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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 123, pp. 1--9.\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 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/123\hfil Four-point boundary-value problems]
{Solvability of a four-point boundary-value problem  for 
fourth-order ordinary \\ differential equations}

\author[J. Ge, C. Bai \hfil EJDE-2007/123\hfilneg]
{Jing Ge, Chuanzhi Bai} 

\address{Jing Ge  \newline
Department of  Mathematics \\
 Huaiyin Teachers College\\
  Huaian, Jiangsi 223300,  China}
  \email{gejing0512@163.com}

\address{Chuanzhi Bai \newline
Department of  Mathematics \\
 Huaiyin Teachers College\\
  Huaian, Jiangsi 223300,  China}
\email{czbai8@sohu.com}

\thanks{Submitted May 10, 2007. Published September 21, 2007.}
\thanks{Supported by grants NNSFC-10771212 and  06KJB110010.}
\subjclass[2000]{34B10, 34B15}
\keywords{Four-point boundary-value problem; existence of solutions;
\hfill\break\indent fixed point theorem}

\begin{abstract}
 In this paper we investigate the existence of solutions of a class
 of  four-point boundary-value problems for fourth-order ordinary
 differential equations. Our analysis relies on  a fixed point
 theorem due to Krasnoselskii and Zabreiko.
\end{abstract}

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


\section{Introduction}

In recent years, boundary-value problems for second and higher
order differential equations have been extensively studied. The
monographs of Agarwal \cite{a1} and Agarwal, O'Regan, and Wong
\cite{a2} contain excellent surveys of known results. Recently an
increasing interest in studying the existence of solutions and
positive solutions to boundary-value problems for higher order
differential equations is observed; see for example \cite{a3, b1,
g1, g2, g3, h1}.


Very recently, Zhang, Chen and L\"u \cite{z1} by
using the upper and lower solution method investigated the fourth
order nonlinear ordinary differential equation
\begin{equation}
u^{(4)}(t) = f(t, u(t), u''(t)), \quad 0 < t < 1,
\label{e1.1}
\end{equation}
with the four-point boundary conditions
\begin{equation}
\begin{gathered}
u(0) = u(1) = 0, \\
 a u''(\xi_1) - b
u'''(\xi_1) = 0, \quad  c u''(\xi_2) + d u'''(\xi_2) = 0,
\end{gathered}
\label{e1.2}
\end{equation}
  where $a, b, c, d$ are nonnegative constants satisfying $a d + b c
+ a c(\xi_2 - \xi_1) > 0$, $0 \leq \xi_1 < \xi_2 \leq 1$ and $f \in
C([0, 1] \times \mathbb{R} \times \mathbb{R})$.  They proved the following
Lemma (a key lemma):

\begin{lemma}[{\cite[Lemma 2.2]{z1}}] \label{lem1.1}
 Suppose $a, b, c, d, \xi_1, \xi_2$ are nonnegative constants satisfying
$0 \leq \xi_1 < \xi_2 \leq 1$, $b - a \xi_1 \geq 0$,
$d - c + c \xi_2 \geq 0$ and
$\delta = a d + b c + a c (\xi_2 - \xi_1) \not= 0$. If
$u(t) \in C^4[0, 1]$ satisfies
\begin{itemize}
\item $u^{(4)}(t) \geq 0$, $t \in (0, 1)$,

\item $u(0) \geq 0$, $u(1) \geq 0$,

\item $a u''(\xi_1) - bu'''(\xi_1) \leq  0$,
$ c u''(\xi_2) + d u'''(\xi_2) \leq 0$,
\end{itemize}
then $u(t) \geq 0$ and $u''(t) \leq 0$ for $t \in [0,1]$.
\end{lemma}

Unfortunately this Lemma is wrong as shown below.

