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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 120, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2011/??\hfil Solvability of a second-order DE]
{Solvability of a second-order multi-point boundary-value
problems at resonance on a half-line with dim\,ker L=2 }

\author[W. Jiang, B. Wang, Z. Wang\hfil EJDE-2011/??\hfilneg]
{Weihua Jiang, Bin Wang, Zhenji Wang}  % in alphabetical order

\address{Weihua Jiang  \newline
College of Sciences, Hebei University of Science and Technology,
Shijiazhuang, 050018, Hebei, China}
\email{weihuajiang@hebust.edu.cn}

\address{Bin Wang \newline
Department of Basic Courses,
Hebei Professional and Technological College of Chemical and
Pharmaceutical Engineering, Shijiazhuang, 050026, Hebei,  China}
\email{wb@hebcpc.cn}


\address{Zhenji Wang \newline
Department of Basic Courses,
Hebei Professional and Technological College of Chemical and
Pharmaceutical Engineering, Shijiazhuang, 050026, Hebei,  China}
\email{wzj@hebcpc.cn}

\thanks{Submitted June 30, 2011. Published September 19, 2011.}
\subjclass[2000]{34B10, 34B15}
\keywords{Resonance; Fredholm operator; multi-point boundary-value problem; 
\hfill\break\indent coincidence degree theory}

\begin{abstract}
 We show the existence of solutions for a second-order multi-point
 boundary-value problem at resonance on a half-line, where the
 dimension of the kernel of the differential operator is 2.
 Our main tools are the coincidence degree theory due to Mawhin,
 suitable operators, and algebraic methods.
 Our results are illustrated with an example.
\end{abstract}

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


\section{Introduction}

 In this article, we show the existence of
solutions for the boundary-value problem
\begin{gather}
x''(t)=f(t,x(t),x'(t))+e(t),\quad t\in (0,+\infty),\label{e1.1}\\
x(0)=\sum_{i=1}^{m}\alpha_ix(\xi_i),\quad
\lim_{t\to +\infty}x'(t)=\sum_{j=1}^{n}\beta_jx'(\eta_j),\label{e1.2}
\end{gather}
where $f:[0,+\infty)\times R^2\to R$,  $e\in L^1[0,+\infty)$,
$0<\xi_1<\xi_2<\dots<\xi_m<+\infty$,
$0<\eta_1<\eta_2<\dots<\eta_n<+\infty$, $m\geq 2$, $n\geq 1$.

 Multi-point boundary value problems of
ordinary differential equations arise in a variety of different
areas of Applied Mathematics and Physics. For example, the
vibrations of a guy wire of a uniform cross-section being composed
of $N$ parts of different densities can be set up as a multi-point
boundary-value problem (see \cite{m5}); many problems in the theory of
elastic stability can be handled by the method of multi-point
problems(see \cite{t1}). Bridges of small size are often designed with
two supported points, which leads to a standard two-point boundary
condition and bridges of large size are sometimes contrived with
multi-point supports, which corresponds to a multi-point boundary
condition (see \cite{z3}).

Boundary-value problem \eqref{e1.1}-\eqref{e1.2} is called a
problem at resonance if $Lx:=x''(t)=0$ has non-trivial solutions
under the boundary condition \eqref{e1.2}; i.e., when
 $\dim\ker L\geq 1$.

On the finite interval [0,1], the first-order, second-order and
high-order multi-point boundary-value problems at resonance have
been studied by many authors (see
\cite{d1,f1,f2,g1,g2,g3,g4,k1,l3,l4,l5,l7,m1,m2,m3,n1,p1}),
where $\dim\ker L=1$. In \cite{k2,z1,z2}, the second-order multi-point
boundary-value problems at resonance have been discussed when
$\dim \ker L=2$ on the finite interval [0,1]. Recently, the
boundary-value problems at resonance on the infinite interval with
$\dim \ker L=1$ has been investigated by many authors, see
\cite{k3,l2,y1,l6}and references cited therein.
Although the existing literature on
solutions of multi-point boundary-value problems is quite wide, to
the best of our knowledge, there is few paper to investigate the
resonance case with $\dim \ker L =2$ on the infinite interval.

Motivated by the above results, by constructing the suitable
operators and getting help from the algebraic methods, we will
show the existence of solutions for the second-order multi-point
boundary-value problem at resonance on a half-line with
$\dim \ker L= 2$, which brings many difficulties.
 And we give an example to
illustrate our results. Some methods used in this paper are new
and they can be used to solve the $nth$-order boundary-value
problems at resonance with $1<\dim \ker L\leq n$.

