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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 139, 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}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/139\hfil Solvability]
{Solvability of a three-point nonlinear boundary-value problem}

\author[A. Guezane-Lakoud, S. Kelaiaia \hfil EJDE-2010/139\hfilneg]
{Assia Guezane-Lakoud, Smail Kelaiaia}  % in alphabetical order

\address{Assia Guezane-Lakoud \newline
Department of Mathematics\\
Faculty of Sciences \\
University Badji Mokhtar\\
B.P. 12, 23000, Annaba, Algeria}
\email{a\_guezane@yahoo.fr}

\address{Smail Kelaiaia \newline
Department of Mathematics\\
Faculty of Sciences \\
University Badji Mokhtar\\
B.P. 12, 23000, Annaba, Algeria}
\email{kelaiaiasmail@yahoo.fr}

\thanks{Submitted March 19, 2010. Published September 27, 2010.}
\subjclass[2000]{34B10, 34B15}
\keywords{Fixed point theorem; three-point boundary-value problem;
\hfill\break\indent non trivial solution}

\begin{abstract}
 Using the Leray Schauder nonlinear alternative, we prove the
 existence of a nontrivial solution for the three-point
 boundary-value problem
 \begin{gather*}
 u''+f(t,u)= 0,\quad 0<t<1 \\
 u(0)= \alpha u'(0),\quad u(1)=\beta u'(\eta ),
 \end{gather*}
 where $\eta \in (0,1)$, $\alpha ,\beta \in \mathbb{R}$,
 $f\in C([0,1] \times\mathbb{R},\mathbb{R})$.
 Some examples are given to illustrate the results obtained.
\end{abstract}

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

\section{Introduction and preliminaries}

During the previous  years, many authors have studied three-point
boundary-value problems (BVP) for second order differential
equations. Such problems have potential applications in physics,
biology, chemistry, etc. For example, a second-order three-point
(BVP) is used as a model for the membrane response of a spherical
cap \cite{p1} in nonlinear diffusion generated by nonlinear sources
and in chemical reactor theory.

In this article, we investigate the existence of a nontrivial
solution for the  second-order three-point boundary-value problem
\begin{gather}
u''+f(t,u)=0,\quad 0<t<1,  \label{e1} \\
u(0)=\alpha u'(0),\quad
u(1)=\beta u'(\eta )\label{e2}
\end{gather}
where $\eta \in (0,1)$, $\alpha $, $\beta \in\mathbb{R}$,
$f\in C([0,1] \times\mathbb{R},\mathbb{R})$.
We do not assume any monotonicity condition on the nonlinearity
$f$, and the parameters $\alpha $ and $\beta $ belong to
$\mathbb{R}$, so our conditions are more general than the conditions
found in the literature. Such problem arises in the study of the
equilibrium states of a heated bar. In this situation two controllers
at $t=0$ and $t=\eta $ alter the heat according to the temperatures
detected by a sensor at $t=1$.

This study is motivated by Il'in and Moiseev's results
\cite{i1}, on
similar boundary value problems for certain linear ordinary
differential  equations. Many of the results involving nonlocal
boundary value problems are studied in
\cite{f1,g1,g2,i1,i2,k1,m1,m2,p1,s1,s2}.
 In \cite{s2} the author
used the Leray-Shauder nonlinear alternative to establish some
results on the existence of solutions for the equation \eqref{e1}
subject to the  conditions
\[
u(0)=0,\quad u(1)=\alpha u(\eta )
\]
Gupta \cite{g1} studied certain three-point boundary-value problem with
the above nonlocal conditions. Similar boundary value problem with
the conditions
\[
u(0)=\alpha u'(0),\quad u(1)=\beta u(\eta )
\]
is considered in \cite{s1}. Under some conditions on the nonlinearity
of $f$ and by using Leray Schauder nonlinear alternative, we
establish the existence of nontrivial solution of the BVP
\eqref{e1}-\eqref{e2}. Some examples  illustrate our results.

This article is organized as follows. First, we list some
preliminary material to be used later. Then in Section 2, we
present and prove our main results which consist in existence
theorems and corollaries. We end our work with some illustrating
examples.

