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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 62, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/62\hfil Boundary and initial value problems]
{Boundary and initial value problems for second-order neutral
functional differential equations}
\author[H. H. Le, T. P. N. Le\hfil EJDE-2006/62\hfilneg]
{Hoan Hoa Le, Thi Phuong Ngoc Le}  

\address{Hoan Hoa Le \newline
Department of Mathematics, Ho Chi Minh City University of  Education,
280 An Duong Vuong Str., Dist. 5, Ho Chi Minh City, Vietnam}

\address{Thi Phuong Ngoc Le \newline
Nha Trang Educational College, 01 Nguyen Chanh Str., Nha Trang City,
Vietnam}
\email{phuongngoccdsp@dng.vnn.vn\quad lephuongngoc@netcenter-vn.net}

\date{}
\thanks{Submitted February 20, 2006. Published May 11, 2006.}
\subjclass[2000]{34M50, 34K40}
\keywords{Three-point boundary-value problem; topological degree;
\hfill\break\indent
 Leray-Schauder nonlinear alternative; Contraction mapping principle}

\begin{abstract}
 In this paper, we consider the three-point boundary-value problem
 for the second order neutral functional differential equation
 $$
 u''+ f(t,u_t, u'(t))= 0, \quad  0 \leq t\leq 1,
 $$
 with the three-point boundary condition
 $u_0= \phi$, $u(1) = u(\eta)$. Under suitable
 assumptions on the function $f$ we prove the existence,
 uniqueness and continuous dependence of solutions.
 As an application of the methods used, we study the existence
 of solutions for the same equation with a ``mixed" boundary
 condition
 $u_0 = \phi, u(1) = \alpha [u'(\eta) - u'(0)]$,
 or with an initial condition $ u_0 = \phi, u'(0) =0$.
 For the initial-value problem, the uniqueness and
 continuous dependence of solutions are also considered.
 Furthermore, the paper shows that the solution
 set of the initial-value problem is nonempty, compact and
 connected. Our approach is based on the fixed point theory.
\end{abstract}

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


\section{Introduction}

Let $ C= C ([-r,0]; \mathbb {R})$, with $ r >0 $ is a fixed constant,
be the Banach space of all continuous  functions
$ \phi : [-r,0] \to \mathbb {R}$, with the sup-norm
$ \| \phi \| = \sup \lbrace |\phi (\theta)| : -r \leq \theta \leq 0\rbrace$.
 For any continuous function $ u:[-r,1] \to \mathbb {R}$ and for any
$ t \in [0,1]$, we denote by $ u_t$ the element of $ C$ defined by
 $ u_t(\theta) = u( t+\theta), \theta \in [-r,0]$.
In this paper, we consider the  second-order neutral
functional differential equation
\begin{equation} \label{eE}
 u''+f(t,u_t, u'(t))= 0, \quad 0 \leq t\leq 1,
\end{equation}
where $ f : [0,1] \times C \times \mathbb {R} \to  \mathbb {R}$
is continuous, with one of the following boundary conditions
\begin{gather}
u_0= \phi,\quad  u(1) = u(\eta) \label{BC1} \\
u_0 = \phi, \quad u(1) = \alpha [ u'(\eta) - u'(0)], \label{BC2}
\end{gather}
or with the initial conditions
\begin{equation}
u_0= \phi, \quad u'(0) = 0, \label{IC3}
\end{equation}
where $\phi \in C$, $0 < \eta <1$, and $\alpha \in \mathbb {R}$.

The boundary-value problems for ordinary differential equation
and for neutral functional differential equations have been
studied by several authors by using the Leray-Schauder continuation
theorem, Leray-Schauder nonlinear alternative, topological
transversality method. We refer the reader for example to
\cite{h1,m1,n1,o1,y1,z2} and references therein.

In \cite{n1}, the author proved the existence of solution for the  neutral FDE
\begin{gather*}
\frac {d} {dt} [x'(t)-g(t,x_t)] = f(t,x_t, x'(t)),\quad
 0 \leq t \leq 1, \\
x_0 = \phi, x(1) = \eta,
\end{gather*}
where $ f : [0, 1] \times C \times \mathbb {R}^n \to \mathbb {R} ^n, g
: [0,1] \times C  \to \mathbb {R} ^n $ are continuous functions,
$ \phi \in C, \eta \in \mathbb {R}^n$.
In \cite{z2}, the existence, uniqueness and continuous dependence on a
 real parameter $ \alpha $ of the solution for the following problem
were established
\begin{gather*}
 \big ( \Lambda (t) x'(t)) ' = f(t,x_t, x'(t)), \quad 0 \leq t \leq T, \\
 x_0 = \phi, \quad A x(T)+ B x'(T) = v,
\end{gather*}
where $ \Lambda (t)$ is an $n \times n$ continuous matrix defined on
$ [0, T]$, $ A$ and $ B$ are $ n \times n$ constant matrices,
$v\in \mathbb {R}^n, \phi \in C = C \big ([-r,0]; \mathbb {R}^n \big )$.

Recently in \cite{m1,y1}, the authors studied the boundary-value problem
$$
u'' + f(t,u) = 0, \quad 0 < t < 1,
$$
where $ f : [0,1] \times \mathbb {R} \to \mathbb {R} $ is continuous,
with one of the following boundary conditions
$$
u(0) = 0, \quad u(1) = \alpha u(\eta),
$$
or
$$
u'(0) = 0, \quad u(1) = \alpha u'(\eta).
$$
In the base of the above papers, we shall consider the problems for
FDEs \eqref{eE}, \eqref{BC1}; \eqref{eE}, \eqref{BC2}
and \eqref{eE}, \eqref{IC3}.
This paper is organized as follows.
In section 2, we present some preliminaries.
 By using Leray-Schauder nonlinear alternative, the existence theorems
 of boundary-value problem \eqref{eE}-\eqref{BC1} are given in section 3.
Furthermore, the uniqueness, based on the contraction mapping principle,
and continuous dependence of solution are established.
In sections 4; 5, as an application of the methods which are used in
 the proofs of section 3, we also study the existence of solution for
the equation \eqref{eE} with a "mixed" bonundary condition \eqref{BC2}
 or with an initial condition \eqref{IC3}. For the initial value
 problem \eqref{eE}-\eqref{IC3}, the uniqueness and continuous dependence
of solution are also considered. From the results, based on the
topological degree theory of compact vector fields, the paper shows
that the solution set of the initial value problem is nonempty,
compact and connected.

\section{Preliminaries}
We denote by $C[0,1]$ and $C^1[0,1]$, respectively, the Banach
spaces of continuous real functions and continuously differentiable
real functions on $[0,1]$, with the norms:
\begin{gather*}
\| u \|_0 = \sup \{ {| {u(t)}|:0 \le t \le 1} \},\\
\| u \|_1 = \max \{ {\| u \|_0 ,\| {u}' \|_0 } \},
\end{gather*}
where $\| {u}' \|_0 =\sup \{ {| {{u}'(t)}|: 0 \le t \le 1}\}$,
and by $L^1[0,1]$ the space of all real functions
$x(t)$ such that $| {x(t)}|$ is Lebesgue integrable on
$[0,1]$.
The proofs of our theorems are based on the following theorems result.

\begin{theorem}[Nonlinear Alternative of Leray-Schauder] \label{thm2.1}
Let $E$ be a Banach space and $\Omega $ be a bounded open subset of $E$,
$0\in \Omega $, $T:\overline {\Omega }\to E$ be a completely
continuous operator. Then, either there exists
$x\in \partial \Omega $ such that $Tx=\lambda x$ for some $\lambda >1$,
or there exists a fixed point $x\in \Omega $.
\end{theorem}

The proof of the theorem above can be found in
 \cite[Theorem 2.10]{o1}.

\begin{theorem}[\cite{k1}] \label{thm2.2}
Let $(E,\vert \cdot \vert)$  be a real Banach space, $D$ be a bounded
open subset of $E$ with boundary $\partial D$, closure $\overline{D}$
and $T:\overline{D}\to E$ be a completely continuous operator.
Assume that $T$ satisfies the follows conditions:
\begin{itemize}
\item[(i)] $T$ has no fixed points on $\partial D$ and
 $\gamma (I-T,D)\neq 0$.
\item[(ii)] For each $\varepsilon >0$, there is a completely continuous operator
$T_{\varepsilon }$  such that
$\vert T_{\varepsilon}(x)-T(x)\vert <\varepsilon $, for all
$x\in \overline{D}$, and such that for each $h$  with
$\vert h\vert <\varepsilon $, the equation
$x=T_{\varepsilon }(x)+h$ has at most one solution in $\overline{D}$.
\end{itemize}
Then the set of fixed points of $T$ is nonempty, compact and
connected.
\end{theorem}

The proof of the theorem above  can be found in \cite[theorem 48.2]{k1}.
We remark that condition (i) is equivalent to the following condition.
\begin{itemize}
\item[(\~i)] $T$ has no fixed points on $\partial D$
and $\deg (I-T,D,0)\neq 0$.
\end{itemize}
\noindent
Because of this, if a completely continuous operator $T$ is defined on
$\overline{D}$ and has no fixed points on $\partial D$, then the
rotation $\gamma (I-T,D)$ coincides with the Leray-Schauder degree
of $I-T $ on $D$ with respect to the origin,  see
\cite[section 20.2]{k1}.

\begin{theorem}[\cite{d1}] \label{thm2.3}
Let $E,F$ be Banach spaces, $D$ be an open subset of $E$ and
$f:D \to F$ be continuous. Then for each $\varepsilon > 0$, there is
a mapping $f_\varepsilon :D \to F$ that is locally Lipschitz such that
$$
| {f(x) - f_\varepsilon (x)}| < \varepsilon ,\quad \forall  x \in D,
$$
and $f_\varepsilon (D)$ is a subset of the closed convex  hull of $f(D)$.
\end{theorem}

The proof of the above theorem can be found in \cite[p. 53]{d1}.
We will need the following lemmas later. The proofs of these
lemmas are not difficult and we omit them.

