\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 114, 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 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/114\hfil Conjugate boundary-value problems]
{Solvability of (k,n-k) conjugate boundary-value problems at resonance}

\author[W. Jiang, J. Qiu \hfil EJDE-2012/114\hfilneg]
{Weihua Jiang, Jiqing Qiu}  % 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{Jiqing Qiu \newline
College of Sciences, Hebei University of Science and Technology,
Shijiazhuang, 050018, Hebei,  China}
\email{qiujiqing@263.net}

\thanks{Submitted May 1, 2012. Published July 5, 2012.}
\subjclass[2000]{35B34, 34B10, 34B15}
\keywords{Resonance; Fredholm operator; boundary-value problem}
 
\begin{abstract}
 Using the coincidence degree theory due to Mawhin and constructing
 suitable operators, we prove the existence of solutions for $(k,n-k)$
 conjugate boundary-value problems at resonance.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

The existence of solutions for $(k,n-k)$ conjugate
boundary-value problems at non-resonance has been studied in many
papers (see \cite{1,2,3,4,7,8,11,12,13,14,17,22,26,27,28,31,32,33}).
For example, using fixed point theorem in a cone, Jiang \cite{13} obtained
the existence of positive solutions for $(k,n-k)$ conjugate boundary-value
problem
\begin{gather*}
(-1)^{n-k}y^{(n)}(t)=f(t,y(t)),\quad 0<t<1,\\
y^{(i)}(0)=y^{(j)}(1)=0,\quad 0\leq i\leq k-1, \; 0\leq j\leq n-k-1,
\end{gather*}
where $f(t,y)$ may be singular at $y=0$, $t=0$, $t=1$. By using fixed
point index theory, Zhang and Sun \cite{33} studied the existence of
positive solutions for the problem
$$
(-1)^{n-k}\varphi^{(n)}(x)=h(x)f(\varphi(x)),\quad 0<x<1,\;
n\geq 2,\; 1\leq k\leq n-1,
$$
subject to the boundary conditions
$$
\varphi(0)=\sum_{i=1}^{m-2}a_i\varphi(\xi_i),\quad
\varphi^{(i)}(0)=\varphi^{(j)}(1)=0,\quad 1\leq i\leq k-1,\;
0\leq j\leq n-k-1,
$$
and
$$
\varphi(1)=\sum_{i=1}^{m-2}a_i\varphi(\xi_i),\quad
\varphi^{(i)}(0)=\varphi^{(j)}(1)=0,\;
0\leq i\leq k-1,\quad 1\leq j\leq n-k-1,
$$
respectively. Solvability of boundary-value problems
at resonance has been investigated
by many authors (see \cite{5,6,9,10,15,16,18,19,20,21,23,25,29,30,34}).
For example, in \cite{5}, using the coincidence degree
theory due to Mawhin, Du, Lin and Ge investigated the existence of
solutions for the $(n-1,1)$ boundary-value problems at resonance
\begin{gather*}
x^{(n)}(t)=f(t,x(t),x'(t),\dots,x^{(n-1)}(t))+e(t),
\quad\text{a.e. } t\in (0,1),\\
x(0)=\sum_{i=1}^{m-2}\alpha_ix(\xi_i),\quad
x'(0)=x''(0)=\dots=x^{(n-2)}(0)=0,\quad
x(1)=x(\eta).
\end{gather*}
Motivated by the  results in \cite{5,13,33}, in this
paper, we discuss the existence of solutions for the $(k,n-k)$
conjugate boundary-value problem at resonance
\begin{gather}
(-1)^{n-k}y^{(n)}(t)=f(t,y(t),y'(t),\dots,y^{(n-1)}(t))+\varepsilon(t),
 \quad \text{a.e. }t\in [0,1], \label{e1.1}\\
\begin{gathered}
y^{(i)}(0)=y^{(j)}(1)=0,\quad 0\leq i\leq k-1,\;
0\leq j\leq n-k-2,\\
y^{(n-1)}(1)=\sum_{i=1}^{m}\alpha_iy^{(n-1)}(\xi_i),
\end{gathered}\label{e1.2}
\end{gather}
where $1\leq k\leq n-1$, $0<\xi_1<\xi_2<\dots<\xi_{m}<1$.

