\documentclass[twoside]{article}

\usepackage{amssymb, amsmath}
\pagestyle{myheadings}
\setcounter{page}{47}

\markboth{\hfil A priori estimate for multi-point boundary-value problems\hfil}%
{\hfil Chaitan P. Gupta \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc 16th  Conference on Applied Mathematics, Univ. of Central Oklahoma},
\newline
Electronic Journal of Differential Equations, Conf. 07, 2001, pp. 47--59.
\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu (login: ftp)}
 \vspace{\bigskipamount} \\
%
  A new a priori estimate for multi-point boundary-value problems
%
\thanks{ {\em Mathematics Subject Classifications:} 34B10, 34B15, 34G20.
\hfil\break\indent
{\em Key words:} Three-point boundary-value problem, $m$-point boundary-value
 problem, \hfil\break\indent
 a-priori estimates, Leray-Schauder Continuation theorem,
 Caratheodory's conditions.
\hfil\break\indent
\copyright 2001 Southwest Texas State University. \hfil\break\indent
Published July 20, 2001.} }

\date{}
\author{ Chaitan P. Gupta }
\maketitle

\begin{abstract}
Let $f:[0,1]\times \mathbb{R}^2\to \mathbb{R}$ be a function satisfying
Caratheodory's conditions and $e(t)\in L^{1}[0,1]$. Let $0<\xi
_1<\xi_2<\dots <\xi_{m-2}<1$ and $a_i\in \mathbb{R}$ for
$i=1,2,\dots ,m-2$ be given. A priori estimates of the
form
$$
\|x\|_{\infty }\leq C\| x''\|_1, \quad \|x'\|_{\infty }\leq
C\|x''\|_1, $$
are needed to obtain the existence
of a solution for the multi-point bound\-ary-value problem
\begin{gather*}
x''(t)=f(t,x(t),x'(t))+e(t),\quad 0<t<1, \\
x(0)=0,\quad x(1)=\sum_{i=1}^{m-2}a_ix(\xi_i),
\end{gather*}
using Leray Schauder continuation theorem. The purpose of this
paper is to obtain a new a priori estimate of the form $\| x\|
_{\infty }\leq C\| x''\|_1$. This new estimate then enables us to
obtain a new existence theorem. Further, we obtain a new a priori estimate
of the form $\| x\|_{\infty }\leq C\| x''\|_1$
for the three-point boundary-value problem
\begin{gather*}
x''(t)=f(t,x(t),x'(t))+e(t),\quad 0<t<1, \\
x'(0)=0,\quad x(1)=\alpha x(\eta ),
\end{gather*}
where $\eta \in (0,1)$ and $\alpha \in \mathbb{R}$ are given.
The estimate obtained for the three-point boundary-value problem
turns out to be sharper than the one obtained by particularizing
the $m$-point boundary value estimate to the three-point case.
\end{abstract}

\newtheorem{theorem}{Theorem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}


\section{Introduction}

Let $f:[0,1]\times \mathbb{R}^2\to \mathbb{R}$ be a function satisfying
Caratheodory's conditions and $e(t)\in L^{1}[0,1]$. Let
$0<\xi_1<\xi_2<\dots <\xi_{m-2}<1$ and $a_i\in \mathbb{R}$
for $i=1,2,\dots ,m-2$ be given. Let us consider the
problem of existence of a solution for the multi-point boundary-value problem
\begin{equation}\begin{gathered}
x''(t)=f(t,x(t),x'(t))+e(t),0<t<1, \\
x(0)=0,x(1)=\sum_{i=1}^{m-2}a_ix(\xi_i). \end{gathered}\label{mbvp}
\end{equation}
In \cite {gt3} the author and Sergei Trofimchuk had studied this problem earlier
and obtained existence results using the Leray-Schauder continuation theorem.
Now, to apply the Leray-Schauder continuation theorem requires a priori
estimates of the form
\[
\|x\|_{\infty }\leq C\| x''\|_1, \quad
\|x'\|_{\infty }\leq C\| x''\|_1.
\]
For a function $x(t)\in W^{2,1}(0,1)$ with $x(0)=0$, $x(1)=%
\sum_{i=1}^{m-2}a_ix(\xi_i)$, and $\sum_{i=1}^{m-2}a_i\xi_i\neq 1$,
Gupta and Trofimchuk obtained the a priori estimate
\[
\| x'\|_{\infty }\leq \frac{1}{1-\tau }\| x''\|_1,
\]
where, $0\leq \tau <1$ is suitable constant defined by $a_i$,
and $\xi_i $, $i=1,2,\dots ,m-2$. Using, then the
estimate $\| x\|_{\infty }\leq \| x'\|_{\infty }$, for
functions $x(t)\in W^{2,1}(0,1)$ with $x(0)=0$, they obtained the
estimate
\[
\| x\|_{\infty }\leq \frac{1}{1-\tau }\| x''\|_1.
\]
The purpose of this paper is to obtain a new and sharper estimate
$\|
x\|_{\infty }\leq C\| x''\|_1$ for
$x(t)\in W^{2,1}(0,1)$ with $x(0)=0$, $x(1)=\sum_{i=1}^{m-2}a_ix(\xi_i)$,
and \break
$\sum_{i=1}^{m-2}a_i\xi_i\neq 1$. This new estimate then
enables us to obtain a new existence theorem for the above
boundary-value problem. Further, we obtain  a new a priori estimate of the
form $\| x\|_{\infty }\leq C\| x''\|_1$ for the three-point
boundary-value problem
\begin{equation}\begin{gathered}
x''(t)=f(t,x(t),x'(t))+e(t),\quad 0<t<1,\\
x'(0)=0,\quad x(1)=\alpha x(\eta ), \end{gathered} \label{3bp}
\end{equation}
where $\eta \in (0,1)$ and $\alpha \in \mathbb{R}$ are given.
The estimate obtained for the three-point boundary-value problem
turns out to be sharper than the one obtained by particularizing
the $m$-point boundary-value estimate to the three-point case.
These a priori estimates have been motivated by the results of
\cite{gt2}.