\subsection*{Counterexample to \cite[Lemma 2.2]{z1}}
Let $u(t) = \frac{1}{3} t^4 +
\frac{1}{4} t^3 - \frac{4}{3} t^2 + \frac{3}{4} t$ which belongs to $C^4[0, 1]$,
$\xi_1 = \frac{1}{10}$, $\xi_2 = \frac{1}{8}$, $a, b, c, d$ be
nonnegative constants satisfying $b \geq  \frac{1}{10} a = a \xi_1$,
$ d = \frac{15}{16} c > \frac{7}{8} c = (1 - \xi_2) c$ and
$\delta = a d + b c + \frac{1}{40} a c \not= 0$. Then we have
\begin{gather*}
u^{(4)}(t) = 8 > 0, \quad  t \in (0, 1), \\
 u(0) = 0, \quad  u(1) = 0, \\
\begin{aligned}
 a u''(\xi_1) - b u'''(\xi_1)
 & = a \Big[4 t^2 + \frac{3}{2} t -
\frac{8}{3}\Big]_{t = 1/10} - b \Big[8 t + \frac{3}{2}\Big]_{t= 1/10} \\
 & = - 2\frac{143}{300} a - 2\frac{3}{10} b  \leq 0,
\end{aligned}
\end{gather*}
 and
\begin{align*}
c u''(\xi_2) + d u'''(\xi_2)
&= c \Big[4 t^2 + \frac{3}{2} t - \frac{8}{3}\Big]_{t = 1/8} + d
\Big[8 t  + \frac{3}{2}\Big]_{t = 1/8} \\
&= - \frac{29}{12} c + \frac{5}{2} d  = - \frac{29}{12} c +
\frac{5}{2} \cdot \frac{15}{16} c = - \frac{7}{96} c \leq 0.
\end{align*}
 But
$$
u\big(\frac{8}{9}\big) = - 0.0031 < 0;
$$
 that is, \cite[Lemma 2.2]{z1} is incorrect.

So the conclusions of \cite{z1} should be reconsidered.  The aim
of this paper is to investigate  the existence of solutions of the
BVP \eqref{e1.1}-\eqref{e1.2} by using a fixed point theorem due
to Krasnoselskii and Zabreiko in \cite{k1}.

\section{Main result}

First, we give some lemmas which are needed in our discussion of
the main results.

\begin{lemma} \label{lem2.1}
Suppose $a, b, c, d, \xi_1, \xi_2$
are nonnegative constants satisfying $0 \leq \xi_1 < \xi_2 \leq 1$
and $\delta = a d + b c + a c (\xi_2 - \xi_1) \not= 0$. If  $h \in
C[0, 1]$, then the boundary-value problem
\begin{gather}
  v''(t) = h(t), \quad   t \in [0, 1], \label{e2.1} \\
 a v(\xi_1) -  b v'(\xi_1) = 0, \quad  c v(\xi_2) + d
v'(\xi_2) = 0, \label{e2.2}
\end{gather}
has a unique solution
\begin{equation}
 v(t) = \int_{\xi_1}^t (t - s) h(s) ds + \frac{1}{\delta}
\int_{\xi_1}^{\xi_2} (a(\xi_1 - t) - b)(c(\xi_2 - s) + d)h(s) ds.
\label{e2.3}
\end{equation}
\end{lemma}

\begin{proof}
By \eqref{e2.1},  it is easy to know that
\begin{equation}
 v(t) = C_1 + C_2 t + \int_0^t (t - s) h(s) ds,
\label{e2.4}
\end{equation}
  where $C_1, C_2$ are any two constants. Substituting \eqref{e2.4} into
boundary  conditions \eqref{e2.2}, by a routine calculation, we get
\begin{gather}
 C_1 = \int_0^{\xi_1} s h(s) ds +
\frac{1}{\delta} (a \xi_1 - b) \int_{\xi_1}^{\xi_2} (c(\xi_2 - s) +
d) h(s) ds,
\label{e2.5}
\\
C_2 = - \int_0^{\xi_1} h(s) ds - \frac{a}{\delta}
\int_{\xi_1}^{\xi_2} (c(\xi_2 - s) + d) h(s) ds.
 \label{e2.6}
\end{gather}
Substituting  \eqref{e2.5} and \eqref{e2.6} into \eqref{e2.4}, we
obtain \eqref{e2.3} which implies lemma.
\end{proof}


