\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 23, 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  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/23\hfil Existence of positive solutions]
{Existence of positive solutions for fractional
differential equations with Riemann-Liouville
left-hand and right-hand fractional derivatives}

\author[Shuqin Zhang\hfil EJDE-2004/23\hfilneg]
{Shuqin Zhang} 

\address{Shuqin Zhang \hfill\break
Department of Mathematical Sciences, Tsinghua University,\\
Beijing, 100084 China}
\email{szhang@math.tsinghua.edu.cn\quad shuqinzhang\_1971@hotmail.com}


\date{}
\thanks{Submitted June 15, 2003. Published February 16, 2004.}
\thanks{Supported by the Postdoctoral Foundation of China.}
\subjclass[2000]{34B15}
\keywords{Existence of positive solution, fractional differential equation,
\hfill\break\indent fixed point, cone}


\begin{abstract}
 Combining  properties of Riemann-Liouville fractional calculus 
 and fixed point theorems, we obtain three existence results 
 of one positive solution and of multiple positive solutions 
 for initial value problems with fractional differential equations. 
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{prop}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}
Let  $s$ be a real number and $n = [s]+1$ where $[s]$
the integer part of $s$. For a function $f:[a,b] \to \mathbb{R}$
the expressions
\begin{gather*}
D_{a+}^sf(x)=\frac 1{\Gamma(n-s)}(\frac d{dx})^n
\int_a^x\frac{f(t)}{(x-t)^{s-n+1}}dt\\
D_{b-}^sf(x)=\frac {(-1)^n}{\Gamma(n-s)}(\frac d{dx})^n
\int_x^b\frac{f(t)}{(t-x)^{s-n+1}}dt
\end{gather*}
are called, respectively, the
Riemann-Liouville left-hand and right-hand fractional
derivative of order $s$.  If $s$ is an integer,
the derivative of order $s$ is understood in the sense of usual
differentiation:
$$
D_{a+}^s=(\frac d{dx})^s,\quad D_{b-}^s=(-1)^s(\frac d{dx})^s\quad
s=1,2,3,\dots
$$
Here we consider the initial-value problem
\begin{equation} \label{e1}
 \begin{aligned}
 D_{1-}^\alpha D_{0+}^\delta u(t)
   =g(t,u(t))\quad 0< t< 1 , 0<\alpha,\delta<1\\
t^{1-\delta}u(t)\big|_{t=0}=a\geq 0,\quad
(1-t)^{1-\alpha}D_{0+}^\delta u(t)|_{t=0}=b\geq 0,
\end{aligned}
\end{equation}
where $D_{1-}^\alpha,D_{0+}^\delta $ are the Riemann-Liouville
right-hand and left-hand fractional derivatives.

For $x >0$, the expressions
\begin{gather*}
I_{a+}^sf(x)= \frac 1{\Gamma(s)}\int_a^x\frac{f(t)}{(x-t)^{1-s}}dt,\quad x>a,\\
I_{b-}^sf(x)= \frac 1{\Gamma(s)}\int_x^b\frac{f(t)}{(t-x)^{1-s}}dt,\quad x<b
\end{gather*}
are called, respectively, the  Riemann-Liouville left-hand
and  right-hand fractional integral of order $s$; see \cite{b}


\begin{prop}[{\cite[theorem 2.4]{b}}] \label{prop1}
If $s>0$ then $D_{a+}^sI_{a+}^sf(x)=f(x)$
for any $f\in L_1(a,b)$, while
\begin{equation} \label{e2}
I_{a+}^sD_{a+}^sf(x)=f(x)
 \end{equation} is satisfied for $f\in
I_{a+}^s(L_1(a,b))$ with
\[
I_{a+}^s(L_1(a,b))=\{g(x): g(x)=I_{a+}^s\varphi(x),\quad \varphi\in L_1(a,b)\}
\]
If  $f, D_{a+}^s f\in L_1(a,b)$, then \eqref{e2} is not true in
general. However
$$
I_{a+}^sD_{a+}^sf(x)=f(x)-\Sigma_{k=0}^{n-1}\frac{(x-a)^{s-k-1}}{\Gamma(s-k)}
f_{n-s}^{(s-k-1)}(a)
$$
where  $n=[s]+1, f_{n-s}(x)=I_{a+}^{n-s}f(x)$. In particular
 for $0<Re s<1$, we have
$$
I_{a+}^sD_{a+}^sf(x)=f(x)-\frac{f_{1-s}(a)}{\Gamma(s)}(x-a)^{s-1}.
$$
\end{prop}

\begin{remark}\label{rem1} \rm
Similar results hold for right-hand fractional derivatives.
\end{remark}

The following theorems play major role in this article.

\begin{theorem}[\cite{k}] \label{theoA}
Let $X$ be a Banach space, and let
$P\subset X$ be a cone in $X$. If $\Omega_1, \Omega_2$ are
open subsets of $X$ with $0\in \Omega_1\subset
{\overline{\Omega}_1}\subset{\Omega _2}$, and let $S:P\rightarrow
P$ be a completely continuous operator such that either
\begin{enumerate}
\item $\|Sw\|\leq \|w\|$, $w\in P\cap {\partial \Omega_1}$,
  $\|Sw\|\geq \|w\|$,  $w\in P\cap {\partial \Omega_2}$,\quad
or
\item $\|Sw\|\geq \|w\|$, $w\in P\cap {\partial \Omega_1}$,
$\|Sw\|\leq \|w\|\quad w\in P\cap {\partial \Omega_2}$
\end{enumerate}
then $S$ has a fixed point in
$P\cap {\overline{\Omega}_2\backslash{\Omega_1}}$.
\end{theorem}

\begin{theorem}[\cite{j}]\label{theoB}
Let $K$ be a cone and $K_c=\{y\in K|\|y\|\leq c\}$, and
$A:\overline{K_c}\rightarrow \overline{K_c}$
be completely continuous and $\alpha $ be a nonnegative continuous
concave function on $K$ such that $\alpha(y)\leq\|y\|$, for all
$y\in\overline{K_c}$. If there exist $0<a<b<d\leq c$ such
that
\begin{itemize}
\item[(C1)] $\{y\in K(\alpha,b,d\}|\alpha(y)>b\}\neq\emptyset$ and
$\alpha(Ay)>b$, for all $y\in K\{\alpha,b,d\}$,