Let $E=C[0,1] $, with the supremum norm
$\| y(t)\|=\sup_{t\in [0,1] }| y(t)| $, for all  $y\in E$. Now we
state two preliminary results.

\begin{lemma} \label{lem1}
Let $y\in E$.  If $\beta (\eta +\alpha )\neq
\alpha +1$, then the three-point BVP
\begin{gather*}
u''(t)+y(t)=0,\quad 0<t<1 \\
u(0)=\alpha u'(0),\quad u(1)=\beta u'(\eta )
\end{gather*}
has a unique solution
\[
u(t)=-\int_{0}^{t}(t-s)y(s)ds+\frac{t+\alpha }{1+\alpha -\beta }
\int_{0}^{1}(1-s)y(s)ds-\beta \frac{t+\alpha }{1+\alpha -\beta }
\int_{0}^{\eta }y(s)ds.
\]
\end{lemma}

\begin{proof}
 Rewriting the differential equation as $u''(t)
=-y(t)$,  and integrating  twice, we obtain
\begin{equation}
u(t)=-\int_{0}^{t}(t-s)y(s) ds+C_1t+C.  \label{e3}
\end{equation}
Then $u'(0) = C_1$,
\begin{gather*}
u'(\eta ) = -\int_{0}^{\eta }y(s)ds+C_1, \\
u(1) = -\int_{0}^{1}(1-s)y(s)ds+C_1+C_2, \\
u(0) = C_2=\alpha C_1.
\end{gather*}
Combining these with the second equality in condition \eqref{e2},
we obtain
\[
-\int_{0}^{1}(1-s)y(s)ds+C_1(1+\alpha )=-\beta
\int_{0}^{\eta }y(s)ds+C_1\beta ,
\]
which is equivalent to
\[
C_1(1+\alpha -\beta )=\int_{0}^{1}(1-s)
y(s)ds-\beta \int_{0}^{\eta }y(s)ds.
\]
Since $1+\alpha -\beta \neq 0$,
\begin{gather*}
C_1 = \frac{1}{1+\alpha -\beta }(\int_{0}^{1}(1-s)
y(s)ds-\beta \int_{0}^{\eta }y(s)ds). \\
C_2 = \alpha C_1.
\end{gather*}
Substituting $C_1$ and $C_2$ by their values in \eqref{e3}, we
obtain the solution in  the statement of the lemma.
This completes the proof.
\end{proof}

We define the integral operator $T:E\to E$, by
\begin{equation} \label{e4}
\begin{aligned}
Tu(t) &= -\int_{0}^{t}(t-s)f(s,u(s))ds
+\frac{t+\alpha }{1+\alpha -\beta }
\int_{0}^{1}(1-s)f(s,u(s))ds\\
&\quad -\beta \frac{t+\alpha }{1+\alpha -\beta }
\int_{0}^{\eta }f(s,u(s))ds
\end{aligned}
\end{equation}
 By Lemma \ref{lem1},  BVP \eqref{e1}-\eqref{e2} has a solution
if and only if the operator $T$ has a fixed point in $E$. By
Ascoli Arzela theorem we prove that $T$ is a completely continuous
operator. Now we cite the Leray Schauder nonlinear alternative.

\begin{lemma}[\cite{d1}] \label{lem2}
 Let $F$ be a Banach space and $\Omega $
 a bounded open subset of $F$,
$0\in \Omega $. $T:\overline{\Omega }\to F$  be a completely
continuous operator. Then,
either there exists $x\in \partial \Omega $, $\lambda >1$
such that $T(x)=\lambda x$, or there exists a fixed
point $x^{\ast }\in \overline{\Omega }$.
\end{lemma}

\section{Main Results}

In this section, we present and prove our main results.

