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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 61, 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}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/61\hfil Solutions to third-order multi-point BVPs]
{Solutions to third-order multi-point
boundary-value problems at resonance with three dimensional kernels}

\author[S. Li, J. Yin, Z. Du \hfil EJDE-2014/61\hfilneg]
{Shuang Li, Jian Yin, Zengji Du}  % in alphabetical order

\address{Shuang Li \newline
School of Management,
China University of Mining and Technology,
Xuzhou, Jiangsu 221116,  China}
\email{lishuangchina@163.com}

\address{Jian Yin \newline
School of Mathematics and Statistics, Jiangsu Normal University,
Xuzhou, Jiangsu 221116, China}
\email{yinjian719@163.com}

\address{Zengji Du (corresponding author)\newline
School of Mathematics and Statistics, Jiangsu Normal University,
Xuzhou, Jiangsu 221116, China}
\email{duzengji@163.com, Fax 0086-516-83403299}

\thanks{Submitted  October 13, 2013. Published March 5, 2014.}
\subjclass[2000]{34B15}
\keywords{Multi-point boundary-value problem; coincidence degree theory; 
\hfill\break\indent resonance}

\begin{abstract}
 In this article, we consider the boundary-value problem
 \begin{gather*}
 x'''(t)=f(t, x(t), x'(t),x''(t)), \quad t\in (0,1),\\
 x''(0)=\sum_{i=1}^{m}\alpha_i x''(\xi_i), \quad
 x'(0)=\sum_{k=1}^{l}\gamma_k x'(\sigma_{k}),\quad
 x(1)=\sum_{j=1}^{n}\beta_jx(\eta_j),
 \end{gather*}
 where $f: [0, 1]\times \mathbb{R}^3\to \mathbb{R}$ is a Carath\'eodory
 function, and the kernel to the linear operator has dimension three.
 Under some resonance conditions, by using the coincidence
 degree theorem, we show the existence of solutions. An example
 is given to illustrate our results. 
\end{abstract}

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

\section{Introduction}

This concerns the  third-order
nonlinear differential equation
\begin{equation}
x'''(t)=f(t, x(t), x'(t),x''(t)),\quad t\in(0,1), \label{e1.1}
\end{equation}
with the  boundary conditions
\begin{equation}
x''(0)=\sum_{i=1}^{m}\alpha_i x''(\xi_i),\quad
x'(0)=\sum_{k=1}^{l}\gamma_k x'(\sigma_k),\quad
x(1)=\sum_{j=1}^{n}\beta_jx(\eta_j),\label{e1.2}
\end{equation}
where $f: [0, 1]\times \mathbb{R}^3\to \mathbb{R}$ is a Carath\'eodory
function, $0<\xi_1<\dots<\xi_m<1$,
$0<\sigma_1<\dots<\sigma_l<1$, $0<\eta_1<\dots<\eta_n<1$,
$\alpha_{i}, \gamma_{k},\beta_j\in\mathbb{R}$
($i=1,\dots, m$; $k=1,\dots,l$; $j=1,\dots, n$) and
$\sigma_1>\{\xi_1,\dots,\xi_m\}$.


The existence of solutions for multi-point boundary-value
problems at resonance case has been extensively studied by many
authors \cite{d1,d2,d3,g1,l1,l2,l3,l4,x1}.
When the linear equation $Lx=x'''=0$ with the boundary
conditions \eqref{e1.2} has a non-trivial solution, i.e.
$\dim\ker L\geq1$. we say that boundary value problem (BVP for
short) \eqref{e1.1} and \eqref{e1.2} is a resonance problem.

The case of $\dim\ker L=1$ has been discussed by many authors
\cite{d1,d2,g1,l2}.
For the case of $\dim\ker L=2$, there are some results in \cite{d3,l3,l4,x1}.,
Lin, Du and Meng \cite{l3} discussed the
 third-order multi-point boundary value-problem with $\dim\ker L=2$,
\begin{gather}
x'''(t)=f(t, x(t), x'(t),x''(t)),\quad  t\in(0,1), \label{e1.3}\\
x''(0)=\sum_{i=1}^{m}\alpha_i x''(\xi_i),\quad
 x'(0)=0, \quad x(1)=\sum_{j=1}^{n}\beta_jx(\eta_j).\label{e1.4}
\end{gather}
Zhao, Liang and Ren  \cite{z1} studied the  nonlinear
third-order boundary-value problems with $\dim\ker L=3$,
\begin{gather*}
x'''(t)=f(t, x(t), x'(t),x''(t))+e(t),\quad t\in(0,1),\\
x'(0)=\sum_{j=1}^{n}\alpha_jx'(\xi_j),\quad
x''(0)=x''(\eta),\quad x''(1)=x''(\zeta).
\end{gather*}
In \cite{z1}, the results are obtained under the assumption that
\begin{equation}
M=\begin{vmatrix}
m_{11}&m_{12}&m_{13}\\m_{21}&m_{22}&m_{23}\\
m_{31}&m_{32}&m_{33}\\
\end{vmatrix} \neq 0. \label{e1.5}
\end{equation}
This condition  is difficult to verify and looks superfluous.

 Inspired by the above results, in this paper, we study
\eqref{e1.1}-\eqref{e1.2} at resonance and show that the 
assumption
\begin{align*}
&\Lambda(p,q,r)\\
&=\begin{vmatrix}
(p+1)(p+2)\sum_{i=1}^m\alpha_i\xi_i^p 
 &(p+2)\sum_{k=1}^l\gamma_k\sigma_k^{p+1}
 &2(1-\sum_{j=1}^n\beta_j\eta_j^{p+2})\\
(q+1)(q+2)\sum_{i=1}^m\alpha_i\xi_i^q 
 &(q+2)\sum_{k=1}^l\gamma_k\sigma_k^{q+1}
 &2(1-\sum_{j=1}^n\beta_j\eta_j^{q+2})\\
(r+1)(r+2)\sum_{i=1}^m\alpha_i\xi_i^r
 &(r+2)\sum_{k=1}^l\gamma_k\sigma_k^{r+1}
 &2(1-\sum_{j=1}^n\beta_j\eta_j^{r+2})
\end{vmatrix}\neq0 
\end{align*}
can replace \eqref{e1.5} from \cite{l1,z1}, by using Lemma 
\ref{lem3.1} below. 
If there exists $\sigma_k(k=1,2,\dots,l)$ satisfying 
$x'(0)=\sum_{k=1}^{l}\gamma_k x'(\sigma_k)=0$,
then BVP \eqref{e1.3}-\eqref{e1.4} is a special case of
 BVP \eqref{e1.1}-\eqref{e1.2}.

  The remaining part of this article is organized as follows: 
In section 2, we will state some definitions and lemmas which would be
useful in the proving of main results of this paper. 
In section 3, by applying Mawhin coincidence degree theory, we obtain some
sufficient conditions which guarantee the existence of solution for
BVP \eqref{e1.1}-\eqref{e1.2} at resonance case. 
In the last section, an example is given to illustrate our results.

