\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 89, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/89\hfil Multiple sign-changing solutions]
{Multiple sign-changing solutions for some m-point boundary-value problems}

\author[X. Xu\hfil EJDE-2004/89\hfilneg]
{Xian Xu}

\address{Xian Xu \hfill\break
 Department of  Mathematics,  Xuzhou Normal University, Xuzhou,
Jiangsu, 221116, China}
\email{xuxian68@163.com}

\date{}
\thanks{Submitted April 26, 2004. Published July 5, 2004.}
\subjclass[2000]{34B15, 34B25}
\keywords{m-point boundary-value problem; sign-changing solution;
\hfill\break\indent fixed point index}

\begin{abstract}
 In this paper, we show existence results for multiple 
 sign-changing solutions for m-point boundary-value problems. 
 We use fixed point index and Leray-Schauder degree methods.
\end{abstract}

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

\section{Introduction}
 In this paper, we consider the  second-order
multi-point boundary-value problem
\begin{equation}
 \begin{gathered}
 y''(t)+f(y)=0, \quad 0\leq t\leq 1,\\
 y(0)=0,\quad y(1)=\sum_{i=1}^{m-2}\alpha_i y(\eta_i),
 \end{gathered}
 \label{e1.1}
\end{equation}
 where $0<\alpha_i$, $i=1,2,\dots, m-2$, $0<\eta_1<\eta_2<\dots<\eta_{m-2}<1$,
 $f\in C(\mathbb{R},\mathbb{R})$.

The multi-point boundary-value problems for ordinary differential
equations arise  in different areas of applied mathematics and
physics.  For examples, the vibrations of a guy wire of  uniform
cross-section and composed of $N$ parts of different densities can
be set up as a multi-point boundary-value problem (see \cite{m4}),
many problems in the theory of elastic stability can be handled as
multi-point problems (see \cite{t1}). Recently, there is much
attention focused on the existence
  of nontrivial or positive solutions of the nonlinear multi-point
   boundary-value problems(see \cite{f1,g1,g2,l1,m1, m3,r1,w1,x1,x2,y1} and the
references therein). For example, Ruyun Ma \cite{m1} considered the
m-point boundary-value problem
\begin{equation}
\begin{gathered}
 u''(t)+a(t)f(u)=0,\quad t\in (0,1),\\
 u'(0)=\sum_{i=1}^{m-2}b_iu'(\xi_i),\quad u(1)=\sum_{i=1}^{m-2}a_iu(\xi_i),
 \end{gathered} \label{e1.2}
\end{equation}
where $f\in C(\mathbb{R}^+,\mathbb{R}^+)$, $\xi_i\in (0,1)$ with
$0<\xi_1<\xi_2<\dots<\xi_{m-2}<1$, $a_i,\ b_i\in \mathbb{R}^+$ with
$0<\sum_{i=1}^{m-2}a_i<1$, and
$0<\sum_{i=1}^{m-2}b_i<1$.
Set
$$
 f_0=\lim_{u\to 0^+}\frac{f(u)}{u},\quad
 f_{\infty}=\lim_{u\to+\infty}\frac{f(u)}{u}.
$$
 Then $f_0=0$ and $f_{\infty}=\infty$ correspond to the
 super-linear case, and $f_0=\infty$ and $f_{\infty}=0$ correspond to the
 sub-linear case. By applying the fixed point theorem in
 cones, Ruyun Ma \cite{m1} showed that the m-point boundary value
 problem  \eqref{e1.2} has at least one positive solution if $f$ is
 either super-linear or sub-linear.

 In this paper, we shall study the cases $f_0$, $f_\infty\not\in\{0,+\infty\}$.
In these cases, the m-point boundary-value problem
\eqref{e1.1} may have sign-changing solutions.  Quite recently, the
existence and qualitative properties of sign-changing solutions
for elliptic boundary-value problems have been extensively
studied. To the author's knowledge, however, there were fewer
papers
  considered the sign-changing solutions for multi-point boundary
  value problems. The purpose of this paper is to give some existence
  results  for multiple sign-changing solution for m-point boundary value
  problem \eqref{e1.1}. We shall
 follow the idea employed in \cite{l2} by Liu. To show the main result
 in this paper we need to study the the spectrum properties of
  the linear operator related the m-point boundary-value problem
\eqref{e1.1}.  Gupta and Sergej Trofimchuk \cite{g1} studied the
problem of  existence of solutions for the three-point boundary-value problem
\begin{equation}
\begin{gathered}
 x''(t)=f(t, x(t), x'(t)),\quad t\in (0,1),\\
 x(0)=0,\quad x(1)=\alpha x(\eta),
 \end{gathered}\label{e1.3}
\end{equation}
where $\alpha\in \mathbb{R}$, $\alpha\leq 1$ and $\eta\in(0,1)$ are
given. Using the spectrum radius of some related linear operators,
the authors proved some existence results for nontrivial
solutions of the three-point  boundary-value problem \eqref{e1.3}.

  We shall organize this paper as follows. In $\S2$ some
  preliminary results are given including the study of the
  eigenvalues of the linear operator $A'(\theta)$ and $A'(\infty)$. In $\S3$
  by using the fixed point index and Leray-Shauder degree method, we will prove the main
   result.

\section{Preliminary Lemmas}

From \cite[Theorem 2.3.1]{d1}, we have the following definition.
Let $X$ be a retract of real Banach space $E$, $U$ be a relatively
bounded open subset of $X$, $A:D\mapsto X$ be completely
continuous operator. The integer $i(A, U, X)$ be defined by
$$
i(A, U, X)=\deg(I-A\cdot r, B(\theta, R)\cap r^{-1}(U),\theta),
$$
where $r:E\mapsto X$ is an arbitrary retraction and
$R>0$ such that $B(\theta, R)\supset U$. Then the integer $i(A, U,X)$ be
called the fixed point index of $A$ on $U$ with respect to
$X$.

Set
$$
\beta_0=\lim_{x\to 0}\frac{f(x)}{x},\quad  \beta_1=\lim_{|x|\to \infty}\frac{f(x)}{x}.
$$
 Let us list some conditions to be used in this paper.
\begin{itemize}
\item[(H0)] Assume that the sequence of positive solutions of the equation
$$\sin
 \sqrt{x}=\sum_{i=1}^{m-2}\alpha_i \sin\eta_i\sqrt{x}$$
is  $\lambda_1<\lambda_2<\dots<\lambda_n<\lambda_{n+1}<\dots$.

