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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 155, pp. 1--12.\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-2010/155\hfil Existence and uniqueness]
{Existence and uniqueness of a positive solution for a third-order
three-point boundary-value problem}

\author[A. P. Palamides, N. M. Stavrakakis \hfil EJDE-2010/155\hfilneg]
{Alex P. Palamides, Nikolaos M. Stavrakakis}  % in alphabetical order

\address{Alex P. Palamides \newline
Technological Educational Institute of Piraeus, Department
Electronic Computer Systems Engineering, Athens, Greece}
\email{palamid@teipir.gr}

\address{Nikolaos M. Stavrakakis \newline
Department of Mathematics, National Technical University, Zografou
Campus, 157 80 Athens, Greece}
\email{nikolas@central.ntua.gr}

\thanks{Submitted May 10, 2010. Published October 28, 2010.}
\subjclass[2000]{34B10, 34B18, 34B15, 34G20}
\keywords{Three point singular boundary value problem;
positive solutions; \hfill\break\indent
third order differential equation;
existence; uniqueness; fixed points in cones; Green's functions}

\begin{abstract}
 In this work we study a third-order three-point boundary-value
 problem (BVP). We derive sufficient conditions that guarantee
 the positivity of the solution of the corresponding linear BVP
 Then, based on the classical Guo-Krasnosel'skii's  fixed point
 theorem, we obtain positive solutions to the nonlinear BVP.
 Additional hypotheses guarantee the uniqueness of the solution.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}

\section{Introduction}

In this article, we are concerned with a certain class of third-order
differential equations, known as the three-point boundary-value problem
(BVP), given by
\begin{equation}
\begin{gathered}
u'''(t)=-f(t,u(t)),\quad 0<t<1, \\
u(0)-qu'(0)=0,\quad u'(\eta)=0,\quad u(1)=0,
\end{gathered}  \label{E}
\end{equation}
where
\begin{equation}
q\geq \frac{1}{2\eta }(1-2\eta ),\ \ 0<\eta <1/2.
\label{H1}
\end{equation}
The problem \eqref{E} consists of a new set of boundary conditions
but it is closely related with several boundary conditions.

 Recently, Sun, Cho and O'Regan \cite{SCO} proved the existence of
positive solutions to the third-order boundary-value problem
\begin{gather*}
z'''+q(t)f(t,z)=0,\quad 0<t<1, \\
z(0)=z'(0)=z(1)=0,
\end{gather*}
mainly under a local (at $z=0$) monotone condition and sublinearity
(at $ z=+\infty $) of the nonlinearity. In that paper,
they constructed the corresponding Green function and then
applied the Krasnosel'ski\u{\i}'s fixed point theorem.

 Lately there have been several papers on third-order boundary
value problems. Hopkins and Kosmatov \cite{hopkins},
 Infante and Webb \cite{We}, Li \cite{li}, Liu et al \cite{liu,liu2},
Guo et al \cite{guo} and Kang et al \cite{minghe} have all
considered third-order problems. Graef and Yang
\cite{gy2} and Wong \cite{wong} considered three-point focal problems.
Also Henderson et al \cite{GHY, GHY1} studied higher order boundary
value problem.

 Anderson et al  \cite{AAK} proved the existence of Green's
function and found an explicit formula for it, associated with the
homogeneous BVP:
\begin{gather*}
x'''(t)=0,\quad 0\leq t\leq 1 \\
ax(0)-bx'(0)=0,\quad \gamma x(\eta)-\delta x'(\eta )=0,
\quad x''(1)=0.
\end{gather*}
Finally, Palamides and Smyrlis \cite{PS} proved the existence
of positive solutions for the general nonlinear boundary-value problem
\begin{gather*}
x'''(t)=a(t)F(t,x(t),x'(t),x''(t)),\quad 0<t<1, \\
x(0)=x'(\eta )=x''(1)=0.
\end{gather*}
In this article, mainly motivated by the above mentioned papers
we consider a new set of boundary conditions and assume
similar hypothesis as in \cite{SCO}. Initially we construct the
Green's function to the homogeneous BVP corresponding to \eqref{E}
and then we derive the sufficient condition \eqref{H1} which
guarantees that the Green's function is positive. Then, based on
the Guo-Krasnosel'skii's fixed point theorem, we obtain a positive
solution to the nonlinear BVP \eqref{E}, under superlinear
(or sublinear) type growth-rate on the nonlinearity.
Straightforwardly, we also conclude existence of a
negative solution of the above BVP, under the hypothesis of negativity
of the nonlinearity. Finally, we give conditions under which the
existing solution is unique.

