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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 14, 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{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/14\hfil Multiple sign-changing solutions]
{Multiple sign-changing solutions for sub-linear impulsive
three-point boundary-value problems}

\author[G. Bao, X. Xu\hfil EJDE-2010/14\hfilneg]
{Gui Bao, Xian Xu}  % in alphabetical order

\address{Department of Mathematics,
Xuzhou Normal University, Xuzhou, Jiangsu, 221116,  China}
\email[Gui Bao]{baoguigui@163.com}
\email[Xian Xu]{xuxian68@163.com}

\thanks{Submitted August 26, 2009. Published January 21, 2010.}
\subjclass[2000]{34B15, 34B25}
\keywords{Impulsive three-point boundary-value problem; \hfill\break\indent
 Leray-Schauder degree; sign-changing solution; strict upper
 and lower solutions}

\begin{abstract}
 In this article, we study the existence of sign-changing solutions
 for some second-order impulsive boundary-value problem with
 a sub-linear condition at infinity. To obtain the results we
 use the Leray-Schauder degree and the upper and lower
 solution method.
\end{abstract}

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

\section{Introduction}

This article concerns the impulsive differential
equation
\begin{equation}
\begin{gathered}
y''(t )+f(t,y(t),y'(t))=0 ,  \quad  t\in J, \; t\neq t_k ,\\
\Delta y'|_{t=t_k}= \bar{I}_k (y(t_k)), \quad k = 1,2,\dots,m,\\
 y(0)=0  ,  \quad y(1)=\alpha y(\eta) ,
\end{gathered}\label{e1.1}
\end{equation}
where $J = [0,1]$, $f \in C[J\times \mathbb{R}^2,\mathbb{R}^1]$,
$\bar{I}_k \in C[ \mathbb{R}^{1},\mathbb{R}^{1}]$,
$k = 1,2,\dots,m$, $0\leq \alpha<1$,
$0=t_0<t_1<t_2<\dots<t_m<\eta<t_{m+1}=1$.

The theory of impulsive differential equations describes
processes which experience a sudden change of their state at certain
moments. Processes with such a character arise naturally and often,
for example, phenomena studied in physics, chemical technology,
population dynamics, biotechnology and economics. For an
introduction of the basic theory of impulsive differential equation,
we refer the reader to \cite{l1}.

In recent years, there have been many papers studying the
existence of sign-changing solutions to some boundary-value problems,
see \cite{d1,m1,r1,s1,s2,z2} and the references therein. However, to the
authors¡¯ best knowledge, there are few papers that considered the
sign-changing solutions for the impulsive boundary-value problems.
Usually, to show the existence of sign-changing solutions one
employs the variational method and the Leray-Schauder degree method.
However, a suitable variational structure for impulsive
boundary-value problems is yet unknown.
In \cite{p1,x2,x3}, authors computed the
algebraic multiplicities of the linear problems corresponding to the
discussed boundary-value problems, but we know that the algebraic
multiplicities of impulsive boundary-value problem are not easy to
compute. Thus, there are many difficulties in studying the
sign-changing solutions for the impulsive boundary-value problem
\eqref{e1.1} by the method mentioned
above.

In this paper, we consider the sign-changing solutions for
the impulsive three-point boundary-value problem \eqref{e1.1} by the
Leray-Schauder degree and strict upper and lower solution method. We
assume a sub-linear condition at infinity, and we construct another
pair of strict upper and lower solutions by conditions of $f$ and
$\bar{I}_k$.
 We will show a result of at least four sign-changing solutions,
two positive solutions and two negative solutions for \eqref{e1.1}.
Moreover, we will give a description of the exact locations
of them.


\section{Preliminaries}

 Let $J'= J\setminus\{ t_1,t_2,\dots,t_m\}$,
$PC^1[J,\mathbb{R}^1]=\{ x:J\to \mathbb{R}^1$,
$x'$ is continuous at $t \neq t_k$, $x'$ is left continuous
at $t=t_k$, $x'(t_{k}^{+})$  exists\}.
 For  $x\in PC^1[J,\mathbb{R}^1]$, let
$$
\|x\|_{PC^1} =\max \{ \| x \| ,\| x' \| \},
$$
where $\|x \|= \sup_{t\in J}|x(t)|$ and
$\|x' \|= \sup_{t\in J}|x'(t)|$. Then $PC^1[J,\mathbb{R}^1]$
is a real Banach space with norm  $\|\cdot\|_{PC^1}$.
  Let $x,y \in C[J,\mathbb{R}^1]$. Define $\prec$ as
 follows
$$
 x \prec y ~\text{if $x(t)<y(t)$ for all $ t\in  J$}.
$$