\item[(H1)] $0<\sum_{i=1}^{m-2}\alpha_i<1$, $f\in C(\mathbb{R}, \mathbb{R})$,
  $f(0)=0$, $xf(x)> 0$
 for all $x\in \mathbb{R}\backslash\{0\} $.


\item[(H2)] There exist  positive integers $n_0$  and $n_1$ such that
$$
\lambda_{2n_0}<\beta_0<\lambda_{2n_0+1},\quad
\lambda_{2n_1}<\beta_1<\lambda_{2n_1+1}.
$$

\item[(H3)] There exists $C_0>0$ such that
 $$
 |f(x)|<\frac{2(1-\sum_{i=1}^{m-2}\alpha_i\eta_i)}
 {5-\sum_{i=1}^{m-2}\alpha_i\eta_i}C_0,
$$
for all $x$ with $|x|\leq C_0$.
\end{itemize}

  The main result of this paper is the following.

\begin{theorem} \label{thm2.1}
 Suppose that (H0)--(H3) hold. Then the m-point boundary-value problem \eqref{e1.1}
 has at least two sign-changing solutions. Moreover, the  m-point boundary-value
 problem \eqref{e1.1} also has at least
 two positive solutions and two negative solutions.
\end{theorem}

Before giving the proof of Theorem \ref{thm2.1}, we  list
some  preliminary lemmas.
Let
\begin{gather*}
E=\{x\in C^1[0,1] : x(0)=0,\; x(1)=\sum_{i=1}^{m-2}\alpha_i x(\eta_i)\}\\
P=\{x\in E: x(t)\geq 0 \mbox{ for }  t\in [0,1]\}.
\end{gather*}
For $x\in E$, let
$\|x\|=\|x\|_0+\|x'\|_0$,
where $\|x\|_0=\max_{t\in[0,1]}|x(t)|$ and
 $\|x'\|_0=\max_{t\in[0,1]}|x'(t)|$. It is easy to show that $E$ is a
 Banach space with
the norm $\|\cdot\|$ and  $P$ is a  cone of $E$. Let the operators
$K$, $F$ and $A$ be defined by
\begin{equation}
\begin{aligned}
(Kx)(t)&= \frac{t}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}\int^1_0(1-s)x(s)ds
-\int^t_0(t-s)x(s)ds\\
&\quad -\frac{t}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}
\sum_{i=1}^{m-2}\alpha_i\int^{\eta_i}_0(\eta_i-s)x(s)ds,\quad
t\in [0,1],\;x\in E,
\end{aligned}
\label{e2.1)}
\end{equation}
$(Fx)(t)=f(x(t))$ for $t\in [0,1]$, $x\in E$
and $A=KF$.

From \cite[Lemma 2.3.1]{d1}, we get the following Lemma.

\begin{lemma} \label{lm1}
 Let $\theta\in \Omega$
and $A:P\cap\bar{\Omega}\mapsto P$ be condensing. Suppose that
$$
Ax\neq \mu x, \quad  \forall x\in P\cap\partial\Omega,\; \mu\geq 1.
$$
Then $i(A, P\cap\Omega, P)=1$.
\end{lemma}

 From \cite[Corollary 2, p.p.146]{d2}, we have the following Lemma.

\begin{lemma} \label{lm2}
Let $\Omega$ be a open set in $E$ and
$\theta\in \Omega$, $A:\bar{\Omega}\mapsto E$ be completely
continuous. Suppose that
$$ \|Ax\|\leq\|x\|, \quad Ax\neq x, \;  \forall x\in \partial \Omega.
$$
Then $\deg(I-A, \Omega, \theta)=1$.
\end{lemma}

\noindent{\bf Remark}\;
 Obviously, Lemma \ref{lm2} can also be directly
obtained by the normality and homotopic invariance  property of
Leray-Schauder degree.

The following Lemma can be easily obtained.

\begin{lemma} \label{lm3}
Suppose that $\sum_{i=1}^{m-2} \alpha_i\eta_i<1$. If $u\in
 C[0,1]$, then $y\in C^2[0,1]$ is a solution the
 m-point boundary-value problem
 \begin{gather*}
 y''(t)+u(t)=0,\quad 0\leq t\leq 1,\\
 y(0)=0,\quad  y(1)=\sum_{i=1}^{m-2} \alpha_i y(\eta_i)
 \end{gather*}
if and only if $y\in C[0,1]$ is a solution of the
 integral equation
 $y(t)=(Ku)(t), t\in [0,1]$.
\end{lemma}

\noindent{\bf Remark} %2
By Lemma \ref{lm3} we can easily show that $A:E\mapsto  E$ is a completely continuous
operator.

\begin{lemma} \label{lm4}
Suppose that (H1) and (H2) hold.
 Then the operator $A$ is Fr\'echet differentiable at
 $\theta$ and $\infty$. Moreover, $A'(\theta)=\beta_0 K$,
 and $A'(\infty)=\beta_1 K.$
\end{lemma}

