\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 111, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}


\begin{document}

\title[\hfilneg EJDE-2004/111\hfil Nontrivial solution for a three-point BVP]
{Nontrivial solution for a three-point boundary-value problem}

\author[Y.-P. Sun \hfil EJDE-2004/111\hfilneg]
{Yong-Ping Sun}

\address{Yong-Ping Sun \hfill\break
Department of Fundamental Courses \\
Hangzhou Radio and TV University \\
Hangzhou, Zhejiang 310012, China}
\email{syp@mail.hzrtvu.edu.cn}

\date{}
\thanks{Submitted June 15, 2004. Published September 22, 2004.}
\thanks{Supported by the Education Department of Zhejiang
Province of China(20040495)}
\subjclass[2000]{34B10, 34B15}
\keywords{Three-point boundary value problem; Nontrivial solution;
\hfill\break\indent
Leray-Schauder nonlinear alternative}

\begin{abstract}
 In this paper, we study the existence of nontrivial
 solutions for the second-order three-point boundary-value problem
 \begin{gather*}
 u''+f(t,u)=0,\quad  0<t<1, \\
 u'(0)=0,\quad u(1)=\alpha u'(\eta).
 \end{gather*}
 where $\eta \in (0,1)$, $\alpha \in \mathbb{R}$,
 $f\in C([0,1]\times \mathbb{R},\mathbb{R})$.
 Under certain growth conditions on the nonlinearity  $f$
 and by using  Leray-Schauder nonlinear alternative,
 sufficient conditions for the existence of nontrivial solution
 are obtained. We illustrate the results obtained with some examples.
\end{abstract}

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

\section{Introduction}

This paper, we  prove existence results for the following
second-order three-point boundary value problem (BVP):
\begin{equation}\begin{gathered}
u''+f(t,u)=0,\quad  0<t<1, \\
u'(0)=0,\quad u(1)=\alpha u'(\eta).
\end{gathered} \label{e1.1}
\end{equation}
where $\eta \in (0,1)$, $\alpha \in \mathbb{R}$,
 $f\in C([0,1]\times \mathbb{R}, \mathbb{R})$.

The study of three-point BVP for certain nonlinear ordinary
differential equations was initiated by Gupta \cite{g1}. Over the ten
yeas, three-point boundary value problems have been extensively
studied by many authors, for example, see
\cite{f1,f2,g2,g3,h1,i1,i2,l1,l2,m1,m2,m3,m4,w1},
 and references therein. But in the existing literature on the
BVP \eqref{e1.1} is few.
Most of them studied the following three-point BVP
\begin{gather*}
u''+f(t,u)=0,\quad  0<t<1, \\
 u(0)=0,\quad u(1)=\alpha u(\eta).
\end{gather*}
(for example, Gupta \cite{g2}, Ma \cite{m2}, Liu \cite{l1}, Webb \cite{w1},
He and Ge \cite{h1}) or
\begin{gather*}
 u''+f(t,u)=0,\quad  0<t<1, \\
 u'(0)=0,\quad u(1)=\alpha u(\eta).
\end{gather*}
(for example, Liu \cite{l2} and Webb \cite{w1}).
 In a recent paper,  Infante \cite{i2}
 investigated the BVP \eqref{e1.1} for the first time.
 The aim of the present paper is to establish some simple criteria of
 the existence of nontrivial solution for the BVP \eqref{e1.1}.
 Note that we do not require any monotonicity and nonnegativity on
$f$. The results we obtained are new.

The paper is organized as follows. In Section 2, we present two
lemmas that will be used to prove the main results. In Section 3,
we obtain some existence results for nontrivial solution of the
BVP \eqref{e1.1}. Finally, in Section 4, as an application, we give some
examples to illustrate the results we obtained.

\section{Preliminaries}

 Let $E=C[0,1]$, with supremum norm
 $\|y\|=\sup_{t\in [0,1]}|y(t)|$ for any $y\in E$. A solution $u(t)$ of
the BVP \eqref{e1.1} is called nontrivial solution if $u(t)\not\equiv 0$.
In arriving at our results, we need to state two preliminary
results.

\begin{lemma}\label{lem1}
Let $y\in C[0,1]$, then the three-point BVP
\begin{gather*}
u''+y(t)=0,\quad 0<t<1,\\
 u'(0)=0,\quad u(1)=\alpha u'(\eta).
 \end{gather*}
has a unique solution
$$
u(t)=\int_0^1(1-s)y(s)ds-\int_0^t(t-s)y(s)ds
-\alpha \int_0^\eta y(s)ds.
$$
\end{lemma}

The proof of this lemma is easy, and we omit it.

