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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 149, pp. 1--10.\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-2013/149\hfil Existence of three positive solutions]
{Existence of three positive solutions for an $m$-point boundary-value
 problem on time scales}

\author[A. Dogan \hfil EJDE-2013/149\hfilneg]
{Abdulkadir Dogan}  % in alphabetical order

\address{Abdulkadir Dogan  \newline
Department of  Applied Mathematics,
Faculty of Computer Sciences,
Abdullah Gul University, Kayseri, 38039 Turkey\newline
Tel: +90 352 224 88 00\quad Fax:+90 352 338 88 28}
\email{abdulkadir.dogan@agu.edu.tr}

\thanks{Submitted March 29, 2013. Published June 27, 2013.}
\subjclass[2000]{34B15, 34B16, 34B18, 39A10}
\keywords{Time scales; boundary value problem; positive solutions;
\hfill\break\indent fixed point theorem}

\begin{abstract}
 We study an $m$-point boundary-value problem on time scales.
 By using a fixed point  theorem, we prove the existence of at least
 three positive solutions, under suitable growth conditions imposed
 on the nonlinear term. An example is given to illustrate
 our results.
\end{abstract}

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

\section{Introduction}

The theory of dynamic equation on time scales (or measure chains) was
initiated by Stefan Hilger in his Ph. D. thesis
in 1988 \cite{Hilger} (supervised by Bernd Aulbach) as a means of
unifying structure for the study of differential equations in the
continuous case and study of finite difference equations in the discrete case.
In recent years, it has found a considerable amount of interest and
attracted the attention of many researchers; see for example
 \cite{RPAgarwal,APeterson,Avery,AGraef,AKong,Sang,WLi,Sun,Zhang}.
It is still a new area, and research in this area is rapidly growing.
The study of time scales has led to several important applications, e.g.,
in the study of insect population models, heat transfer, neural networks,
phytoremediation of metals, wound healing, and epidemic models
\cite{Peter,Jones,Sped,Thomas}.

Throughout the remainder of this article, let $\mathbb{T}$ be a closed nonempty
subset of $R$, and let $\mathbb{T}$
have the subspace topology on $R$. In some of the current literature,
$\mathbb{T}$ is called a time scale. For convenience,
we make the blanket assumption that $0,T$ are points in $\mathbb{T}$.

Sang and Xi \cite{YSang} considered the following $p$-Laplacian dynamic
equation on time scales
\begin{gather*}
(\phi_p(u^\Delta(t)))^\nabla+a(t)f(t,u(t))=0,\quad t\in[0,T]_{\mathbb{T}}, \\
 \phi_p(u^\Delta(0))=\sum_{i=1}^{m-2} a_i \phi_p(u^\Delta(\xi_i)), \quad
 u(T)= \sum_{i=1}^{m-2} b_i u(\xi_i),
\end{gather*}
where $\phi_p(s)=| s | ^{p-2} s$, $ p>1$.

He \cite{Zhe} studied the existence of at least two positive solutions
by way of a new double fixed-point theorem for the equation
\begin{gather*}
 [\varphi_p(u^\Delta(t))]^\nabla   +a(t)f(u(t))=0,\quad  t\in[0,T]_{\mathbb{T}}, \\
 u(0)-B_0(u^{\Delta}(\eta))=0  \quad  u^\Delta(T)=0,\text{or }\\
u^{\Delta}(0)=0,  \quad  u(T)+B_1(u^{\Delta}(\eta))=0,
\end{gather*}
where $\varphi_p(s)=| s | ^{p-2} s$, $p>1$, $\eta \in(0,\rho(t))_{\mathbb{T}}$.

Anderson et al \cite{Avery} showed the existence of at least one
solution for the corresponding
boundary-value problem
\begin{gather*}
 [g(u^\Delta(t))]^\nabla   +c(t)f(u(t))=0,\quad  t\in (a,b),  \\
 u(a)-B_0(u^{\Delta}(\upsilon))=0,  \quad  u^\Delta(b)=0,
\end{gather*}
where $g(z)=| z | ^{p-2} z$, $p>1$, and  $\upsilon \in(a, b)\subset \mathbb{T}$.

