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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 52, 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 2010 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2010/52\hfil Multiplicity of positive solutions]
{Multiplicity of positive solutions for four-point
boundary value problems of impulsive differential equations with
$p$-Laplacian}

\author[L. Shen, X. Liu, Z. Lu \hfil EJDE-2010/52\hfilneg]
{Li Shen, Xiping Liu, Zhenhua Lu} 

\address{Li Shen \newline
College of Science, University of Shanghai for Science and Technology\\
Shanghai 200093, China} 
\email{eric0shen@gmail.com}

\address{Xiping Liu \newline
College of Science, University of Shanghai for Science and Technology\\
Shanghai 200093, China}
\email{xipingliu@163.com}

\address{Zhenhua Lu\newline
College of Science, University of Shanghai for Science and Technology\\
Shanghai 200093, China}
\email{ehuanglu@163.com}



\thanks{Submitted November 16, 2009. Published April 14, 2010.}
\thanks{Supported by grant 10ZZ93 from Innovation Program
of Shanghai Municipal \hfill\break\indent Education Commission}
\subjclass[2000]{34A37, 34B37}
\keywords{$p$-Laplacian;  impulsive; positive solutions;  boundary
value problem}

\begin{abstract}
 Using a fixed-point theorem in cones, we obtain
 sufficient conditions for the multiplicity of positive
 solutions for four-point boundary value problems of third-order
 impulsive differential equations with $p$-Laplacian.
\end{abstract}

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

\section{Introduction}

Recently, there has been much attention focused on the theory of
impulsive differential equation as it is widely used in various
areas such as mechanics, electromagnetism, chemistry. A lot of
theories have been established to solve these problems, see
\cite{Lak}, \cite{Fu} and the references therein. Guo \cite{G1}
obtained the existence of solutions, via cone theory, for
second-order impulsive differential equation
\begin{gather*}
 x''=f(t,x,Tx), \quad  t\geq  0, \; t\neq t_{k}\; k=1,2,3,\dots,  \\
 \Delta x|_{t=t_{k}} =I_{k}(x(t_{k})), \quad k =1,2,3,\dots,\\
 \Delta x'|_{t=t_{k}} =\overline{I}_{k}(x(t_{k})), \quad  k = 1,2,3,\dots,\\
 x(0)=x_{0}, \quad x'(0)=x_{0}^{*}.
\end{gather*}

In  \cite{AR}, using Leggett-Williams fixed point theorem,
authors studied the multiplicity result for second order impulsive
differential equations
\begin{gather*}
  y''+\phi(t)f(y(t))=0, \quad t\in (0,1)\setminus \{t_1,t_2,\dots,t_m\},  \\
 \Delta y(t_{k}) =I_{k}(y(t_{k}^{-})), \;k =1,2,3,\dots,m,\\
 \Delta y'(t_{k}) =J_{k}(y(t_{k}^{-})), \;k =1,2,3,\dots,m,\\
  y(0)=y(1)=0.
\end{gather*}
Kaufmann \cite{Kaufmanna} studied a second-order nonlinear
differential equation on an unbounded domain with solutions subject
to impulsive conditions and the Sturm-Liouville type boundary
conditions.
In  \cite{J1}-\cite{JL}, the authors studied positive
solutions of multiple points boundary value problems for second
order impulsive differential equations.

All the works above concern  boundary value problems
with second-order impulsive equations, and there are just a
few works that consider multiplicity of positive solutions for
third-order impulsive equations with $p$-Laplacian.