\begin{definition} \label{def2.1}\rm
 A function $u \in PC^1[J,\mathbb{R}^1]\cap  C^2[J',\mathbb{R}^1]$
is called a strict lower solution of \eqref{e1.1}, if
\begin{equation}
\begin{gathered}
u''(t)+f(t,u(t),u'(t))>0,  \quad  t\neq t_k,\\
\Delta u'|_{t=t_k}> \bar{I}_k (u(t_k)), \quad k = 1,2,\dots,m,\\
u(0)< 0 ,  \quad    u(1)-\alpha  u(\eta)< 0 .
\end{gathered} \label{e2.1}
\end{equation}
A function $v\in PC^1[J,\mathbb{R}^1]\cap C^2[J',\mathbb{R}^1]$
is called a strict upper solution of \eqref{e1.1}, if
\begin{equation}
\begin{gathered}
v''(t)+f(t,v(t),v'(t))<0, \quad t\neq t_k,\\
\Delta v'|_{t=t_k}< \bar{I}_k (v(t_k)), \quad k = 1,2,\dots,m,\\
v(0)> 0 ,  \quad    v(1)-\alpha  v(\eta)> 0 .\\
\end{gathered} \label{e2.2}
\end{equation}
\end{definition}

Let us introduce the following constants:
\begin{equation}
\begin{gathered}
\beta= \limsup_{|x|+|y|\to \infty}\max_{t\in J}\frac{|f(t,x,y)|}{|x|+|y|},\\
\bar{\beta_k}=\limsup_{|x|\to \infty}\frac{|\bar{I}_k(x)|}{|x|},
\quad  k=1,2,\dots m, \\
\gamma =\frac{4}{1-\alpha\eta}(2\beta +\sum_{k=1}^{m}\bar{\beta}_k).
\end{gathered}\label{e2.3}
\end{equation}
To state the main results in this paper we need the following
assumptions:
\begin{itemize}
\item[(H1)] For each $k\in \{1,2,\dots,m\}$, $\bar{I}_k(0)=0$ and
$$
\lim_{x\to 0}\frac{\bar{I}_k(x)}{x}=d_0>0.
$$
\item[(H2)] $f:[0,1]\times \mathbb{R}^2 \to \mathbb{R}^1$ is continuous,
$f(t, 0, 0)=0$ and
$$
\lim_{x\to 0}\frac{f(t,x,y)}{x}=d_1<0,
$$
uniformly for $t\in[0,1]$.
\end{itemize}
 From \cite[Lemma 5.4.1]{g1}, we have the following result.

\begin{lemma} \label{lem2.1}
 $H\subset PC^1[J,\mathbb{R}^1]$ is a relatively compact set
if and only if for any $x\in H$, $x(t)$ and $x'(t)$ are uniformly
bounded on $J$ and equicontinuous at any $ J_k (k=1,2,\dots,m)$,
where
$J_1=[0,t_1]$, $J_i=(t_{i-1},t_i]$, $ i=2,3,\dots,m+1$.
\end{lemma}

Now we define the operator $A: PC^1[J,\mathbb{R}^1]\to
PC^1[J,\mathbb{R}^1]$ as follows:
\begin{align*}
&(Ax)(t)\\
&=\frac{t}{1-\alpha\eta}\int_0^1(1-s)f(s,x(s),x'(s))ds
 -\frac{\alpha
t}{1-\alpha\eta}\int_0^{\eta}(\eta-s)f(s,x(s),x'(s))ds \\
&\quad - \int_0^t(t-s)f(s,x(s),x'(s))ds
+\sum_{0<t_k<t}[\bar{I}_k(x(t_k))(t-t_k)]\\
&\quad - \frac{t}{1-\alpha\eta}\sum_{k=1}^m
\{[1-t_k-\alpha(\eta-t_k)]\bar{I}_k(x(t_k))\},
\quad x\in PC^1[J,\mathbb{R}^1].
\end{align*}
 From Lemma \ref{lem2.1}, we know $A:PC^1[J,\mathbb{R}^1]\to
PC^1[J,\mathbb{R}^1]$ is a completely continuous operator.
The following Lemma  can be easily obtained.

 \begin{lemma} \label{lem2.2}
$y\in PC^1[J,\mathbb{R}^1]$ is a solution of \eqref{e1.1} if and
only if $ y(t) = Ay(t) $  for $t\in [0,1]$
\end{lemma}

\begin{theorem} \label{thm2.1}
 Assume that $u_1$ and $u_2$ are two strict lower
solutions of \eqref{e1.1}, $0\leq \gamma<1$, then there exists $R_0>0$
large enough such that
$$
\deg(I-A,\Omega,\theta)=1,
$$
where $\Omega= \{ x \in B(\theta,R_0) :  \sigma_1\prec x \}$,
$\sigma_1 (t)= \sup_{t\in J}\{u_1(t),u_2(t)\}$.
\end{theorem}