As far as we know, this is the first paper to study the existence
of solutions for $(k,n-k)$ boundary-value problems at resonance
with $1\leq k\leq n-1$.

 In this paper, we assume the following conditions:
\begin{itemize}
\item[(H1)] $0<\xi_1<\xi_2<\dots<\xi_{m}<1$,
$\sum_{i=1}^{m}\alpha_i=1$,
$\sum_{i=1}^{m}\alpha_i\xi_i\neq 1$.

\item[(H2)] $\varepsilon(t)\in L^\infty[0,1]$,
$f:[0,1]\times \mathbb{R}^n \to \mathbb{R}$ satisfies
Carath\'aodory conditions; i.e., $f(\cdot,x)$ is
measurable for each fixed $x\in \mathbb{R}^n$, $f(t,\cdot)$ is
continuous for a.e. $t\in[0,1]$, and for each $r>0$, there exists
$\Phi_r\in L^\infty[0,1]$ such that
$|f(t,x_1,x_2,\dots,x_n)|\leq\Phi_r(t)$ for all
$|x_i|\leq r$, $i=1,2,\dots,n$, a.e. $t\in[0,1]$.
\end{itemize}

\section{Preliminaries}\label{s2}

First, we introduce some notation and state a
theorem to be used later. For more details see \cite{24}.

 Let $X$ and $Y$ be real Banach spaces and
$L:\operatorname{dom}L\subset X\to Y$ 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
$$
L\big|_{\operatorname{dom}L\cap \ker P}:\operatorname{dom}L\cap \ker
 P\to \operatorname{Im}L
$$
is invertible. We denote the inverse by $K_{P}$.

 Assume that $\Omega$ 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{24}] \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)] $\deg(QN|_{\ker L},\Omega\cap \ker L,0)\neq 0$, 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}

Take $X=C^{n-1}[0,1]$ with norm
$\|u\|=\max\{\|u\|_{\infty},\|u'\|_{\infty},\dots,\|u^{(n-1)}\|_{\infty}\}$,
where $\|u\|_{\infty}=\max_{t\in[0,1]}|u(t)|$, $Y=L^1[0,1]$
with norm $\|x\|_1=\int_0^1|x(t)|dt$. Define the operator
$Ly(t)=(-1)^{n-k}y^{(n)}(t)$ with
\begin{align*}
\operatorname{dom}L=\{&y\in X: y^{(n)}\in Y,\;
y^{(i)}(0)=y^{(j)}(1)=0,\; 0\leq i\leq k-1, \\
 & 0\leq j\leq n-k-2,\; y^{(n-1)}(1)=\sum_{i=1}^{m}\alpha_iy^{(n-1)}(\xi_i)\}.
\end{align*}
Let $N:X\to Y$ be defined as
$$
Ny(t)=f(t,y(t),y'(t),\dots,y^{(n-1)}(t))+\varepsilon(t),\quad t\in [0,1].
$$
Then problem \eqref{e1.1}, \eqref{e1.2} becomes  $Ly=Ny$.

\section{Main results}

By Cramer's rule, we can get the following lemmas.