Motivated by all the works above, we concentrate on getting multiple
positive solutions for four-point boundary value problems of
third-order impulsive differential equations with $p$-Laplacian
\begin{equation}\label{eq1.1}
\begin{gathered}
 (\phi_p(u''(t)))'=f(t,u(t),u'(t)), \quad
  t \in (0,1) \backslash \{t_1,t_2,\dots,t_m\},  \\
 \Delta u''(t)|_{t=t_{k}}=0, \quad k =1,2,\dots,m,\\
 \Delta u'(t)|_{t=t_{k}} =I_{k}(u(t_{k})), \quad k =1,2,\dots,m,\\
 \Delta u(t)|_{t=t_{k}} =J_{k}(u(t_{k})), \quad k = 1,2,\dots,m,\\
 u''(0)=0,\quad u'(0)=\alpha u'(\xi)+\beta u'(\eta),\quad
u(1)=\delta u(0),
\end{gathered}
\end{equation}
 where $\phi_p$ is $p$-Laplacian operator
$$
  \phi_p(s)=|s|^{p-2}s, p>1,\quad
 (\phi_p)^{-1}=\phi_q,\quad \frac{1}{p}+\frac{1}{q}=1,
$$
 $t_{k}$, $k = 0,1,2,\dots,m,m+1$, are constants which satisfy
$$
  0=t_0<t_1<t_2<\dots<t_k<\dots<t_m<t_{m+1}=1,
$$
$\Delta u|_{t=t_{k}}=u(t_k ^{+})-u(t_k ^{-})$, in which $u(t_k
 ^{+})$ ($u(t_k
 ^{-})$ respectively) denote the right limit (left limit respectively)
of $u(t)$ at  $t=t_k$,
and
$\alpha,\beta>0$,
$\alpha+\beta<1$;
$0<\xi$, $\eta<1$; $\xi,\eta\neq t_{k}$ ($k = 1,2,\dots,m$);
$\delta>1$;
$f\in C([0,1]\times[0,+\infty)\times\mathbb{R},[0,+\infty))$,
$I_k, J_k\in C([0,+\infty),[0,+\infty))$.


\section{Preliminaries}

Let $J=[0,1]\backslash\{t_1,t_2,\dots,t_m\}$,
$ PC[0,1]=\{ u: [0,1]\to R,\; u$ is continuous at $t\neq t_k$,
$u(t_k^+),\; u(t_k^-)$ exist, and $u(t_k^-)=u(t_k),
k=1,2,\dots,m\}$, $PC^1[0,1]=\{ u\in PC[0,1]\;|\ u' $ is
continuous at $t\neq t_k$,
  $u'(t_k^+),\; u'(t_k^-)$ exist, $k=1,2,\dots,m\}$,
with the norm
$$
  \|u\|_{PC}=\sup_{t\in J}|u(t)|,\quad
  \|u\|_{PC^1}=\mathrel{\max _{{t\in J}}}
  \{\|u\|_{PC}, \|u'\|_{PC}\}.
$$
Obviously $PC[0,1]$ and $PC^1[0,1]$ are Banach spaces.

