can be found in
the literature on the existence of multiple positive solutions for
the third-order Sturm-Liouville boundary value problem \eqref{e1.1},
\eqref{e1.2}. The purpose here is to fill this gap in the
literature. Motivated by the works  \cite{a1} and \cite{h1}, we
shall establish the existence of at least two or at least three
positive solutions to third-order Sturm-Liouville boundary value
problem with $p$-Laplacian \eqref{e1.1}, \eqref{e1.2} by using fixed
point theorems in cones.

By a positive solution of \eqref{e1.1} and \eqref{e1.2} we
understand a function
$u(t)\in C^2[0,1]$ which is positive on $0<t<1$ and satisfies the
differential equation \eqref{e1.1} and the boundary conditions
\eqref{e1.2}.

In this article, we use the following assumptions:
\begin{itemize}
\item[(A1)] $\rho:=\gamma\beta+\alpha\gamma+\alpha\delta>0$,
$0<\sigma:=\min\big\{\frac{4\delta+\gamma}{4(\delta+\gamma)},\frac
{\alpha+4\beta}{4(\alpha+\beta)}\big\}<1$.

\item[(A2)] $G(t,s)$ is the Green's function of the
 differential equation $u''(t)=0$, $t\in  (0,1)$ with respect to
 the boundary value condition \eqref{e1.2}, i.e.,
$$
  G(t,s)=  \begin{cases}
\frac 1\rho (\gamma+\delta-\gamma t)(\beta +\alpha
  s), &0\leq s\leq t\leq 1.\\
\frac 1\rho (\beta+\alpha
t)(\gamma+\delta-\gamma s), & 0\leq t\leq s\leq 1.
\end{cases}
$$
Evidently  $ G(t,s)\leq G(s,s)$, $0\leq t,s\leq 1$.

\item[(A3)] $f\in C([0,1]\times [0,\infty);[0,\infty))$.

\end{itemize}
For convenience, we denote
\begin{gather*}
 \zeta(a)=\max\{f(t,u):0\leq t\leq 1,\ 0\leq u\leq a\},\\
 \psi(b)=\min\{f(t,u):\frac 14\leq t\leq \frac 34,\ b\leq u\leq {\frac
 b{\sigma^2}}\}.
\end{gather*}
 Where $\sigma$ is given as in (A1).
Our main results are the following.

\begin{theorem} \label{thm1.1}
 Assume  {(A1)--(A3)}, and that
there exist constants $0<a<b$ such that
\begin{gather} \label{ei}
 \zeta(a)<(ma)^{p-1}\\
\label{eii} \psi(b)\geq (lb)^{p-1}.
\end{gather}
Then  the boundary value problem \eqref{e1.1}, \eqref{e1.2} has at
least two positive solutions $u_1, u_2$ satisfying
$\|u_1\|<a$, $\min_{t\in[\frac 14,\frac 34]}u_2(t)<b$ and
$\|u_2\|>a$, where
\begin{gather*}
  m=\Big(\int^1_0 G(s,s)ds\Big)^{-1}
=\frac {6\rho}{\alpha\gamma+3\alpha\delta+3\beta\gamma+6\beta\delta},\\
 l= \frac 2{\sigma {4}^{1-q}}
\Big(\int^{3/4}_{1/4}G(\frac 12,s)ds\Big)^{-1}=
\frac 2{\sigma {4}^{1-q}}\cdot \frac {32
\rho}{3\alpha\gamma+7\alpha\delta+7\beta\gamma+16\beta\delta}.
\end{gather*}
\end{theorem}

\begin{theorem} \label{thm1.2}
Assume {\rm (A1)--(A3)} and that there exist constants $a,b,c$
such that $0<a<b<\sigma^2c$  implies
\begin{gather} \label{ei'}
  \zeta(a)<(ma)^{p-1},\\
\label{eii'}  \psi(b)\geq (lb)^{p-1},\\
\label{eiii'} \zeta(c)\leq (mc)^{p-1}.
\end{gather}
Then  the boundary value problem \eqref{e1.1}, \eqref{e1.2} has at
least three positive solutions $u_1, u_2$ and $u_3$ with
$\|u_1\|<a$, $\min_{t\in[\frac 14,\frac 34]}u_2(t)>b$,
$\|u_3\|>a$ and $\min_{t\in[\frac 14,\frac 34]}u_3(t)<b$, where
$\sigma$ is given as in (A1) and $m,l$ are given as in Theorem
\ref{thm1.1}.
\end{theorem}

The proofs of theorems are based upon the Leggett-Williams
fixed-point theorems \cite{l1}. These theorems have been useful
technique for proving  the existence of three or two solutions for
boundary value problems of differential and difference equations,
see \cite{a1,a2,h1}.