\section{Preliminaries}


 Now, some notation and an abstract existence result \cite{m1} are
introduced.
  Let $Y, Z$ be real Banach spaces and let $L:\operatorname{dom}L\subset
Y\to Z$ be a linear operator which is a Fredholm map of
index zero and $P: Y\to Y$, $Q: Z\to Z$ be continuous
projectors such that $\operatorname{Im}P = \ker L$, 
$\ker Q = \operatorname{Im}L$ and $Y =\ker L \oplus \ker P$, 
$Z =\operatorname{Im}L\oplus \operatorname{Im}Q$.
 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 
of that map by $K_P$. Let $\Omega$ be an open bounded subset of $Y$
such that $\operatorname{dom}L\cap \Omega\neq\emptyset$, the map $N:Y\to Z$ 
is said to be $L$-compact on $\bar{\Omega}$ if the map $QN(\bar{\Omega})$ is
bounded and $K_P(I-Q)N:\bar{\Omega}\to Y$ is compact.

\begin{lemma}[{\cite[Theorem IV]{m1}}]  \label{lem2.1}
Let $L$ be a Fredholm map of index zero and let $N$ be $L$-compact on
$\bar{\Omega}$. Assume that the following conditions are satisfied:
\begin{itemize}
\item[(i)] $Lx\neq \lambda Nx$ for every $(x,\lambda)\in
((\operatorname{dom}L\backslash\ker L)\cap\partial\Omega)\times [0,1]$;

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

\item[(iii)] $\deg (QN|_{\ker L},\Omega\cap\ker L,0)\neq 0$, where 
$Q:Z\to Z$ is a continuous projector as above with $\operatorname{Im}L=\ker Q$.
\end{itemize}
Then the abstract equation $Lx=Nx$ has at least one solution in
$\operatorname{dom}L\cap\bar{\Omega}$.
\end{lemma}

We use the classical spaces $C[0,1]$, $C^1[0,1]$, $C^2[0,1]$ and $L^1[0,1]$. 
For $x$ in $C^2[0,1]$, we use the norms
$\|x\|_{\infty}=\max_{t\in (0, 1)}\{|x(t)|\}$ and 
\[
\|x\|=\max\{\|x\|_{\infty}, \|x'\|_{\infty}, \|x''\|_{\infty}\}. 
\]
For $L^1[0,1]$ we denote the norm by $\|\cdot\|_1$.
We define the Sobolev space
\begin{align*}
W^{3, 1}(0, 1)=\big\{&x: [0, 1]\to R: x, x', x'' 
\text{ are  absolutely  continuous on $[0, 1]$}\\
&\text{with }  x'''\in L^{1}[0, 1]\big\}.
\end{align*}
Let $Y=C^2[0,1]$, $Z=L^1[0,1]$, and define the linear operator 
$L: \operatorname{dom}L\subset Y\to Z$ as $Lx=x'''$, $x\in \operatorname{dom}L$,
where
$$
\operatorname{dom}L=\{x\in W^{3, 1}(0, 1):x \text{ satisfies boundary conditions 
\eqref{e1.2}}\}.
$$
We define $N: Y\to Z$ as
$$
Nx=f(t, x(t), x'(t),x''(t)), \quad t\in (0,1), 
$$
then \eqref{e1.1}-\eqref{e1.2} can be written as $Lx=Nx$.

The resonance conditions of \eqref{e1.1}-\eqref{e1.2} are as
follows
\begin{equation} \label{eC}
\begin{gathered}
\sum_{i=1}^{m}\alpha_i=1,\quad \sum_{k=1}^l\gamma_k=1,\quad \sum_{j=1}^n\beta_j=1,\\
\sum_{j=1}^n\beta_j\eta_j=1,\quad \sum_{j=1}^n\beta_j\eta_j^2=1,\quad
\sum_{k=1}^l\gamma_k\sigma_k=0.
\end{gathered}
\end{equation}
  Define the following symbols:
\begin{gather*}
\begin{aligned}
&\Lambda(p,q,r)\\
&=\begin{vmatrix}
(p+1)(p+2)\sum_{i=1}^m\alpha_i\xi_i^p &(p+2)\sum_{k=1}^l\gamma_k\sigma_k^{p+1}
 &2(1-\sum_{j=1}^n\beta_j\eta_j^{p+2})\\
(q+1)(q+2)\sum_{i=1}^m\alpha_i\xi_i^q&(q+2)\sum_{k=1}^l\gamma_k\sigma_k^{q+1}
 &2(1-\sum_{j=1}^n\beta_j\eta_j^{q+2})\\
(r+1)(r+2)\sum_{i=1}^m\alpha_i\xi_i^r&(r+2)\sum_{k=1}^l\gamma_k\sigma_k^{r+1}
 &2(1-\sum_{j=1}^n\beta_j\eta_j^{r+2})
\end{vmatrix},
\end{aligned}
\\
M_{11}=\begin{vmatrix}
(q+2)\sum_{k=1}^l\gamma_k\sigma_k^{q+1}&2(1-\sum_{j=1}^n\beta_j\eta_j^{q+2})\\
(r+2)\sum_{k=1}^l\gamma_k\sigma_k^{r+1}&2(1-\sum_{j=1}^n\beta_j\eta_j^{r+2})
\end{vmatrix},
\\
M_{12}=-\begin{vmatrix}
(q+1)(q+2)\sum_{i=1}^m\alpha_i\xi_i^q&2(1-\sum_{j=1}^n\beta_j\eta_j^{q+2})\\
(r+1)(r+2)\sum_{i=1}^m\alpha_i\xi_i^r&2(1-\sum_{j=1}^n\beta_j\eta_j^{r+2})
\end{vmatrix},
\\
M_{13}=\begin{vmatrix}
(q+1)(q+2)\sum_{i=1}^m\alpha_i\xi_i^q&(q+2)\sum_{k=1}^l\gamma_k\sigma_k^{q+1}\\
(r+1)(r+2)\sum_{i=1}^m\alpha_i\xi_i^r&(r+2)\sum_{k=1}^l\gamma_k\sigma_k^{r+1}
\end{vmatrix},
\\
M_{21}=-\begin{vmatrix}
(p+2)\sum_{k=1}^l\gamma_k\sigma_k^{p+1}&2(1-\sum_{j=1}^n\beta_j\eta_j^{p+2})\\
(r+2)\sum_{k=1}^l\gamma_k\sigma_k^{r+1}&2(1-\sum_{j=1}^n\beta_j\eta_j^{r+2})
\end{vmatrix},
\\
M_{22}=\begin{vmatrix}
(p+1)(p+2)\sum_{i=1}^m\alpha_i\xi_i^p&2(1-\sum_{j=1}^n\beta_j\eta_j^{p+2})\\
(r+1)(r+2)\sum_{i=1}^m\alpha_i\xi_i^r&2(1-\sum_{j=1}^n\beta_j\eta_j^{r+2})
\end{vmatrix},
\\
M_{23}=-\begin{vmatrix}
(p+1)(p+2)\sum_{i=1}^m\alpha_i\xi_i^p&(p+2)\sum_{k=1}^l\gamma_k\sigma_k^{p+1}\\
(r+1)(r+2)\sum_{i=1}^m\alpha_i\xi_i^r&(r+2)\sum_{k=1}^l\gamma_k\sigma_k^{r+1}
\end{vmatrix},
\\
M_{31}=\begin{vmatrix}
(p+2)\sum_{k=1}^l\gamma_k\sigma_k^{p+1}&2(1-\sum_{j=1}^n\beta_j\eta_j^{p+2})\\
(q+2)\sum_{k=1}^l\gamma_k\sigma_k^{q+1}&2(1-\sum_{j=1}^n\beta_j\eta_j^{q+2})
\end{vmatrix},
\\
M_{32}=-\begin{vmatrix}
(p+1)(p+2)\sum_{i=1}^m\alpha_i\xi_i^p&2(1-\sum_{j=1}^n\beta_j\eta_j^{p+2})\\
(q+1)(q+2)\sum_{i=1}^m\alpha_i\xi_i^q&2(1-\sum_{j=1}^n\beta_j\eta_j^{q+2})
\end{vmatrix},
\\
M_{33}=\begin{vmatrix}
(p+1)(p+2)\sum_{i=1}^m\alpha_i\xi_i^p&(p+2)\sum_{k=1}^l\gamma_k\sigma_k^{p+1}\\
(q+1)(q+2)\sum_{i=1}^m\alpha_i\xi_i^q&(q+2)\sum_{k=1}^l\gamma_k\sigma_k^{q+1}
\end{vmatrix}
\end{gather*}