\begin{lemma} \label{lem2.1}
$u\in PC^1[0,1]\bigcap C^3[J] $ is a solution of \eqref{eq1.1}
if and only if
\begin{equation} \label{eq2.1}
  \begin{aligned}
      u(t)&=
     u(0)+u'(0)t+\int_0^t(t-s)\phi_q\Big(\int_0^s
      f(r,u(r),u'(r))dr\Big)ds \\
    &\quad +\sum_{t_k<t}(t-t_k)I_k(u(t_k))+\sum_{t_k<t}J_k(u(t_k)),
  \end{aligned}
\end{equation}
where
\begin{equation} \label{eq2.2}
  \begin{aligned}
      u(0)=
& \frac{\alpha\int_0^{\xi}\phi_q(\int_0^sf(r,u(r),u'(r))dr)ds
+\beta\int_0^{\eta}\phi_q(\int_0^s
      f(r,u(r),u'(r))dr)ds}{(\delta-1)(1-\alpha-\beta)} \\
    & +\frac{\alpha\sum_{t_k<\xi}I_k(u(t_k))+\beta\sum_{t_k<\eta}I_k(u(t_k))}
     {(\delta-1)(1-\alpha-\beta)} \\
&\quad +\frac{\int_0^1\int_0^s\phi_q(\int_0^r
     f(w,u(w),u'(w))dw)\,dr\,ds}{\delta-1} \\
&\quad +\frac{1}{\delta-1}\sum_{k=1}^{m}\big((1-t_k)I_k(u(t_k)+J_k(u(t_k))\big),
  \end{aligned}
\end{equation}
\begin{equation} \label{eq2.3}
  \begin{aligned}
     u'(0)&=
    \frac{\alpha\int_0^{\xi}\phi_q(\int_0^s
     f(r,u(r),u'(r))dr)ds+\beta\int_0^{\eta}\phi_q(\int_0^s
     f(r,u(r),u'(r))dr)ds}{1-\alpha-\beta} \\
   &\quad +\frac{\alpha\sum_{t_k<\xi}I_k(u(t_k))+\beta\sum_{t_k<\eta}I_k(u(t_k))}
     {1-\alpha-\beta}.
  \end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
 Suppose $u\in PC^1[0,1]\bigcap C^3[J] $ is a
solution of  \eqref{eq1.1}, for all $k=1,2,\dots,m$, from
Lagrange's mean value theorem we have
$$
  u(t_k)-u(t_k-h)=u'(\xi_k)h,\quad
  0<h<t_k-t_{k-1},\; \xi_k\in(t_k-h,t_k),
$$
because $u'(t_k^-)$ exists, we get
$$
  u'_-(t_k)=\lim_{h\to 0^+}\frac{u(t_k)-u(t_k-h)}{h}
  =\lim_{\xi_k\to t_k^-}u'(\xi_k)=u'(t_k^-).
$$
Let $u'(t_k)=u_-'(t_k)=u'(t_k^-)$, $k=1,2,\dots,m$.
We use Lagrange's mean value theorem again and obtain
$$
  u'(t_k)-u'(t_k-h)=u''(\eta_k)h,\quad 0<h<t_k-t_{k-1},\;
 \eta_k\in(t_k-h,t_k),
$$
we can get $u''_-(t_k)$ exists from $\Delta
u''(t)|_{t=t_{k}}=u''(t_k^+)-u''(t_k^-)=0$, and
$$
  u''_-(t_k)=\lim_{h\to 0^+ }\frac{u'(t_k)-u'(t_k-h)}{h}
  =\lim_{\xi_k\to t_k^-}u''(\xi_k)=u''(t_k^-).
$$
Let $u''(t_k)=u''(t_k^-)$, $k=1,2,\dots,m$.
Integrating the differential equation  \eqref{eq1.1} we have
$$
  \phi_p(u''(t))-\phi_p(u''(0))=\int_0^tf(s,u(s),u'(s))ds,\quad
0  \leq t \leq  t_1 .
$$
By $u''(0)=0$, we have
$$
  u''(t)=\phi_q(\int_0^tf(s,u(s),u'(s))ds);
$$
that is,
$$
  u''(t)=\phi_q(\int_0^tf(s,u(s),u'(s))ds),
$$
and
$$
  u''(t_1)=\phi_q(\int_0^{t_1}f(s,u(s),u'(s))ds).
$$
Since $\Delta u''(t)|_{t=t_{1}}=u''(t_1^+)-u''(t_1^-)=0$, for $t_1<
t \leq t_2$, we obtain
$$
  u''(t)=\phi_q(\int_0^tf(s,u(s),u'(s))ds).
$$
Similarly, by $\Delta u''(t)|_{t=t_{k}}=u''(t_k^+)-u''(t_k^-)=0,\
k=1,2,\dots,m$, we can show for all $t \in [0,1]$,
\begin{equation}\label{eq2.4}
   u''(t)=\phi_q(\int_0^tf(s,u(s),u'(s))ds).
\end{equation}


For each $t \in (0,1)$, there exist $0\leq t_k<t_{k+1}\leq 1$, such
that $t_k< t\leq t_{k+1}$, by integrating both sides of
\eqref{eq2.4}, we obtain
  \begin{gather*}
    u'(t_1^-)-u'(0)=\int_0^{t_1}\phi_q(\int_0^sf(r,u(r),u'(r))dr)ds, \\
    u'(t_2^-)-u'(t_1^+)=\int_{t_1}^{t_2}\phi_q(\int_0^sf(r,u(r),u'(r))dr)ds, \\
    \dots \\
    u'(t_k^-)-u'(t_{k-1}^+)=\int_{t_{k-1}}^{t_k}\phi_q(\int_0^sf(r,u(r),u'(r))dr)ds, \\
    u'(t)-u'(t_k^+)=\int_{t_k}^{t}\phi_q(\int_0^sf(r,u(r),u'(r))dr)ds.
  \end{gather*}
Hence,
$$
  u'(t)=u'(0)+\int_0^{t}\phi_q(\int_0^sf(r,u(r),u'(r))dr)ds
+\sum_{t_k<t}I_k(u(t_k)).
$$
We have
  \begin{gather*}
     \alpha u'(\xi)=\alpha u'(0)+\alpha \int_0^{\xi}\phi_q(\int_0^sf(r,u(r),u'(r))dr)ds
     +\alpha \sum_{t_k<\xi}I_k(u(t_k)), \\
     \beta u'(\eta)=\beta u'(0)+\beta \int_0^{\eta}\phi_q(\int_0^sf(r,u(r),u'(r))dr)ds+\beta
      \sum_{t_k<\eta}I_k(u(t_k)).
  \end{gather*}
It follows that
\begin{align*}
     u'(0)&=
  \frac{\alpha\int_0^{\xi}\phi_q(\int_0^s
     f(r,u(r),u'(r))dr)ds+\beta\int_0^{\eta}\phi_q(\int_0^s
     f(r,u(r),u'(r))dr)ds}{1-\alpha-\beta} \\
&\quad +\frac{\alpha\sum_{t_k<\xi}I_k(u(t_k))+\beta\sum_{t_k<\eta}I_k(u(t_k))}
     {1-\alpha-\beta}
\end{align*}
from $u'(0)=\alpha u'(\xi)+\beta u'(\eta)$.