\begin{lemma} \label{lem3.1}
 For given $u\in Y$, the system of linear equations
\begin{equation}
\begin{gathered}
 \frac{x_k}{k!}+\frac{x_{k+1}}{(k+1)!}+\dots+\frac{x_{n-2}}{(n-2)!}
 +\frac{(-1)^{n-k}}{(n-1)!}\int_0^1(1-s)^{n-1}u(s)ds=0\\
\frac{x_k}{(k-1)!}+\frac{x_{k+1}}{k!}+\dots+\frac{x_{n-2}}{(n-3)!}
 +\frac{(-1)^{n-k}}{(n-2)!}\int_0^1(1-s)^{n-2}u(s)ds=0\\
\dots\\
\begin{aligned}
&\frac{x_k} {[k-(n-k-2)]!}+\frac{x_{k+1}}{[k+1-(n-k-2)]!} +\dots+
\frac{x_{n-2}}{[n-2-(n-k-2)]!}\\
&+ \frac{(-1)^{n-k}}{[n-1-(n-k-2)]!}\int_0^1(1-s)^{k+1}u(s)ds=0
\end{aligned}
\end{gathered}\label{e3.1}
\end{equation}
has an only one solution, $(x_k,x_{k+1},\dots,x_{n-2})$ with
\begin{align*}
x_m&=\int_0^1\frac{(-1)^{n-k-1}m!}{(m-k)!(k-1)!(n-m-2)!}
 \sum_{i=0}^{m-k}(-1)^{m-k-i}\frac{C_{m-k}^i}{m-i}\\
&\quad \times \Big[\sum_{j=0}^{n-m-2}(-1)^{j}C_{n-m-2}^j
 \frac{(1-s)^{n-1-i-j}}{n-1-i-j}\Big]u(s)ds,\quad
 m=k,k+1,\dots,n-2.
\end{align*}
\end{lemma}

\begin{lemma} \label{lem3.2}
The system of linear equations
\begin{equation}
\begin{gathered}
 \frac{x_k}{k!}+\frac{x_{k+1}}{(k+1)!}+\dots+\frac{x_{n-2}}{(n-2)!}+\frac{1}{(n-1)!}=0
\\
\frac{x_k}{(k-1)!}+\frac{x_{k+1}}{k!}+\dots+\frac{x_{n-2}}{(n-3)!}+\frac{1}{(n-2)!}=0\\
\dots\\
\begin{aligned}
&\frac{x_k} {[k-(n-k-2)]!}+\frac{x_{k+1}}{[k+1-(n-k-2)]!} +\dots \\
&+\frac{x_{n-2}}{[n-2-(n-k-2)]!}+ \frac{1}{[n-1-(n-k-2)]!}=0
\end{aligned}
\end{gathered} \label{e3.2}
\end{equation}
 has an only one solution,  $(x_k,x_{k+1},\dots,x_{n-2})$ with
\begin{align*}
x_m&=-\frac{m!}{(m-k)!(k-1)!(n-m-2)!}\sum_{i=0}^{m-k}(-1)^{m-k-i}
 \frac{C_{m-k}^i}{m-i}\\
&\quad \times \Big(\sum_{j=0}^{n-m-2}(-1)^{j}C_{n-m-2}^j\frac{1}{n-1-i-j}\Big),
\quad m=k,k+1,\dots,n-2.
\end{align*}
\end{lemma}

Let $(B_k(u),B_{k+1}(u),\dots,B_{n-2}(u))$
 denote the only solution of \eqref{e3.1}, and let
$(A_k,A_{k+1},\dots,A_{n-2})$ denote the only solution
 of \eqref{e3.2}, and  let $A_{n-1}=1$.

 In order to obtain our main results, we firstly
present and prove the following lemmas.

\begin{lemma} \label{lem3.3}
Suppose {\rm (H1)} holds, then $L:\operatorname{dom}L\subset
X\to Y$ is a Fredholm operator of index zero and the
linear continuous projector $Q:Y\to Y$ can be defined as
$$
Qu=\frac{1}{1-\sum_{i=1}^{m}\alpha_i\xi_i}
\sum_{i=1}^{m}\alpha_i\int_{\xi_i}^1u(s)ds,
$$
and the linear operator $K_P:\operatorname{Im}L\to \operatorname{dom}L\cap \ker P$
can be written as
$$
K_Pu=\sum_{i=k}^{n-2}\frac{B_i(u)}{i!}t^{i}
+\frac{(-1)^{n-k}}{(n-1)!}\int_0^t(t-s)^{n-1}u(s)ds.
$$
\end{lemma}

