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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 64, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/64\hfil Positive solutions]
{Positive solutions for a nonlocal multi-point boundary-value problem
of fractional and second order}

\author[A. M. A. El-Sayed, E. O. Bin-Taher \hfil EJDE-2013/64\hfilneg]
{Ahmed M. A. El-Sayed, Ebtisam O. Bin-Taher}  % in alphabetical order

\address{Ahmed M. A. El-Sayed \newline
Faculty of Science, Alexandria University, Alexandria, Egypt}
\email{amasayed@hotmail.com}

\address{Ebtisam O. Bin-Taher \newline
Faculty of Science, Hadhramout Univeristy of Science and
Technology, Hadhramout, Yemen}
\email{ebtsamsam@yahoo.com}

\thanks{Submitted March 19, 2012. Published March 5, 2013.}
\subjclass[2000]{34B10, 26A33}
\keywords{Fractional calculus; boundary value problem; nonlocal condition;
\hfill\break\indent  integral condition; positive solution}

\begin{abstract}
 In this article we study the existence of positive solutions for
 the nonlocal multi-point boundary-value problem
 \begin{gather*}
 u''(t)+f(t, ^{c}D^{\alpha}u(t))=0, \quad   \alpha \in(0, 1), \text{ a.e. }
 t\in(0, 1), \\
  u(0)=0, \quad u(1)=\sum_{k=1}^m a_k u(\tau_k), \quad
 \tau_k\in(a, b)\subset (0, 1).
 \end{gather*}
 We also consider the corresponding integral condition, and the
 two special cases  $\alpha = 0 $ and $ \alpha = 1$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

Problems with non-local conditions have been extensively studied
by several authors in the previous two decades; see for example
\cite{B0}-\cite{B2},  \cite{ee}-\cite{sk},
and the references therein.
In this work we show the existence of at least one solution for the
 nonlocal multi-point boundary-value problem consisting of
 second and fractional-orders differential equation
\begin{equation}\label{e1.1}
 u''(t)+f(t, ^{c}D^{\alpha}u(t))=0, \quad \alpha \in(0, 1), \text{ a.e. }
   t\in(0, 1)
\end{equation}
with the nonlocal conditions
\begin{equation}\label{e1.2}
 u(0)=0, \quad u(1)=\sum_{k=1}^m a_k u(\tau_k), \quad
 \tau_k\in(a, b)\subset (0, 1).
\end{equation}
Also we deduce the same results for the two differential equations
\begin{equation}\label{*}
u''(t)+f(t, u(t))=0,  \quad   \text{a.e. }   t\in(0, 1)
\end{equation}
and
\begin{equation}\label{**}%
u''(t)+f(t, u'(t))=0,  \quad    \text{a.e. }   t\in(0, 1).
\end{equation}
with the nonlocal conditions \eqref{e1.2}.
Also we study  problems \eqref{e1.1}, \eqref{*} and \eqref{**} with an
integral condition.

\section{Preliminaries}

Let $ L^1(I) $ denote the class of Lebesgue integrable functions on
the interval $ I = [a,b]$,where $0\leq a<b<\infty$ and let $ \Gamma(.) $ denote
 the gamma function.

\begin{definition}[\cite{P2}]\label{def1.1.1}\rm
The fractional-order integral of the function $f\in L^1[a,b]$ of
order $\beta > 0$ is defined by
\[
I_a^\beta f(t) = \int_a^t \frac{(t - s)^{\beta - 1}}{\Gamma(\beta)} f(s) ds,
\]
\end{definition}

\begin{definition}[\cite{P1,P2}]\label{def1.1.2} \rm
The Caputo fractional-order derivative of order
$\alpha \in (0,1]$ of the absolutely continuous function $f(t)$ is
defined by
\[
D_a^\alpha f(t)=I_a^{1 - \alpha} \frac{d}{dt} f(t)
=\int_a^t \frac{(t - s)^{-\alpha}}{\Gamma(1 - \alpha)} \frac{d}{ds} f(s) ds.
\]
\end{definition}

\begin{theorem}[Schauder fixed point theorem \cite{KD}]\label{thm13}
Let $E$ be a Banach space and $Q$ be a convex subset of
$E$, and $T:Q\to Q$ a compact, continuous map.
 Then $T$ has at least one fixed point in $Q$.
\end{theorem}

\begin{theorem}[Kolmogorov compactness criterion \cite{JD}]\label{thm17}
Let $\Omega \subseteq L^p (0,1)$, $1 \leq p < \infty$. If
\begin{itemize}
\item[(i)] $\Omega$ is bounded in $L^p (0,1)$, and
\item[(ii)] $u_h \to u$ as $h \to 0$ uniformly with respect to
$u \in \Omega$, then $\Omega$ is relatively compact in $L^p (0,1)$,
where
\[
u_h(t) = \frac{1}{h} \int_t^{t+h} u(s) ds.
\]
\end{itemize}
\end{theorem}