Similarly, we get the results as follows with the method above
\begin{align*}
      u(t)&=
     u(0)+u'(0)t+\int_0^t(t-s)\phi_q\Big(\int_0^s
      f(r,u(r),u'(r))dr\Big)ds \\
&\quad  +\sum_{t_k<t}(t-t_k)I_k(u(t_k))+\sum_{t_k<t}J_k(u(t_k)).
\end{align*}
Note that by the boundary condition $u(1)=\delta u(0)$,
\begin{align*}
    u(0)
&=  \frac{\alpha\int_0^{\xi}\phi_q(\int_0^sf(r,u(r),u'(r))dr)ds
   +\beta\int_0^{\eta}\phi_q(\int_0^s
    f(r,u(r),u'(r))dr)ds}{(\delta-1)(1-\alpha-\beta)} \\
&\quad +\frac{\alpha\sum_{t_k<\xi}I_k(u(t_k))+\beta\sum_{t_k<\eta}I_k(u(t_k))}
   {(\delta-1)(1-\alpha-\beta)}\\
&\quad +\frac{\int_0^1\int_0^s\phi_q(\int_0^r
   f(w,u(w),u'(w))dw)\,dr\,ds}{\delta-1} \\
&\quad +\frac{1}{\delta-1}\sum_{k=1}^{m}[(1-t_k)I_k(u(t_k)+J_k(u(t_k))].
\end{align*}
On the other hand, let $u\in  PC^1[0,1]\bigcap C^3[J] $ be a
solution of \eqref{eq2.1}, differentiate \eqref{eq2.1} when $t\neq
t_k$, we have
$$
  u''(t)=\phi_q(\int_0^t f(s,u(s),u'(s))ds);
$$
that is,
$$
  \phi_p(u''(t))=\int_0^t f(s,u(s),u'(s))ds\,.
$$
Differentiating again,
$$
  (\phi_p(u''(t)))'=f(t,u(t),u'(t)).
$$
By \eqref{eq2.1}, we can easily get
  \begin{gather*}
      \Delta u''(t)|_{t=t_{k}}=0, \quad k =1,2,\dots,m,\\
      \Delta u'(t)|_{t=t_{k}} =I_{k}(u(t_{k})), \quad k =1,2,\dots,m,\\
      \Delta u(t)|_{t=t_{k}} =J_{k}(u(t_{k})), \quad k = 1,2,\dots,m,\\
      u''(0)=0,\quad u'(0)=\alpha u'(\xi)+\beta u'(\eta),\quad
 u(1)=\delta      u(0).
  \end{gather*}
\end{proof}

Next, we give the Bai-Ge fixed point theorem  which is used in the
proof of our main result. Let $E$ be a Banach space, $P\subset E$ be
a cone, $\theta$, $\psi:P\to[0,+\infty)$ be nonnegative convex
functions which satisfy
\begin{equation}\label{eq2.5}
  \|x\|\leq k \max \{\theta(x),\psi(x)\},\quad \text{for all }
  x \in P,
\end{equation}
where $k$ is a positive constant.
\begin{equation}\label{eq2.6}
   \Omega=\{x\in P: \theta(x)<r,\psi(x)<L\}\neq\phi, \mbox{ where}
   \ r>0,L>0.
\end{equation}

Let $r>a>0, L>0$ be constants, $\theta$, $\psi: P \to[0,+\infty)$ be
two nonnegative continuous convex functions which satisfy
\eqref{eq2.5} and \eqref{eq2.6}, and $\gamma$ be a nonnegative
concave function on $P$. We define convex sets as follows
  \begin{gather*}
    P(\theta,r;\psi,L)=\{x\in P: \theta(x)<r, \psi(x)<L\}, \\
    \overline{P}(\theta,r;\psi,L)=\{x\in P: \theta(x)\leq r, \psi(x)\leq L\}, \\
    P(\theta,r;\psi,L;\gamma,a)=\{x\in P: \theta(x)<r, \psi(x)<L,\gamma (x)>a\}, \\
    \overline{P}(\theta,r;\psi,L;\gamma,a)=\{x\in P: \theta(x)\leq r,
     \psi(x)\leq L,\gamma(x)\geq a\}.
  \end{gather*}