\begin{proof}
For any $\varepsilon>0$, by (H2) there
 exists $\delta>0$ such that for any $0<|x|<\delta$,
 $$
\big|\frac{f(x)}{x}-\beta_0\big|<\varepsilon,
$$
that is $|f(x)-\beta_0x|<\varepsilon |x|$, for all $0\leq|x|<\delta$.
Then, for any $x\in E$ with $\|x\|<\delta$, we have
 \begin{align*}
&|(Ax-A\theta-\beta_0 Kx)(t)|\\
&=|(K(Fx-\beta_0x))(t)|\\
&\leq\frac{1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}\int^1_0(1-s)\max_{s\in
  [0,1]}|f(x(s))-\beta_0x(s)|ds\\
&\quad +\int^1_0(1-s)\max_{s\in  [0,1]}|f(x(s))-\beta_0x(s)|ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}\sum_{i=1}^{m-2}
\alpha_i\int^{\eta_i}_0(\eta_i-s)\max_{s\in  [0,1]}|f(x(s))-\beta_0x(s)|ds\\
&\leq  \Big [\frac{1}{2(1-\sum_{i=1}^{m-2}\alpha_i\eta_i)}+
  \frac{1}{2}+\frac{\sum_{i=1}^{m-2}\alpha_i\eta_i^2}{2(1-
  \sum_{i=1}^{m-2}\alpha_i\eta_i)}\Big ]\|x\|_0\varepsilon\\
&\leq \frac{1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}\|x\|\varepsilon,t\in [0,1].
\end{align*}
This implies
\begin{equation}
\|Ax-A\theta-\beta_0 Kx\|_0\leq \frac{1}{1-\sum_{i=1}^{m-2}
\alpha_i\eta_i}\|x\|\varepsilon,\quad
x\in E, \; \|x\|<\delta.\label{e2.1}
\end{equation}
 Similarly, we can show that
for any $x\in E$, $\|x\|<\delta$,
$$
|(Ax-A\theta-\beta_0 Kx)'(t)|\leq
\frac{3-\sum_{i=1}^{m-2}\alpha_i\eta_i}{2(1-\sum_{i=1}^{m-2}\alpha_i\eta_i)}\|x\|
\varepsilon, \quad t\in [0,1]
$$
and so
\begin{equation}
\|(Ax-A\theta-\beta_0 Kx)'\|_0\leq
\frac{3-\sum_{i=1}^{m-2}\alpha_i\eta_i}{2(1-\sum_{i=1}^{m-2}\alpha_i\eta_i)}
\|x\|\varepsilon,\ x\in E, \ \|x\|<\delta.\label{e2.2}
\end{equation}
By \eqref{e2.1} and \eqref{e2.2}, we have
\begin{align*}
\|Ax-A\theta-\beta_0 Kx\|
&=\|Ax-A\theta-\beta_0 Kx\|_0+\|(Ax-A\theta-\beta_0 Kx)'\|_0\\
&\leq\frac{5-\sum_{i=1}^{m-2}\alpha_i\eta_i}{2(1
-\sum_{i=1}^{m-2}\alpha_i\eta_i)}\|x\|\varepsilon
\end{align*}
Consequently,
$$
\lim_{\|x\|\to 0}\frac{\|Ax-A\theta-\beta_0 Kx\|}{\|x\|}=0.
$$
This means that $A$ is Fr\'echet differentiable at $\theta$,
and $A'(\theta)=\beta_0K$.

For each $\varepsilon>0$, by (H2), there exists $R>0$ such that
$$ |f(x)-\beta_1x|<\varepsilon |x|$$
for $|x|>R$. Let $b=\max_{|x|\leq R}|f(x)-\beta_1 x|$.
Then we have for any $x\in \mathbb{R}$,
$$ |f(x)-\beta_1 x|\leq \varepsilon |x|+b.
$$
Consequently,
 \begin{align*}
&|(Ax-\beta_1 Kx)(t)|\\
&=|(K(Fx-\beta_1x))(t)|\\
&\leq\frac{1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}\int^1_0(1-s)\max_{s\in
  [0,1]}|f(x(s))-\beta_1x(s)|ds\\
&\quad +\int^1_0(1-s)\max_{s\in
  [0,1]}|f(x(s))-\beta_1x(s)|ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}
  \sum_{i=1}^{m-2}\alpha_i\int^{\eta_i}_0(\eta_i-s)\max_{s\in
  [0,1]}|f(x(s))-\beta_1x(s)|ds\\
&\leq  \Big[\frac{1}{2(1-\sum_{i=1}^{m-2}\alpha_i\eta_i)}+
  \frac{1}{2}+\frac{\sum_{i=1}^{m-2}\alpha_i\eta_i^2}{2(1-\sum_{i=1}^{m-2}
  \alpha_i\eta_i)}\Big](\varepsilon\|x\|_0+b)\\
&\leq \frac{1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}(\varepsilon\|x\|+b),\quad t\in [0,1].
\end{align*}
This implies
\begin{equation}
\|Ax-\beta_1 Kx\|_0\leq
\frac{1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}(\varepsilon\|x\|+b),\quad
x\in E.\label{e2.3}
\end{equation}
Similarly, we can show that
\begin{equation}
\|(Ax-\beta_1 Kx)'\|_0\leq
\frac{3-\sum_{i=1}^{m-2}\alpha_i\eta_i}{2(1-\sum_{i=1}^{m-2}\alpha_i\eta_i)}
(\varepsilon\|x\|+b),\quad
x\in E.\label{e2.4}
\end{equation}
By \eqref{e2.3} and \eqref{e2.4}, we have
\begin{align*}
\|Ax-\beta_1 Kx\|&=\|Ax-\beta_1
Kx\|_0+\|(Ax-\beta_1 Kx)'\|_0\\
&\leq\frac{5-\sum_{i=1}^{m-2}\alpha_i\eta_i}{2(1-\sum_{i=1}^{m-2}
\alpha_i\eta_i)}(\varepsilon\|x\|+b).
\end{align*}
Consequently,
$$
\lim_{\|x\|\to \infty}\frac{\|Ax-\beta_1 Kx\|}{\|x\|}=0\,.
$$
This means that $A$ is Fr\'echet
differentiable at $\infty$, and $A'(\infty)=\beta_1K$. The
proof is complete.
\end{proof}

\begin{lemma} \label{lm5}
Suppose that (H0) and (H1) hold.
Let $\beta$ be a positive number. Then the sequence of positive
eigenvalues of the operator $\beta K$ is
$$
\frac{\beta}{\lambda_1}>\frac{\beta}{\lambda_2}>\dots>\frac{\beta}{\lambda_n}\dots.
$$
Moreover, the positive eigenvalues $\frac{\beta}{\lambda_n}$
$(n=1,2,\dots)$ have algebraic multiplicity one.
\end{lemma}