 In recent years,  much attention has been paid  to the existence of positive
solutions of boundary value problems (BVPs)
on time scales for  $p(t)\equiv 1$ and $\varphi(u)=| u | ^{p-2} u$, $p>1$;
see \cite{Agarwal,Avery,Zhe, Erk,Wtli,HRSun,WLi} and the references therein.
The key condition used in the above papers is
 the oddness of a $p$-Laplacian operator.  Nevertheless, we
define a new operator which improves and generalizes a $p$-Laplacian operator
for some $p>1$, and $\varphi$ is not necessary odd. In addition, there are
not many results concerning increasing homeomorphism and positive
homomorphism on time scales; see \cite{SLiang,Liang}.

Motivated by works mentioned above, in this paper,
we study the existence of at least three positive solutions
to the following $p$-Laplacian multipoint BVP on time scales
\begin{gather} \label{e1.1}
[\varphi( p(t)u^\Delta(t))]^\nabla+a(t)f(u(t))=0,\quad
 t\in[0,T]_{{\mathbb{T}}^K\cap{\mathbb{T}}_K }, \\
u(0)=\sum_{i=1}^{m-2}a_i u(\xi_i), \quad
 u^\Delta(T)=0, \quad \label{e1.2}
\end{gather}
where $\varphi:\mathbb{R}\to\mathbb{R}$ is an increasing homeomorphism
and positive homomorphism and $\varphi(0)=0$,
$p \in C([0,T]_{\mathbb{T}},(0,+\infty))$ and $\xi_i \in [0,T]_{\mathbb{T}}  $
with $0<\xi_1<\xi_2<\dots <\xi_{m-2}<T$,
$0 < \sum_{i=1}^{m-2}a_i <1$ $a:\mathbb{T} \to  [0,+\infty)$ is ld-continuous
and not identically zero on any closed subinterval of $[0,T]_{\mathbb{T}}$.
The usual notation and terminology for time scales as can be found
in \cite{Bohner,Peter}, will be used here.

A projection $\varphi: R\to R$ is called an increasing
homeomorphism and homomorphism if the following
conditions are satisfied:
\begin{itemize}
\item[(i)] if $x \leq y$, then $ \varphi(x)\leq \phi(y),  \forall x,y \in R$;
\item[(ii)]  $\varphi$ is a continuous bijection and its inverse mapping 
is also continuous;
\item[(iii)] $\varphi(xy)=\varphi(x)\varphi(y),  \forall x,y \in R$.
\end{itemize}

 If  the above conditions hold, then it implies that $\varphi$ 
is homogeneous and generates a $p$-Laplacian operator. 
It is well known that  the $p$-Laplacian operator is odd. 
Nevertheless, the operator which we defined above is not necessarily odd.

Throughout this article we assume that the following conditions are satisfied:
\begin{itemize}
\item[(H1)]   $f:\mathbb{R}^{+}\to  \mathbb{R}^{+}  $   is continuous and
            $0 <\sum_{i=1}^{m-2}a_i<1 $;

\item[(H2)]  $p \in C([0,T]_{\mathbb{T}},[0,\infty)) $and nondecreasing on
$  [0,T]_{\mathbb{T}}$;

\item[(H3)]  $a:\mathbb{T} \to[0,\infty) $  is ld-continuous and not identical
zero on any closed subinterval of    $[0,T]_{\mathbb{T}}$.
\end{itemize}

The rest of article is arranged as follows.
In Section 2, we state some definitions, notation,
lemmas and prove several preliminary results.
 The main theorem on the existence of at least three positive solutions
and its proof are presented in Section 3.
In last section 4, we give an example to demonstrate our results.

\section{Preliminaries}

In this section, we provide some background materials from theory of cones
in Banach spaces. The following
definitions can be found in the book by Deimling \cite{Deim} and
in the book by Guo and Lakshmikantham \cite{Guo}.