\section{Preliminaries}

In this section we summarize some basic concepts and results which
are taken from  Guo and Lakshmikantham \cite{g2},  and from Leggett
and Williams \cite{l1}.

\begin{definition} \label{def2.1}\rm
Let $ E$ be a real Banach space and $P$ be a nonempty, convex closed
set in $E$. We say that $P$ is a cone if it satisfies the following
properties: (i) $\lambda u\in P $ for $u\in P,\lambda\geq 0$; (ii)
$u,-u\in P$ implies $u=\theta(\theta$ denotes the null element of $
E)$.
\end{definition}

If $P\subset E $ is a cone, we denote the order induced by $P$
on $E$ by $\leq$. For $ u,v\in P$, we write $u\leq v$ if
$ v-u\in P$.

\begin{definition} \label{def2.2} \rm
The map $\varphi$ is said to be a nonnegative continuous concave
functional on $P$ provided that
$\varphi: P\to [0,\infty)$ is continuous and
$\varphi (tx+(1-t)y)\geq t\varphi (x)+(1-t)\varphi (y)$
for all $ x,y\in P$ and $0\leq t\leq1$.
\end{definition}

\begin{definition} \label{def2.3} \rm
Let $0<a<b$ be given and let $\varphi$ be a
nonnegative continuous concave functional on the cone
$P$. Define the convex sets $P_r,\bar{P_r}$ and $P(\varphi,a,b)$ by
$P_r=\{y\in P: \|y\|<r\},\ \ \bar{P_r}=\{y\in P: \|y\|\leq r\}$,
$P(\varphi,a,b)=\{y\in P: a\leq\varphi (y),\|y\|\leq b\}$.
\end{definition}

\begin{theorem}[Leggett-Williams \cite{l1}] \label{thm2.4}
Let $T:\bar{P_c}\to \bar{P_c}$ be a completely continuous
operator and let $\varphi$ be a nonnegative continuous concave
functional on $P$ such that $\varphi  (y)\leq \|y\|$ for all
$y\in \bar{P_c}$. Suppose that there exist $0<a<b<d\leq c$ such that
\begin{itemize}
\item[(a')]  $ \{y\in P(\varphi,b,d): \varphi (y)>b\}\neq\emptyset$
and $\varphi (Ty)>b$ for $y\in P(\varphi,b,d)$;

\item[(b')] $\|Ty\|<a$ for $ \|y\|\leq a$;

\item[(c')] $\varphi (Ty)>b$ for $y\in P(\varphi,b,c)$ with $\|Ty\|>d$.
\end{itemize}
Then $T$ has at least three fixed points $y_1, y_2, y_3$ in
$\bar{P_c}$ satisfying $\|y_1\|<a, \varphi(y_2)>b, \|y_3\|>a$ and
$\varphi(y_3)<b$
\end{theorem}

\begin{theorem}[\cite{l1}] \label{thm2.5}
Let $T:\bar{P_c}\to P$ be a
completely continuous operator and let $\varphi$ be a nonnegative
continuous concave functional on $P$ such that
$\varphi (y)\leq \|y\|$ for all $y\in \bar{P_c}$.
 Suppose that there exist $0<a<b< c$ such that
\begin{itemize}
\item[(a'')] $\{y\in P(\varphi,b,c): \varphi (y)>b\}\neq\emptyset$,
and $\varphi (Ty)>b$ for $y\in P(\varphi,b,c)$;

\item[(b'')] $\|Ty\|<a$ for $ \|y\|\leq a$;

\item[(c'')] $\varphi (Ty)>\frac bc \|Ty\|$ for
$y\in \bar{ P_c}$ with $\|Ty\|>c.$
\end{itemize}
 Then $T$ has at least two fixed points $y_1, y_2$ in $\bar{P_c}$
satisfying $\|y_1\|<a$, $\|y_2\|>a$ and $\varphi(y_2)<b$.
\end{theorem}