\begin{proof}
Let $\bar{\lambda}$ be a positive eigenvalue of the linear
operator $\beta K$, and $y\in E\backslash\{\theta\}$ be an
eigenfunction corresponding to the eigenvalue $\bar{\lambda}$.  By
Lemma \ref{lm3}, we have
\begin{equation}
\begin{gathered}
 y''(t)+\frac{\beta}{\bar{\lambda}}y(t)=0,0\leq t\leq 1,\\
y(0)=0,\quad y(1)=\sum_{i=1}^{m-2}\alpha_i y(\eta_i).
\end{gathered} \label{e2.5}
\end{equation}
The auxiliary equation of the differential equation \eqref{e2.5} has
roots $\pm \sqrt{\frac{\beta}{\bar{\lambda}}}i$. Thus the
general solution of \eqref{e2.5} is of the form
$$
y(t)=C_1 \cos t\sqrt{\frac{\beta}{\bar{\lambda}}}+C_2
\sin t\sqrt{\frac{\beta}{\bar{\lambda}}},\quad t\in [0,1].
$$
Applying the condition $y(0)=0$, we obtain that $C_1=0$, and so the general
solution can be reduce to
$$
y(t)=C_2 \sin t\sqrt{\frac{\beta}{\bar{\lambda}}},\quad  t\in [0,1].
$$
Applying the second condition
$y(1)=\sum_{i=1}^{m-2}\alpha_i y(\eta_i)$, we obtain that
$$
\sin\sqrt{\frac{\beta}{\bar{\lambda}}}=
\sum_{i=1}^{m-2}\alpha_i\sin\eta_i\sqrt{\frac{\beta}{\bar{\lambda}}}.
$$
 Since the positive solutions of the equation
$\sin \sqrt{x}=\sum_{i=1}^{m-2}\alpha_i\sin \eta_i\sqrt{x}$ are
$0<\lambda_1<\lambda_2<\dots$,
then $\bar{\lambda}$ is one of the values
$$
\frac{\beta}{\lambda_1}>\frac{\beta}{\lambda_2}>\dots>\frac{\beta}{\lambda_n}\dots
$$
and the eigenfunction corresponding to the eigenvalue
$\frac{\beta}{\lambda_n}$ is
   $$
y_n(t)=C \sin t\sqrt{\lambda_n}, \quad t\in [0,1],
$$
 where $C$ is a nonzero constant. By ordinary method, we can show that any two
eigenfunctions  corresponding to the same
 eigenvalue $\frac{\beta}{\lambda_n}$ are merely nonzero constant multiples
of each other.
 Consequently,
\begin{equation}
\dim \ker(\frac{\beta}{\lambda_n}I- \beta K)=\dim \ker(I-\lambda_n K)=1.
\label{e2.6}
\end{equation}
Now we  show that
\begin{equation}
\ker(I-\lambda_nK)=\ker (I-\lambda_nK)^2.\label{e2.7}
\end{equation}
 Obviously, we  need to show only that
$$ \ker (I-\lambda_nK)^2\subset \ker(I-\lambda_nK).
$$
For any $y\in \ker (I-\lambda_nK)^2$,  $(I-\lambda_nK)y$ is an
 eigenfunction of linear operator $\beta K$ corresponding to the
 eigenvalue $\frac{\beta}{\lambda_n}$ if $(I-\lambda_nK)y\not =\theta$. Then
 there exists nonzero constant $\gamma$ such that
 $$
(I-\lambda_nK)y=\gamma \sin t\sqrt{\lambda_n}, \quad t\in [0,1].
$$
 By direct computation, we have
\begin{equation}
\begin{gathered}
y''(t)+\lambda_ny =-\lambda_n \gamma \sin t\sqrt{\lambda_n}, \quad t\in [0,1],\\
y(0)=0,\quad y(1)=\sum_{i=1}^{m-2}\alpha_i y(\eta_i).
\end{gathered} \label{e2.8}
\end{equation}
It is easy to see that the general solutions of \eqref{e2.8} is of the form
 \begin{align*}
 y(t)&=C_1\cos t\sqrt{\lambda_n}+C_2\sin t\sqrt{\lambda_n}+
 (\frac{\gamma t\sqrt{\lambda_n}}{2}-\frac{\gamma}{4}\sin
 2t\sqrt{\lambda_n})\cos t \sqrt{\lambda_n}\\
&\quad +\frac{\gamma}{4}\cos
 2t\sqrt{\lambda_n}\cdot \sin t\sqrt{\lambda_n},\quad t\in [0,1],
 \end {align*}
 where $C_1$, $C_2$ are two nonzero constants. Applying the condition  $y(0)=0$,
we obtain that  $C_1=0$. Since
$\sin\sqrt{\lambda_n}=\sum_{i=1}^{m-2}\alpha_i\sin \eta_i\sqrt{\lambda_n}$, then
we have
\begin{equation}
 \begin{aligned}
 y(1)&=C_2\sin \sqrt{\lambda_n}+
 (\frac{\gamma \sqrt{\lambda_n}}{2}-\frac{\gamma}{4}\sin
 2\sqrt{\lambda_n})\cos \sqrt{\lambda_n}+\frac{\gamma}{4}\cos
 2\sqrt{\lambda_n}\cdot \sin \sqrt{\lambda_n}\\
&=\sum_{i=1}^{m-2}\alpha_iC_2 \sin \eta_i\sqrt{\lambda_n}+
 \frac{\gamma\sqrt{\lambda_n}}{2}\cos\sqrt{\lambda_n}-
 \frac{\gamma}{2}\sum_{i=1}^{m-2}
 \alpha_i\sin\eta_i\sqrt{\lambda_n}\cos^2\sqrt{\lambda_n}\\
&\quad + \frac{\gamma}{4}\sum_{i=1}^{m-2}\alpha_i\cos2
 \sqrt{\lambda_n}\sin\eta_i\sqrt{\lambda_n},
\end{aligned}
 \label{e2.9}
\end{equation}
 and
\begin{equation}
 \begin{aligned}
  \sum_{i=1}^{m-2}\alpha_i y(\eta_i)
&= \sum_{i=1}^{m-2}\alpha_i C_2\sin \eta_i\sqrt{\lambda_n}+
 \sum_{i=1}^{m-2}(\frac{\gamma \alpha_i\eta_i\sqrt{\lambda_n}}{2}
 -\frac{\gamma\alpha_i}{4}\sin
 2\eta_i\sqrt{\lambda_n})\cos \eta_i\sqrt{\lambda_n}\\
&\quad +\sum_{i=1}^{m-2}\frac{\gamma\alpha_i}{4}\cos
 2\eta_i\sqrt{\lambda_n}\cdot \sin \eta_i\sqrt{\lambda_n}.\
\end{aligned}  \label{e2.10}
\end{equation}
Since $y(1)=\sum_{i=1}^{m-2}\alpha_i y(\eta_i)$, by \eqref{e2.9}
and \eqref{e2.10}, we have
$$
\cos\sqrt{\lambda_n}=\sum_{i=1}^{m-2}\alpha_i\eta_i\cos\eta_i\sqrt{\lambda_n}.
$$
By the Schwarz inequality, we obtain
\begin{align*}
1-\sin ^2\sqrt{\lambda_n}
&=(\sum_{i=1}^{m-2}\alpha_i\eta_i\cos\eta_i\sqrt{\lambda_n})^2\\
&\leq (\sum_{i=1}^{m-2}\eta_i^2)(\sum_{i=1}^{m-2}\alpha_i^2\cos^2\eta_i\sqrt{\lambda_n})\\
&=(\sum_{i=1}^{m-2}\eta_i^2)(\sum_{i=1}^{m-2}\alpha_i^2)-(\sum_{i=1}^{m-2}
\eta_i^2)(\sum_{i=1}^{m-2}\alpha_i^2\sin^2\eta_i\sqrt{\lambda_n}).
\end{align*}
Applying  the condition $\sin
\sqrt{\lambda_n}=\sum_{i=1}^{m-2}\alpha_i\sin\eta_i\sqrt{\lambda_n}$,
we obtain
\begin{align*}
1&\leq
(\sum_{i=1}^{m-2}\eta_i^2)(\sum_{i=1}^{m-2}\alpha_i^2)+(\sum_{i=1}^{m-2}\alpha_i\sin\eta_i\sqrt{\lambda_n})^2-(\sum_{i=1}^{m-2}\eta_i^2)
(\sum_{i=1}^{m-2}\alpha_i^2\sin^2\eta_i\sqrt{\lambda_n})\\
&=(\sum_{i=1}^{m-2}\eta_i^2)(\sum_{i=1}^{m-2}\alpha_i^2)
+(1-(\sum_{i=1}^{m-2}\eta_i^2))(
\sum_{i=1}^{m-2}\alpha_i^2\sin^2\eta_i\sqrt{\lambda_n})
\\
&+\sum_{i\not =j}\alpha_i\alpha_j\sin\eta_i\sqrt{\lambda_n}\sin\eta_j\sqrt{\lambda_n}\\
&\leq(\sum_{i=1}^{m-2}\eta_i^2)(\sum_{i=1}^{m-2}\alpha_i^2)+
(1-(\sum_{i=1}^{m-2}\eta_i^2))(\sum_{i=1}^{m-2}\alpha_i^2)
+\sum_{i\neq j}\alpha_i\alpha_j\\
&=(\sum_{i=1}^{m-2}\alpha_i)^2,
\end{align*}
which is  a contradiction of $\sum_{i=1}^{m-2}\alpha_i<1$.
Thus, \eqref{e2.7} holds. It follows from \eqref{e2.6} and \eqref{e2.7} that the
algebraic  multiplicity of the eigenvalue
$\frac{\beta}{\lambda_n}$ is 1. The proof is complete.
\end{proof}