\section{Constructing the Green's function}

 Consider first the homogeneous third-order boundary-value
problem
\begin{equation}
\begin{gathered}
u'''(t)=0,\quad 0<t<1, \\
u(0)-qu'(0)=0,\quad u'(\eta )=0,\quad u(1)=0.
\end{gathered}  \label{E0}
\end{equation}

\begin{lemma}\label{Le1}
If $\eta \neq 1/(2(1+q))$, the boundary value problem
\eqref{E} has the unique solution
\[
u(t)=0,\quad 0\leq t\leq 1.
\]
\end{lemma}

\begin{proof}
The general solution of the BVP \eqref{E0} has the form
$u(t)=at^{2}+bt+c$. The conditions at $t=\eta$ and
$t=1$ imply that $2a\eta +b=0$ and $a+b+c=0$.
Moreover the condition at $t=0$ yields $c-qb=0$. Hence
we immediately obtain the expected result.
\end{proof}


 Consider now the inhomogeneous BVP
\begin{equation}
\begin{gathered}
u'''(t)=-1,\quad 0\leq t\leq 1, \\
u(0)-qu'(0)=0,\quad u'(\eta)=0,\quad u(1)=0.
\end{gathered} \label{E1}
\end{equation}

\begin{lemma}\label{Le2}
Assume that $\eta \neq 1/(2(1+q)$. Then, the Green's
function of the BVP \eqref{E1} is given by: for
$s<\eta$,
\[
G(t,s)=\begin{cases}
u_1^{\ast }(t,s), & 0\leq t\leq s \\
v_1^{\ast }(t,s), & s\leq t\leq 1
\end{cases}
\]
and  for $s>\eta$,
\[
G(t,s)=\begin{cases}
u_2^{\ast }(t,s),& 0\leq t\leq s \\
v_2^{\ast }(t,s),& s\leq t\leq 1,
\end{cases}
\]
where
\begin{gather*}
\begin{aligned}
u_1^{\ast }(t,s)
&=C_0\Big((1-2\eta -2q\eta +s^{2}+2qs)t^{2}-2(s-2s\eta +s^{2}\eta)t\\
&\quad -2q(s-2s\eta +s^{2}\eta )\Big)
\end{aligned}\\
v_1^{\ast }(t,s)=C_0\Big((s^{2}+2qs)t^{2}
-2(2qs\eta +s^{2}\eta )t
 -(s^{2}+2qs-4qs\eta -2s^{2}\eta )\Big),
\\
u_2^{\ast }(t,s)=C_0\Big(
(1+s^{2}-2s)t^{2}-2(\eta -2s\eta +s^{2}\eta )t
 -2q(\eta -2s\eta +s^{2}\eta )\Big),
\\
\begin{aligned}
v_2^{\ast }(t,s)
&=C_0\Big( (2\eta +2q\eta +s^{2}-2s)t^{2}-2(\eta -s+2qs\eta
 +s^{2}\eta )t\\
&\quad  -2q\eta +s^{2}-4qs\eta -2s^{2}\eta\Big),
\end{aligned}\\
C_0=-\frac{1}{2(-1+2\eta +2q\eta )}.
\end{gather*}
\end{lemma}

\begin{proof}
To obtain the solution of \eqref{E1}, we  proceed by cases on
the two branches of the solution, via the above Green's function
$G(t,s)$:

 If $t<\eta $,
\begin{equation}
\begin{aligned}
u_1(t)
&=C_0\Big( \int_0^t\Big(
(s^{2}+2qs)t^{2}-2(2qs\eta +s^{2}\eta )t
-(s^{2}+2qs-4qs\eta -2s^{2}\eta ) \Big)ds \\
&\quad +\int_{t}^{\eta }\Big(
(1-2\eta -2q\eta +s^{2}+2qs)t^{2}-2(s-2s\eta +s^{2}\eta )t\\
&\quad -2q(s-2s\eta +s^{2}\eta )\Big)ds \\
&\quad +\int_{\eta }^{1}\Big(
(1+s^{2}-2s)t^{2} -2(\eta -2s\eta +s^{2}\eta )t
-2q(\eta -2s\eta +s^{2}\eta )\Big)ds\Big)
\end{aligned} \label{14}
\end{equation}
Hence an easy computation ensures that
$u_1(0)-qu_1'(0)=0$  and
$u_1'(\eta )=0$.


