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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 240, 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 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/240\hfil Existence of solutions]
{Existence of solutions for four-point resonance boundary-value
 problems on time scales}

\author[N. Wang, H. Zhou, L. Yang \hfil EJDE-2015/240\hfilneg]
{Ning Wang, Hui Zhou, Liu Yang}

\address{Ning Wang \newline
College of Mathematics and Statistics,
Institute of Applied Mathematics,
Hefei Normal University, Hefei 230061, China}
\email{Ningw@hftc.edu.cn}

\address{Hui Zhou \newline
College of Mathematics and Statistics,
Institute of Applied Mathematics,
Hefei Normal University, Hefei 230061, China}
\email{zhouhui0309@126.com}

\address{Liu Yang \newline
College of Mathematics and Statistics,
Institute of Applied Mathematics,
Hefei Normal University, Hefei 230061, China}
\email{yliu722@163.com}

\thanks{Submitted March 6, 2015. Published September 21, 2015.}
\subjclass[2010]{34N05, 34B10}
\keywords{Boundary-value problem; time scale; coincidence degree}

\begin{abstract}
 By using Mawhin's continuation theorem, we prove the existence of solutions
 for a class of multi-point boundary-value problem under different resonance
 conditions on time scales.
\end{abstract}

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

\section{Introduction}

Consider multi-point boundary-value problem on time scales
\begin{equation}
x^{\Delta\nabla}(t)=f(t,x(t),x^{\Delta}(t))+e(t),t\in (0,1)\cap T,\label{e1.1}
\end{equation}
subject to boundary condition
\begin{equation}
x(0)=\alpha x(\eta),x(1)=\beta x(\xi),\label{e1.2}
\end{equation}
where $T$ is a time scale such that $0,1\in T,\eta,\xi\in(0,1)\cap T,$
$f: T\times R^{2}\to \mathbb{R}$ is a continuous function
and $e(t)\in L^{1}[0,1].\alpha,\beta\in R$ hold
\begin{equation}
\alpha=1,\beta=1\label{e1.3}
\end{equation}
or
\begin{equation}
\alpha\eta(1-\beta)+(1-\alpha)(1-\beta\xi)=0.\label{e1.4}
\end{equation}
  By condition \eqref{e1.3} or \eqref{e1.4},
the differential operator in \eqref{e1.1} is not invertible,
which is called problem at resonance. The study on multi-point
nonlocal boundary-value problems for linear second-order ordinary
differential equations
was initiated by Il'in and Moiseev \cite{i1,i2}. Since then some
existence results were obtained for general boundary-value
problems by several authors. We refer the reader to some recent results,
such as \cite{b2,g1,l1,m1,m2} at non-resonance, and \cite{f1,r1,r2,r3}
at resonance and reference therein.

 For problem \eqref{e1.1}, \eqref{e1.2} in the continuous setting,
by using upper and lower solution method
Rach\"unkov\'a \cite{r1,r2,r3}  obtained  excellent
results about the problem
\begin{equation}
x''(t)=f(t,x(t),x'(t))+e(t),\quad t\in (0,1),\label{e1.5}
\end{equation}
subject to boundary condition
\begin{equation}
x(0)=x(\eta),x(1)=x(\xi)\label{e1.6}
\end{equation}
Bai \cite{b1} developed the upper and lower solution method and obtained
existence and multiplicity results for the resonance problem
\begin{equation}
x''(t)=f(t,x(t),x'(t))+e(t),\quad t\in (0,1),\label{e1.7}
\end{equation}
subject to boundary condition
\begin{equation}
x(0)=\alpha x(\eta),\quad x(1)=\beta x(\xi)\label{e1.8}
\end{equation}
where condition \eqref{e1.4} is satisfied. But the existence results for
resonance cases on time scales is rare (see \cite{k1}). Motivated by
works above, we apply coincidence degree theory \cite{m3} to establish
existence results for resonance problems \eqref{e1.1}--\eqref{e1.4}.
Considering that the main tools we used are different from 
\cite{b1,r1,r2,r3},
the results we establish follows are  new even in the continuous setting.

 The article is organized as follows. Some basic definitions and conclusions
 on time scales and the main tools we used are introduced in section 2.
In section 3, we discuss the existence of solutions under condition \eqref{e1.3}
and the existence results of case \eqref{e1.4} is considered in section 4.