Define the integral operator $T: E\to E$ by
\begin{equation} \label{e2.1}
\begin{aligned}
Tu(t)&=\int_0^1(1-s)f(s,u(s))ds-\int_0^t(t-s)f(s,u(s))ds\\
&\quad-\alpha\int_0^\eta f(s,u(s))ds,\quad t\in [0,1].
\end{aligned}
\end{equation}
By Lemma \ref{lem1}, the BVP \eqref{e1.1} has a solution if and only if
 the operator $T$ has a fixed point in $E$. So we only need to seek a
fixed point of $T$ in $E$. By Ascoli-Arzela Theorem, we can prove
that $T$ is a completely continuous operator. The key tool in our
approach is the following Leray-Schauder nonlinear alternative
(See \cite{d1}).

\begin{lemma} \label{lem2}
Let $E$ be Banach space and $\Omega$
 be a bounded open subset of $E$, $0\in \Omega$
 $T:\overline{\Omega}\rightarrow E$ 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{thm1}
Suppose $f(t,0)\not\equiv 0$, 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),\ a.e.\ (t,x)\in [0,1]\times \mathbb{R},\\
2\int_0^1(1-s)k(s)ds+ |\alpha|\int_0^\eta k(s)ds <1.
\end{gather*}
Then the BVP \eqref{e1.1} has at least one nontrivial solution
$u^\ast \in C[0,1]$.
\end{theorem}

\begin{proof}  Let
\begin{gather*}
A=2\int_0^1(1-s)k(s)ds+ |\alpha |\int_0^\eta k(s)ds,\\
B=2\int_0^1(1-s)h(s)ds+
 |\alpha |\int_0^\eta h(s)ds.
\end{gather*}
Then $A<1$. Since $f(t,0)\not\equiv 0$, there exists an interval
$[\sigma,\tau]\subset [0,1]$ such that $\min_{\sigma \leq
t\leq\tau}|f(t,0)|>0$. On the other hand, from $h(t)\geq
|f(t,0)|$, a.e. $t\in [0,1]$, we know that $B>0$. Let
$m=B(1-A)^{-1}$, $\Omega=\{u\in C[0,1]:\|u\|< m\}$. Suppose $u\in
\partial \Omega$, $\lambda >1$ such that $Tu=\lambda u$, then
\begin{align*}
\lambda m&=\lambda \|u\|=\|Tu\|=\max_{0\leq t\leq 1}|(Tu)(t)|\\
&\leq\int_0^1(1-s)|f(s,u(s))|ds+
\max_{0\leq t\leq 1}\int_0^t(t-s)|f(s,u(s))|ds\\
&\quad +|\alpha |\int_0^\eta |f(s,u(s))|ds\\
&\leq 2\int_0^1(1-s)|f(s,u(s))|ds
+|\alpha |\int_0^\eta |f(s,u(s))|ds\\
&\leq \Big[2\int_0^1(1-s)k(s)|u(s)|ds
+|\alpha |\int_0^\eta k(s)|u(s)|ds\Big]\\
&\quad+\Big[2\int_0^1(1-s)h(s)ds
+|\alpha |\int_0^\eta h(s)ds\Big]\\
&\leq A\|u\|+B=Am+B.
\end{align*}
Therefore,
$$
\lambda\leq
A+\frac{B}{m}=A+\frac{B}{B(1-A)^{-1}}=A+(1-A)=1,
$$
this contradicts $\lambda>1$. By Lemma \ref{lem2}, $T$ has a fixed point
$u^\ast\in \overline{\Omega}$. In view of  $f(t,0)\not\equiv 0$,
the BVP \eqref{e1.1} has a nontrivial solution $u^\ast\in C[0,1]$.
This completes the proof.
\end{proof}

\begin{theorem}\label{thm2}
Suppose $f(t,0)\not\equiv 0$, and there exist
nonnegative functions $k,h\in L^1[0,1]$ such that
$$
|f(t,x)|\leq k(t)|x|+h(t),\ a.e.\ (t,x)\in [0,1]\times \mathbb{R}.
$$
If one of the following conditions is fulfilled:
\begin{itemize}
\item[(1)] There exists constant $p>1$ such that
$$\int_0^1k^p(s)ds<\big[\frac{(1+q)^{1/q}}{2 +|\alpha|\ [\eta(1+q)]^{1/q} } \big]^p,
\quad  \big( \frac{1}{p} +\frac{1}{q}=1 \big).
$$