\begin{definition} \label{def2.1} \rm
 Let $E$  be a real Banach space. A nonempty, closed, convex set $P \subset E$ is
a cone if it satisfies the following two conditions:
\begin{itemize}
\item[(i)]  $x \in P$, $ \lambda \geq 0$ imply   $\lambda x\in P$;
\item[(ii)]  $x \in P$, $ -x\in P$ imply   $x=0$.
\end{itemize}
Every cone $P \subset E$ induces an ordering in $E$ given by
$x\leq y$ if and only if $y-x \in P$.
\end{definition}

\begin{definition} \label{def2.2} \rm
We say the map $\alpha$ is a nonnegative continuous convex functional
 on a cone $P$ of a real
Banach space $E$ if $\alpha: P \to [0,\infty)$ is continuous and
$$
\alpha (tx+(1-t)y) \leq t\alpha(x)+(1-t)\alpha(y)
$$
for all $x,y \in P$ and $t\in [0,1]$.
\end{definition}

\begin{definition} \label{def2.3} \rm
 Given a nonnegative continuous functional $\gamma$ on a cone $P$ of $E$,
for each $d>0$ we define the set
$$
P(\gamma, d)=\{x \in P:\gamma(x)<d\}.
$$
\end{definition}

Let the Banach space $E=C_{ld}([0,T]_{\mathbb{T}},\mathrm{R})$ with norm
$\| u\|= \sup_{t\in [0,T]_{\mathbb{T}}} | u(t)|$ and define the cone
$P \subset E$ by
$$
P= \{u\in E  |  u(t)\text{ is a concave and nonnegative nondecreasing function on }
 [0,T]_{{\mathbb{T}}^K\cap{\mathbb{T}}_K }   \}.
$$

\begin{lemma} \label{lem2.1}
  If  $\sum_{i=1}^{m-2}\alpha_i \ne 1$
then for  $h\in C_{ld}[0,T]_{\mathbb{T}}$,
\begin{gather}
[\varphi( p(t)u^\Delta(t))]^\nabla+h(t)=0,\quad
t\in[0,T]_{{\mathbb{T}}^K\cap{\mathbb{T}}_K },
  \label{e2.1}\\
u(0)=\sum_{i=1}^{m-2}a_i u(\xi_i), \quad  u^\Delta(T)=0 \quad
\label{e2.2}
\end{gather}
has the unique solution
\begin{equation}
\begin{aligned}
u(t)&=\int_0^t \frac{1}{p(s)}\varphi^{-1}
\Big( \int_s^T h(\tau) \nabla \tau \Big)\Delta s\\
&\quad +\frac {\sum_{i=1}^{m-2} a_i }{1-\sum_{i=1}^{m-2} a_i}
 \int_0^{\xi_i} \frac{1}{p(s)}\varphi^{-1}
\Big( \int_s^T h(\tau) \nabla \tau \Big)\Delta s.
\end{aligned}  \label{e2.3}
\end{equation}
\end{lemma}

\begin{proof}
 Let  $u$  be as in \eqref{e2.3},  taking the delta derivative of \eqref{e2.3},
we have
$$
u^\Delta(t)= \frac{1}{p(t)}\varphi^{-1}
\Big( \int_t^T h(\tau) \nabla \tau \Big),
$$
moreover, we get
$$
\varphi (p(t) u^\Delta(t))= \int_t^T h(\tau) \nabla \tau,
$$
taking the nabla derivative of this expression yields
$[\varphi (p(t) u^\Delta(t))]^\nabla = -h(t)$. Routine calculations verify that
$u$ satisfies the boundary value conditions in \eqref{e2.2}, so that $u$
given in \eqref{e2.3} is a solution of \eqref{e2.1} and \eqref{e2.2}.
It is easy to see that the BVP
$$
[\varphi (p(t) u^\Delta(t))]^\nabla =0,\quad
u(0)=\sum_{i=1}^{m-2} a_i u(\xi_i), \quad
u^\Delta(T)=0
$$
has only the trivial solution. Thus $u$
in  \eqref{e2.3} is the unique solution of \eqref{e2.1} and \eqref{e2.2}.
\end{proof}

\begin{lemma} \label{lem2.2}
If  $u\in P$, then
$$
u(t)\geq\frac{t}{T} \|u\|,\quad  t\in [0,T]_{\mathbb{T}},
$$
where
$$
\|u\|=\sup_{ t\in [0,T]_{\mathbb{T}}}| u(t)|.
  $$
\end{lemma}

\begin{proof}
Since $u^{\Delta \nabla}(t)\leq 0$, it follows that $u^{\Delta}(t)$
 is nonincreasing.
Thus, for  $ 0< t < T$,
\begin{gather*}
u(t)-u(0)=\int_0^t u^{\Delta}(s) \Delta s\geq t u^{\Delta}(t),\\
u(T)-u(t)=\int_t^T u^{\Delta}(s) \Delta s\leq (T-t) u^{\Delta}(t)
\end{gather*}
from which we have
$$
u(t)\geq \frac{t u(T)+(T-t)u(0)}{T} \geq \frac{t}{T} u(T)= \frac{t}{T}\|u\|.
$$
The proof is complete.
\end{proof}