\begin{lemma} \label{lm6}
Suppose that (H0) and (H1) hold and $y\in P\backslash\{\theta\}$ is a solution
of the boundary-value problem \eqref{e1.1}. Then $y\in \stackrel{\circ}{P}$.
\end{lemma}

\begin{proof} Since $y''(t)=-f(y(t))\leq 0$ for
$t\in [0,1]$, then $y$ is a concave function on $[0,1]$. For all
$i\in\{1,2,\dots, m-2\}$, we have from the concavity of $y$ that
$$
 y(t)\leq \frac{y(1)-y(\eta_i)}{1-\eta_i}(t-1)+y(1), \quad t\in [0,\eta_1]
$$
that is
 $ y(t)(1-\eta_i)\leq (y(1)-y(\eta_i))(t-1)+y(1)(1-\eta_i)$,
 $t\in [0,\eta_1]$.
 This together with the boundary condition
$y(1)=\sum_{i=1}^{m-2}\alpha_iy(\eta_i)$ implies
\begin{equation}
\begin{aligned}
y(t)&\leq
y(1)\frac{\sum_{i=1}^{m-2}\alpha_i(1-\eta_i)+
(1-\sum_{i=1}^{m-2}\alpha_i)(1-t)}{\sum_{i=1}^{m-2}\alpha_i(1-\eta_i)} \\
&\leq y(1)\frac{\sum_{i=1}^{m-2}\alpha_i(1-\eta_i)+
(1-\sum_{i=1}^{m-2}\alpha_i)}{\sum_{i=1}^{m-2}
\alpha_i(1-\eta_i)}\\
 &= y(1)\frac{1-\sum_{i=1}^{m-2}\alpha_i\eta_i
}{\sum_{i=1}^{m-2}\alpha_i(1-\eta_i)},\quad   t\in [0,\eta_1].
\end{aligned} \label{e2.11}
\end{equation}
 From the concavity of $y$ and this inequality, we have
\begin{equation}
y(t)\leq \frac{y(\eta_1)}{\eta_1}t\leq
\frac{y(\eta_1)}{\eta_1}\leq
y(1)\frac{1-\sum_{i=1}^{m-2}\alpha_i\eta_i
}{\sum_{i=1}^{m-2}\alpha_i(1-\eta_i)\eta_1} , t\in
[\eta_1,1].  \label{e2.12)}
\end{equation}
 From  this inequality and \eqref{e2.11} it follows that
$$
y(1)\geq\frac{\sum_{i=1}^{m-2}\alpha_i(1-\eta_i)
\eta_1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}\|y\|_0.
$$
Since
$y$ is a concave function on [0,1], we have
\begin{equation}
y(t)\geq (y(1)-y(0))t=y(1)t\geq\frac{\sum_{i=1}^{m-2}
\alpha_i(1-\eta_i)
\eta_1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}\|y\|_0t, \quad
 t\in [0,1].\label{e2.13}
\end{equation}
Consequently,
$$
y'(0)=\lim_{t\to 0}\frac{y(t)}{t}\geq
\frac{\sum_{i=1}^{m-2}\alpha_i(1-\eta_i)
\eta_1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}\|y\|_0>0.
$$
Then there exist $\varepsilon>0$ and $\tau_1>0$ such that
\begin{equation}
y'(t)>\tau_1,\forall t\in [0,\varepsilon].\label{e2.14}
\end{equation}
By \eqref{e2.13}, there exists $\tau_2>0$ such that
\begin{equation}
y(t)>\tau_2,\quad \forall t\in [\varepsilon,1]\label{e2.15}
\end{equation}
Let $\tau=\min\{\tau_1,\tau_2\}$. Then by \eqref{e2.14} and \eqref{e2.15}, we
obtain
$u(t)\geq 0$, $t\in[0,1]$ for any $u\in E$ with $\|u-y\|<\tau$.
Therefore, $B(y,\tau)\subset P$ and $y\in \stackrel{\circ}{P}$, where
$B(y,\tau)=\{x\in E:\|x-y\|<\tau\}$.
The proof is complete.
\end{proof}