\begin{lemma} [\cite{m1}] \label{lem2.4}
For $y \in C[0,1]$, the problem
\begin{gather*}
 {u}'' + y(t) = 0,\quad t \in (0,1), \\
 u(0) = 0,\quad u(1) = u(\eta ),
 \end{gather*}
with $\eta \in (0,1)$, has a unique solution
$$
u(t) = - \int_0^t {(t - s)y(s)ds - \frac{t}{1 - \eta }}
\int_0^\eta {(\eta - s)y(s)ds}
 + \frac{t}{1 - \eta }\int_0^1 (1 - s)y(s)ds,
$$
 $t \in [0,1]$.
\end{lemma}

\begin{lemma} \label{lem2.5}
For $ y \in C[0,1]$,  the ``mixed" boundary-value problem
\begin{gather*}
 {u}'' + y(t) = 0,\quad t \in (0,1), \\
 u(0) = 0,\quad u(1) = \alpha ({u}'(\eta ) - {u}'(0)),
 \end{gather*}
with $\eta \in (0,1)$ and $\alpha \in \mathbb{R}$, has a unique solution
$$
u(t) = - \int_0^t (t - s)y(s)ds - \alpha t \int_0^\eta
{y(s)ds}
 + t\int_0^1 {(1 - s)y(s)ds,} \quad t \in [0,1].
$$
\end{lemma}

\begin{lemma} \label{lem2.6}
For $ y \in C[0,1]$,  the initial-value problem
\begin{gather*}
 {u}'' + y(t) = 0,\quad 0 < t \le 1, \\
 u(0) = 0,\quad u'(0) = 0,
\end{gather*}
has a unique solution
$$
u(t) = - \int_0^t {(t - s)y(s)ds} ,\quad t \in [0,1].
$$
\end{lemma}

\section{Main Results}

In this section, we present our existence results for the
boundary-value problem \eqref{eE}-\eqref{BC1}.


\begin{theorem} \label{thm3.1}
Let $ f:[0,1]\times C\times \mathbb{R}\to \mathbb{R}$ be a continuous
function. Assume that there exist nonnegative functions
$p$, $q$, $r\in L^{1}[0,1]$ such that
\begin{itemize}
\item[(H1)] $|f(t,u,v)|\leq p(t)\|u\|+ q(t)|v|+ r(t)$,
 for all $(t,u,v)\in [0,1]\times C\times \mathbb{R}$

\item[(H2)] $\frac{2-\eta }{1-\eta }\int_0^1 (1-s)p(s)ds
+\frac{1}{1-\eta }\int_0^{\eta } (\eta -s)p(s)ds <1$,

\item[(H3)] $\int_0^1 [ p(s)+q(s)]ds  + \frac{1}{1-\eta }
\int_0^1 (1-s)[p(s)+q(s)]ds +\frac{1}{1-\eta }
\int_0^{\eta } (\eta -s) [p(s)+q(s)]ds < 1$.

\end{itemize}
Then the boundary-value problem \eqref{eE}-\eqref{BC1} has at
least one solution.
\end{theorem}

\begin{proof}
{\bf Step 1}. Consider first the case $\phi (0)= 0$. Put
$$
C_{0}= \{u\in C^{1}[0,1]:u(0)=0\}.
$$
Then $C_{0}$ is the subspace of $C^{1}[0,1]$. We note that
for all $u\in C_{0}$, $ u(t)=\int_0^t u'(s)ds $, so
\begin{equation}\label{e3.1}
\| u\|_{0}\leq \|u'\| _{0}.
\end{equation}
For a function $ u \in C_{0}$, we define the function
$\widehat{u}:[-r,1]\to \mathbb{R}$
by
$$ \widehat{u}(t)= \begin{cases}
\phi (t), & t\in [-r,0],\\
u(t), &  t\in [0,1].
\end{cases}
$$
We also note that
\begin{equation} \label{e3.2}
\|\widehat{u}_{t}\|^{k}\leq \max \{\|u\| _{0}^{k},\|\phi \|^{k}\}\leq
\|u\| _{0}^{k}+ \|\phi \|^{k}, \forall t\in [0,1], k\geq 0.
\end{equation}
Define the integral operator $ T:C_{0}\to C^{1}[0,1]$ by
\begin{equation} \label{e3.3}
\begin{split}
Tu(t)=& -\int_0^t (t-s)f(s,\widehat{u}_{s},u'(s))ds -
\frac{t}{1-\eta }\int_0^{\eta} (\eta -s)f(s,\widehat{u}
_{s},u'(s))ds \\
& +\frac{t}{1-\eta }\int_0^1(1-s)f(s,\widehat{u}_{s},u'(s))ds,\quad
t\in [0,1].
\end{split}
\end{equation}
By Lemma \ref{lem2.4}, it is obvious that $\overline{u}$ is a solution
of the boundary-value problem \eqref{eE}-\eqref{BC1} if and only
if the operator $T$ has a fixed point $u \in C_{0}$, where
$$
\overline{u}(t)=\begin{cases}
\phi (t), & t\in [-r,0],\\
u(t), & t\in [0,1].
\end{cases}
$$
Using (H1) and \eqref{e3.2}, for all $u \in C_{0}$, for all
 $ t\in [0,1]$, we obtain
\begin{equation*}
\begin{split}
|Tu(t)|&\leq  \int_0^1 (1-s)[p(s)\|\widehat{u}_{s}\| + q(s)|u'(s)|+ r(s)] ds \\
&\quad + \frac{1}{1-\eta }\int_0^{\eta}(\eta -s)[ p(s)\|\widehat{u}_{s}\| + q(s)|u'(s)|+ r(s)] ds \\
&\quad  + \frac{1}{1-\eta }\int_0^1 (1-s)[p(s)\|\widehat{u}_{s}\| + q(s)|u'(s)|+ r(s)] ds \\
&\leq A_1\|u\|_0+ B_1 \|u'\|_0 + C_1,
\end{split}
\end{equation*}
where
\begin{gather*}
A_{1}=\frac{2-\eta }{1-\eta }\int_0^1(1-s)p(s)ds+
\frac{1}{1-\eta }\int_0^{\eta}(\eta -s)p(s)ds, \\
B_{1}=\frac{2-\eta }{1-\eta }\int_0^1(1-s)q(s)
+ \frac{1}{1-\eta }\int_0^{\eta}(\eta -s)q(s)ds, \\
\begin{aligned}
C_{1}&=\Big (\frac{2-\eta }{1-\eta }\int_0^1(1-s)p(s)ds+
\frac{1}{1-\eta }\int_0^{\eta}(\eta -s)p(s)ds \Big)\|\phi\|\\
& \quad+ \frac{2-\eta }{1-\eta }\int_0^1(1-s)r(s)ds+
\frac{1}{1-\eta }\int_0^{\eta}(\eta -s)r(s)ds.
\end{aligned}
\end{gather*}
Hence
\begin{equation}\label{e3.4}
\|Tu\|_0 \leq A_1\|u\|_0+ B_1 \|u'\|_0 + C_1,\quad \forall  u \in C_{0}.
\end{equation}
On the other hand,
\begin{equation}\label{e3.5}
\begin{split}
(Tu)'(t)= & - \int_0^t f(s,\widehat{u}_{s},u'(s))ds
- \frac{1}{1-\eta }\int_0^{\eta}(\eta
-s)f(s,\widehat{u}_{s},u'(s))ds \\
& +\frac{1}{1-\eta }\int_0^1 (1-s)f(s,\widehat{u}_{s},u'(s))ds,\quad
 t\in [0,1].
\end{split}
\end{equation}
Similarly, it follows from (H1) and \eqref{e3.2} that
\begin{equation} \label{e3.6}
\|(Tu)'\|_0\leq A_2\|u\|_0+ B_2 \|u'\|_0 + C_2, \quad
\forall u \in C_{0},
\end{equation}
where
\begin{gather*}
 A_{2}= \int_0^1 p(s)ds +\frac{1}{1-\eta }\int_0^1(1-s)p(s)ds+
\frac{1}{1-\eta }\int_0^{\eta}(\eta -s)p(s)ds, \\
B_{2}=\int_0^1 q(s)ds + \frac{1}{1-\eta }\int_0^1(1-s)q(s)+ \frac{1}{1-\eta }\int_0^{\eta}(\eta -s)q(s)ds, \\
\begin{aligned}
 C_{2}&=\Big (\int_0^1 p(s)ds + \frac{1}{1-\eta }\int_0^1(1-s)p(s)ds +
\frac{1}{1-\eta }\int_0^{\eta}(\eta -s)p(s)ds \Big)\|\phi\|\\
& \quad + \int_0^1 r(s)ds + \frac{1}{1-\eta }\int_0^1(1-s)r(s)ds+
\frac{1}{1-\eta }\int_0^{\eta}(\eta -s)r(s)ds.
\end{aligned}
\end{gather*}
Put
\begin{equation}\label{e3.7}
A= \max \lbrace A_1, A_2 + B_2 \rbrace.
\end{equation}
 From (H2)-(H3), it follows that $A_1<1$, $A_2+B_2 < 1$, so $A <1$.
We now choose a constant $B > 0$ such that
\begin{equation}\label{e3.8}
B \geq \max \lbrace  \frac {B_1 C_2} {1- A_2 -B_2}+ C_1, C_2 \rbrace,
\end{equation}
and put
\begin{equation}\label{e3.9}
m = \frac {B}{1-A}, \quad
\Omega  =  \lbrace  u \in C_0 : \|u\|_1 < m \rbrace.
\end{equation}
Then $\Omega $ be a bounded open subset of $C_0 $, $ 0\in \Omega $, and
$ \partial \Omega = \{u\in C_{0}: \|u\|_1= m \}$.
We shall show that $ T:\overline{\Omega }=\Omega \cup \partial \Omega
\to C^1[0,1]$ has a fixed point $u\in \Omega $ by applying Theorem
\ref{thm2.1}.

