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

\AtBeginDocument{{\noindent\small \emph{
Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 25, pp. 1--6.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/25\hfil Existence of solutions]
{Existence of solutions for some third-order boundary-value problems}

\author[Z. Bai\hfil EJDE-2008/25\hfilneg]
{Zhanbing Bai}

\address{Zhanbing Bai \newline
 Institute of Mathematics,
 Shandong University of Science and Technology\\
 Qingdao 266510,  China}
\email{zhanbingbai@163.com}


\thanks{Submitted September 21, 2007. Published February 22, 2008.}
\thanks{Supported by grant 10626033 from the
Tianyuan Youth Grant Office of China} 
\subjclass[2000]{34B15}
\keywords{Third-order boundary-value problem; lower and upper
solutions; \hfill\break\indent fixed-point theorem}

\begin{abstract}
 In this paper concerns the third-order boundary-value problem
 \begin{gather*}
 u'''(t)+ f(t, u(t),u'(t), u''(t))=0, \quad 0 < t < 1, \\
 r_1 u(0) - r_2 u' (0)= r_3 u(1) + r_4 u'(1)= u''(0)=0.
 \end{gather*}
 By placing certain restrictions on the nonlinear term $f$,
 we prove the existence of at least one solution to the
 boundary-value problem with the use of lower and upper solution
 method and of  Schauder fixed-point theorem.
 The construction of lower or upper solutions is also presented.
\end{abstract}

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

\section{Introduction}

Recently, third-order boundary-value problems have been considered
in many papers. Some problems of regulation and control of some
actions by a control level or by a signal reduce to solving the
third-order equations. Other applications of  third-order
differential equations are encountered in the control of a flying
apparatus in cosmic space, the deflection of sandwich beam, and the
study of draining and coating flows. For details, see
the references in this article and the references therein.

As it is pointed out by Anderson {\it et al.} \cite{a1}, a large part of
the literature on solution to higher-order boundary-value problems
seems to be traced to Krasnosel'skii's work on nonlinear operator
equations as well as other fixed-point theorem such as
Leggett-Williams' fixed-point theorem.

The method of upper and lower solution is extensively developed for
lower order equations with linear and nonlinear boundary conditions.
But there are only a few applications to higher-order ordinary
differential equations. For applications to higher-order ODEs, we
refer the reader to Ehme \cite{e1}, Klaasen \cite{k1} and the references
therein. Specially, in Cabada \cite{c1} and Yao \cite{y1},
the lower and upper
solution method is employed to acquire existence results about some
third-order boundary-value problems with some monotonic or
quasi-monotonic nonlinear term $f$ which is no dependence on any
lower-order derivatives. On the other hand, to my best knowledge,
there are few papers referred to lower and upper solutions of
third-order equation consider the relationship between the property
of nonlinear term and the construction of lower and upper solutions.

The purpose of this paper is to study the existence of solution for
two class nonlinear third-order boundary-value problems
\begin{gather}
 u'''(t)+ f(t,u(t),u''(t))=0, \quad 0 \leq t \leq 1, \label{e1}\\
 r_1 u(0) - r_2 u' (0)=r_3 u(1) + r_4 u' (1)= u''(0)=0, \label{e2}
\end{gather}
and
\begin{gather}
 u'''(t)+ f(t,u(t),u'(t),u''(t))=0, \quad  0 \leq t \leq 1, \label{e3} \\
u(0) = u'(1)= u''(0)=0.\label{e4}
\end{gather}
The method used here is not based on the Krasnosel'skii's
fixed-point theorem or monotonic operator theory; rather, it is
based on Schauder fixed-point theorem, the appropriate integral
transvestites and lower and upper solution method. The
construction of lower or upper solution is also presented.