In the rest of this section we  assume that (A1)-(A3)
hold. Let $E=C[0,1]$ and $C^+[0,1]=\{x\in E| x(t)\geq 0$,
$t\in [0,1]\}$. Define an operator $T$ by
$$
 (Tu)(t)=\int^1_0G(t,v)\phi_q \Big(\int^v_0 f(s,u(s))ds\Big)dv,\quad
 \forall u\in C^+[0,1].
$$
 From (A2) and (A3), we can easily get
$(Tu)(t)\geq 0$, $t\in [0,1]$ for $u\in C^+[0,1]$.


\begin{remark} \label{rmk2.6}\rm
Suppose that $u\in C^+[0,1]$ satisfies of the
operator equation, $Tu=u$. We can obtain
\begin{gather*}
\begin{aligned}
u'(t)&= -\frac \gamma\rho\int^t_0(\beta+\alpha v)\phi_q
\Big(\int^v_0 f(s,u(s))ds\Big)dv\\
&\quad +\frac \alpha\rho\int^1_t(\gamma+\delta-\gamma v)\phi_q
\Big(\int^v_0 f(s,u(s))ds\Big)dv,
\end{aligned}\\
 u''(t)=(Tu)''(t)=-\phi_q \Big(\int^t_0 f(s,u(s))ds\Big).
\end{gather*}
  So we have
$$
 \phi_p(u''(t))=-\int^t_0 f(s,u(s))ds,
$$
  and in consequence,
$(\phi_p(u''(t)))'=-f(t,u(t))$.
 Moreover, it is clear  that
\begin{align*}
\alpha u(0)-\beta u'(0)
&= \alpha\int^1_0G(0,v)\phi_q \Big(\int^v_0
f(s,u(s))ds\Big)dv\\
&\quad -\beta\cdot \frac \alpha\rho \int^1_0(\gamma+\delta-\gamma
v)\phi_q \Big(\int^v_0 f(s,u(s))ds\Big)dv=0;
\end{align*}
\begin{align*}
\gamma u(1)+\delta u'(1)
&= \gamma\int^1_0G(1,v)\phi_q \Big(\int^v_0
f(s,u(s))ds\Big)dv\\
&\quad +\big(-\frac \gamma\rho\big)\cdot \delta\int^1_0(\beta+\alpha
v)\phi_q \Big(\int^v_0 f(s,u(s))ds\Big)dv=0.
\end{align*}
Further, $u''(0)=0$, that is to say, all the fixed points of
operator $T$ are the solutions for the problem
\eqref{e1.1}, \eqref{e1.2}.
\end{remark}

\begin{lemma}[\cite{l2}] \label{lem2.7}
Suppose that $G(t,s)$ is defined as in
(A2). Then
\begin{gather*}
\frac {G(t,s)}{G(s,s)}\leq 1\quad \mbox{for } t\in [0,1],\; s\in [0,1],\\
\frac {G(t,s)}{G(s,s)}\geq \sigma\quad \mbox{for }
t\in [\frac 14,\frac 34],\; s\in [0,1].
\end{gather*}
\end{lemma}

\begin{lemma} \label{lem2.8}
The operator $T:C^+[0,1]\to C^+[0,1]$ is completely
continuous; i.e., $T$ is continuous and compact.
\end{lemma}