\section{Preliminaries}

First we present some basic definitions on time scales (see \cite{a1}).
A time scale $T$ is a closed nonempty subset of $\mathbb{R}$.
For $t<\sup T$ and $r>\inf T$, we define the forward jump operator
$\sigma$ and the backward jump operator $\rho$ respectively by
$$
\sigma(t)=\inf\{\tau\in T|\tau>t\},\quad
\rho(t)=\sup\{\tau\in T|\tau<t\},
$$
for all $t\in T$. If $\sigma(t)>t$, t is said to be right scattered, and
if $\sigma(t)=t$, $t$ is said to be right dense.
If $\rho(t)<t$, $t$ is said to be left scattered,
and if $\rho(t)=t$, $t$ is said to be left dense.
 A function $f$ is left-dense continuous, if $f$ is continuous at each
left dense point in T and its right-sided
limits exists at each right dense points in $T$.

  For $u:T\to \mathbb{R}$ and $t\in T$, we define the delta-derivative of
$u(t)$, $u^{\Delta}(t)$, to be the number (when it exists),
with the property that for each $\varepsilon>0$, there is a neighborhood,
$U$, of $t$ such that for all $s\in U$,
$$
|u(\sigma(t))-u(s)-u^{\Delta}(t)(\sigma(t)-s)|\leq \varepsilon|\sigma(t)-s|.
$$
  For $u:T\to \mathbb{R}$ and $t\in T$, we define the nabla-derivative of
$u(t),u^{\nabla}(t)$, to be the number (when it exists),
with the property that for each $\varepsilon>0$, there is a neighborhood,
$U$, of $t$ such that for all $s\in U$,
$$
|u(\rho(t))-u(s)-u^{\nabla}(t)(\rho(t)-s)|\leq \varepsilon|\sigma(t)-s|.
$$
 Then we recall some notations and an abstract existence result briefly.

 Let $X$, $Y$ be real Banach spaces and let
$L:\operatorname{dom}L\subset X\to Y$ be a Fredholm operator with index zero,
here $\operatorname{dom} L$ denotes the domain of $L$.
This means that $\operatorname{Im}L$ is closed in $Y$ and dim
$\ker L=\dim(Y/\operatorname{Im}L)<+\infty$. Consider the supplementary
subspaces $X_1$ and $Y_1$ such that $X=\ker L\oplus X_1$ and
$Y=\operatorname{Im}L\oplus Y_1$ and let $P:X\to \ker L$ and
$Q:Y\to Y_1$ be the natural projections.
Clearly, $\ker L\cap (\operatorname{dom}L\cap X_1)=\{0\}$,
thus the restriction $L_{p}:=L|_{\operatorname{dom}L\cap X_1}$ is invertible.
 Denote by $K$ the inverse of $L_{p}$.

 Let $\Omega$ be an open bounded subset of $X$ with
$\operatorname{dom}L\cap\Omega\neq\emptyset$.
A map $N:\overline{\Omega}\to Y$ is said to be $L$-compact in
$\overline{\Omega}$ if $QN(\overline{\Omega})$ is bounded and the operator
$K(I-Q)N:\overline{\Omega} \to X$ is compact.
We first give the famous Mawhin continuation theorem.

\begin{lemma}[Mawhin \cite{m3}] \label{lem2.1}
 Suppose that $X$ and $Y$ are Banach spaces, and
$L:\operatorname{dom}L\subset X\to Y$ is a Fredholm operator
with index zero.Furthermore, $\Omega\subset X$ is an open bounded
set and $N:\overline{\Omega}\to Y$ is $L$-compact
on $\overline{\Omega}$. If
\begin{itemize}
\item[(1)] $Lx\neq \lambda Nx$ for all
$x\in \partial \Omega\cap \operatorname{dom}L,\lambda\in (0,1)$;

\item[(2)] $Nx\not\in \operatorname{Im}L$ for all $x\in \partial \Omega\cap \ker L$;

\item[(3)] $\deg\{JQN, \Omega\cap \ker L,0\}\neq 0$, where
$J:\ker L\to \operatorname{Im}Q$ is an isomorphism,
\end{itemize}
then the equation $Lx=Nx$ has a solution in
$\overline{\Omega}\cap \operatorname{dom}L$.
\end{lemma}