\noindent(a) First, $T$ is continuous. Indeed, for each
$u_{0}\in \overline{\Omega }$, let $\{u_{n}\}$ be a sequence in
$\overline{\Omega }$ such that $ \lim_{n \to \infty } u_{n}=u_{0}$.
For all $t\in [0,1]$, from \eqref{e3.3}, we get
\begin{equation*}
\begin{split}
Tu_n(t)-Tu_0(t)= & - \int_0^t(t-s)\Big [ f( s,(\widehat{u}_n)_s,u'_{n}(s)) - f( s,(\widehat{u}_0)_{s},u'_0(s)) \Big ] ds \\
& -\frac {t}{1-\eta }\int_0^{\eta } (\eta -s)\Big [ f( s,(\widehat{u}_n)_{s},u'_{n}(s)) - f( s,(\widehat{u}_0)_s,u'_0(s)) \Big] ds \\
& + \frac {t}{1-\eta }\int_0^1 (1-s)\Big [ f( s,(\widehat{u}_n)_{s},u'_{n}(s)) -f( s,(\widehat{u}_0)_{s},u'_{0}(s)) \Big ] ds.
\end{split}
\end{equation*}
Put $D= \{(\widehat{u}_n)_s : s \in [0,1],\, n = 0,1,2,\dots\}$, then $D$
is compact in $C$.
Since $ f :[0,1]\times C\times \mathbb{R} \to \mathbb{R}$  is continuous,
$f$ is uniformly continuous on the compact subset
$[0,1]\times D\times [-m,m]$. This implies that, for all
$\varepsilon >0$, there exists $\delta >0$ such that for each
$(s_1,\phi _1,\nu _1)$,
$(s_2,\phi _2,\nu _2)\in [0,1]\times D\times [-m,m]$,
\begin{gather*}
|s_1-s_2| <\delta,\quad
\|\phi_1-\phi_2\| <\delta,\quad
|\nu_1-\nu_2| <\delta \\
\Rightarrow \quad
 |f(s_1,\phi _1,\nu _1)-f(s_2,\phi _2,\nu _2)|
 <\frac{\varepsilon }{2\beta },
\end{gather*}
with
$\beta =1+\frac{2}{1-\eta }> 0$.
Since $\lim_{n\to \infty } u_{n}=u_{0}$ in
$\overline{\Omega }$, with respect to
$\Vert \cdot\Vert _{1}$, there exists
$n_{0}$ such that for all $n\geq n_{0}$,
$$
\|(\widehat{u}_n)_{s}-(\widehat{u}_0)_{s}\| <\delta,
|u'_n(s)-u'_0(s)| <\delta, \quad
\forall s\in [0,1].
$$
On the other hand, for all $ s\in [0,1]$,
$\big ( s,(\widehat{u}_n)_s,u'_n(s)\big ),\big( s,(\widehat{u}_0)_s,u'_0(s))  \in [0,1]\times D\times [-m,m]$, therefore, for all $
n \geq n_0$,
\begin{equation*}
\begin{split}
|Tu_n(t)-Tu_0(t)| & \leq (1+\frac {2}{1-\eta}) \int_0^1
 \Big | f( s,(\widehat{u}_n)_s,u'_{n}(s))
 - f( s,(\widehat{u}_0)_{s},u'_0(s)) \Big | ds \\
& < (1+\frac {2}{1-\eta})\frac{\varepsilon }{2\beta }
=\frac{\varepsilon }{2},\quad \forall  t\in [0,1].
\end{split}
\end{equation*}
Similarly
$$
|(Tu_n)'(t)-(Tu_0)'(t)|  < \frac{\varepsilon }{2},\quad
\forall t\in [0,1].
$$
This implies that for all $ n\geq n_{0}$,
$$
\|Tu_{n}-Tu_{0}\|_1 = \max \Big
\lbrace \|Tu_{n}-Tu_{0}\|_{0},\|( Tu_{n})'-( Tu_{0})'\|_{0}\Big \rbrace
\leq \frac{\varepsilon }{2}<\varepsilon.
$$
(b) Next, we show that $T(\overline{\Omega })$ is relatively compact.
Let $\{Tu_n\}$ be a bounded sequence of $T(\overline{\Omega })$,
corresponding $\{u_n\} \subset \overline{\Omega }$, we shall show
that $\{Tu_{n}\}$ contains a convergence subsequence in $C^{1}[0,1]$,
with respect to $\|.\|_1$. The proof of this fact is obtained as follows.
For all $n$, it follows from $\eqref{e3.4}, \eqref{e3.6}, \eqref{e3.9}$ that
\begin{gather*}
\|Tu_{n}\|_{0}\leq A_{1}\| u_{n}\|_{0}+ B_{1}\|u'_n\|_{0}+ C_{1}\leq A_1m
+ B_1 m + C_1,\\
\|(Tu_{n})'\|_{0}\leq A_{2}\| u_{n}\|_{0}+ B_{2}\|u'_n\|_{0}
+ C_{2}\leq A_2m + B_2 m + C_2.
\end{gather*}
Hence, the sequences $\{Tu_{n}\},\{(Tu_{n})'\}$
are uniformly bounded. On the other hand, combining
\eqref{e3.3}, \eqref{e3.5}, \eqref{e3.9} and (H1), for all $n$,
for all $t_1, t_2\in [0,1]$,  we have
\begin{align*}
&|Tu_n(t_1)-Tu_n(t_2)|\\
&\leq \Big | \int_{t_1}^{t_2}(1-s)[(m+\|\phi\|) p(s)+m q(s) +r(s)]ds
\Big |\\
&\quad + \frac{1}{1-\eta } \Big ( \int_0^{\eta}(\eta-s)[(m+\|\phi\|) p(s)
+ m q(s) +r(s)] ds \Big ) |t_1 - t_2|\\
&\quad + \frac{1}{1-\eta } \Big ( \int_0^1 [(m+\|\phi\|) p(s)
+ m q(s) +r(s)] ds \Big ) |t_1 - t_2|\\
& \leq K_1 |t_1 - t_2|,
\end{align*}
\begin{align*}
|(Tu_n)'(t_1)-(Tu_n)'(t_2)| &\leq \Big| \int_{t_1}^{t_2}[(m+\|\phi\|) p(s)
+m q(s) +r(s)]ds \Big |\\
& \leq K_2 |t_1 - t_2|,
\end{align*}
where $K_{1}$, $K_{2}$ are independent of $t_1$, $t_2$ and $n$.
This implies that the sequences $\{Tu_n \}, \{( Tu_n)'\} $ are
equi-continuous.
By using the Ascoli-Arzela theorem, we have $\{Tu_{n}\}$,
$\{(Tu_n)'\}$ are relatively compact in $C[0,1]$. Therefore,
 there exists a subsequence $\{u_{n_k}\} \subset \{u_n \}$, such that
$$
 Tu_{n_k}\to u \quad \text {and}\quad ( Tu_{n_k})'\to v,\quad
  \text {as} \quad k \to \infty,
$$
with respect to $\|.\|_0$. Then $u$ is differentiable and $u'= v$,
so $Tu_{n_k}\to u$, as $ k\to \infty $, in $C^1[0,1]$, with respect
to $\|.\|_1$. Thus $T$ is completely continuous.

\noindent(c) Finally, suppose that there exists
$u^* \in \partial \Omega $,
such that $T(u^*)=\lambda u^*$, for some $\lambda >1$.
Then, we have the following set is bounded
$$
\lbrace u^*\in \partial \Omega : T(u^*)= \lambda u^*, \lambda >1 \rbrace.
$$
Indeed, it follows from \eqref{e3.6} that
\begin{equation}\label{e3.10}
\|(u^*)'\|_0 = \frac {1}{\lambda}\|(Tu^*)'\|_0
\leq \|(Tu^*)'\|_0 \leq A_2\|u^*\|_0 + B_2 \|(u^*)'\|_0 + C_2 .
\end{equation}
Combining \eqref{e3.1}, \eqref{e3.10}, we get
$$
( 1- A_{2}-B_{2}) \|(u^*)'\|_0 \leq C_2.
$$
Since $A_{2}+B_{2}<1$, this implies that
\begin{equation}\label{e3.11}
\|(u^*)'\|_0  \leq M,
\end{equation}
where $ M =C_2/(1-A_2-B_2) $ is a constant.
Thus, combining \eqref{e3.1}, \eqref{e3.4}, \eqref{e3.6}-\eqref{e3.8},
 \eqref{e3.10} and \eqref{e3.11}, we obtain
\begin{equation}\label{e3.12}
\begin{gathered}
\begin{aligned}
\|Tu^*\|_0  &\leq A_1\|u^*\|_0 + B_1 \|(u^*)'\|_0 + C_1 \\
 &\leq A_1\|u^*\|_0 + B_1 M + C_1 \\
 &\leq A\|u^*\|_0 + B,
\end{aligned}\\
\begin{aligned}
\|(Tu^*)'\|_0 &\leq A_2|u^*\|_1 + B_2\|u^*\|_1 + C_2\\
 &\leq A\|u^*\|_1 + B.
\end{aligned}
\end{gathered}
\end{equation}
Consequently
$$
\lambda \|u^*\|_1 =\|Tu^*\|_1 \leq A\|u^*\|_1 +B,
$$
which implies
$$
 \lambda m \leq Am+B \quad\text{or}\quad
  \lambda \leq A + \frac{B}{m}, \quad \text{i.e. } \lambda \leq 1,
$$
this contradicts $\lambda >1$. The proof of step 1 is complete.