Let us define the mapping $A$ from $P$ to $E$ by the formula
\begin{equation}
\begin{aligned}
(Au)(t)
&=\int_0^t \frac{1}{p(s)}\varphi^{-1}
\Big( \int_s^T a(\tau) f(u(\tau) \nabla \tau \Big)\Delta s\\
&\quad +\frac {\sum_{i=1}^{m-2} a_i }{1-\sum_{i=1}^{m-2} a_i}
 \int_0^{\xi_i} \frac{1}{p(s)}\varphi^{-1}
\Big( \int_s^T a(\tau) f(u(\tau) \nabla \tau \Big)\Delta s,  \quad u\in P.
\end{aligned}\label{e2.4}
\end{equation}

\begin{lemma} \label{lem2.3}
The mapping $A:P\to P$  is completely continuous.
\end{lemma}

\begin{proof}  For each $u \in P$, we have $(Au)(t)\geq 0$, for all
$t\in [0,T]_{\mathbb{T}}$.
Taking the delta derivative  of \eqref{e2.4}, we have
$$
(Au)^\Delta(t)=\frac{1}{p(t)}\varphi^{-1}
\Big( \int_t^T a(\tau) f(u(\tau) )\nabla \tau \Big).
$$
Clearly, $(Au)^\Delta(t)$ is a continuous function and  $(Au)^\Delta(t)\geq 0$,
that is  $(Au)(t)$   is decreasing on   $[0,T]_{\mathbb{T}}$.
\begin{itemize}
\item[(i)]
 If $t\in[0,T]_{{\mathbb{T}}^K\cap{\mathbb{T}}_K }$ is left scattered, we have
$$
(Au)^{\Delta \nabla}(t)=\frac{(Au)^{\Delta}(\rho(t))-(Au)^{\Delta}(t)}{\rho(t)-t}
 \leq 0.
$$
\item[(ii)]   If $t\in[0,T]_{{\mathbb{T}}^K\cap{\mathbb{T}}_K }$ is a left dense,
we have
$$
(Au)^{\Delta \nabla}(t) =\lim_{s\to t}\frac{(Au)^{\Delta}(t)
-(Au)^{\Delta}(s)}{t-s}\leq 0.
$$
\end{itemize}
By (i) and (ii), we have $(Au)^{\Delta \nabla} (t) \leq 0$,
$t\in[0,T]_{{\mathbb{T}}^K\cap{\mathbb{T}}_K }$; i.e., $(Au)$
is concave on  $[0,T]_{\mathbb{T}}$. This implies that $Au\in P$ and
$A: P \to P$. With standard argument
one may show that $A: P \to P$ is completely continuous.
\end{proof}

The following fixed point theorem is fundamental  for the proofs
of our main results.


\begin{theorem}[\cite{Ren}] \label{thm2.1}
 Let $P$ be a cone in a  Banach space $E$. Let  $\alpha,\beta$ and $\gamma$
be three increasing, nonnegative  and continuous  functionals on $P$,
satisfying for some  $c>0$ and $M>0$ such that
$$
\gamma(x)\leq \beta(x)\leq \alpha(x),   \|u \| \leq M \gamma(x)
$$
for all $x\in \overline{P(\gamma, c)}$. Suppose there exists a
completely continuous operator $T:\overline{P(\gamma, c)}\to P$
and $0<a<b<c$ such that
 \begin{itemize}
\item[(S1)]  $\gamma(Tx)<c$, for all    $ x\in \partial P(\gamma,c)$;
\item[(S2)]  $\beta(Tx)>b$,  for all   $x\in \partial P(\beta,b)$;
\item[(S3)]  $P(\alpha,a)\ne \emptyset$,    and   $\alpha(Tx)<a$,
for all   $x\in \partial P(\alpha,a)$.
\end{itemize}
Then  $T$  has at least three  fixed points
 $x_1, x_2, x_3  \in \overline{P(\gamma, c)}$ such that
 $$
0\leq \alpha(x_1) < a <\alpha(x_2),   \beta(x_2)< b <\beta(x_3),
   \gamma(x_3) < c.
$$
\end{theorem}