\begin{theorem} \label{thm3}
We assume that $f(t,0)\neq 0$, $1+\alpha
\neq \beta $  and there exist nonnegative functions
$k,h\in L^{1} [0,1] $ such that
\begin{gather}
| f(t,x)| \leq k(t)| x| +h(t),\quad \text{a.e. }(t,x)\in [0,1
] \times\mathbb{R},  \label{e5}
\\
(1+\frac{| 1+\alpha | }{| 1+\alpha
-\beta | })\int_{0}^{1}(1-s)k(s)ds+| \beta
\frac{1+\alpha }{1+\alpha -\beta }| \int_{0}^{\eta
}k(s)ds<1 \label{e6}
\end{gather}
Then  BVP \eqref{e1}-\eqref{e2} has at least
one nontrivial solution  $u^{\ast }\in C[0,1] $.
\end{theorem}

\begin{proof}
Setting
\begin{gather*}
M=(1+\frac{| 1+\alpha | }{| 1+\alpha
-\beta | })\int_{0}^{1}(1-s)k(s)ds+|
\beta \frac{1+\alpha }{1+\alpha -\beta }| \int_{0}^{\eta }k(s)ds,
\\
N=(1+\frac{| 1+\alpha | }{| 1+\alpha
-\beta | })\int_{0}^{1}(1-s)h(s)
ds+| \beta \frac{1+\alpha }{1+\alpha -\beta }|
\int_{0}^{\eta }(\eta -s)h(s)ds.
\end{gather*}
By \eqref{e6}, we have $M<1$. Since $f(t,0)\neq 0$, then
there exists an interval $[\sigma ,\tau ] \subset [0,1] $
such that $\underset{\sigma \leq t\leq r}{\min }| f(t,0)| >0$
 and as $h(t)\geq |f(t,0)| $, for all $t\in [0,1] $ then
$N>0$. Let $m=\frac{N}{(1-M)}$, $\Omega =\{ u\in C[0,1] :\| u\| <m\}$.
Assume that $u\in \partial \Omega $, $\lambda >1$ such
$Tu=\lambda u$, then
\begin{align*}
\lambda m&=\lambda \|u\| =\| Tu\|
=\max_{0\leq t\leq 1}| (Tu)(t)|\\
&\leq \| u\| \Big[(1+\frac{| 1+\alpha| }{| 1+\alpha -\beta | })
\int_{0}^{1}(1-s)k(s)ds+| \beta \frac{1+\alpha }{
1+\alpha -\beta }| \int_{0}^{\eta }k(s)ds \\
&\quad +  (1+\frac{| 1+\alpha | }{|
1+\alpha -\beta | })\int_{0}^{1}(1-s)
h(s)ds+| \beta \frac{1+\alpha }{1+\alpha -\beta }|
\int_{0}^{\eta }h(s)ds\Big] \\
&= M\| u\| +N
\end{align*}
From this we obtain
\[
\lambda \leq M+\frac{N}{m}=M+\frac{N}{N(1-M)^{-1}}=M+(1-M)=1,
\]
this contradicts  $\lambda >1$. By Lemma \ref{lem2} we
conclude that operator $T$ has a fixed point
$u^{\ast }\in \overline{\Omega }$ and then
 BVP \eqref{e1}-\eqref{e2} has a
nontrivial solution $u^{\ast }\in C[0,1]$.
\end{proof}

\begin{theorem} \label{thm4}
Assume  \eqref{e5} and one of the following four
conditions:
\begin{itemize}
\item[(1)] There exists a constant $p>1$ such that
\begin{equation}
\int_{0}^{1}k^{p}(s)ds<\Big[\frac{(1+q)^{1/q}}{1+|
\frac{ 1+\alpha }{1+\alpha -\beta }| +(| \beta
\frac{ 1+\alpha }{1+\alpha -\beta }| )(\eta
(1+q))^{1/q}} \Big] ^{p}\quad
(\frac{1}{p}+\frac{1}{q}=1); \label{e7}
\end{equation}

\item[(2)] There exists constant $\mu >-1$ such that
\begin{equation}
k(s)<\frac{(\mu +1)(\mu +2)}{1+| \frac{ 1+\alpha
}{1+\alpha -\beta }| +(| \beta \frac{ 1+\alpha
}{1+\alpha -\beta }| )(\mu +2)\eta ^{\mu +1}}s^{\mu }
\label{e8}
\end{equation}
and
\[
\operatorname{meas}\big\{ s\in [0,1] :
k(s)<\frac{(\mu +1)(
\mu +2)}{1+| \frac{1+\alpha }{1+\alpha -\beta }|
+| \beta \frac{1+\alpha }{1+\alpha -\beta }| (\mu
+2)\eta ^{\mu +1}}s^{\mu }\big\} >0;
\]