\section{Existence results under condition \eqref{e1.3}}

In this section  problem \eqref{e1.1}, \eqref{e1.2} is considered under
the assumption that $\alpha=1$ and $\beta=1$.
We introduce the space
$$
X=\{x:[0,1]\to \mathbb{R}:x^{\Delta}\in AC[0,1],x^{\Delta \nabla}\in L^{1}[0,1]\}
$$
endowed with the norm $\|x\|=\sup\{\|x\|_0,\|x^{\Delta}\|_0\}$,
where $\|x\|_0=\sup_{t\in[0,1]}|x(t)|$.
Let $Y=L^{1}[0,1]$ with the norm
$$
\|x\|_1=\int^{1}_0|x(t)|\nabla t.
$$
Let a linear mapping $L:\operatorname{dom} L\subset X\to Y$ with
$$
\operatorname{dom} L=\{x\in X:x(0)=x(\eta),x(1)=x(\xi)\}
$$
be defined by $Lx=x^{\Delta \nabla}(t)$.
Define the mapping $N:X\to Y$ by
$$
Nx(t)=f(t,x(t),x^{\Delta}(t))+e(t)
$$
Then problem \eqref{e1.1}, \eqref{e1.2} can be written as $Lx=Nx$, here
$L$ is a linear operator.

\begin{lemma} \label{lem3.1}
If $\beta=1$ and $\alpha=1$ then
\begin{itemize}
\item[(1)]
\begin{equation}
\operatorname{Im}L=\{y(t)| \int^{\eta}_0(\xi -s)y(s)\nabla s
+\int^{\xi}_{\eta}\eta(\xi -1)y(s)\nabla s
-\int^{1}_0\eta(1-s)y(s)\nabla s=0\};\label{e3.1}
\end{equation}

\item[(2)] $L:\operatorname{dom}L\subset X\to Y$ is a Fredholm operator
 with index zero;

\item[(3)] Define projector operator $P:X\to \ker L$ as $Px=x(0)$,
then the generalized inverse of operator
$L$, $K_{P}:\operatorname{Im}L\to \operatorname{dom}L\cap \ker P$
can be written as
\begin{equation}
K_{P}(y)=- \frac{t}{\eta}\int^{\eta}_0(\eta-s)y(s)\nabla s
+\int^{t}_0(t-s)y(s) \nabla s,\label{e3.2}
\end{equation}
satisfying $\|K_{P}(y(t))\|_1\leq 2\|y\|_1$.
\end{itemize}
\end{lemma}

\begin{proof} (1) First we show that
$$
\operatorname{Im}L=\{y(t)\in Y| \int^{\eta}_0(\xi -s)y(s)\nabla s
+\int^{\xi}_{\eta}\eta(\xi -1)y(s)\nabla s
-\int^{1}_0\eta(1-s)y(s)\nabla s=0\}.
$$
First suppose $y(t)\in \operatorname{Im}L$, then there exists $x(t)$
such that $Lx=y$. That is
$$
x(t)=\int^{t}_0(t-s)y(s)\nabla s+x^{\Delta}(0)t+x(0).
$$
Condition $x(0)=x(\eta)$, $x(1)=x(\xi)$ imply that
$$
\int^{\eta}_0(\xi -s)y(s)\nabla s+\int^{\xi}_{\eta}\eta(\xi -1)y(s)\nabla s
-\int^{1}_0\eta(1-s)\nabla s=0.
$$
Next we suppose that
\[
y(t)\in\{y(t)| \int^{\eta}_0(\xi -s)y(s)\nabla s
 +\int^{\xi}_{\eta}\eta(\xi -1)y(s)\nabla s
-\int^{1}_0\eta(1-s)\nabla s=0\}.
\]
Let $x(t)\in X$, where
$$
x(t)=\int^{t}_0(t-s)y(s)\nabla s-\frac{t}{\eta}\int^{\eta}_0(\eta-s)y(s) \nabla s\,.
$$
Then $Lx=x^{\Delta \nabla}=y(t)$.
because
\[
y(t)\in\{y(t)| \int^{\eta}_0(\xi -s)y(s)\nabla s
+\int^{\xi}_{\eta}\eta(\xi -1)y(s)\nabla s
-\int^{1}_0\eta(1-s)y(s)\nabla s=0\},
\]
 by a simple computation we have
$x(0)=x(\eta),x(1)=x(\xi)$.
Thus $y(t)\in \operatorname{Im}L$. Summing up two steps above we obtain 
$$
\operatorname{Im}L=\{y(t)\in Y| \int^{\eta}_0(\xi -s)y(s)\nabla s
+\int^{\xi}_{\eta}\eta(\xi -1)y(s)\nabla s
-\int^{1}_0\eta(1-s)y(s)\nabla s=0\}.
$$