\section{Main results}

We define the increasing, nonnegative, continuous functionals:
\begin{gather*}
\gamma (u)=\max_{t\in[0,\xi_{1}]_{\mathbb{T}} } u(t)=u(\xi_1),\\
\beta (u)=\min_{t\in[\xi_{1},\xi_{m-2}]_{\mathbb{T}} } u(t)=u(\xi_1),\\
\alpha (u)=\max_{t\in[0,\xi_{m-2}]_{\mathbb{T}} } u(t)=u(\xi_{m-2}).
\end{gather*}
Clearly for every  $u\in P$
$$
\gamma (u)\leq \beta (u) \leq  \alpha (u).
$$
Moreover, for each  $u\in P$, Lemma \ref{lem2.2} implies
$\gamma (u)=u(\xi_1)\geq\frac{\xi_1}{T}\|u\|$.
That is, $\|u\|\leq \frac{T}{\xi_1}\gamma(u)$ for all $u\in P$.

For simplicity, we use the following symbols:
\begin{gather*}
 \lambda_1  =  \frac{1}{p(0)}
 \Big(\xi_1+\frac { \sum_{i=1}^{m-2} a_i T} {1-\sum_{i=1}^{m-2} a_i}\Big)
 \varphi ^{-1}\Big( \int_{0}^T a(\tau) \nabla \tau \Big), \\
\lambda_2  =  \frac{1}{p(T)}
 \Big(\xi_1+\frac { \sum_{i=1}^{m-2} a_i T} {1-\sum_{i=1}^{m-2} a_i}\Big)
 \varphi ^{-1}\Big( \int_{\xi_{m-2}}^T a(\tau) \nabla \tau \Big), \\
\lambda_3  =  \frac{1}{p(0)}
\Big(\xi_{m-2}+\frac { \sum_{i=1}^{m-2} a_i T} {1-\sum_{i=1}^{m-2} a_i}\Big)
 \varphi ^{-1}\Big( \int_{0}^T a(\tau) \nabla \tau \Big).
\end{gather*}

\begin{theorem} \label{thm3.1}
 Suppose that  conditions {\rm (H1), (H2), (H3)} are satisfied. Let
$0< a < \frac{\xi_1}{T}b < b <\frac{\lambda_2}{\lambda_1}c$,
 and suppose that $f$ satisfies the following conditions:
\begin{itemize}
\item[(i)] $ f(u) < \varphi (c/\lambda_1)$  for all  $u\in [0,Tc/\xi_1]$;
\item[(ii)] $ f(u) > \varphi (b/\lambda_2)$  for all  $u\in [b,Tb/\xi_1]$;
\item[(iii)] $ f(u) < \varphi (a/\lambda_3)$ for all $u\in [0,Ta/\xi_1]$.
\end{itemize}
Then there exist  at least three positive solutions  $u_1,u_2,  u_3$
of \eqref{e1.1} and\eqref{e1.2} such that
$$
0 \leq \alpha(u_1) < a <\alpha(u_2), \quad
  \beta(u_2)< b <\beta(u_3), \quad
   \gamma(u_3) < c.
$$
\end{theorem}

\begin{proof}
 Define the completely continuous operator $A$ by \eqref{e2.4}.
Let $u \in \partial P(\gamma,c)$,
then $(Au)(t)\geq 0$ for  $t\in[0,T]_{\mathbb{T}}$. By Lemma \ref{lem2.3} we know that
$A: \overline{ P(\gamma,c)}\to P$.