\noindent{\bf Step 2}. The case $\phi (0)\neq 0$.
By the transformation $v = u - \phi (0)$, the boundary-value problem
\eqref{eE}-\eqref{BC1} reduces to the  boundary-value problem
\begin{gather*}
v''+ f(t, v_t+ \phi(0), v'(t))= 0, \quad
 0 \leq t\leq 1, \quad \label{eEtilde}\\
v_0 = \phi -\phi(0) \equiv \widetilde \phi, \quad v(1) = v(\eta),
\label{BC1tilde}
\end{gather*}
with  $\widetilde {\phi }\in C$ and $\widetilde{\phi }(0)=0$.
By step 1, this boundary-value problem has at least one
solution. Step 2 follows and Theorem \ref{thm3.1} is proved.
\end{proof}

\begin{theorem} \label{thm3.2}
Let $ f:[0,1]\times C\times \mathbb{R}\to \mathbb{R}$ be a continuous
function. Assume that  there exist nonnegative functions
$p$, $q$, $r\in L^{1}[0,1]$ and reals constants $k,l \in [0,1]$
such that (H2) holds and
\begin{itemize}
\item[(\~H1)] $|f(t,u,v)|\leq p(t)\|u\|^k+ q(t)|v|^l+ r(t)$,
 for all $(t,u,v)\in [0,1]\times C\times \mathbb{R}$,
\item[(\~H3)] $Q(k)A_2+Q(l) B_2 <1$,
\end{itemize}
where
\begin{gather*}
 A_{2}= \int_0^1 p(s)ds +\frac{1}{1-\eta }\int_0^1(1-s)p(s)ds+
\frac{1}{1-\eta }\int_0^{\eta}(\eta -s)p(s)ds, \\
B_{2}=\int_0^1 q(s)ds + \frac{1}{1-\eta }\int_0^1(1-s)q(s)+ \frac{1}{1-\eta }\int_0^{\eta}(\eta -s)q(s)ds,
\end{gather*}
and
$$ Q(\mu)=
\begin{cases}
0, & 0 \leq \mu <1,\\
1, & \mu =1.
\end{cases}
$$
Then the boundary-value problem $\eqref{eE}-\eqref{BC1}$ has at
least one solution.
\end{theorem}

\begin{proof}
It is obvious that the Theorem \ref{thm3.1} is a special case of this
theorem with $k=l=1$.
Here, we consider only the case $\phi (0)=0$ and let the subspace
$C_{0}$, the function $\widehat{u}$ and the operator $T$ be defined
as in Theorem \ref{thm3.1}.
Using (\~H1) and \eqref{e3.2}, for all $u\in C_{0}$ and all $t\in
[0,1]$, we have
\begin{equation*}
\begin{split}
|Tu(t)|&\leq  \int_0^1 (1-s)[p(s)\|\widehat{u}_{s}\|^k + q(s)|u'(s)|^l
 + r(s)] ds \\
&\quad + \frac{1}{1-\eta }\int_0^{\eta}(\eta -s)
 [ p(s)\|\widehat{u}_{s}\|^k + q(s)|u'(s)|^l+ r(s)] ds \\
&\quad + \frac{1}{1-\eta }\int_0^1 (1-s)[p(s)\|\widehat{u}_{s}\|^k
 + q(s)|u'(s)|^l+ r(s)] ds \\
&\leq A_1\|u\|_0^k+ B_1 \|u'\|_0^l + C_3,
\end{split}
\end{equation*}
where $A_{1}$ and $B_{1}$ as in Theorem \ref{thm3.1}, and
\begin{equation*}
\begin{split}
C_{3}=& \Big (\frac{2-\eta }{1-\eta }\int_0^1(1-s)p(s)ds+
\frac{1}{1-\eta }\int_0^{\eta}(\eta -s)p(s)ds \Big)\|\phi\|^k\\
& + \frac{2-\eta }{1-\eta }\int_0^1(1-s)r(s)ds+
\frac{1}{1-\eta }\int_0^{\eta}(\eta -s)r(s)ds.
\end{split}
\end{equation*}
It follows that for all $u\in C_{0}$,
\begin{equation}\label{e3.13}
\|Tu\|_0 \leq A_1\|u\|_0^k+ B_1 \|u'\|_0^l + C_3.
\end{equation}
Similarly, for all $u\in C_{0}$, we  obtain
\begin{equation}\label{e3.14}
\begin{split}
\|(Tu)'\|_0& \leq A_2\|u\|_0^k+ B_2 \|u'\|_0^l + C_4\\
&\leq A_2\|u'\|_0^k+ B_2 \|u'\|_0^l + C_4,
\end{split}
\end{equation}
where $A_{2}$ and $B_{2}$ are as above and
\begin{equation*}
\begin{split}
C_{4}= & \Big (\int_0^1 p(s)ds + \frac{1}{1-\eta }\int_0^1(1-s)p(s)ds +
\frac{1}{1-\eta }\int_0^{\eta}(\eta -s)p(s)ds \Big)\|\phi\|^k\\
& + \int_0^1 r(s)ds + \frac{1}{1-\eta }\int_0^1(1-s)r(s)ds+
\frac{1}{1-\eta }\int_0^{\eta}(\eta -s)r(s)ds.
\end{split}
\end{equation*}
Clearly, as the proof of the Theorem \ref{thm3.1}, if we show the boundedness
of the following set
\begin{equation}\label{e3.15}
\lbrace u^{*}\in \partial \Omega :T(u^{*})=\lambda u^{*}, \lambda >1
 \rbrace,
\end{equation}
then, combining the assume (H2), the proof of Theorem \ref{thm3.2} will be
completely. That is proved as follows.

Suppose that there exists $u^* \in \partial \Omega $ such that
$T(u^{*})= \lambda u^{*}$ for some $\lambda >1$.
We consider three cases. \\
{\it Case 1}: $0\leq k < 1$,  $ 0 \leq l<1$.
If $\|( u^*)'\|_{0} > 1$, then from (3.14), we have
\begin{equation}\label{e3.16}
\|( Tu^*)'\|_{0}\leq (A_2+ B_2)\|( u^*)'\|_{0}^h +C_4,
\end{equation}
where  $h= \max \{ k,l \}$.
It follows that
\begin{equation}\label{e3.17}
\|( u^*)'\|_{0}= \frac{1}{\lambda }\|( Tu^*)'\|_{0}\leq \|(Tu^*)'\|_{0}
\leq ( A_{2}+B_{2}) \|( u^*)'\|_{0}^{h}+ C_{4}.
\end{equation}
Here, let us note that if $ K\geq 0$, $H>0$, $0\leq \beta <2$ are given
constants, then there exists a constant $C>0$ such that
\begin{equation}\label{e3.18}
Kx^{\beta }\leq \frac{Hx^{2}}{2}+ C, \quad \forall x\geq 0.
\end{equation}
Hence, with $ x=\sqrt{\|( u^*)'\|_0}$, $K =A_{2}+ B_{2}$,
$\beta =2h, H=1$, the inequality \eqref{e3.18} implies that
$$
( A_{2} + B_{2})\|( u^*)'\|_{0}^{h} + C_{4}\leq
\frac{1}{2}\|( u^*)'\|_{0}+ C_{4}+C.
$$
Combining the above inequalities,
$$
\|( u^*)'\|_{0}\leq \frac{1}{2}\|( u^*)'\|_{0} + C_{4} + C
\quad\text{or}\quad
 \|( u^*)'\|_{0} \leq 2C_{4}+2C.
$$
We can choose $C$ such that $2C_{4}+ 2C > 1$; therefore,
$$
\|( u^*)'\|_{0} \leq 2C_{4}+2C,
$$
although $\|( u^*)'\|_{0}\leq 1$ or $\|( u^*)'\|_{0}>1$.
Thus, in case 1, there exists a positive constant
$\widetilde{M}= 2C_{4}+2C$, such that
\begin{equation}\label{e3.19}
\|( u^*)'\|_{0}\leq \widetilde{M}.
\end{equation}
{\it Case 2}: $ k = 1$, $0\leq l <1$. From \eqref{e3.14}, we have
$$
\|( Tu^*)'\|_{0}\leq A_2 \|( u^*)'\|_{0} + B_2 \|( u^*)'\|_{0}^l +C_2,
$$
where $C_{4}=C_{2}$, since $k = 1$.
So we have
$$
(1-A_{2})\|( u^*)'\|_{0}\leq B_{2}\|( u^*)'\|_{0}^{l}+C_{2}.
$$
Clearly, from (\~H3), $A_{2}<1$.
Using  \eqref{e3.18} again, with
$x = \sqrt{\|( u^*)'\|_0}$, $K = B_{2}$,  $\beta =2l$,
$H = 1- A_{2}$, we get
$$
B_{2}\|( u^*)'\|_{0}^{l}+C_{2}\leq \frac{1}{2}(1-A_{2})\|( u^*)'\|_{0}
+C_{2}+\widetilde{C},
$$
and so
$$
(1-A_{2})\|( u^*)'\|_{0}\leq \frac{1}{2}(1-A_{2})\|( u^*)'\|_{0}
+C_{2}+\widetilde{C}
\Leftrightarrow \|( u^*)'\|_{0} \leq \frac{2C_{2}+2\widetilde{C}}{1-A_{2}},
$$
where $\widetilde{C}$ is a positive constant.
We deduce that \eqref{e3.19}  also holds in the second case,
in which $\widetilde{M}=\frac{2C_{2}+2\widetilde{C}}{1-A_{2}}$.