 By \cite[Lemmas 2.3.7,  2.3.8]{d1}, we have the following Lemma.

\begin{lemma} \label{lm7}
Let $A:P\mapsto P$ be completely continuous,
Suppose that $A$ is differentiable at $\theta$ and
$\infty$ along $P$ and 1 is not an eigenvalue of
$A'_+(\theta)$ and  $A'_+(\infty)$ corresponding to a
positive eigenfunction.
\begin{itemize}
\item[(1)] If  $A'_+(\theta)$ has a positive eigenfunction
corresponding to an eigenvalue greater than 1,  and
$A\theta=\theta$. Then there exists $\tau>0$ such that
$i(A, P\cap B(\theta, r), P)=0$
for any $0<r<\tau$.


\item[(2)] If  $A'_+(\infty)$ has a positive eigenfunction
which corresponds to an eigenvalue greater than 1. Then there exists
$\varsigma>0$ such that
$i(A, P\cap B(\theta, R), P)=0$
for any $R>\varsigma$.
\end{itemize}
\end{lemma}

\begin{lemma} \label{lm8}
Suppose that (H0)--(H3) hold. Then
\begin{itemize}
\item[(1)] There exists $C_0>r_0>0$ such that for any $0<r\leq r_0$,
$$i(A, P\cap B(\theta,r),P)=0,\quad
i(A, -P\cap B(\theta,r),-P)=0$$

\item[(2)] There exists $R_0>C_0$ such that for any $R\geq R_0$,
$$i(A, P\cap B(\theta,R),P)=0,\quad
i(A, -P\cap B(\theta,R),-P)=0.$$
\end{itemize}
\end{lemma}

\begin{proof}  We  prove only  conclusion (1). The same
way, conclusion (2) can be proved. First we claim that
$K(P)\subset P$ and $K(-P)\subset -P$. Let $x\in P$ be fixed and
$y= Kx$. Obviously, $y\in C^1[0,1]$. By direct computation, we
have
\begin{equation}
\begin{aligned}
y(1)&=\frac{1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}
(\sum_{i=1}^{m-2}\eta_i\int^1_0(1-s) x(s)ds
-\sum_{i=1}^{m-2}\alpha_i\int^{\eta_i}_0(\eta_i-s) x(s)ds)\\
&\geq \frac{1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}\sum_{i=1}^{m-2}\alpha_i
\int^{\eta_i}_0(1-\eta_i)sx(s)\,ds
\geq 0\,.
\end{aligned}
 \label{e2.16}
\end{equation}
It follows from Lemma \ref{lm3} that
\begin{gather}
y''(t)=-x(t)\leq 0,\quad \forall t\in [0,1].\label{e2.17}\\
y(0)=0, \quad  y(1)=\sum_{i=1}^{m-2}\alpha_i y(\eta_i)\label{e2.18}
\end{gather}
By \eqref{e2.17}, we see that $y$ is a concave function on [0,1]. Then
the boundary condition \eqref{e2.16} and \eqref{e2.18} mean that $y(t)\geq 0$
for $t\in [0,1]$.  Therefore, $y\in P$, and so $K(P)\subset P$,
$K(-P)\subset (-P)$. Since $xf(x)> 0$ for $x\in \mathbb{R}\backslash\{0\}$, then we see that $A(P)\subset P$ and
$A(-P)\subset (-P)$.

It follows from Lemmas \ref{lm4} and \ref{lm5} that $A'_+(\theta)=\beta_0 K$,
$\beta_0/\lambda_1$ $(>1)$ is an eigenvalue of
the linear operator $\beta_0 K$ and the eigenfunction
corresponding to $\frac{\beta_0}{\lambda_1}$ is
$$
y(t)=C\sin t\sqrt{\lambda_1},\quad t\in [0,1],
$$
where $C$ is an arbitrary positive  constant and $\lambda_1$ is
the smallest positive solution of the equation
$\sin\sqrt{x}=\sum_{i=1}^{m-2}\alpha_i \sin
\eta_i\sqrt{x}$.
Since
$$
\lim_{x\to 0}\frac{\sin\sqrt{x}-\sum_{i=1}^{m-2}\alpha_i \sin
\eta_i\sqrt{x}}{\sqrt{x}}=1-\sum_{i=1}^{m-2}\alpha_i\eta_i>0\,,
$$
there exists $\delta_0\in (0,1)$ small enough such that
$$
\frac{\sin\sqrt{\delta_0}-\sum_{i=1}^{m-2}\alpha_i \sin
\eta_i\sqrt{\delta_0}}{\sqrt{\delta_0}}
\geq\frac{1}{4}(1-\sum_{i=1}^{m-2}\alpha_i\eta_i)>0.
$$
On the other hand,
$$
\sin\sqrt{\pi^2}-\sum_{i=1}^{m-2}\alpha_i
\sin \eta_i\sqrt{\pi^2}=-\sum_{i=1}^{m-2}\alpha_i \sin
\eta_i\pi<0.
$$
Then, by the intermediate-value principle,
$\lambda_1\in (\delta_0, \pi^2)$. Consequently,
$$
y(t)=C\sin t\sqrt{\lambda_1}\geq 0,\quad  t\in [0,1].
$$
It follows from Lemma \ref{lm7} that there exists $\tau_0>0$ such that
$i(A, P\cap B(\theta, r),P)=0$
for any $0<r\leq \tau_0$.