(2) We claim that $L$ is a Fredholm operator with index zero.
It is easy to see that $\ker L=R$. Next we prove that
$Y=\operatorname{Im}L\oplus \ker L$. Suppose $y(t)\in Y$,
define the projector operator $Q$ as
$$
Q(y)= \frac{ \int^{\eta}_0(\xi -s)y(s)\nabla s
 +\int^{\xi}_{\eta}\eta(\xi -1)y(s)\nabla s
-\int^{1}_0\eta(1-s)y(s)\nabla s}{ \int^{\eta}_0(\xi -s)\nabla s
 +\int^{\xi}_{\eta}\eta(\xi -1)\nabla s -\int^{1}_0\eta(1-s)\nabla s}
$$
Let $y^{\ast}=y(t)-Q(y(t))$, by a simple computation,
$y^{\ast}\in \operatorname{Im}L$. Hence $Y=\operatorname{Im}L+\ker L$,
furthermore considering $\operatorname{Im}L\cap \ker L=\{0\}$,
 we have $Y=\operatorname{Im}L\oplus \ker L$. Thus
$\dim \ker L=\operatorname{codim} \operatorname{Im}L$,
which means $L$ is a Fredholm operator with index zero.

(3) Define the projector operator $P:X\to \ker L$ as
$Px=x(0)$,  for $y(t)\in \operatorname{Im}L$,
 $$
(LK_{P})(y(t))=y(t),
$$
and for $x(t)\in \operatorname{dom}L\cap \ker P$, we know
\begin{align*}
(K_{P}L)(x(t))
&=K_{P}(x^{\Delta \nabla}(t))\\
&=-\frac{t}{\eta}\int^{\eta}_0(\eta-s)x^{\Delta \nabla}(s)\nabla s
+\int^{t}_0\int^{s}_0x^{\Delta \nabla}(\tau)\nabla\tau \nabla s \\
&=x(t).
\end{align*}
This shows that $K_{P}=(L_{\operatorname{dom}L\cap \ker P})^{-1}$.
Furthermore from the definition of the norms in the $X,Y$,
\[
\|K_{P}(y(t))\|_1\leq 2\|y\|_1.
\]
The above arguments complete the proof.
\end{proof}

\begin{theorem} \label{thm3.1}
Let $f:[0,1]\times R^{2}\to \mathbb{R}$ be a continuous function.
Assume that following conditions are satisfied:
\begin{itemize}
\item[(C1)]  There exist functions $a, b,c, r\in L^{1}[0,1]$ and constant
$\theta\in [0,1)$ such that for all $(x,y)\in R^{2}$,
either
\begin{equation}
|f(t,x,y)|\leq a(t)|x|+b(t)|y|+c(t)|y(t)|^{\theta}+r(t),\label{e3.3}
\end{equation}
or 
\begin{equation}
|f(t,x,y)|\leq a(t)|x|+b(t)|y|+c(t)|x(t)|^{\theta}+r(t).\label{e3.4}
\end{equation}

\item[(C2)] There exists a constant $M>0$ such that for
$x\in \operatorname{dom} L,|x(\rho(t))|>M$, for all $t\in [0,1]$,
\begin{equation}
\begin{aligned}
&\int^{\eta}_0(\xi -s)(f(s,x,x^{\Delta})+e(s))\nabla s+\int^{\xi}_{\eta}\eta(\xi -1)
(f(s,x,x^{\Delta})+e(s))\nabla s\\
&-\int^{1}_0\eta(1-s)(f(s,x,x^{\Delta})+e(s))\nabla s\neq 0
\end{aligned}\label{e3.5}
\end{equation}

\item[(C3)] There exists $M^{\ast}>0$ such that for $d\in R$,
if $|d|>M^{\ast}$, then either
\begin{equation}
d\times \int^{\eta}_0(f(t,d,0)+e(t))\nabla t>0\label{e3.6}
\end{equation}
or else
\begin{equation}
d\times \int^{\eta}_0(f(t,d,0)+e(t))\nabla t<0\label{e3.7}
\end{equation}
Then for each $e\in L^{1}[0,1]$, resonance problem \eqref{e1.1}, \eqref{e1.2} with
$\alpha=1$ and $\beta=1$ has at least one solution provided that
$$
\|a\|_1+\|b\|_1<\frac{1}{3}.
$$
\end{itemize}
\end{theorem}