\begin{proof}
 If we let $I_k=0$ in the proof of \cite[Theorem 2.1]{x1},
we can easily get this theorem by slight modification. But for the
completeness of this paper we will give details of the proof of this
theorem.
 For $0\leq \gamma<1$, we  take $\beta'>\beta$,
 $\bar{\beta}_k'>{\bar{\beta}}_k$,
$(k =1,2,\dots,m)$ with
\begin{equation}
\gamma ':= \frac{4}{1-\alpha\eta}
(2\beta'+\sum_{k=1}^{m}\bar{\beta'}_k) <1.\label{e2.4}
\end{equation}
 From the definition of $\beta $, there exists $N>0$, such that
$$
|f(t,x,y)|<\beta'(|x|+|y|), \quad  \forall t \in J,|x|+|y|\geq N,
$$
and so
\begin{equation}
|f(t,x,y)|\leq \beta'(|x|+|y|)+ M,\quad  \forall t \in J,  \; x,y \in
\mathbb{R}^1, \label{e2.5}
\end{equation}
where $M=\sup_{(t,x,y)\in J\times \mathbb{R}^2,\,
|x|+|y|\leq  N} |f(t,x,y)|$. Similarly, we
have
\begin{equation}
|\bar{I}_k(x)|\leq  \bar{\beta'}_k|x|+ \overline{M}_k , \quad
\forall x\in \mathbb{R}^1 ,\label{e2.6}
\end{equation}
where $\overline{M}_k$ is a positive constant. Take
\begin{equation}
R_0 > \max\{\|u_1\|_{PC^1},\|u_2\|_{PC^1},
 \frac{1}{1-\gamma'}\frac{4}{1-\alpha\eta}
(M+\sum_{k=1}^{m}\overline{M}_k)\}.\label{e2.7}
\end{equation}
Let $\sigma_1 (t)=\sup_{t\in J}\{u_1(t),u_2(t)\}$
 for all $t\in J$. Then $\sigma_1\in PC[J,\mathbb{R}^1]$.
Now we define  $h_1:J \times \mathbb{R}^2 \to \mathbb{R}^1$,
$\bar{J}_{k,1}:\mathbb{R}^1\to \mathbb{R}^1$,
$(k=1,2,\dots,m)$ as  follows:
\begin{equation}
h_1(t,x,y)=\begin{cases}
 f(t,\sigma_1(t),y),  &  x<\sigma_1(t),\\
 f(t,x,y),   &  x\geq \sigma_1(t),
 \end{cases}\label{e2.8}
\end{equation}
\begin{equation}
\bar{J}_{k,1}(x)=\begin{cases}
\bar{I}_k(\sigma_1(t_k)),  &  x<\sigma_1(t_k),\\
 \bar{I}_k(x),   &    x\geq\sigma_1(t_k).
 \end{cases}\label{e2.9}
\end{equation}
Define the nonlinear operator
$A_1:PC^1[J,\mathbb{R}^1]\to PC^1[J,\mathbb{R}^1]$ as follows:
\begin{align*}
&(A_1x)(t)\\
&= \frac{t}{1-\alpha\eta}\int_0^1(1-s)h_1(s,x(s),x'(s))ds
 -\frac{\alpha
t}{1-\alpha\eta}\int_0^{\eta}(\eta-s)h_1(s,x(s),x'(s))ds \\
&\quad - \int_0^t(t-s)h_1(s,x(s),x'(s))ds
+\sum_{0<t_k<t}[\bar{J}_{k,1}(x(t_k))(t-t_k)]\\
&\quad -\frac{t}{1-\alpha\eta}
 \sum_{k=1}^m\{[1-t_k-\alpha(\eta-t_k)]\bar{J}_{k,1}(x(t_k)\},
\quad \forall t\in J.
\end{align*}
Clearly, $A_1:PC^1[J,\mathbb{R}^1]\to PC^1[J,\mathbb{R}^1] $  is a
completely continuous operator.
Let
$$
B(\theta,R_0)=\{x\in PC^1[J,\mathbb{R}^1]:\|x\|_{PC^1}<R_0\}.
$$
For any $x\in \overline{B}(\theta,R_0)$,
by \eqref{e2.5}-\eqref{e2.9}, we have for all