 For $\eta \leq t\leq 1$,
\begin{align*}
u_2(t)
&=C_0\Big(
\int_0^{\eta }\Big(
(s^{2}+2qs)t^{2}-2(+2qs\eta +s^{2}\eta )t
 -(s^{2}+2qs-4qs\eta -2s^{2}\eta)\Big)ds\\
&\quad
+\int_{\eta }^t\Big(
(2\eta +2q\eta +s^{2}-2s)t^{2}
-2(\eta -s+2qs\eta +s^{2}\eta )t\\
&\quad -(2q\eta +s^{2}-4qs\eta -2s^{2}\eta)\Big)ds \\
&\quad +\int_{t}^{1}\Big(
(1+s^{2}-2s)t^{2}-2(\eta -2s\eta +s^{2}\eta )t
-2q(\eta -2s\eta +s^{2}\eta )\Big)ds
\Big)
\end{align*}
By another calculation, we may obtain that
$u_2'(\eta )=0$  and $u_2(1)=0$.
Furthermore,
\[
u_1'''(t)=-1\quad\text{and}\quad
u_1(t)=u_2(t)=u(t),\quad 0\leq t\leq 1.
\]
Hence the obtained function $u(t)$, $0\leq t\leq 1$, is a
solution of  \eqref{E1}.
\end{proof}


\begin{lemma}\label{Le00}
Assume hypothesis \eqref{H1}. Then the Green function is
nonnegative.
\end{lemma}

\begin{proof}
By  condition \eqref{H1}, we have $C_0\leq 0$. Then
\[
4s\eta -2s-2s^{2}\eta -t^{2}(2\eta -2s)=-2s(-2\eta +s\eta
+1)-t^{2}(2\eta -2s)\leq 0,
\]
since by the definition of $u_1^{\ast }(t,s)$,
 $-2\eta +1\geq 0$ and $s\leq \eta $. Consequently
\begin{align*}
u_1^{\ast }(t,s)
&=C_0\Big(
(4s\eta -2s-2s^{2}\eta -t^{2}(2\eta -2s))q\\ \ \ \
&\quad  +(t^{2}(s^{2}-2\eta +1)-2t(s-2s\eta +s^{2}\eta ))
\Big)\\
&\geq C_0\Big(
(4s\eta -2s-2s^{2}\eta -t^{2}(2\eta -2s))(-
\frac{1}{2\eta }(2\eta -1))\\
&\quad +(t^{2}(s^{2}-2\eta +1)-2t(s-2s\eta +s^{2}\eta ))\Big).
\end{align*}
That is,
\begin{equation}
u_1^{\ast }(t,s)\geq C_0s(t-1)(t-2\eta
+1)\frac{-2\eta +s\eta +1}{\eta }\geq 0.  \label{15}
\end{equation}
 Similarly,
\begin{align*}
u_2^{\ast }(t,s)
&=C_0((4\eta s-2\eta s^{2}-2\eta )q+(t^{2}(s^{2}-2s+1)-2t(\eta
s^{2}-2\eta s+\eta )) ) \\
& \geq C_0\Big((4\eta s-2\eta s^{2}-2\eta )(-\frac{1}{
2\eta }(2\eta -1))+(t^{2}(s^{2}-2s+1)\\
&\quad -2t(\eta s^{2}-2\eta s+\eta )) \Big)
\end{align*}
Hence
\begin{equation}
u_2^{\ast }(t,s)\geq C_0(s-1)^{2}(
t-1)(t-2\eta +1)\geq 0.  \label{16}
\end{equation}
 In the same way, we may verify that
\begin{equation}
\begin{gathered}
v_1^{\ast }(t,s)\geq C_0s(t-1)(t-2\eta
+1)\frac{-2\eta +s\eta +1}{\eta }\geq 0,   \\
v_2^{\ast }(t,s)\geq C_0(s-1)^{2}(t-1)(t-2\eta +1)\geq 0.
\end{gathered}  \label{17}
\end{equation}
\end{proof}

 For convenience, we set: For $s \leq \eta$,
\[
\frac{\partial }{\partial t}G(t,s)
= 2C_0\times\begin{cases}
(s^{2}+2qs+1-2\eta -2q\eta ) t -(s-2s\eta +s^{2}\eta ),
& t\leq s
\\
(s^{2}+2qs)t-(2q\eta s+s^{2}\eta ),
&s\leq t\leq 1
\end{cases}
\]
and for $s >\eta$,
\[
\frac{\partial }{\partial t}G(t,s)
=2C_0\times \begin{cases}
(s^{2}-2s+1) t-(\eta s^{2}-2\eta s+\eta),
& 0\leq t\leq s \\
(s^{2}-2s+2\eta +2q\eta ) t-(\eta -s+2qs\eta +s^{2}\eta ),
&  s\leq t\leq 1
\end{cases}
\]
Moreover, for $s <\eta$,
\[
\frac{\partial ^{2}}{\partial t^{2}}G(t,s)
=2C_0\times\begin{cases}
s^{2}+2qs+1-2\eta -2q\eta  , & 0\leq t\leq s \\
 s^{2}+2qs, & s\leq t\leq 1
\end{cases}
\]
and  for $s >\eta$,
\[
\frac{\partial ^{2}}{\partial t^{2}}G(t,s)
=2C_0\times\begin{cases}
s^{2}-2s+1 , &0\leq t\leq s \\
s^{2}-2s+2\eta +2q\eta , & s\leq t\leq 1.
\end{cases}
\]
Consider the Banach space $C=C([ 0,1] ,\mathbb{R})$
of continuous maps, equipped with the standard norm
\[
\|y\|=\max \{ |y(t)|:0\leq t\leq 1\} ,
\]
$0<\theta \leq \eta <1/2$ and let
\begin{align*}
K_0=\Big\{&
y\in C:y(t)\geq 0,\; t\in [0,1],\; y''(t)\leq 0,\;
t\in [\theta ,1-\theta ],  \\
&\max_{0\leq t\leq 1}y(t)=y(\eta )\text{  and  }
y(1)=0\Big\} .
\end{align*}
It is obvious that $K_0$ is a cone in $C$. We define furthermore the
subcone
\[
K=\big\{ y\in K_0:\min_{t\in [\theta ,1-\theta ]}y(
t)\geq \theta \|y\|\big\}
\]