\begin{proof}
 Firstly, we show that $T:C^+[0,1]\to C^+[0,1]$
is continuous. From Remark \ref{rmk2.6}, we know that
$T:C^+[0,1]\to C^+[0,1]$.
 Suppose $\{u_n\}\subset C^+[0,1],\ u_n\to
\bar{u}(n\to \infty)$. Then $\bar{u}\in C^+[0,1]$ and there
exists a constant $M_0>0$ such that
$\|u_n\|\leq M_0$,
$\|\bar{u}\|\leq M_0$. Let
$M_1=\max\{f(t,u)| t\in [0,1],\ u\in [0,M_0]\}$.
Then for $t\in [0,1]$ we have
\begin{align*}
|Tu_n(t)-T\bar{u}(t)|
&\leq  \int^1_0G(t,v)\big|\phi_q
\Big(\int^v_0 f(s,u_n(s))ds\Big)-\phi_q \Big(\int^v_0
f(s,\bar{u}(s))ds \Big)\big|dv\\
&\leq  \int^1_0G(v,v)\big|\phi_q \Big(\int^v_0
f(s,u_n(s))ds\Big)-\phi_q
\Big(\int^v_0 f(s,\bar{u}(s))ds\Big)\big|dv\\
&\leq  \int^1_0 2\phi_q(M_1)G(v,v)dv.
\end{align*}
Note that $f(t,u)$ is continuous.   We know that
$\phi_q(\int^v_0 f(s,u)ds)$
is continuous in $u$ on $[0,\infty)$. Then for
for each $\varepsilon>0$, there exists $\delta_1>0$, such that
$|u_1-u_2|<\delta_1$ and
we
 $$
 \big|\phi_q \Big(\int^v_0 f(s,u_1(s))ds\Big)-\phi_q
\Big(\int^v_0 f(s,u_2(s))ds\Big)\big|<\frac
\varepsilon{G(v,v)}.
$$
  In view of $u_n(s)\to \bar{u}(s)$, as
$n\to \infty$, there exists a natural number $N>0$, for
$n>N$ with $|u_n(s)-\bar{u}(s)|<\delta_1$, we have
$$
 \big|\phi_q \Big(\int^v_0 f(s,u_n(s))ds\Big)-\phi_q \Big(\int^v_0
f(s,\bar{u}(s))ds\Big)\big|<\frac \varepsilon{G(v,v)}.
$$
Thus
for $\varepsilon>0$, there exists $N>0$, such that when $n>N$,
$$
 G(v,v)\big|\phi_q \Big(\int^v_0 f(s,u_n(s))ds\Big) -\phi_q
\Big(\int^v_0 f(s,\bar{u}(s))ds\big)\big|< \varepsilon,\quad
\text{a.e. } [0,1].
$$
An application of Lebesgue's dominated convergence theorem implies
$$
 |Tu_n(t)-T\bar{u}(t)|\to 0 (\text{as }n\to\infty),\;
t\in [0,1].
$$
So operator $T:C^+[0,1]\to C^+[0,1]$ is continuous.

Next we prove that  $T$ is compact. Let $\Omega\subset C^+[0,1]$ be
a bounded set. Then there exists $R>0$ such that
$\Omega\subset \{u\in C^+[0,1]| \|u\|\leq R\}$.
Set $M=\max\{f(t,u)|t\in [0,1],\,u\in \Omega\}$.
For any $u\in \Omega$, we have
$$
 |(Tu)(t)|=\big|\int^1_0G(t,v)\phi_q \Big(\int^v_0 f(s,u(s))ds\Big)dv
\big|\leq \int^1_0G(v,v)\phi_q (M)dv,
$$
which implies that $T(\Omega)$ is uniformly bounded.

Furthermore, for any $u\in \Omega$ and $t\in [0,1]$, we have
\begin{align*}
|(Tu)'(t)|& = \Big|-\frac \gamma\rho\int^t_0(\beta+\alpha v)\phi_q
\Big(\int^v_0 f(s,u(s))ds\Big)dv\\
&\quad + \frac \alpha\rho\int^1_t(\gamma+\delta-\gamma v)\phi_q
\Big(\int^v_0 f(s,u(s))ds\Big)dv\Big|\\
&\leq  \phi_q(M)\Big[\frac \gamma\rho\int^t_0(\beta+\alpha v)
dv+\frac \alpha\rho\int^1_t(\gamma+\delta-\gamma v)
dv\Big]\\
&= \phi_q(M)t\leq \phi_q(M).
\end{align*}
Hence $\|(Tu)'\|\leq \phi_q(M)$. So we can easily prove that
$T(\Omega)$ is equicontinuous. The Arzela-Ascoli Theorem guarantee
that $T(\Omega)$ is relatively compact and therefore that $T$ is compact.
\end{proof}