This paper is organized as follows.
In Section 2, some necessary backgrounds will be stated
and some lemmas be proved. In Section 3, the main results
will be given and proved. In Section 4, an example is given
to illustrate our results.

In this article, we will assume the following conditions:

\begin{itemize}
\item[(C1)] $f:[0,+\infty)\times R^2\to R$ is a
$S$-Carath\'eodory function; i.e.,
\begin{itemize}
\item[(i)] $f(t,\cdot)$ is continuous on $R^2$ for a.e.
$t\in[0,+\infty)$.

\item[(ii)] $f(\cdot,x)$ is Lebesgue measurable on $[0,+\infty)$
for each $x\in\mathbb{R}^2$.

\item[(iii)]  For each $r>0$, there exists a function
$\varphi_r\in L^1[0,+\infty)$, $\varphi_r(t)\geq 0$,
$t\in[0,+\infty)$ satisfying $\int_0^{+\infty}s\varphi_r(s)ds<+\infty$
 such that
$$
|f(t,x)|\leq \varphi_r(t),\quad \text{a. e. }t\in
[0,+\infty),\;||x||<r.
$$
\end{itemize}

\item[(C2)] $\sum_{i=1}^m\alpha_i=\sum_{j=1}^n\beta_j=1$,
$\sum_{i=1}^m\alpha_i\xi_i=0$.

\item[(C3)] $\Delta=
\left|\begin{matrix} Q_1e^{-t}& Q_2e^{-t}\\
Q_1te^{-t}& Q_2te^{-t}
\end{matrix}\right|
:=\left|\begin{matrix} a_{11} & a_{12} \\
a_{21}& a_{22}
\end{matrix}\right| \neq 0$,
where
$$
Q_1y=\sum_{i=1}^{m}\alpha_i\int_{0}^{\xi_i}(\xi_i-s)y(s)ds,\quad
Q_2y=\sum_{j=1}^{n}\beta_j\int_{\eta_j}^{+\infty}y(s)ds.
$$
\end{itemize}

\section{Preliminary }

 To obtain our results, we introduce some
notation and two theorems.

 Let $X$ and $Y$ be real Banach spaces and let
$L:\operatorname{dom}(L)\subset X\to Y$ be a Fredholm operator with
index zero, $P:X\to X$, $Q:Y\to Y$ be projectors
such that
$$
\operatorname{Im}P=\ker L,\quad
\ker Q=\operatorname{Im}L,\quad
X=\ker L\oplus \ker P,\quad
Y=\operatorname{Im}L\oplus \operatorname{Im}Q.
$$
It follows that
$$
L|_{\operatorname{dom} L\cap \ker P}:
\operatorname{dom}L\cap \ker P\to \operatorname{Im}L
$$
is invertible. We denote the inverse by $K_{P}$.

 If $\Omega$ is an open bounded subset of
$X$, $\operatorname{dom}L\cap\overline{\Omega}\neq\emptyset$, the map
$N:X\to Y$ will be called $L$-compact on
$\overline{\Omega}$ if $QN(\overline{\Omega})$ is bounded and
$K_{P}(I-Q)N:\overline{\Omega}\to X$ is compact.

\begin{theorem}[cite{m4}] \label{thm2.1}
 Let $L:\operatorname{dom}L\subset X\to Y$ be a
Fredholm operator of index zero and $N:X\to Y$
$L$-compact on $\overline{\Omega}$. Assume that the following
conditions are satisfied:
\begin{itemize}
\item[(1)] $Lx\neq \lambda Nx$ for every
$(x,\lambda)\in [(\operatorname{dom}L\setminus
\ker L)\cap\partial\Omega]\times(0,1)$;

\item[(2)] $Nx\notin \operatorname{Im}L$ for every
 $x\in \ker L\cap\partial\Omega$;

\item[(3)] $deg(QN|_{\ker L},~\Omega\cap \ker L,0)\neq0$, where
$Q:Y\to Y$ is a projection such that $\operatorname{Im}L=\ker Q$.