\section{Preliminaries}

In this section, we consider  \eqref{e1}--\eqref{e2}, under the
assumption that  $f:[0, 1] \times R^2 \to R$ is continuous, $ r_1,
r_2, r_3, r_4 \geq 0 $ and $\rho := r_2r_3 + r_1r_3 + r_1r_4 >0$. We
give some lemmas which indicate some restrictions on the nonlinear
term and let us construct lower or upper solutions.

\begin{definition} \label{def2.1} \rm
 We call $\alpha$, $\beta \in C^2[0, 1]\bigcap C^3(0, 1) $ lower
and upper solutions of Problem \eqref{e1}--\eqref{e2}, respectively, if
\begin{gather*}
 \alpha'''(t)+ f(t,\alpha(t),\alpha''(t)) \geq 0,\quad 0 < t < 1, \\
 r_1\alpha(0) - r_2\alpha' (0)= r_3\alpha(1)
+ r_4\alpha'(1)=0, \quad \alpha''(0) \geq 0;
\end{gather*}
\begin{gather*}
\beta'''(t)+ f(t,\beta(t),\beta''(t))\leq 0, \quad  0 < t < 1,\\
 r_1\beta(0) - r_2\beta' (0)= r_3\beta(1) + r_4\beta'(1)=0, \quad
 \beta''(0) \leq 0.
\end{gather*}
\end{definition}

Denote by $G(t, s)$ the Green's function of
\begin{equation}
\begin{gathered}
 -u''(t) = 0, \quad 0 < t < 1, \\
 r_1u(0)-r_2u' (0)= r_3u(1) + r_4u'(1)=0,
\end{gathered} \label{e5}
\end{equation}
then
\begin{equation}
G(t, s) = \begin{cases}
 \frac{1}{\rho}x(t)y(s), & 0 \leq s < t \leq 1, \\
\frac{1}{\rho}x(s)y(t), & 0 \leq t < s \leq 1,
\end{cases} \label{e6}
\end{equation}
where $x(t):= r_3+r_4-r_3t$, $y(t) := r_2+r_1t $, for
$ t \in [0, 1]$, $\rho = r_2r_3 + r_1r_3 + r_1r_4$. Clearly,
$G(t, s) \geq 0$ for $0 \leq  t, s \leq 1$.

The following Lemma comes from Lian \cite{l1} with small modification.

\begin{lemma}[\cite{l1}] \label{lem2.1}
For  $G(t, s)$ defined by \eqref{e6}, the following holds:
\begin{itemize}
\item[(R1)] $\frac{G(t, s)}{G(s, s)} \leq 1$  for
$t,s \in (0, 1)$,

\item[(R2)] $\frac{G(t, s)}{G(s, s)} \geq
C=:\min \{ \frac{r_4}{r_3+r_4}, \frac{r_2}{r_1+r_2}\} \geq 0$,
 for $t, s \in (0,1)$.
\end{itemize}
\end{lemma}

 Let $\eta = \int_0^1 G(s, s)s\,ds >0$. Then we have the following results.


\begin{lemma} \label{lem2.2}
If there exists a constant $M \geq 0$ such that
$$
f(t,s,r) \leq M, \quad\text{for } 0\leq t \leq 1,\;
\eta M C\leq s \leq \eta M, \; - M \leq r \leq 0,
$$
 then Problem \eqref{e1}-\eqref{e2} has an upper solution.
\end{lemma}

\begin{proof} Setting $v(t) = -u''(t)$, Problem
\eqref{e1}--\eqref{e2} is equivalent to
\begin{gather}
v'(t) = f(t, (Av)(t), -v(t)), \quad 0 < t < 1,\label{e7} \\
v(0)= 0, \quad \label{e8}
\end{gather}
 where $(Av)(t) = \int_0^1G(t,s)v(s)ds $ and $G(t, s) $
is defined by \eqref{e6}.
It is clear that the restriction on $f$ guarantee that
$\psi (t) = Mt$ satisfies
\begin{gather*}
\psi'(t) - f(t, (A\psi)(t), -\psi(t))\geq 0, \quad 0<t<1, \\
\psi(0) \geq 0.
\end{gather*}
 This shows that $\beta(t)= (A\psi)(t)$ is an upper solution of
Problem \eqref{e1}--\eqref{e2}.
\end{proof}

\begin{lemma} \label{lem2.3}
If there exists a constant $N \leq 0 $ such that
$$
f(t,s,r) \geq N, \quad \text{for } 0\leq t \leq 1, \; \eta N \leq s \leq
\eta NC, \; 0 \leq r \leq -N,
$$
 then Problem \eqref{e1}--\eqref{e2} has a lower solution.
\end{lemma}

\begin{proof}
Setting $\varphi(t)=Nt, \alpha(t) = (A \varphi)(t)$,
we can complete the proof as in Lemma \ref{lem2.2}.
\end{proof}

\begin{remark} \label{rmk2.1}
In fact, we can write the upper solution
$\beta (t) $ and lower solution $\alpha(t)$ explicitly.
Particularly, if $r_1, r_3 \not= 0$ and $r_2 =r_4=0$, then
$\beta(t) = -\frac{M}{6}t^3 + \frac{M}{6}t$,
$\alpha(t) = - \frac{N}{6}t^3 +\frac{N}{6}t $
are upper and lower solutions of Problem \eqref{e1}$,
\eqref{e2}$.
\end{remark}