\section{A priori estimates}

We begin this section by first describing an estimate obtained by
Gupta and Trofimchuk. Let $a_i\in \mathbb{R}$, $\xi_i\in (0,1)$,
$i=1,2,\dots ,m-2$, $0<\xi_1<\xi_2<\dots
<\xi_{m-2}<1$, with $\sum_{i=1}^{m-2}a_i\xi_i\neq 1$, be
given. Let $x(t)\in W^{2,1}(0,1)$ be such that $x(0)=0$,
$x(1)=\sum_{i=1}^{m-2}a_ix(\xi_i)$.  Let us write the
condition $x(1)=\sum_{i=1}^{m-2}a_ix(\xi_i)$ in symmetric form
$\sum_{i=1}^{m-1}a_ix(\xi_i)=0$ by setting $a_{m-1}=-1$ and
$\xi_{m-1}=1$. Then the assumption $\sum_{i=1}^{m-2}a_i\xi_i\neq 1$
is equivalent to $\sum_{i=1}^{m-1}a_i\xi_i\neq 0$. Let us, define,
for $i$, $j=1,2,\dots ,m-1$,
\begin{eqnarray*}
\sigma_{ij} &=&a_i(\xi_i-\xi_{j}) \quad \text{for }i\neq j \,,\\
\sigma_{jj} &=&(\sum_{i=1}^{m-1}a_i)\xi_{j}\,.
\end{eqnarray*}
We observe that
\[
\sum_{i=1}^{m-1}\sigma_{ij}=\sum_{i=1}^{m-1}a_i\xi_i\neq 0\text{, for }%
j=1,2,\dots ,m-1\,.
\]
For $a\in \mathbb{R}$, setting $a_{+}=\max (a,0)$ and $a_{-}=\max
(-a,0)$ so that $a=a_{+}-a_{-}$, $|a|=a_{+}+a_{-}$, we see
that
\begin{equation}
\sum_{i=1}^{m-1}(\sigma_{ij})_{+}\neq \sum_{i=1}^{m-1}(\sigma_{ij})_{-}.
\label{eq0}
\end{equation}
We, next, define
\[
\sigma_{+}^{j}=\sum_{i=1}^{m-1}(\sigma_{ij})_{+},\sigma
_{-}^{j}=\sum_{i=1}^{m-1}(\sigma_{ij})_{-}\quad \text{for }j=1,2,\dots ,m-1\,,
\]
and
\begin{equation}
\tau =\min \{\frac{\sigma_{+}^{j}}{\sigma_{-}^{j}},\frac{\sigma
_{-}^{j}}{\sigma_{+}^{j}} : j=1,2,\dots,m-1\}.  \label{tau}
\end{equation}
We, note, that $0\leq \tau <1$ in view of (\ref{eq0}).

\begin{proposition} \label{Prop1}
 Let $a_i\in \mathbb{R}$, $\xi_i\in (0,1)$, $i=1,2,\dots ,m-2$,
$0<\xi_1<\xi_2<\dots <\xi_{m-2}<1$,
with $\sum_{i=1}^{m-2}a_i\xi_i\neq 1$, be given. Then for
$x(t)\in W^{2,1}(0,1)$ with $x(0)=0$,
$x(1)=\sum_{i=1}^{m-2}a_ix(\xi _i)$ we have
\begin{equation}
\| x\|_{\infty }\leq \frac{1}{1-\tau }\| x''\|_1, \label{GTest}
\end{equation}
where $\tau $ is as given in (\ref{tau}).
\end{proposition}

We refer the reader to \cite{gt3} for a proof of this
proposition.

\begin{theorem} \label{Th1}
 Let $a_i\in \mathbb{R}$, $\xi_i\in (0,1)$, $i=1,2,\dots ,m-2$,
$0<\xi_1<\xi_2<\dots <\xi_{m-2}<1$,
with $\sum_{i=1}^{m-2}a_i\xi_i\neq 1$, $\sum_{i=1}^{m-2}a_i\neq 1$, be
given. Then for $x(t)\in W^{2,1}(0,1)$ with $x(0)=0$, $x(1)=%
\sum_{i=1}^{m-2}a_ix(\xi_i)$ we have
\begin{equation}
\| x\|_{\infty }\leq C\| x''\|_1, \label{est1}
\end{equation}
where
\[
C=\min \{\frac{1}{1-\tau }\text{,  }C_1\},
\]
with $\tau $ as defined in (\ref{tau}),
\[
C_1=\max \{C_2,\frac{1}{1-\tau }\sum_{i=1}^{m-2}|\frac{%
a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}|\},
\]
and $C_2$ as defined below in (\ref{est4}).
\end{theorem}