\item[(3)] The function $k(s)$ satisfies
\begin{equation}
k(s)<\frac{| 1+\alpha -\beta | }{|
1+\alpha | +| 1+\alpha -\beta |
+| \beta ( 1+\alpha )| \eta }  \label{e9}
\end{equation}
and
\[
\operatorname{meas}\big\{ s\in [0,1]:k(s)<\frac{| 1+\alpha -\beta
| }{| 1+\alpha | +| 1+\alpha -\beta
| +| \beta (1+\alpha )| \eta }
\big\} >0;
\]

\item[(4)] The function $f(t,x)$ satisfies
\begin{equation}
\omega =\limsup_{| x| \to \infty} \max_{t\in [0,1] }
| \frac{f(t,x)}{x}| <\frac{1}{2}(\frac{| 1+\alpha -\beta
| }{| 1+\alpha | +| 1+\alpha
-\beta | +| \beta (1+\alpha )| \eta }). \label{e10}
\end{equation}

\end{itemize}
Then BVP \eqref{e1}-\eqref{e2} has at least
one nontrivial solution $u^{\ast }\in C[0,1]$.
\end{theorem}

\begin{proof}
Let $M$ be defined as in the proof of Theorem \ref{thm3}.
To prove Theorem \ref{thm4}, we only
need to prove that $M<1$.

(1) By using H\"{o}lder inequality, we obtain
\begin{align*}
M &\leq \big(1+| \frac{1+\alpha }{1+\alpha -\beta }|\big)
\Big(\int_{0}^{1}k^{p}(s)ds\Big)^{1/p}
\Big(\int_{0}^{1}(1-s)^{q}ds\Big)^{1/q}\\
&\quad + | \beta \frac{1+\alpha }{1+\alpha -\beta }|
\Big(\int_{0}^{\eta }k^{p}(s)ds\Big)^{1/p}
\Big(\int_{0}^{\eta}ds\Big)^{1/q}\,.
\end{align*}
Then
\begin{align*}
M &\leq \Big(\int_{0}^{1}k^{p}(s)ds\Big)^{1/p}
\Big[\big(1+| \frac{1+\alpha }{1+\alpha -\beta }| \big)
\Big(\int_{0}^{1}(1-s)^{q}ds\Big)^{1/q}\\
&\quad + |\beta \frac{1+\alpha }{1+\alpha -\beta }|
\Big(\int_{0}^{\eta }ds\Big)^{1/q}\Big].
\end{align*}
Integrating, it yields
\begin{align*}
M&<\frac{(1+q)^{1/q}}{(1+| \frac{1+\alpha }{1+\alpha
-\beta }| )+|\beta \frac{1+\alpha }{1+\alpha -\beta }|(\eta
(1+q))^{1/q}}\\
&\quad \times
\Big[(1+\frac{|1+\alpha |}{|1+\alpha -\beta |})
(\frac{1}{1+q})^{1/q}+|\beta \frac{1+\alpha }{1+\alpha -\beta }|
\eta ^{1/q}\Big] =1.
\end{align*}

(2) Using the same reasoning as in the proof of the first
statement we obtain
\begin{align*}
M &<\frac{(\mu +1)(\mu +2)}{1+| \frac{
1+\alpha }{1+\alpha -\beta }| +(| \beta \frac{
1+\alpha }{1+\alpha -\beta }| )(\mu +2)\eta
^{\mu +1}}\\
&\quad \times
\Big[\big(1+| \frac{1+\alpha }{1+\alpha -\beta }|
\big)\int_{0}^{1}(1-s)s^{\mu }ds+| \beta \frac{1+\alpha }{
1+\alpha -\beta }| \int_{0}^{\eta }s^{\mu }ds\Big] \\
&=\frac{(\mu +1)(\mu +2)}{1+| \frac{
1+\alpha }{1+\alpha -\beta }| +| \beta \frac{1+\alpha }{
1+\alpha -\beta }| (\mu +2)\eta ^{\mu +1}}
\big(1+| \frac{1+\alpha }{1+\alpha -\beta }| \big)
\big(\frac{1}{(\mu +1)(\mu +2)}\big)\\
&\quad + \big| \beta \frac{1+\alpha }{1+\alpha -\beta }\big|
\big(\frac{\eta ^{\mu +1}}{\mu +1}\big)= 1
\end{align*}