\end{itemize}
Then the equation $Lx=Nx$ has at least one solution in
$\operatorname{dom}L\cap\overline\Omega$.
\end{theorem}

Let
$$
X=\{x\in C^1[0,+\infty):\lim_{t\to
+\infty}\frac{|x(t)|}{1+t}~{\rm and
}~\lim_{t\to +\infty} x'(t)~{\rm exist}\}
$$
with norm
$\|x\|=\max\{\|x\|_0,~\|x'\|_\infty\}$, where
$$
\|x\|_0=\sup_{t\in[0,\infty)}\frac{|x(t)|}{1+t},\quad
\|x\|_\infty=\sup_{t\in[0,+\infty)}|x(t)|.
$$
It is easy to prove that $(X,\|\cdot\|)$ is a Banach space.

\begin{theorem}[\cite{a1}] \label{thm2.2}
 Let $M\subset X$. Then $M$ is relatively
compact if the following conditions hold:
\begin{itemize}
\item[(a)] $M$ is bounded in $X$;

\item[(b)] the functions belonging to $M$ are equi-continuous on any
compact interval of $R^+$;

\item[(c)] the functions from $M$ are equi-convergent at $+\infty$.
\end{itemize}
\end{theorem}

Let $Y=L^1[0,+\infty)$ with the norm
$\|y\|_1=\int_{0}^{+\infty}|y(s)|ds$.
Define $Lx=x''$, with domain
$$
\operatorname{dom}L
=\{x\in X:x''\in L^1[0,+\infty),x(0)=\sum_{i=1}^{m}\alpha_ix(\xi_i),
\lim_{t\to+\infty}x'(t)=\sum_{j=1}^{n}\beta_jx'(\eta_j)\}.
$$
Obviously, $\ker L=\{a+bt:a,b\in\mathbb{R}\}$. Now, we will prove that
$$
\operatorname{Im}L=\{y\in Y: Q_1y=Q_2y=0\}.
$$
In fact, if $Lx=y$, then $y\in Y$ and
$$
x(t)=x(0)+x'(0)t+\int_{0}^{t}(t-s)y(s)ds.
$$
It follows from \eqref{e1.2} that $Q_1y=Q_2y=0$.

On the other hand, assume $y\in Y$ satisfying $Q_1y=Q_2y=0$. Take
$$
x(t)=\int_0^t(t-s)y(s)ds.
$$
Then $x\in X,$ $x''(t)=y(t)$ and $x$ satisfies \eqref{e1.2}. So,
$x\in \operatorname{dom}L$; i.e., $y\in \operatorname{Im}L$.

Define operators $T_1,~T_2:Y\to Y$ as follows:
$$
T_1y=\frac{1}{\Delta}(\Delta_{11}Q_1y+\Delta_{12}Q_2y)e^{-t},\quad
T_2y=\frac{1}{\Delta}(\Delta_{21}Q_1y+\Delta_{22}Q_2y)e^{-t},
$$
where $\Delta_{ij}$ is the algebraic cofactor of $a_{ij}$.
Define the operator $Q:Y\to Y$ by
$$
Qy=T_1y+(T_2y)\cdot t.
$$
By a simple calculation, we obtain
$T_1(T_1y)=T_1y$,
$T_1(T_2y t)=0,~T_2(T_1y)=0$,
$T_2(T_2yt)=T_2y$. So,
$Q^2y=Qy$; i.e., $Q:Y\to Y$ is a linear projector.
Obviously, $Q$ is continuous.

For $y\in Y$, $y=(y-Qy)+Qy$, we have $Qy\in \operatorname{Im}Q$
and $Q(y-Qy)=0$. It follows from $Q(y-Qy)=0$, the definitions
of $Q$, $T_1$, $T_2$ and
condition (C3), that $Q_1(y-Qy)=Q_2(y-Qy)=0$; i.e.,
$y-Qy\in \operatorname{Im}L$. So,
$Y=\operatorname{Im}L+\operatorname{Im}Q$.
Take $y\in \operatorname{Im}L\cap \operatorname{Im}Q$, then
$y=Qy=0$; i.e., $Y=\operatorname{Im}L\oplus \operatorname{Im}Q$,
So, we have
$\dim \ker L=\operatorname{codim}\operatorname{Im}L=2$, thus $L$
is a Fredholm operator with index zero.

Define the continuous projection $P:X\to \ker L$ by
$$
(Px)(t)=x(0)+x'(0)t,\quad t\in[0,+\infty).
$$
Then $X=\ker L\oplus \ker P$.