\section{Main results}

Consider the fractional-order functional integral equation
\begin{equation}\label{equa3.1}
\begin{aligned}
y(t)&=f\Big(t,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)}
\big\{A\int_0^{1}(1-s) y(s) ds-A\sum_{k=1}^ma_k
\int_0^{\tau_k}(\tau_k-s)y(s) ds\big\}\\
&\quad -\int_0^t\frac{(t-s)^{1-\alpha}}{\Gamma(2-\alpha)}y(s) ds\Big).
\end{aligned}
\end{equation}

 The function $y$ is called a solution of the fractional-order functional
 integral equation \eqref{equa3.1}, if $y$ belongs to $L^{1}[0,1]$ and
 satisfies  \eqref{equa3.1}.

We consider the following assumptions:
\begin{itemize}
\item[(H1)] $f:[0,1]\times R\to R^{+}$ be a function with the
 following properties:
\begin{itemize}
\item[(a)] $u\to f(t,u)$ is continuous for almost all $t \in [0,1]$,
\item[(b)] $t\to f(t,u)$ is measurable for all $u\in R$,
\end{itemize}

\item[(H2)]there exists an integrable function $ a\in L^1[0,1]$ and
constant $b,$ such that
\[
|f(t,u)| \leq a(t)+b |u|,  \textrm{a.e}  t\in[0,1],
\]
\end{itemize}

\begin{theorem} \label{thm3.1}  % Theorem 3.1
Let the {\rm (H1), (H2)} be satisfied. If
\begin{equation}
 B=\frac{b}{\Gamma(3-\alpha)} < 1,
\end{equation}
then  \eqref{equa3.1}  has at least one positive solution $y\in L^1[0,1]$.
\end{theorem}

\begin{proof}
Define the operator $T$ associated with  \eqref{equa3.1} by
\begin{align*}
Ty(t)&=f\Big(t,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)}
 \big\{A\int_0^{1}(1-s) y(s) ds
 - A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds\big\} \\
&\quad - \int_0^t\frac{(t-s)^{1-\alpha}}{\Gamma(2-\alpha)}y(s) ds\Big)
\end{align*}
Let $Q^{+}_{r} = \{y\in R^{+}:\|y\|<r, r>0\}$,
\[
r=\frac{\|a\|}{1 - B(1+A+A\sum_{k=1}^m a_k)}.
\]
Let $y$ be an arbitrary element in $Q^{+}_{r}$, then from assumptions
 (H1) and (H2), we obtain
\begin{align*}
\|Ty\|_{L^1}
&=\int_0^1 |Ty(t)| dt\\
&=\int_0^1 \Big|f\Big(t,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)}
\big\{A\int_0^{1}(1-s) y(s) ds
 - A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds\big\} \\
&\quad - I^{2-{\alpha}}y(t)\Big)\Big|dt
\\
&\leq \int_0^1|a(t)|dt+b \int_0^{1}
 \frac{t^{1-\alpha}}{\Gamma(2-\alpha)}\,dt
 \Big\{A\int_0^{1}(1-s) |y(s)| ds\\
&\quad +A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) |y(s)| ds\Big\}
  +b\int_0^1\int_0^t\frac{(t-s)^{1-\alpha}}{\Gamma(2-\alpha)}|y(s)| \,ds\,dt
  \\
&=\|a\|_{L^1}+b \int_0^{1} \frac{t^{1-\alpha}}{\Gamma(2-\alpha)}\,dt
 \Big\{A\int_0^{1} |y(s)| ds+A\sum_{k=1}^m a_k\int_0^{1} |y(s)| ds\Big\}
 \\
&\quad+ b\int_0^1\int_s^1\frac{(t-s)^{1-\alpha}}{\Gamma(2-\alpha)}dt|y(s)| ds
\\
&\leq \|a\|_{L^1}+b (A+A\sum_{k=1}^m a_k)\|y\|_{L^1}
 \int_0^{1}  \frac{t^{1-\alpha}}{\Gamma(2-\alpha)}dt\\
&\quad +b\int_0^1\frac{(t-s)^{2-\alpha}}{(2-\alpha)
 \Gamma(2-\alpha)}|_{s}^{1}|y(s)| \,ds
\\
&\leq \|a\|_{L^1}+\frac{b (A+A\sum_{k=1}^m a_k)}{\Gamma(3-\alpha)}
 \|y\|_{L^1}+b\int_0^1\frac{1}{\Gamma(3-\alpha)}|y(s)| \,ds\\
&\leq \|a\|_{L^1}+\frac{b (A+A\sum_{k=1}^m a_k)}{\Gamma(3-\alpha)}
 \|y\|_{L^1}+\frac{b}{\Gamma(3-\alpha)}\|y\|_{L^1}\\
&\leq \|a\|_{L^1}+\frac{b (1+A+A\sum_{k=1}^m a_k)}{\Gamma(3-\alpha)}
 \|y\|_{L^1}= r\,,
\end{align*}
which implies that the operator $T$ maps $Q^{+}_{r}$ into it self.
Assumption (H1) implies that $T$ is continuous.