\begin{proof}
  By simple calculations, we obtain that
$$
\ker L=\big\{y:y=c\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}t^{i}\Big),\; c\in \mathbb{R}\big\}.
$$
Define linear operator $P:X\to X$ as follows
$$
Py(t)=\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}t^{i}\Big)y^{(n-1)}(0).
$$
Obviously, $\operatorname{Im}P=\ker L$ and $P^2y=Py$.
For any $y\in X$, it follows
from $y=(y-Py)+Py$ that $X=\ker P+\ker L$. By simple calculation, we
can get that $\ker L\cap \ker P=\{0\}$. So, we have
\begin{equation}
X=\ker L\oplus \ker P.\label{e3.3}
\end{equation}
We will show that
$$
\operatorname{Im}L=\big\{u\in Y: \sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1u(s)ds=0 \big\}.
$$
In fact, if $u\in \operatorname{Im}L$, there exists $y\in \operatorname{dom}L$
 such that $u=Ly\in Y$. So, we have
$$
y=\sum_{i=k}^{n-1}\frac{c_i}{i!}t^{i}+\frac{(-1)^{n-k}}{(n-1)!}
\int_0^t(t-s)^{n-1}u(s)ds.
$$
Since $\sum_{i=1}^{m}\alpha_i=1$ and
$y^{(n-1)}(1)=\sum_{i=1}^{m}\alpha_iy^{(n-1)}(\xi_i)$, we have
$$
\int_0^1u(s)ds=\sum_{i=1}^{m}\alpha_i\int_0^{\xi_i}u(s)ds;
$$
i.e., $\sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1u(s)ds=0$.

 On the other hand, if $u\in Y$ satisfies
$\sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1u(s)ds=0$, we take
$$
y=\sum_{i=k}^{n-2}\frac{B_i(u)}{i!}t^{i}
+\frac{(-1)^{n-k}}{(n-1)!}\int_0^t(t-s)^{n-1}u(s)ds.
$$
Obviously, $Ly=u$ and $y^{(n-1)}(1)=\sum_{i=1}^{m}\alpha_iy^{(n-1)}(\xi_i)$. By
Lemma \ref{lem3.1}, we  obtain that
 $y\in \operatorname{dom}L$; i.e., $u\in \operatorname{Im}L$.

Define operator $Q:Y\to Y$ as follows
$$
Qu=\frac{1}{1-\sum_{i=1}^m\alpha_i\xi_i}
\Big(\sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1u(s)ds\Big).
$$
Obviously, $Q^2y=Qy$ and $\operatorname{Im}L=\ker Q$. For $y\in Y$, set
$y=(y-Qy)+Qy$. Then
$y-Qy\in \ker Q=\operatorname{Im}L$, $Qy\in \operatorname{Im}Q$. It follows
from $\ker Q= \operatorname{Im}L$ and $Q^2y=Qy$ that
 $\operatorname{Im}Q\cap \operatorname{Im}L=\{0\}$. So we
have
$$
Y=\operatorname{Im}L\oplus \operatorname{Im}Q.
$$
This, together with \eqref{e3.3}, means that $L$ is a Fredholm operator
of index zero.

Define operator $K_P:Y\to X $ as follows
$$
K_Pu=\sum_{i=k}^{n-2}\frac{B_i(u)}{i!}t^{i}
+\frac{(-1)^{n-k}}{(n-1)!}\int_0^t(t-s)^{n-1}u(s)ds.
$$
Now we  show that $K_P(\operatorname{Im}L)\subset \operatorname{dom}L\cap \ker P$.
Take $u\in \operatorname{Im}L$. Obviously, $(K_P(u))^{(n-1)}(0)=0$. This implies
that $K_P(u)\in \ker P$. It is easy to see that
$(K_P(u))^{(i)}(0)=0,~0\leq i\leq k-1$. It follows from Lemma \ref{lem3.1}
that $(K_P(u))^{(j)}(1)=0$, $0\leq j\leq n-k-2$. 
From  $u\in \operatorname{Im}L$, we obtain
$$
(K_P(u))^{(n-1)}(1)=\sum_{i=1}^m\alpha_i(K_P(u))^{(n-1)}(\xi_i).
$$
So, $K_P(u)\in \operatorname{dom}L$.