\section{Main results}

In this section, we give some existence results for Problems
 \eqref{e1}--\eqref{e2} and \eqref{e3}--\eqref{e4}.

\begin{theorem} \label{thm3.1}
Suppose there are two constants $M \geq 0 \geq N$,
$ M \geq |N| $ such that
\begin{gather}
f(t,s,r) \leq M, \quad\text{for } 0\leq t \leq 1,\;  \eta MC \leq s \leq
\eta M, \; - M \leq r \leq 0, \label{e9} \\
f(t,s,r) \geq N, \quad \text{for } 0\leq t \leq 1, \;  \eta N \leq s \leq
\eta NC, \; 0 \leq r \leq -N. \label{e10}
\end{gather}
 If $f(t, s, r)$ is increasing in $s$, then Problem \eqref{e1}--\eqref{e2}
 has a solution $u(t)$ such that $\alpha(t) \leq u(t) \leq \beta (t)$,
$\beta''(t) \leq u''(t) \leq \alpha''(t)$, $t \in [0, 1] $,
 where
\[
\beta(t) =M\int_0^1G(t, s)s\,ds,\quad
\alpha(t) = N\int_0^1G(t, s)s\,ds\,.
\]
\end{theorem}

\begin{proof}
 From Lemmas \ref{lem2.2} and \ref{lem2.3} combined, conditions \eqref{e9} and
\eqref{e10} yield that  \eqref{e1}--\eqref{e2} has a lower solution
$\alpha (t)$ and an upper solution $\beta (t)$.
Set $\varphi =-\alpha''$, $\psi = - \beta'' $, then
$\varphi(t)=Nt \leq Mt= \psi(t)$. On the other hand,
$M \geq 0 \ge N$ implies
$\alpha''(t) \geq \beta''(t)$, for $0<t<1$. Taking into account
boundary condition \eqref{e2} and the fact that the Green's function
$G(t, s)\geq 0$ yield $\alpha(t) \leq \beta(t)$,
for $0 \leq t \leq 1$.

Now, consider the  truncated problem
\begin{equation}
Lv = Fv, \quad v \in \mathop{\rm dom}(L), \label{e11}
\end{equation}
where $L: \mathop{\rm dom}L=\{ v \in C^1(0, 1)\bigcap C[0, 1]: v(0)=0\}
\to C[0, 1] $ is a derivative operator such that
$(Lv)(t)= v' (t), t \in (0, 1) $, and $F: C[0, 1] \to C[0, 1]$ is
a continuous operator defined as
\begin{equation}
(Fv)(t)= f\big(t,A[v(t)]_{\varphi(t)}^{\psi(t)},
-[v(t)]_{\varphi(t)}^{\psi(t)}\big), \label{e12}
\end{equation}
where $[v]_{\varphi}^{\psi}= \min\{\psi, \max\{v, \varphi\}\}$.

Firstly, we prove that if $v^* $ is a solution of \eqref{e11},
then $\varphi(t) \leq v^*(t) \leq \psi(t)$, $t \in [0, 1]$.
Consequently, $v^*(t)$ is a solution of \eqref{e7}--\eqref{e8}.
Furthermore, $u^*(t) = (Av^*)(t)$ is a solution of
 \eqref{e1}--\eqref{e2} satisfying $\alpha(t) \leq u^*(t) \leq \beta(t)$.