\begin{lemma}[\cite{BG}] \label{lem2.2}
Let $E$ be Banach space, $P\subset E$ be a cone
 and $r_2\geq d>b>r_1>0, L_2\geq L_1>0$ be constants. Assume $\theta, \psi: P \to[0,+\infty)$
are nonnegative continuous convex functions which satisfy
\eqref{eq2.5} and \eqref{eq2.6}. $\gamma$ is a nonnegative concave
function on $P$ such that for all $x$ in
$\overline{P}(\theta,r_2;\psi,L_2)$ satisfies
$\gamma(x)\leq\theta(x)$.
$T:\overline{P}(\theta,r_2;\psi,L_2)\to\overline{P}(\theta,r_2;\psi,L_2)$
is a completely continuous operator. Suppose
\begin{itemize}
\item[(C1)] $\{x\in \overline{P}(\theta,d;\psi,L_2;\gamma,b):
   \gamma(x)>b\}\neq\phi$,  and $\gamma(Tx)>b$, for \\
   $x\in \overline{P}(\theta,d;\psi,L_2;\gamma,b)$;
\item[(C2)] $\theta(Tx)<r_1,\  \psi(Tx)<L_1$, for $x\in
   \overline{P}(\theta,r_1;\psi,L_1)$;
\item[(C3)] $\gamma(Tx)>b$, for $x\in
    \overline{P}(\theta,r_2;\psi,L_2;\gamma,b)$ with $\theta (Tx)>d$.
\end{itemize}
Then $T$ has at least three fixed points $x_1,x_2,x_3$ in
$\overline{P}(\theta,r_2;\psi,L_2)$. Further,
  \begin{gather*}
    x_1 \in \overline{P}(\theta,r_1;\psi,L_1),\; \ x_2\in\{\overline{P}(\theta,r_2;\psi,L_2;\gamma,b)
    : \gamma(x)>b\},\\
    x_3\in \overline{P}(\theta,r_2;\psi,L_2)\backslash
 \big(\overline{P}(\theta,r_1;\psi,L_1)
    \cup\overline{P}(\theta,r_2;\psi,L_2;\gamma,b)\big).
\end{gather*}
\end{lemma}

 \section{Main results}


Let closed  cone $P$ be defined by
$$
  P=\{ u\in PC^1[0,1]: u(t)\geq 0\}.
$$
Define operator $T:P\to PC^1[0,1]$ by
\begin{align*}
    Tu(t)
  &=  u(0)+u'(0)t+\int _0^t(t-s)\phi_q\Big(\int_0^s
    f(r,u(r),u'(r))dr\Big)ds\\
  &\quad +\sum_{t_k<t}(t-t_k)I_k(u(t_k))+\sum_{t_k<t}J_k(u(t_k)),
  \; t\in[0,1],
\end{align*}
which $u(0)$, $u'(0)$ are defined in \eqref{eq2.2}, \eqref{eq2.3}.

 The nonnegative continuous convex functions $\theta,\psi$, and
nonnegative continuous concave function $\gamma$ are defined by
$$
  \theta(u)=\sup_{0\leq t\leq 1}u(t),\quad
  \psi(u)=\sup_{0\leq t\leq 1}|u'(t)|,\quad
  \gamma(u)=\min_{t\in [a_m,b_m]}u(t),
$$
for all $u\in P$, where $a_m=\frac{3t_m+t_{m+1}}{4}$,
$b_m=\frac{t_m+3t_{m+1}}{4}$.
Let
\begin{gather*}
  l=\frac{\delta-1}{\int_{a_m}^{b_m}(b_m-r)\phi_q(r-a_m)ds}
  =\frac{2^{q+1}q(q+1)(\delta-1)}{(1-t_m)^{q+1}},
\\
  I_u^R=\max \{I_1(u),I_2(u),\dots,I_m(u)\},\quad u\in [0,R],
\\
  M_1=\frac{1-\alpha-\beta}{\int_0^1\phi_q(s)ds+m(1-\alpha-\beta)
 +x\alpha+y\beta}
     =\frac{1-\alpha-\beta}{1/q+m(1-\alpha-\beta)+x\alpha+y\beta},
\end{gather*}
where $x$ and $y$ satisfy $t_x<\xi<t_{x+1}$, $t_y<\eta<t_{y+1}$.