Now we  prove that $K_P$ is the inverse of $L|_{\operatorname{dom}L\cap
\ker P}$.
Obviously, $LK_Pu=u$, for $u\in \operatorname{Im}L$. On the other hand, for $y\in
\operatorname{dom}L\cap \ker P$, we have
\begin{align*}
K_PLy(t)
&=\sum_{i=k}^{n-2}\frac{B_i(Ly)}{i!}t^{i}+\frac{(-1)^{n-k}}{(n-1)!}
 \int_0^t(t-s)^{n-1}(-1)^{n-k}y^{(n)}(s)ds\\
&=\sum_{i=k}^{n-2}\big(\frac{B_i(Ly)-y^{(i)}(0)}{i!}\big)t^i+y(t).
\end{align*}
Since $y$ and $K_PLy\in \operatorname{dom}L$, we have
$(K_PLy)^{(j)}(1)=y^{(j)}(1)=0,~0\leq j\leq n-k-2$. This means
that
$(B_k(Ly)-y^{(k)}(0),~B_{k+1}(Ly)-y^{(k+1)}(0),\dots,B_{n-2}(Ly)-y^{(n-2)}(0))$
is the only zero solution of the system of linear equations
\begin{gather*}
 \frac{x_k}{k!}+\frac{x_{k+1}}{(k+1)!}+\dots+\frac{x_{n-2}}{(n-2)!}=0\\
  \frac{x_k}{(k-1)!}+\frac{x_{k+1}}{k!}+\dots+\frac{x_{n-2}}{(n-3)!}=0\\
 \dots\\
\frac{x_k} {[k-(n-k-2)]!}+\frac{x_{k+1}}{[k+1-(n-k-2)]!} +\dots+
\frac{x_{n-2}}{[n-2-(n-k-2)]!}=0.
\end{gather*}
 So, we have $K_PLy=y$, for $y\in \operatorname{dom}L\cap \ker P$.
Thus, $K_P=(L|_{\operatorname{dom}L\cap \ker P})^{-1}$. The proof is
complete.
\end{proof}

\begin{lemma} \label{lem3.4}
Assume $\Omega\subset X$ is an open bounded
subset and $\operatorname{dom}L\cap \overline{\Omega}\neq \emptyset$, then $N$ is
$L$-compact on $\overline{\Omega}$.
\end{lemma}

\begin{proof}
 Obviously, $QN(\overline{\Omega})$ is bounded. Now we
will show that $K_P(I-Q)N:\overline{\Omega}\to X$ is
compact.

It follows from (H2) that there exists constant $M_0>0$ such
that $|(I-Q)Ny|\leq M_0$; a.e., $t\in [0,1]$,
$y\in \overline{\Omega}$. Thus, $K_P(I-Q)N(\overline{\Omega})$ is
bounded. By (H2) and Lebesgue Dominated Convergence theorem, we
get that $K_P(I-Q)N:\overline{\Omega}\to X$ is continuous.
Since $\big\{\int_0^t(t-s)^j(I-Q)Ny(s)ds,~y\in
\overline{\Omega}\big\}$, $j=0,1\dots,n-1$ are equi-continuous,
and $t^j$, $j=0,1\dots,n-1$ are uniformly continuous on $[0,1]$,
using Ascoli-Arzela theorem, we obtain that
$K_P(I-Q)N:\overline{\Omega}\to X$ is compact. The proof
is complete.
\end{proof}