\begin{remark} \label{rmk2.2} \rm
 Let $\xi_1 = 0$, $\xi_2 = 1$, then \eqref{e2.3} reduces to
 $$  v(t) = -  \int_0^1 G(t, s) h(s) ds,  $$
 where
\[
G(t, s) = \frac{1}{\delta} \begin{cases}
(as + b)(d + c(1 - t)), & 0 \leq s \leq t \leq 1,\\
(at + b)(d + c(1 - s)), & 0 \leq t < s \leq 1.
 \end{cases}
 \]
\end{remark}

\begin{remark} \label{rmk2.3} \rm
 Let
 \begin{equation} \label{e2.7}
\begin{gathered}
  R(t) = \frac{1}{\delta} ((a(t - \xi_1)
+ b)x_3 + (c(\xi_2 - t) + d) x_2),
\\
 G_1(t, s) = \begin{cases}
s(1 - t), & 0 \leq s \leq t \leq 1, \\
t(1 - s), & 0 \leq t < s \leq 1,
\end{cases}
\\
 G_2(t, s) = \frac{1}{\delta}
  \begin{cases}
(a(s - \xi_1) + b)(d + c(\xi_2 - t)), & \xi_1 \leq s \leq t \leq \xi_2,\\
(a(t - \xi_1) + b)(d + c(\xi_2 - s)), & \xi_1 \leq t < s \leq
\xi_2.
 \end{cases}
\end{gathered}
\end{equation}
In \cite[Lemma 2.2]{z1} it is claimed that
 \begin{equation}
  u(t) = t x_1 + (1 - t) x_0 - \int_0^1
G_1(t, \xi) R(\xi) d \xi + \int_0^1 G_1(t, \xi) \int_{\xi_1}^{\xi_2}
G_2(\xi, s) h(s) ds d \xi, \label{e2.8}
\end{equation}
  is the solution of the  boundary-value problem
\begin{gather*}
 u^{(4)}(t) = h(t), \quad  0 < t < 1,\\
 u(0) =  x_0, \quad  u(1) = x_1, \\
 a u''(\xi_1) - b u'''(\xi_1) = x_2,
\quad  c u''(\xi_2) + d
u'''(\xi_2) = x_3.
\end{gather*}
  However \eqref{e2.8} is wrong. Indeed, by
Lemma \ref{lem2.1}, \eqref{e2.8} should be
replaced by
 $$
u(t) = t x_1 + (1 - t) x_0 - \int_0^1
G_1(t, \xi) R(\xi) d \xi - \int_0^1 G_1(t, \eta) v(\eta) d \eta,
$$
where
$$
v(\eta) = \int_{\xi_1}^{\eta} (\eta - s) h(s) ds +
\frac{1}{\delta} \int_{\xi_1}^{\xi_2} (a(\xi_1 - \eta) - b)(c(\xi_2
- s) + d)h(s) ds.
$$
\end{remark}

\begin{remark} \label{rmk2.4} \rm
 In  \cite[Theorem 3.1]{z1}, the operator $T : C[0, 1] \to C[0,1]$
is defined as
 $$
T u(t) = \int_0^1
G_1(t, \eta) \int_{\xi_1}^{\xi_2} G_2(\eta, s) f(s, u(s),
u''(s)) ds d \eta,
$$
  where $G_1(t, s)$ and $G_2(t, s)$ are as in Remark \ref{rmk2.2}.
By \ref{lem2.1} and Remark \ref{rmk2.2}, the definition of $T$ is
incorrect. In fact, the operator $T$ should be defined as
\begin{align*}
T u(t) & = \int_0^1 G_1(t, \eta) \int_{\xi_1}^{\eta} (s - \eta) f(s,
u(s), u''(s)) ds d \eta\\
& \quad  + \frac{1}{\delta} \int_0^1 G_1(t, \eta)
\int_{\xi_1}^{\xi_2} (b - a(\xi_1 - \eta))(c(\xi_2 - s) + d) f(s,
u(s), u''(s)) ds d \eta.
 \end{align*}
 \end{remark}