\item[(C2)] $\|Ay\|<a$, for $\|y\|\leq a$

\item[(C3)] $\alpha(Ay)>b$, for $y\in K\{\alpha,b,c\}$ with
$\|Ay\|>d$,
\end{itemize}
then $A$ has at least three fixed points $y_1,y_2,y_3$ satisfying
$$
\|y_1\|<a,\quad b<\alpha(y_2),\quad \|y_3\|>a \quad
\mbox{with}\quad\alpha(y_3)<b. $$
\end{theorem}

 \section{Main Results}
  Let $X=\{u\in C(0,1) : t^{1-\delta}(1-t)^{1-\alpha}u\in C[0,1]\}$ be
the Banach space endowed with the
 norm
 $$
 \|u\|=\max_{0\leq t\leq 1}t^{1-\delta}(1-t)^{1-\alpha}|u(t)|.
 $$
Let $K$ be the cone $K=\{u\in X; u(t)\geq 0,\; 0\leq t\leq 1\}$.
Applying $I_{1-}^\alpha $ to the first equation in \eqref{e1}
it follows that
$(1-t)^{\alpha-1}D_{0+}^\delta u(t)|_{t=0}=b$ that
 \begin{equation} \label{e3}
  D_{0+}^\delta u(t)=b(1-t)^{\alpha-1}+I_{1-}^\alpha
g(t,u(t))
\end{equation}
From this equation, Proposition \ref{prop1} and the condition
$t^{1-\delta}u(t)|_{t=0}=a$, we have
\begin{align*}
u(t)&= at^{\delta-1}+I_{0+}^\delta(b(1-t)^{\alpha-1} +I_{1-}^\alpha g(t,u(t)))\\
&=at^{\delta-1}+\frac 1{\Gamma(\delta)}\int_0^t(t-s)^{\delta-1}b(1-s)^{\alpha-1}ds \\
& \quad +\frac 1{\Gamma(\delta)}\int_0^t(t-s)^{\delta-1}I_{1-}^\alpha g(s,u(s))ds\\
&= at^{\delta-1}+\frac
b{\Gamma(\delta)}\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&+\quad \frac1{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}
\int_s^1(\tau-s)^{\alpha-1} g(\tau,u(\tau))d\tau\,ds
\end{align*}
Defining $T: X\rightarrow X$ by
\begin{align*}
Tu(t)&=at^{\delta-1}+\frac
b{\Gamma(\delta)}\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad+\frac 1{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}
\int_s^1(\tau-s)^{\alpha-1} g(\tau,u(\tau))d\tau\,ds,
\end{align*}
we see that $u$ is a solution to \eqref{e1} if and only if $u$ is a fixed
point of $T$.

\begin{lemma}  \label{lm1}
If $g$ is continuous, then $T$ is a completely continuous operator.
\end{lemma}

\begin{proof} Since $g$ is continuous, $T$ transforms $X$ into $X$.
Let $M=\{u\in X;
\|u\|\leq l, l>0\}$. For $u \in M$,
\begin{align*}
&t^{1-\delta}(1-t)^{1-\alpha}|T u(t)|\\
&=|a(1-t)^{1-\alpha}+\frac {bt^{1-\delta}(1-t)^{1-\alpha}}
{\Gamma(\delta)}\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&+\quad \frac {t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)
\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
g(\tau,u(\tau))d\tau\,ds|\\
&\leq a+\frac {bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-t)^{\alpha-1}ds\\
&+\quad \frac {t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)
\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
|g(\tau,u(\tau))|d\tau\,ds\\
&\leq a+\frac{b}{\Gamma(1+\delta)}+\frac{L}{\Gamma(\delta)\Gamma(\alpha)}
\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} d\tau\,ds\\
&= a+\frac{b}{\Gamma(1+\delta)}+\frac {L}{\Gamma(\delta)\Gamma(1+\alpha)}
\int_0^t(t-s)^{\delta-1}(1-s)^\alpha ds\\
&\leq a+\frac{b}{\Gamma(1+\delta)}+\frac{L}{\Gamma(\delta)
\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1}ds\\
&= a+\frac{b}{\Gamma(1+\delta)}+\frac
{L}{\Gamma(1+\delta)\Gamma(1+\alpha)}t^\delta\\
&\leq a+\frac{b}{\Gamma(1+\delta)}+\frac {L}{\Gamma(1+\delta)\Gamma(1+\alpha)}
\end{align*}
where
$$
L=\max_{0\leq t\leq 1,\|u\|\leq l}|g(t,u(t))|+1\,.
$$
So $T(M)$ is bounded.