\paragraph{Proof}
Let $\xi_{m-1}=1$, $a_{m-1}=-1$ so that the
condition $x(1)=\sum_{i=1}^{m-2}a_ix(\xi_i)$ $ $ may be
written in the symmetric form $\sum_{i=1}^{m-1}a_ix(\xi_i)=0$
and $\sum_{i=1}^{m-1}a_i\neq 0$. Since $x(t)\in W^{2,1}(0,1)$
there exists a
$c\in \lbrack 0,1]$ such that $\| x\|_{\infty }=|x(c)|$%
. We may assume that $x(c)>0$, by replacing $x(t)$ by $-x(t)$, if
necessary. Next, since $x(0)=0$, we see that $c\in (0,1]$. In
case, $c\in (0,1)$ we must have $x'(c)=0$. Applying, now,
the Taylor's formula with integral remainder after the second term
at each $\xi_i$, $i=1,2,\dots ,m-1$, to get
\begin{equation}
x(\xi_i)=x(c)+r_i,  \label{eq1}
\end{equation}
where
\begin{equation}
r_i=\int_{c}^{\xi_i}(\xi_i-s)x''(s)ds\leq 0, \label{eq2}
\end{equation}
$i=1,2,\dots ,m-1$. Multiplying the equation
(\ref{eq1}) by $a_i$, $i=1,2,\dots ,m-1$, and adding
the resulting equations we obtain
\begin{equation}
0=\sum_{i=1}^{m-1}a_ix(\xi
_i)=\sum_{i=1}^{m-1}a_ix(c)+\sum_{i=1}^{m-1}a_ir_i\text{.
} \label{eq3}
\end{equation}
Now, equations (\ref{eq2}), (\ref{eq3}) imply that
\begin{gather}
0<x(c)=-\frac{1}{\sum_{i=1}^{m-1}a_i}\sum_{i=1}^{m-1}a_ir_i=-%
\sum_{i=1}^{m-1}(\frac{a_i}{\sum_{i=1}^{m-1}a_i})\int_{c}^{\xi
_i}(\xi
_i-s)x''(s)ds  \nonumber \\
\leq
\sum_{i=1}^{m-1}(\frac{a_i}{\sum_{i=1}^{m-1}a_i})_{+}|\int_{c}^{\xi_i}(\xi_i-s)x''(s)ds|.
\label{est2}
\end{gather}
We, next, observe that
\[
|\int_{c}^{\xi_i}(\xi_i-s)x''(s)ds|\leq |\xi
_i-c|\int_{c}^{\xi_i}|x''(s)|ds
\leq |\xi_i-c|\int_0^{1}|x''(s)|ds,
\]
for $i=1,2,\dots ,m-1$. We thus see from (\ref{eq2})
that
\begin{multline}
\| x\|_{\infty }=x(c)\leq \sum_{i=1}^{m-1}(\frac{a_i}{%
\sum_{i=1}^{m-1}a_i})_{+}|\int_{c}^{\xi_i}(\xi
_i-s)x''(s)ds|  \\
\leq \sum_{i=1}^{m-1}(\frac{a_i}{\sum_{i=1}^{m-1}a_i})_{+}|\xi
_i-c|\int_0^{1}|x''(s)|ds  \\
\leq \max_{u\in \lbrack 0,1]}(\sum_{i=1}^{m-1}(\frac{a_i}{
\sum_{i=1}^{m-1}a_i})_{+}|\xi_i-u|)\int_0^{1}|x''(s)|ds.  \label{est3}
\end{multline}
Since, now,
$\sum_{i=1}^{m-1}(\frac{a_i}{\sum_{i=1}^{m-1}a_i})_{+}|\xi
_i-u|$ is a piecewise linear function, its maximum value is
attained at one of the points, $0$, $\xi_{j}$, $j=1,2,\dots ,m-1 $.
Accordingly, we get
\begin{align}
&\max_{u\in \lbrack 0,1]}(\sum_{i=1}^{m-1}(\frac{a_i}{\sum_{i=1}^{m-1}a_i}
)_{+}|\xi_i-u|)  \nonumber \\
&=\max \left\{
\begin{array}{c}
\sum_{i=1}^{m-1}\xi_i(\frac{a_i}{\sum_{i=1}^{m-1}a_i})_{+}, \\
\sum_{i=1,i\neq
j}^{m-1}(\frac{a_i}{\sum_{i=1}^{m-1}a_i})_{+}|\xi_i-\xi
_{j}|,j=1,2,\dots ,m-1,
\end{array}
\right\}  \label{est4} \\
&=\max \left\{
\begin{array}{c}
\sum_{i=1}^{m-2}\xi_i(\frac{a_i}{1-\sum_{i=1}^{m-2}a_i})_{-}+(\frac{1
}{1-\sum_{i=1}^{m-2}a_i})_{+}, \\
\sum_{i=1,i\neq j}^{m-2}(\frac{a_i}{1-\sum_{i=1}^{m-2}a_i})_{-}
|\xi_i-\xi_{j}|+(\frac{1}{1-\sum_{i=1}^{m-2}a_i})_{+}(1-\xi_{j}),  \\
j=1,2,\dots ,m-2, \\
\sum_{i=1}^{m-2}(\frac{a_i}{1-\sum_{i=1}^{m-2}a_i})_{-}(1-\xi_i)
\end{array}
\right\} \equiv C_2.  \nonumber
\end{align}
Accordingly, when $x(c)=\| x\|_{\infty }$ with $c\in (0,1)$ we see
that
\begin{equation}
\| x\|_{\infty }\leq C_2\| x''\|_1. \label{est5}
\end{equation}
Let, now, $c=1$ so that $\| x\|_{\infty }=x(1)$. We, then, see
that there exists a $\lambda_i$, for each $i=1,2,\dots ,m-2$, such that
\begin{equation}
x(1)-x(\xi_i)=(1-\xi_i)x'(\lambda_i). \label{eq4}
\end{equation}
It follows from equations (\ref{eq4}) that
\[
(\sum_{i=1}^{m-2}a_i-1)x(1) =\sum_{i=1}^{m-2}a_i(x(1)-x(\xi_i))
=\sum_{i=1}^{m-2}a_i(1-\xi_i)x'(\lambda_i).
\]
Accordingly, we get
\begin{eqnarray}
\|x\|_{\infty }&=&x(1)
=\sum_{i=1}^{m-2}\frac{a_i(1-\xi_i)}{\sum_{i=1}^{m-2}a_i-1}%
x'(\lambda_i)  \nonumber \\
&\leq &\sum_{i=1}^{m-2}|\frac{a_i(1-\xi_i)}{\sum_{i=1}^{m-2}a_i-1}%
|\| x'\|_{\infty }  \nonumber \\
&\leq &(\frac{1}{1-\tau }\sum_{i=1}^{m-2}|\frac{a_i(1-\xi_i)}{%
\sum_{i=1}^{m-2}a_i-1}|)\| x''\|_1.  \label{est6}
\end{eqnarray}
Thus from estimates (\ref{est5}), (\ref{est6}) we obtain
\begin{equation}
\|x\|_{\infty }\leq \max \{C_2,\frac{1}{1-\tau }
\sum_{i=1}^{m-2}|\frac{a_i(1-\xi_i)}{\sum_{i=1}^{m-2}a_i-1}|\}\| x''\|_1
\equiv C_1\| x''\|_1. \label{est7}
\end{equation}
The estimate (\ref{est1}) is now immediate since $\| x\|_{\infty
}\leq $ $\frac{1}{1-\tau }\| x''\|_1 $, from Proposition 1.  This
completes the proof of Theorem \ref{Th1}. \quad $\Box$