Define the operator
$K_P:\operatorname{Im}L\to \operatorname{dom}L\cap \ker P$ by
$$
K_Py=\int_0^t(t-s)y(s)ds.
$$
Then $K_P$ is the inverse operator of
$L|_{\operatorname{dom}L\cap \ker P}$ and
\begin{equation}
\|K_Py\|\leq \|y\|_1. \label{e2.1}
\end{equation}
In fact, for $x\in \operatorname{dom}L\cap
\ker P$, $K_PL(x)=\int_0^t(t-s)x''(s)ds=x(t)$. On the other hand, for
$y\in \operatorname{Im}L, LK_P(y)=(\int_0^t(t-s)y(s)ds)''=y(t)$.
By
\begin{gather*}
\frac{|K_Py|}{1+t}\leq \int_0^{+\infty}|y(s)|ds=\|y\|_1, \\
|(K_Py)'(t)|=|\int_0^ty(s)ds|\leq \|y\|_1,
\end{gather*}
we obtain \eqref{e2.1}.

Let the nonlinear operator $N:X\to Y$ be defined by
$$
Nx=f(t,x(t),x'(t))+e(t),\quad t\in[0,+\infty).
$$
Then  problem \eqref{e1.1}--\eqref{e1.2} is equivalent  to
$$
Lx=Nx,\quad x\in \operatorname{dom}L.
$$

\begin{lemma} \label{lem2.1}
Suppose that $\Omega$ is an open bounded subset
of $X$ such that $\operatorname{dom}L\cap\overline{\Omega}\neq\Phi$.
Then $N$ is $L$-compact on $\overline{\Omega}$.
\end{lemma}

\begin{proof}
 Since $\Omega$ is bounded, there exists a constant
$r>0$ such that $\|x\|\leq r$ for any $x\in \overline{\Omega}$.
For $x\in \overline{\Omega}$, by (C1), we obtain
\begin{align*}
|Q_1Nx|
&=|\sum_{i=1}^{m}\alpha_i\int_{0}^{\xi_i}(\xi_i-s)[f(s,x(s),x'(s))
 +e(s)]ds|\\
&\leq\sum_{i=1}^{m}|\alpha_i\xi_i|\int_{0}^{+\infty}\varphi_r(s)
 +|e(s)|ds:=l_1
\end{align*}
and
\begin{align*}
|Q_2Nx|
&=|\sum_{j=1}^{n}\beta_j\int_{\eta_j}^{+\infty}f(s,x(s),x'(s))
+e(s)ds| \\
&\leq\sum_{j=1}^{n}|\beta_j|\cdot\int_{0}^{+\infty}\varphi_r(s)
+|e(s)|ds:=l_2.
\end{align*}
Thus,
\begin{equation}
\begin{split}
\|QNx\|_1
&=\int_{0}^{+\infty}|QNx(s)|ds\\
&\leq\int_{0}^{+\infty}|T_1Nx(s)|ds+\int_{0}^{+\infty}|T_2Nx(s)|sds\\
&\leq \frac{1}{|\Delta|}[|\Delta_{11}|\cdot|Q_1Nx|
 +|\Delta_{12}|\cdot|Q_2Nx|]\\
&\quad +\frac{1}{|\Delta|}[|\Delta_{21}|\cdot|Q_1Nx|
 +|\Delta_{22}|\cdot|Q_2Nx|]\\
&\leq\frac{1}{|\Delta|}[(|\Delta_{11}|+|\Delta_{21}|)l_1
+(|\Delta_{12}|
+|\Delta_{22}|)l_2].
\end{split} \label{e2.2}
\end{equation}
So, $QN(\overline{\Omega})$ is bounded.
Now, we will prove that $K_P(I-Q)N(\overline{\Omega})$ is compact.

(a). Obviously, $K_P(I-Q)N:\overline{\Omega}\to Y$ is
continuous. For $x\in\overline{\Omega}$, since
\begin{equation}
\|Nx\|_1=\int_{0}^{+\infty}|f(s,x(s),x'(s))
+e(s)|ds\leq\int_{0}^{+\infty}\varphi_r(s)
+|e(s)|ds:=l_3, \label{e2.3}
\end{equation}
\begin{align*}
\frac{|K_P(I-Q)N x(t)|}{1+t}
&=\frac{1}{1+t}|\int_0^t(t-s)(I-Q)Nx(s)ds|\\
&\leq\int_{0}^{+\infty}|Nx(s)|+|QNx(s)|ds\\
&=\|Nx\|_1+\|QNx\|_1,
\end{align*}
and
\begin{align*}
|[K_P(I-Q)N x]'(t)|
&=|\int_0^t(I-Q)Nx(s)ds|\\
&\leq\int_{0}^{+\infty}|Nx(s)|+|QNx(s)|ds\\
&=\|Nx\|_1+\|QNx\|_1,
\end{align*}
by \eqref{e2.2} and \eqref{e2.3}, we obtain that
$K_P(I-Q)N(\overline{\Omega})$ is bounded.