The following well-known fixed point theorem \cite{k1} will play
an important role in the proof of our theorem.

\begin{lemma} \label{lem2.5}
 Let $X$ be a Banach space, and $F : X \to X$ be completely continuous.
Assume that $A : X \to X$ is a
bounded linear operator such that 1 is not an eigenvalue of $A$ and
$$
\lim _{\|x\| \to \infty} \frac{\|F(x) - A(x)\|}{\|x\|} =0.
$$
 Then $F$ has a fixed point in $X$.
\end{lemma}

Let $X = C^2[0, 1]$ be endowed with the norm by
 $$
\|u\|_0 = \max \{\|u\|, \|u^{\prime \prime}\|\},
$$
  where $\|u\| = \max_{0 \leq t \leq 1} |u(t)|$.

We are now in a position to present and prove our main result.
Let

 \begin{itemize}
\item[(H1)]  $a, b, c, d, \xi_1, \xi_2$
are nonnegative constants satisfying $0 \leq \xi_1 < \xi_2 \leq 1$,
$b - a \xi_1 \geq 0$  and  $\delta = ad + bc + ac(\xi_2 - \xi_1)
\not= 0$,

\item[(H2)]  $f(t, u, v) = p(t) g(u) + q(t) h(v)$, where
$g, h : \mathbb{R} \to \mathbb{R}$ are continuous with
  $$
\lim _{u \to \infty} \frac{g(u)}{u} = \lambda,
\quad  \lim _{v \to \infty} \frac{h(v)}{v} = \mu,
$$
 where  $p, q \in C[0, 1]$. Moreover, there exists some
$t_0 \in [0, 1]$ such that $p(t_0) g(0) + q(t_0) h(0) \not= 0$,  and there
exists a continuous nonnegative function $w : [0, 1] \to
\mathbb{R}^+$ such that $|p(s)| + |q(s)| \leq w(s)$ for each
$s \in [0, 1]$.
\end{itemize}

\begin{theorem} \label{thm2.6}
Assume {\rm (H1)--(H2)}.  If
 $ \max \{|\lambda|, |\mu|\} < \min\big\{\frac{1}{L_1}, \frac{1}{L_2}\big\}$,
 where
\begin{align*}
 L_1 & = \frac{1}{12} \Big[\int_0^{\xi_1} \tau^3 (2-\tau) w(\tau)
d\tau + \int_{\xi_1}^1 (1-\tau)^3(1+\tau) w(\tau) d\tau
\\
 & \quad  + \frac{2(b-a \xi_1) +
a}{\delta} \int_{\xi_1}^{\xi_2} (c(\xi_2 - \tau) + d) w(\tau)d\tau
\Big],
\end{align*}
 and
 $$
L_2 = \int_{\xi_1}^1 (1-\tau) w(\tau)
d\tau + \frac{1}{\delta} \int_{\xi_1}^{\xi_2} (b +
a(1-\xi_1))(c(\xi_2 - \tau) + d) w(\tau)d\tau,
$$
  then BVP \eqref{e1.1} and \eqref{e1.2} has at least one nontrivial
solution $u \in C^2[0, 1]$.
\end{theorem}