To obtain our main results, we need the following conditions.
\begin{itemize}
\item[(H3)] There exists a constant $M>0$ such that if
$|y^{(n-1)}(t)|>M$, $t\in [\xi_m,1]$ then
$$
\sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1\big[f(s,y(s),y'(s),
\dots,y^{(n-1)}(s))+\varepsilon(s)\big]ds\neq 0.
$$

\item[(H4)] There exist functions $g,h,\psi_i\in L^1[0,1]$,
$i=1,2,\dots,n$, with $\sum_{i=1}^{n}\|\psi_i\|_1<1/2$, $\theta\in[0,1)$,
some $1\leq j\leq n$ such that
$$
|f(t,x_1,x_2,\dots,x_n)|\leq g(t)+\sum_{i=1}^{n}\psi_i(t)|x_i|+h(t)|x_j|^\theta.
$$

\item[(H5)] There exists a constant $c_0>0$ such that, if $|c|>c_0$,
one of the following two conditions holds
\begin{gather}
c\sum_{i=1}^m\alpha_i\int_{\xi_i}^1
\Big[f\Big(s,c\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}s^i\Big),
 c \Big(\sum_{i=k}^{n-1}\frac{A_i}{(i-1)!}s^{i-1}\Big),\dots,
 c\Big)+\varepsilon(s)\Big]ds>0,\label{e3.4}
\\
c\sum_{i=1}^m\alpha_i\int_{\xi_i}^1
\Big[f\Big(s,c\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}s^i\Big),
 c\Big(\sum_{i=k}^{n-1}\frac{A_i}{(i-1)!}s^{i-1}\Big),\dots,
 c\Big)+\varepsilon(s)\Big]ds<0.\label{e3.5}
\end{gather}
\end{itemize}

\begin{lemma} \label{lem3.5}
Assume {\rm (H1)--(H4)}. Then the set
$$\Omega_1=\big\{y\in \operatorname{dom}L\setminus \ker L:
Ly=\lambda Ny,\;\lambda\in (0,1)\big\}
$$
is bounded.
\end{lemma}

\begin{proof} Take $y\in \Omega_1$. Since $Ny\in \operatorname{Im}L$,
we have
\begin{equation}
\sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1\left[f(s,y(s),y'(s),
\dots,y^{(n-1)}(s))+\varepsilon(s)\right]ds=0.\label{e3.6}
\end{equation}
Since $Ly=\lambda Ny$ and $y\in \operatorname{dom}L$, it follows that
\begin{equation}
y(t)=\sum_{i=k}^{n-1}\frac{c_i}{i!}t^{i}+\frac{(-1)^{n-k}}{(n-1)!}
\lambda\int_0^t(t-s)^{n-1}\left[f(s,y(s),y'(s),\dots,y^{(n-1)}(s))
+\varepsilon(s)\right]ds,\label{e3.7}
\end{equation}
where $c_i$, $i=k,k+1,\dots,n-1$ satisfy
\begin{gather*}
\sum_{i=k}^{n-1}\frac{c_i}{i!}
=-\frac{(-1)^{n-k}}{(n-1)!}\lambda\int_0^1(1-s)^{n-1}
 \big[f(s,y(s),y'(s),\dots,y^{(n-1)}(s))+\varepsilon(s)\big]ds\\
\sum_{i=k}^{n-1}\frac{c_i}{(i-1)!}=-
\frac{(-1)^{n-k}}{(n-2)!}\lambda\int_0^1(1-s)^{n-2}
 \big[f(s,y(s),y'(s),\dots,y^{(n-1)}(s))+\varepsilon(s)\big]ds\\
\dots\\
\begin{aligned}
\sum_{i=k}^{n-1}\frac{c_i}{[i-(n-k-2)]!}
&=-\frac{(-1)^{n-k}}{[n-1-(n-k-2)]!}\lambda\int_0^1(1-s)^{k+1}\\
&\quad \times \big[f(s,y(s),y'(s),\dots,y^{(n-1)}(s))+\varepsilon(s)\big]ds.
\end{aligned}
\end{gather*}
 It follows from $y^{(i)}(0)=y^{(j)}(1)=0$,
$0\leq i\leq k-1$, $0\leq j\leq n-k-2$ that there exists at least one point
$\delta_i\in [0,1]$ such that
$y^{(i)}(\delta_i)=0$, $i=0,1,\dots,n-2$. So, we have
$$
y^{(i)}(t)=\int_{\delta_i}^ty^{(i+1)}(s)ds,\quad i=0,1,\dots,n-2.
$$
Therefore,
\begin{equation}
\|y^{(i)}\|_\infty\leq \|y^{(i+1)}\|_1\leq
\|y^{(i+1)}\|_\infty,\quad i=0,1,\dots,n-2.\label{e3.8}
\end{equation}