\begin{remark} \label{Remark1}\rm
Let $\eta \in (0,1)$, $\alpha \in \mathbb{R}$ with
$\alpha \eta \neq 1$ be given.  It was proved earlier by
Gupta and Trofimchuk for $x(t)\in W^{2,1}(0,1)$ with $x(0)=0$,
$x(1)=\alpha x(\eta )$ that
$$\begin{gathered}
\|x\|_{\infty }\leq \| x''\|
_1\quad \text{if }\alpha \leq 1, \\
\|x\|_{\infty }\leq \frac{1-\eta }{1-\alpha \eta }
\|x''\|_1\quad \text{if }\alpha \eta <1\text{ and }\alpha >1, \\
\|x\|_{\infty }\leq \frac{\alpha -1}{\alpha \eta -1}
\| x''\|_1\quad \text{if }\alpha >1 \text{ and }\alpha \eta >1,
\end{gathered}$$
so that
$$\begin{gathered}
\tau =0\quad \text{if }\alpha \leq 1, \\
\frac{1}{1-\tau } =\frac{1-\eta }{1-\alpha \eta }\quad \text{if }\alpha >1%
\text{ and }\alpha \eta <1, \\
\frac{1}{1-\tau } =\frac{\alpha -1}{\alpha \eta -1}\quad \text{if }\alpha >1
\text{ and }\alpha \eta >1\,.
\end{gathered}$$
\end{remark}

\begin{remark}\label{Remark2}\rm
Let us note that for $x(t)\in W^{2,1}(0,1)$ with
$x(0)=0$, $x(1)=\alpha x(\eta )$ the constant $C_2$ defined in
(\ref{est4}) is given by
\[
C_2=\max \{\eta (\frac{\alpha }{1-\alpha })_{-}+(\frac{1}{1-\alpha })_{+}
,(\frac{1}{1-\alpha })_{+}(1-\eta ),(\frac{\alpha }{1-\alpha
})_{-}(1-\eta )\}.
\]
It follows that
\[
C_2=\left\{
\begin{array}{ll}
\max \{\frac{1+|\alpha |\eta }{1+|\alpha |},\frac{%
|\alpha |(1-\eta )}{1+|\alpha |}\} &\text{for }\alpha \leq 0, \\
\frac{1}{1-\alpha } &\text{for }0\leq \alpha <1, \\
\max \{\frac{\alpha \eta }{\alpha -1},\frac{\alpha (1-\eta
)}{\alpha -1}\} &\text{for }\alpha >1\,.
\end{array}
\right.
\]
Next, we see from the definition of $C_1$ in (\ref{est7}) and
(\ref {Remark1}) that
\[
C_1=\left\{
\begin{array}{ll}
\max \{\frac{1+|\alpha |\eta }{1+|\alpha |},\frac{
|\alpha |(1-\eta )}{1+|\alpha |}\} &\text{for }\alpha \leq 0, \\[2pt]
\frac{1}{1-\alpha } &\text{for }0\leq \alpha <1, \\[2pt]
\max \{\frac{\alpha \eta }{\alpha -1},\frac{\alpha (1-\eta )^2}{
(\alpha -1)(1-\alpha \eta )}\}& \text{for }\alpha \eta <1
\text{ and }\alpha >1, \\[3pt]
\max \{\frac{\alpha \eta }{\alpha -1},\frac{\alpha (1-\eta )}{
(\alpha \eta -1)}\} &\text{for }\alpha \eta >1\text{ and }\alpha >1\,.
\end{array}
\right.
\]
Finally, we see that for $x(t)\in W^{2,1}(0,1)$ with $x(0)=0$,
$x(1)=\alpha x(\eta )$ we have
\begin{equation}
\| x\|_{\infty }\leq C\| x''\|_1, \label{est8}
\end{equation}
where $C=\min \{\frac{1}{1-\tau }$,  $C_1\}$ is given by
\[
C=\left\{
\begin{array}{ll}
\max \{\frac{1+|\alpha |\eta }{1+|\alpha |},\frac{
|\alpha |(1-\eta )}{1+|\alpha |}\} &\text{for }\alpha \leq 0, \\[2pt]
1 &\text{for }0\leq \alpha <1, \\[2pt]
\min \{\frac{1-\eta }{1-\alpha \eta },\max \{\frac{\alpha \eta }{
\alpha -1},\frac{\alpha (1-\eta )^2}{(\alpha -1)(1-\alpha \eta )}
\}\} &\text{for }\alpha \eta <1\text{ and }\alpha >1, \\[3pt]
\min \{\frac{\alpha -1}{\alpha \eta -1},\max \{\frac{\alpha \eta }{
\alpha -1},\frac{\alpha (1-\eta )}{(\alpha \eta -1)}\}\}
& \text{for }\alpha \eta >1\text{ and }\alpha >1.
\end{array}
\right.
\]
The following theorem gives a better estimate than (\ref{est8})
for an $x(t)\in W^{2,1}(0,1)$ with $x(0)=0$, $x(1)=\alpha x(\eta
)$.
\end{remark}

\begin{theorem}\label{Th2}
 Let $\alpha \in \mathbb{R}$ and $\eta \in (0,1)$ with $\alpha \neq 1$,
$\alpha \eta \neq 1$, be given. Then for $x(t)\in W^{2,1}(0,1)$ with $x(0)=0$,
$x(1)=\alpha x(\eta )$ we have
\[
\| x\|_{\infty }\leq M\|x''\|_1
\]
where
$$M=\left\{\begin{array}{ll}
\max \{\frac{1+|\alpha |\eta }{1+|\alpha |},
\frac{|\alpha |(1-\eta )}{1+|\alpha |}\} &\text{if }\alpha\leq -1. \\
\frac{1-\alpha \eta }{1-\alpha } &\text{if  }-1\leq \alpha <0, \\
1 &\text{if  }0\leq \alpha <1, \\
\max \{\frac{\eta }{2},\frac{\alpha (1-\eta )}{\alpha -1},
\frac{\alpha \eta (1-\eta )}{1-\alpha \eta }\} &\text{if  }\alpha>1
\text{ and }\alpha \eta <1, \\
\max \{\frac{\eta }{2},\frac{\alpha \eta -1}{\alpha -1},
\frac{\alpha \eta (1-\eta )}{\alpha \eta -1}\} &\text{if  }\alpha >1
\text{ and }\alpha \eta >1\,.
\end{array}\right.
$$
\end{theorem}