 Now, we  show that all the conditions of Theorem \ref{thm2.1} are satisfied. To verify
(S1) of Theorem \ref{thm2.1} holds, we choose $u \in \partial P(\gamma,c)$. Then
$\gamma (u)=\max_{t\in[0,\xi_{1}]_{\mathrm{T}} } u(t)=u(\xi_1)=c$.
If we recall that $\|u\|\leq \frac{T}{\xi_1}\gamma(u)=\frac{T}{\xi_1}c$.
Therefore
$$
0 \leq u(t)\leq\frac{T}{\xi_1}c,   \quad\text{for all }
t\in[0,T]_{\mathbb{T}}.
$$
As a consequence of (i),
$$
f(u(s))<\varphi (c/\lambda_1),  \quad \text{for }  s\in[0,T]_{\mathbb{T}}.
$$
Since $Au \in P$, we have
\begin{align*}
 \gamma(Au)
&=  (Au)(\xi_1)\\
&=  \int_0^{\xi_1} \frac{1}{p(s)}\varphi^{-1}
 \Big( \int_s^T a(\tau) f(u(\tau)) \nabla \tau \Big)\Delta s\\
&\quad  +\frac {\sum_{i=1}^{m-2} a_i }{1-\sum_{i=1}^{m-2} a_i}
\int_0^{\xi_i} \frac{1}{p(s)}\varphi^{-1}
 \Big( \int_s^T a(\tau) f(u(\tau) )\nabla \tau \Big)\Delta s\\
& \leq  \frac{1}{p(0)}\int_0^{\xi_1} \varphi^{-1}
 \Big( \int_0^T a(\tau) f(u(\tau)) \nabla \tau \Big)\Delta s\\
&\quad  +  \frac {\frac{1}{p(0)}    \sum_{i=1}^{m-2} a_i }{1-\sum_{i=1}^{m-2} a_i}
\int_0^{\xi_i} \varphi^{-1}
 \Big( \int_0^T a(\tau) f(u(\tau)) \nabla \tau \Big)\Delta s\\
& <  \frac{1}{p(0)}
 \Big(\xi_1+\frac {\sum_{i=1}^{m-2} a_i T}{1-\sum_{i=1}^{m-2} a_i}\Big)
\varphi^{-1} \Big( \int_0^T a(\tau) \nabla \tau \Big)\frac{c}{\lambda_1}=c.
\end{align*}
Thus,  (S1) of Theorem \ref{thm2.1} is satisfied.

Secondly, we prove that (S2) of Theorem \ref{thm2.1} is fulfilled. For this, we choose
$u \in \partial P(\beta,b)$. Then
$\beta (u)=\min_{t\in[\xi_1,\xi_{m-2}]_{\mathbb{T}} } u(t)=u(\xi_1)=b$.
This means $u(t)\geq b,  t\in[\xi_i,T]_{\mathbb{T}}$ and since $u \in P$, we have
$ b \leq u(t) \leq \|u\|=u(T)$   for   $t\in[\xi_1,T]_{\mathbb{T}}$.
Note that $\|u\| \leq \frac{T}{\xi_1}\gamma(u)=
\frac{T}{\xi_1}\beta(u)=\frac{T}{\xi_1}b$ for all   $u \in P$.
Therefore,
$$
b \leq u(t)\leq\frac{T}{\xi_1}b,  \quad \text{for all }
  t\in[\xi_1,T]_{\mathbb{T}}.
$$
From  (ii), we have
$$
f(u(s))>\varphi \left(\frac{b}{\lambda_2}\right ), \quad  \text{for }
 s\in[\xi_1,T]_{\mathbb{T}},
$$
and so
\begin{align*}
 \beta(Au) &=  (Au)(\xi_1)\\
&=  \int_0^{\xi_1} \frac{1}{p(s)}\varphi^{-1}
\Big( \int_s^T a(\tau) f(u(\tau)) \nabla \tau \Big)\Delta s\\
&\quad  +\frac {\sum_{i=1}^{m-2} a_i }{1-\sum_{i=1}^{m-2} a_i}
\int_0^{\xi_i} \frac{1}{p(s)}\varphi^{-1}
\Big( \int_s^T a(\tau) f(u(\tau) )\nabla \tau \Big)\Delta s\\
& >  \frac{1}{p(T)}\int_0^{\xi_1} \varphi^{-1}
 \Big( \int_{\xi_{m-2}}^T a(\tau) f(u(\tau)) \nabla \tau \Big)\Delta s\\
&\quad  +  \frac {\frac{1}{p(T)}
\sum_{i=1}^{m-2} a_i }{1-\sum_{i=1}^{m-2} a_i}
\int_0^{\xi_i} \varphi^{-1}
 \Big( \int_ {\xi_{m-2}}^T a(\tau) f(u(\tau)) \nabla \tau \Big)\Delta s\\
& >  \frac{1}{p(T)} 
\Big(\xi_1+\frac {\sum_{i=1}^{m-2} a_i T}{1-\sum_{i=1}^{m-2} a_i}\Big)
\varphi^{-1} \Big( \int_ {\xi_{m-2}}^T a(\tau) \nabla \tau \Big)
 \frac{b}{\lambda_2}=b.
\end{align*}
Thus, (S2) of Theorem \ref{thm2.1} is satisfied.