Similarly, we can show that there exists $\tau_1>0$ such that
$i(A, -P\cap B(\theta, r),-P)=0$ for any $0<r\leq \tau_1$. Let
$r_0=\min\{\tau_0,\tau_1\}$. Then the conclusion (1) holds and the
the proof is complete.
\end{proof}


From \cite[Theorems 21.6, 21.2]{k1}, we have the following two
lemmas.

\begin{lemma}  \label{lm9}
Let $A$ be a completely continuous operator, let $x_0\in E$ be a fixed
point of $A$ and assume that $A$ is defined in a neighborhood of $x_0$ and
$Fr\acute{e}chet$ differentiable at $x_0$. If $1$ is not an
eigenvalue of the linear operator $A'(x_0)$, then $x_0$ is
an isolated singular point of the completely continuous vector
field $I-A$ and for small enough $r>0$
$$
\deg (I-A, B(x_0, r), \theta)=(-1)^k,
$$
where $k$ is the sum of the algebraic multiplicities of the real
eigenvalues of $A'(x_0)$ in $(1, +\infty)$.
\end{lemma}

\begin{lemma} \label{lm10}
Let $A$ be a completely continuous operator which
is defined on all $E$. Assume that 1 is not an eigenvalue of the
asymptotic derivative. The  completely continuous vector field
$I-A$ is then nonsingular on spheres $S_\rho=\{x|\|x\|=\rho\}$ of
sufficiently large radius $\rho$ and
$$
\deg(I-A, B(\theta, \rho), \theta)=(-1)^k,
$$
where $k$ is the sum of the algebraic multiplicities of the real
eigenvalues of $A'(\infty)$ in $(1, +\infty)$.
\end{lemma}

 \section {Proof of main Theorem}

\begin{proof}[Proof of Theorem \ref{thm2.1}]
From Lemma \ref{lm3}, a function  $y$ is a solution of the
boundary-value problem \eqref{e1.1} if and only if $y$ is a fixed
point of the operator $A$. By (H3), we have for any $x\in E$,
$\|x\|=C_0$,
\begin{align*}
&|(Ax)(t)|\\
&\leq\frac{1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}\int^1_0(1-s)\max_{s\in
  [0,1]}|f(x(s))|ds
 +\int^1_0(1-s)\max_{s\in  [0,1]}|f(x(s))|ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}\alpha_i\eta_i}
  \sum_{i=1}^{m-2}\alpha_i\int^{\eta_i}_0(\eta_i-s)\max_{s\in
  [0,1]}|f(x(s))|ds\\
&<\frac{2(1-\sum_{i=1}^{m-2}\alpha_i\eta_i)}{5-\sum_{i=1}^{m-2}\alpha_i\eta_i}
  \Big(\frac{1}{2(1-\sum_{i=1}^{m-2}\alpha_i\eta_i)}
  +\frac{1}{2}+\frac{\sum_{i=1}^{m-2}\alpha_i\eta_i}
  {2(1-\sum_{i=1}^{m-2}\alpha_i\eta_i)}\Big)C_0\\
&\leq \frac{2C_0}{5-\sum_{i=1}^{m-2}\alpha_i\eta_i},\quad t\in [0,1]\,.
\end{align*}
Therefore,
\begin{equation}
\|Ax\|_0<\frac{2C_0}{5-\sum_{i=1}^{m-2}\alpha_i\eta_i} .\label{e3.1}
\end{equation}
Similarly, we can show that for any $x\in E$, with $\|x\|=C_0$,
\begin{equation}
\|(Ax)'\|_0<\frac{3-\sum_{i=1}^{m-2}\alpha_i\eta_i}
{5-\sum_{i=1}^{m-2}\alpha_i\eta_i}C_0.\label{e3.2}
\end{equation}
It follows from \eqref{e3.1} and \eqref{e3.2} that
$\|Ax\|<C_0$, for all $\| x\|=C_0$.
 Then, by Lemmas \ref{lm1} and \ref{lm2} we have
\begin{equation}
i(A, P\cap B(\theta, C_0), P)=1,\label{e3.3}\end{equation}
\begin{equation}i(A,-P\cap B(\theta, C_0), -P)=1,\label{e3.4}\end{equation}
\begin{equation}\deg(I-A, B(\theta, C_0),\theta)=1.\label{e3.5}\end{equation}
By (H2) and Lemma \ref{lm5}, the eigenvalues of the operator
$A'(\theta)=\beta_0 K$ which are large than 1 are
$$
\frac{\beta_0}{\lambda_1},\quad \frac{\beta_0}{\lambda_2},\quad
\frac{\beta_0}{\lambda_3}, \dots,\frac{\beta_0}{\lambda_{2n_0}}.
$$
Therefore, by Lemmas \ref{lm5}, \ref{lm8}, and \ref{lm9}, there exists
$0<r_1<r_0$ such that
\begin{equation}
\deg(I-A, B(\theta, r_1),\theta)=(-1)^{2n_0}=1\,.\label{e3.6}
\end{equation}
Similarly, by Lemmas \ref{lm5}, \ref{lm8} and \ref{lm10}, we have for some
$R_1\geq R_0$,
\begin{equation}
\deg(I-A, B(\theta, R_1),\theta)=1\,.\label{e3.7}
\end{equation}
By Lemma \ref{lm8}, we have
\begin{equation}i(A, P\cap B(\theta, r_1), P)=0,\label{e3.8}\end{equation}
\begin{equation}i(A, -P\cap B(\theta, r_1), -P)=0,\label{e3.9}\end{equation}
\begin{equation}i(A, P\cap B(\theta, R_1), P)=0,\label{e3.10}\end{equation}
\begin{equation}i(A, -P\cap B(\theta, R_1), -P)=0.\label{e3.11}\end{equation}
Then, by \eqref{e3.3}, \eqref{e3.8} and \eqref{e3.10}, we have
\begin{gather}
i(A, P\cap ( B(\theta, R_1)\backslash\overline{B(\theta, C_0)}),
P)=0-1=-1,\label{e3.12} \\
i(A, P\cap (B(\theta, C_0)\backslash\overline{B(\theta, r_1)}),
P)=1-0=1.\label{e3.13}
\end{gather}
Therefore, the operator $A$ has at least
two fixed points  $x_1\in P\cap (B(\theta,R_1)\backslash\overline{B(\theta, C_0)})$
and $x_2\in P\cap (B(\theta, C_0)\backslash\overline{B(\theta, r_1)})$,
respectively. Obviously, $x_1$ and $x_2$ are positive solutions of
the boundary-value problem \eqref{e1.1}.

Similarly, by \eqref{e3.4}, \eqref{e3.9} and \eqref{e3.11}, we have
\begin{gather}
i(A, -P\cap (B(\theta, R_1)\backslash\overline{B(\theta, C_0)}),
-P)=-1,\label{e3.14}\\
i(A, -P\cap (B(\theta, C_0)\backslash\overline{B(\theta, r_1)}),
-P)=1.\label{e3.15}
\end{gather}
Therefore, the operator $A$ has at least two fixed points  $x_3\in
(-P)\cap (B(\theta, C_0)\backslash\overline{B(\theta, r_1)})$ and
$x_4\in (-P)\cap (B(\theta, R_1)\backslash\overline{B(\theta,
C_0)})$, respectively. Obviously, $x_3$ and $x_4$ are negative
solutions of the boundary-value problem \eqref{e1.1}.