In fact, if $\varphi(t) \not\leq v^*(t)$, there exists
$\bar t \in (0, 1)$ such that $v^*(\bar t) < \varphi(\bar t)$.
As $v^*(0) =0 = -\alpha''(0)=\varphi(0)$, by the continuity of $v^*$
and $\varphi$, there exist $t_1 \in [0, \bar t), t_2 \in (\bar t,1]$
such that $v^*(t_1)= \varphi(t_1) $ and
$v^*(t) < \varphi(t)$, for $t \in (t_1, t_2)$. Therefore,
for $t \in [t_1, t_2]$,
\begin{align*}
 (Fv^*)(t)&= f\big(t,A[v^*(t)]_{\varphi(t)}^{\psi(t)},
-[v^*(t)]_{\varphi(t)}^{\psi(t)}\big)\\
&=f\big(t,A[v^*(t)]_{\varphi(t)}^{\psi(t)},-\varphi(t)\big).
\end{align*}
 Let $p(t) = v^*(t) -\varphi(t)$, $t\in [t_1, t_2]$,
taking into account the monotonicity of $f$ and $A$, one has
\begin{gather*}
\varphi'(t) = - \alpha'''(t)
\leq f(t, \alpha(t), \alpha''(t)) = f(t,
A\varphi(t),-\varphi(t)), \quad t \in [0, 1],\\
{v^*}'(t) = (Fv^*)(t)=f\big(t,A[v^*(t)]_{\varphi(t)}^{\psi(t)},
-\varphi(t)\big), \quad
t \in [t_1, t_2],
\end{gather*}
 so $p'(t) \geq 0$, $t \in [t_1, t_2]$. Thus, $p(t_1) =0$
implies $p(t) \geq 0$, $t \in [t_1,t_2]$. Namely,
$v^*(t) \geq \varphi(t)$, $t \in [t_1, t_2]$. It is
a contradiction. Thus, $\varphi(t) \leq v^*(t) $ for
$t \in [0,1]$.
Analogously, we can prove $v^*(t) \leq \psi(t)$, $t \in [0, 1]$.

Secondly, we prove operator equation \eqref{e11} has a solution. Let
$T: C[0, 1] \to C[0, 1] $ by
$$
(Tv)(t) = \int_{0}^t (Fv)(s)ds, \quad t \in [0, 1].
$$
It is clear that $ T $ is a continuous operator and the fixed point
of $ T $ is a solution of Problem \eqref{e11}.
Set $B_M=\{\omega \in C[0, 1]: \Vert \omega \Vert \leq M \}$,
taking into account that
$|N| = \Vert \varphi \Vert \leq \Vert \psi \Vert = M$,
so we deduce that $\varphi, \psi \in B_M $. Now from condition
\eqref{e9}, one has
\begin{gather*}
\Vert Tv \Vert \leq M, \quad v \in B_M,\\
|(Tv)(t)- (Tv)(s)| = \big|\int_s^t(Fv)(r)dr\big| \leq M |t-s|.
\end{gather*}
That is,  $ \{T(B_M) \} $ is equi-continuous and bounded
uniformly. From Arzela-Ascoli theorem, we assert that
$T: B_M \to B_M$ is a completely continuous operator,
furthermore, Schauder fixed-point theorem guarantee that $T$ has a
fixed point  $ v^* \in B_M$. Therefore,
 $$
u^*(t) = \int_0^1G(t, s)v^*(s)ds
$$
 is a solution of Problem \eqref{e1}--\eqref{e2} in
$C^2[0, 1] \bigcap C^3(0, 1) $ such that
$\alpha(t) \leq u^*(t) \leq \beta(t)$,
$\ \beta''(t) \le u^*(t) \le \alpha''(t), \quad t \in [0, 1]$.
\end{proof}

 From the above proof, we should have noticed that the existence
of $M$ is very important. It make us not only acquire an upper solution of
Problem \eqref{e1}--\eqref{e2}, but guarantee that
$ T: B_M \to B_M $ is a completely continuous operator.
On the other hand, the truncated problem used here is different
from other earlier literature.
As the second order derivatives of upper and lower
solutions are employed to make truncating, we take the operator
$F$ instead of the place of the traditional truncated function.