\begin{theorem} \label{thm3.1}
Suppose there exist constants $r_2\geq d\geq \delta b>b>r_1>0$,
$L_2\geq L_1>0$ such that
$$
  r_2\geq\frac{bl\delta(m+1/q)}{(\delta-1)(1-\alpha-\beta)}, \quad
L_2\geq  \frac{bl\delta (m+1/q)}{1-\alpha-\beta},
$$
and the following conditions hold
\begin{itemize}
\item[(H1)]
    $f(t,u,v)<\phi_p(\min\{\frac{\delta-1}{\delta}M_1r_1,M_1L_1\})$,
    $(t,u,v)\in [0,1]\times [0,r_1]\times [-L_1,L_1]$;
\item[(H2)]
    $\phi_p(bl)<f(t,u,v)$, $(t,u,v)\in [a_m,b_m]
 \times [b,d]\times [-L_2,L_2]$;
\item[(H3)]
    $f(t,u,v)<\phi_p(\min\{\frac{\delta-1}{\delta}M_1r_2,M_1L_2\})$,
    $(t,u,v)\in [0,1)\times [0,r_2]\times [-L_2,L_2]$;
\item[(H4)]
    $t_kI_k(u)>J_k(u)$ for $u\in[0,r_2]$,
$I_u^{r_1}<\min\{\frac{\delta-1}{\delta}M_1r_1,M_1L_1\}$,\\
     $I_u^{r_2}<\min\{\frac{\delta-1}{\delta}M_1r_2,M_1L_2\}$.
\end{itemize}
Then boundary-value problem \eqref{eq1.1} has at least three
positive solutions $u_1,u_2,u_3\in
\overline{P}(\theta,r_2;\psi,L_2)$ which satisfy
\begin{gather*}
  \sup_{0\leq t\leq 1}u_1(t)\leq r_1,\quad
  \sup_{0\leq t\leq 1}|u_1'(t)| \leq L_1; \\
 b<\min_{t\in[a_m,b_m]}u_2(t)\leq\sup_{0\leq t\leq 1}u_2(t)\leq r_2, \quad
 \sup_{0\leq t\leq 1}|u_2'(t)| \leq L_2;\\
 \sup_{0\leq t\leq 1}u_3(t)\leq\delta d,\quad
 \sup_{0\leq t\leq  1}|u_3'(t)|\leq L_2.
\end{gather*}
\end{theorem}

\begin{proof}
We need to prove
$\frac{\delta-1}{\delta}M_1r_1\geq bl$ in order to make sure that
the theorem makes sense, since we have $r_2\geq bl\delta
\frac{ m+1/q}{(\delta-1)(1-\alpha-\beta)}$, and
\begin{align*}
     \frac{\delta-1}{\delta}M_1r_1
  &  =\frac{(\delta-1)(1-\alpha-\beta)}{ \delta(1/q
     +m(1-\alpha-\beta)+x\alpha+y\beta)}r_1 \\
  &  \geq\frac{\delta-1}{\delta}\times
      \frac{1-\alpha-\beta}{1/q+m}\times\frac{m+1/q}{
     (\delta-1)(1-\alpha-\beta)}bl\delta\\
  &  =bl.
\end{align*}

Similarly, we have $M_1L_2\geq M_1r_2(\delta -1)\geq bl\delta>bl$,
so there has no contradiction among conditions.
It is easy to see that  \eqref{eq1.1} has
a solution if and only if
\begin{align*}
    Tu(t)&=
    u(0)+u'(0)t+\int _0^t(t-s)\phi_q\left(\int_0^s
    f(r,u(r),u'(r))dr\right)ds\\
  &\quad +\sum_{t_k<t}(t-t_k)I_k(u(t_k))+\sum_{t_k<t}J_k(u(t_k)),
  \quad t\in[0,1]
\end{align*}
has a fixed point.

Next, we will check the conditions (C1), (C2) and (C3) of Lemma
\ref{lem2.2} are satisfied for the operator $T$.

Obviously, we can get $Tu(t)\geq 0$, $(Tu)'(t)\geq 0$, for all
$t\in [0,1]$ and $u\in P$, that also means $Tu$ is a monotone increasing
function.