\section{Main results}


\begin{lemma} \label{lem3.1}
Assume condition \eqref{eC} holds, then
there exist $p\in\{1,2,\dots,n\}$, $q\in{Z^+}$, $q\geq{p+1}$ and
$r\in{Z^+}$  large enough number, such that
$\Lambda(p,q,r)\neq0$.
\end{lemma}

\begin{proof}
Clearly there exists $p\in\{1,2,\dots,n\}$
such that $\sum_{i=1}^m\alpha_i\xi_i^{p+2}\neq0$. Otherwise, we
have
$\sum_{i=1}^m\alpha_i\xi_i^{p+2}=0$ and
$p\in\{1,2,\dots,n\}$, then
$$
\sum_{i=1}^m\alpha_i\xi_i^j(1-\xi_i)=0,\quad j=1,2,\dots,{n-1};
$$
i.e.,
\[ 
\begin{pmatrix}
\xi_1(1-\xi_1)&\dots&\xi_m(1-\xi_m)\\
\vdots&\ddots&\vdots\\
\xi_1^{n-1}(1-\xi_1)&\dots&\xi_m^{n-1}(1-\xi_m)\\
\end{pmatrix}
\begin{pmatrix} \alpha_1\\ \vdots\\ \alpha_m
\end{pmatrix}
= \begin{pmatrix} 0\\ \vdots\\ 0
\end{pmatrix}.
\]
Because 
\[
 \begin{vmatrix}
\xi_1(1-\xi_1)&\dots&\xi_m(1-\xi_m)\\
\vdots&\ddots&\vdots\\
\xi_1^{n-1}(1-\xi_1)&\dots&\xi_m^{n-1}(1-\xi_m)\\
\end{vmatrix}
=\prod_{i=1}^m\xi_i(1-\xi_i)\prod_{1\leq i\leq
j\leq{m-1}}(\xi_j-\xi_i)
\neq 0.
\]
So we have $\alpha_i=0$, $i\in\{1,2,\dots,m\}$. Which is a
contradiction to $\sum_{i=1}^m\alpha_i=1$ of condition \eqref{eC}.

  Similarly, there exists $q\in\{1,2,\dots,{n+1}\}$, such that
$\sum_{i=1}^m\alpha_i\xi_i^q\neq0$. And for each 
$s\in Z,s\geq0$, there exists $k_s\in\{sn+1,\dots,(s+1)n\}$ such that
$$
\sum_{ k=1}^l\gamma_k\sigma_k^{k_s+1}\neq0,\quad
\sum_{i=1}^m\alpha_i\xi_i^{k_s}\neq0.
$$
Set
$$
S=\Big\{{k_s}\in Z:\sum_{k=1}^l\gamma_k\sigma_k^{q+1}=
\frac{(q+1)\sum_{i=1}^m\alpha_i\xi_i^q
\sum_{k=1}^l\gamma_k\sigma_k^{k_s+1}}{(k_s+1)\sum_{i=1}^m\alpha_i\xi_i^{k_s}}\Big\},
$$
then $S$ is a finite set. If else, there exists a monotone
sequence
$\{k_{s_t}\}$, $t=1,2,\dots$, $k_{s_t}\leq k_{s_{t+1}}$,
such that
$$
\sum_{k=1}^l\gamma_k\sigma_k^{q+1}=
\frac{(q+1)\sum_{i=1}^m\alpha_i\xi_i^q
\sum_{k=1}^l\gamma_k\sigma_k^{k_{s_t}+1}}{(k_{s_t}+1)
\sum_{i=1}^m\alpha_i\xi_i^{k_{s_t}}}.
$$
For $\sigma_1>\{\xi_1,\dots,\xi_m\}$,
$$
\sum_{k=1}^l\gamma_k\sigma_k^{q+1}=
\lim_{k_{s_t}\to\infty}\frac{(q+1)
\sum_{i=1}^m\alpha_i\xi_i^q\sum_{k=1}^l\gamma_k
\sigma_k^{k_{s_t}+1}}{(k_{s_t}+1)\sum_{i=1}^m\alpha_i\xi_i^{k_{s_t}}}=\infty,
$$
which is a contradiction. Then 
$$
\begin{vmatrix}
(p+1)(p+2)\sum_{i=1}^m\alpha_i\xi_i^p&(p+2)\sum_{k=1}^l\gamma_k\sigma_k^{p+1}\\
(q+1)(q+2)\sum_{i=1}^m\alpha_i\xi_i^q&(q+2)\sum_{k=1}^l\gamma_k\sigma_k^{q+1}
\end{vmatrix}
\neq0
$$ 
Thus, we have
\begin{align*}
&\lim_{r\to\infty}\Lambda(p,q,r)\\
&=\begin{vmatrix}
(p+1)(p+2)\sum_{i=1}^m\alpha_i\xi_i^p&(p+2)\sum_{k=1}^l\gamma_k\sigma_k^{p+1}
 &2(1-\sum_{j=1}^n\beta_j\eta_j^{p+2})\\
(q+1)(q+2)\sum_{i=1}^m\alpha_i\xi_i^q&(q+2)\sum_{k=1}^l\gamma_k\sigma_k^{q+1}
 &2(1-\sum_{j=1}^n\beta_j\eta_j^{q+2})\\
 0&0&2\\
\end{vmatrix}\\
&=2\begin{vmatrix}
 (p+1)(p+2)\sum_{i=1}^m\alpha_i\xi_i^p&(p+2)\sum_{k=1}^l\gamma_k\sigma_k^{p+1}\\
 (q+1)(q+2)\sum_{i=1}^m\alpha_i\xi_i^q&(q+2)\sum_{k=1}^l\gamma_k\sigma_k^{q+1}
\end{vmatrix} \neq0.
\end{align*}
 So there exist
$p\in\{1,2,\dots,n\}$, $q\in{Z^+}$, $q\geq{p+1}$ and $r\in{Z^+}$ 
large enough number, such that $\Lambda(p,q,r)\neq0$. 
\end{proof}