Let us see that $\overline {T(M)}$ is equicontinuous. For
$u\in M, \varepsilon>0, t_1,t_2\in[0,1],t_1<t_2$, let
$$
\eta<\big\{\frac \varepsilon{4(a+1)},
\big(\frac{\varepsilon\Gamma(1+\delta)}{4(2b+1)}\big)^{1/\delta},
\big(\frac{\varepsilon\Gamma(1+\alpha)\Gamma(1+\delta)}{8L}\big)
^{1/\delta},
\frac{\varepsilon\Gamma(1+\delta)\Gamma(1+\alpha)}{4(b\Gamma(1+\alpha)+L)}
\big\}
$$
For $t_2-t_1\leq\max\{t_2-t_1,
t_2^{1-\delta}-t_1^{1-\delta},
(1-t_1)^{1-\alpha}-(1-t_2)^{1-\alpha}\}<\eta$, we have
\begin{align*}
&|t_2^{1-\delta}(1-t_2)^{1-\alpha}|T u(t_2)|-t_1^{1-\delta}(1-t_1)
  ^{1-\alpha}|T u(t_1)||\\
&=|a(1-t_2)^{1-\alpha}-a(1-t_1)^{1-\alpha}+\frac
{bt_2^{1-\delta}(1-t_2)^{1-\alpha}}{\Gamma(\delta)}\int_0^{t_2}(t_2-s)
^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad -\frac{bt_1^{1-\delta}(1-t_1)^{1-\alpha}}{\Gamma(\delta)}
\int_0^{t_1}(t_1-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad +\frac{t_2^{1-\delta}(1-t_2)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}
\int_0^{t_2}(t_2-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}g(\tau,u(\tau))d\tau\,ds\\
&\quad -\frac{t_1^{1-\delta}(1-t_1)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}
\int_0^{t_1}(t_1-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}g(\tau,u(\tau))d\tau\,ds|\\
&\leq \frac{bt_1^{1-\delta}(1-t_1)^{1-\alpha}}{\Gamma(\delta)}\int_0^{t_1}
((t_1-s)^{\delta-1}-(t_2-s)^{\delta-1})(1-t_1)^{\alpha-1}ds\\
&\quad +\frac{bt_2^{1-\delta}(1-t_2)^{1-\alpha}}{\Gamma(\delta)}
\int_{t_1}^{t_2}(t_2-s)^{\delta-1} (1-t_2)^{\alpha-1}ds+a((1-t_1)^{1-\alpha}
-(1-t_2)^{1-\alpha})\\
&\quad +\frac {Lt_1^{1-\delta}(1-t_1)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}
\int_0^{t_1}((t_1-s)^{\delta-1}-(t_2-s)^{\delta-1})\int_s^1(\tau-s)
^{\alpha-1}d\tau\,ds\\
&\quad + \frac{Lt_2^{1-\delta}(1-t_2)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}
\int_{t_1}^{t_2}(t_2-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}d\tau\,ds\\
&\quad +\frac{b(t_2^{1-\delta}(1-t_2)^{1-\alpha}-t_1^{1-\delta}(1-t_1)
^{1-\alpha})} {\Gamma(\delta)}\int_0^{t_1}(t_2-s)^{\delta-1}
(1-t_2)^{\alpha-1}ds\\
&\quad+\frac {L(t_2^{1-\delta}(1-t_2)^{1-\alpha}-t_1^{1-\delta}(1-t_1)^{1-\alpha})}
{\Gamma(\delta)\Gamma(\alpha)}\int_0^{t_1}(t_2-s)^{\delta-1}
\int_s^1(\tau-s)^{\alpha-1}d\tau\,ds\\
&\leq \frac {bt_1^{1-\delta}}{\Gamma(1+\delta)}(t_1^\delta+(t_2-t_1)
^\delta-t_2^\delta)\\
&\quad +\frac {bt_2^{1-\delta}}{\Gamma(1+\delta)}(t_2-t_1)^\delta+
a((1-t_1)^{1-\alpha}-(1-t_2)^{1-\alpha})\\
&\quad +\frac {Lt_1^{1-\delta}(1-t_1)^{1-\alpha}}{\Gamma(\delta)\Gamma(1+\alpha)}
\int_0^{t_1}((t_1-s)^{\delta-1}-(t_2-s)^{\delta-1})ds\\
&+\quad \frac{Lt_2^{1-\delta}(1-t_2)^{1-\alpha}}{\Gamma(\delta)\Gamma(1+\alpha)}
\int_{t_1}^{t_2}(t_2-s)^{\delta-1} ds\\
&\quad +\frac{b(t_2^{1-\delta}(1-t_2)^{1-\alpha}-t_1^{1-\delta}(1-t_1)
^{1-\alpha})} {\Gamma(1+\delta)}t_2^\delta (1-t_2)^{\alpha-1}\\
&\quad + \frac {L(t_2^{1-\delta}(1-t_2)^{1-\alpha}-t_1^{1-\delta}(1-t_1)^{1-\alpha})}
{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^{t_1}(t_2-s)^{\delta-1}ds\\
&\leq \frac{bt_1^{1-\delta}}{\Gamma(1+\delta)}(t_2-t_1)^\delta
+\frac {bt_2^{1-\delta}}{\Gamma(1+\delta)}(t_2-t_1)^\delta\\
&+\quad a((1-t_1)^{1-\alpha}-(1-t_2)^{1-\alpha})
+\frac {Lt_1^{1-\delta}(1-t_1)^{1-\alpha}}{\Gamma(1+\delta)\Gamma(1+\alpha)}
(t_1^\delta+(t_2-t_1)^\delta-t_2^\delta)\\
&\quad +\frac{Lt_2^{1-\delta}(1-t_2)^{1-\alpha}}{\Gamma(1+\delta)\Gamma(1+\alpha)}
(t_2-t_1)^\delta+\frac {b(t_2^{1-\delta}-t_1^{1-\delta})}
{\Gamma(1+\delta)}+\frac {L(t_2^{1-\delta}-t_1^{1-\delta})}
{\Gamma(1+\delta)\Gamma(1+\alpha)}\\
&\leq \frac {b}{\Gamma(1+\delta)}(t_2-t_1)^\delta
+\frac{b}{\Gamma(1+\delta)}(t_2-t_1)^\delta
+a((1-t_1)^{1-\alpha}-(1-t_2)^{1-\alpha})\\
&\quad +\frac {L}{\Gamma(1+\delta)\Gamma(1+\alpha)}
(t_2-t_1)^\delta+\frac {L}{\Gamma(1+\delta)\Gamma(1+\alpha)}(t_2-t_1)^\delta\\
&\quad +\frac {b(t_2^{1-\delta}-t_1^{1-\delta})}
{\Gamma(1+\delta)}+\frac {L(t_2^{1-\delta}-t_1^{1-\delta})}
{\Gamma(1+\delta)\Gamma(1+\alpha)}\\
&< \frac {2b+1}{\Gamma(1+\delta)}\eta^\delta+ (a+1)\eta
+\frac {2L}{\Gamma(1+\delta)\Gamma(1+\alpha)} \eta^\delta\\
&\quad +(\frac {b} {\Gamma(1+\delta)}+\frac {L}
{\Gamma(1+\delta)\Gamma(1+\alpha)})\eta\\
&<\frac \varepsilon 4+\frac \varepsilon 4+\frac \varepsilon 4+\frac \varepsilon 4\\
&=\varepsilon
\end{align*}
By Arzela-Azcoli's theorem $\overline {TM}$ is equicontinuous, so
the operator $T$ is completely continuous.
\end{proof}