\begin{proof}  We divide the proof into two steps.
\smallskip

\noindent\textbf{Step 1.} Let
$$
\Omega_1=\{x\in \operatorname{dom}\ L\setminus \ker L:Lx=\lambda Nx\}
\quad \text{for some } \lambda\in [0,1],
$$
Then $\Omega_1$ is bounded.
Suppose that $x\in \Omega_1$, $Lx=\lambda Nx$, thus
$\lambda\neq 0$, $Nx\in \operatorname{Im}L=\ker Q$, hence
\begin{equation}
\begin{aligned}
&\int^{\eta}_0(\xi -s)(f(s,x,x^{\Delta})+e(s))\nabla s
 +\int^{\xi}_{\eta}\eta(\xi -1)(f(s,x,x^{\Delta})+e(s))\nabla s\\
&-\int^{1}_0\eta(1-s)(f(s,x,x^{\Delta})+e(s))\nabla s= 0
\end{aligned}\label{e3.8}
\end{equation}
Then \eqref{e3.8} and condition (C2) imply that there exists $t_0\in T$
such that $|x(t_0)|<M$.
In view of $x(0)=x(\rho(t_0))- \int_0^{t_0}x^{\Delta}(s)\nabla s$, we obtain that
\begin{equation}
|x(0)|<M+\|x^{\Delta}\|_0,\quad t\in T.\label{e3.9}
\end{equation}
For $x(1)=x(\eta)$, there exists $t_1\in(\eta,1)\cap T$ such that
$x^{\Delta}(t_1)=0$. Then from
$$
x^{\Delta}(t)=x^{\Delta}(t_1)+\int^{t}_{t_1}x^{\Delta \nabla}(s)\nabla s,
$$
we have
\begin{equation}
\|x^{\Delta}(t)\|_0\leq \|Nx\|_1\label{e3.10}
\end{equation}
Hence, from \eqref{e3.9} and \eqref{e3.10}, we have
$$
|x(0)|\leq M+\|Nx\|_1.
$$
Again for $x\in \Omega_1,x\in \operatorname{dom}L\setminus \ker L$,
then $(I-P)x\in \operatorname{dom}L\cap \ker P,\ LPx=0$,
thus from Lemma \ref{lem3.1}, we have
\begin{equation}
\|(I-P)x\|=\|K_{P}L(I-P)x\|\leq 2\|L(I-P)x\|_1=2\|Lx\|_1
\leq 2\|Nx\|_1.\label{e3.11}
\end{equation}
Then
$$
\|x\|\leq \|Px\|+\|(I-P)\|\leq M+3\|Nx\|_1.
$$
If assumption \eqref{e3.3} holds, we obtain
\begin{align*}
\|x\|&\leq M+3\|Nx\|_1= M+3|f(t,x(t),x^{\Delta}(t))+e(t)|_1\\
&\leq M+3(\|a\|_1|x|+\|b\|_1|x^{\Delta}|
 +\|c\|_1|x^{\Delta}|^{\theta}+\|r\|_1+\|e\|_1) \\
&\leq 3\|a\|_1\|x\|_0+3[\|b\|_1\|x^{\Delta}\|_0+\|c\|_1\|x^{\Delta}\|_0^{\theta}
+\|r\|_1+\|e\|_1+\frac{M}{3}].
\end{align*}
Thus
$$
\|x\|_0\leq \frac{3}{1-3\|a\|_1}[\|b\|_1\|x^{\Delta}\|_0
+\|c\|_1\|x^{\Delta}\|_0^{\theta}
+\|r\|_1+\|e\|_1+\frac{M}{3}].
$$
Then
\begin{align*}
\|x^{\Delta}\|_0
&\leq \|x\|
 \leq [\frac{3\|a\|_1}{1-3\|a\|_1}+1]
3[\|b\|_1\|x^{\Delta}\|_0+\|c\|_1\|x^{\Delta}\|_0^{\theta}
+\|r\|_1+\|e\|_1+\frac{M}{3}]\\
&=3\frac{\|b\|_1}{1-3\|a\|_1}\|x^{\Delta}\|_0+\frac{3}{1-3\|a\|_1}
[\|c\|_1\|x^{\Delta}\|_0^{\theta}
+\|r\|_1+\|e\|_1+\frac{M}{3}]
\end{align*}
By a simple computation,
$$
\|x^{\Delta}\|_0
\leq\frac{3\|c\|_1}{1-3\|a\|_1-3\|b\|_1}\|x^{\Delta}\|_0^{\theta}
+\frac{3}{1-3\|a\|_1-3\|b\|_1}[\|r\|_1+\|e\|_1+\frac{M}{3}].
$$
Since $\theta \in [0,1)$ and $\|a\|_1+\|b\|_1<1/3$, there exists
positive constant $M_1$ such that
$\|x^{\Delta}\|_0\leq M_1$.
Then from \eqref{e3.9}, there exists positive constant $M_2$ such that
$\|x\|_0\leq M_1$.
Hence $\|x\|=\max\{\|x\|_0,\|x^{\Delta}\|_0\}\leq \max\{M_1,M_2\}$,
which means that $\Omega_1$ is bounded. If \eqref{e3.4} hold, similar
to the above argument, we can prove $\Omega_1$ is also bounded.
\smallskip