{\it Case 3}: $0\leq k<1$, $l = 1$. We conclude from the hypothesis
(\~H3) that  $B_{2}<1$, hence that it is similar to the above cases,
\eqref{e3.19}  also holds.
Therefore, Theorem \ref{thm3.2} is proved.
\end{proof}

Now, we present the uniqueness of the solution of the boundary-value
problem \eqref{eE}-\eqref{BC1}.

\begin{theorem} \label{thm3.3}
Let $ f:[0,1]\times C\times \mathbb{R}\to \mathbb{R}$ be continuous
function and satisfy on $[0,1]\times C\times \mathbb{R}$ the Lipschitz
condition
$$
|f(t,u,v)-f(t,\widetilde{u},\widetilde{v})|
\leq \theta ( \|u-\widetilde{u}\|+ |v-\widetilde{v}|),
$$
for some positive constant $\theta $.
If $2( 1+\frac{2}{1-\eta }) \theta <1$,
then there exists a unique solution of \eqref{eE}-\eqref{BC1}.
\end{theorem}

\begin{proof}
Let $S$  be the space of continuous functions $ u:[-r,1] \to \mathbb{R}$
such that $u$ is continuously differentiable on $[0,1]$ and $u_{0}=\phi $.
We define
\begin{equation}\label{e3.20}
d(u,v)= \max \big \lbrace \max_{0\leq t\leq 1} |u(t)-v(t)|,
\max_{0\leq t\leq 1} |u'(t)-v'(t)|\big \rbrace.
\end{equation}
Then $S$ is a completely metrizable space with the distance function $d$.
By Lemma \ref{lem2.4}, for each $u\in S$, the problem
\begin{equation}\label{e3.21}
\begin{gathered}
x''+f(t,u_t, u'(t))= 0, \quad  0 \leq t\leq 1, \\
x(0)= \phi (0),\quad  x(1) = x(\eta),
\end{gathered}
\end{equation}
has a unique solution  on $[0,1]$ which is defined as
\begin{equation*}
\begin{split}
x(t)=& \phi (0)-\int_0^t (t-s)f(s,u_{s},u'(s))ds - \frac{t}{1-\eta }\int_0^{\eta}(\eta
-s)f(s,u_{s},u'(s))ds \\
& + \frac{t}{1-\eta }\int_0^1 (1-s)f(s,u_{s},u'(s))ds, \quad t \in [0,1].
\end{split}
\end{equation*}
We define $\widetilde{u}\in S$, by $\widetilde{u}(t)= x(t)$ on $[0,1]$
and $\widetilde{u}_{0}=\phi $.
Therefore, the mapping $P: S \to S$ is defined by
$$
P(u)=\widetilde{u}, \quad u\in S.
$$
For any $u, v\in S$, we put $w =\widetilde{u}-\widetilde{v}$. Then $w$
satisfies
\begin{equation}\label{e3.22}
\begin{gathered}
w''+ f(t,u_t, u'(t))- f(t,v_{t},v'(t))=0, \quad 0 \leq t\leq 1, \\
w_0= 0, \quad w(1) = w(\eta).
\end{gathered}
\end{equation}
It follows that for all $t\in [0,1]$, we have
\begin{equation}\label{e3.23}
\begin{split}
|w(t)| & \leq \int_0^1 |f(s,u_s, u'(s))- f(t,v_{s},v'(s))|ds\\
&\quad + \frac{1}{1-\eta }\int_0^{\eta} |f(s,u_s, u'(s))- f(t,v_{s},v'(s))|ds\\
&\quad + \frac{1}{1-\eta }\int_0^1 |f(s,u_s, u'(s))- f(t,v_{s},v'(s))|ds\\
&\leq K\theta \int_0^1 \big ( \|u_s-v_s\|+ |u'(s)-v'(s)|\big )ds\\
& \leq K\theta \Big (\max_{0\leq t\leq 1} |u(t)-v(t)|
+ \max_{0\leq t\leq 1} |u'(t)-v'(t)|\Big ),
\end{split}
\end{equation}
where $K = 1+\frac{2}{1-\eta }$.
Similarly,
\begin{equation}\label{e3.24}
\begin{split}
|w'(t)| &\leq K \int_0^1 |f(s,u_s, u'(s))- f(t,v_{s},v'(s))|ds\\
& \leq K\theta \Big (\max_{0\leq t\leq 1} |u(t)-v(t)|
+ \max_{0\leq t\leq 1} |u'(t)-v'(t)|\Big ).
\end{split}
\end{equation}
By the definition of $d$, we have
\begin{equation*}
\begin{split}
 d(\widetilde{u},\widetilde{v})
& = \max \big \lbrace \max_{0\leq t\leq 1}|\widetilde u(t)
   - \widetilde v(t)|, \max_{0\leq t\leq 1}|\widetilde u'(t)
   -\widetilde v'(t)|\big \rbrace \\
& \leq K\theta \Big (\max_{0\leq t\leq 1} |u(t)-v(t)|
   + \max_{0\leq t\leq 1}|u'(t)-v'(t)|\Big )\\
&\leq 2K\theta d(u,v).
\end{split}
\end{equation*}
Since $2K \theta =2( 1+\frac{2}{1-\eta }) \theta <1$, we deduce that $P$
is the contraction mapping. Therefore there exists a unique $u\in S$
such that $P(u)=u$. This implies that $u$ is the unique solution of the
boundary-value problem \eqref{eE}-\eqref{BC1}.
Then Theorem \ref{thm3.3} is proved.
\end{proof}

We remark that Theorem \ref{thm3.3} remains valid if we consider the
boundary-value problem
\begin{equation}\label{e3.25}
\begin{gathered}
u'' + f(t,u_t,u'(t), \lambda ), \quad 0 \leq t \leq 1,\\
u_0 = \phi, \quad u(1) = u(\eta),
\end{gathered}
\end{equation}
where $\lambda $ is a real parameter and
\begin{equation}\label{e3.26}
|f(t,u,v,\lambda )-f(t,\widetilde{u},\widetilde{v},\lambda)|
 \leq \theta ( \|u-\widetilde{u}\| + |v -\widetilde{v}|),
\end{equation}
on $[0,1]\times C\times \mathbb{R}\times \mathbb{R}$ for some positive constant $\theta $, with
\begin{equation}\label{e3.27}
2 (1+\frac{2}{1-\eta }) \theta <1.
\end{equation}
In other words, by  Theorem \ref{thm3.3}, if \eqref{e3.26} , \eqref{e3.27}
hold then the boundary-value problem
\eqref{e3.25} has a unique solution $u(t)=u(t,\lambda )$ for each
$\lambda $. We will show that the solution of \eqref{e3.25} depends
continuously on the parameter $\lambda $ if
\begin{equation}\label{e3.28}
|f(t,u,v,\lambda _{1})-f(t,u,v,\lambda _{2})|\leq L | \lambda _{1}
-\lambda _{2}|,
\end{equation}
for some positive constant $L$, for all $\lambda _{1}$, $\lambda _{2}$.

\begin{theorem} \label{thm3.4}
Let $ f:[0,1]\times C\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$
be a continuous function. If \eqref{e3.26} -\eqref{e3.28} hold
then the solution of \eqref{e3.25} depends continuously on $\lambda$.
\end{theorem}

\begin{proof} Let $u(t)=u(t,\lambda _{1})$ and $v(t)=v(t,\lambda _{2})$
 be solutions of \eqref{e3.25}  with $\lambda =\lambda _{1}$ and
$\lambda =\lambda _{2}$, respectively. It follows from
\eqref{e3.23}, \eqref{e3.24} and \eqref{e3.28} that for all
$t\in [0,1]$,
\begin{equation*}
\begin{split}
|u(t)-v(t)| & \leq K \int_0^1 |f(s,u_s, u'(s), \lambda_1)- f(t,v_{s},v'(s), \lambda_2)|ds\\
&\leq K \int_0^1 |f(s,u_s, u'(s), \lambda_1)- f(t,v_{s},v'(s), \lambda_1)|ds\\
&\quad + K \int_0^1 |f(s,v_s, v'(s), \lambda_1)- f(t,v_{s},v'(s), \lambda_2)|ds\\
& \leq  K \theta \Big (\max_{0\leq t\leq 1} |u(t)-v(t)|
+ \max_{0\leq t\leq 1} |u'(t)-v'(t)|\Big )+KL |\lambda_1 -\lambda_2|,
\end{split}
\end{equation*}
\begin{equation*}
|u'(t)-v'(t)|\leq  K \theta \Big (\max_{0\leq t\leq 1} |u(t)-v(t)|
+ \max_{0\leq t\leq 1} |u'(t)-v'(t)|\Big )+ K L |\lambda_1 -\lambda_2|,
\end{equation*}
where $K = 1+\frac{2}{1-\eta }$. Thus, in the completely metrizable
space $(S,d)$ which is defined as above, we have
\begin{equation*}
\begin{split}
 d(u,v) & = \max \big \lbrace \max_{0\leq t\leq 1}{\max }| u(t)- v(t)|,
  \max_{0\leq t\leq 1}{\max }| u'(t)- v'(t)| \big \rbrace \\
& \leq K\theta \Big (\max_{0\leq t\leq 1} |u(t)-v(t)|
+ \max_{0\leq t\leq 1} |u'(t)-v'(t)|\Big )+ K L |\lambda_1 -\lambda_2|\\
&\leq 2K \theta d(u,v)+ KL |\lambda_1 -\lambda_2|.
\end{split}
\end{equation*}
By \eqref{e3.27}, we have $2K \theta <1$, so
$$
d(u,v) \leq \frac{KL}{1-2K \theta }|\lambda _{1}-\lambda_{2}|.
$$
Thus, the solution of \eqref{e3.25} depends continuously on the
parameter $\lambda$.
The proof of Theorem \ref{thm3.4} is complete.
\end{proof}

\section {Application for the ``mixed" boundary value problem}

Now, we present our existence results for the solution to the
boundary-value problem \eqref{eE}-\eqref{BC2}.
Based on lemma \ref{lem2.5}, the proofs for the
following theorems are similar to that of the section 3.