\begin{proof}
Define an operator $F : C^2[0, 1] \to C^2[0, 1]$  by
\begin{equation}
\begin{aligned}
  F u(t)
 & := \int_0^1 G_1(t, s) \int_{\xi_1}^s (\tau - s)
[p(\tau)
g(u(\tau)) + q(\tau) h(u''(\tau)] d \tau ds \\
& \quad + \frac{1}{\delta} \int_0^1 G_1(t, s) \int_{\xi_1}^{\xi_2} (b -
a(\xi_1 - s))(c(\xi_2 - \tau) + d) \\
&\quad \times \big[p(\tau) g(u(\tau)) + q(\tau)
h(u''(\tau)\big] d \tau ds,
\end{aligned} \label{e2.9}
\end{equation}
  where $G_1(t, s)$ is as in \eqref{e2.7}. Then by Lemma \ref{lem2.1}
  and Remark \ref{rmk2.4},    we easily know that the fixed points of
   $F$ are the solutions to the boundary-value problem \eqref{e1.1}
    and \eqref{e1.2}. It is well known that the operator
  $F$ is a completely continuous operator. Now, we consider
  the following boundary-value problem
\begin{equation}
\begin{gathered}
u^{(4)}(t) = \lambda p(t)u(t) + \mu q(t)u''(t), \quad  0 < t < 1\\
u(0) = u(1) = 0, \\
a u''(\xi_1) - b u'''(\xi_1) = 0, \quad  c u''(\xi_2) + d
u'''(\xi_2) = 0.
\end{gathered} \label{e2.10}
\end{equation}

Define
\begin{equation}
\begin{aligned}
 A u(t)  &:= \int_0^1 G_1(t, s) \int_{\xi_1}^s (\tau - s) [\lambda
p(\tau) u(\tau) + \mu q(\tau) u''(\tau) d \eta ds
\\
 &\quad  + \frac{1}{\delta} \int_0^1 G_1(t,
s) \int_{\xi_1}^{\xi_2} (b - a(\xi_1 - s))(c(\xi_2 - \tau) + d) \\
&\quad [\lambda p(\tau) u(\tau) + \mu q(\tau) u''(\tau)] d \eta ds.
\end{aligned} \label{e2.11}
\end{equation}
  Obviously,  $A$ is a bounded linear operator.
Furthermore, the fixed point of $A$ is a solution of the BVP
\eqref{e2.10}  and conversely.