\noindent\textbf{Step 2}
The set $\Omega_2=\{x\in \ker L:Nx\in \operatorname{Im}L\}$ is bounded.
The fact that $x\in \Omega_2$ implies that $x=d$, $N(x)=f(t,d,0)+e(t)$
and $QNx=0$, thus
\begin{equation}
\begin{aligned}
Q(Nx)
&= \Big( \int^{\eta}_0(\xi -s)(f(s,d,0)+e(s))\nabla s
 +\int^{\xi}_{\eta}\eta(\xi -1)(f(s,d,0)+e(s))\nabla s\\
&\quad -\int^{1}_0\eta(1-s)(f(s,d,0)+e(s))\nabla s\Big)\\
&\quad \div\Big( \int^{\eta}_0(\xi -s)\nabla s
 +\int^{\xi}_{\eta}\eta(\xi -1)\nabla s
-\int^{1}_0\eta(1-s)\nabla s\Big)
=0
 \end{aligned}\label{e3.12}
\end{equation}
So $|d|<M^{\ast}$, thus $x=d$ is bounded. The proof
of step 2 is complete.
\smallskip

 Let $\Omega=\{x\in X:\|x\|<N_1\}$, where  $N_1>\max\{M_1,M_2,M^{\ast}\}$. Then
$\overline{\Omega}_1\subset\Omega$ and $\overline{\Omega}_2\subset\Omega$.
From argument above, it is obviously that
conditions (1), (2) of Lemma \ref{lem2.1} are satisfied.
Furthermore, by using Ascolli-Arezela theorem, it is easy to see that
$K_{P}(I-Q)N:\overline{\Omega}\to Y$ is compact, thus $N$ is $L$-compact
on $\overline{\Omega}$.

  Next, we claim that condition (3) of Lemma \ref{lem2.1} is also satisfied.
If the first part of condition (C3) is satisfied,
define the isomorphism $J:\operatorname{Im}Q\to \ker L$ by
$J(a)=a$ and let
$$
H(\lambda,x)=\lambda x+(1-\lambda)JQNx,(\lambda,x)\in \Omega\times[0,1].
$$
By a simple calculation, we obtain,
for $(\lambda,x)\in \partial (\Omega\cap \ker L)\times [0,1]$,
$$
xH(\lambda,x)=\lambda x^{2}+(1-\lambda)xQNx>0.
$$
Thus $H(\lambda,x)\neq 0$.

 If the second part of condition (C3) is satisfied, define
$$
H(\lambda,x)=-\lambda x+(1-\lambda)JQNx,(\lambda,x)\in \Omega\times[0,1],
$$
where $J:\operatorname{Im}Q\to \ker L$ defined by $J(a)=-a$,
similar with above, we can obtain that $H(\lambda,x)\neq 0$.
Thus
\begin{align*}
\deg (JQN,\Omega\cap \ker L,0)
&=\deg(H(x,0),\Omega\cap \ker L,0)\\
&=\deg (H(x,1),\Omega\cap \ker L,0)\\
&=\deg (I,\Omega\cap \ker L,0)\neq 0.
\end{align*}
Then by Lemma \ref{lem2.1}, $Lx=Nx$ has at least one solution in
$\operatorname{dom}L\cap \Omega$, which means resonance
problem \eqref{e1.1}, \eqref{e1.2} has at
least one solution. The proof is complete.
\end{proof}

\section{Existence result under condition \eqref{e1.4}}

This section we consider problem \eqref{e1.1},\eqref{e1.2} under the assumption that
$$
\alpha\eta(1-\beta)+(1-\alpha)(1-\beta\xi)=0.
$$
The normed spaces $X,Y$ and operators $L,N$ are defined as in Section 3.

\begin{lemma} \label{lem4.1}
 If $\alpha\eta(1-\beta)+(1-\alpha)(1-\beta\xi)=0$, then
\begin{itemize}
\item[(1)]
\begin{align*}
\operatorname{Im}L
&=\{y(t)| \alpha(1-\beta)\int^{\eta}_0(\eta -s)y(s)
\nabla s+(1-\alpha)\beta\int^{\xi}_0(\xi -s)y(s)\nabla s\\
&\quad -(1-\alpha)\int^{1}_0(1-s)y(s)\nabla s=0\};
\end{align*}

\item[(2)]  $L:\operatorname{dom}L\subset X\to Y$ is a Fredholm
operator with index zero.