$t\in J$,
 $$
|h_1(t,x(t),x'(t))| \leq  \beta' \sup_{t\in J}\{|x(t)|,|u_1(t)|,
|u_2(t)|\} + \beta' |x'(t)| + M \leq  2\beta' R_0 + M,
$$
and for $k= 1,2,\dots,m $,
$$
|\bar{J}_{k,1}(x(t_k))|\leq
\bar{\beta}'_k \max\{ |x(t_k)|,|u_1(t_k)|,|u_2(t_k)|\} +
\overline{M}_k \leq  \bar{\beta}'_k R_0 + \overline{M}_k.
$$
Then
\begin{equation}
\begin{aligned}
&|A_1x(t)| \\
&\leq [\frac{1}{1-\alpha\eta}\int_0^1(1-s)ds +\frac{\alpha
}{1-\alpha\eta}\int_0^{\eta}(\eta-s)ds+\int_0^1(1-s)ds]  (2\beta'
R_0 + M ) \\
&\quad+ \frac{1}{1-\alpha\eta}
\sum_{k=1}^m(\bar{\beta}'_k R_0 +
\overline{M}_k)+\sum_{k=1}^m(\bar{\beta}'_k R_0 +
\overline{M}_k)\\&\leq \frac{2}{1-\alpha\eta}(2\beta' R_0 + M
)+\sum_{k=1}^m(\frac{1}{1-\alpha\eta}+1) \bar{\beta}'_k
R_0+\sum_{k=1}^m(\frac{1}{1-\alpha\eta}+1)\overline{M}_k\\
&\leq \frac{2}{1-\alpha\eta}(2\beta'+\sum_{k=1}^m\bar{\beta}'_k)R_0+
\frac{2}{1-\alpha\eta}(M+\sum_{k=1}^m\overline{M}_k).
\end{aligned} \label{e2.10}
\end{equation}
Also we have
\begin{equation}
\begin{aligned}
|(A_1x)'(t)|&\leq  [\frac{1}{1-\alpha\eta}\int_0^1(1-s)ds
+\frac{\alpha }{1-\alpha\eta}\int_0^{\eta}(\eta-s)ds+1]  (2\beta'
R_0 + M ) \\
&\quad  +  \frac{1}{1-\alpha\eta}
\sum_{k=1}^m(\bar{\beta}'_k R_0 +
\overline{M}_k)+\sum_{k=1}^m(\bar{\beta}'_k R_0 +
\overline{M}_k)\\
&\leq \frac{2}{1-\alpha\eta}(2\beta'+\sum_{k=1}^m\bar{\beta}'_k)R_0+
\frac{2}{1-\alpha\eta}(M+\sum_{k=1}^m\overline{M}_k).
\end{aligned} \label{e2.11}
\end{equation}
Thus
$$
\|A_1x\|_{PC^1}\leq \frac{4}{1-\alpha\eta}(2\beta'
+\sum_{k=1}^m\bar{\beta}'_k)R_0+
\frac{4}{1-\alpha\eta}(M +\sum_{k=1}^m\overline{M}_k)<R_0 .
$$
Then $A_1(\overline{B}(\theta,R_0))\subset B(\theta,R_0) $.
Hence
\begin{equation}
\deg(I-A_1,B(\theta,R_0),\theta)=1.\label{e2.12}
\end{equation}
Now we prove that $x_0\in \Omega $ whenever
$x_0\in \overline{B}(\theta,R_0)$
with $x_0=A_1x_0$. By Lemma \ref{lem2.2},   we have
\begin{equation}
\begin{gathered}
x_0''(t)+h_1(t,x_0(t),x_0'(t))=0,  \quad t\in J , \;  t\neq t_k,\\
\Delta x_0'|_{t=t_k}= \bar{J}_{k,1} (x_0(t_k)), \quad k = 1,2,\dots,m,\\
x_0(0)= 0 ,\quad    x_0(1)-\alpha  x_0(\eta)= 0,
\end{gathered}\label{e2.13}
\end{equation}
for any $x_0\in \overline{B}(\theta,R_0)$ with
$x_0=A_1x_0$. We need to prove
\begin{equation}
\sigma_1\prec x_0. \label{e2.14}
\end{equation}
Let $\omega(t)=\sigma_1(t)-x_0(t)$ for all $t\in J$. Then
$\omega\in PC[J,\mathbb{R}^1]$.   If \eqref{e2.14} is not true,
then $\sup_{t\in J}\omega(t)\geq 0 $. We have several cases to
consider.