\begin{lemma}\label{Le3}
For any $y\in K_0$,
\[
\min_{t\in [\theta ,1-\theta ]}y(t)\geq \theta
\|y\|=\theta y(\eta ).
\]
\end{lemma}

\begin{proof}
Since $y\in K_0$, $y(t)\geq 0$ for $0\leq t\leq 1$ and
moreover it is concave downward on the interval
$[\theta ,1-\theta ]$. Thus for any
$t_1,t_2\in [\theta ,1-\theta ]$
and $\lambda \in [0,1]$,
\[
y(\lambda t_1+(1-\lambda )t_2)\geq \lambda
y(t_1)+(1-\lambda )y(t_2).
\]
Therefore,
\[
y(t)\geq \|y\|\min_{t\in [\theta ,1-\theta ]
}\big\{ \frac{t}{\eta },\frac{1-t}{1-\eta }\big\}
 \geq \|y\|\min_{t\in [\theta ,1-\theta ]}\{ t,1-t\}
\geq \theta \|y\|.
\]
\end{proof}

 The next result is very useful.

\begin{proposition} \label{Pr1}
Assume  condition \eqref{H1} holds and let $y:[0,1
]\to [ 0,+\infty ))$ be a continuous map.
 Then the BVP
\begin{equation}
\begin{gathered}
u'''(t)=-y(t),\quad 0\leq t\leq 1, \\
u(0)-qu'(0)=0,\quad u'(\eta)=0,\quad u(1)=0.
\end{gathered} \label{Ey}
\end{equation}
admits the unique positive solution $u\in K$, where
\[
u(t)=\int_0^{1}G(t,s)y(s)ds.
\]
\end{proposition}


\begin{proof}
We notice firstly that $u(t)\geq 0$, $0\leq t\leq 1$. Indeed,
this fact follows directly by the nonnegativity of the Green's
function (see Lemma \ref{Le00}). On the other hand, for
$0\leq t\leq \eta$, we have
\begin{align*}
u'(t)&= \int_0^{1}\frac{\partial }{\partial t}G(t,s)y(s)ds \\
&= \int_0^t2C_0((s^{2}+2qs)t- (\eta s^{2}+2q\eta s))y(s)ds\; \\
&\quad +\int_{t}^{\eta }2C_0\Big(
(s^{2}+2qs+1-2\eta -2q\eta ) t-(s-2s\eta +s^{2}\eta )
\Big)y(s)ds  \\
&\quad +\int_{\eta }^{1}2C_0\big((
s^{2}-2s+1) t-(\eta s^{2}-2\eta s+\eta )
\big)y(s)ds.
\end{align*}
Consequently,
\begin{align*}
u'(\eta )&=\int_0^{\eta }(2C_0((
s^{2}+2qs)\eta -( \eta s^{2}+2q\eta s)))y(s)ds  \\
&\quad  +\int_{\eta }^{1}(
2C_0((s^{2}-2s+) \eta -(\eta s^{2}-2\eta s+\eta )))y(s)ds
\\
&=\int_0^{1}2C_0[(s^{2}+2qs)\eta -(
\eta s^{2}+2q\eta s)]y(s)ds  \\
& =\int_0^{1}0y(s) ds=0.
\end{align*}
Similarly, we may prove that
$u(0)-qu'(0)=0$ and $u(1)=0$.
Furthermore,
\begin{align*}
u''(t)
&= \int_0^{1}\frac{\partial ^{2}}{
\partial t^{2}}G(t,s)y(s)ds \\
&= \int_0^t2C_0(s^{2}+2qs)y(s)ds \\
&\quad +\int_{t}^{\eta }2C_0(s^{2}+2qs+1-2\eta -2q\eta )
y(s)ds \\
&\quad +\int_{\eta }^{1}2C_0(s^{2}-2s+1)y(s)ds.
\end{align*}
Hence, recalling that $C_0=1/2(1-2\eta -2q\eta )$,
\[
u'''(t)= 2C_0(t^{2}+2qt)y(t)
-2C_0(t^{2}+2qt+1-2\eta -2q\eta )y(t)
= -y(t).
\]