\item[(2)] There exists a constant $\mu >-1$ such that
\begin{gather*}
k(s)\leq \frac{(1+\mu)(2+\mu)} {2+|\alpha
|(2+\mu)\eta^{1+\mu}}s^\mu ,\quad\mbox{a.e. }s\in [0,1],\\
\mathop{\rm meas}
\big\{s\in [0,1]: k(s)< \frac{(1+\mu)(2+\mu)}{2
+|\alpha|(2+\mu)\eta^{1+\mu}}s^\mu \big\}>0.
\end{gather*}

\item[(3)] There exists a constant $\mu >-1$ such that
\begin{gather*}
k(s)\leq \frac{(1+\mu)(2+\mu)}{2(1+\mu) +|\alpha
|(2+\mu)}(1-s)^\mu ,\mbox{a.e. } s\in [0,1],\\
\mathop{\rm meas}
\big\{s\in [0,1]: k(s)< \frac{(1+\mu)(2+\mu)}{2(1+\mu) +|\alpha
|(2+\mu)}(1-s)^\mu  \big\}>0.
\end{gather*}

\item[(4)]  $k$ satisfies
\begin{gather*}
k(s)\leq \frac{1}{1+|\alpha |\eta} ,\quad\mbox{a.e. } s\in [0,1],\\
\mathop{\rm meas} \big\{s\in [0,1]: k(s)<
\frac{1}{1+|\alpha |\eta} \big\}>0.
\end{gather*}

\item[(5)] $f$ satisfies
$$
\Lambda:=\limsup_{|x|\to\infty}\max_{t\in
[0,1]}\big|\frac{f(t,x)}{x}\big|<\frac{1}{1+|\alpha |\eta}.
$$
\end{itemize}
 Then the BVP \eqref{e1.1} has at least one nontrivial solution
$u^\ast \in C[0,1]$.
\end{theorem}

\begin{proof}
Let $A$ be given in Theorem \ref{thm1}. In view
of Theorem \ref{thm1}, we only need to prove $A<1$.

\noindent (1)\ By using the H\"{o}lder inequality, we have
\begin{align*}
A&\leq\Big[\int_0^1k^p(s)ds\Big]^{1/p}
\Big\{2\Big[\int_0^1(1-s)^qds\Big]^{1/q}+|\alpha|
\Big[\int_0^\eta 1^qds\Big]^{1/q}\Big\}\\
&\leq\Big[\int_0^1k^p(s)ds\Big]^{1/p}
\Big[2\big(\frac{1}{1+q}\big)^{1/q}+|\alpha|\ \eta^{1/q}\Big]\\
&<\frac{(1+q)^{1/q}}{2 +|\alpha|\, [\eta(1+q)]^{1/q} } \cdot
\frac{2 +|\alpha|\, [\eta(1+q)]^{1/q} }{(1+q)^{1/q}} =1.
\end{align*}
(2)\ In this case, we have
\begin{align*}
A&<\frac{(1+\mu)(2+\mu)} {2+|\alpha |(2+\mu)\eta^{1+\mu}}
\Big[2\int_0^1(1-s)s^\mu ds+|\alpha| \int_0^\eta s^\mu ds\Big]\\
&\leq\frac{(1+\mu)(2+\mu)} {2+|\alpha |(2+\mu)\eta^{1+\mu}}
\Big[\frac{2}{(1+\mu)(2+\mu)}
+|\alpha|\cdot\frac{\eta^{1+\mu}}{1+\mu}\Big]\\
&=\frac{(1+\mu)(2+\mu)} {2+|\alpha |(2+\mu)\eta^{1+\mu}}\cdot
\frac {2+|\alpha |(2+\mu)\eta^{1+\mu}}{(1+\mu)(2+\mu)}=1.
\end{align*}
(3)\ In this case, we have
\begin{align*}
A&<\frac{(1+\mu)(2+\mu)}{2(1+\mu) +|\alpha |(2+\mu)}
\Big[2 \int_0^1(1-s)^{1+\mu} ds+|\alpha|
\int_0^\eta(1-s)^\mu ds\Big]\\
&=\frac{(1+\mu)(2+\mu)}{2(1+\mu) +|\alpha |(2+\mu)}
\Big[\frac{2}{2+\mu}+|\alpha|
\cdot\frac{1-(1-\eta)^{1+\mu}}{1+\mu}\Big]\\
&\leq\frac{(1+\mu)(2+\mu)}{2(1+\mu) +|\alpha |(2+\mu)}
\Big[\frac{2}{2+\mu}+ |\alpha|\cdot\frac{1}{1+\mu}\Big]\\
&=\frac{(1+\mu)(2+\mu)}{2(1+\mu) +|\alpha |(2+\mu)}\cdot
\frac{2(1+\mu) +|\alpha |(2+\mu)}{(1+\mu)(2+\mu)}=1.
\end{align*}
(4)\ In this case, we have
\begin{align*}
A&<\frac{1}{1+|\alpha |\eta}
\Big[2\int_0^1(1-s)ds+|\alpha|\int_0^\eta ds\Big]\\
&=\frac{1}{1+|\alpha |\eta}\left(1 +|\alpha|\eta\right)=1.
\end{align*}
(5)\ Let $\varepsilon=\frac{1}{2}(\frac{1}{1+|\alpha
|\eta}-\Lambda)$, then there exists $c>0$ such that
$$
|f(t,x)|\leq\big(\frac{1}{1+|\alpha |\eta}-\varepsilon\big)|x|,
\quad (t,x)\in [0,1]\times \mathbb{R}\setminus(-c,c).
 $$
Set $M=\max\{|f(t,x)|: (t,x)\in [0,1]\times [-c,c]\}$, then
$$
|f(t,x)|\leq\big(\frac{1}{1+|\alpha |\eta}-\varepsilon\big)|x|+M, \quad
 (t,x)\in [0,1]\times \mathbb{R}.
$$
Set $k(s)=\frac{1}{1+|\alpha |\eta}-\varepsilon$, $h(s)=M$, then
(4) holds. This completes the proof.
\end{proof}