(b). For any $T\in[0,+\infty)$, we will prove that functions
belonging to $K_P(I-Q)N(\overline{\Omega})$ are equi-continuous on
$[0,T]$.
In fact, for $x\in\overline{\Omega}$, we have
\begin{gather}
|Nx(s)|\leq\varphi_r(s)+|e(s)|.\quad s\in[0,\infty),\label{e2.4}\\
|QNx(s)|\leq\frac{1}{|\Delta|}[(|\Delta_{11}|l_1+|\Delta_{12}|l_2)
+(|\Delta_{21}|l_1+|\Delta_{22}|l_2)s]e^{-s}. \label{e2.5}
\end{gather}
For any $t_1,t_2\in[0,T]$, $t_1<t_2$, we have
\begin{align*}
&\big|\frac{K_P(I-Q)N x(t_1)}{1+t_1}-\frac{K_P(I-Q)N x(t_2)}{1+t_2}\big|
\\
&=|\frac{\int_{0}^{t_1}(t_1-s)(I-Q)Nx(s)ds}{1+t_1}
 -\frac{\int_{0}^{t_2}(t_2-s)(I-Q)Nx(s)ds}{1+t_2}|\\
&\leq|\frac{t_1}{1+t_1}\int_{0}^{t_1}(I-Q)Nx(s)ds
 -\frac{t_2}{1+t_2}\int_{0}^{t_2}(I-Q)Nx(s)ds|\\
&\quad +|\frac{1}{1+t_1}\int_{0}^{t_1}s(I-Q)Nx(s)ds
 -\frac{1}{1+t_2}\int_{0}^{t_2}s(I-Q)Nx(s)ds|\\
&\leq|\frac{t_1}{1+t_1}-\frac{t_2}{1+t_2}|\cdot
 \int_{0}^{+\infty}|Nx(s)|+|QNx(s)|ds+\int_{t_1}^{t_2}|Nx(s)|
 +|QNx(s)|ds\\
&\quad +|\frac{1}{1+t_1}-\frac{1}{1+t_2}|\cdot
 T\int_{0}^{+\infty}|Nx(s)|+|QNx(s)|ds\\
&\quad +T\cdot\int_{t_1}^{t_2}|Nx(s)|
 +|QNx(s)|ds\\
&=(|\frac{t_1}{1+t_1}-\frac{t_2}{1+t_2}|
 +|\frac{1}{1+t_1}-\frac{1}{1+t_2}|\cdot T)(\|Nx\|_1+\|QNx\|_1)\\
&\quad +(1+T)\int_{t_1}^{t_2}|Nx(s)|+|QNx(s)|ds.
\end{align*}
and
\begin{align*}
|[K_P(I-Q)N x]'(t_1)-[K_P(I-Q)N x]'(t_2)|
&=|\int_{t_1}^{t_2}(I-Q)Nx(s)ds|\\
&\leq\int_{t_1}^{t_2}|Nx(s)|+|QNx(s)|ds.
\end{align*}
By \eqref{e2.2}--\eqref{e2.5}, the continuity of
$\frac{t}{1+t}$ and $\frac{1}{1+t}$ and the absolute continuity
of integral, we obtain that functions from
$K_P(I-Q)N(\overline{\Omega})$ are
equi-continuous on $[0,T]$.

(c). Now, we will show that functions in
$K_P(I-Q)N(\overline{\Omega})$ are equi-convergent at $+\infty$.
For $x\in\overline{\Omega}$, we have
\begin{gather*}
\lim_{t\to+\infty}\frac{K_P(I-Q)N x(t)}{1+t}
 =\int_{0}^{+\infty}(I-Q)Nx(s)ds.\\
\lim_{t\to+\infty}[K_P(I-Q)N x]'(t)=\int_{0}^{+\infty}(I-Q)Nx(s)ds.
\end{gather*}
By
\begin{align*}
&\big|\frac{K_P(I-Q)N x(t)}{1+t}-\int_{0}^{+\infty}(I-Q)Nx(s)ds\big|\\
&\leq|\frac{t}{1+t}\int_{0}^{t}(I-Q)Nx(s)ds
 -\int_{0}^{+\infty}(I-Q)Nx(s)ds\big|\\
&\quad +\frac{1}{1+t}\int_{0}^{t}|s(I-Q)Nx(s)|ds\\
&\leq\int_{t}^{+\infty}|(I-Q)Nx(s)|ds\\
&\quad +\frac{1}{1+t}\Big[\int_{0}^{+\infty}|(I-Q)Nx(s)|ds
 +\int_{0}^{+\infty}|s(I-Q)Nx(s)|ds\Big]\\
&\leq\int_{t}^{+\infty}|Nx(s)|+|QNx(s)|ds
 +\frac{1}{1+t}\int_{0}^{+\infty}(1+s)[|Nx(s)|+|QNx(s)|]ds,
\end{align*}
and
\[
|[K_P(I-Q)N x]'(t)-\int_{0}^{+\infty}(I-Q)Nx(s)ds|
\leq\int_{t}^{+\infty}|Nx(s)|+|QNx(s)|ds,
\]
From \eqref{e2.4} and \eqref{e2.5}, we can get that functions from
$K_P(I-Q)N(\overline{\Omega})$ are equi-continuous at $+\infty$.
By Theorem \ref{thm2.2}, we obtain that $K_P(I-Q)N(\overline{\Omega})$ is
compact. Therefore, $N$ is $L-$compact on $\overline{\Omega}$.
\end{proof}