\begin{theorem}\label{theo1}
If $g$ is continuous,
$a+b\neq 0$, and there exists $0<\mu\leq 1$ such that
\begin{equation}\label{h2}\
\lim_{|u|\rightarrow \infty}\frac{g(t,u(t))}{|u|^\mu}=0,
\end{equation}
then problem \eqref{e1} has one positive solution.
\end{theorem}

\begin{proof}
 As pointed out above, we only need
 to prove the existence of fixed point of operator  $T$ in $K$.
 It follows from the Lemma \ref{lm1} that
$T: K\rightarrow K$ is a completely continuous operator.
 From \eqref{h2}, there exists $N>0$, such that for
$0< \varepsilon<\frac {\Gamma(1+\mu(\alpha-1)+\alpha)
\Gamma(1+\mu(\delta-1)+\delta)}
 {4\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}$,
$$
 g(t,u(t))\leq \varepsilon |u|^\mu, \quad \mbox{for $t\in [0,1],|u|\geq N$}
$$
 So we have
$$g(t,u(t))\leq \varepsilon |u|^\mu+c, \quad \mbox{for $t\in[0,1],u\in [0,+\infty)$}
$$
where
$$c=\max_{0\leq t\leq 1,|u|\leq N}|g(t,u(t))|+1.
$$
Let $\Omega_1=\{u\in K; \|u\|<R_1\}$,
 where $R_1>\{1,4a,\frac {4b}{\Gamma(1+\delta)},\frac
 {4c}{\Gamma(1+\delta)\Gamma(1+\alpha)}\}$, for
$u\in \partial{\Omega_1}$, we have
\begin{align*}
&t^{1-\delta}(1-t)^{1-\alpha}|T u(t)|\\
&=|a(1-t)^{1-\alpha}+\frac{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&+\quad \frac{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}
\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}g(\tau,u(\tau))d\tau\,ds|\\
&\leq a+\frac{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-t)^{\alpha-1}ds\\
&\quad +\frac{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}
\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}|g(\tau,u(\tau))|d\tau\,ds\\
&\leq a+\frac b{\Gamma(1+\delta)}+\frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}
\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
(\varepsilon |u(\tau)|^\mu+c)d\tau\,ds\\
&\leq \frac{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}
\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
(\varepsilon \tau^{\mu(\delta-1)}(1-\tau)^{\mu(\alpha-1)}\|u\|^\mu+c)d\tau\,ds\\
&+a+\frac{b}{\Gamma(1+\delta)}\\
&\leq\frac{t^{1-\delta}(1-t)^{1-\alpha}\varepsilon
\|u\|^\mu}{\Gamma(\delta)\Gamma(\alpha)}
\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
s^{\mu(\delta-1)}(1-\tau)^{\mu(\alpha-1)} d\tau\,ds\\
&+\quad \frac c{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}
\int_s^1(\tau-s)^{\alpha-1}d\tau\,ds+a+\frac{b}{\Gamma(1+\delta)}\\
&=\frac {t^{1-\delta}(1-t)^{1-\alpha}\varepsilon
\|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\alpha+
\mu(\alpha-1))}\int_0^t(t-s)^{\delta-1}
s^{\mu(\delta-1)}(1-s)^{\mu(\alpha-1)+\alpha} ds\\
&\quad +\frac c{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1}(1-s)^\alpha
ds +a+\frac{b}{\Gamma(1+\delta)}
\end{align*}
If $\mu(\alpha-1)+\alpha<0$, then the first equality of above
becomes
\begin{align*}
&\frac {t^{1-\delta}(1-t)^{1-\alpha}\varepsilon
\|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\alpha+\mu(\alpha-1))}\int_0^t(t-s)^{\delta-1}
s^{\mu(\delta-1)}(1-s)^{\mu(\alpha-1)+\alpha} ds \\
&\leq \frac {t^{1-\delta}(1-t)^{1-\alpha}\varepsilon
\|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\mu(\alpha-1)+\alpha)}\int_0^t(t-s)^{\delta-1}
s^{\mu(\delta-1)}(1-t)^{\mu(\alpha-1)+\alpha} ds \\
&= \frac {t^{1-\delta}(1-t)^{1+\mu(\alpha-1)}\varepsilon
\|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\mu(\alpha-1)+\alpha)}