\begin{corollary}\label{c1}
Suppose $f(t,0)\not\equiv 0$, and there exist two nonnegative functions
$k,h\in L^1[0,1]$ such that
$$
|f(t,x)|\leq k(t)|x|+h(t),\ a.e.\ (t,x)\in [0,1]\times \mathbb{R}.
$$
If one of the following conditions holds
\begin{itemize}
\item[(1)] There exists a constant $p>1$ such that
$$
\int_0^1k^p(s)ds<\Big[\frac{(1+q)^{1/q}}{2 +|\alpha|\ (1+q)^{1/q} } \Big]^p,\quad
 \big( \frac{1}{p} +\frac{1}{q}=1 \big).
$$
\item[(2)] There exists a constant $\mu >-1$ such that
\begin{gather*}
k(s)\leq \frac{(1+\mu)(2+\mu)} {2+|\alpha |(2+\mu)}s^\mu ,\quad\mbox{a.e. }
s\in [0,1],\\
\mathop{\rm meas}\big\{s\in [0,1]: k(s)< \frac{(1+\mu)(2+\mu)}{2
+|\alpha|(2+\mu)}s^\mu \big\}>0.
\end{gather*}

\item[(3)] $k$ satisfies
\begin{gather*}
k(s)\leq \frac{1}{1+|\alpha |} ,\quad\mbox{a.e. } s\in [0,1],\\
\mathop{\rm meas} \big\{s\in [0,1]: k(s)<
\frac{1}{1+|\alpha|} \big\}>0.
\end{gather*}

\item[(4)] $f$ satisfies
$$
\Lambda=:\limsup_{|x|\to\infty}\max_{t\in
[0,1]}\big|\frac{f(t,x)}{x}\big|<\frac{1}{1+|\alpha|}.
$$
\end{itemize}
 Then the BVP \eqref{e1.1} has at least one nontrivial solution
$u^\ast \in C[0,1]$.
\end{corollary}

\begin{proof} In this case, we have
\[
A=\ 2\int_0^1(1-s)k(s)ds+|\alpha|\int_0^\eta k(s)ds
\leq 2\int_0^1(1-s)k(s)ds+|\alpha|\int_0^1k(s)ds.
\]
The rest of the proof is the same as in Theorem \ref{thm2}. \end{proof}

\begin{corollary}\label{c2}
Suppose $f(t,0)\not\equiv 0$,
and there exist two nonnegative functions $k,h\in L^1[0,1]$ such
that
$$
|f(t,x)|\leq k(t)|x|+h(t),\quad\mbox{a.e. } (t,x)\in [0,1]\times \mathbb{R}.
$$
If one of the following conditions is holds.
\begin{itemize}
\item[(1)] There exists a constant $p>1$ such that
$$
\int_0^1k^p(s)ds<\Big[\frac{1}{2+|\alpha|}\cdot\big(
\frac{1+q}{2^{1+q}-1}\big)^{1/q}\Big]^p,\quad
 \big( \frac{1}{p} +\frac{1}{q}=1 \big).
$$