\item[(3)] Define projector operator $P:X\to \ker L$ as
$Px=x(0)$,
then the generalized inverse of operator
$L$, $K_{P}:\operatorname{Im}L\to \operatorname{dom}L\cap \ker P$ can be written as
$$
K_{P}(y)=\int^{t}_0(t-s)y(s)\nabla s+\frac{t}{1-\beta\xi}
[\beta\int^{\xi}_0(\xi-s)y(s) \nabla s-\int^{1}_0(1-s)y(s) \nabla s],
$$
satisfying
\[
\|K_{P}(y(t))\|_1\leq (1+|\frac{\beta+1}{1-\beta\xi}|)\|y\|_1.
\]
\end{itemize}
\end{lemma}

\begin{proof}
(1) First we show that
\begin{align*}
\operatorname{Im}L
&=\Big\{y(t)| \alpha(1-\beta)\int^{\eta}_0(\eta -s)y(s)\nabla s
+(1-\alpha)\beta\int^{\xi}_0(\xi -s)y(s)\nabla s\\
&\quad -(1-\alpha)\int^{1}_0(1-s)y(s)\nabla s=0\Big\}
\end{align*}
First suppose $y(t)\in \operatorname{Im}L$, then there exists $x(t)$
such that $Lx=y$. That is
$$
x(t)=\int^{t}_0(t-s)y(s)\nabla s+u^{\Delta}(0)t+u(0).
$$
Then boundary condition
$x(0)=\alpha x(\eta),x(1)=\beta x(\xi)$ together with
$\alpha\eta(1-\beta)+(1-\alpha)(1-\beta\xi)=0$ imply that
$$
\alpha(1-\beta)\int^{\eta}_0(\eta -s)y(s)\nabla s
+(1-\alpha)\beta\int^{\xi}_0(\xi -s)y(s)\nabla s
-(1-\alpha)\int^{1}_0(1-s)y(s)\nabla s=0.
$$
Next we assume that
\begin{align*}
y(t)&\in\Big\{y(t)| \alpha(1-\beta)\int^{\eta}_0(\eta -s)y(s)\nabla s
+(1-\alpha)\beta\int^{\xi}_0(\xi -s)y(s)\nabla s \\
&\quad -(1-\alpha)\int^{1}_0(1-s)y(s)\nabla s=0\Big\}.
\end{align*}
Let
$x(t)\in X$, where
$$
x(t)=\int^{t}_0(t-s)y(s)\nabla s+\frac{t}{1-\beta\xi}
[\beta\int^{\xi}_0(\xi-s)y(s) \nabla s-\int^{1}_0(1-s)y(s) \nabla s]\,.
$$
Then $Lx=x^{\Delta \nabla}=y(t)$.
Since
\begin{align*}
y(t)&\in\Big\{ y(t)| \alpha(1-\beta)\int^{\eta}_0(\eta -s)y(s)\nabla s
+(1-\alpha)\beta\int^{\xi}_0(\xi -s)y(s)\nabla s\\
&\quad -(1-\alpha)\int^{1}_0(1-s)y(s)\nabla s=0\Big\},
\end{align*}
by a simple computation we have
$$
x(0)=\alpha x(\eta),\quad x(1)=\beta x(\xi)
$$
Thus $y(t)\in \operatorname{Im}L$. Summing up the two steps above we obtain that:
(1)
\begin{align*}
\operatorname{Im}L
&=\{y(t)| \alpha(1-\beta)\int^{\eta}_0(\eta -s)y(s)\nabla s
+(1-\alpha)\beta\int^{\xi}_0(\xi -s)y(s)\nabla s\\
&\quad -(1-\alpha)\int^{1}_0(1-s)y(s)\nabla s=0\}.
\end{align*}

(2) Next we claim that $L$ is a Fredholm operator with index zero.
It is easy to see that $\ker L=R$. Next we prove that
$Y=\operatorname{Im}L\oplus \ker L$.
Suppose $y(t)\in Y$, define the projector operator $Q$ as
\begin{align*}
Q(y)&= \Big( \alpha(1-\beta)\int^{\eta}_0(\eta -s)y(s)\nabla s
 +(1-\alpha)\beta\int^{\xi}_0(\xi -s)y(s)\nabla s\\
&\quad -(1-\alpha)\int^{1}_0(1-s)y(s)\nabla s\Big)\\
&\div \Big(\alpha(1-\beta)\int^{\eta}_0(\eta -s)\nabla s
 +(1-\alpha)\beta\int^{\xi}_0(\xi -s)\nabla s
-(1-\alpha)\int^{1}_0(1-s)\nabla s\Big)
\end{align*}
Let $y^{\ast}=y(t)-Q(y(t))$, by a simple computation,
$y^{\ast}\in \operatorname{Im}L$.Hence $Y=\operatorname{Im}L+\ker L$,
furthermore considering $\operatorname{Im}L\cap \ker L=\{0\}$, we have
$Y=\operatorname{Im}L\oplus \ker L$. Thus
$$
\dim \ker L=\operatorname{codim} \operatorname{Im}L,
$$
which means L is a Fredholm operator with index zero.