(1) $\omega(0)=\sup_{t\in J}\omega(t)\geq 0$. In
this case,
$$
0\leq \omega(0)=\sigma_1(0)-x_0(0)=\sigma_1(0)
=\max\{u_1(0),u_2(0)\}<0,
$$
which is  a  contradiction.

 (2) $\omega(1)=\sup_{t\in J}\omega(t)\geq 0$. Assume without
loss of generality that $\sigma_1(1)=u_1(1)$. Then
$$
0\leq \omega(1)=u_1(1)-x_0(1)<\alpha u_1(\eta)-\alpha
x_0(\eta)\leq \alpha\omega(\eta)\leq \alpha\omega(1),
$$
which is  a  contradiction.

 (3) There exists $k_0\in \{1,2,\dots,m,m+1\}$ and
$\tau_0\in (t_{k_0-1},t_{k_0})$
 such that $\omega(\tau_0)=\sup_{t\in J}\omega(t)\geq  0$.
We may assume $\sigma_1(\tau_0)=u_1(\tau_0) $. We have
 two subcases:
(3A) $u_2(\tau_0)<u_1(\tau_0)$, and
(3B) $u_2(\tau_0)=u_1(\tau_0)$.

For case (3A), we take  $\delta_0>0$ small  enough  such  that
 $[\tau_0-\delta_0,\tau_0+\delta_0]\subset(t_{k_0-1},t_{k_0})$ and
 $\sigma_1(t)=u_1(t)$ for all $t\in [\tau_0-\delta_0,\tau_0+\delta_0]$.
Then $\omega(t)=u_1(t)-x_0(t)$ for all
$t\in [\tau_0-\delta_0,\tau_0+\delta_0] $. Thus,
$\omega\in C^2[\tau_0-\delta_0,\tau_0+\delta_0]$ and
 $\omega(\tau_0)$ is a local maximum of $\omega$ in
 $[\tau_0-\delta_0,\tau_0+\delta_0]$.
Therefore $\omega'(\tau_0)=0$ , $\omega''(\tau_0)\leq 0$ and so
\begin{align*}
0&\geq \omega''(\tau_0)=u_1''(\tau_0)-x_0''(\tau_0)\\
&=u_1''(\tau_0)+h_1(\tau_0,x_0(\tau_0),x_0'(\tau_0))\\
&= u_1''(\tau_0)+f(\tau_0,u_1(\tau_0),u_1'(\tau_0))>0,
\end{align*}
which is  a  contradiction.

For case (3B), let $\omega_1(t)=u_2(t)-x_0(t)$  for all
$t\in (t_{k_0-1},t_{k_0})$. For  $t'\in (t_{k_0-1},t_{k_0})$, we
have
\begin{align*}
\omega_1(\tau_0)
&=  u_2(\tau_0)-x_0(\tau_0)\\
&=\sigma_1(\tau_0)-x_0(\tau_0)
 = \omega(\tau_0)\\
&\geq\omega(t')=\sigma_1(t')-x_0(t')\\
&\geq u_2(t')-x_0(t')=\omega_1(t').
\end{align*}
Then $\omega_1(\tau_0)$ is a local maximum of $\omega_1$ in
$(t_{k_0-1},t_{k_0})$. Thus  $\omega_1'(\tau_0)=0$,
$\omega_1''(\tau_0)\leq  0$. Therefore
\begin{align*}
0&\geq \omega_1''(\tau_0)
 =u_2''(\tau_0)-x_0''(\tau_0)\\
&=u_2''(\tau_0)+h_1(\tau_0,x_0(\tau_0),x_0'(\tau_0))\\
&= u_2''(\tau_0)+f(\tau_0,u_2(\tau_0),u_2'(\tau_0))>0,
\end{align*}
which is  a  contradiction.

(4)  There exists $k_0\in\{1,2,\dots,m\}$  such that
$\omega(t_{k_0})=\sup_{t \in J}\omega(t)\geq 0$. We take
$\delta_0>0$ small enough such that $\omega(t_{k_0})$ is a local
maximum of $\omega(t)$ in $[t_{k_0}-\delta_0,t_{k_0}+\delta_0]$,
then we have $\omega'(t_{k_0})\geq0$ and
$\omega'(t^+_{k_0})\leq 0$. Thus,
\begin{align*}
0&\geq \omega'(t_{k_0}^+)=u_1'(t_{k_0}^+)-x_0'(t_{k_0}^+)\\
&>[u_1'(t_{k_0})+\bar{I}_{k_0}(u_1(t_{k_0}))]
 -[x_0'(t_{k_0})+\bar{J}_{k_0,1}(x_0(t_{k_0}))]\\
&= u_1'(t_{k_0})-x_0'(t_{k_0})\\
&=\omega'(t_{k_0})\geq0,
\end{align*}
which is  a  contradiction.

 From the discussion of cases (1)-(4), we see that \eqref{e2.14} holds.
Since $\Omega=\{x\in B(\theta,R_0)|\sigma_1\prec x \}$, it follows that
$\Omega\subset PC^1[J,\mathbb{R}^1] $ is an open set.
 We see from \eqref{e2.12}
\eqref{e2.14} and the properties of topological degree that
$$
\deg(I-A_1,\Omega,\theta)=1.
$$
Notice that $A_1x=Ax$ for all $x\in \overline{\Omega}$,
and so we have
$$
\deg(I-A,\Omega,\theta)=1 .
$$
This completes the proof.
\end{proof}


\begin{corollary} \label{coro2.1}
Assume that $u_1$ is a strict lower solution
of \eqref{e1.1}, $0\leq \gamma<1$, then there exists $R_0>0$ large
enough such that
$$
\deg(I-A,\Omega,\theta)=1,
$$
where $\Omega= \{ x \in B(\theta,R_0) : u_1 \prec x \}$.
\end{corollary}

Also we have the following Theorems.