(3) we have
\[
M = (1+\frac{| 1+\alpha | }{| 1+\alpha
-\beta | })\int_{0}^{1}(1-s)k(s)ds
+| \beta \frac{1+\alpha }{1+\alpha -\beta }|
\int_{0}^{\eta }k(s)ds.
\]
Then
\begin{align*}
M &<\big(\frac{| 1+\alpha -\beta | }{| (
1+\alpha )| +| 1+\alpha -\beta |
+| \beta (1+\alpha )| \eta }\big)\\
&\quad \times
\big(1+| \frac{1+\alpha }{1+\alpha -\beta }| \big)
\int_{0}^{1}(1-s)ds+| \beta \frac{1+\alpha }{1+\alpha -\beta }
| \int_{0}^{\eta }ds\\
&=\frac{1}{2}(\tfrac{| 1+\alpha -\beta | }{
| (1+\alpha )| +| 1+\alpha -\beta
| +| \beta (1+\alpha )| \eta }
)(1+| \tfrac{1+\alpha }{1+\alpha -\beta }|
)+| \beta \tfrac{1+\alpha }{1+\alpha -\beta }|
\eta =1
\end{align*}

(4) From $\omega =\limsup_{| x| \to \infty } \max_{t\in [0,1] }
| \frac{ f(t,x)}{x}| $ we deduce that there exists $c>0$ such
that for $| x| >c$ we have
\[
| f(t,x)| \leq (\omega +\varepsilon )| x| \quad \forall \varepsilon >0.
\]
Set
\[
h(t)=\max \{ | f(t,x)| :(
t,x)\in [0,1] \times (-c,c)\}.
\]
Then for $(t,x)\in [0,1] \times\mathbb{R}$,
with $\varepsilon =\omega$, we obtain
\begin{align*}
| f(t,x)|
&\leq 2\omega |x| +h(t)\\
&\leq \frac{| 1+\alpha -\beta | }{| (1+\alpha
)| +| 1+\alpha -\beta | +|
\beta (1+\alpha )| \eta }| x| +h(t).
\end{align*}
Setting
\[
k(t)<\frac{| 1+\alpha -\beta | }{|
(1+\alpha )| +| 1+\alpha -\beta |
+| \beta (1+\alpha )| \eta },
\]
by applying the above statement we complete the proof.
\end{proof}

\begin{corollary} \label{coro5}
Assume the conditions of Theorem \ref{thm4}, and  one of the following
four conditions:
\begin{itemize}
\item[(1)]  There exists a constant $p>1$ such that
\[
\int_{0}^{1}k^{p}(s)ds<\Big[\frac{(1+q)^{1/q}}{1+| \frac{
1+\alpha }{1+\alpha -\beta }| +| \beta \frac{1+\alpha }{
1+\alpha -\beta }| (1+q)^{1/q}}\Big] ^{p},\quad
(\frac{1}{p}+\frac{1}{q}=1).
\]

\item[(2)] There exists a constant $\mu >-1$ such that
\begin{gather*}
k(s)<\frac{(\mu +1)(\mu +2)}{1+| \frac{
1+\alpha }{1+\alpha -\beta }| +(| \beta \frac{
1+\alpha }{1+\alpha -\beta }| )(\mu +2)}s^{\mu},
\\
\operatorname{meas}\big\{ s\in [0,1] : k(s)<\frac{(\mu +1)(
\mu +2)}{1+| \frac{1+\alpha }{1+\alpha -\beta }|
+| \beta \frac{1+\alpha }{1+\alpha -\beta }| (\mu
+2)}s^{\mu }\big\} >0
\end{gather*}