\paragraph{Proof}
For $\alpha \leq 0$ we see from Theorem \ref{Th1}
and remark \ref{Remark2} that
\[
M=\max \{\frac{1+|\alpha |\eta }{1+|\alpha |},
\frac{|\alpha |(1-\eta )}{1+|\alpha |}\}.
\]
This implies, in particular, for $\alpha \leq -1$ that $M=\max \{\frac{
1+|\alpha |\eta }{1+|\alpha |}$, $\frac{|\alpha |(1-\eta )}{1+|\alpha |}\}$.
Note that for $-1\leq \alpha <0$,
\[
\frac{1-\alpha \eta }{1-\alpha }=\frac{1+\eta |\alpha |}{1+|\alpha |}\geq
\frac{|\alpha |(1+\eta )}{1+|\alpha |}>%
\frac{|\alpha |(1-\eta )}{1+|\alpha |}
\]
and so we again see from Theorem \ref{Th1} and Remark
\ref{Remark2} that
\[
M=\left\{\begin{array}{ll}
\frac{1-\alpha \eta }{1-\alpha }&\text{if }-1\leq \alpha <0\\
1 &\text{if } 0\leq \alpha <1\,.\end{array}\right.
\]

Finally, we consider the case $\alpha >1$. Let $ x(\eta )=z$
so that $x(1)=\alpha z$. We may assume without loss of
generality that $z\geq 0$, replacing $x(t)$ by $-x(t)$ if
necessary. Suppose, now, $\| x\|_{\infty }=1$ so that there exists
a $c\in \lbrack 0,1]$ such that either $x(c)=1$ or $x(c)=-1$. We
consider all possible cases of the location for $c$.

(i) Suppose that $c\in (0,\eta ] $ and $x(c)=1$. Then $ x'(c)=0$, $c\neq \eta $.
Now, by mean value theorem there exist
$\nu_1\in \lbrack c,\eta ], \nu_2\in \lbrack \eta ,1]$ such that
\[
x'(\nu_1)=\frac{x(\eta )-x(c)}{\eta -c}=-\frac{1-z}{\eta -c},\quad
x'(\nu_2)=\frac{x(1)-x(\eta )}{1-\eta }=\frac{\alpha z-z}{1-\eta }.
\]
We note that $x'(\nu_1)\leq 0$, $x'(\nu_2)\geq 0$ since $0\leq z\leq 1$
and $\alpha >1$. It follows that
\begin{align*}
\int_0^{1}|x''(s)|ds \geq& |\int_{c}^{\nu_1}x''(s)ds|+|\int_{\nu
_1}^{\nu_2}x''(s)ds|\\
=&2|x'(\nu_1)|+x'(\nu_2)
=2\frac{1-z}{\eta -c}+\frac{\alpha z-z}{1-\eta }\\
\geq& \min_{c\in\lbrack 0,\eta ),z\in \lbrack 0,\frac{1}{\alpha }]}
\{2\frac{1-z}{\eta -c} +\frac{\alpha z-z}{1-\eta }\}\\
\geq& \min_{c\in \lbrack 0,\eta )}\{\frac{2}{\eta -c},\frac{2(\alpha -1)}
{\alpha (\eta -c)}+\frac{\alpha -1}{\alpha (1-\eta )}\}\\
\geq& \min\{\frac{2}{\eta },\frac{\alpha -1}{\alpha (1-\eta )}\}.
\end{align*}