Finally, by the nonnegativity of the solution $u(t)$ and the
boundary conditions $u(0)-qu'(0)=0$ and $u(1)=0$, we may assume
that $u''(0) \leq 0$. Otherwise, if $u''(0)>0$,  we get
$u'(t)>0$, in a right neighborhood of $0$, due to the
differential equation
$u'''(t)=-y(t)$, $0\leq t\leq 1$, and since $u'(\eta )=0$).
Hence there is a $\theta \in [ 0,\eta )$, such that $u''(\theta )=0$.
 Thus in both the cases, we conclude that
\[
u''(t)\leq 0,\quad \theta \leq t\leq 1-\theta .
\]
Consequently, in view of Lemma \ref{Le3}, we obtain that $u\in K$.
\end{proof}


\begin{corollary}\label{Co}
 Assume that hypotheses of Proposition \ref{Pr1} are
satisfied. Consider the BVP
\begin{equation}
\begin{gathered}
u'''(t)=y(t),\quad 0\leq t\leq 1, \\
u(0)-qu'(0)=0,\quad u'(\eta)=0,\quad u(1)=0.
\end{gathered}  \label{E1*}
\end{equation}
Then, the map
\[
u(t)=-\int_0^{1}G(t,s)y(s)ds
\]
is clearly a non-positive solution of \eqref{E1*}.
\end{corollary}

\section{Main Results}

In this section we prove the existence of at least one
positive solution of  \eqref{E}. We assume that
\begin{equation}
f\in C([0,1]\times [ 0,+\infty ),[0,+\infty))  \label{H2}
\end{equation}
In view of Proposition \ref{Pr1}, we consider the positive solution
$u_1(t)$ of \eqref{E1} and set
\begin{gather*}
A_0 = \max \{ u_1(t):0\leq t\leq 1\}
=\max_{0\leq t\leq 1}\Big(\int_0^{1}G(t,s)ds\Big),\\
B_0 = \max \{ u_1(t):\theta \leq t\leq 1-\theta\}
=\max_{\theta \leq t\leq 1-\theta }
\Big(\int_{\theta }^{1-\theta}G(t,s)ds\Big).
\end{gather*}
In view of Lemma \ref{Le00}, we get $A_0\geq B_0>0$.
We define the operator
\[
\mathcal{T}u(t)=\int_0^{1}G(t,s)f(s,u(s))ds.
\]
Obviously, BVP \eqref{E} has a solution $u=u(t)$, if and
only if $u$ is a fixed point of $T$. Moreover,
recalling that the operator
$\mathcal{T}:K\to C([0,1])$
is called \textit{completely continuous},
if it is continuous and maps bounded sets into precompact sets
we state the next well-known result \cite{Su}.

\begin{proposition}
Assume that \eqref{H1}-\eqref{H2} hold. Then $\mathcal{T}:K\to K$
is a completely continuous operator.
\end{proposition}

\begin{proof}
It is sufficient to show that $\mathcal{T}(K)\subset K$. This is
easily derived from Lemma \ref{Le3} and Proposition \ref{Pr1}, due to
assumption \eqref{H1} and the definition of the cone $K$.
\end{proof}


We will employ the following fixed point theorem due
to Krasnosel'skii \cite{Kr}.

\begin{theorem} \label{Th1} Let $E$ be a Banach space, $K\subseteq E$
be a cone and suppose that $\Omega _1$, $\Omega _2$ are bounded
open balls of $E$ centered at the origin with
$\overline{\Omega }_1\subset \Omega _2$. Furthermore,
suppose that $\mathcal{T}:K\cap (\overline{\Omega }_2
\setminus \Omega_1)\to K$ is a completely continuous operator
such that either
:$\| \mathcal{T}u\| \leq \| u\| $, $u\in K\cap \partial
\Omega _1$ and $\| \mathcal{T}u\| \geq \| u\| $, $u\in K\cap
\partial \Omega _2$;
 or $\| \mathcal{T}u\| \geq \| u\| $, $u\in K\cap \partial \Omega _1$
and $\| \mathcal{T}u\| \leq \|u\|$, $u\in K\cap \partial \Omega _2$
holds.
Then $\mathcal{T}$  admits a fixed point in $K\cap (\overline{\Omega
}_2\setminus \Omega _1)$.
\end{theorem}

Now we are ready to formulate and prove our main result.