\section{Main results}

The following theorem is our main result.

\begin{theorem} \label{thm3.1}
Assume that  {\rm (C1)--(C3)} and the following conditions hold:
\begin{itemize}
\item[(H1)] There exist functions
$\alpha(t),\beta(t),\gamma(t),\delta(t) \in
L^1[0,+\infty)$, and $\theta\in[0,1)$ such that either
$$
|f(t,u,v)|\leq\alpha(t)+\beta(t)\frac{|u|}{1+t}+\gamma(t)|v|
+\delta(t)(\frac{|u|}{1+t})^\theta
$$
or
$$
|f(t,u,v)|\leq\alpha(t)+\beta(t)\frac{|u|}{1+t}+\gamma(t)|v|
+\delta(t)|v|^\theta;
$$

\item[(H2)] There exist constants $A>0,~B>0$ such that, if $|x(t)|>A$
for every $t\in[0,B]$ or $|x'(t)|>A$ for every $t\in[0,+\infty)$,
then
either $Q_1Nx\neq0$ or $Q_2Nx\neq0$,
where
$\|\beta\|_1+\|\gamma\|_1<\frac{1}{2+B}$;

\item[(H3)] There exists a constant $C>0$ such that, if $|a|>C$ or
$|b|>C$, then either
\begin{itemize}
\item[(1)] $ aQ_1N(a+bt)+bQ_2N(a+bt)<0$, or

\item[(2)] $aQ_1N(a+bt)+bQ_2N(a+bt)>0$.
\end{itemize}
\end{itemize}
Then the boundary-value problem \eqref{e1.1}--\eqref{e1.2}
has at least one solution in $X$.
\end{theorem}


\begin{proof} We divide the proof into four steps.

\textbf{Step1.} Let
$$
\Omega_1=\{x\in \operatorname{dom}L\setminus \ker L:Lx=\lambda Nx,
\text{ for some }\lambda\in[0,1]\}.
$$
We will prove that $\Omega_1$ is bounded.
In fact, $x\in \Omega_1$ means $\lambda\neq 0$ and
$Nx\in \operatorname{Im}L$.
Thus
$$
Q_1Nx=Q_2Nx=0.
$$
By (H2), there exist $t_0\in[0,B]$, $t_1\in[0,+\infty)$ such that
$$
|x(t_0)|\leq A,\quad |x'(t_1)|\leq A.
$$
So,
$$
|x'(t)|=|x'(t_1)-\int_{t}^{t_1}x''(s)ds|
\leq A+\int_{t}^{t_1}|Nx(s)|ds\leq A+\|Nx\|_1;
$$
i.e., $\|x'\|_{\infty}\leq A+\|Nx\|_1$.
Considering
\begin{align*}
|x(0)|
&=|x(t_0)-\int_{0}^{t_0}x'(s)ds|
 \leq A+|\int_{0}^{t_0}x'(s)ds|\\
&\leq A+\|x'\|_{\infty}\cdot B
 \leq A(1+B)+B\cdot\|Nx\|_1,
\end{align*}
we have
$$
\|Px\|\leq |x(0)|+|x'(0)|\leq A(2+B)+(1+B)\|Nx\|_1.
$$
By $LPx=0$, \eqref{e2.1} and (H1), we obtain
\begin{align*}
\|x\|
&=\|Px+(I-P)x\|\leq \|Px\|+\|K_PL(I-P)x\|\\
&\leq\|Px\|+\|Lx\|_1\leq \|Px\|+\|Nx\|_1\\
&\leq(2+B)(A+\|Nx\|_1)\\
&\leq(2+B)(A+\|\alpha\|_1+\|\beta\|_1\cdot\|x\|
 +\|\gamma\|_1\cdot\|x\|+\|\delta\|_1\cdot\|x\|^{\theta}+\|e\|_1).
\end{align*}
So,
$$
\|x\|\leq\frac{2+B}{1-(2+B)(\|\beta\|_1+\|\gamma\|_1)}
(A+\|\alpha\|_1+\|e\|_1+\|\delta\|_1\cdot\|x\|^{\theta}).
$$
It follows from $\theta\in[0,1)$ that $\Omega_1$ is bounded.