We now assert that 1 is not an eigenvalue of $A$.
In fact, if $\lambda = 0$ and $\mu = 0$, then the BVP \eqref{e2.10}
has no nontrivial solution.
If $\lambda \not= 0$ or $\mu \not= 0$, suppose the BVP \eqref{e2.10}
has a nontrivial solution $u$ and $\|u\|_0 > 0$, then
\begin{align*}
 |A u(t)|
 & \leq \int_0^1 G_1(t, s) \int_{\xi_1}^s |(\tau - s)
  [\lambda p(\tau) u(\tau) + \mu q(\tau) u''(\tau)]| d \tau ds \\
& \quad + \frac{1}{\delta} \int_0^1 G_1(t, s) \int_{\xi_1}^{\xi_2}
  \big|(b - a(\xi_1 - s))(c(\xi_2 - \tau) + d) \\
&\quad\times [\lambda p(\tau) u(\tau) + \mu q(\tau) u''(\tau)] \big| d \tau ds\\
 & \leq \int_0^1 s(1-s)
\int_{\xi_1}^s (s - \tau) [|\lambda| |p(\tau)| |u(\tau)| + |\mu|
|q(\tau)| |u''(\tau)|] d \tau ds \\
 & \quad + \frac{1}{\delta} \int_0^1 s(1-s)
\int_{\xi_1}^{\xi_2} (b - a(\xi_1 - s))(c(\xi_2 - \tau) + d) \\
&\quad\times [|\lambda| |p(\tau)||u(\tau)| + |\mu||q(\tau)||u''(\tau)|] d \tau
ds \\
 &  \leq  \Big[\int_0^1 s(1-s)
\int_{\xi_1}^s (s - \tau) [|\lambda||p(\tau)| + |\mu||q(\tau)|] d
\tau ds  + \frac{1}{\delta} \int_0^1 s(1-s) \\
&\quad\times
\int_{\xi_1}^{\xi_2} (b - a(\xi_1 - s))(c(\xi_2 - \tau) + d)
[|\lambda||p(\tau)| + |\mu||q(\tau)|] d \tau ds\Big]\|u\|_0 \\
 &  = \frac{1}{12} \Big[\int_0^{\xi_1}
\tau^3 (2-\tau) (|\lambda||p(\tau)| + |\mu||q(\tau)|) d\tau \\
&\quad + \int_{\xi_1}^1 (1-\tau)^3(1+\tau)(|\lambda||p(\tau)|
 + |\mu||q(\tau)|) d\tau \\
& \quad  + \frac{2(b-a \xi_1) + a}{\delta} \int_{\xi_1}^{\xi_2}
 (c(\xi_2 - \tau) + d) (|\lambda||p(\tau)|
 +|\mu||q(\tau)|) d\tau \Big]\|u\|_0 \\
 & \leq \max \{|\lambda|, |\mu|\} \frac{1}{12} \Big[\int_0^{\xi_1}
\tau^3 (2-\tau) w(\tau) d\tau + \int_{\xi_1}^1 (1-\tau)^3(1+\tau)
w(\tau) d\tau  \\
& \quad + \frac{2(b-a \xi_1) + a}{\delta} \int_{\xi_1}^{\xi_2}
(c(\xi_2 - \tau) + d) w(\tau) d\tau \Big]\|u\|_0,  \quad t \in
[0, 1],
\end{align*}
  which implies that
 $$
|A u(t)| \leq \max \{|\lambda|, |\mu|\}
L_1 \|u\|_0 <  \frac{1}{L_1} L_1 \|u\|_0 = \|u\|_0.
 $$
  On the other hand, we have
  \begin{align*}
 |(Au)''(t)|
 & = \big|\int_{\xi_1}^t
(s-t) [\lambda p(s) u(s) + \mu q(s) u''(s)] ds  \\
 & \quad   + \frac{1}{\delta}
\int_{\xi_1}^{\xi_2} (b - a(\xi_1 - t))(c(\xi_2 - s) + d) [\lambda
p(s) u(s) + \mu q(s) u''(s)] ds\big| \\
 &  \leq \Big[\int_{\xi_1}^1 (1-s) (|\lambda||p(s)| +
|\mu||q(s)|) ds \\
 & \quad   +
\frac{1}{\delta} \int_{\xi_1}^{\xi_2} (b + a(1-\xi_1))(c(\xi_2 - s)
+ d) (|\lambda||p(s)| + |\mu||q(s)|) ds\Big]\|u\|_0  \\
 &   \leq \max \{|\lambda|, |\mu|\} L_2
\|u\|_0 < \frac{1}{L_2} L_2 \|u\|_0 = \|u\|_0, \quad t \in [0, 1].
\end{align*}
  Then  $\|Au\|_0 < \|u\|_0$. This contradiction means
that BVP \eqref{e2.10} has no nontrivial solution. Hence, 1 is not
an eigenvalue of $A$.