\item[(3)] The function $k(s)$ satisfies
\begin{gather*}
k(s)<\frac{| 1+\alpha -\beta | }{| (
1+\alpha )| +| 1+\alpha -\beta |
+| \beta (1+\alpha )| \eta },
\\
\operatorname{meas}\left\{ s\in [0,1] ,k(s)<\frac{| 1+\alpha -\beta
| }{| (1+\alpha )| +|
1+\alpha -\beta | +| \beta (1+\alpha )
| \eta }\right\} >0
\end{gather*}

\item[(4)] The function $f(t,x)$ satisfies
\begin{align*}
\omega &= \limsup_{| x| \to \infty }
\max_{t\in [0,1]} | \frac{f(t,x)}{x}| \\
&< 1/2(\frac{| 1+\alpha -\beta | }{| (
1+\alpha )| +| 1+\alpha -\beta |
+| \beta (1+\alpha )| })
\end{align*}
\end{itemize}
Then the BVP \eqref{e1}-\eqref{e2} has at least
one nontrivial solution $u^{\ast }\in C[0,1]$.
\end{corollary}

\begin{proof}
Taking into account
\begin{align*}
M &= \big(1+\frac{| 1+\alpha | }{| 1+\alpha
-\beta | }\big)\int_{0}^{1}(1-s)k(s)ds+|
\beta \frac{1+\alpha }{1+\alpha -\beta }| \int_{0}^{\eta }k(s)ds \\
&\leq \big(1+\frac{| 1+\alpha | }{| 1+\alpha
-\beta | }\big)\int_{0}^{1}(1-s)k(s)ds+|
\beta \frac{1+\alpha }{1+\alpha -\beta }| \int_{0}^{1}k(s)ds.
\end{align*}
 the proof follows.
\end{proof}

\section{Examples}

To illustrate our results, we give the following examples.

\begin{example} \label{exa6} \rm
Consider the three-point BVP
\begin{equation}
\begin{gathered}
u''+\frac{u}{2}(t\sin \sqrt{u}-e^{-t}\cos u)
+te^{t}=0,\quad 0<t<1 \\
u(0)=\frac{1}{2}u'(0),\quad
u(1)=\frac{1}{2}u'(\tfrac{1}{2}).
\end{gathered} \label{e11}
\end{equation}
Where $\alpha =\beta =\eta =1/2$, $+\alpha -\beta =1/2\neq 0$
and
\[
f(t,x)=\frac{x}{2}(t\sin \sqrt{x}-e^{-t}\cos x)+te^{t},
\]
It is easy to see that
\[
| f(t,x)| \leq \frac{1}{2}(
t+e^{-t})| x| +te^{t},\quad (t,x)\in [0,1] \times\mathbb{R},
\]
set
\[
k(t)=\frac{1}{2}(t+e^{-t})\geq 0,\quad h(t)=te^{t}\geq 0,
\]
we have $k$, $h\in L^{1}[0,1] $, $f(t,0)=te^{t}\neq 0$ and
\begin{align*}
M&=\frac{5}{2}\int_{0}^{1}(1-s)k(s)ds+\frac{3}{4}\int_{0}^{\frac{
1}{2}}k(s)ds\\
&=\frac{5}{4}\int_{0}^{1}(1-s)(s+e^{-s})ds+\frac{3}{8
}\int_{0}^{1/2}(s+e^{-s})ds=0.86261.
\end{align*}
Then, by Theorem \ref{thm3}, the BVP \eqref{e11} has at least
one nontrivial solution $u^{\ast }$ in $C[0,1]$.
\end{example}