Now let $\Omega$ be a bounded subset of
$Q^{+}_{r}$, then $T(\Omega)$ is bounded in $L^1[0,1]$; i.e., condition
(i) of Theorem \ref{thm17} is satisfied.
Let $y\in \Omega$. Then
\begin{align*}
\|(Ty)_h-Ty\|
&=\int_0^1|(Ty)_h(t)-(Ty)(t)| dt\\\\
&= \int_0^1 |\frac{1}{h} \int_t^{t+h} (Ty)(s) ds - (Ty)(t)| dt\\\\
&\leq \int_0^1 \Big(\frac{1}{h} \int_t^{t+h} |(Ty)(s) -(Ty)(t)| ds\Big) dt\\
&\leq \int_0^1\frac{1}{h}\int_t^{t+h}
\Big|f\Big(s,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)}
\big\{A\int_0^{1}(1-s)y(s) ds \\
&\quad -A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s)y(s) ds\big\}
-I^{2-{\alpha}}y(t)\Big)\\
&\quad - f\Big(t,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)}
\big\{A\int_0^{1}(1-s) y(s) ds\\
&\quad -A\sum_{k=1}^m a_k
\int_0^{\tau_k}(\tau_k-s) y(s) ds\Big\}
 -I^{2-{\alpha}}y(t)\Big)\Big| \,ds\,dt.
\end{align*}
then assumption (H2) implies that $f\in L^1(0,1)$ and
\begin{align*}
&\frac{1}{h}\int_t^{t+h}
\Big|f\Big(s,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)}
 \big\{A\int_0^{1}(1-s) y(s) ds - A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s)
 y(s) ds\big\}\\
&- I^{2-{\alpha}}y(t)\Big)
- f\Big(t,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)}
\big\{A\int_0^{1}(1-s) y(s) ds
- A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds\big\}\\
&- I^{2-{\alpha}}y(t)\Big)\Big| \,ds
\to 0
\end{align*}
Therefore, by Theorem \ref{thm17}, we have
that $T(\Omega)$ is relatively compact; that is, $T$ is a compact operator.
Then the operator $T$ has a fixed point $Q^{+}_{r}$, which proves
the existence of positive solution $y\in L^{1}[0,1]$  for \eqref{equa3.1}.
\end{proof}

For the existence of solutions to the nonlocal problem \eqref{e1.1}--\eqref{e1.2},
we have the following theorem.

\begin{theorem} \label{thm3.2}
Under the  assumptions of Theorem \ref{thm3.1}, if $0<\sum_{k=1}^m a_k \tau_k<1$,
 then nonlocal problem \eqref{e1.1}--\eqref{e1.2}
has at least one positive solution $ u\in C[0,1]$, with
$u' \in AC[0,1]$.
\end{theorem}