Firstly, we have $\theta(u)\leq r_2$, $\psi(u)\leq L_2$ for all $u
\in \overline{P}(\theta,r_2;\psi,L_2)$. By the condition (H4)
$t_kI_k(u)>J_k(u)$ and
$I_u^{r_2}<\min\{\frac{\delta-1}{\delta}M_1r_2,M_1L_2\}$, we get
$$
  \sum_{k=1}^{m}\Big((1-t_k)I_k(u(t_k))+J_k(u(t_k))\Big)\leq
  \sum_{k=1}^{m}I_k(u(t_k))\leq m\frac{\delta-1}{\delta}M_1r_2.
$$
By  condition (H3),
$f(t,u,v)<\phi_p(\frac{\delta-1}{\delta}M_1r_2)$, we obtain
$$
  \phi_q\Big(\int_0^s f(t,u(r),u'(r))dr\Big)\leq
  \frac{\delta-1}{\delta}M_1r_2\phi_q(s).
$$
Hence,
$$
  u(0)\leq\frac{ \alpha/q+\beta/q+x\alpha+y\beta}{\delta(1-\alpha-\beta)}M_1r_2
  +\frac{1}{q\delta}M_1r_2+\frac{m}{\delta}M_1r_2.
$$
Similarly,
$$
  u'(0)\leq\frac{(\delta-1)(\alpha+\beta+q(x\alpha +y\beta))}
  {q\delta(1-\alpha-\beta)}M_1r_2.
$$
Therefore, we can show that
\begin{align*}
    \theta(Tu)
&= \sup_{0\leq t\leq1}(u(0)+u'(0)t+\int _0^t(t-s)\phi_q\Big(\int_0^s
    f(r,u(r),u'(r))dr\Big)ds\\
  &\quad +\sum_{t_k<t}(t-t_k)I_k(u(t_k))+\sum_{t_k<t}J_k(u(t_k)))\\
  & \leq
   \frac{ \alpha/q+\beta/q+x\alpha+y\beta}{\delta(1-\alpha-\beta)}M_1r_2
    +\frac{1}{q\delta}M_1r_2+\frac{m}{\delta}M_1r_2\\
  &\quad +\frac{(\delta-1)(\alpha+\beta+q(x\alpha +y\beta))}{q\delta(1-\alpha-\beta)}M_1r_2
    +\frac{(\delta-1)(1+mq)}{q\delta}M_1r_2\\
  & = \frac{1/q+m(1-\alpha-\beta)+x\alpha+y\beta}{1-\alpha-\beta}M_1r_2.
\end{align*}
Since
$M_1=\frac{1-\alpha-\beta}{1/q+m(1-\alpha-\beta)+x\alpha+y\beta}$,
we have $\theta (Tu)\leq r_2$.

Similarly, we have
\begin{align*}
    \psi(Tu)&=
   \sup_{0\leq t\leq 1}|(Tu)'(t)|\\
  & =  \sup_{0\leq t\leq 1}|u'(0)+\int_0^t\phi_q(\int_0^sf(r,u(r),u'(r))dr)ds
    +\sum_{t_k<t}I_k(u(t_k))| \\
 & \leq \frac{1/q+m(1-\alpha-\beta)+x\alpha+y\beta}{1-\alpha-\beta}M_1L_2=L_2.
\end{align*}
Therefore, $T:\overline{P}(\theta,r_2;\psi,L_2)\to
\overline{P}(\theta,r_2;\psi,L_2)$, and it is easy to see that $T$
is a completely continuous operator.

The proof of the condition (C2) in Lemma \ref{lem2.2} is similar
to the one above.

 To check condition (C1) of Lemma \ref{lem2.2}, we choose $u_0=d$.
It is easy to see that
$u_0\in \overline{P}(\theta,d;\psi,L_2)$ and $\gamma(u)=d>b$,
so $\{x\in \overline{P}(\theta,d;\psi,L_2;\gamma,b):
\gamma(x)>b\}\neq\phi$.