(ii) Let, now, $c\in (0,\eta ]$, $ x(c)=-1$. Then since $x'(c)=0$,
$c\neq \eta $, we again see from mean value theorem that
there exist $\nu_{3}\in \lbrack c,\eta ], \nu _{4}\in \lbrack \eta,1]$
such that
\[
 x'(\nu_{3})=\frac{x(\eta )-x(c)}{\eta -c}=\frac{z+1}{\eta -c},
 \quad x'(\nu_{4})=\frac{x(1)-x(\eta )}{1-\eta }=\frac{\alpha z-z}{1-\eta }.
\]
Again we note that $x'(\nu_{3})>0$, $x'(\nu_{4})\geq 0$ since $0\leq z\leq 1$
and $\alpha >1$ and we have
\begin{equation}\begin{aligned}
\int_0^{1}|x''(s)|ds\geq& |\int_{c}^{\nu_{3}}x''(s)ds|
+|\int_{\nu_{3}}^{\nu_{4}}x''(s)ds|\\
=&x'(\nu_{3})+|x'(\nu_{4})-x'(\nu_{3})|
=\frac{1+z}{\eta -c}+|\frac{\alpha z-z}{1-\eta }-\frac{1+z}{\eta-c}|.
\end{aligned} \label{EQ7}
 \end{equation}
Let $F(z,c)=\frac{1+z}{\eta -c}+|\frac{\alpha z-z}{1-\eta }-\frac{1+z}{
\eta -c}|$. We need to estimate $\min_{c\in \lbrack 0,\eta ),z\in \lbrack 0,
\frac{1}{\alpha }]}F(z,c)$. We note that
\begin{gather*}
F(0,c) =\frac{2}{\eta -c}\geq \frac{2}{\eta }\quad \text{for }c\in \lbrack
0,\eta ), \\
F(\frac{1}{\alpha },c) =\frac{\alpha +1}{\alpha (\eta
-c)}+|\frac{\alpha -1}{\alpha (1-\eta )}-\frac{\alpha +1}{\alpha
(\eta -c)}|\geq \frac{\alpha -1}{\alpha (1-\eta )}\quad
\text{for }c\in \lbrack 0,\eta ).
\end{gather*}
Let $z_0$ be such that  $\frac{\alpha z_0-z_0}{1-\eta }
-\frac{1+z_0}{\eta -c}=0$ so that $z_0=\frac{1-\eta }{\alpha \eta
-1-c(\alpha -1)}$. It is easy to see that $z_0\in \lbrack
0,\frac{1}{\alpha }]$ if $\eta >\frac{\alpha +1}{2\alpha }$  and
$c\in (0,\frac{2\alpha \eta
-\alpha -1}{\alpha -1})$. In this case we get $F(z_0,c)=\frac{\alpha -1}{%
\alpha \eta -1-c(\alpha -1)}\geq \frac{\alpha -1}{\alpha \eta
-1}$.
Accordingly we see that $F(z,c)\geq \min \{\frac{2}{\eta },\frac{\alpha -1%
}{\alpha (1-\eta )}\}$ if $\alpha \eta \leq 1$ and $F(z,c)\geq \min \{\frac{%
2}{\eta },\frac{\alpha -1}{\alpha (1-\eta )},\frac{\alpha
-1}{\alpha \eta -1}\}$ if $\alpha \eta >1$. We thus have from
(\ref{EQ7}) that
\begin{align*}
\int_0^{1}|x''(s)|ds\geq &|\int_{c}^{\nu
_{3}}x''(s)ds|+|\int_{\nu_{3}}^{\nu_{4}}x''(s)ds|=x'(\nu_{3})+|x'(\nu_{4})
-x'(\nu_{3})| \\
=&\frac{1+z}{\eta -c}+|\frac{\alpha z-z}{1-\eta }-\frac{1+z}{\eta -c}| \\
\geq& \left\{
\begin{array}{cc}
\min \{\frac{2}{\eta },\frac{\alpha -1}{\alpha (1-\eta )}\}, &
\text{if }\alpha \eta \leq 1, \\
\min \{\frac{2}{\eta },\frac{\alpha -1}{\alpha (1-\eta )},\frac{\alpha -1%
}{\alpha \eta -1}\}, & \text{ if }\alpha \eta >1.
\end{array}\right.
\end{align*}

(iii) Next, suppose that $c\in (\eta ,1)$, $ x(c)=1$. Again,
$x'(c)=0$ and we have from mean value theorem that there
exist $\nu_{5}\in \lbrack \eta ,c], \nu_{6}\in \lbrack c,1]$ such
that
\[
 x'(\nu_{5})=\frac{x(c)-x(\eta )}{c-\eta
}=\frac{1-z}{c-\eta }, \quad
x'(\nu_{6})=\frac{x(1)-x(c)}{1-c}=\frac{\alpha z-1}{1-c}.
\]
Note that $x'(\nu_{5})\geq 0$, $x'(\nu_{6})\leq
0$ since $x(1)=\alpha z\leq 1$. Accordingly, we obtain
\begin{equation}\begin{aligned}
\int_0^{1}|x''(s)|ds\geq &|\int_0^{\nu
_{5}}x''(s)ds|+|\int_{\nu_{5}}^{\nu_{6}}x''(s)ds| \\
=& x'(\nu_{5})+|x'(\nu_{6})-x'(\nu
_{5})|=2x'(\nu_{5})+|x'(\nu_{6})|  \\
=& 2\frac{1-z}{c-\eta }+\frac{1-\alpha z}{1-c}\geq \frac{2(\alpha
-1)}{\alpha (1-\eta )},\quad \text{since } 0\leq z\leq \frac{1}{\alpha}.
\end{aligned}\label{eq6}
\end{equation}

(iv) Next, suppose that $c\in (\eta ,1)$, $ x(c)=-1$. Again,
$x'(c)=0$ and we have from mean value theorem that there
exist $\nu_{7}\in \lbrack \eta ,c], \nu_{8}\in \lbrack c,1]$ such
that
\[
 x'(\nu_{7})=\frac{x(c)-x(\eta )}{c-\eta }=\frac{-1-z}{c-\eta }, \quad
x'(\nu_{8})=\frac{x(1)-x(c)}{1-c}=\frac{\alpha z+1}{1-c}.
\]
Note that $x'(\nu_{7})\leq 0$, $x'(\nu_{8})\geq0$. Accordingly, we obtain
\begin{align*}
\int_0^{1}|x''(s)|ds\geq &|\int_0^{\nu_{7}}x''(s)ds|
+|\int_{\nu_{7}}^{\nu_{8}}x''(s)ds| \\
=&|x'(\nu_{7})|+|x'(\nu_{8})-x'(\nu_{7})|=2|x'(\nu_{7})|+x'(\nu_{8}) \\
=&2\frac{1+z}{c-\eta }+\frac{1+\alpha z}{1-c}\geq \frac{2}{c-\eta }
+\frac{1}{1-c}\\
\geq& \frac{2}{1-\eta }\geq \frac{2(\alpha -1)}{\alpha (1-\eta)}.
\end{align*}