\begin{theorem} \label{Th2}
Assume that \eqref{H1}-\eqref{H2} hold and there exist
positive constants $r\neq R$ such that
\begin{gather}
 |f(t,x)|\leq \frac{r}{A_0},\quad (t,x)\in [0,1]\times [0,r];
  \label{A1}\\
 | f(t,x)|\geq \frac{R}{B_0},\quad (t,x)\in [0,1]
\times [ \theta R,R].   \label{A2}
\end{gather}
Then the boundary value problem \eqref{E} admits a positive solution
$u=u(t)$, $0\leq t\leq 1$, such that
\[
\min \{ r,R\} \leq \|u\|\leq \max \{ r,R\} .
\]
Moreover, the obtained solution $u=u(t)$, $0\leq t\leq 1$ is
concave downward.
\end{theorem}

\begin{proof}
Assuming first that $r<R$, we consider the open balls
\[
\Omega _1=\{ u\in C([0,1]):\|u\|<r\},\quad
\Omega _2=\{ u\in C([0,1]):\ \|u\|<R\} .
\]
Let $u\in K\cap \partial \Omega _1$ be any function.
By noticing the sign of nonlinearity, the assumption \eqref{A1} yields
\begin{align*}
\| \mathcal{T}u\|
&= \max_{0\leq t\leq 1}|\int_0^{1}G(t,s)f(s,u(s))ds| \\
&\leq \max_{0\leq t\leq 1}\Big(\int_0^{1}G(t,s)\frac{r}{A_0}ds\Big)
=r=\| u\| .
\end{align*}
Therefore, the first part of the assumption of Theorem \ref{Th2}, is
fulfilled.
Similarly, for every $u\in K\cap \partial \Omega _2$, in view
of Lemma \ref{Le3}, it obvious that $\theta R\leq u(s)\leq R$,
$\theta \leq s\leq 1-\theta $. Thus the assumption \eqref{A2} implies
\begin{align*}
\| \mathcal{T}u\|
&= \max_{0\leq t\leq 1}|\int_0^{1}G( t,s)f(s,u(s))ds| \\
&\geq \max_{0\leq t\leq 1}|\int_{\theta }^{1-\theta }G(t,s)
f(s,u(s))ds| \\
&\geq \max_{\theta \leq t\leq 1-\theta }
\Big(\int_{\theta }^{1-\theta }G(t,s)\frac{R}{B_0}ds\Big)\\
&=R=\| u\| .
\end{align*}
Therefore,
$\| \mathcal{T}u\| \geq \| u\|$, for $u\in K\cap \partial \Omega _2$.

Finally, we may apply Theorem \ref{Th1}, to obtain a solution
$u=u(t)$, $0\leq t\leq 1$ of  BVP \eqref{E}.
Additionally by the definition of $K\subset K_0$ and the fact
that $u\in K$, we conclude that $u(t)$ is a positive solution.
Noticing that $u\in K\cap (\overline{\Omega }_2\setminus \Omega _1)$,
it is obvious that
\[
r\leq \|u\|\leq R.
\]

We assume now that $r>R$. We consider the open balls
\[
\Omega _1=\{ u\in C([0,1]): \|u\|<R\},\quad
\Omega _2=\{ u\in C([0,1]):\ \|u\|<r\} .
\]
and let $u\in K \cap \partial \Omega _1$. By Lemma \ref{Le3}, we have
\[
\min_{t\in [\theta ,1-\theta ]}u(t)\geq \theta
\|u\|=\theta R.
\]
Then from assumption \eqref{A2}, we conclude that
\begin{align*}
\| \mathcal{T}u\| &= \max_{0\leq t\leq 1}\big|\int_0^{1}G(
t,s)f(s,u(s))ds\big| \\
&\geq \max_{0\leq t\leq 1}\Big(\int_0^{1}G(t,s)\frac{R}{
B_0}ds\Big)\\
&\geq \max_{\theta \leq t\leq 1-\theta }\Big(\int_{\theta }^{1-\theta
}G(t,s)\frac{R}{B_0}ds\Big)\\
&=R=\| u\| .
\end{align*}
Similarly, if $u\in K\cap \partial \Omega _2$, then
$0\leq u(s)\leq r$, $0\leq s\leq 1$.
 Thus \eqref{A1} implies
\begin{align*}
\| \mathcal{T}u\|
&= \max_{0\leq t\leq 1}|\int_0^{1}G(t,s)f(s,u(s))ds| \\
&\leq \max_{0\leq t\leq 1}\Big(\int_0^{1}G(t,s)\frac{r}{A_0}ds\Big)\\
&=r=\| u\| .
\end{align*}
Therefore, the existence result follows.
\end{proof}