Finally, we prove that
 $$
\lim _{\|u\|_0  \to \infty} \frac{\|Fu - Au\|_0}{\|u\|_0} = 0.
$$
According to $\lim _{u \to \infty} \frac{g(u)}{u} = \lambda$
and $\lim _{v \to \infty} \frac{h(v)}{v} = \mu$,  for any
$\varepsilon > 0$, there must be $R > 0$ such that
 $$
|g(u) - \lambda u| < \varepsilon |u|, \quad
|h(v) - \mu v| < \varepsilon |v|, \quad |u|, |v| > R.
$$
Set $R^* = \max \{\max_{|u| \leq R} |g(u)|, \max_{|v| \leq R}
|h(v)|\}$ and select $M > 0$ such that
$R^* + \max \{|\lambda|, |\mu|\} < \varepsilon M$.  Denote
\begin{gather*}
E_1 = \{t \in [0, 1] : |u(t)| \leq R, \ |v(t)| > R\},  \\
E_2 = \{t \in [0, 1] : |u(t)| > R, \ |v(t)| \leq R\}, \\
E_3 = \{t \in [0, 1] : \max \{|u(t)|, |v(t)|\} \leq R\}, \\
E_4 = \{t \in [0, 1] : \min \{|u(t)|, |v(t)|\} > R\}.
\end{gather*}
 Thus for any $u \in C^2[0, 1]$ with $\|u\|_0 > M$,  when
$t \in E_1$, we have
 $$
|g(u(t)) - \lambda u(t)| \leq |g((u(t))| + |\lambda|
|u(t)| \leq R^* + |\lambda| R < \varepsilon M < \varepsilon
\|u\|_0,
$$
and
$$
  |h(v(t)) - \mu v(t)| < \varepsilon |v(t)| \leq
\varepsilon \|v\|_0.
$$
  Similarly, we conclude that for any  $u \in C^2[0, 1]$ with
$\|u\|_0 > M$, when $t \in E_i$ ($i = 2, 3, 4$), we also have that
 $$
|g(u(t)) - \lambda u(t)| <  \varepsilon \|u\|_0,
\quad |h(v(t)) - \mu v(t)| <  \varepsilon \|v\|_0.
$$
  Hence, we get
\begin{equation}
\begin{aligned}
& |Fu(t) - Au(t)| \\
& = \big| \int_0^1 G_1(t, s) \int_{\xi_1}^s (\tau - s) (
 p(\tau)[g(u(\tau)) - \lambda u(\tau)] + q(\tau) [h(u''(\tau)) -
 \mu u''(\tau)]) d \tau ds
\\
 & \quad  + \frac{1}{\delta} \int_0^1
G_1(t, s)  \int_{\xi_1}^{\xi_2} (b - a(\xi_1 - s))(c(\xi_2 - \tau)
 + d)  \\
& \quad \times (p(\tau)[g(u(\tau)) - \lambda u(\tau)] +
 q(\tau)[h(u''(\tau)) - \mu u''(\tau)]) d \tau ds\big| \\
&  \leq \Big[\int_0^1 G_1(s, s)
\int_{\xi_1}^s (s - \tau) (|p(\tau)| + |q(\tau)|)d \tau ds
 \\
 & \quad + \frac{1}{\delta} \int_0^1
G_1(s, s) \int_{\xi_1}^{\xi_2} (b - a(\xi_1 - s))(c(\xi_2 - \tau)
+ d) (|p(\tau)| + |q(\tau)|) d \tau ds \Big] \varepsilon \|u\|_0
\\
 &  \leq \frac{1}{12}
\Big[\int_0^{\xi_1} \tau^3 (2-\tau) w(\tau) d\tau +
\int_{\xi_1}^1 (1-\tau)^3(1+\tau) w(\tau) d\tau  \\
 & \quad + \frac{2(b-a \xi_1) +
a}{\delta} \int_{\xi_1}^{\xi_2} (c(\xi_2 - \tau) + d) w(\tau)
d\tau \Big]\varepsilon \|u\|_0. \\
 & = \varepsilon L_1 \|u\|_0.
\end{aligned} \label{e2.12}
\end{equation}
 On the other hand, we have
 \begin{align*}
 & |(Fu - Au)''(t)| \\
 &  = \big|\int_{\xi_1}^t (s - t) (p(s)[g(u(s)) - \lambda u(s)] +
q(s)[h(u''(s)) - \mu u''(s)]) ds  \\
 & \quad   + \frac{1}{\delta}
\int_{\xi_1}^{\xi_2} (b - a(\xi_1 - t))(c(\xi_2 - s) + d) \\
& \quad   \times (p(s)[g(u(s)) - \lambda u(s)] +
q(s)[h(u''(s)) - \mu u''(s)]) ds \big| \\
 &  \leq \Big[\int_{\xi_1}^1 (1-s) w(s)
ds  + \frac{1}{\delta} \int_{\xi_1}^{\xi_2} (b +
a(1-\xi_1))(c(\xi_2 - s) + d) w(s) ds\Big] \varepsilon \|u\|_0
 \\
 & = \varepsilon L_2 \|u\|_0.
\end{align*}
  Combining the above inequality with \eqref{e2.12}, we have
 $$
\lim _{\|u\|_0 \to \infty} \frac{\|Fu - Au\|_0}{\|u\|_0} = 0.
$$
 Lemma \ref{lem2.5}  now guarantees
that BVP \eqref{e1.1} and \eqref{e1.2}  has a solution
$u^* \in C^2[0, 1]$. Obviously, $u^* \not= 0$ when
$p(t_0)g(0) + q(t_0)h(0) \not= 0$ for some $t_0 \in [0, 1]$.
In fact, if $u^* = 0$, then
$(0)^{(4)} = p(t_0)g(0) + q(t_0)h(0) \not= 0$ will lead to a
contradiction. This completes the proof.
\end{proof}