\item[(2)] There exists a constant $\mu >-1$ such that
\begin{gather*}
k(s)\leq \frac{(1+\mu)(2+\mu)}{(2+|\alpha|)(\mu+3)}s^\mu ,\quad\mbox{a.e. }
s\in [0,1],\\
\mathop{\rm meas} \big\{s\in [0,1]: k(s)<
\frac{(1+\mu)(2+\mu)}{(2+|\alpha|)(\mu+3)}s^\mu \big\}>0.
\end{gather*}

\item[(3)] There exists a constant $\mu >-2$ such that
\begin{gather*}
k(s)\leq \frac{(2+\mu)}{(2+|\alpha|)(2^{2+\mu}-1)}(2-s)^\mu ,\quad
\mbox{a.e. } s\in [0,1],\\
\mathop{\rm meas} \big\{ s\in [0,1]: k(s)<
\frac{(2+\mu)}{(2+|\alpha|)(2^{2+\mu}-1)} (2-s)^\mu \big\}>0.
\end{gather*}
\end{itemize}
Then the BVP \eqref{e1.1} has at least one nontrivial solution $u^\ast
\in C[0,1]$.
 \end{corollary}

\begin{proof} In this case,
\begin{align*}
A&= 2\int_0^1(1-s)k(s)ds+|\alpha|\int_0^\eta k(s)ds\\
&\leq 2\int_0^1(1-s)k(s)ds+|\alpha|\int_0^1k(s)ds\\
&\leq (2+|\alpha|)\int_0^1(2-s)k(s)ds.
\end{align*}
(1)\ Using the H\"{o}lder inequality,
\begin{align*}
A&\leq (2+|\alpha|)\int_0^1(2-s)k(s)ds\\
&\leq(2+|\alpha|)\Big[\int_0^1k^p(s)ds\Big]^{1/p}
\Big[\int_0^1(2-s)^qds\Big]^{1/q}\\
&< (2+|\alpha|)\cdot\frac{1}{2+|\alpha|}\big(
\frac{1+q}{2^{1+q}-1}\big)^{1/q}\cdot\big(
\frac{2^{1+q}-1}{1+q}\big)^{1/q}=1.
\end{align*}
(2)\ In this case, we have
\begin{align*}
A&\leq (2+|\alpha|)\int_0^1(2-s)k(s)ds\\
&< (2+|\alpha|)\cdot \frac{(1+\mu)(2+\mu)}{(2+|\alpha|)
(\mu+3)}\int_0^1(2-s)s^{\mu}ds\\
&=\frac{(1+\mu)(2+\mu)}{\mu+3}\cdot
\frac{\mu+3}{(1+\mu)(2+\mu)}=1.
\end{align*}
(3)\ In this case,
\begin{align*}
A&\leq (2+|\alpha|)\int_0^1(2-s)k(s)ds\\
&<(2+|\alpha|)\cdot \frac{(2+\mu)}{(2+|\alpha|)
(2^{2+\mu}-1)}\int_0^1(2-s)^{1+\mu}ds\\
&=\frac{2+\mu}{2^{2+\mu}-1}\cdot \frac{2^{2+\mu}-1}{2+\mu}=1.
\end{align*}
The proof is complete.\end{proof}

\section{Examples}

 In this section, in order to illustrate
our results, we consider some examples.