\begin{theorem} \label{thm2.2}
Assume that $v_1$ and $v_2$ are two strict upper
solutions of \eqref{e1.1}, $0\leq \gamma<1$, then there exists $R_0>0$
large enough such that
$$
\deg(I-A,\Omega,\theta)=1,
$$
 where $\Omega$=$ \{ x \in B(\theta,R_0):  x\prec\sigma_2  \}$,
$\sigma_2 (t)=\inf_{t\in J}\{v_1(t),v_2(t)\}$.
\end{theorem}

\begin{corollary} \label{coro2.2}
Assume that $v_1$ is a strict upper solution of
\eqref{e1.1}, $0\leq \gamma<1$, then there exists $R_0>0$ large enough
such that
$$
\deg(I-A,\Omega,\theta)=1,
$$
where $\Omega= \{ x \in B(\theta,R_0): x\prec v_1 \}$.
\end{corollary}

 \begin{theorem} \label{thm2.3}
Assume that $u_1$ is a strict lower solution and $v_1$ is a strict upper
solution of \eqref{e1.1}, $0\leq \gamma<1$, then there exist $R_0>0$
large enough such that
$$
\deg(I-A,\Omega,\theta)=1,
$$
where $\Omega=\{x\in B(\theta,R_0):u_1\prec x\prec v_1 \}$.
\end{theorem}

\section{Main Results}

\begin{theorem} \label{thm3.1}
Assume that {\rm (H1), (H2)} are satisfied,
$0\leq \gamma<1$  and \eqref{e1.1} has a strict lower solution $u_1$
 and a strict upper solution $v_1$, such that
$u_1\prec v_1$ and $u_1$, $v_1$ are sign-changing on $[0,1]$. Then
\eqref{e1.1} has at least four sign-changing
solutions, two positive solutions and two negative solutions.
\end{theorem}

\begin{proof}
 From (H2), there exists $0<\varepsilon_0<R_0$ such that
$$
f(t,-\varepsilon,0)>0, \quad f(t,\varepsilon,0)<0, \quad
\forall t\in[0,1], \; \forall\varepsilon\in(0,\varepsilon_0).
$$
Let $u_{1,i}(t)=-1/i$, $v_{1,j}(t)=\frac{1}{j}$,
$i,j=1,2,\dots$. Then there exists a natural number
$n_0>\frac{1}{\varepsilon_0}$ such that
 $$
u_{1,i}\not\preceq v_1, \quad u_1 \not\preceq v_{1,j},
$$
for each $i,j\geq n_0$.
Since $u_{1,i}(t)=-\frac{1}{i}<0$, it follows that
$u_{1,i}(t_k)=-1/i<0$, $k=1,2,3,\dots ,m$. By (H1) and
(H2), we can easily show that
\begin{gather*}
u_{1,i}''(t)+f(t,u_{1,i}(t),u_{1,i}'(t))>0,  \quad t\neq t_k,\\
\Delta u_{1,i}'|_{t=t_k}> \bar{I}_k (u_{1,i}(t_k)), \quad
k = 1,2,\dots,m,\\
u_{1,i}(0)< 0 ,  \quad    u_{1,i}(1)-\alpha  u_{1,i}(\eta)< 0 .
\end{gather*}
 So, $u_{1,i}(t)$ is a strict lower solution of \eqref{e1.1}.
Similarly, we know $v_{1,j}$ is a strict upper solution of
\eqref{e1.1}.