\begin{theorem} \label{thm4.1}
Let $ f:[0,1]\times C\times \mathbb{R}\to \mathbb{R}$ be a
continuous function and assume there exist nonnegative functions
$p$, $q$, $r\in L^{1}[0,1]$ such that
\begin{itemize}
\item[(M1)] $|f(t,u,v)|\leq p(t)\|u\|+ q(t)|v|+ r(t)$, for all
   $(t,u,v)\in [0,1]\times C\times \mathbb{R}$
\item[(M2)] $2 \int_0^1 (1-s)p(s)ds + |\alpha|\int_0^{\eta} p(s)ds <1,$
\item[(M3)]
$\int_0^1 (2-s) [p(s)+q(s)]ds  + |\alpha|\int_0^{\eta} [p(s)+q(s)]ds < 1$.
\end{itemize}
Then the boundary-value problem \eqref{eE}-\eqref{BC2} has at least
one solution.
\end{theorem}

\begin{proof}
We first consider the case $\phi (0)=0$ and let the
 subspace $C_{0}$, the functions $\widehat{u}$ be defined as in
Theorem \ref{thm3.1}. Define the integral operator $T:C_{0}\to C^{1}[0,1]$ by
\begin{equation}\label{e4.1}
\begin{split}
Tu(t)=& -\int_0^t (t-s)f(s,\widehat{u}_{s},u'(s))ds -\alpha t \int_0^{\eta}f(s,\widehat{u}
_{s},u'(s))ds \\
& + t \int_0^1(1-s)f(s,\widehat{u}_{s},u'(s))ds, t\in [0,1].
\end{split}
\end{equation}
Using (M1) and \eqref{e3.2}, it follows that
\begin{equation}\label{e4.2}
\|Tu\|_0 \leq  a_1 \|u\|_0 + b_1\|u'\|_0 + c_1, \quad\forall u \in C_0,
\end{equation}
where
\begin{gather*}
a_1 =2 \int_0^1(1-s)p(s)ds + |\alpha|\int_0^{\eta}p(s)ds,\\
b_1 =2 \int_0^1(1-s)q(s)ds + |\alpha|\int_0^{\eta}q(s)ds, \\
\begin{aligned}
c_1 &= \Big(2 \int_0^1(1-s)p(s)ds + |\alpha|\int_0^{\eta}p(s)ds \Big)
\|\phi\|\\
&\quad + 2 \int_0^1(1-s)r(s)ds + |\alpha|\int_0^{\eta}r(s)ds.
\end{aligned}
\end{gather*}
Also using (M1) and \eqref{e3.2}, we obtain
\begin{equation}\label{e4.3}
\|(Tu)'\|_0 \leq  a_2 \|u\|_0 + b_2\|u'\|_0 + c_2,
\quad \forall  u \in C_0,
\end{equation}
where
\begin{gather*}
 a_2 = \int_0^1(2-s)p(s)ds + |\alpha|\int_0^{\eta}p(s)ds,\\
 b_2 = \int_0^1(2-s)q(s)ds + |\alpha|\int_0^{\eta}q(s)ds, \\
\begin{aligned}
 c_2 =& \Big( \int_0^1(2-s)p(s)ds + |\alpha|\int_0^{\eta}p(s)ds \Big)
\|\phi\|\\
&+  \int_0^1(2-s)r(s)ds + |\alpha|\int_0^{\eta}r(s)ds.
\end{aligned}
\end{gather*}
As in the proof of the theorems \ref{thm3.1}, \ref{thm3.2}, we conclude from
\eqref{e4.2}, \eqref{e4.1} and (M3) that the following set is bounded
\begin{equation}\label{e4.4}
\lbrace u^{*}\in \partial \Omega :T(u^*)=\lambda u^*, \lambda >1 \rbrace.
\end{equation}
Hence that, combining the assumption (M2) and the continuity of $f$,
 $T$ has a fixed point $u\in C_0$.
In the case $\phi (0)\neq 0$, by the transformation $v = u-\phi (0)$,
we can rewrite the boundary-value problem \eqref{eE}-\eqref{BC2}
  in the form
\begin{gather*}
v''+f(t,v_t+ \phi(0), v'(t))= 0, \quad 0 \leq t\leq 1, \\
v_0 = \phi -\phi(0) \equiv \widetilde \phi,\quad
v(1) = \alpha [ v'(\eta) - v'(0)]-\phi(0),
\end{gather*}
in which $\widetilde{\phi }\in C$ and $\widetilde{\phi }(0)=0$.
Here, we also consider the subspace $C_{0}$ and for a function
$v\in C_{0}$, we define the function $\widehat{v}:[-r,1]\to \mathbb{R}$ by
\begin{equation*}
\widehat{v}(t)= \begin{cases}
\widetilde{\phi }(t), & t\in [-r,0],\\
v(t), & t \in [0,1].
\end{cases}
\end{equation*}
Consider the operator $\widetilde{T}: C_{0}\to C^{1}[0,1]$ defined by
\begin{align*}
\widetilde{T}v(t)
=&-\int_0^t(t-s)f(s,\widehat{v}_s+\phi (0),v'(s))ds
 - \alpha t \int_0^{\eta}f(s,\widehat{v}_{s}+\phi (0),v'(s))ds \\
& -\phi (0)t + t\int_0^1(1-s)f(s,\widehat{v}_s+ \phi(0),v'(s))ds,
\quad t\in [0,1].
\end{align*}
Then, we can prove in a similar manner as above that $\widetilde{T}$
 has a fixed point $v\in C_0$.
This completes the proof of Theorem \ref{thm4.1}.
\end{proof}

\begin{theorem} \label{thm4.2}
Let $ f:[0,1]\times C\times \mathbb{R}\to \mathbb{R}$ be a continuous
function. Suppose that there exist nonnegative functions
$p$, $q$, $r\in L^{1}[0,1]$ and reals constants $k,l \in [0,1]$
 such that (M2) holds and
\begin{itemize}
\item[(\~M1)] $|f(t,u,v)|\leq p(t)\|u\|^k+ q(t)|v|^l+ r(t)$, for all
$(t,u,v)\in [0,1]\times C\times \mathbb{R}$
\item[(\~M3)] $Q(k)a_2 + Q(l) b_2 <1$,
\end{itemize}
where
\begin{gather*}
 a_{2}= \int_0^1(2-s) p(s)ds +|\alpha|\int_0^{\eta}p(s)ds, \\
 b_{2}=\int_0^1(2-s) q(s)ds +|\alpha|\int_0^{\eta}q(s)ds,
\end{gather*}
and the function $Q(\mu)$ is defined as in the Theorem \ref{thm3.2}.
Then the boundary-value problem \eqref{eE}-\eqref{BC2} has at least
one solution.
\end{theorem}

The proof for the above theorem is similar to that of the
Theorem \ref{thm3.2} and
is omitted.

\section{Application for the initial value problem}

First, by the same method as in section 3, combining  Lemma \ref{lem2.6},
we also establish the following results for the existence, uniqueness,
continuous dependence on a real parameter of  the solution to the
IVP \eqref{eE}-\eqref{IC3}.

\begin{theorem} \label{thm5.1}
Let $ f:[0,1]\times C\times \mathbb{R}\to \mathbb{R}$ be continuous
function and there exist nonnegative functions $p$, $q$, $r\in L^{1}[0,1]$
such that
\begin{itemize}
\item[(I1)] $|f(t,u,v)|\leq p(t)\|u\|+ q(t)|v|+ r(t)$, for all
$(t,u,v)\in [0,1]\times C\times \mathbb{R}$,
\item[(I2)]
$\int_0^1 p(s)ds  + \int_0^1 q(s)ds <1$.
\end{itemize}
Then the \eqref{eE}-\eqref{IC3} has at least one solution.
\end{theorem}

We remark that the above theorem may be a special case of
\cite[Corollary 4.2]{n1} which is stated there without proving.

\begin{proof}[Proof of the Theorem \ref{thm5.1}]
Here, we consider only the case  $\phi (0)= 0$ and let the
subspace $C_0$, and the function $\widehat{u}$ defined as
in Theorem \ref{thm3.1}.
Define the integral operator $ T: C_{0}\to C^{1}[0,1]$ by
\begin{equation}\label{e5.1}
Tu(t)= -\int_0^t (t-s)f(s,\widehat{u}_{s},u'(s))ds, \quad  t\in [0,1].
\end{equation}
Using (I1), \eqref{e3.2} and \eqref{e5.1}, for all $u \in C_{0}$, we obtain
\begin{equation}\label{e5.2}
\|Tu\|_0 \leq \widetilde A_1\|u\|_0+ \widetilde B_1 \|u'\|_0
+  \widetilde C_1,
\end{equation}
where
\begin{gather*}
 \widetilde A_{1}= \int_0^1(1-s)p(s)ds,\quad
 \widetilde B_{1}= \int_0^1(1-s)q(s)ds, \\
\widetilde C_{1}= \|\phi\| \int_0^1 (1-s)p(s)ds + \int_0^1(1-s)r(s)ds,
\end{gather*}
and
\begin{equation}\label{e5.3}
\|(Tu)'\|_0\leq \widetilde A_2\|u\|_0+ \widetilde B_2 \|u'\|_0
+ \widetilde C_2,
\end{equation}
where
\begin{gather*}
 \widetilde A_{2}= \int_0^1 p(s)ds,\quad
 \widetilde B_{2}= \int_0^1 q(s)ds, \\
 \widetilde C_{2}=\|\phi\| \int_0^1 p(s)ds + \int_0^1 r(s)ds.
\end{gather*}
It is easy to see that
$$
 \widetilde A_1 \leq \widetilde A_2,\quad
 \widetilde B_1 \leq \widetilde B_2,\quad
 \widetilde C_1 \leq \widetilde C_2.
$$
This implies from (I2) and \eqref{e5.2}, \eqref{e5.3} that the
following set is bounded
\begin{equation}\label{e5.4}
\lbrace u^*\in \partial \Omega : T(u^*)= \lambda u^*, \lambda >1 \rbrace.
\end{equation}
Choose the constants $ \widetilde A, \widetilde B,\widetilde m$ as follows
\begin{equation}\label{e5.5}
\widetilde A = \max \lbrace \widetilde A_1, \widetilde A_2
+ \widetilde B_2 \rbrace = \widetilde A_2+ \widetilde B_2,
\end{equation}
by (I2), we have $\widetilde A_2+ \widetilde B_2 <1$, so
$\widetilde A <1$,
\begin{equation}\label{e5.6}
 \widetilde B > \max \lbrace  \frac {\widetilde B_1 \widetilde C_2}
{1- \widetilde A}+ \widetilde C_1, \widetilde C_2 \rbrace,
\end{equation}
clearly, $\widetilde B > 0$. Put
\begin{equation}\label{e5.7}
\Omega  =  \lbrace  u \in C_0 : \|u\|_1 < \widetilde m \rbrace,
\quad \text {with } \widetilde m = \frac {\widetilde B}{1-\widetilde A}.
\end{equation}
Clearly, $\Omega $ is a bounded open subset of $C_0 $, $ 0\in \Omega $,
and $ \partial \Omega = \{u\in C_{0}: \|u\|_1= \widetilde m \}$.
Then, we can prove that the operator
$T:\overline{\Omega }=\Omega \cup
\partial \Omega \to C^{1}[0,1]$
is completely continuous and there
is not $u^* \in \partial \Omega $ such that
$T(u^*)= \lambda u^*$, for some $\lambda >1$.
By using theorem \ref{thm2.1}, $T$  has a fixed point $u\in \Omega $. The proof of
Theorem \ref{thm5.1} is complete.
\end{proof}