\begin{lemma} \label{lem3.2}
If condition \eqref{eC} holds,
then $L:\operatorname{dom}L\subset Y\to Z$ is a Fredholm operator of
index zero. Furthermore, the continuous projector operator
$Q:Z\to Z$ can be defined by
$$
Q(y)=(T_1y(t))t^{p-1}+(T_2y(t))t^{q-1}+(T_3y(t))t^{r-1},
$$
where  $p,q,r$ are given by Lemma \ref{lem3.1} and
\begin{gather*}
T_1y=\frac{p(p+1)(p+2)}{\Lambda(p,q,r)}[M_{11}Q_1y+M_{12}Q_2y+M_{13}Q_3y],\\
T_2y=\frac{q(q+1)(q+2)}{\Lambda(p,q,r)}[M_{21}Q_1y+M_{22}Q_2y+M_{23}Q_3y],\\
T_3y=\frac{r(r+1)(r+2)}{\Lambda(p,q,r)}[M_{31}Q_1y+M_{32}Q_2y+M_{33}Q_3y],\\
Q_1y=\sum_{i=1}^m\int_0^{\xi_i}y(s)ds,
Q_2y=\sum_{k=1}^l\gamma_k\int_0^{\sigma_k}(\sigma_k-s)y(s)ds,\\
Q_3y=\int_0^1(1-s)^2y(s)ds-\sum_{j=1}^n\beta_j\int_0^{\eta_j}(\eta_j-s)^2y(s)ds.
\end{gather*}
The linear operator $K_P:\operatorname{Im}L\to \operatorname{dom}L\cap
\ker P$ can be written as
$$
K_Py(t)=\frac{1}{2}\int_0^t(t-s)^2y(s)ds,~y\in \operatorname{Im}L.
$$ 
Furthermore
$\|K_Py\|\leq\|y\|_1$, $y\in \operatorname{Im}L$.
\end{lemma}

\begin{proof}
  It is clear that 
$$
\ker L=\{x\in \operatorname{dom}L:x=a+bt+ct^2,\, a,b,c\in\mathbb{R}\}.
$$ 
Now we show that
\begin{equation} \label{e3.1}
\operatorname{Im}L=\{y\in Z:Q_1y=Q_2y=Q_3y=0\}.
\end{equation}
Since the problem
\begin{equation}
x'''=y \label{e3.2}
\end{equation}
has a solution $x(t)$, such that boundary conditions \eqref{e1.2} hold; i.e.,
$$
x''(0)=\sum_{i=1}^{m}\alpha_i x''(\xi_i),\quad
x'(0)=\sum_{k=1}^{l}\gamma_kx'(\sigma_k), \quad
x(1)=\sum_{j=1}^{n}\beta_jx(\eta_j),
$$
if and only if
\begin{equation}
Q_1y=Q_2y=Q_3y=0. \label{e3.3}
\end{equation}
In fact, if \eqref{e3.2} has a solution $x(t)$ that satisfies the boundary
conditions \eqref{e1.2}, then we have
$$
x(t)=x(0)+x'(0)t+\frac{1}{2}x''(0)t^2+\frac{1}{2}\int_0^t(t-s)^2y(s)ds.
$$
According to condition \eqref{eC}, we have $Q_1y=Q_2y=Q_3y=0$.