\begin{example} \label{ex2.7} \rm
 Consider the  fourth-order four-point boundary-value problem
\begin{equation}
 \begin{gathered}
 u^{(4)}(t) = \frac{t \sin 2\pi t}{t^2 + 1}u(t)
  - \frac{1}{2}t e^{\cos t} \cos u''(t),
  \quad 0 < t < 1,\\
u(0) = u(1) = 0, \\
 u''(1/3) - u'''(1/3) = 0,
\quad  u''(2/3) + u'''(2/3) = 0.
\end{gathered}
\label{e2.13}
\end{equation}
To show \eqref{e2.13} has at least one nontrivial solution we
apply Theorem \ref{thm2.6} with $p(t) = \frac{t \sin 2\pi t}{t^2 +
1}$, $q(t) = \frac{1}{2}t e^{\cos t}$, $g(u) = u$, $h(u) = \cos
u$,  $a = b = c = d = 1$, $\xi_1 = 1/3$ and $\xi_2 = 2/3$. Clearly
(H1) is satisfied. Obviously,
 $$ p(t_0) g(0) + q(t_0) h(0) =
\frac{1}{2} t_0 e^{\cos t_0} \not= 0, \quad   t_0 \in (0, 1].
$$
  Since $|p(s)| + |q(s)| \leq (\frac{e}{2} + 1) s :=
w(s)$ for each $s \in [0, 1]$, we have
\begin{align*}
L_1 & = \frac{\frac{e}{2}+1}{12} \Big[\int_0^{1/3} \tau^4
(2-\tau) d\tau + (e + 1) \int_{1/3}^1 (1-\tau)^3(1+\tau) \tau
d\tau  \\
 & \quad   +  \int_{1/3}^{2/3}
(\frac{5}{3} - \tau) \tau d\tau \Big],
\end{align*}
 $$
L_2 = \big(\frac{e}{2}+1\big)\Big[\int_{1/3}^1 \tau (1-\tau) d\tau +
\frac{5}{7} \int_{1/3}^{2/3}\big(\frac{5}{3} - \tau\big) \tau
d\tau \Big].
$$
By simple calculation we easily know that
 $$
L_1 < L_2 < \frac{1}{3}\big(\frac{e}{2}+1\big) < 1.
$$
 Notice
 $$  \lambda = \lim _{u \to \infty} \frac{g(u)}{u} = 1, \quad
   \mu = \lim _{u \to \infty}\frac{h(u)}{u} = 0,
 $$
we have
 $$  \max \{\lambda, \mu\} < 1 < \min
\{\frac{1}{L_1}, \frac{1}{L_2}\}.
$$
 So (H2) is satisfied. Thus, Theorem \ref{thm2.6} now
guarantees that  BVP \eqref{e2.13} has at least one nontrivial
solution $u \in C^2[0, 1]$.
\end{example}





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