\begin{corollary} \label{Co1}
Assume \eqref{H1}-\eqref{H2} and in addition we suppose either:
 The nonlinearity is superlinear at both points $x=0$
and $x=+\infty$; i.e.,
\begin{equation}
\lim_{x\to 0+}\max_{0\leq t\leq 1}\frac{f(t,x)}{x}=0+\quad
\text{and}\quad
\lim_{x\to +\infty }\min_{0\leq t\leq 1}\frac{f(t,x)}{x} =+\infty ;
\label{A3}
\end{equation}
or
 the nonlinearity is sublinear at both points $x=0$
and $x=+\infty$, i.e.,
\begin{equation}
\lim_{x\to 0+}\min_{0\leq t\leq 1}\frac{f(t,x)}{x}=+\infty \quad
\text{and}\quad
\lim_{x\to +\infty }\max_{0\leq t\leq 1}\frac{f(t,x)}{x}
=0+. \label{A4}
\end{equation}
Then  boundary value problem \eqref{E} admits a positive, concave
downward solution $u=u(t)$, $0\leq t\leq 1$.
\end{corollary}

\begin{proof}
By the superlinearity of $f$, there exists an $r>0$ such that
$\frac{f(t,x)}{x}\leq \frac{1}{A_0}$, for all $(t,x)\in [0,1]
\times [0,r]$ and this yields assumption \eqref{A1} of previous
Theorem \ref{Th2}. Similarly by the superlinearity at $+\infty $,
we get an $R>r$ such that $\frac{f(t,x)}{x}\geq \frac{1}{\theta B_0}$,
for all $(t,x)\in [0,1]\times [ \theta R,R]$. Hence
Theorem \ref{Th2} is applicable.
On the other hand, when the nonlinearity is sublinear, we examine the
following cases:
(a) If $f$ is bounded, say by $M>0$, we may choose any $R\geq A_0M$
and then we obtain
\begin{equation}
\begin{aligned}
\| \mathcal{T}u\|
&\leq \max_{0\leq t\leq 1}|\int_0^{1}G(t,s)f(s,u(s))ds| \\
&\leq \max_{0\leq t\leq 1}\Big(\int_0^{1}G(t,s)Mds\Big) \\
&=MA_0\leq R=\| u\| ,
\end{aligned} \label{33}
\end{equation}
for $u\in K$ with $\|u\|=R$.

(b) If $f$ is unbounded, let also an $R$ be large enough such that
\[
\frac{|f(t,R)|}{R}\leq \frac{1}{A_0}\quad \text{and}\quad
|f(t,u)|\leq |f(t,R)|,\quad  (t,u)\in [0,1]\times [0,R].
\]
Therefore,
\[
|f(t,u)|\leq |f(t,R)|\leq \frac{R}{A_0},\quad
 (t,u)\in [0,1]\times [0,R].
\]
Consequently,
\begin{align*}
\| \mathcal{T}u\|
&=\max_{0\leq t\leq 1}|\int_0^{1}G(t,s)f(s,u(s))ds| \\
&\leq \max_{0\leq t\leq 1}\Big(\int_0^{1}G(t,s)\frac{R}{
A_0}ds\Big) \\
&\leq \frac{R}{A_0}A_0=\| u\|
\end{align*}
for  $u\in K$ with $\|u\|=R$.
Moreover, by the sublinearity of $f$ at $u=0$, there exists
an $r<R$ such that for any $u\in K$, $\|u\|=r$
(then we know that $r\geq u(s)\geq \theta \|u\|=\theta r$,
$\theta \leq s\leq 1-\theta$)
\[
|f(s,u(s))|\geq \frac{u(s)}{\theta
B_0}\geq \frac{\theta r}{\theta B_0}=\frac{r}{B_0},\quad
 (s,u(s))\in [\theta ,1-\theta ]\times [\theta r,r].
\]
Hence, for any $u\in K$ such that $\|u\|=r$, we have
\begin{align*}
\| \mathcal{T}u\|
& =\max_{0\leq t\leq 1}|\int_0^{1}G(t,s)f(s,u(s))ds| \\
& \geq \max_{\theta \leq t\leq 1-\theta }|\int_{\theta
}^{1-\theta }G(t,s)f(s,u(s))ds| \\
&\geq \max_{\theta \leq t\leq 1-\theta }\Big(\int_{\theta }^{1-\theta
}G(t,s)\frac{r}{B_0}ds\Big)\\
&=r\geq \| u\|.
\end{align*}
This clearly completes the proof.
\end{proof}

\begin{remark} \label{rmk1} \rm
We notice that the  positive solution, $u=u(t)$, obtained above
 satisfies the properties
\[
u'(0)>0,\quad u''(0)\leq 0,\quad
u'(1)<0, \quad u''(1)\leq 0.
\]
Furthermore the map $u''(t)$, $0\leq t\leq 1$ is
non-increasing.
\end{remark}