On the other hand, we let
$$
x(t)=a+bt+ct^2+\frac{1}{2}\int_0^t(t-s)^2y(s)ds,
$$
where $a, b, c$ are arbitrary constants. If \eqref{e3.3} holds, then $x(t)$
is a solution of \eqref{e3.2} and \eqref{e1.2}. Hence \eqref{e3.1} holds.
  By Lemma \ref{lem3.1}, there exists
$p\in\{1,2,\dots,n\},~q\in{Z^+},~q\geq{p+1}$ and $r\in{Z^+}$ is a
large enough number, such that $\Lambda(p,q,r)\neq0$. Set
\begin{gather*}
T_1y=\frac{p(p+1)(p+2)}{\Lambda(p,q,r)}[M_{11}Q_1y+M_{12}Q_2y+M_{13}Q_3y],\\
T_2y=\frac{q(q+1)(q+2)}{\Lambda(p,q,r)}[M_{21}Q_1y+M_{22}Q_2y+M_{23}Q_3y],\\
T_3y=\frac{r(r+1)(r+2)}{\Lambda(p,q,r)}[M_{31}Q_1y+M_{32}Q_2y+M_{33}Q_3y].
\end{gather*}
And we define
$$
Q(y)=(T_1y(t))t^{p-1}+(T_2y(t))t^{q-1}+(T_3y(t))t^{r-1},
$$ 
then $\dim\operatorname{Im}Q=3$.
So we have
\begin{align*}
&T_1((T_1y)t^{p-1})\\
&= \frac{p(p+1)(p+2)}{\Lambda(p,q,r)}[M_{11}Q_1((T_1y)t^{p-1})
 +M_{12}Q_2((T_1y)t^{p-1})+M_{13}Q_3((T_1y)t^{p-1})]\\
&= \frac{p(p+1)(p+2)}{\Lambda(p,q,r)}[M_{11}Q_1(t^{p-1})+M_{12}Q_2(t^{p-1})
 +M_{13}Q_3(t^{p-1})](T_1y)\\
&= \frac{1}{\Lambda(p,q,r)}[M_{11}(p+1)(p+2)\sum_{i=1}^m\alpha_i\xi_i^p
 +M_{12}(p+2)\sum_{k=1}^l\gamma_k\sigma_k^{p+1}\\
&\quad +2M_{13}(1-\sum_{j=1}^n\beta_j\eta_j^{p+2})](T_1y)\\
&= T_1y,
\end{align*}
\begin{align*}
&T_1((T_2y)t^{q-1})\\
&= \frac{q(q+1)(q+2)}{\Lambda(p,q,r)}[M_{11}Q_1((T_2y)t^{q-1})
 +M_{12}Q_2((T_2y)t^{q-1})+M_{13}Q_3((T_2y)t^{q-1})]\\
&= \frac{q(q+1)(q+2)}{\Lambda(p,q,r)}[M_{11}Q_1(t^{q-1})+M_{12}Q_2(t^{q-1})
 +M_{13}Q_3(t^{q-1})](T_2y)\\
&= \frac{1}{\Lambda(p,q,r)}[M_{11}(q+1)(q+2)\sum_{i=1}^m\alpha_i\xi_i^q
 +M_{12}(q+2)\sum_{k=1}^l\gamma_k\sigma_k^{q+1}\\
&\quad +2M_{13}(1-\sum_{j=1}^n\beta_j\eta_j^{q+2})](T_2y)
= 0,
\end{align*}
\begin{align*}
&T_1((T_3y)t^{r-1})\\
&= \frac{r(r+1)(r+2)}{\Lambda(p,q,r)}[M_{11}Q_1((T_3y)t^{r-1})
 +M_{12}Q_2((T_3y)t^{r-1})+M_{13}Q_3((T_3y)t^{r-1})]\\
&= \frac{r(r+1)(r+2)}{\Lambda(p,q,r)}[M_{11}Q_1(t^{r-1})+M_{12}Q_2(t^{r-1})
 +M_{13}Q_3(t^{r-1})](T_3y)\\
&= \frac{1}{\Lambda(p,q,r)}[M_{11}(r+1)(r+2)\sum_{i=1}^m\alpha_i\xi_i^r
 +M_{12}(r+2)\sum_{k=1}^l\gamma_k\sigma_k^{r+1}\\
&\quad +2M_{13}(1-\sum_{j=1}^n\beta_j\eta_j^{r+2})](T_3y)
= 0.
\end{align*}
Similarly, we obtain
\begin{gather*}
T_2((T_1y)t^{p-1})=0, \quad T_2((T_2y)t^{q-1})=T_2y,\quad
T_2((T_3y)t^{r-1})=0,\\
T_3((T_1y)t^{p-1})=0, \quad T_3((T_2y)t^{q-1})=0, \quad
T_3((T_3y)t^{r-1})=T_3y.
\end{gather*}
So we have
\begin{align*}
Q^{2}y
&= Q(T_1y(t))t^{p-1}+(T_2y(t))t^{q-1}+(T_3y(t))t^{r-1})\\
&= T_1((T_1y(t))t^{p-1}+(T_2y(t))t^{q-1}+(T_3y(t))t^{r-1})t^{p-1}\\
&\quad +T_2((T_1y(t))t^{p-1}+(T_2y(t))t^{q-1}+(T_3y(t))t^{r-1})t^{q-1}\\
&\quad +T_3((T_1y(t))t^{p-1}+(T_2y(t))t^{q-1}+(T_3y(t))t^{r-1})t^{r-1}\\
&= (T_1y(t))t^{p-1}+(T_2y(t))t^{q-1}+(T_3y(t))t^{r-1}\\
&=  Qy .
\end{align*}
Thus, $Q$ is a well defined operator.

  Now we need to show that $\ker Q=\operatorname{Im}L$. 
If $y\in \ker Q$, then $Qy=0$. By
the definition of $Qy$, we have
\begin{gather*}
M_{11}Q_1y+M_{12}Q_2y+M_{13}Q_3y=0,\\
M_{21}Q_1y+M_{22}Q_2y+M_{23}Q_3y=0,\\
M_{31}Q_1y+M_{32}Q_2y+M_{33}Q_3y=0.
\end{gather*}
 Because of
\[ 
\begin{vmatrix}
M_{11} & M_{12} & M_{13}\\
M_{21} & M_{22} & M_{23}\\
M_{31} & M_{32} & M_{33}
\end{vmatrix}
=\Lambda^{2}(p,q,r)\neq 0,
\]
we have $Q_1y=Q_2y=Q_3y=0$, i.e. $y\in ImL$.

  If $y\in \operatorname{Im}L$, we have $Q_1y=Q_2y=Q_3y=0$. From the
definition of $Qy$, it is obvious that $Qy=0$, thus $y\in KerQ$.
Hence, $\ker Q=\operatorname{Im}L$.

  For $y\in Z$, let $y=(y-Qy)+Qy$, since $Q(y-Qy)=Qy-Q^2y=0$, we know
$y-Qy\in \ker Q=\operatorname{Im}L$, and we have $Qy\in \operatorname{Im}Q$.
Thus $Z=\operatorname{Im}L+\operatorname{Im}Q$. And for any 
$y\in \operatorname{Im}L\cap \operatorname{Im}Q$, from $y\in \operatorname{Im}Q$,
 there exists constants
$a,b,c\in\mathbb{R}$, such that 
$y(t)=at^{p-1}+bt^{q-1}+ct^{r-1}$. From
$y\in \operatorname{Im}L$, we obtain
\begin{equation} \label{e3.4}
\begin{gathered}
\frac{a}{p}\sum_{i=1}^m\alpha_i\xi_i^p+\frac{b}{q}\sum_{i=1}^m\alpha_i\xi_i^q
+\frac{c}{r}\sum_{i=1}^m\alpha_i\xi_i^r=0,
\\
\frac{a}{p(p+1)}\sum_{k=1}^l\gamma_k\sigma_k^{p+1}
+\frac{b}{q(q+1)}\sum_{k=1}^l\gamma_k\sigma_k^{q+1}
+\frac{c}{r(r+1)}\sum_{k=1}^l\gamma_k\sigma_k^{r+1}=0,
\\
\begin{aligned}
&a(\frac{1}{p}-\frac{2}{p+1}+\frac{1}{p+2})(1-\sum_{j=1}^n\beta_j\eta_j^{p+2})
+b(\frac{1}{q}-\frac{2}{q+1}+\frac{1}{q+2})(1-\sum_{j=1}^n\beta_j\eta_j^{q+2})\\
&+c(\frac{1}{r}-\frac{2}{r+1}+\frac{1}{r+2})(1-\sum_{j=1}^n\beta_j\eta_j^{r+2})=0.
\end{aligned}
\end{gathered}
\end{equation}
In view of
\begin{align*}
&\begin{vmatrix}
\frac{1}{p}\sum_{i=1}^m\alpha_i\xi_i^p
 &\frac{1}{q}\sum_{i=1}^m\alpha_i\xi_i^q&\frac{1}{r}\sum_{i=1}^m\alpha_i\xi_i^r,\\
\frac{1}{p(p+1)}\sum_{k=1}^l\gamma_k\sigma_k^{p+1}
 &\frac{1}{q(q+1)}\sum_{k=1}^l\gamma_k\sigma_k^{q+1}
 &\frac{1}{r(r+1)}\sum_{k=1}^l\gamma_k\sigma_k^{r+1}\\
A_{31} &A_{32} &A_{33}
\end{vmatrix}
\\
&=\frac{1}{p(p+1)(p+2)q(q+1)(q+2)r(r+1)(r+2)}\Lambda(p,q,r)\neq0,
\end{align*}
where the entries of the third row are
\begin{gather*}
A_{31}= (\frac{1}{p}-\frac{2}{p+1}+\frac{1}{p+2})(1-\sum_{j=1}^n\beta_j\eta_j^{p+2})\\
A_{32}=(\frac{1}{q}-\frac{2}{q+1}+\frac{1}{q+2})(1-\sum_{j=1}^n\beta_j\eta_j^{q+2})\\
A_{33}=(\frac{1}{r}-\frac{2}{r+1}+\frac{1}{r+2})(1-\sum_{j=1}^n\beta_j\eta_j^{r+2})\,.
\end{gather*}