\begin{proof}
 For the problem \eqref{e1.1}-\eqref{e1.2}, let $- y(t) = u''(t)$. Then
\begin{equation}\label{equa5}
u(t) = tu'(0) - I^{2}y(t),
\end{equation}
where $y$ is the solution of the fractional-order functional integral
equation \eqref{equa3.1}.
Letting $t=\tau_k$ in  \eqref{equa5}, we obtain
\[
u(\tau_k)=- \int_0^{\tau_k}(\tau_k-s) y(s) ds+\tau_k u'(0)
\]
and
\begin{equation}\label{e*}
\sum_{k=1}^m a_ku(\tau_k)
=- \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds
 + u'(0)\sum_{k=1}^m a_k \tau_k
\end{equation}
From equation \eqref{e1.2} and \eqref{e*}, we obtain
\[
- \int_0^{1}(1-s) y(s) ds+u'(0)
=- \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds
+u'(0)\sum_{k=1}^m a_k \tau_k\,.
\]
Then
\begin{gather*}
u'(0) =A \Big(\int_0^{1}(1-s) y(s) ds
- \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds\Big), \\
 A=(1 - \sum_{k=1}^m a_k \tau_k)^{-1}.
\end{gather*}
Then
\begin{equation}\label{e6}
\begin{aligned}
u(t) &=A t\int_0^{1}(1-s) y(s) ds
 - A t \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds
  - \int_0^{t}(t-s)y(s) ds \\
&=t\Big\{\int_0^{1}(1-s) y(s) ds+A \sum_{k=1}^m a_k
\tau_k\int_0^{1}(1-s)y(s) ds\\
&\quad - A\sum_{k=1}^m a_k \int_0^{\tau_k}(\tau_k-s)y(s) ds\Big\}
- \int_0^{t}(t-s)y(s) ds.
\end{aligned}
\end{equation}
where $y$ is the solution of the fractional-order functional integral
equation \eqref{equa3.1}.
Hence, by Theorem \ref{thm3.1}, Equation \eqref{e6} has  at least one solution
$ u \in C(0,1)$.

Now, from equation \eqref{e6}, we have
\begin{gather*}
u(0)=\lim_{t\to 0^{+}}u(t)=0,\\
\begin{aligned}
u(1)&=\lim_{t\to 1^{-}}u(t)\\
 &=A \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds
- A\int_0^{1}(1-s) y(s) ds- \int_0^{1}(1-s)y(s) ds
\end{aligned}
\end{gather*}
from which we deduce that  \eqref{e6} has at least one positive solution
$ u \in C[0,1]$.
Now,
\begin{align*}
\sum_{k=1}^m a_k \int_0^{\tau_k}(\tau_k-s)y(s) ds
&<\sum_{k=1}^m a_k\tau_k \int_0^{1}(1-\frac{s}{\tau_k})y(s) ds\\
&<\sum_{k=1}^m a_k\tau_k \int_0^{1}(1-s)y(s) ds
\end{align*}
and
\[
\int_0^{t}(t-s)y(s) ds<t \int_0^{1}(1-\frac{s}{t})y(s) ds
<\int_0^{1}(1-s)y(s) ds\,.
\]
Then the solution of \eqref{e6} is positive.

To complete the proof, we show that  \eqref{e6} satisfies the nonlocal
 problem  \eqref{e1.1}--\eqref{e1.2}.
Differentiating \eqref{e6},  we obtain
\[
\frac{d^{2}u}{dt^{2}}=- y(t),
\]
and
\begin{align*}
D^{\alpha} u(t)
&=I^{1-{\alpha}} \frac{d}{dt}u(t)\\
&=- I^{1-{\alpha}}\Big(A \int_0^{1}(1-s) y(s) ds
- A \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds
- \int_0^{t}y(s) ds\Big)
\\
& = \frac{t^{1-\alpha}}{\Gamma(2-\alpha)}
\Big\{A\int_0^{1}(1-s) y(s) ds
 - A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds\Big\}\\
&\quad - \int_0^t\frac{(t-s)^{1-\alpha}}{\Gamma(2-\alpha)}y(s) ds,
\end{align*}
where
\[
 u''(t)+f(t, D^{\alpha}u(t))=0
\]
and
\begin{align*}
&\sum_{k=1}^m a_ku(\tau_k)\\
&= A \sum_{k=1}^m a_k\tau_k \sum_{k=1}^m a_k
 \int_0^{\tau_k}(\tau_k-s) y(s) ds\\
&\quad - A \sum_{k=1}^m a_k \tau_k\int_0^{1}(1-s) y(s) ds
 - \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s)y(s) ds
 \\
&= \Big(A \sum_{k=1}^m a_k\tau_k - 1\Big)
\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds
 - A \sum_{k=1}^m a_k \tau_k\int_0^{1}(1-s) y(s) ds
 \\
&=(A+1 - 1)\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds
 - (A+1)\int_0^{1}(1-s) y(s) ds
 \\
&= A \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds
 - A\int_0^{1}(1-s) y(s) ds- \int_0^{1}(1-s) y(s) ds\\
& = u(1)\,.
\end{align*}
This complete the proof of  the equivalent between
problem \eqref{e1.1}--\eqref{e1.2} and the integral equation \eqref{e6}.
This also implies that there exists at least one positive solution
$ u\in C[0,1] $ of the nonlocal problem  \eqref{e1.1}--\eqref{e1.2}.
\end{proof}