Finally we prove that (S3) of Theorem \ref{thm2.1} is also satisfied. 
We note that $u(t)=a/2$,  $t\in[0,T]_{\mathbb{T}}$ is a member of $P(\alpha,a)$ and
$\alpha (u)=\frac{a}{2}<a$. Therefore  $ P(\alpha,a)\ne \emptyset$. Now let
$ u \in P(\alpha,a)$. Then
$\alpha (u)=\max_{t\in[0,\xi_{m-2}]_{\mathbb{T}} } u(t)=u(\xi_{m-2})=a$.
This implies that $0 \leq u(t)\leq a$ for  $t\in[0,\xi_{m-2}]_{\mathbb{T}}$.
Recalling that $\|u\| \leq \frac{T}{\xi_{m-2}}\gamma(u)\leq
\frac{T}{\xi_{m-2}}\alpha(u)\leq\frac{T}{\xi_1}a$ for all $ u \in P$, we have
$$
0 \leq u(t)\leq\frac{T}{\xi_1}a, \quad   \text{for all } t\in[0,T]_{\mathbb{T}}.
$$
From  assumption (iii), we obtain
$$
f(u(s))<\varphi \Big(\frac{a}{\lambda_3}\Big), \quad  \text{for }
 s\in[0,T]_{\mathbb{T}},
$$
and so
\begin{align*}
 \alpha(Au) &=  (Au)(\xi_{m-2})\\
&=  \int_0^{\xi_{m-2}} \frac{1}{p(s)}\varphi^{-1}
\Big( \int_s^T a(\tau) f(u(\tau)) \nabla \tau \Big)\Delta s\\
 &\quad  +\frac {\sum_{i=1}^{m-2} a_i }{1-\sum_{i=1}^{m-2} a_i}
\int_0^{\xi_i} \frac{1}{p(s)}\varphi^{-1}
 \Big( \int_s^T a(\tau) f(u(\tau) )\nabla \tau \Big)\Delta s\\
& \leq  \frac{1}{p(0)}\int_0^{\xi_{m-2}} \varphi^{-1}
 \Big( \int_0^T a(\tau) f(u(\tau)) \nabla \tau \Big)\Delta s\\
&\quad  +  \frac {\frac{1}{p(0)}
 \sum_{i=1}^{m-2} a_i }{1-\sum_{i=1}^{m-2} a_i}
\int_0^{T} \varphi^{-1} \Big( \int_0^T a(\tau) f(u(\tau)) \nabla \tau \Big)
 \Delta s\\
& <  \frac{1}{p(0)} \Big(\xi_{m-2}
 +\frac {\sum_{i=1}^{m-2} a_i T}{1-\sum_{i=1}^{m-2} a_i}\Big)
\varphi^{-1} \Big( \int_0^T a(\tau) \nabla \tau \Big)\frac{a}{\lambda_3}=a.
\end{align*}
Then condition (S3) of Theorem \ref{thm2.1} is satisfied.
So Theorem \ref{thm2.1} implies that $A$ has a least
three fixed points which are positive solutions $u_1,u_2,u_3$
belonging to $\overline{P(\gamma,c)}$ of \eqref{e1.1} and \eqref{e1.2} such that
$$
0 \leq \alpha(u_1) < a <\alpha(u_2), \quad
 \beta(u_2)< b <\beta(u_3), \quad    \gamma(u_3) < c.
$$
The proof  is complete.
\end{proof}

