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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 43, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/43\hfil Smallest eigenvalues]
{Existence and comparison of smallest eigenvalues for a fractional\\
boundary-value problem}

\author[P. W. Eloe, J. T. Neugebauer \hfil EJDE-2014/43\hfilneg]
{Paul W. Eloe, Jeffrey T. Neugebauer}  % in alphabetical order

\address{Paul W. Eloe \newline 
Department of Mathematics, University of Dayton,
Dayton, OH 45469, USA}
\email{peloe1@udayton.edu}

\address{Jeffrey T. Neugebauer \newline 
Department of Mathematics and Statistics,
Eastern Kentucky University, Richmond, KY  40475, USA}
\email{jeffrey.neugebauer@eku.edu}

\thanks{Submitted August 22, 2013. Published February 10, 2014.}
\subjclass[2000]{26A33}
\keywords{Fractional boundary value problem; smallest eigenvalues;
 \hfill\break\indent $u_0$-positive operator}

\begin{abstract}
 The theory of $u_0$-positive operators with respect to a cone in a
 Banach space is applied to the fractional linear differential equations
 $$
 D_{0+}^{\alpha} u+\lambda_1p(t)u=0\quad\text{and}\quad 
 D_{0+}^{\alpha} u+\lambda_2q(t)u=0, 
 $$
 $0< t< 1$, with each satisfying the boundary conditions $u(0)=u(1)=0$.
 The existence of smallest positive eigenvalues is established,
 and a comparison theorem for smallest positive eigenvalues is obtained.
\end{abstract}

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

\section{Introduction}

We consider the eigenvalue problems
\begin{gather}\label{1}
D_{0+}^{\alpha} u+\lambda_1p(t)u=0, \quad 0<t<1,\\
\label{2}
D_{0+}^{\alpha} u+\lambda_2q(t)u=0, \quad 0<t<1,
\end{gather}
satisfying the boundary conditions
\begin{equation}\label{3}
u(0)=u(1)=0,
\end{equation}
where $1<\alpha\leq 2$ is a real number, $D_{0+}^{\alpha}$ is the standard
 Riemann-Liouville derivative, and $p(t)$ and $q(t)$ are continuous 
nonnegative functions on $[0,1]$, where neither $p(t)$ nor $q(t)$ 
vanishes identically on any nondegenerate compact subinterval of $[0,1]$.

The Krein Rutman theory \cite{KR62} has been employed extensively 
to establish the existence of and compare smallest eigenvalues 
of boundary value problems 
for differential equations, difference equations,
and dynamic equations on time scales.  
For some examples, see \cite{CDHY98,DEH99,EH89,EH92,GT67,HP90,Hof03,Neu11,Tra86}
 and the references therein.  A standard approach to show the existence 
of smallest eigenvalues is to apply the theory of $u_0$-positive
operators \cite{K64}.  Operators are defined whose eigenvalues are reciprocals 
of the eigenvalues of the original boundary value problems.
These operators are constructed by using the corresponding Green's function; 
the $u_0$-positivity of these operators are obtained by showing the operator
maps nonzero elements of a cone into the interior of that cone.  
Sign properties of the Green's function are employed to map the cone into 
the cone and higher order derivatives of the Green's functions are employed 
to map elements to the interior of the cone.  The theory of $u_0$-positivity,
as developed by Krasnosel'skii \cite{K64}, gives the existence of largest 
eigenvalues of the operator, with the corresponding eigenfunction existing 
in a cone.

In this article, we apply the standard approach, described above, 
to a boundary value problem for a fractional differential equation.
We are not aware of any previous application of $u_0$-positive operators 
to fractional differential equations.  Fixed point theory is now commonly 
applied to boundary value problems for fractional equations;
see, for example, the bibliography found in \cite{ABH10}.  
In many of these applications, the common Banach space to employ is $C[0,1];$ 
this space is not appropriate for applications of $u_0$-positivity to 
\eqref{1}, \eqref{2} or \eqref{1}, \eqref{3}, since the corresponding Green's 
function, $G(t,s)$, has unbounded slope at $t=0$.  The primary contribution
 of this article then is to consider an appropriate Banach space and cone, 
with nonempty interior, so that theory of $u_0$-positive operators does apply. 
 The motivation for the Banach space used here is found in \cite[Theorem 2.5]{D04}
 or \cite[Theorem 3.1]{SKM93}.  The particular approach to construct the 
Banach space and cone is modeled after \cite{Neu11}.  
For other work on eigenvalue problems of fractional differential equations, 
see \cite{A12,HG12,WZ12,ZLWW12}.

In Section 2, we state the preliminary definitions and theorems.  
In Section 3, we define the appropriate Banach space and establish 
the existence of and compare smallest eigenvalues of \eqref{1}, \eqref{2}
 and \eqref{1}, \eqref{3}.