\textbf{Step2.} Set $\Omega_2=\{x\in \ker L:Nx\in \operatorname{Im}L\}$.
Then $\Omega_2$ is bounded.
In fact, $x\in\Omega_2$ implies $x=a+bt$ and
$Q_1N(a+bt)=Q_2N(a+bt)=0$. By $(H_3)$, we obtain $|a|\leq C,~|b|\leq
C$. So, $\Omega_2$ is bounded.

\textbf{Step3.} Define the isomorphism $J:\ker L\to \operatorname{Im}Q$
by
$$
J(a+bt)=\frac{1}{\Delta}[\Delta_{11}a+\Delta_{12}b+(\Delta_{21}a
+\Delta_{22}b)t]e^{-t}.
$$
Assume (H3)(1) holds. Let
$$
\Omega_3=\{x\in \ker L:-\lambda Jx+(1-\lambda)QNx=0,
\text{ for some }\lambda\in[0,1]\}.
$$
Then $\Omega_3$ is bounded.

In fact, $x\in \Omega_3$ means that there exist constants
$a,b\in\mathbb{R}$, $\lambda\in[0,1]$ such that $x=a+bt$ and $\lambda
Jx=(1-\lambda)QNx$. If $\lambda=0$, then $QNx=0$. So,
\begin{gather*}
\Delta_{11}Q_1Nx+\Delta_{12}Q_2Nx=0,\\
\Delta_{21}Q_1Nx+\Delta_{22}Q_2Nx=0.
\end{gather*}
It follows from $\Delta\neq0$ that $Q_1Nx=Q_2Nx=0$. By (H3),
we obtain $|a|\leq C$, $|b|\leq C$.

If $\lambda=1$, we can similarly get $a=b=0$.
For $\lambda\in(0,1)$, by $\lambda Jx=(1-\lambda)QNx$, we obtain
\begin{gather*}
\lambda\Delta_{11}a+\lambda\Delta_{12}b
 =(1-\lambda)\Delta_{11}Q_1N(a+bt)+(1-\lambda)\Delta_{12}Q_2N(a+bt),\\
\lambda\Delta_{21}a+\lambda\Delta_{22}b
 =(1-\lambda)\Delta_{21}Q_1N(a+bt)+(1-\lambda)\Delta_{22}Q_2N(a+bt).
\end{gather*}
It follows from $\Delta\neq 0$ that
\begin{gather*}
\lambda a=(1-\lambda)Q_1N(a+bt),\\
\lambda b=(1-\lambda)Q_2N(a+bt).
\end{gather*}
If $|a|>C$, $|b|>C$, by (H3)(1), we obtain
$$
\lambda(a^2+b^2)=(1-\lambda)[aQ_1N(a+bt)+bQ_2N(a+bt)]<0,
$$
a contradiction. So, $\Omega_3$ is bounded.

\begin{remark} \label{rmk3.1} \rm
If (H3)(2) holds, take
$$
\Omega_3=\{x\in \ker L:\lambda Jx+(1-\lambda)QNx=0,
\text{ for some }\lambda\in[0,1]\}.
$$
We can similarly prove that $\Omega_3$ is bounded.
\end{remark}

\textbf{Step4.} Take an open bounded set
$\Omega\supset\bigcup_{i=1}^{3}\overline{\Omega_i}\bigcup\{0\}$.
We will prove that \eqref{e1.1}--\eqref{e1.2}
 has at least one solution in
$\operatorname{dom}L\cap\overline{\Omega}$.

By Step1 and Step2, we obtain
\begin{itemize}
 \item[(1)] $Lx\neq \lambda Nx,$ for every
 $(x,\lambda)\in[(\operatorname{dom}L\setminus \ker L)
 \cap\partial\Omega]\times(0,1)$;

\item[(2)] $Nx\not\in \operatorname{Im}L$, for every
$x\in \ker L\cap\partial\Omega$.
\end{itemize}
Now we will show that
\begin{itemize}
\item[(3)] $\deg(QN|_{\ker L}, \Omega\cap \ker L,~0)\neq0$.