\begin{theorem} \label{thm5.2}
Let $ f:[0,1]\times C\times \mathbb{R}\to \mathbb{R}$ be continuous
function. Assume that  there exist nonnegative functions
$p$, $q$, $r\in L^{1}[0,1]$ and reals constants $k,l \in [0,1]$  such that
\begin{itemize}
\item[(\~I1)] $|f(t,u,v)|\leq p(t)\|u\|^k+ q(t)|v|^l+ r(t)$,
 for all $(t,u,v)\in [0,1]\times C\times \mathbb{R}$
\item[(\~I2)] $Q(k)\int_0^1 p(s)ds + Q(l) \int_0^1 q(s)ds  <1$,
\end{itemize}
where the function $Q(\mu)$ is defined as in the Theorem \ref{thm3.2}.
Then \eqref{eE}-\eqref{IC3} has at least one solution.
\end{theorem}

\begin{theorem} \label{thm5.3}
Let $ f:[0,1]\times C\times \mathbb{R}\to \mathbb{R}$ be continuous
function and satisfy on $[0,1]\times C\times \mathbb{R}$
the Lipschitz condition
$$
|f(t,u,v)-f(t,\widetilde{u},\widetilde{v})|
\leq \theta ( \|u-\widetilde{u}\|+ |v-\widetilde{v}|),
$$
for some positive constant $\theta $.
If $2 \theta < 1$, then there exists a unique solution to
 \eqref{eE}-\eqref{IC3}.
\end{theorem}

Now, we consider the problem
\begin{equation}\label{e5.8}
\begin{gathered}
u'' + f(t,u_t,u'(t), \lambda ), \quad 0 \leq t \leq 1,\\
u_0 = \phi, \quad u'(0) = 0,
\end{gathered}
\end{equation}
where $\lambda $ is a real parameter and
\begin{equation}\label{e5.9}
|f(t,u,v,\lambda )-f(t,\widetilde{u},\widetilde{v},\lambda)|
 \leq \theta ( \|u-\widetilde{u}\| + |v -\widetilde{v}|),
\end{equation}
on $[0,1]\times C\times \mathbb{R}\times \mathbb{R}$ for some positive
constant $\theta $, with
\begin{gather}\label{e5.10}
2 \theta <1, \\ \label{e5.11}
|f(t,u,v,\lambda _{1})-f(t,u,v,\lambda _{2})|
\leq L | \lambda _{1}-\lambda _{2}|,
\end{gather}
for some positive constant $L$, for all $\lambda _{1}$, $\lambda _{2}$.

\begin{theorem} \label{thm5.4}
Let $ f:[0,1]\times C\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$
be continuous function. If \eqref{e5.9}-\eqref{e5.11} hold, then
the solution to \eqref{e5.8} depends continuously on $\lambda$.
\end{theorem}

The proofs of Theorems \ref{thm5.2}--\ref{thm5.4} are similar to that
of Theorems \ref{thm3.2}--\ref{thm3.4}, respectively, let us omit them.

Next, we shall show that the solution set of \eqref{eE}-\eqref{IC3}
 is nonempty, compact and connected. To this end, we need  the
following result.

\begin{proposition}  \label{prop5.5}
Let $f: [0,1] \times C \times \mathbb{R} \to \mathbb{R}$ be continuous
and locally Lipschitz with respect to $ C \times \mathbb{R}$,
 i.e. for every  $(t_0,u_0,v_0) \in [0,1] \times C \times \mathbb{R}$,
there exist positive constants $\delta,\rho, \sigma $ and
$\theta \geq 0 $ such that
$$
|f(t,u,v)-f(t,\widetilde u,\widetilde v)|
 \leq \theta ( \| u -\widetilde{u}\|+ |v-\widetilde{v}|),
$$
for some positive constant $\theta$, for all
$ t \in [0,1]$, $(u,v), (\widetilde{u},\widetilde{v})
\in C \times \mathbb{R}$, with
$$
|t-t_{0}| \leq \delta ,\quad
\|u-u_{0}\|\leq \rho,\quad
\| \widetilde{u}-u_{0}\|\leq \rho, \quad
\ |v-v_{0}| \leq \sigma, | \widetilde{v}-v_{0}| \leq \sigma.
$$
Then \eqref{eE}-\eqref{IC3} has at most a solution.
\end{proposition}

\begin{proof}
Suppose that \eqref{eE}-\eqref{IC3} have two solutions $u(t), v(t)$
on $[-r,1]$. Then
$$
u(t)=v(t), \quad \text{for all } t\in [-r,0].
$$
We shall show that  $u(t)=v(t)$, for all $t \in [-r,1]$.
Put
\begin{equation}\label{e5.12}
b= \max \big \lbrace \tau :u(t)=v(t), \forall t\in [-r,\tau] \big \rbrace.
\end{equation}
Clearly, $b\geq 0$. Thus $ 0 \leq b \leq 1$.
We suppose by contradiction that $ b < 1$. Since $f$ is locally lipschitz,
for $(b, u_{b}, u'(b))\in [0,1] \times C \times \mathbb{R}$,
there exist real numbers $\delta, \rho , \sigma $ and $\theta \geq 0$
such that
\begin{equation*}
|f(t,\widetilde{u}_{1},\widetilde{v}_{1})-f(t,\widetilde{u}_{2},
\widetilde{v}_{2})| \leq \theta \big ( \| \widetilde{u}_{1}
-\widetilde{u}_{2}\| + |\widetilde{v}_{1}-\widetilde{v}_{2}|\big),
\end{equation*}
for all $t\in [0,1]$, $(\widetilde{u}_{1},\widetilde{v}_{1})$,
$(\widetilde{u}_{2},\widetilde{v}_{2})\in C\times \mathbb{R}$,
 with $| t-b| \leq \delta$,
$$
\|\widetilde{u}_{1}-u_{b}\| \leq \rho,\quad
\|\widetilde{u}_{2}-u_{b}\|\leq \rho, \quad
|\widetilde{v}_{1}-u'(b)|\leq \sigma,\quad
|\widetilde{v}_{2}-u'(b)|\leq \sigma.
$$
Note that $u_{b}= v_{b}$, $ u'(b)=v'(b)$  and $\,\ b+ \delta \leq 1$.

For each fixed $u \in C([-r,1];\mathbb{R})$ which is continuously
differentiable on $[0,1]$, since the mappings
$$
s \mapsto u_{s},\quad  s \mapsto u'(s) \quad \text{with } s \in [0,1],
$$
are continuous, so there exists $\delta' >0$ with $\delta' <\delta $
and $ 2\theta \delta'<1$, such that
$$
\| u_{s}-u_{b}\|\leq \rho, \quad
\|v_{s}-u_{b}\| \leq \rho, \quad
 |u'(s)-u'(b)|\leq \sigma,\quad
 |v'(s)-u'(b)|\leq \sigma,
$$
for all $s \in [b,b+\delta']$.

Let $S_b$ be the space of continuous functions
$ x :[-r,b + \delta'] \to \mathbb{R}$
which are continuously differentiable on $[b, b + \delta']$ with
$ x_b= u_b$. We define
$$
d_b(x,y)= \max \Big \lbrace \max_{b \leq t \leq b+ \delta'}
|x(t)-y(t)|,  \max_{b \leq t\leq b+\delta'}|x'(t)-y'(t)| \Big \rbrace.
$$
Then $S_{b}$ is a completely metrizable space with the distance
function $d_{b}$.
It is easy to see that  $\overline{u}= u |_{[-r,b+\delta']} \in S_{b}$
and  $\overline{v}= v  |_{[-r,b+\delta']} \in S_{b}$.
Put $w= \overline{u}- \overline{v}$, then $w$ satisfies
\begin{equation}\label{e5.13}
\begin{gathered}
w'' + f(t,\overline{u}_{t},\overline{u}'(t))
- f(t,\overline{v}_{t},\overline{v}'(t))= 0, \quad
b\leq t \leq  b+ \delta',\\
w_{b}=0, \quad w'(b)=0.
\end{gathered}
\end{equation}
It follows that for all $t\in [b,b+\delta']$, we have
\begin{align*}
|w(t)|& \leq \int_b^t (1-s)|f(s,\overline{u}_{s},\overline{u}'(s))
  -f(s,\overline{v}_{s},\overline{v}'(s))| ds \\
&\leq \theta \int_b^t \big( \|\overline{u}_{s}-\overline{v}_{s}\|
  + |\overline{u}'(s)-\overline{v}'(s)| \big) ds \\
&\leq \theta \delta '\Big ( \max_{b\leq t\leq b+\delta '} |\overline{u}(t)
-\overline{v}(t)|
+ \max_{b \leq t \leq  b+\delta'} |\overline{u}'(t)- \overline{v}'(t)|\Big ).
\end{align*}
Similarly,
\begin{align*}
|w'(t)|& \leq \int_b^t |f(s,\overline{u}_{s},\overline{u}'(s))
  -f(s,\overline{v}_{s},\overline{v}'(s))| ds \\
&\leq \theta \delta'\Big ( \max_{b\leq t\leq b+\delta'}
|\overline{u}(t)-\overline{v}(t)| + \max_{b\leq t\leq b+\delta'}
|\overline{u}'(t)-\overline{v}'(t)|\Big ).
\end{align*}
By the definition of the distance $d_{b}$, we have
\begin{align*}
d_{b}(\overline{u},\overline{v})
& = \max \Big \lbrace \max_{b\leq t\leq b+\delta'}
|\overline{u}(t)-\overline{v}(t)|, \max_{b\leq t\leq b+\delta'}
 |\overline{u}'(t)- \overline{v}'(t)| \Big \rbrace\\
& \leq \theta \delta'\Big ( \max_{b\leq t\leq b+\delta'}
|\overline{u}(t)-\overline{v}(t)| + \max_{b\leq t\leq b+\delta'}
|\overline{u}'(t)-\overline{v}'(t)|\Big )\\
& \leq 2 \theta \delta' d_{b}(\overline{u},\overline{v}).
\end{align*}
Since  $2\theta \delta'<1$, we deduce that
$d_{b}(\overline{u},\overline{v})= 0$ i.e. $\overline{u}=\overline{v}$.
Therefore,
$$
u(t)=v(t), \quad \forall  t \in [-r, b+\delta'].
$$
This leads to a contradiction with the definition of $b$ in \eqref{e5.12}.
Then the proof is complete.
\end{proof}