(3) Define the projector operator $P:X\to \ker L$ as $Px=x(0)$,
for $y(t)\in \operatorname{Im}L$, we have
$$
(LK_{P})(y(t))=y(t),
$$
and for $x(t)\in \operatorname{dom}L\cap \ker P$, we know
\begin{align*}
&(K_{P}L)(x(t))\\
&=K_{P}(x^{\Delta \nabla}(t)) \\
&=\int^{t}_0(t-s)x^{\Delta \nabla} \nabla s+\frac{t}{1-\beta\xi}
[\beta\int^{\xi}_0(\xi-s)x^{\Delta \nabla} \nabla s
 -\int^{1}_0(1-s)x^{\Delta \nabla} \nabla s] \\
&=x(t)
\end{align*}
These shows that $K_{P}=(L_{\operatorname{dom}L\cap \ker P})^{-1}$.
Furthermore from the definition of the norms in the $X,\ Y$, we have
$$
\|K_{P}(y(t))\|_1\leq (1+|\frac{\beta+1}{1-\beta\xi}|)\|y\|_1.
$$
\end{proof}

\begin{theorem} \label{thm4.1}
Let $f:[0,1]\times R^{2}\to \mathbb{R}$ be a continuous function.
Condition $C_1, C_2$ are satisfied and:
\begin{itemize}
\item[(C5)] There exists a constant $M>0$ such that for
$x\in \operatorname{dom}L,|x(\rho(t))|>M$, for all $t\in [0,1]$,
\begin{equation}
\begin{aligned}
&\alpha(1-\beta)\int^{\eta}_0(\eta -s)(f(s,x,x^{\Delta})+e(s))\nabla s \\
&+ (1-\alpha)\beta\int^{\xi}_0(\xi -s)(f(s,x,x^{\Delta})+e(s))\nabla s \\
&-(1-\alpha)\int^{1}_0(1-s)(f(s,x,x^{\Delta})+e(s))\nabla s
\neq 0.
\end{aligned}\label{e4.5}
\end{equation}
\end{itemize}
Then for each $e\in L^{1}[0,1]$, resonance problem \eqref{e1.1}, \eqref{e1.2} with
$\alpha\eta(1-\beta)+(1-\alpha)(1-\beta\xi)=0$ has at least one solution
provided that
$$
\|a\|_1+\|b\|_1<\frac{1}{2+|\frac{\beta+1}{1-\beta\xi}|}.
$$
\end{theorem}

The proof is similar to that of Theory 3.1 and it is omitted here.

\section{An example}

In this section we give an example to illustrate the main results of this article.
Let $T=\{0\}\cup\{\frac{1}{2^{n+1}}\}\cup[\frac{1}{2},1]$, $(n=1,2,\dots)$.
We consider the four-point boundary-value problem
\begin{equation}
x^{\Delta\nabla}(t)=\frac{1}{18}x(t)+\frac{1}{18}x^{\Delta}(t)
+\frac{1}{12}\sin(x^{\Delta}(t))^{1/5},\quad t\in (0,1)\cap T,\label{e5.1}
\end{equation}
subject to the boundary condition
\begin{equation}
x(0)= x(\frac{1}{3}),\quad x(1)=x(\frac{1}{2}),\label{e5.2}
\end{equation}
It is easy to see that $\alpha=\beta=1$,
$\eta=1/3$, $\xi=1/2$, $a(t)=b(t)=\frac{1}{18}$,
$c(t)=1/12$, $\|a(t)\|+\|b(t)\|<1/3$ and 
$$
f(t,\ x ,\ y)\leq \frac{1}{18}|x|+\frac{1}{18}|y|
+\frac{1}{12}|y|^{1/5},\quad
 f(t,\ x ,\ y)\geq \frac{1}{18}|x|-\frac{1}{18}|y|-\frac{1}{12}.
$$
Then all conditions of Theorem \ref{thm3.1} hold. Hence, the problem has at least
one nontrival solution at resonance.

\subsection*{Acknowledgements}
 The authors would like to thank the anonymous referees for their careful
reading of this manuscript and for suggesting useful changes.
 This work was sponsored by the NSFC (11201109), Anhui
Provincial Natural Science Foundation (1408085QA07), the Higher
School Natural Science Project of Anhui Province (KJ2014A200) and
the outstanding talents plan of Anhui High school.

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\end{document}