\begin{remark} \label{rmk3.1} \rm
 If the nonlinear term $ f $ satisfies
$$
\bar f_\infty = \limsup_{u \to + \infty} \max_{0 \leq t \leq 1}
\frac{f(t, \eta u, -u)}{u} \leq 1,
$$
 then there exists $M>0$ such that $\beta(t) = \int_0^1G(t, s)Ms\,ds$
is an upper solution of \eqref{e1}--\eqref{e2}.
If the nonlinear term $ f $ satisfies
$$
\underline f_\infty = \liminf_{u \to - \infty} \min_{0 \leq t \leq 1}
\frac{f(t, \eta u, -u)}{u} \leq 1,
$$
 then there exists $N<0$ such that $\alpha(t) = \int_0^1G(t, s)Ns\,ds$
is a lower solution of \eqref{e1}--\eqref{e2}.
\end{remark}

\begin{example} \label{exa1} \rm
Consider the  problem
\begin{gather}
 u'''(t) + \frac{1}{3} [t + \ln (1+u(t))-u''(t)]=0, \quad 0<t<1
\label{e13}\\
r_1 u(0) - r_2 u' (0)= r_3 u(1) + r_4 u'(1) = u''(0) = 0.  \label{e14}
\end{gather}
Let $f(t, s, r) = \frac{1}{3} [t+\ln(1+s) - r]$, it is
easy to check that $M=1, N=0 $ satisfy condition \eqref{e9}, \eqref{e10},
respectively, therefore,
$$
\alpha(t)=0, \quad \beta(t) = \int_0^1G(t, s)s\,ds
$$
are lower and upper solutions of \eqref{e13}--\eqref{e14}.
By Theorem \ref{thm3.1}, Problem \eqref{e13}--\eqref{e14} has a positive
solution $u^* $ such that $0 \leq u^*(t) \leq \int_0^1G(t, s)s\,ds$.
\end{example}


We now give some sufficient conditions with which there is at
least one solution to \eqref{e3}--\eqref{e4}. In the following we
assume  $f:[0, 1] \times R^3 \to R$ is continuous. It
is not difficult to see that  \eqref{e3}--\eqref{e4} is equivalent to
\begin{gather}
v'(t) = f(t, (Av)(t),(Bv)(t),-v(t)), \quad 0 <t<1 \label{e15}\\
v(0)=0,  \label{e16}
\end{gather}
 where $(Av)(t)= \int_0^1G(t,s)v(s)ds$, $(Bv)(t)= \int_t^1v(s)ds $.
Lemmas similar to Lemmas \ref{lem2.2} and \ref{lem2.3}
 can be obtained analogously
and so are omitted. An argument similar to the one in
Theorem \ref{thm3.1} provides the following result about
Problem \eqref{e3}--\eqref{e4}.

\begin{theorem} \label{thm3.2}
Suppose there are two constants $M \geq 0 \geq N$, $ M \geq |N| $
such that
\begin{gather}
f(t,s,l,r) \leq M, \quad \text{for } t \in [0, 1], \;
 \frac{s}{\eta}, 2l \in [0, M],\;  r \in [-M, 0], \label{e17}\\
f(t,s,l,r) \geq N, \quad \text{for } t \in [0, 1],\;
s \in [\eta N, 0], \; 2l, r \in[0, -N]. \label{e18}
\end{gather}
 If $f(t,s, l, r)$ is increasing in each of $s$ and $l$, then
\eqref{e3}--\eqref{e4} has a solution $u(t)$ such that
$\alpha(t) \leq u(t) \leq \beta (t)$,
$\beta''(t) \leq u''(t) \leq \alpha''(t)$, $t \in [0, 1] $, where
$\beta(t) = M\int_0^1G(t, s)s\,ds$, $\alpha(t) = N\int_0^1G(t, s)s\,ds$,
$\eta=1/3$.
\end{theorem}

\begin{example} \label{exa2}\rm
 Consider the  problem
\begin{gather}
u'''(t) + \frac{1}{4} [t+ e^{u(t)} + (u'(t))^2+u''(t)]=0, \quad
 0<t<1, \label{e19}\\
u(0) = u'(1) = u''(0) = 0.  \label{e20}
\end{gather}
We can check that $M=2$, $N=0 $ satisfy \eqref{e17}, \eqref{e18},
respectively, therefore,
$$
\alpha(t)=0, \quad \beta(t) = 2\int_0^1G(t, s)s\,ds
$$
are lower and upper solutions of  \eqref{e19}--\eqref{e20}.
By Theorem \ref{thm3.2}, Problem \eqref{e19}--\eqref{e20} has a positive
solution $u^* $ such that $0 \leq u^*(t) \leq 2\int_0^1G(t, s)s\,ds$.
\end{example}

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\end{document}