\begin{corollary} \label{Co0}
Under the assumptions of Theorem \ref{Th2} or Corollary \ref{Co1},
the  BVP
\begin{equation}
\begin{gathered}
u'''(t)=f(t,u(t)),\quad 0<t<1, \\
u(0)-qu'(0)=0,\quad u'(\eta)=0,\quad u(1)=0
\end{gathered}  \label{34}
\end{equation}
admits a negative and concave upward solution $u=u(t)$.
Here again the map $u''(t)$, $0\leq t\leq 1$ is non-increasing.
\end{corollary}

\begin{proof}
Now the above negative solution satisfies
\[
u'(0)<0,\quad u''(0)>0,\quad
u'(1)>0,\quad u''(1) \geq 0.
\]
\end{proof}


\begin{corollary}\label{Co2}
Under the assumptions of Theorem \ref{Th2} or Corollary\ \ref{Co1},
 BVP \eqref{34} admits a negative, concave upward solution
$u=u(t)$, $0\leq t\leq 1$.
\end{corollary}

\begin{proof}
Let $u=u_1(t)$, $0\leq t\leq 1$ be a solution of  BVP
\eqref{E}. Then the function
\[
u=-u_1(t),\quad 0\leq t\leq 1
\]
is obviously the desired solution of \eqref{34}. We notice that
\[
u'(0)<0,\quad u''(0)>0,\quad
u'(1)>0,\quad u''(1) \geq 0.
\]
Moreover, the map $u''(t)$, $0\leq t\leq 1$ is
nondecreasing.
\end{proof}

\begin{corollary} \label{Co3}
Under the assumptions of Theorem \ref{Th2} or Corollary\ \ref{Co1},
BVP \eqref{34} admits a positive and concave downward solution
$u=u(t)$, $0\leq t\leq 1$.
\end{corollary}

\begin{proof}
Obviously the desired solution is given by
\[
u(t)=\int_0^{1}[-G(t,s)]f(s,u(s))ds.
\]
Here also the map $u''(t)$, $0\leq t\leq 1$ is
nondecreasing.
\end{proof}

\begin{example} \label{exa1} \rm
Consider the boundary value problem
\begin{gather*}
u'''(t)=-\sqrt[3]{u(t)+(u(t))^{2}},\quad  0<t\leq 1 \\
u(0)=u'(0),\quad u'(3/10)=u(1)=0.
\end{gather*}
The nonlinearity $f(t,u)=\sqrt[3]{u+t}$ is sublinear. Thus,
Corollary \ref{Co1} guarantees the existence of a positive and concave
downwards solution to the above BVP.
\end{example}

\section{Uniqueness of solution}

\begin{theorem} \label{thm3}
Under the assumptions of Theorem \ref{Th2} or Corollary \ref{Co1},
  BVP \eqref{E} admits a unique solution, provided that the map
$f(t,.):[0,+\infty )\to [ 0,+\infty )$ is nondecreasing for
every $ t\in [0,1]$ and moreover
\[
|f(t,u_2)-f(t,u_1)|\leq L|u_2-u_1|,\quad
(t,u_{i})\in [0,1]\times [ 0,+\infty ),
\]
where
\[
\frac{1}{L}>\frac{1}{6}\eta (\eta -1)^{2}
\frac{2q+\eta +q\eta }{2\eta +2q\eta -1}.
\]
\end{theorem}

\begin{proof}
Let $w_{i}(t)$, $i=1,2$ be two  solutions of  BVP (
\eqref{E}, Since the Green's function $G(t,s)$ is positive, we
have
\begin{align*}
w_2(t)-w_1(t)
&=\int_0^{1}G(t,s) [f(s,w_2(s))-f(s,w_1(s))]ds  \\
&\leq \int_0^{1}G(t,s)|f(s,w_2(s))-f(s,w_1(s))|ds \\
&\leq \int_0^{1}G(t,s)L| w_2(s)-w_1(s)|ds \\
&\leq \int_0^{1}G(t,s) L\|w_2-w_1\|ds\\
&=L\|w_2-w_1\|\int_0^{1}G(t,s)ds \\
&=L\|w_2-w_1\|u_1(t),\quad 0\leq t\leq 1,
\end{align*}
where $u_1(\ t)$,  is the unique positive
solution of BVP \eqref{E1}. Consequently, we obtain the contradiction
\begin{align*}
\|w_2-w_1\|
&\leq L\|w_2-w_1\|\|u_1\|
\leq L\|w_2-w_1\|u_1(\eta )\\
&= L\|w_2-w_1\|\frac{1}{6}\eta (\eta -1)^{2}\frac{2q+\eta
+q\eta }{2\eta +2q\eta -1}\\
&<\|w_2-w_1\|.
\end{align*}
\end{proof}

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\end{document}