By \eqref{e3.6} and (H3),  there exists $t_0\in [\xi_m,1]$
such that $|y^{(n-1)}(t_0)|\leq M$. This, together with \eqref{e3.7},
implies
\begin{equation}
|c_{n-1}|\leq M+\int_0^1\left|f(s,y(s),y'(s),\dots,y^{(n-1)}(s))\right|ds
+\|\varepsilon\|_1.\label{e3.9}
\end{equation}
It follows from \eqref{e3.7}-\eqref{e3.9} and (H4) that
\begin{align*}
\|y^{(n-1)}\|_\infty
&\leq M+2\int_0^1\big|f(s,y(s),y'(s),\dots,y^{(n-1)}(s))\big|ds
 +2\|\varepsilon\|_1\\
&\leq M+2[\|g\|_1+\sum_{i=1}^{n}\|\psi_i\|_1\|y^{(i-1)}\|_\infty
 +\|h\|_1\|y^{(j-1)}\|_\infty^\theta]+2\|\varepsilon\|_1\\
&\leq M+2\|g\|_1+2\sum_{i=1}^{n}\|\psi_i\|_1\|y^{(n-1)}\|_\infty
 +2\|h\|_1\|y^{(n-1)}\|_\infty^\theta+2\|\varepsilon\|_1.
\end{align*}
So, we obtain
$$
\|y^{(n-1)}\|_\infty
\leq\frac{M+2\|g\|_1+2\|\varepsilon\|_1}{1-2\sum_{i=1}^{n}\|\psi_i\|_1}+
\frac{2\|h\|_1}{1-2\sum_{i=1}^{n}\|\psi_i\|_1}\|y^{(n-1)}\|_\infty^\theta.
$$
Then $\theta\in [0,1)$ implies that
$\{\|y^{(n-1)}\|_\infty|: y\in \Omega_1\}$ is bounded. Considering of \eqref{e3.8},
 we obtain that $\Omega_1$ is bounded.
\end{proof}

\begin{lemma} \label{lem3.6}
Assume {\rm (H1), (H2), (H5)}. Then the set
$$
\Omega_2=\{y:y\in \ker L,\;Ny\in \operatorname{Im}L\}
$$
is bounded.
\end{lemma}

\begin{proof}
Take $y\in \Omega_2$, then $y(t)=
c\big(\sum_{i=k}^{n-1}\frac{A_i}{i!}t^i\big)$, $c\in \mathbb{R}$
and $Ny\in \operatorname{Im}L$. So, we have
$$
c \sum_{i=1}^m\alpha_i\int_{\xi_i}^1\Big[f\Big(s,c\Big(\sum_{i=k}^{n-1}
\frac{A_i}{i!}s^i\Big),c\Big(\sum_{i=k}^{n-1}\frac{A_i}{(i-1)!}s^{i-1}\Big),
\dots,c\Big)
+\varepsilon(s)\Big]ds=0.
$$
By (H5), we obtain that $|c|\leq c_0$. So, $\Omega_2$ is
bounded.
\end{proof}