Let
$$
S=\{x|x=Ax, x\in P\cap (B(\theta, R_1)\backslash\overline{B(\theta,
C_0)})\}.
$$
It follows from Lemma \ref{lm6} that $S\subset \stackrel{\circ}{P}$. Therefore,
for any $x\in S$, there exists $\delta_x>0$ such that
$B(x,\delta_x)\subset P\cap (B(\theta,R_1)\backslash\overline{B(\theta, C_0)})$.
Let $O_1=\bigcup_{x\in S}B(x, \delta_x)$. Then, we have
$O_1\subset P\cap (B(\theta, R_1)\backslash\overline{B(\theta,C_0)})$.
By \eqref{e3.12} and the excision property of the fixed point
index, we have
\begin{equation}
i(A, O_1, P)=-1.\label{e3.16}
\end{equation}
By the definition of the fixed point index, we have
\begin{equation}
i(A, O_1, P)=\deg(I-A\cdot r, B(\theta, \bar{R})\cap
r^{-1}(O_1),\theta),\label{e3.17}
\end{equation}
where $r:E\mapsto P$ is an
arbitrary retraction and $\bar{R}$ is a large enough positive
number such that $O_1\subset B(\theta, \bar{R})$.  Now, we assume
that $y^*\in B(\theta,\bar{R})\cap r^{-1}(O_1)$ such that
$y^*=A\cdot r(y^*)$. Since $r:E\mapsto P$ and $A:P\mapsto P$, then
$y^*\in P$, and so $y^*=ry^*\in O_1$. Therefore,  $y^*\in O_1$
whenever $y^*\in B(\theta, \bar{R})\cap r^{-1}(O_1)$ is a fixed
point  of the operator $A\cdot
r$. Then, by the excision property of the degree we have
\begin{equation}
\deg(I-A\cdot r, B(\theta, \bar{R})\cap r^{-1}(O_1),\theta)
=\deg (I-A, O_1,\theta).\label{e3.18}
\end{equation}
By \eqref{e3.16}-\eqref{e3.18}, we have
\begin{equation}
\deg (I-A,
O_1,\theta)=-1.\label{e3.19}
\end{equation}
 Similarly, by \eqref{e3.13}-\eqref{e3.15}, we can
show that there exist open sets $O_2,\ O_3$ and $O_4$ such that
\begin{gather}
O_2\subset P\cap(B(\theta, C_0)\backslash\overline{B(\theta, r_1)}),\nonumber\\
O_3\subset -P\cap(B(\theta, C_0)\backslash\overline{B(\theta, r_1)}),\nonumber \\
O_4\subset -P\cap(B(\theta, R_1)\backslash\overline{B(\theta, C_0)}),\nonumber\\
\deg (I-A, O_2, \theta)=1,\label{e3.20}\\
\deg (I-A, O_3, \theta)=1,\label{e3.21} \\
\deg (I-A, O_4, \theta)=-1.\label{e3.22}
\end{gather}
It follows from \eqref{e3.5}, \eqref{e3.6}, \eqref{e3.20} and \eqref{e3.21} that
$$
\deg(I-A, B(\theta, C_0)\backslash(\overline{O_2}\cup\overline{O_3}
\cup\overline{B(\theta, r_1)}), \theta)=1-1-1-1=-2.
$$
This implies that $A$ has at least one fixed point $x_5\in B(\theta,
C_0)\backslash(\overline{O_2}\cup\overline{O_3}\cup\overline{B(\theta,
r_1)})$.
Similarly, by \eqref{e3.5}, \eqref{e3.7}, \eqref{e3.19} and \eqref{e3.22},
$$
\deg(I-A, B(\theta,R_1)\backslash(\overline{O_1}\cup\overline{O_4}
\cup\overline{B(\theta, C_0)}), \theta)=1-1+1+1=2.
$$
This implies  that $A$ has at least one fixed point $x_6\in
B(\theta,
R_1)\backslash(\overline{O_1}\cup\overline{O_4}\cup\overline{B(\theta,
C_0)}) $. Obviously, $x_5$ and $x_6$ are two distinct
sign-changing solutions of the boundary-value problem
\eqref{e1.1}. The proof is complete.
\end{proof}

By the method used in the proof of
Theorem \ref{thm2.1}, it is easy to show  the following four corollaries.

\begin{corollary} \label{coro1}
 Suppose that (H0), (H1) and (H3) hold, and that there exists positive integer
$n_0$ such that $\lambda_{2n_0}<\beta_0<\lambda_{2n_0+1}$.
  Then the boundary-value problem \eqref{e1.1} has at least one sign-changing solution.
 Moreover, the boundary-value problem \eqref{e1.1} has at least one positive solution and one
 negative solution.
\end{corollary}

\begin{corollary} \label{coro2}
Suppose that (H0), (H1) and (H3) hold, and that  there
exists positive integer $n_1$ such that
 $\lambda_{2n_1}<\beta_1<\lambda_{2n_1+1}$.
 Then the conclusion of Corollary \ref{coro1} holds.
\end{corollary}

\begin{corollary} \label{coro3}
Suppose that (H0) and (H1) hold,
$\beta_0>\lambda_1$, $\beta_1<\lambda_1$ (or $\beta_0<\lambda_1$,
$\beta_1>\lambda_1$).
    Then the boundary-value problem \eqref{e1.1} has at least one positive
solution and    one  negative solution.
\end{corollary}

\begin{corollary} \label{coro4}
Suppose that (H0), (H1) and (H3) hold,
  $\beta_0>\lambda_1$, $\beta_1>\lambda_1$.
Then the boundary-value problem \eqref{e1.1} has at least two positive
solutions and    two negative solutions.
\end{corollary}

\noindent{\bf Remark.}\;  %3.
In Theorem \ref{thm2.1}, we show not only the existence
of multiple sign-changing solutions, but also the existence of
multiple positive solutions and negative solutions. Obviously, we
can employ this method to show the existence of sign-changing
solutions for other nonlinear boundary-value problems.

\subsection*{Acknowledgment}
The author wishes to express his sincere gratitude to the
anonymous referees for their helpful suggestions in improving the paper.

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