\marginpar{I split the above matrix, please check it}

  Therefore \eqref{e3.4} has an unique solution $a=b=c=0$, which implies
$\operatorname{Im}Q\cap\operatorname{Im}L=\{0\}$ and
$Z=\operatorname{Im}L\oplus \operatorname{Im}Q$. Since 
$\dim\ker L=\text{dimIm}Q=\text{codimIm}L=3$,
thus $L$ is a Fredholm map of index zero.

Let $P:Y\to Y$ be defined by
$$
Px(t)=x(0)+x'(0)t+\frac{1}{2}x''(0)t^2,\quad t\in(0,1).
$$ 
then, the generalized inverse operator 
$K_P:\operatorname{Im}L\to{\operatorname{dom}L\cap \ker P}$ be defined by
$$
K_P(y(t))=\frac{1}{2}\int_0^t(t-s)^2y(s)ds,\quad y\in \operatorname{Im}L.
$$ 
If $y\in \operatorname{Im}L$, we have $(LK_P)(y(t))=(K_Py)'''=y(t)$.
If $x(t)\in{\operatorname{dom}L\cap \ker P}$, we have
\begin{align*}
(K_PL)x(t)
&= (K_P)(x'''(t))\\
&= \frac{1}{2}\int_0^t(t-s)^2x'''(s)ds\\
&= x(t)-[x(0)+x'(0)t+\frac{1}{2}x''(0)t^2]\\
&= x(t)-Px(t).
\end{align*}
Because $x\in {\operatorname{dom}L\cap \ker P}$, we know that
$Px(t)=0$. Thus $(K_PL)(x(t))=x(t)$.
It is obviously that $\|K_Py\|\leq\|y\|_1$. 
\end{proof}

For the next theorem we use the assumptions:
\begin{itemize}
\item[(H1)] There exists 
$\alpha(t), \beta(t), \gamma(t), \theta(t)\in L^1[0,1]$, 
such that
for all $(x_1,x_2,x_3)$ in $\mathbb{R}^3$, $t\in(0,1)$,
$$
|f(t,x_1,x_2,x_3)|\leq\alpha(t)|x_1|+\beta(t)|x_2|+\gamma(t)|x_3|+\theta(t).
$$

\item[(H2)] There exists a constant $A>0$ such that for
$x(t)\in\operatorname{dom}L$, if $|x(t)|>A$ or $|x'(t)|>A$ or $|x''(t)|>A$
for all $t\in (0,1)$, then
$Q_1N(x(t))\neq0$ or $Q_2N(x(t))\neq0$ or $Q_3N(x(t))\neq0$.

\item[(H3)] There exists a constant $B>0$ such that for $a,b,c\in\mathbb{R}$,
if $|a|>B$, $|b|>B$, $|c|>B$, then either
\begin{equation}
Q_1N(a+bt+ct^2)+Q_2N(a+bt+ct^2)+Q_3N(a+bt+ct^2)>0, \label{e3.5}
\end{equation}
or
\begin{equation}
Q_1N(a+bt+ct^2)+Q_2N(a+bt+ct^2)+Q_3N(a+bt+ct^2)<0. \label{e3.6}
\end{equation}
\end{itemize}

\begin{theorem} \label{thm3.1}
Let the conditions \eqref{eC}, {\rm (H1), (H2), (H3)} hold.
Then BVP \eqref{e1.1}-\eqref{e1.2} have at least one solution in
$C^2[0,1]$, provided that $\|\alpha\|_1+\|\beta\|_1+\|\gamma\|_1<1$.
\end{theorem}

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

\noindent \textbf{Step 1:} Let
$$
\Omega_1=\{x\in\operatorname{dom}L\backslash \ker L:Lx=\lambda
Nx, \lambda\in[0,1]\}.
$$ 
Then $\Omega_1$ is bounded.
Suppose that $x\in\Omega_1$, we have $Lx=\lambda Nx$. 
Thus $\lambda\neq0,~Nx\in \operatorname{Im}L=\ker Q$, hence
$$
Q_1(Nx(t))=Q_2(Nx(t))=Q_3(Nx(t))=0.
$$ 
From (H2), there exists
$t_1, t_2, t_3\in(0,1)$, such that 
$|x(t_1)|\leq A,~ |x'(t_2)|\leq A$, $|x''(t_3)|\leq A$. Since $x, x', x''$ 
are absolutely continuous for all $t\in(0,1)$, and
$$
x'(t)=x'(t_2)+\int_{t_2}^tx''(s)ds,\quad
x''(t)=x''(t_3)+\int_{t_3}^tx'''(s)ds.
$$
Thus
$$
\|x''\|_\infty\leq A+\|x'''\|_1,~ \|x'\|_\infty\leq 2A+\|x'''\|_1.
$$ 
And
\begin{align*}
\|x\|_{\infty}
&\leq \|(I-P)x\|_{\infty}+\|Px\|_{\infty}=\|Px\|_{\infty}+\|K_PL(I-P)x\|_{\infty}\\
&= \|Px\|_{\infty}+\|K_PL x \|_{\infty}\leq \|Px\|_{\infty}+\|K_P\|L x\|_1\\
&\leq\|Px\|_{\infty}+\|L x\|_1=\|Px\|_{\infty}+\|x'''\|_1.
\end{align*}
  From $(H_1)$, we obtain
\begin{align*}
\|x'''\|_1
&=\|Lx\|_1 \leq \|Nx\|_1\\
&\leq \|\alpha\|_1\|x\|_\infty+\|\beta\|_1\|x'\|_\infty
 +\|\gamma\|_1\|x''\|_\infty+\|\theta\|_1\\
&\leq (\|\alpha\|_1+\|\beta\|_1+\|\gamma\|_1)\|x'''\|_1
 +(2\|\beta\|_1+\|\gamma\|_1)A+\|\alpha\|_1\|Px\|_{\infty}+\|\theta\|_1.
\end{align*}
Then
$$
\|x'''\|_1\leq\frac{1}{1-(\|\alpha\|_1+\|\beta\|_1
+\|\gamma\|_1)}[(2\|\beta\|_1+\|\gamma\|_1)A
+\|\alpha\|_1\|Px\|_{\infty}+\|\theta\|_1].
$$
So there exists a constant $M_1>0$ such that $\|x\|\leq M_1$. Hence
we show that $\Omega_1$ is bounded.
\smallskip