\section{Proofs of main results}

In this section, we prove the existence of multiplicity results. Let
$E=C[0,1]$ be endowed with the maximum norm $\|y\|=\max_{t\in[0,1]}|
y(t)|$, and the ordering $x\leq y$ if $x(t)\leq y(t)$ for all
$t\in[0,1]$. Define the cone $P\subset E$ by
$$
 P=\{u\in C^+[0,1] :
 \min_{t\in [\frac 14,\frac 34]}u(t)\geq \sigma\|u\|\},
$$
where $\sigma$ is given as in (A1). Next we
show that $T(P)\subset P$. For any $u\in P$ and $ t\in [0,1]$, from
Lemma \ref{lem2.7} we have
$$
 Tu(t)=\int^1_0G(t,v)\phi_q \Big(\int^v_0 f(s,u(s))ds\Big)dv
\leq \int^1_0G(v,v)\phi_q \Big(\int^v_0 f(s,u(s))ds\Big)dv.
$$
  Consequently,
$$
 \|Tu\|\leq \int^1_0G(v,v)\phi_q \Big(\int^v_0
f(s,u(s))ds\Big)dv.
$$
  Further, for $u\in P$ and $t\in [\frac 14,\frac 34]$,
from Lemma \ref{lem2.7} we obtain
\begin{align*}
\min_{t\in [\frac 14, \frac 34]}Tu(t)
&= \min_{t\in [\frac 14,
\frac 34]}\int^1_0G(t,v)\phi_q \Big(\int^v_0 f(s,u(s))ds\Big)dv\\
&\geq  \sigma \int^1_0G(v,v)\phi_q \Big(\int^v_0
f(s,u(s))ds\Big)dv
\geq \sigma\|Tu\|.
\end{align*}
 From Lemma \ref{lem2.8}, we know that $T:P\to P$ is completely continuous. Let
$\varphi :P \to [0,\infty) $ be the nonnegative continuous
concave functional defined by
$$
  \varphi (u)=\min_{t\in [\frac 14,\frac 34]} u(t) , \quad  u\in P.
$$
Evidently, for each $u\in P$, we have $ \varphi (u)\leq \|u\|$. We
are now in a position to proving the main results.