\begin{example} \label{ex4.1} \rm
Consider the three-point BVP
\begin{equation} \label{e4.1}
\begin{gathered}
u''+(t-t^2)|u|\sin u-t^2u+t^3-2\sin t=0,\quad  0<t<1, \\
 u'(0)=0,\quad  u(1)=4u'(1/2).
\end{gathered}
\end{equation}
 Set $\alpha=4$, $\eta=\frac{1}{2}$, and
\begin{gather*}
f(t,x) = (t-t^2)|x|\sin x-t^2x + t^3-2\sin t,\\
k(t) = t,\quad h(t)=t^3+2\sin t.
\end{gather*}
It is easy to prove that $k,h\in L^1[0,1] $ are nonnegative functions and
\begin{eqnarray*}
f(t,x)\leq k(t)|x|+ h(t), \quad (t, x)\in [0,1]\times \mathbb{R}.
\end{eqnarray*}
and
\[
 A= 2\int_0^1(1-s)k(s)ds+|\alpha|\int_0^\eta k(s)ds=\frac{5}{6}<1.
\]
Hence, by Theorem \ref{thm1}, the BVP \eqref{e4.1} has at least one nontrivial
solution $u^*$ in $C[0, 1]$.
\end{example}

\begin{example} \label{ex4.2} \rm
Consider the three-point BVP
\begin{equation} \label{e4.2}
\begin{gathered}
u''+\frac{2\sqrt{t}u^3}{3+u^4}e^{-|sin (u^2-t)|}+3e^t-2\sin t=0,\quad  0<t<1, \\
u'(0)=0,\quad u(1)=\sqrt{3}u'(1/4).
\end{gathered}
\end{equation}
 Set $\alpha=\sqrt{3}$, $\eta=\frac{1}{4}$, and
\begin{gather*}
f(t,x)= \frac{2\sqrt{t}x^3}{3+x^4}e^{-|\sin (x^2-t)|}+3e^t-2\sin t,\\
k(t)= \sqrt{\frac{t}{3}},\quad h(t)=3e^t+2\sin t.
\end{gather*}
It is easy to prove that $k, h\in L^1[0,1] $ are nonnegative
functions and
\[
f(t,x)\leq k(t)|x|+ h(t),\quad (t, x)\in [0,1]\times \mathbb{R}.
\]
 Let $p=q=2$, then
\begin{gather*}
 \int_0^1k^p(s)ds=\int_0^1\frac{1}{3}s\,ds=\frac{1}{6},\\
 \Big[\frac{(1+q)^{1/q}}{2+|\alpha|[\eta(1+q)]^{1/q}}\Big]^p=\frac{12}{49}.
\end{gather*}
Therefore,
$$
\int_0^1k^p(s)ds<\Big[\frac{(1+q)^{1/q}}{2+|\alpha|[\eta(1+q)]^{1/q}}\Big]^p.
$$
 Thus, by Theorem \ref{thm2} (1), the BVP \eqref{e4.2} has at least one
nontrivial solution $u^*$ in $C[0, 1]$.
\end{example}

\begin{example} \label{ex4.3}\rm
Consider the three-point BVP
\begin{equation} \label{e4.3}
\begin{gathered}
u''+\frac{u^2e^{-t}}{3(1+u^2)(1+2e^u)\sqrt{t}}
+\frac{1}{7\sqrt{t}}u+e^t-\sqrt{t}\cos t=0,\quad  0<t<1, \\
u'(0)=0,\quad u(1)=\frac{1}{3}u'(1/4).
\end{gathered}
\end{equation}
 Set $\alpha=\frac{1}{3}$, $\eta=\frac{1}{4}$, and
\begin{gather*}
f(t,x)=\frac{x^2e^{-t}}{3(1+x^2)(1+2e^x)\sqrt{t}}+\frac{1}{7\sqrt{t}}x
+e^t-\sqrt{t}\cos t,\\
k(t)= \frac{1}{6\sqrt{t}}+\frac{1}{7\sqrt{t}},\quad
h(t)=e^t+\sqrt{t}\cos t.
\end{gather*}
It is easy to prove that $k, h\in L^1[0,1] $ are nonnegative functions, and
$$
f(t,x)\leq k(t)|x|+ h(t),\quad\mbox{a.e. } (t, x)\in [0,1]\times \mathbb{R},
$$
Let $\mu=-\frac{1}{2}$, then
$$
\frac{(1+\mu)(2+\mu)}{2+|\alpha|(2+\mu)\eta^{1+\mu}}=\frac{1}{3}.
$$
Therefore,
\begin{align*}
k(s)&= \frac{1}{6\sqrt{s}}+\frac{1}{7\sqrt{s}}< \frac{1}{6\sqrt{s}}
+\frac{1}{6\sqrt{s}}=\frac{1}{3}s^{-\frac{1}{2}}\\
&=\frac{(1+\mu)(2+\mu)}{2+|\alpha|(2+\mu)\eta^{1+\mu}}s^{\mu},\quad
\mbox{a.e. } s\in [0,1],
\end{align*}
$$
\mathop{\rm meas}
\big\{s\in [0,1]: k(s)< \frac{(1+\mu)(2+\mu)}{2+|\alpha|(2+\mu)
\eta^{1+\mu}}s^{\mu} \big\}=1>0.
$$
 Hence, by Theorem \ref{thm2} (2), the BVP \eqref{e4.3} has at least one
nontrivial solution $u^*\in C[0, 1]$.
\end{example}