\begin{example} \label{exa7} \rm
Consider the three-point BVP
\begin{equation}
\begin{gathered}
u''+\frac{u^{4}}{2\sqrt {t^{3}+1}(1+u^{3})}
+\cos e^{t}(1-\sin t)=0,\quad 0<t<1, \\
u(0)=-2u'(0),\quad u(1)
=2u'(1/3)
\end{gathered} \label{e12}
\end{equation}
then
\begin{gather*}
f(t,x)= \frac{x^{4}}{2\sqrt{t^{3}+1}(1+x^{3})}
+\cos e^{t}(1-\sin t),\\
\begin{aligned}
| f(t,x)| &\leq \frac{1}{4\sqrt {t^{3}+1}
}| x| +\cos e^{t}(1-\sin t)\\
&= k(t)| x| +h(t),
\end{aligned}
\end{gather*}
where
\[
k(t)=\frac{1}{4\sqrt {t^{3}+1}}\geq 0,\quad
h(t)=\cos e^{t}(1-\sin t)\geq 0
\]
\begin{align*}
M &= \frac{4}{3}\int_{0}^{1}(1-s)k(s)ds+\frac{2}{3}
\int_{0}^{\eta }k(s)ds \\
&= \frac13 \int_{0}^{1}(1-s)\frac{1}{\sqrt {s^{3}+1}}
ds+\frac{2}{16} \int_{0}^{1/3}\frac{1}{\sqrt {s^{3}+1}}ds=0.201\,41<1
\end{align*}
Applying the third statement of Theorem \ref{thm4} we obtain
\[
\max_{s}k(s)=\frac{1}{4}<\frac{| 1+\alpha -\beta | }{
| (1+\alpha )| +| 1+\alpha -\beta
| +| \beta (1+\alpha )| \eta }
\frac{9}{14}=0.642\,86\,,
\]
\begin{align*}
\operatorname{meas}\{ s\in [0,1]:k(s)
&<\frac{| 1+\alpha -\beta
| }{| (1+\alpha )| +|1+\alpha -\beta | +| \beta (1+\alpha )
| \eta }\} \\
&= \operatorname{meas}\{ s\in [0,1]: k(s)<\frac{9}{14}=0.642\,86\}\\
&=\operatorname{meas}[0,1] >0\,.
\end{align*}
Hence, by Theorem \ref{thm4},  BVP \eqref{e12} has at least one nontrivial
solution $u^{\ast }$ in $C[0,1]$.
\end{example}

\begin{example} \label{exa8} \rm
Consider the three-point BVP
\begin{equation}
\begin{gathered}
u''+\frac{u^{3}}{2(1+e^{-t^{2}})^{4}(1+u^{2})}-\sqrt{t}\sin t=0 \\
u(0)=\frac{1}{3}u'(0),\quad
u(1)=\frac{-1}{6}u'(\frac{1}{4}).
\end{gathered}  \label{e13}
\end{equation}
We have
\begin{gather*}
f(t,x)=\frac{x^{3}}{2(1+e^{-t^{2}})^{4}(
1+x^{2})}-\sqrt{t}\sin t,
\\
| f(t,x)| \leq | x| \frac{(1+e^{-t^{2}})^{-4}}{2}+\sqrt{t}\sin t
= k(t)| x| +h(t)
\\
| \frac{f(t,x)}{x}|
\leq \frac{(1+e^{-t^{2}})^{-4}}{2}+\frac{\sqrt{t}\sin t}{| x| }
\leq 0.142+\frac{1}{| x|}.
\end{gather*}
Applying the fourth statement in Theorem \ref{thm4} we obtain
\begin{align*}
\omega &= \limsup_{| x| \to \infty }
\max_{t\in [0,1] } | \frac{f(t,x)}{x}|
=\limsup_{| x| \to\infty } (0.142+\frac{1}{| x| })\\
&=0.142< \frac{1}{2}(\frac{\frac{13}{9}}{\frac{4}{3}
+\frac{13}{9}+\frac{1}{27}})=0.25658
\end{align*}
Hence,  BVP \eqref{e13} has at least one nontrivial solution
$u^{\ast }$ in $C[0,1]$.
\end{example}