\int_0^t(t-s)^{\delta-1}
s^{\mu(\delta-1)} ds \\
&=\frac {t^{1+\mu(\delta-1)}(1-t)^{1+\mu(\alpha-1)}\varepsilon
\|u\|^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha)
\Gamma(1+\mu(\delta-1)+\delta)}\\
&\leq \frac {\varepsilon
R_1^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha)
\Gamma(1+\mu(\delta-1)+\delta)}\,.
\end{align*}
If $\mu(\alpha-1)+\alpha\geq 0$, then the first equality of above
becomes
\begin{align*}
&\frac {t^{1-\delta}(1-t)^{1-\alpha}\varepsilon
\|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\alpha+\mu(\alpha-1))}\int_0^t(t-s)^{\delta-1}
s^{\mu(\delta-1)}(1-s)^{\mu(\alpha-1)+\alpha} ds \\
&\leq\frac {t^{1-\delta}(1-t)^{1-\alpha}\varepsilon
\|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\mu(\alpha-1)+\alpha)}\int_0^t(t-s)^{\delta-1}
s^{\mu(\delta-1)} ds \\
&=\frac {t^{1+\mu(\delta-1)}(1-t)^{1-\alpha}\varepsilon
\|u\|^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha)
\Gamma(1+\mu(\delta-1)+\delta)}\\
&\leq \frac {\varepsilon
R_1^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha)
\Gamma(1+\mu(\delta-1)+\delta)}\,.
\end{align*}
 Therefore, we obtain
 \begin{align*}
t^{1-\delta}(1-t)^{1-\alpha}|T u(t)|
 &\leq\frac {\varepsilon
\|u\|^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha)
\Gamma(1+\mu(\delta-1)+\delta)}\\
&\quad +\frac c{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1}ds+a
 +\frac{b}{\Gamma(1+\delta)}\\
&\leq \frac {\varepsilon
R_1^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha)
\Gamma(1+\mu(\delta-1)+\delta)}\\
&\quad +\frac c{\Gamma(1+\delta)\Gamma(1+\alpha)}t^\delta+a
+\frac{b}{\Gamma(1+\delta)}\\
&\leq \frac {\varepsilon R_1\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}
{\Gamma(1+\mu(\alpha-1)+\alpha) \Gamma(1+\mu(\delta-1)+\delta)}\\
&\quad +\frac c{\Gamma(1+\delta)\Gamma(1+\alpha)}+a+\frac{b}{\Gamma(1+\delta)}\\
&\leq \frac {R_1}4+\frac {R_1}4+\frac {R_1}4+\frac {R_1}4
=R_1
\end{align*}
and $\|Tu\|\leq R_1=\|u\|$. Taking
$$ \Omega_2=\{u\in K; \|u\|<R_2\} $$
 where $R_2<\{a(\frac 12)^{\delta-1}+\frac
b{\Gamma(1+\delta)}(\frac 12)^\delta, R_1\}$, then for $u\in
 \partial{\Omega_2}$, we obtain
\begin{align*}
T u(\frac 12)
=&a(\frac 12)^{\delta-1}+\frac b{\Gamma(\delta)}\int_0^{\frac 12
}(\frac 12-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad +\frac 1{\Gamma(\delta)\Gamma(\alpha)}\int_0^{\frac 12}(\frac 12
-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}g(\tau,u(\tau))d\tau\,ds\\
&\geq a(\frac 12)^{\delta-1}+\frac
b{\Gamma(\delta)}\int_0^{\frac 12}(\frac 12-s)^{\delta-1}ds\\
&= a(\frac 12)^{\delta-1}+\frac b{\Gamma(1+\delta)}(\frac 12)^\delta\\
&\geq R_2\,.
\end{align*}
Therefore, $\|Tu\|\geq R_2=\|u\|$.
 Theorem \ref{theoA} implies that operator $T$ has one fixed point $u^*(t)\in
 {\overline {\Omega_1}\backslash {\Omega_2}}$, then $u^*(t)$ is one
 positive solution of problem \eqref{e1}.
\end{proof}

\begin{theorem}\label{theo2} If $g$ is continuous, and
there exists constant $c_1,c_2>0$ and $1\leq
\lambda<\min\{\frac \alpha {1-\alpha},\frac 1{1-\delta}\}$ with
$\alpha\geq\frac 12$ or $1\leq\lambda<\min\{\frac
1{1-\alpha},\frac 1{1-\delta}\}$ with $0<\alpha\leq\frac 12$ such
that
\begin{equation}\label{h3}
g(t,u(t))\leq c_1+c_2|u(t)|^\lambda, \quad \mbox{for all $t\in[0,1],
u\in [0,+\infty)$}
\end{equation}
then problem \eqref{e1} has at least one solution.
\end{theorem}