\noindent \textbf{Step 2:}
Let $\Omega_2=\{x\in \ker L:Nx\in \operatorname{Im}L\}$. Then $\Omega_2$ is bounded.
 Since $x\in\Omega_2$, $x\in \ker L=\{x\in
\operatorname{dom}L:x=a+bt+ct^2,\, a,b,c\in\mathbb{R}\}$, and $QNx=0$, thus,
$$
Q_1(N(a+bt+ct^2))=Q_2(N(a+bt+ct^2))=Q_3(N(a+bt+ct^2))=0.
$$
 From (H3),
$$
\|x\|\leq|a|+|b|+|c|\leq3B.
$$ 
So $\Omega_2$ is bounded.
\smallskip

\noindent \textbf{Step 3:} Let 
$$
\Omega_3=\{x\in \ker L:\lambda Jx+(1-\lambda)QNx=0,\; \lambda\in[0,1]\}.
$$ 
Here
$J:\ker L\to \operatorname{Im}Q$ is the linear isomorphism
given by
$$
J(a+bt+ct^2)=\frac{1}{\Lambda(p,q,r)}(a_1t^{p-1}+b_1t^{q-1}+c_1t^{r-1}),\quad
a,b,c\in \mathbb{R}
$$ 
where
\begin{gather*}
a_1=p(p+1)(p+2)(M_{11}|a|+M_{12}|b|+M_{13}|c|),\\
b_1=q(q+1)(q+2)(M_{21}|a|+M_{22}|b|+M_{23}|c|),\\
c_1=r(r+1)(r+2)(M_{31}|a|+M_{32}|b|+M_{33}|c|).
\end{gather*}
Then $\Omega_3$ is bounded.

Set 
\begin{gather*}
B_1=\frac{p(p+1)(p+2)}{\Lambda(p,q,r)},\quad
B_2=\frac{q(q+1)(q+2)}{\Lambda(p,q,r)},\quad
B_3=\frac{r(r+1)(r+2)}{\Lambda(p,q,r)},\\
X_1=\lambda|a|+(1-\lambda)Q_1N(a+bt+ct^2),\quad
X_2=\lambda|b|+(1-\lambda)Q_2N(a+bt+ct^2),\\
X_3=\lambda|c|+(1-\lambda)Q_3N(a+bt+ct^2).
\end{gather*}
Since $x(t)=a+bt+ct^2\in\Omega_3$, then we have 
$\lambda Jx+(1-\lambda)QNx=0$; i.e.,
\begin{gather*}
B_1M_{11}X_1+B_1M_{12}X_2+B_1M_{13}X_3=0,\\
B_2M_{21}X_1+B_2M_{22}X_2+B_2M_{23}X_3=0,\\
B_3M_{31}X_1+B_3M_{32}X_2+B_3M_{33}X_3=0.
\end{gather*}
Because 
\begin{align*}
\begin{vmatrix}
B_1M_{11}&B_1M_{12}&B_1M_{13}\\
B_2M_{21}&B_2M_{22}&B_2M_{23}\\
B_3M_{31}&B_3M_{32}&B_3M_{33}
\end{vmatrix} 
&= (B_1B_2B_3)
\begin{vmatrix}
M_{11}&M_{12}&M_{13}\\
M_{21}&M_{22}&M_{23}\\
M_{31}&M_{32}&M_{33}
\end{vmatrix}\\
&= \frac{p(p+1)(p+2)q(q+1)(q+2)r(r+1)(r+2)}{\Lambda(p,q,r)}
\neq 0.
\end{align*}
Then $X_1=X_2=X_3=0$. 
If $\lambda=1$, then $|a|=|b|=|c|=0$.
If $\lambda\neq1$ and $|a|>B$ or $|b|>B$ or $|c|>B$ and \eqref{e3.5} hold,
then 
\begin{align*}
\lambda(|a|+|b|+|c|)
&=-(1-\lambda)\big[Q_1(N(a+bt+ct^2))+Q_2(N(a+bt+ct^2))\\
&\quad +Q_3(N(a+bt+ct^2))\big]<0,
\end{align*}
which contradicts  $\lambda(|a|+|b|+|c|)>0$. Thus
$$
\|x\|\leq|a|+|b|+|c|\leq3B.
$$
So $\Omega_3$ is bounded.

  If \eqref{e3.6} holds, we  let
$$
\Omega_3=\{x\in \ker L:-\lambda
Jx+(1-\lambda)QNx=0,~\lambda\in[0,1]\}.
$$ 
Similarly, we can proof that $\Omega_3$ is bounded.
\smallskip

\noindent \textbf{Step 4:}
 In the following, we shall prove that all
the conditions of Lemma \ref{lem2.1} are satisfied. Let $\Omega\subset Y$ be
a bounded open set such that
$\cup_{i=1}^3\overline{\Omega_i}\subset\Omega$. 
By the Ascoli-Arzela theorem, we can show that
$K_P(I-Q)N:\overline{\Omega}\to Y$ is compact, thus 
$N$ is $L-$compact on $\overline{\Omega}$.
Then by the above argument, we obtain 
\begin{itemize}
\item[(i)] $Lx\neq\lambda Nx$ for each
$(x,\lambda)\in((\operatorname{dom}L\backslash
\ker L)\cap\partial\Omega)\times[0,1]$;

\item[(ii)] $Nx\notin \operatorname{Im}L$ for each $x\in
\ker L\cap\partial\Omega$.
\end{itemize}

  At last we  prove that (iii) of the Lemma \ref{lem2.1} is
satisfied. Let $H(x,\lambda)=\pm\lambda Jx+(1-\lambda)QNx$.
According to the above argument, we have $H(x,\lambda)\neq0$, for
$x\in \ker L\cap\partial\Omega$. Thus, by the homotopy property
of degree, we get
\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) \neq 0.
\end{align*}
Then by Lemma \ref{lem2.1}, $Lx=Nx$ has at least one solution in
$\operatorname{dom}L\cap\overline{\Omega}$, so that BVP \eqref{e1.1}-\eqref{e1.2} 
has at least one solution in $C^2[0,1]$. 
\end{proof}