\begin{proof}[Proof of Theorem \ref{thm1.1}]
 It is easy to see that $T:\bar{P_{\frac b{\sigma^2}}}\to P$ is
completely continuous and $0<a<b<\frac b{\sigma^2}$. Choose
$u(t)=\frac b{\sigma^2}$, then
$$
 u\in P\big(\varphi,b,\frac b{\sigma^2}\big),\quad
 \varphi (u)=\frac b{\sigma^2}>b.
$$
So $\{u\in P(\varphi,b,\frac b{\sigma^2}): \varphi (u)>b\}\neq \emptyset$.
Hence, if $u\in P(\varphi,b,\frac b{\sigma^2})$, then
$b\leq u(t)\leq\frac b{\sigma^2}$ for $t\in [\frac 14,\frac 34]$.
Thus for
$t\in [\frac 14,\frac 34]$, from assumption \eqref{eii}, we have
$$
 f(t,u(t))\geq \psi(b)\geq (lb)^{p-1},\ t\in [\frac 14,\frac 34].
$$
Hence
\begin{align*}
Tu(\frac 12)
&= \int^1_0G(\frac 12,v)\phi_q
\Big(\int^v_0 f(s,u(s)\Big)ds)dv\\
& \geq \int^{3/4}_{1/4}G(\frac 12,v)\phi_q
\Big(\int^v_0 f(s,u(s))ds\Big)dv\\
&\geq \int^{3/4}_{1/4} G(\frac 12,v)lb v^{q-1}dv \\
&\geq (\frac 14)^{q-1} lb\int^{3/4}_{1/4}G(\frac 12,v) dv
= \frac {2b}{\sigma}>\frac b\sigma.
\end{align*}
Consequently,
$$
 \min_{t\in [\frac 14,\frac 34]}Tu(t)\geq \sigma\|Tu\|>
 \sigma\times \frac b{\sigma}=b\ \ \mbox{for}\ b\leq u(t)
\leq \frac b{\sigma^2},\ t\in [\frac 14,\frac 34].
$$
That is,
 $$
 \varphi(Tu)>b, \forall\ u\in P\big(\varphi,b,\frac b{\sigma^2}\big).
$$
Therefore, condition (a'') of Theorem \ref{thm2.5} is satisfied.
 Now if $u\in \bar{P_a}$, then $\|u\|\leq a$. By assumption \eqref{ei}, we have
 $f(t,u(t))\leq \zeta(a)<(ma)^{p-1}$, $t\in[0,1]$. Consequently,
\begin{align*}
\|Tu\|&= \max_{t\in [0,1]}|Tu(t)|=\max_{t\in
[0,1]}\int^1_0G(t,v)\phi_q \Big(\int^v_0 f(s,u(s))ds\Big)dv\\
&<  ma\max_{t\in [0,1]}\int^1_0 G(t,v)dv
\leq ma \int^1_0 G(v,v)dv=a.
\end{align*}
This shows that $T:\bar{P_a}\to P_a$. That is, $\|Tu\|< a$ for $u\in
\bar{P_a}$. This shows that condition (b'') of Theorem \ref{thm2.5}
is satisfied. Finally, we show that (c'') of Theorem \ref{thm2.5}
also holds. Assume that $u\in \bar {P_{\frac b{\sigma^2}}}$ with
$\|Tu\|>\frac b{\sigma^2}$, then by the definition of cone $P$, we
have
$$
 \varphi (Tu)=\min_{t\in [\frac 14,\frac 34]}Tu(t)\geq \sigma\|Tu\|
>\sigma^2\|Tu\|
=b/ {\frac b{\sigma^2}}\|Tu\|.
$$
  So condition (c'') of Theorem \ref{thm2.5}
is satisfied. Thus using Theorem \ref{thm2.5}, $T$ has at least two fixed
points. That is to say, problem \eqref{e1.1},\eqref{e1.2} has
at least two
positive solutions $u_1, u_2$ in $\bar{P_{\frac b{\sigma^2}}}$
satisfying $\|u_1\|<a$,
$\min_{t\in[\frac 14,\frac 34]}u_2(t)<b$ and $\|u_2\|>a$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
  It follows from the conditions
\eqref{ei'}-\eqref{eiii'} in Theorem \ref{thm1.2} that
$a<b<\frac b{\sigma^2}<c$. Using the same arguments as in the
proof of Theorem \ref{thm1.1}, we have:
$T:\bar{P_c}\to \bar{P_c}$
is a completely continuous operator and
$T:\bar{P_a}\to P_a$. Also
$$
\{u\in P\big(\varphi,b,\frac b{\sigma^2}\big): \varphi(u)>b\}\neq
\emptyset,\quad
\varphi(Tu)>b \; \forall u\in
P\big(\varphi,b,\frac b{\sigma^2}\big).
$$
Moreover, for $u\in P(\varphi,b,c)$ and $\|Tu\|>\frac b{\sigma^2}$,
we have
$$
  \varphi (Tu)=\min_{t\in [\frac 14,\frac 34]}Tu(t)
\geq \sigma\|Tu\|>\frac b\sigma>b.
$$
So all the conditions of Theorem \ref{thm2.4} are satisfied.  Thus using
Theorem \ref{thm2.4}, $T$ has at least three fixed points. That is to say,
the boundary value problem \eqref{e1.1},\eqref{e1.2} has at least
three positive solutions $u_1, u_2\, u_3$ with
$\|u_1\|<a$,
$\min_{t\in[\frac 14,\frac 34]}u_2(t)>b$,
$\|u_3\|>a$ and $\min_{t\in[\frac 14,\frac 34]}u_3(t)<b$.
\end{proof}

\begin{corollary} \label{coro3.1}
 Assume {\rm (A1)--(A3)} and that
 there exist constants $0<a_j<b_j<\sigma^2a_{j+1}$
$(j=1, 2,\dots, n-1)$, $n\in N$ such that
\begin{itemize}
\item[(B1)] $\zeta(a_j)<(ma_j)^{p-1}$, $1\leq j\leq n$.