\begin{proof}
 As in Theorem \ref{theo1}, we only need to consider
existence of fixed point of operator $T$. By Lemma \ref{lm1}, $T$
is a completely continuous operator. We will make use of the
Schauder Fixed Point Theorem to prove this theorem.
  Let $0<R<1$, and
  $$
 B_R=\{u\in C([0,\gamma],[0,+\infty));\|u-(at^{\delta-1}+
 \frac b{\Gamma(\delta)}\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1} ds\|\leq R\}
$$
 be a convex bounded and closed subset of the Banach space
 $C[0,\gamma]$, where
\begin{align*}
 \gamma<\min
  \big\{&1,\big(\frac{R\Gamma(1+\delta)\Gamma(1+\alpha)}{2c_1}\big)^{1/\delta},\\
&\big(\frac{\Gamma(1+\delta+\lambda(\delta-1))\Gamma(1+\alpha+\lambda(\alpha-1))}{2c_2
\Gamma(1+\lambda(\delta-1))\Gamma(1+\lambda(\alpha-1))}\big)
^{\frac1{1+\lambda(\delta-1)} }\big\}
\end{align*}
Note that for all $u\in B_R$,
  \begin{align*}
&t^{1-\delta}(1-t)^{1-\alpha} |Tu(t)-(at^{\delta-1}+\frac b{\Gamma(\delta)}
 \int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1} ds)|\\
&=\frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
g(\tau,u(\tau))d\tau\,ds\\
&\leq\frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
(c_1+c_2|u(\tau)|^\lambda )d\tau\, ds\\
&\leq\frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}
\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}\\
&\quad\times (c_1+c_2\tau^{\lambda(\delta-1)}(1-\tau)^{\lambda(\alpha-1)}
\|u\|^\lambda )d\tau\, ds\\
&\leq \frac
{c_1t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(1+\alpha)}
\int_0^t(t-s)^{\delta-1}(1-s)^\alpha ds\\
&\quad+\frac {c_2\|u\|^\lambda
t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}
\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}s^{\lambda(\delta-1)}
(1-\tau)^{\lambda(\alpha-1)}d\tau\,ds\\
&=\frac
{c_1t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(1+\alpha)}
\int_0^t(t-s)^{\delta-1}(1-s)^\alpha ds\\
&\quad +\frac {c_2\|u\|^\lambda
t^{1-\delta}(1-t)^{1-\alpha}\Gamma(1+\lambda(\alpha-1))}
{\Gamma(\delta)\Gamma(1+\alpha+\lambda(\alpha-1))}
\int_0^t(t-s)^{\delta-1}s^{\lambda(\delta-1)}
(1-s)^{\lambda(\alpha-1)+\alpha}ds
\end{align*}
If $1\leq \lambda<\min\{\frac\alpha{1-\alpha},\frac
1{1-\delta}\}$, then for the second formula of above becomes
\begin{align*}
&\frac {c_2\|u\|^\lambda
t^{1-\delta}(1-t)^{1-\alpha}\Gamma(1+\lambda(\alpha-1))}
{\Gamma(\delta)\Gamma(1+\alpha+\lambda(\alpha-1))}
\int_0^t(t-s)^{\delta-1}s^{\lambda(\delta-1)}
(1-s)^{\lambda(\alpha-1)+\alpha}ds\\
&\leq \frac {c_2\|u\|^\lambda
t^{1-\delta}(1-t)^{1-\alpha}\Gamma(1+\lambda(\alpha-1))}
{\Gamma(\delta)\Gamma(1+\alpha+\lambda(\alpha-1))}
\int_0^t(t-s)^{\delta-1}s^{\lambda(\delta-1)} ds\\
&=\frac {c_2\|u\|^\lambda
(1-t)^{1-\alpha}\Gamma(1+\lambda(\alpha-1))\Gamma(1+\lambda(\delta-1))}
{\Gamma(1+\delta+\lambda(\delta-1))\Gamma(1+\alpha+\lambda(\alpha-1))}
t^{\lambda(\delta-1)+1}\\
&\leq \frac {c_2\|u\|^\lambda
\Gamma(1+\lambda(\alpha-1))\Gamma(1+\lambda(\delta-1))}
{\Gamma(1+\delta+\lambda(\delta-1))\Gamma(1+\alpha+\lambda(\alpha-1))}
\gamma^{\lambda(\delta-1)+1}
\end{align*}
Similarly, if $1\leq \lambda<\min\{\frac 1{1-\alpha},\frac
1{1-\delta}\}$, then for the second formula of above becomes
\begin{align*}
&\frac {c_2\|u\|^\lambda
t^{1-\delta}(1-t)^{1-\alpha}\Gamma(1+\lambda(\alpha-1))}
{\Gamma(\delta)\Gamma(1+\alpha+\lambda(\alpha-1))}
\int_0^t(t-s)^{\delta-1}s^{\lambda(\delta-1)}
(1-s)^{\lambda(\alpha-1)+\alpha}ds\\
&\leq\frac {c_2\|u\|^\lambda
t^{1-\delta}(1-t)^{1-\alpha}\Gamma(1+\lambda(\alpha-1))}
{\Gamma(\delta)\Gamma(1+\alpha+\lambda(\alpha-1))}
\int_0^t(t-s)^{\delta-1}s^{\lambda(\delta-1)}(1-t)^{\lambda(\alpha-1)+\alpha} ds\\
&=\frac {c_2\|u\|^\lambda
(1-t)^{1+\lambda(\alpha-1)}\Gamma(1+\lambda(\alpha-1))\Gamma(1+\lambda(\delta-1))}
{\Gamma(1+\delta+\lambda(\delta-1))\Gamma(1+\alpha+\lambda(\alpha-1))}
t^{\lambda(\delta-1)+1}\\
&\leq\frac {c_2\|u\|^\lambda
\Gamma(1+\lambda(\alpha-1))\Gamma(1+\lambda(\delta-1))}
{\Gamma(1+\delta+\lambda(\delta-1))\Gamma(1+\alpha+\lambda(\alpha-1))}
\gamma^{\lambda(\delta-1)+1}
\end{align*}
Therefore, we have
\begin{align*}
&t^{1-\delta}(1-t)^{1-\alpha} |Tu(t)-(at^{\delta-1}+\frac b{\Gamma(\delta)}
 \int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}
   ds)|\\
&=\frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
g(\tau,u(\tau))d\tau\,ds\\
 &\leq\frac
{c_1}{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1}ds+\frac
{c_2\|u\|^\lambda
\Gamma(1+\lambda(\alpha-1))\Gamma(1+\lambda(\delta-1))}
{\Gamma(1+\delta+\lambda(\delta-1))\Gamma(1+\alpha+\lambda(\alpha-1))}
\gamma^{\lambda(\delta-1)+1}\\
&=\frac {c_1}{\Gamma(1+\delta)\Gamma(1+\alpha)}t^\delta+\frac
{c_2\|u\|^\lambda
\Gamma(1+\lambda(\alpha-1))\Gamma(1+\lambda(\delta-1))}
{\Gamma(1+\delta+\lambda(\delta-1))\Gamma(1+\alpha+\lambda(\alpha-1))}\gamma^{\lambda(\delta-1)+1}\\
&\leq\frac
{c_1}{\Gamma(1+\delta)\Gamma(1+\alpha)}\gamma^\delta+\frac
{c_2R\Gamma(1+\lambda(\alpha-1))\Gamma(1+\lambda(\delta-1))}
{\Gamma(1+\delta+\lambda(\delta-1))\Gamma(1+\alpha+\lambda(\alpha-1))}\gamma^{\lambda(\delta-1)+1}\\
&<\frac R2+\frac R2 =R
\end{align*}
Therefore, $T(B_R)\subset B_R$.Now by the  Schauder Fixed Point Theorem
there exists  $u^*(t)\in B_R$ such that
$Tu^*(t)=u^*(t)$, and this completes the proof.
\end{proof}