For  $u\in \overline{P}(\theta,d;\psi,L_2;\gamma,b)$, we have
$b\leq u(t)\leq d$, $|u'(t)|\leq L_2$ for all $t \in [a_m,b_m]$.
Since $Tu$ is a monotone increasing function, and $(Tu)(t)\geq 0$,
$t\in [0,1]$, we have
\begin{align*}
    \gamma (Tu)
&=  \min_{t\in [a_m,b_m]}\big(u(0)+u'(0)t+\int _0^t(t-s)\phi_q(\int_0^s
    f(r,u(r),u'(r))dr)ds\\
 &\quad   +\sum_{t_k<t}[(t-t_k)I_k(u(t_k)+\sum_{t_k<t}J_k(u(t_k))\big)\\
 &= Tu(a_m).
\end{align*}
By (H2) and $u(0),u'(0)$ defined before, we
have
$$
\phi_p(bl)<f(t,u,v),\quad
 (t,u,v)\in [a_m,b_m]\times [b,d]\times [-L_2,L_2],$$
\begin{align*}
    Tu(a_m)
&= u(0)+u'(0)a_m+\int_0^{a_m}(a_m-s)\phi_q(\int_0^s f(r,u(r),u'(r))dr)ds\\
  &\quad +\sum_{t_k<a_m}(a_m-t_k)I_k(u(t_k))+\sum_{t_k<a_m}J_k(u(t_k))\\
  &\geq   u(0) \\
  & \geq \frac{1}{\delta-1}\int_0^1\int_0^s\phi_q(\int_0^r f(w,u(w),u'(w))dw)\,dr\,ds\\
  & > \frac{1}{\delta-1}\int _{a_m}^{b_m}ds\int_{a_m}^s\phi_q(\int_{a_m}^r \phi_p (bl)dw)dr\\
  & = \frac{bl}{ \delta-1}\int _{a_m}^{b_m}ds\int_{a_m}^s\phi_q(r-a_m)dr\\
  & = \frac{bl}{\delta-1}\int_{a_m}^{b_m}(b_m-r)\phi_q(r-a_m)dr
   = b.
\end{align*}
Thus $\gamma(Tu)>b $ and the condition (C1) of Lemma \ref{lem2.2}
also holds.

Finally  to prove (C3) of Lemma \ref{lem2.2}, we check
$\gamma(Tu)>b$ to be satisfied for all
$u\in\overline{P}(\theta,r_2;\psi,L_2;\gamma,b)$ with $\theta
(Tu)>d$. Since $Tu$ is a nonnegative monotone increasing function,
we can get
\begin{gather*}
  \theta (Tu)=\sup_{0\leq t\leq 1}Tu(t)=Tu(1),\\
  \gamma(Tu)=\min_{t\in [a_m,b_m]}Tu(t)=Tu(a_m), \\
  Tu(a_m)\geq Tu(0)=\frac {1}{\delta}Tu(1)>\frac{d}{\delta}
  \geq b;
\end{gather*}
that is,
$\gamma (Tu)>b$.

 We have checked Lemma \ref{lem2.2} to make sure all the
conditions are satisfied with the work we have done in the section above.
Then $T$
has at least three fixed points $u_1,u_2,u_3$ in
$\overline{P}(\theta,r_2;\psi,L_2)$. Further,
\begin{gather*}
  u_1 \in  \overline{P}(\theta,r_1;\psi,L_1),\quad
  u_2\in\{\overline{P}(\theta,r_2;\psi,L_2;\gamma,b): \gamma(x)>b\},\\
  u_3\in\overline{P}(\theta,r_2;\psi,L_2)\backslash\{\overline{P}
  (\theta,r_1;\psi,L_1)\cup\overline{P}(\theta,r_2;\psi,L_2;\gamma,b)\}.
\end{gather*}
Therefore  \eqref{eq1.1} has at least
three positive solutions $u_1,u_2,u_3$. From the boundary conditions we
have $u_3(1)=\delta u_3(0)$ and $u_3$ is a monotone increasing
function, so we have
$$
  b>\gamma(u_3)=\min_{a_m\leq t\leq b_m}
u_3(t)=u_3(a_m)\geq u_3(0)
=\frac{1}{\delta}u_3(1)=\frac{1}{\delta}\theta(u_3),
$$
so $\theta(u_3)\leq \delta b$, that means
$\sup_{0\leq t\leq 1} u_3(t)\leq \delta d$, and $u_1,u_2,u_3$ satisfy
\begin{gather*}
  \sup_{0\leq t\leq 1}u_1(t)\leq r_1, \quad
 \sup_{0\leq t\leq 1}|u_1'(t)| \leq L_1; \\
 b<\min_{t\in[a_m,b_m]}u_2(t)\leq \sup_{0\leq t\leq 1}u_2(t)\leq r_2, \quad
 \sup_{0\leq t\leq 1}|u_2'(t)| \leq L_2; \\
\sup_{0\leq t\leq 1}u_3(t)\leq\delta d,\quad
\sup_{0\leq t\leq1}|u_3'(t)|\leq L_2.
\end{gather*}
\end{proof}

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\end{document}