Take $u_{1,n_0}$ and $v_{1,n_0}$, let
\begin{gather*}
O_1 = \{x \in B(\theta,R_0)| u_1\prec x\}, \quad
O_2 = \{x \in B(\theta,R_0)|x\prec v_1\},\\
O_3 = \{ x\in B(\theta,R_0)|u_{1,n_0}\prec x\} ,\quad
O_4 = \{ x\in B(\theta,R_0)|x\prec v_{1,n_0}\},\\
\Omega_1=O_1\setminus (\overline{O_1\cap O_2})
 \cup (\overline{O_1\cap O_3}),\quad
\Omega_2=O_2\setminus (\overline{O_1\cap O_2})\cup
 (\overline{O_2\cap O_4}),\\
\Omega_3=O_3\setminus (\overline{O_1\cap O_3})\cup
 (\overline{O_3\cap O_4}), \quad
\Omega_4=B(\theta,R_0)\setminus
(\overline{O}_1\cup\overline{O}_4\cup\overline{\Omega}_2
 \cup\overline{\Omega}_3).
\end{gather*}
 From Theorems \ref{thm2.1}-\ref{thm2.3} and
Corollaries \ref{coro2.1}-\ref{coro2.2}, we have
\begin{gather}
\deg(I-A,O_1,\theta)=1, \label{e3.1}\\
\deg(I-A,O_2,\theta)=1, \label{e3.2}\\
\deg(I-A,O_1\cap O_2,\theta)=1,\label{e3.3}\\
\deg(I-A,O_1\cap O_3,\theta)=1,\label{e3.4}\\
\deg(I-A,O_2\cap O_4,\theta)=1.\label{e3.5}
\end{gather}
Thus,
\begin{gather}
\deg(I-A,\Omega_1,\theta)=1-1-1=-1,\label{e3.6}\\
\deg(I-A,\Omega_2,\theta)=1-1-1=-1.\label{e3.7}\end{gather}
 So, there exist $x_1\in O_1\cap O_2$,
$x_2\in \Omega_1$, $x_3\in \Omega_2$, which are sign-changing
solutions of \eqref{e1.1}.
  From Corollaries \ref{coro2.1}-\ref{coro2.2} and Theorem \ref{thm2.3}, we have
\begin{gather}
\deg(I-A,O_3,\theta)=1,\label{e3.8}\\
\deg(I-A,O_4,\theta)=1,\label{e3.9}\\
\deg(I-A,O_3\cap O_4,\theta)=1.\label{e3.10}
\end{gather}
Thus, from \eqref{e3.4}, \eqref{e3.7} and \eqref{e3.9}, we have
\begin{equation}
\deg(I-A,\Omega_3,\theta)=1-1-1=-1.\label{e3.11}
\end{equation}
 From the proof of \eqref{e2.12}, it is easy to get
\begin{gather}
 \deg(I-A_,B(\theta,R_0),\theta)=1.\label{e3.12}
\end{gather}
Then we have from \eqref{e3.1}, \eqref{e3.7}, \eqref{e3.9},
\eqref{e3.11} and \eqref{e3.12} that
$$
\deg(I-A,\Omega_4,\theta)=1-1-1-(-1)-(-1)=1.
$$
So, there exists a fourth sign-changing solution $x_4\in \Omega_4$.
By \eqref{e3.4}, we can get a solution $x_{5,i}\in O_1\cap O_3$ for
$i\geq n_0$.  From $\|x_{5,i}\|=\| Ax_{5,i}\|<R_0$, we know
$\{x_{5,i}\}_{i=n_0}^\infty$ is a bounded set. Notice that $A$ is a
completely continuous operator, then $\{x_{5,i}\}_{i=n_0}^\infty$ is
a relatively compact set. Without loss of generality, assume that
$x_{5,i}\to x_5$ as $i\to\infty$. Then $x_5$ is a
solution of \eqref{e1.1}. Since $u_{1,i}\to 0$ as
$i\to\infty$, then $x_5$ is a positive solution of \eqref{e1.1}.
Similarly, we can get $x_6$, $x_7$ and $x_8$ such that
\begin{gather*}
\theta\prec x_6\prec R_0, \quad u_1\not\prec x_6\not\prec v_{1,n_0}.\\
-R_0\prec x_7\prec v_1,\quad  -R_0\prec x_7\prec\theta,\\
-R_0\prec x_8\prec \theta,\quad  u_{1,n_0}\not\prec x_8\not\prec v_1.
\end{gather*}
It is easy to see that $x_6$ is a positive solution of \eqref{e1.1},
$x_7$ and  $x_8$ are two negative solutions of \eqref{e1.1}.
This completes the proof.
\end{proof}

\begin{remark} \label{rmk3.1}\rm
 Obviously, we can replace the sub-linear condition
$0\leq  \gamma<1$ with a pair of strict upper and lower
solutions, but then we need to introduce a Nagumo condition for
nonlinear item $f$.
\end{remark}

In this paper, we give some existence results for
sign-changing solutions. Up to now, there were few papers that
considered the existence of sign-changing solutions for impulsive
multi-point boundary-value problem. Moreover, we give the exact
positions of them. Therefore, the result of this paper is new.

 The method of this paper is of interest even if there
exists a jump of $x(t)$ at $t=t_k$, $k=1,2,3,\dots,m$ at the same
time.