\begin{lemma} \label{lem3.7}
Assume {\rm (H1), (H2), (H5)}. Then the set
$$
\Omega_3=\{y\in \ker L:\lambda Jy+(1-\lambda)\theta QNy=0,\,\lambda\in [0,1]\}
$$
is bounded, where $J:\ker L\to \operatorname{Im}Q$ is a linear isomorphism
given by
 $$
J\Big(c\sum_{i=k}^{n-1}\frac{A_i}{i!}t^i\Big)
= \frac{c}{1-\sum_{i=1}^{m}\alpha_i\xi_i} ,\quad c \in \mathbb{R}
$$
and
$\theta=\begin{cases}
1 & \text{if \eqref{e3.4} holds},\\
-1,&\text{if \eqref{e3.5} holds}.
\end{cases}$.
\end{lemma}

\begin{proof} For $y\in \Omega_3$, we get
$y=c\big(\sum_{i=k}^{n-1}\frac{A_i}{i!}t^i\big)$ with
$$
\lambda c+(1-\lambda)\theta \sum_{i=1}^m\alpha_i\int_{\xi_i}^1
\Big[f\Big(s,c\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}s^i\Big),
c\Big(\sum_{i=k}^{n-1}\frac{A_i}{(i-1)!}s^{i-1}\Big),\dots,c\Big)
+\varepsilon(s)\Big]ds=0.
$$
If $\lambda=0$, by (H5), we get
$|c|\leq c_0$. If $\lambda=1$, $c=0$. For $\lambda\in (0,1)$, if
$|c|\geq c_0$,
then
\begin{align*}
\lambda c^2&=-(1-\lambda)\theta c\sum_{i=1}^m\alpha_i\int_{\xi_i}^1
\Big[f\Big(s,c\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}s^i\Big),
c\Big(\sum_{i=k}^{n-1}\frac{A_i}{(i-1)!}s^{i-1}\Big),\dots, c\Big) \\
&\quad +\varepsilon(s)\Big]ds<0.
\end{align*}
This is a contradiction. So, $\Omega_3$ is bounded.
\end{proof}


\begin{theorem} \label{thm3.1}
Assume {\rm (H1)--(H5)} Then  problem
\eqref{e1.1}--\eqref{e1.2} has at least one solution in $X$.
\end{theorem}

\begin{proof} 
Let $\Omega\supset\cup_{i=1}^3\overline{\Omega_i}\cup\{0\}$
 be a bounded open subset of $X$. It follows from Lemma \ref{lem3.4} 
that $N$ is $L-$compact on $\overline{\Omega}$. By 
Lemmas \ref{lem3.5} and \ref{lem3.6}, we obtain:
(1) $Ly\neq \lambda Ny$ for every 
$(y,\lambda)\in [(\operatorname{dom}L\setminus
\ker L)\cap\partial\Omega]\times(0,~1)$; and
(2) $Ny\notin \operatorname{Im}L$ for every $y\in \ker L\cap\partial\Omega$.
We need  to prove only
(3) $\deg(QN|_{\ker L},~\Omega\cap \ker L,~0)\neq0$.
To do this, we take
$$
H(y,\lambda)=\lambda Jy+\theta(1-\lambda)QNy.
$$
 According to Lemma \ref{lem3.7}, we know $H(y,\lambda)\neq 0$ for 
$y\in \partial\Omega\cap \ker L$. By the homotopy of degree, we obtain
\begin{align*}
\deg(QN|_{\ker L}, \Omega \cap \ker L,0)
&=\deg(\theta H(\cdot,0),\Omega \cap \ker L,0)\\
&=\deg(\theta H(\cdot,1),\Omega \cap \ker L,0)\\
&=\deg(\theta J, \Omega \cap \ker L,0)\neq 0.
\end{align*}
By Theorem \ref{thm2.1}, we  obtain that $Ly=Ny$ 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 complete.
\end{proof}

\subsection*{Acknowledgments}
This research was supported by grants 11171088 from the Natural Science Foundation
of China, QD201020 from the Doctoral Program Foundation of Hebei
University of Science and Technology, and XL201136 from the
the Foundation of Hebei University of Science and Technology.

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\end{document}