The following result establishes the existence of multiple
positive solutions for the initial value problem \eqref{e1}.
Let
$$f(u)=u(\frac 12),\quad u\in K
$$
Obviously $f$ is a nonnegative  concave function that satisfies
$ f(u)\leq \|u\|$, for $u\in K$
and
\begin{gather*}
 K_c=\{u\in K; \|u\|\leq c\},\\
K(f,d,e)=\{u\in K; d\leq f(u), \|u\|\leq e\}\,.
\end{gather*}

\begin{theorem} \label{thm5}
Let $g$ be continuous. If
$a(\frac 12)^{\delta-1}+\frac b{\Gamma(1+\delta)}
<\overline a<\overline b<d=2\overline b=c$, if
$g$ satisfies:
\begin{itemize}
\item[(H4)] $g(t,u)<(\overline a-a-\frac
b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha)$ for
$0\leq t\leq 1$, $0\leq u\leq\overline a$

\item[(H5)] $g(t,u)<(c-a-\frac b{\Gamma(1+\delta)})\Gamma(1+\delta)
\Gamma(1+\alpha)$ for $0\leq t\leq 1$, $0\leq u\leq c$

\item[(H6)] $g(t,u)>(\frac 12)^{-1-\delta}(\overline b-a(\frac
12)^{\delta-1}-\frac {b(\frac 12)^\delta}
{\Gamma(1+\delta)})\frac{(1+\delta)\Gamma(1+\delta)\Gamma(1+\alpha)}{\delta}$
for $0\leq t\leq 1$, $\overline b\leq u\leq 2\overline b$,
\end{itemize}
then the initial-value problem \eqref{e1} has three positive solutions
$u_1,u_2,u_3$ satisfying
\begin{equation} \label{e6}
\|u_1\|<\overline a,\quad \overline b<f(u_2),\quad \|u_3\|>\overline a
\quad\mbox{with}\quad f(u_3)<\overline b
\end{equation}
\end{theorem}