(v) Finally suppose that $c=1$, so that $x(1)=1=\alpha z$. We then
have that there exists a $\nu_{9}\in (\eta ,1)$ such that
\[
 x'(\nu_{9})=\frac{x(1)-x(\eta )}{1-\eta }
 =\frac{1-\frac{1}{\alpha }}{1-\eta }=\frac{\alpha -1}{\alpha (1-\eta )}.
\]
Also, there exists a $\nu_{10}\in (0,\eta )$ such that
\[
x'(\nu_{10})=\frac{x(\eta )-x(0)}{\eta -0}=\frac{1}{\alpha \eta }.
\]
Thus
\begin{eqnarray*}
\int_0^{1}|x''(s)|ds &\geq &|\int_{\upsilon_{10}}^{\nu
_{9}}x''(s)ds|=|x'(\nu_{9})-x'(\nu
_{10})|\\
&=&|\frac{1-\frac{1}{\alpha }}{1-\eta }-\frac{1}{\alpha \eta
}|=|\frac{\alpha \eta -1}{\alpha \eta (1-\eta )}|.
\end{eqnarray*}
We thus see from (i), (ii), (iii), (iv) and (v) that for $\alpha
>1$, $\| x\|_{\infty }\leq M\|x''\|_1$
with
$$M=\left\{
\begin{array}{ll}
\max \{\frac{\eta }{2},\frac{\alpha (1-\eta )}{\alpha -1},
\frac{\alpha \eta (1-\eta )}{1-\alpha \eta }\}
&\text{if }\alpha \eta \leq 1, \\[3pt]
\max \{\frac{\eta }{2},\frac{\alpha \eta -1}{\alpha -1},
\frac{\alpha \eta (1-\eta )}{\alpha \eta -1}\} &\text{if  }\alpha \eta >1,
\end{array}\right.
$$
since for $\alpha >1$, $\alpha \eta >1$, $\frac{\alpha
\eta (1-\eta )}{\alpha \eta -1}>\frac{\alpha (1-\eta )}{\alpha
-1}$. This completes the present proof. \quad$\Box$

\begin{remark}\rm
 Let $\alpha =4$ and $\eta =\frac{1}{2}$. Let us consider the estimate
\begin{equation}
\| x\|_{\infty }\leq C\| x''\|_1, \label{est9}
\end{equation}
for $x(t)\in W^{2,1}(0,1)$ with $x(0)=0$, $x(1)=4x(\frac{1}{2})$.
Now, the function
\begin{equation}
\varphi (t)=\left\{
\begin{array}{cc}
2t^{3}, & \text{ for }t\in \lbrack 0,\frac{1}{2}], \\
\frac{3t-1}{2}, & \text{ for }t\in \lbrack \frac{1}{2},1],
\end{array}
\right.  \label{phi}
\end{equation}
is such that $\varphi (t)\in W^{2,1}(0,1)$ with $\varphi (0)=0$
and $\varphi (1)=4\varphi (\frac{1}{2})$. Moreover, $\| \varphi \|
_{\infty }=1$ and $\| \varphi ''\|_1=\frac{3}{2}$. It follows that
$C\geq \frac{2}{3}$ in (\ref{est9}). Now, Proposition \ref {Prop1}
and Remark \ref{Remark1} give $C=3$ in (\ref{est9}); while Theorem
\ref{Th1} and Remark \ref{Remark2} give $C=2$ in (\ref{est9}); and
Theorem \ref{Th2} gives $C=1$ in (\ref{est9}). This shows that
Theorem \ref{Th2} gives the best estimate $\| x\| _{\infty }\leq
\| x''\|_1$ for $x(t)\in W^{2,1}(0,1)$ with $x(0)=0$,
$x(1)=4x(\frac{1}{2})$. However, the function $\varphi (t)$
defined in (\ref {phi}) indicates that it may be possible to
improve $C$ in (\ref{est9}). This question remains open at this
time.
\end{remark}

To explore this further we introduce the notion of {\it
approximate best constant} in the following.

\paragraph{Definition}
$B\in \mathbb{R}$ is called ``approximate best constant'' if for every
$\varepsilon >0$ there exists an $\alpha \in \mathbb{R}$ and an $\eta \in
(0,1)$ such that (i) for every $x(t)\in W^{2,1}(0,1)$ with
$x(0)=0$, $x(1)=\alpha x(\eta )$, $\| x\|_{\infty }\leq
(B+\varepsilon )\| x''\|_1$; (ii) there exists a function $\phi
(t)\in W^{2,1}(0,1)$ with $\phi (0)=0$, $\phi (1)=\alpha \phi
(\eta )$, and $\| \phi \|_{\infty }>B\| \phi ''\|_1$.


\begin{theorem} \label{Th3}
For every $k>1$, $1-\frac{1}{k}$ is an approximate best constant.
\end{theorem}

\paragraph{Proof}
For each integer $n>2$, consider the function $\phi_{kn}(t)\in W^{2,1}(0,1)$
defined by
\[
\phi_{kn}(t)=\left\{
\begin{array}{cc}
t^{n}, & \text{ for }t\in \lbrack 0,\frac{1}{k}], \\
\frac{nt}{k^{n-1}}-\frac{n-1}{k^{n}}, & \text{ for }t\in \lbrack
\frac{1}{k},1].
\end{array}
\right.
\]
It is easy to see that $\phi_{kn}(t)\in W^{2,1}(0,1)$, with $\phi
_{kn}(0)=0 $, $\phi_{kn}(1)=\alpha_{kn}\phi_{kn}(\frac{1}{k})$,
where $\alpha_{kn}=n(k-1)+1$, and
\[
\| \phi_{kn}''\|_1=\frac{n}{k^{n-1}}, \quad
\| \phi_{kn}\|_{\infty }=\phi_{kn}(1)=\frac{n(k-1)+1}{%
k^{n}},
\]
so that
\begin{equation}
\| \phi_{kn}\|_{\infty }=\frac{n(k-1)+1}{nk}\| \phi_{kn}''\|_1.
\label{eq7}
\end{equation}
Now, since $\alpha_{kn}\cdot \frac{1}{k}=\frac{n(k-1)+1}{k}=n-\frac{n-1}{%
k}>1$ for $n>2$, we obtain using Theorem \ref{Th2} the
estimate
\begin{equation}\begin{gathered}
\| x\|_{\infty }\leq \frac{n(k-1)+1}{k(n-1)}\| x''\|_1\quad
\text{for }x(t)\in W^{2,1}(0,1) \\
x(0)=0,\quad x(1)=\alpha_{kn}x(\frac{1}{k}).
\end{gathered}\label{est10}
\end{equation}
Let us set
$B_{kn}=\frac{n(k-1)+1}{nk}=1-\frac{1}{k}+\frac{1}{nk}$,
$M_{kn}=\frac{n(k-1)+1}{k(n-1)}=1-\frac{1}{k}+\frac{1}{n-1}$.
We notice that
\[
M_{kn}-B_{kn}=\frac{1}{n-1}-\frac{1}{nk}=\frac{n(k-1)+1}{n(n-1)k}>0,
\]
so that $M_{kn}-B_{kn}>0$.  Also, we note that
\[
\lim_{n\to \infty }B_{kn}=\lim_{n\to \infty }M_{kn}=1-\frac{1}{k}.
\]
Let, now, $\varepsilon >0$ be given.  Choose, $n_0$ such that
$M_{kn_0}<1-\frac{1}{k}+\varepsilon $. It, now, follows from
(\ref{est10}) and (\ref{eq7}) that
\[\begin{gathered}
\| x\|_{\infty }\leq (1-\frac{1}{k}+\varepsilon )\|
x''\|_1\quad \text{for }x(t)\in W^{2,1}(0,1) \\
x(0)=0,\quad x(1)=\alpha_{kn_0}x(\frac{1}{k}),
\end{gathered}\]
and
\[
\| \phi_{kn_0}\|_{\infty }=(1-\frac{1}{k}+\frac{1}{n_0k})
\| \phi_{kn}''\|_1>(1-\frac{1}{k})\| \phi_{kn}''\|_1.
\]
This completes the proof of the Theorem. \quad $\Box$