\begin{example} \label{ex4.4} \rm
Consider the three-point BVP
\begin{equation} \label{e4.4}
\begin{gathered}
u''+\frac{u^2e^{-u^2}}{3(1+u^2)\sqrt[4]{1-t}}-3e^{-t}+\sqrt{\sin t}=0,\quad  0<t<1, \\
 u'(0)=0,\quad  u(1)=-3u'(1/4).
\end{gathered}
\end{equation}
 Set $\eta=\frac{1}{4}, \alpha=-3,$ and
\begin{gather*}
  f(t,x)
 = \frac{x^2e^{-x^2}}{3(1+x^2)\sqrt[4]{1-t}}-3e^{-t}+\sqrt{\sin t},\\
k(t)=\frac{1}{6\sqrt[4]{1-t}},\quad h(t)=3e^{-t}+\sqrt{\sin t}.
\end{gather*}
It is easy to prove that $k, h\in L^1[0,1] $ are nonnegative
functions, and
\[
f(t,x)\leq k(t)|x|+ h(t),\quad\mbox{a.e.} (t, x)\in [0,1]\times \mathbb{R},
\]
Let $\mu=-\frac{1}{4}$. Then
$$
\frac{(1+\mu)(2+\mu)}{2(1+\mu)+|\alpha|(2+\mu)}=\frac{7}{36}.
$$
Therefore,
\begin{gather*}
k(s)= \frac{1}{6\sqrt[4]{1-s}}< \frac{7}{36\sqrt[4]{1-s}}
=\frac{(1+\mu)(2+\mu)}{2(1+\mu)+|\alpha|(2+\mu)}(1-s)^{-\frac{1}{4}},\quad
\mbox{a.e.}  s\in [0,1], \\
\mathop{\rm meas} \big\{s\in [0,1]: k(s)< \frac{(1+\mu)(2+\mu)}{2(1+\mu)
+|\alpha|(2+\mu)}(1-s)^{-\frac{1}{4}} \big\}=1>0.
\end{gather*}
 Hence, by Theorem \ref{thm2} (3), the BVP \eqref{e4.4} has at least one
nontrivial solution $u^*\in C[0, 1]$.
\end{example}

\begin{example} \label{ex4.5} \rm
Consider the three-point BVP
\begin{equation} \label{e4.5}
\begin{gathered}
 u''+\frac{t^2u^2e^{-u^2}}{t^2+u^2}-\cos e^t+3\sin^2 t=0,\quad  0<t<1, \\
 u'(0)=0,\quad  u(1)=3u'(1/3).
 \end{gathered}
\end{equation}
 Set $\eta=\frac{1}{3}$, $\alpha=3$, and
\begin{gather*}
 f(t,x) = \frac{t^2x^2e^{-x^2}}{t^2+x^2}-\cos e^t+3\sin^2 t,\\
 k(t)=\frac{t}{2},\ h(t)=\cos e^t+3\sin^2 t.
\end{gather*}
 Then it is easy to prove that $k, h\in L^1[0,1] $ are nonnegative functions,
and
\begin{gather*}
f(t,x)\leq k(t)|x|+ h(t),\quad\mbox{a.e. } (t, x)\in [0,1]\times \mathbb{R},\\
k(s)=\frac{s}{2} \leq \frac{1}{1+|\alpha|\eta}=\frac{1}{2},\quad
s\in [0,1],\\
\mathop{\rm meas} \big\{s\in [0,1]: k(s)<
\frac{1}{1+|\alpha|\eta} \big\}=1>0.
\end{gather*}
Hence, by Theorem \ref{thm2} (4), the BVP \eqref{e4.5} has at least one
nontrivial solution $u^*\in C[0, 1]$.
\end{example}

\subsection*{Acknowledgment}
The author wishes to express his sincere gratitude to the anonymous
referees for their helpful suggestions in improving the paper.

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\end{document}