\section{Examples}

In  this section, we  give  an example to illustrate our results.
Let
$\mathbb{T}=\{(\frac{2}{3})^{\mathbb{N}_0}  \}
\cup \{1-(\frac{2}{3})^{\mathbb{N}_0}\}$, where
  $\mathbb{N}_0$  denotes the set of all nonnegative integers.
If we choose  $a_1=a_2=1/4$,
$\xi_1=1/3$,  $\xi_2=2/3$,
 $T=1$,  $a(t)\equiv 1$, and  $p(t)\equiv 1  $.
Consider the following BVP on the time scale $\mathbb{T}$:
\begin{gather}
[\varphi( u^\Delta(t))]^\nabla+ f(u(t))=0,\quad
t\in[0,1]_{\mathbb{T}}, \label{e4.1}\\
u(0)=\frac{1}{4}u\Big (\frac{1}{3} \Big )
+\frac{1}{4}u \Big (\frac{2}{3} \Big),\quad
u^{\Delta}(1)=0, \label{e4.2}
\end{gather}
where
\[
 \varphi(u)=  \begin{cases}
\frac{u^5}{1+u^2}, &    u \leq 0,\\
u^2,    &  u>0,
 \end{cases}
\]
and
\[
 f(u)=  \begin{cases}
0.1, &    0\leq u \leq 3,\\
 0.1+\frac{90(u-3)}{4\sqrt{3}-3},    & 3\leq u \leq4\sqrt{3},\\
90.1, & 4\sqrt{3}\leq u.
  \end{cases}
\]
We take $a=1$,  $b=4\sqrt{3}$,  $c=75$.
By simple calculations, we have
\begin{gather*}
 \lambda_1 =  \frac{1}{p(0)}
\Big(\xi_1+\frac { \sum_{i=1}^{m-2} a_i T} {1-\sum_{i=1}^{m-2} a_i}\Big)
 \varphi ^{-1}
\Big( \int_{0}^T a(\tau) \nabla \tau \Big)=\frac{4}{3},
\\
\lambda_2 =  \frac{1}{p(T)}
\Big(\xi_1+\frac { \sum_{i=1}^{m-2} a_i T} {1-\sum_{i=1}^{m-2} a_i}\Big)
 \varphi ^{-1}\Big( \int_{\xi_{m-2}}^T a(\tau) \nabla \tau \Big)
= \frac{4\sqrt{3}}{9}, \\
\lambda_3 =  \frac{1}{p(0)}
\Big(\xi_{m-2}+\frac { \sum_{i=1}^{m-2} a_i T} {1-\sum_{i=1}^{m-2} a_i}\Big)
 \varphi ^{-1}\Big( \int_{0}^T a(\tau) \nabla \tau \Big)=\frac{5}{3}.
\end{gather*}
It is easy to see that
$$
0 < a < \frac{\xi_1}{T} b < b < \frac{\lambda_2}{\lambda_1} c,
$$
and that $f$ satisfies
\begin{gather*}
 f(u)  <  \varphi \Big(\frac{c}{\lambda_1}\Big)
= \Big(\frac{75}{\frac{4}{3}}\Big)^2\thickapprox 3164.0625, \quad
   u\in [0,225];\\
 f(u)  > \varphi \Big(\frac{b}{\lambda_2}\Big)
= \Big(\frac{4\sqrt{3}} {\frac{4\sqrt{3}}{9}}\Big)^2=81, \quad
 u\in [4\sqrt{3},12\sqrt{3}];\\
 f(u) < \varphi \Big(\frac{a}{\lambda_3}\Big)
=\Big(\frac{1}{\frac{5}{3}}\Big)^2=\frac{9}{25}, \quad
   u\in [0,3].
\end{gather*}
By Theorem \ref{thm3.1}, we see that  BVP \eqref{e4.1} and \eqref{e4.2} 
has at least three positive solutions
$u_1, u_2, u_3$ such that
\begin{gather*}
\max_{t\in [0,\frac{2}{3}]_{\mathbb{T}} } \{ u_i(t) \} \leq 75,   \quad
   \text{for }    i=1,2,3; \\
0 \leq \max_{t\in [0,2/3]_{\mathbb{T}} } \{ u_1(t) \} < 1 <
\max_{t\in [0,\frac{2}{3}]_{\mathbb{T}} } \{ u_2(t) \};
\\
 \min_{t\in [ \frac{1}{3},\frac{2}{3}]_{\mathbb{T}} } \{ u_2(t) \}
< 4\sqrt{3} <  \min_{t\in [ \frac{1}{3},\frac{2}{3}]_{\mathbb{T}} } \{ u_3(t) \},
\quad
 \max_{t\in [0,\frac{1}{3}]_{\mathbb{T}} } \{ u_3(t)\} \leq 75.
\end{gather*}

\subsection*{Acknowledgments}

The author would like to thank the anoymous referees and the editor for
their helpful comments and suggestions.
The project is supported by Abdullah Gul University Foundation of Turkey.

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\end{document}