\end{itemize}
Let $H(x,\lambda)=\pm\lambda Jx+(1-\lambda)QNx$. By step 3, we
know that $H(x,\lambda)\neq 0$, for every $(x,\lambda)\in
(\ker L\cap\partial\Omega)\times[0,1]$. Thus, by the homotopy
property of degree, we obtain
\begin{align*}
\deg(QN|_{\ker L},\Omega\cap \ker L,0)
&=\deg(H(\cdot,0),\Omega\cap \ker L,0)  \\
&=\deg(H(\cdot,1),\Omega\cap \ker L,0)\\
& =\deg(\pm J,~\Omega\cap \ker L,0) =\pm 1\neq 0.
\end{align*}

By Theorem \ref{thm2.1}, we can get that $Lx=Nx$ has at least one solution
in $\operatorname{dom}L\cap\overline{\Omega}$; i.e. ,
\eqref{e1.1}--\eqref{e1.2} has at least
one solution in $X$. The prove is completed.
\end{proof}

\section{Example}

Let's consider the  boundary-value problem
\begin{gather}
x''(t)=f(t,x(t),x'(t))+e(t),\quad t\in[0,\infty),\label{e4.1} \\
x(0)=2x(1)-x(2),\quad x'(\infty)=x'(2),\label{e4.2}
\end{gather}
where
\begin{gather*}
f(t,x(t),x'(t))=\begin{cases}
 -e^{-10t}x(0),& 0\leq t\leq 2,\\
e^{-10t}sinx'(t)+e^{-t}\sqrt[3]{x'(t)},& t>2.
\end{cases}
\\
e(t)=\begin{cases}
0, & 0\leq t\leq 2,\\
te^{-t}, & t>2.
\end{cases}
\end{gather*}
Corresponding to  problem \eqref{e1.1}-\eqref{e1.2}, we have
that
$m=2$, $n=1$, $\alpha_1=2$, $\alpha_2=-1$, $\xi_1=1$,
$\xi_2=2$, $\beta_1=1$, $\eta_1=2$.
Obviously, (C1) and (C2) are satisfied. By simple
calculation, we obtain
$a_{11}=-(1-e^{-1})^2$, $a_{21}=6e^{-1}-2-4e^{-2}$,
$a_{12}=e^{-2}$, $a_{22}=3e^{-2}$.
$$
\Delta= \left|\begin{matrix} a_{11} & a_{12} \\
a_{21}& a_{22}
\end{matrix}\right| =e^{-4}-e^{-2}\neq 0\,.
$$
So, (C3) is satisfied.
Take $\alpha(t)=0$, $\theta=\frac{1}{3}$,
\begin{gather*}
\beta(t)=\begin{cases}
(1+t)e^{-10t}, &0\leq t\leq 2,\\
0, & t>2,
\end{cases} \quad
 \gamma(t)=\begin{cases}
0, &0\leq t\leq 2,\\
e^{-10t},& t>2,
\end{cases}\\
\delta(t)=\begin{cases}
0, & 0\leq t\leq 2,\\
e^{-t},& t>2.
\end{cases}
\end{gather*}
Then $f$ satisfies (H1). We can easily get that
$\|\beta\|_1=\frac{1}{10}[\frac{11}{10}-\frac{31}{10}e^{-20}]$,
$\|\gamma\|_1=\frac{1}{10}e^{-20}$.
So, we have $\|\beta\|_1+\|\gamma\|_1<1/5$.

Let $B=2$, $A=e^{-54}/1000$. We get that
$Q_1Nx\neq 0$ if $|x(t)|>A$, for any $t\in[0,2]$ and
$Q_2Nx\neq 0$ if $|x'(t)|>A,$ for any $t\in[0,\infty)$.
This means that (H2) is satisfied.

Set $C=100$. We can easily get that
$$
aQ_1N(a+bt)+bQ_2N(a+bt)>0
$$
if $|a|>C$ or $|b|>C$. So, (H3) is satisfied.

By theorem \ref{thm3.1}, we obtain that problem \eqref{e4.1}--\eqref{e4.2}
has at least one solution.

\subsection*{Acknowledgments}
This work is supported by grants 10875094 and 60874003
from the Natural Science Foundation of China;
08M007, 11171088 and A2009000664 from the Natural Science Foundation of Hebei
Province; (2008153) from the Foundation of Hebei Education
Department; XL200814 from  the Foundation of Hebei University of
Science and Technology.

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