\subsection*{Example}
Our results can be applied to the nonlocal problem
\begin{gather*}
 u''(t)+f(t, D^{\alpha}u(t))=0,\quad  \alpha \in(0, 1), \text{ a.e. }
 t\in(0, 1),\\
u(0) = 0, \quad u(1) = 2 u(1/4) - 3 u(1/2)+3/2 u(3/4).
\end{gather*}

As an application, we have the following corollaries for the
two cases $ \alpha = 1 $ and $ \alpha = 0$.

\begin{corollary} \label{coro3.1}
Under the assumptions of Theorem \ref{thm3.2}, the nonlocal problem
\begin{gather*}
u''(t)+f(t,u'(t))=0 \quad   \text{a.e. }  t\in(0, 1)\\
u(0)=0, \quad u(1)=\sum_{k=1}^m a_k u(\tau_k), \quad
\tau_k\in(a, b)\subset (0, 1), \quad 0<\sum_{k=1}^m a_k \tau_k<1.
\end{gather*}
has at least one positive solution.
\end{corollary}

The proof of the above corollary follows by
letting $\alpha \to 1$ in Theorems \ref{thm3.1} and \ref{thm3.2}
 (see \cite{P1}).


\begin{corollary} \label{coro3.2}
Under  the assumptions of Theorem \ref{thm3.2}, the nonlocal problem
\begin{gather*}
u''(t)+f(t,u(t))=0 \quad   \text{a.e. }  t\in(0, 1)\\
u(0)=0,\quad  u(1)=\sum_{k=1}^m a_k u(\tau_k),\quad
 \tau_k\in(a, b)\subset (0, 1), \quad 0<\sum_{k=1}^m a_k \tau_k<1.
\end{gather*}
has at least one positive solution.
\end{corollary}

The proof of the above corollary follows by
letting $\alpha \to 0$ in Theorems \ref{thm3.1} and \ref{thm3.2}
(see \cite{P1}).


\section{Integral condition}

Let$ u \in C[0,1] $ be the solution of the  problem
\eqref{e1.1}--\eqref{e1.2}.
Let $ a_k=t_j-t_{k-1}$,  $\eta_k\in (t_{k-1}, t_j)$,
$a = t_0 < t_1 < t_2<\dots < t_{n}=b $.
Then  condition \eqref{e1.2} becomes
\[
u(1)=\sum_{k=1}^m (t_j - t_{k-1}) u(\eta_k).
\]
From the continuity of the solution $u$ ,to \eqref{e1.1}--\eqref{e1.2},
 we can obtain
\[
\lim_{m\to\infty}\sum_{k=1}^m (t_j - t_{k-1}) u(\eta_k)
= \int_{a }^{b} u(s) ds.
\]
and  condition \eqref{e1.2} is transformed into the integral condition
\[
u(0)=0, u(1)=\int_{a}^{b} u(s) ds.
\]
Now, we have the following  result.

\begin{theorem} \label{thm4.1}
Under the assumptions of  Theorem \ref{thm3.2}, there exist at least one
positive solution $ u\in AC[0,1] $ to the  problem
\begin{gather*}
u''(t)+f(t, D^{\alpha}x(t))=0, \quad  \alpha\in [0, 1], \text{ a.e. } t\in(0, 1),\\
u(0)=0, \quad u(1)=\int_{a}^{b} u(s)\, ds.
\end{gather*}
\end{theorem}

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\section*{Addendum posted by the editor on December 4, 2013} 

In March 2013, an anonymous reader informed us that the results in this 
article are incorrect:
\begin{quote}
Theorem 3.2 is wrong, the example following it is not valid. 
Corollary 3.3 and Corollary 3.4 are not correct, under the given conditions 
it is possible that NO positive solution exist.
\end{quote}
The authors tried to solve the problem, but the correction was not
satisfactory> The authors were informed, but they have not sent
any new corrections.  
\medskip 

End of addendum.
\end{document}