\section{Applications}

We consider the  boundary value problem
\begin{gather}
x'''(t)=x''(t)+\sin x(t)(1-\cos x'(t)), t\in(0,1),\label{e4.1}\\
x''(0)=2x''(\frac{1}{5})-x''(\frac{1}{4}),\quad 
x'(0) =3x'(\frac{1}{3})-2x'(\frac{1}{2}),\\
x(1) =\frac{9}{5}x(\frac{1}{3})-4x(\frac{1}{2})+\frac{16}{5}x(\frac{3}{4}).
\label{e4.2}
\end{gather}
So 
$f(t,x(t),x'(t),x''(t))=x''(t)+\sin x(t)(1-\cos x'(t))$,
$\alpha_1=2$, $\alpha_2=-1$, $\xi_1=\frac{1}{5}$,
$\xi_2=\frac{1}{4}$, $\gamma_1=3$, $\gamma_2=-2$, $\sigma_1=\frac{1}{3}$,
$\sigma_2=\frac{1}{2}$, $\beta_1=\frac{9}{5}$, $\beta_2=-4$,
$\beta_3=\frac{16}{5}$, $\eta_1=\frac{1}{3}$, $\eta_2=\frac{1}{2}$,
$\eta_3=\frac{3}{4}$.
Then we have $\alpha_1-\alpha_2=1$, $\gamma_1+\gamma_2=1$,
$\beta_1+\beta_2+\beta_3=1$,
$\beta_1\eta_1+\beta_2\eta_2+\beta_3\eta_3=1$,
$\beta_1\eta_1^{2}+\beta_2\eta_2^{2}+\beta_3\eta_3^{2}=1$,
$\gamma_1\sigma_1+\gamma_2\sigma_2=0$. So the condition \eqref{eC}
holds.

  By calculations, we obtain
\begin{gather*}
Q_1y=2\int_{0}^{1/5}y(s)ds,\quad
Q_2y=3\int_{0}^{\frac{1}{3}}(\frac{1}{3}-s)y(s)ds
 -2\int_{0}^{1/2}(\frac{1}{2}-s)y(s)ds,
\\
\begin{aligned}
Q_3y&=\int_{0}^{1}(1-s)^2y(s)ds-\frac{9}{5}
\int_{0}^{\frac{1}{3}}(\frac{1}{3}-s)^2y(s)ds\\
&\quad +4\int_{0}^{1/2}(\frac{1}{2}-s)^2y(s)ds
-\frac{16}{5}\int_{0}^{3/4}(\frac{3}{4}-s)^2y(s)ds,
\end{aligned} \\
\Lambda(1,2,3)=\frac{1225231}{1834400},
\\
M_{11}=-\frac{5731015}{1508544},\quad M_{12}=\frac{45787}{311040},\quad 
M_{13}=-\frac{3429}{38880},
\\
M_{21}=\frac{70758}{279935}, \quad M_{22}=\frac{3089}{9600},\quad 
M_{23}=\frac{8649}{2260},
\\
M_{31}=-\frac{477}{3888},\quad M_{32}=-\frac{1269}{3600}, \quad
M_{33}=-\frac{79}{200}.
\\
T_1y=\frac{6}{\Lambda(p,q,r)}[M_{11}Q_1y+M_{12}Q_2y+M_{13}Q_3y],
\\
T_2y=\frac{24}{\Lambda(p,q,r)}[M_{21}Q_1y+M_{22}Q_2y+M_{23}Q_3y],
\\
T_3y=\frac{60}{\Lambda(p,q,r)}[M_{31}Q_1y+M_{32}Q_2y+M_{33}Q_3y].
\end{gather*}
Define $Qy$ by $Qy=T_1y+(T_2y)t+(T_3y)t^2$, and we take $K_Py(t)$ as
in Lemma \ref{lem3.2}, then Lemma \ref{lem3.2} holds.

   On the other hand, we have 
$$
|f(t,x(t),x'(t),x''(t))|\leq|x''(t)|+2,\quad t\in(0,1).
$$
And let $\alpha(t)=0$, $\beta(t)=0$, $\gamma(t)=1$, $\theta(t)=2$, then
the condition (H1) of Theorem \ref{thm3.1} is satisfied.
    If $x''(t)>8=A$, $t\in(0,1)$, then
\begin{align*}
Q_1y
&= 2\int_{0}^{1/5}(x''(t)+\sin x(t)(1-\cos x'(t)))dt\\
&\quad -\int_{0}^{1/4}(x''(t)+\sin x(t)(1-\cos x'(t)))dt\\
&> 2\int_{0}^{1/5}7dt-\int_{0}^{1/4}10dt
> \frac{9}{10},
\end{align*}
If $x''(t)<-8=-A$, $t\in(0,1)$, then
\begin{align*}
Q_1y
&= 2\int_{0}^{1/5}(x''(t)+\sin x(t)(1-\cos x'(t)))dt\\
&\quad -\int_{0}^{1/4}(x''(t)+\sin x(t)(1-\cos x'(t)))dt\\
&< 2\int_{0}^{1/5}(-6)dt-\int_{0}^{1/4}(-9)dt
<-\frac{3}{20},
\end{align*}
So condition (H2) is satisfied.

If $|a|>16=B$, $|b|>16=B$, $|c|>16=B$, then
\begin{align*}
&Q_1N(a+bt+ct^2)+Q_2N(a+bt+ct^2)+Q_3N(a+bt+ct^2)\\
&= 2\int_{0}^{1/5}(|2c|+\sin (a+bt+ct^2)[1-\cos (b+2ct)])dt\\
&\quad -\int_{0}^{1/4}(|2c|+\sin (a+bt+ct^2)[1-\cos (b+2ct)])dt\\
&\quad +3\int_{0}^{\frac{1}{3}}(\frac{1}{3}-t)(|2c|
 +\sin (a+bt+ct^2)[1-\cos (b+2ct)])dt\\
&\quad -2\int_{0}^{1/2}(\frac{1}{2}-t)(|2c|
 +\sin (a+bt+ct^2)[1-\cos (b+2ct)])dt\\
&\quad +\int_{0}^{1}(1-t)^2(|2c|+\sin (a+bt+ct^2)[1-\cos (b+2ct)])dt\\
&\quad -\frac{9}{5}\int_{0}^{\frac{1}{3}}(\frac{1}{3}-t)^2(|2c|
 +\sin (a+bt+ct^2)[1-\cos (b+2ct)])dt\\
&\quad +4\int_{0}^{1/2}(\frac{1}{2}-t)^2(|2c|+\sin (a+bt+ct^2)
 [1-\cos (b+2ct)])dt\\
&\quad-\frac{16}{5}\int_{0}^{3/4}(\frac{3}{4}-t)^2(|2c|
 +\sin (a+bt+ct^2)[1-\cos (b+2ct)])dt
>0.
\end{align*}
So  condition (H3) is satisfied. Thus all the
conditions of Theorem \ref{thm3.1} are satisfied. Hence BVP 
\eqref{e4.1}-\eqref{e4.2} has
at least one solution in $C^2[0,1]$.


\subsection*{Acknowledgments}
This research was sponsored by the Natural Science Foundation of China
(11071205, 11101349) and by
 PAPD of Jiangsu Higher Education Institutions,
the Humanistic and Social Scientific Research Planning Program 
in Ministry of Education of China (No. 12YJA630063), 
the Social Science Foundation of Jiangsu Province of China (No. 10GLB001).

 The authors express their sincere thanks to  the anonymous reviewer  
for his or her valuable comments and suggestions for improving the quality 
of this article.

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\end{document}