\item[(B2)] $\psi(b_j)\geq (lb_j)^{p-1}$, $1\leq j\leq n-1$,
  where $\sigma$ is given as in (A1) and $m,l$ are
given as in Theorem \ref{thm1.1}.
\end{itemize}
Then the boundary value problem \eqref{e1.1},\eqref{e1.2}
has at least $2n-1$ positive solutions.
\end{corollary}

The proof of the above corollary is an immediate consequence of
Theorem \ref{thm1.2}.

\begin{remark} \label{rmk3.2} \rm
 When $p=2$,  problem \eqref{e1.1}, \eqref{e1.2} is the usual
form of third-order Sturm-Liouville boundary value problem
\begin{gather*}
 u'''(t)+f(t,u(t))=0,\ t\in (0,1),\\
 \alpha u(0)-\beta u'(0)=0,\quad
 \gamma u(1)+\delta u'(1)=0,\quad u''(0)=0.
\end{gather*}
\end{remark}

 Using the same method, we can  present some sufficient
conditions that guarantee the existence of at least two or three
positive solutions for the above boundary value problem.
These results are also new and different from previous results.

\section{An example}

Now we consider an example to illustrate our results. Consider the
third-order Sturm-Liouville boundary value problem, with
$p$-Laplacian,
\begin{gather}
 (\phi_3(u''(t)))'+[\varphi(t)h(u(t))]^2=0, \quad t\in (0,1),\label{e4.1}\\
u(0)- u'(0)=0,\quad u(1)=0,\quad u''(0)=0,\label{e4.2}
\end{gather}
where $\varphi(t)=4t$, $t\in [0,1]$ and
$$
 h(u)=  \begin{cases}
u/2, & 0\leq  u\leq 3/512;\\
 \frac {1021}4u -\frac {3057}{2048}, & \frac 3{512}\leq u\leq \frac
 5{512};\\
 1,&  \frac 5{512} \leq u \leq \frac 5{32};\\
 \frac 8{59}u+\frac {231}{236},  & \frac5{32}\leq u\leq 2;\\
 5u/8,&  u\geq 2.
\end{cases}
$$
In this example, we note that $p=3$, $\alpha=\beta=\gamma=1$,
$\delta=0$. After a simple calculation, we get $q=3/2$,
$\rho=2$, $\sigma=\frac 14<1$, $G(s,s)=\frac 12(1-s^2)$ and
$$
 m=\frac {6\rho}{\alpha\gamma+3\alpha\delta+3\beta\gamma+6\beta\delta}=3,
\quad  l=\frac 2{\sigma {4}^{1-q}}\cdot \frac {32
\rho}{3\alpha\gamma+7\alpha\delta+7\beta\gamma+16\beta\delta}=\frac
{512}5.
$$
  We choose $a=\frac 3{512}$, $b=\frac 5{512}$,
$c=2$. Evidently, $a<b< \sigma^2 c$ and
\begin{itemize}
\item[(i)] for $t\in [0,1]$, $0\leq u\leq \frac 3{512}$, we have
$$
 f(t,u)=[\varphi(t)h(u)]^2\leq \big[4\times \frac 12\times\frac
3{512}\big]^2<(ma)^2.
$$
\item[(ii)] for $t\in [\frac 14,\frac 34]$,
$\frac 5{512}\leq u\leq \frac b{\sigma^2}=\frac 5{32}$, we have
 $$
  f(t,u)=[\varphi(t)h(u)]^2\geq\big[4\times\frac 14\times 1\big]^2
=(lb)^2.
$$

\item[(iii)] for $ t\in [0,1]$, $0\leq u\leq 2$, we have
$$
  f(t,u)=[\varphi(t)h(u)]^2\leq \big[4\times 1\times
\big(\frac 8{59}\times 2+\frac {231}{236}\big)\big]^2\leq (mc)^2.
$$
Thus, $\zeta(a)<(ma)^2$, $\psi(b)\geq (lb)^2$,
$\zeta(c)\leq (mc)^2$.
\end{itemize}
Hence, all the conditions of Theorem \ref{thm1.2} are satisfied. An
application of Theorem \ref{thm1.2} implies that   \eqref{e4.1}, \eqref{e4.2}
has at least three positive solutions $u_1, u_2$ and $u_3$ with
$\|u_1\|<\frac 3{512}$,
$\min_{t\in[\frac 14,\frac 34]}u_2(t)>\frac 5{512}$,
$\|u_3\|>\frac 3{512}$ and
$\min_{t\in[\frac 14,\frac 34]}u_3(t)<\frac 5{512}$.


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\end{document}