\begin{proof} We apply Theorem \ref{theoB}. Since $g$ is continuous,
by Lemma \ref{lm1}, operator $T$ is completely continuous.
Now we choose $u\in \overline {K_c}$, then $\|u\|\leq c$, and
$g(t,u(t))<(c-a-\frac b{\Gamma(1+\delta)})\Gamma(1+\delta)
\Gamma(1+\alpha)$ for  $t\in[0,1]$ by (H5), so we have
\begin{align*}
&t^{1-\delta}(1-t)^{1-\alpha}|T u(t)|\\
&=|a(1-t)^{1-\alpha}+\frac
{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad + \frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
g(\tau,u(\tau))d\tau\,ds|\\
&\leq a(1-t)^{1-\alpha}+\frac
{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad + \frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
|g(\tau,u(\tau))|d\tau\,ds\\
&< a(1-t)^{1-\alpha}+\frac
{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad +\frac {t^{1-\delta}(1-t)^{1-\alpha}(c-a-\frac
b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha)}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}
\int_s^1(\tau-s)^{\alpha-1}
 d\tau\,ds\\
&=a(1-t)^{1-\alpha}+\frac
{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad+ \frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1}
(c-a-\frac b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha)(1-s)^\alpha ds\\
 &\leq a+\frac
{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-t)^{\alpha-1}ds\\
&\quad+\frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1}
(c-a-\frac b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha)ds\\
&\leq a+\frac {bt}{\Gamma(1+\delta)}+t(1-t)^{1-\alpha}
(c-a-\frac b{\Gamma(1+\delta)})\\
&\leq a+\frac {b}{\Gamma(1+\delta)}+c-a-\frac b{\Gamma(1+\delta)}
=c
\end{align*}
That is $\|Tu\|\leq c$. On the other hand, $f(u)\leq\|u\|$ for
$u\in K_c$, and $T:K_c\rightarrow K_c$ is completely continuous by
the above deduction and Lemma \ref{lm1}. Similarly, if $u\in
\overline {K_{\overline a}}$, then $\|u\|\leq \overline a$, and
$g(t,u(t))<(\overline a-a-\frac
b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha)$ for each
$t\in[0,1]$ by (H4), we obtain
\begin{align*}
&t^{1-\delta}(1-t)^{1-\alpha}|T u(t)|\\
&=|a(1-t)^{1-\alpha}+\frac
{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad+\frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
g(\tau,u(\tau))d\tau\,ds|\\
&\leq a(1-t)^{1-\alpha}+\frac
{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad+\frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
|g(\tau,u(\tau))|d\tau\,ds\\
&<a(1-t)^{1-\alpha}+\frac
{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad+\frac {t^{1-\delta}(1-t)^{1-\alpha}(\overline a-a-\frac
b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha)}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
d\tau\,ds\\
&=a(1-t)^{1-\alpha}+\frac
{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad+\frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1}
(\overline a-a-\frac b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha)(1-s)^\alpha ds\\
 &\leq a+\frac
{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)}
\int_0^t(t-s)^{\delta-1}(1-t)^{\alpha-1}ds\\
&\quad+\frac
{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1}
(\overline a-a-\frac b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha) ds\\
&= a+\frac {bt}{\Gamma(1+\delta)}+ {t(1-t)^{1-\alpha}}(\overline
a-a-\frac b{\Gamma(1+\delta)})\\
&\leq a+\frac {b}{\Gamma(1+\delta)}+\overline a-a-\frac
b{\Gamma(1+\delta)}
=\overline a\,.
\end{align*}
So $\|Tu\|<\overline a$ for $\|u\|\leq \overline a$ which proves
the condition (C2) of
Theorem \ref{theoB}.
We note that $u(t)=2\overline b$, $0\leq t\leq 1$ belong to
$K(f,\overline b,2\overline b)$. In fact $f(u)=f(2\overline b
)=2\overline b>\overline b$, so $\{u\in
K(f,\overline b,2\overline b)|f(u)>\overline b\}\neq \emptyset$.
In addition, if we choose $u\in K(f,\overline b,2\overline b)$,
then we have $\overline b<f(u)=u(\frac 12)\leq \|u\|\leq
2\overline b$, and by (H6)
\begin{align*}
f(Tu)&=Tu(\frac 12)\\
&=a(\frac12)^{\delta-1}+\frac
b{\Gamma(\delta)}\int_0^{\frac 12}(\frac 12-s)^{\delta-1}(1-s)^{\alpha-1}ds\\
&\quad+\frac 1{\Gamma(\delta)\Gamma(\alpha)}\int_0^{\frac 12}({\frac
12}-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
g(\tau,u(\tau))d\tau\,ds\\
&>\frac {(\frac 12)^{-1-\delta}(\overline b-a(\frac
12)^{\delta-1}-\frac {b(\frac 12)^\delta}
{\Gamma(1+\delta)})\frac{(1+\delta)\Gamma(1+\delta)
\Gamma(1+\alpha)}{\delta}}{\Gamma(\delta)\Gamma(\alpha)}\\
&\quad\times \int_0^{\frac12}(\frac 12-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}
d\tau\,ds
+a(\frac 12)^{\delta-1}+\frac b{\Gamma(\delta)}\int_0^{\frac 12}
(\frac 12-s)^{\delta-1}ds\\
&=\frac {(\frac 12)^{-1-\delta}(\overline b-a(\frac
12)^{\delta-1}-\frac {b(\frac 12)^\delta}
{\Gamma(1+\delta)})\frac{(1+\delta)\Gamma(1+\delta)}{\delta}}{\Gamma(\delta)}\int_0^{\frac
12}(\frac 12-s)^{\delta-1}
(1-s)^\alpha ds\\
&\quad+a(\frac 12)^{\delta-1}+\frac {b(\frac
12)^\delta}{\Gamma(1+\delta)}\\
&\geq \frac {(\frac 12)^{-1-\delta}(\overline b-a(\frac
12)^{\delta-1}-\frac {b(\frac 12)^\delta}
{\Gamma(1+\delta)})\frac{(1+\delta)\Gamma(1+\delta)}{\delta}}{\Gamma(\delta)}\int_0^{\frac
12}(\frac 12-s)^{\delta-1}(1-s)ds\\
&\quad+a(\frac 12)^{\delta-1}+\frac {b(\frac12)^\delta}{\Gamma(1+\delta)}\\
&> \frac {(\overline b-a(\frac 12)^{\delta-1}-\frac
{b(\frac 12)^\delta}{\Gamma(1+\delta)})(1+\delta)}\delta- \frac
{\overline b-a(\frac 12)^{\delta-1}-\frac {b(\frac12)^\delta}
{\Gamma(1+\delta)}}\delta\\
&\quad+a(\frac 12)^{\delta-1}+\frac {b(\frac
12)^\delta}{\Gamma(1+\delta)}
=\overline b\,.
\end{align*}
So $f(Tu)>\overline b$ for $u\in K(f,\overline b,2\overline b)$
which proves the condition (C1) of
Theorem \ref{theoB}.
Choosing $u\in K(f,\overline b,c)$ such that $\|Tu\|\geq
d=2\overline b$, then $\overline b<f(u)=u(\frac 12)\leq \|u\|\leq
c=d=2\overline b$, therefore by the above deduction that
$f(Tu)=Tu(\frac 12)>\overline b$, which proves the condition
(C3) of Theorem \ref{theoB}.
Thus by Theorem \ref{theo2},  $T$ has three fixed points in $K$, which proves
the theorem.
\end{proof}

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\end{document}