\begin{example} \label{exa9} \rm
Consider the three-point BVP
\begin{equation}
\begin{gathered}
u''+\frac{\sin u}{2\sqrt{4+t}}-t^{2}\cos t+t\cos (t^2)=0,
\quad 0<t<1 \\
u(0)=\frac{-1}{4}u^{'}(0),\quad
u(1)=\frac{-1}{6}u^{'}(\frac{1}{5}),
\end{gathered}  \label{e14}
\end{equation}
where
\begin{gather*}
f(t,x)= \frac{\sin x}{2\sqrt{4+t}}-t^{2}\cos t+t\cos (t^2), \\
| f(t,x)| \leq \frac{|x| }{2\sqrt{4+t}}-t^{2}\cos t+t\cos (t^2)
 = k(t)| x| +h(t),\\
k(t) = \frac{1}{2\sqrt{4+t}}\geq 0,\quad
h(t)=-t^{2}\cos t+t\cos (t^2)\geq 0
\end{gather*}
We see that $k$, $h\in L^{1}[0,1] $,
$f(t,0)=-t^{2}\cos t+t\cos (t^2)\neq 0$.
The first statement of Corollary \ref{coro5} for $p=2$  yields
\begin{align*}
\int_{0}^{1}k^{2}(s)ds
&= \int_{0}^{1}\big(\frac{1}{2(\sqrt{4+t})}\big)^{2}dt
=\frac{1}{4}\int_{0}^{1}\frac{1}{4+t}dt=0.055786 \\
&<\Big[\frac{(1+q)^{1/q}}{1+| \frac{1+\alpha }{1+\alpha
-\beta }| +| \beta \frac{1+\alpha }{1+\alpha -\beta }
| (1+q)^{1/q}}\Big] ^{p}=0.71083.
\end{align*}
Hence,  BVP \eqref{e14} has at least one nontrivial solution
$u^{\ast }$ in $C[0,1]$.
\end{example}

\subsection*{Acknowledgements}
The authors are grateful to the anonymous referees
for their helpful comments and fruitful remarks.
The second author would like to express his thanks to the
Laboratory of advanced material at Annaba university for its
support and hospitality.

\begin{thebibliography}{00}

\bibitem{d1}  K. Deimling; \emph{Non-linear Functional
Analysis}, Springer, Berlin, 1985.

\bibitem{f1} W. Feng, J. R. L. Webb;
 \emph{Solvability of three-point
nonlinear bound-ary value problems at resonance}, Nonlinear Analysis TMA, 30
(6), (1997), 3227-3238.

\bibitem{g1} C. P. Gupta; \emph{Solvability of a three-point
nonlinear boundary value problem for a second order ordinary differential
equations}, J. Math. Anal. Appl. 168 (1992), 540-551.

\bibitem{g2} C. P. Gupta;
\emph{Solvability of a three-point nonnlinear
boundary value problem for a second order differential equation}, J. Math.
Anal. Appl. 168, (1992), 540-551.

\bibitem{i1} V. A. Il'in and E. I. Moiseev;
 \emph{Nonlocal boundary value
problem of the first kind for a Sturm-Liouville operator in its differential
and finite difference aspects}, Differential Equations, 23 (7) (1987),
803--810.

\bibitem{i2} G. Infante and J. R. L. Webb;
 \emph{Three point boundary value
problems with solutions that change sign}, J. Integ. Eqns Appl, 15 (2003),
37-57.

\bibitem{k1} S. A. Krasnoselskii;
\emph{A remark on a second order
three-point boundary value problem}, J. Math. Anal. Appl. 183, (1994),
581-592.

\bibitem{m1} R. Ma;
\emph{Existence theorems for second order
three-point boundary value problems}, J. Math. Anal. Appl. 212 (1997),
545-555.

\bibitem{m2} R. Ma;
 \emph{A Survey On Nonlocal Boundary Value Problems}.
Applied Mathematics E-Notes, 7 (2007), 257-279

\bibitem{p1} P. K. Palamides, G. Infante, P. Pietramala;
 \emph{Nontrivial
solutions of a nonlinear heat flow problem via Sperner's Lemma.}
Applied Mathematics Letters 22 (2009) 1444-1450.

\bibitem{s1} L. Shuhong, Y-P. Sun;
 \emph{Nontrivial solution of a nonlinear
second order three-point boundary-value problem}, Appl. math. J
(2007), 22(1), 37-47.

\bibitem{s2} Y-P. Sun;
 \emph{Nontrivial solution for a three-point
boundary-value problem}, Electron. J. of Differential Equations,
Vol. 2004 (2004), No. 111, 1-10.

\end{thebibliography}

\end{document}