\begin{example} \label{exa3.1} \rm
Let $R_0=100$ and
 $$
u_1(t)=\sin\frac{3}{2}\pi t-\frac{1}{2}, \quad
v_1(t)=\sin\frac{1}{2}\pi t+\frac{1}{2}, \quad
 \forall t\in[0,1].
$$
 Obviously, $u_1(t)$ and $v_1(t)$ are sign-changing on $[0,1]$
and $u_1\prec  v_1$.
Now let the sets $D_1,D_2,D_3$, and $\widetilde{D}_4$ be defined by
\begin{gather*}
D_1=\{(t,u_1(t),u_1'(t)):t\in[0,1]\},\quad
D_2=\{(t,v_1(t),v_1'(t)):t\in[0,1]\},\\
D_3=\{(t,100,0):t\in[0,1]\},\quad
\widetilde{D}_4=\{(t,0,0):t\in[0,1]\}.
\end{gather*}
Then $D_1,D_2,D_3$, and $\widetilde{D}_4$ are four disjoint closed
sets of $\mathbb{R}^3$. Let
$$
r_0=\frac{1}{2}\min\{d(\widetilde{D}_4,D_1),d(\widetilde{D}_4,D_2),
d(\widetilde{D}_4,D_3)\}>0
$$
and
$$
D_4=\{(t,x,y)\in \mathbb{R}^3:d((t,x,y),\widetilde{D}_4)\leq  r_0\}.
$$
Define the function $\widetilde{f}$ by
$$
\widetilde{f}(t,x,y)=\begin{cases}
30, & (t,x,y)\in D_1,\\
-30, & (t,x,y)\in D_2,\\
1, &  (t,x,y)\in D_3,\\
\frac{1}{100}(-x+y),  &  (t,x,y)\in D_4.
 \end{cases}
$$
  From Dugundji's extension theorem, see \cite{g2}, there exists
a continuous function $f:[0,1]\times \mathbb{R}^2\mapsto \mathbb{R}^1$
 such that
$f(t,x,y)=\widetilde{f}(t,x,y)$ while $(t,x,y)\in D_i$ for each
$i=1,2,3,4$, and
$f([0,1]\times \mathbb{R}^2)\subset \widetilde{f}([0,1]\times
\mathbb{R}^2)\subset B(\theta,100)$.
Consider the  impulsive three-point boundary-value problem
\begin{equation}
\begin{gathered}
y''(t )+f(t,y(t),y'(t))=0 ,  \quad   t\in J, \; t\neq t_k ,\\
\Delta y'|_{t=t_k}= \bar{I}_k (y(t_k)),  \quad k = 1,2,\\
 y(0)=0  ,  \quad  y(1)=\alpha y(\eta) ,\\
\end{gathered} \label{e3.13}
\end{equation}
where $t_1=\frac{1}{10}$,
$t_2=\frac{2}{3},\alpha=\frac{1}{2}$,
$\eta=\frac{3}{4}$
and $\bar{I}_k(x)=\frac{1}{50k}x,  k=1,2$.  From the definition of
$u_1(t)$ and $f$ we have
$$
u_1''(t)+f(t,u_1(t),u_1'(t))=-\frac{9}{4}\pi^2\sin\frac{3}{2}\pi t
+\widetilde{f}(t,u_1(t),u_1'(t))>-\frac{9}{4}\pi^2+30>0,
$$
for all $t\in[0,1]$,
\begin{gather*}
\bar{I}_1(u_1(t_1))=\frac{1}{50}(\sin\frac{3}{20}\pi-\frac{1}{2})<0
=\Delta u'_1|_{t=t_1},\\
\bar{I}_2(u_1(t_2))=\frac{1}{100}(\sin\pi-\frac{1}{2})<0
=\Delta u'_1|_{t=t_2},\\
u_1(0)<0, \quad \alpha u_1(\eta)=\frac{1}{2}(\sin\frac{9}{8}\pi
-\frac{1}{2})>-\frac{3}{2}=u_1(1).
\end{gather*}
Then $u_1(t)$ is a strict lower solution of
\eqref{e3.12}. Similarly, $v_1(t)$ is a strict upper solution
of \eqref{e3.12}.
 From
$$
\lim_{x\to 0}\frac{\bar{I}_k(x)}{x}=\frac{1}{50k}>0,
\quad \bar{I}_k(0)=0, \quad k=1,2,
$$
we see that (H1) holds. Next note
$$
\lim_{x\to0}\frac{f(t,x,0)}{x}
=\lim_{x\to0}\frac{\widetilde{f}(t,x,0)}{x}=-\frac{1}{100}<0,f(t,0,0)=0,
$$
uniformly for $t\in[0,1]$, then (H2)  holds. Since
\begin{gather*}
\beta= \limsup_{|x|+|y|\to \infty} \max_{t\in
J}\frac{|f(t,x,y)|}{|x|+|y|}=0, \\
 \bar{\beta_k}=\limsup_{|x|\to \infty}
 \frac{|\bar{I}_k(x)|}{|x|}=\frac{1}{50k}, \quad
k=1,2,
\end{gather*}
it follows that
$$
\gamma =\frac{4}{1-\alpha\eta}(2\beta +\bar{\beta}
_1+\bar{\beta}_2)=\frac{24}{125}<1.
$$
Now all conditions of Theorem \ref{thm3.1} hold.
Therefore, the impulsive boundary-value problem \eqref{e3.2}
has at least four sign-changing solutions,
two positive solutions and two negative solutions.
\end{example}

\subsection*{Acknowledgments}
 This research is supported by grants NSFC10971179 from the
Natural Science Foundation of Jiangsu Education Committee
and 09KJB110008 from the  Qing Lan Project.

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\end{document}