\begin{remark}\rm
We note that $\lim_{k\to \infty }(1-\frac{1}{k})=1$.  In view
of this, it may be conjectured that $1$ may be a best constant in
the sense that there exists an $\alpha \in \mathbb{R}$ and an $\eta \in
(0,1)$ such that for $x(t)\in W^{2,1}(0,1)$ with $x(0)=0$,
$x(1)=\alpha x(\eta )$ one has the estimate
\[
\| x\|_{\infty }\leq \| x''\|_1.
\]
However, since $\lim_{k\to \infty }\alpha_{kn}=\infty $ and
$\lim_{k\to \infty }\frac{1}{k}=0$, it is not clear if such
$\alpha \in \mathbb{R}$ and an $\eta \in (0,1)$ exist.
\end{remark}

\section{Existence theroems}

We state below the existence theorems one obtains using the a
priori estimates obtained above.  We omit the proof of these
theorems as they are similar to the corresponding theorems in
\cite{gt3}.

\begin{theorem}
 Let $f:[0,1]\times \mathbb{R}^2\to \mathbb{R}$ be a function satisfying
Caratheodory's conditions. Assume that there exist functions
$p(t)$, $q(t)$, $r(t)$ in $L^{1}(0,1)$ such that
\[
\left| f(t,x_1,x_2)\right| \leq p(t)\left| x_1\right| +q(t)\left|
x_2\right| +r(t)
\]
for a.e. $t\in \lbrack 0,1]$ and all $(x_1,x_2)\in \mathbb{R}^2$. Let
$a_i\in \mathbb{R}$, $\xi_i\in (0,1)$, $i=1, 2,\dots, m-2$,
$0<\xi_1<\xi_2<\dots <\xi_{m-2}<1$ with
$\sum_{i=1}^{m-2}a_i\xi_i\neq 1$ and $\sum_{i=1}^{m-2}a_i\neq 1$, be given.
Then the multi-point boundary-value problem
\[\begin{gathered}
x''(t) =f(t,x(t),x'(t))+e(t),\quad 0<t<1,\\
x(0) =0,\quad x(1)=\sum_{i=1}^{m-2}a_ix(\xi_i).
\end{gathered}\]
has at least one solution in $C^{1}[0,1]$ provided
\[
C\left\| p(t)\right\|_1+\frac{1}{1-\tau }\left\| q(t)\right\|_1<1,
\]
where $C$ is as given in Theorem \ref{Th1} and $\tau $ as given in
Proposition \ref{Prop1}.
\end{theorem}

\begin{theorem}
Let $f:[0,1]\times \mathbb{R}^2\mapsto \mathbb{R}$ be a function satisfying
Caratheodory's conditions. Assume that there exist functions
$p(t)$, $q(t)$, $r(t)$ such that the functions $p(t)$, $q(t)$,
$r(t)$ are in $L^{1}(0,1)$ and
\[
\left| f(t,x_1,x_2)\right| \leq p(t)\left| x_1\right| +q(t)\left|
x_2\right| +r(t)
\]
for a.e. $t\in \lbrack 0,1]$ and all $(x_1,x_2)\in \mathbb{R}^2$. Let
$\alpha \in \mathbb{R}$, $\eta \in (0,1)$, $\alpha \neq 1$, and $\alpha
\eta \neq 1$ be given. Then, the three-point boundary-value problem
\[\begin{gathered}
x''(t) =f(t,x(t),x'(t))+e(t),\quad 0<t<1,\\
x(0) =0,\quad x(1)=\alpha x(\eta ).
\end{gathered}\]
has at least one solution in $C^{1}[0,1]$ provided
\[
M\left\| p(t)\right\|_1+\frac{1}{1-\tau }\left\| q(t)\right\|_1<1.
\]
where $M$ is as given in Theorem \ref{Th2} and $\tau $ as given in
Proposition \ref{Prop1}.
\end{theorem}

\begin{thebibliography}{9}

\bibitem{gt2}  C. P. Gupta, S. I. Trofimchuk, {\it a priori Estimates for
the Existence of a Solution for a Multi-Point Boundary Value
Problem}, J. of Inequalities and Applications, 5(2000) p. 351-365.

\bibitem{gt3}  C. P. Gupta, S. I. Trofimchuk, {\it Solvability of a
multi-point boundary-value problem and related a priori estimates},
 Canadian Applied Mathematics Quarterly, Vol. 6 (1998) pp. 45-60.
\end{thebibliography}

\noindent\textsc{Chaitan P. Gupta }\\
Department of Mathematics\\
University of Nevada, Reno\\
Reno, NV 89557 USA\\
email: gupta@unr.edu  chaitang@hotmail.com

\end{document}