 From Theorems \ref{thm5.1}, \ref{thm5.2} and Proposition \ref{prop5.5},
 we obtain the following corollary.

\begin{corollary} \label{coro5.6}
Let $ f:[0,1]\times C\times \mathbb{R}\to \mathbb{R}$ be a continuous
function and locally Lipschitz with respect to $C \times \mathbb {R}$.
Assume that  there exist nonnegative functions $p$, $q$, $r\in L^{1}[0,1]$
and reals constants $k,l \in [0,1]$  such that
\begin{itemize}
\item[(\~I1)] $|f(t,u,v)|\leq p(t)\|u\|^k+ q(t)|v|^l+ r(t)$,
  for all $(t,u,v)\in [0,1]\times C\times \mathbb{R}$
\item[(\~I2)] $Q(k)\int_0^1 p(s)ds + Q(l) \int_0^1 q(s)ds  <1$,
\end{itemize}
where the function $Q(\mu)$ is defined as in the Theorem \ref{thm3.2}.
Then \eqref{eE}-\eqref{IC3} has a unique solution.
\end{corollary}

By the above results and applying Theorems \ref{thm2.2}, \ref{thm2.3},
 we have the following theorem.

\begin{theorem} \label{thm5.7}
Let $f:[0,1]\times C\times \mathbb{R}\to \mathbb{R}$
be continuous function and satisfy the conditions (I1)-(I2)
or (\~I1)-(\~I2). Then the solution
set of the IVP \eqref{eE}-\eqref{IC3} is nonempty, compact and connected.
\end{theorem}

\begin{proof}
{\bf Step 1}. {\it The case $\phi (0)=0$.}
We again consider the subspace $C_{0}$, the function $\widehat{u}$ and the
operator $T$, which are defined as in Theorem \ref{thm5.1}
As above,
$T:\overline{\Omega }=\Omega \cup \partial \Omega \to C^{1}[0,1]$ is
completely continuous, where
$$
\Omega = \{ u\in C_{0}:\|u\|_{1}<\widetilde{m}\},\quad
 \widetilde{m}=\frac{\widetilde{B}}{1-\widetilde{A}}.
$$
According to  Theorems \ref{thm5.1}-\ref{thm5.2}, it is obvious
that the fixed
point set of $T$ is nonempty. Furthermore, it is compact and connected.
Indeed, First, for all $u\in \overline{\Omega }$, it follows from
\eqref{e5.2}, \eqref{e5.3}, \eqref{e5.6} and \eqref{e5.7}, that
\begin{gather*}
\|Tu\|_{1}\leq \widetilde{A}~\widetilde{m}+\widetilde{C}_{2},\\
\widetilde{m}=\frac{\widetilde{B}}{1-\widetilde{A}}
>\frac{\widetilde C_2}{1-\widetilde{A}}, \quad \text{i.e.  }
\widetilde{A}\widetilde{m} +\widetilde{C}_{2}<\widetilde{m}.
\end{gather*}
Therefore,
$\|Tu\|_{1}<\widetilde{m}$.
Then we obtain
$$
T(\overline{\Omega })\subset \Omega .
$$
On the other hand, $\Omega $ is convex, so
$$
\deg (I-T,\Omega ,0)\neq 0.
$$
Obviously, $T$  has no fixed points on $\partial \Omega $.

Next, the function $f:[0,1]\times C\times \mathbb{R}\to \mathbb{R}$
is continuous function, by  Theorem \ref{thm2.3}, for each $\varepsilon >0$,
there is a mapping
$f_{\varepsilon }:[0,1]\times C\times \mathbb{R} \to \mathbb{R}$
that is locally Lipschitz with respect to $C\times  \mathbb{R}$, such that
\begin{equation}\label{e5.14}
|f(t,u,v)-f_{\varepsilon }(t,u,v)| \leq \frac {\varepsilon}{2},
\quad \forall (t,u,v)\in [0,1]\times C\times \mathbb{R}.
\end{equation}
Clearly, $f_{\varepsilon }$ is continuous. Moreover, by $f$ satisfies
the conditions (I1)-(I2) or (\~I1)-(\~I2), it follows
from \eqref{e5.14} that  $f_{\varepsilon }$ satisfies the conditions
 (I1)-(I2) or (\~I1)-(\~I2).
Let  $T_{\varepsilon }:\overline{\Omega }\to C^{1}[0,1]$ be defined by
\begin{equation}\label{e5.15}
T_{\varepsilon }u(t)=
-\int_0^t  (t-s)f_{\varepsilon}(s,\widehat{u}_{s},u'(s))ds, \quad
 t\in [0,1].
\end{equation}
It is easy to check that $T_{\varepsilon }$ is completely continuous and
\begin{equation}\label{e5.16}
\|T(u)-T_{\varepsilon }(u)\|_{1}\leq \frac {\varepsilon}{2}< \varepsilon,
\quad \forall u \in \overline{\Omega }.
\end{equation}
Finally, we need prove that for each $h\in \overline{\Omega }$ with
$\|h\|_{1}<\varepsilon$, the equation
\begin{equation}\label{e5.17}
u = T_{\varepsilon }(u) + h,
\end{equation}
has at most one solution. Suppose that $u_{1}$, $u_{2}$ are two
solutions of \eqref{e5.17}. Put
$$
w_{1}=\widehat{u}_{1}-\widehat{h}, \quad
w_{2}=\widehat{u}_{2}-\widehat{h},
$$
where
\begin{equation*}
\widehat{h}(t)= \begin{cases}
\phi (t), & t\in [-r,0], \\
h(t),&  t \in [0,1],
\end{cases}
\qquad
\widehat{u}_{i}(t)= \begin{cases}
\phi (t), & t\in [-r,0], \\
u_i(t), & t \in [0,1],
\end{cases}
\end{equation*}
$i=1, 2$. Then $w_{1}$, $w_{2}$ are two solutions of the  problem
\begin{equation}\label{e5.18}
\begin{gathered}
w'' +  f_{\varepsilon }\big ( t,w_{t}+\widehat{h}_{t},w'(t)+h'(t)\big ) = 0,
\quad  0\leq t\leq 1,\\
w_{0}=0,\quad w'(0)=0.
\end{gathered}
\end{equation}
This implies from Proposition \ref{prop5.5} that the problem \eqref{e5.18}
 has at most one solution, so
$$
w_{1}= w_{2}, \quad \text{i.e. } u_{1}= u_{2}.
$$
It follows that \eqref{e5.17} has at most one solution.

Applying Theorem \ref{thm2.2}, we have the fixed point set of $T$ is nonempty,
compact and connected. Thus, so is the solution set of
\eqref{eE}-\eqref{IC3}. The step 1 is complete.
\smallskip

\noindent{\bf Step 2}. {\it The case $\phi (0)\neq 0$.}
By the transformation $v=u-\phi (0)$, the IVP \eqref{eE}-\eqref{IC3}
can be rewritten in the form
\begin{equation}\label{e5.19}
\begin{gathered}
v'' +  f \big ( t,v_{t}+ \phi (0),v'(t)\big  ) = 0, \quad 0\leq t\leq 1,\\
v_{0}=\phi - \phi(0) \equiv \widetilde{\phi },\quad  v'(0)=0.
\end{gathered}
\end{equation}
in which  $\widetilde{\phi }\in C$ and $\widetilde{\phi }(0)=0$.
By the step 1, we can prove without difficulty that the solution set
of \eqref{e5.19} is nonempty, compact and connected.
In this proof, when $f$ satisfies the conditions
(\~I1)-(\~I2), the inequality \eqref{e3.18} is used again.
Consequently, the solution set of  \eqref{eE}-\eqref{IC3} is nonempty,
compact and connected. Theorem \ref{thm5.7} is proved.
\end{proof}

\subsection*{Acknowledgements}
 The authors wish to express their sincere thanks to the referee
for his/her helpful comments, also to Professor Dung Le and to
Professor Nguyen Thanh Long for their helpful suggestions and remarks.

\begin{thebibliography}{0}

\bibitem{d1} K. Deimling, {\it Nonlinear Functional Analysis},
Springer, NewYork, 1985.

\bibitem{h1} J. Henderson, {\it Boundary Value Problems for Functional
Differential Equations}, World Scientific Publishing, 1995.

\bibitem{k1} M. A. Krasnosel'skii, P. P. Zabreiko;
 {\it Geometrical Methods of Nonlinear Analysis}, Springer-Verlag
Berlin Heidelberg New York Tokyo, 1984.

\bibitem{m1} R. Ma, {\it Positive solutions of a nonlinear
three-point boundary-value problem}, Electronic J. Differential Equations,
Vol. 1998 (1998), No. 34, 1-8.

\bibitem{n1} S. K. Ntouyas; {\it Boundary value problems for neutral
functional differential equations}, Editor Johnny Henderson,
World Scientific,  1995, 239 - 249.

\bibitem{o1} Donal O'Regan; {\it Theory of singular boundary problems},
World Scientific Publishing, 1994.

\bibitem{y1} Yong-Ping Sun; {\it Nontrivial solution for a three-point
boundary-value problem}, Electronic J. Differential Equations, Vol. 2004
(2004),  No. 111, 1-10.

\bibitem{z1} Eberhard Zeidler; {\it Nonlinear Functional Analysis
and its Applications}, Springer-Verlag New York Berlin Heidelberg Tokyo,
Part I.

\bibitem{z2} Bo Zhang, {\it Boundary value problems of second order
functional differential equations}, Editor Johnny Henderson,
World Scientific,  1995, 301- 306.

\end{thebibliography}

\end{document}