\section{Preliminary definitions and theorems}


\begin{definition} \rm
Let $1<\alpha\leq 2$.  The $\alpha$-th Riemann-Liouville fractional 
derivative of the function $u:[0,1]\to\mathbb{R}$, denoted $D_{0+}^{\alpha} u$, is defined as
\[
D_{0+}^{\alpha} u(t)=\frac{1}{\Gamma(2-\alpha)}\frac{d^2}{dt^2}
\int^t_0 (t-s)^{2-\alpha-1}u(s)ds,
\]
provided the right-hand side exists.
\end{definition}

\begin{definition} \rm
Let $\mathcal{B}$ be a Banach space over $\mathbb{R}$. A closed nonempty subset $\mathcal{P}$ of $\mathcal{B}$ 
is said to be a cone provided
\begin{itemize}
\item[(i)] $\alpha u+\beta v \in\mathcal{P}$, for all $u,v\in\mathcal{P}$ and all $\alpha,\beta\geq0$, 
and
\item[(ii)] $u\in\mathcal{P}$ and $-u\in\mathcal{P}$ implies $u=0$.
\end{itemize}
\end{definition}

\begin{definition} \rm
A cone $\mathcal{P}$ is  solid if the interior, $\mathcal{P}^{\circ}$, of $\mathcal{P}$, is nonempty.  
A cone $\mathcal{P}$ is reproducing if $\mathcal{B}=\mathcal{P}-\mathcal{P}$; i.e., given $w\in\mathcal{B}$, 
there exist $u,v\in\mathcal{P}$ such that $w=u-v$.
\end{definition}

\begin{remark} \rm
Krasnosel'skii \cite{K64} showed that every solid cone is reproducing.
\end{remark}

Cones give rise to partial orders on Banach spaces and to partial 
orders on bounded linear operators on Banach spaces in a natural way.

\begin{definition} \rm
Let $\mathcal{P}$ be a cone in a real Banach space $\mathcal{B}$.  
If $u,v\in\mathcal{B}$, we say $u\leq v$ with respect to $\mathcal{P}$ if $v-u\in\mathcal{P}$.  
If both $M,N:\mathcal{B}\to\mathcal{B}$ are bounded linear operators, we say $M\leq N$ 
with respect to $\mathcal{P}$ if $Mu\leq Nu$ for all $u\in\mathcal{P}$.
\end{definition}

\begin{definition}\rm
A bounded linear operator $M:\mathcal{B}\to\mathcal{B}$ is $u_0$-positive with respect
to $\mathcal{P}$ if there exists $u_0\in\mathcal{P}\backslash\{0\}$ such that for each
 $u\in\mathcal{P}\backslash\{0\}$, there exist $k_1(u)>0$ and $k_2(u)>0$ such that 
$k_1u_0\leq Mu\leq k_2u_0$ with respect to $\mathcal{P}$.
\end{definition}

The following two results are fundamental to our comparison results and 
are attributed to Krasnosel'skii \cite{K64}.  
The proof of Theorem \ref{TheoremOne} can be found in Krasnosel'skii's 
book \cite{K64}, and the proof of Theorem \ref{TheoremTwo} is provided 
by Keener and Travis \cite{KT78} as an extension of Krasonel'skii's results.

\begin{theorem}\label{TheoremOne}
Let $\mathcal{B}$ be a real Banach space and let $\mathcal{P}\subset\mathcal{B}$ be a reproducing cone.  
Let $L:\mathcal{B}\to\mathcal{B}$ be a compact, $u_0$-positive, linear operator.
Then $L$ has an essentially unique eigenvector in $\mathcal{P}$, and the 
corresponding eigenvalue is simple, positive, and larger than the 
absolute value of any other eigenvalue.
\end{theorem}

\begin{theorem}\label{TheoremTwo}
Let $\mathcal{B}$ be a real Banach space and $\mathcal{P}\subset\mathcal{B}$ be a cone. 
 Let both $M,N:\mathcal{B}\to\mathcal{B}$ be bounded, linear operators and assume 
that at least one of the operators is $u_0$-positive.
If $M\leq N$, $Mu_1\geq\lambda_1u_1$ for some $u_1\in\mathcal{P}$ and
some $\lambda_1>0$, and $Nu_2\leq \lambda_2u_2$ for some
$u_2\in\mathcal{P}$ and some $\lambda_2>0$, then $\lambda_1\leq \lambda_2$.
 Furthermore, $\lambda_1=\lambda_2$ implies $u_1$ is a scalar multiple
of $u_2$.
\end{theorem}

\section{Comparison of smallest eigenvalues}

In \cite{BL05}, Bai and L\:u showed the Green's function for $-D_{0+}^{\alpha} u(t)=0$ 
satisfying \eqref{3} is 
\begin{equation}
G(t,s)=\begin{cases}
\frac{[t(1-s)]^{\alpha-1}-(t-s)^{\alpha-1}}{\Gamma(\alpha)}, 
 & 0\leq s \leq t\leq 1,\\[4pt]
\frac{[t(1-s)]^{\alpha-1}}{\Gamma(\alpha)}, 
 & 0\leq t\leq s\leq 1.
\end{cases}
\end{equation}

Define the Banach Space
\[
\mathcal{B}= \{ u: u=t^{\alpha -1}v, v\in C^{1}[0,1], v(1)=0\},
\]
with the norm
\[
\|u\|=|v'|_0,
\]
where $|v'|_0=\sup_{t\in [0,1]} |v'(t)|$ denotes the usual supremum norm.

Note that for $v\in C^1[0,1]$, $v(1)=0$, $0\leq t\leq1$,
$$
|v(t)|=|v(t)-v(1)|=\big|\int_1^tv'(s)ds\big|\le (1-t)|v'|\le \|u\|.
$$
Therefore, $|v|_0\le \|u\|=|v'|_0$ and
$$
|u|_0=|t^{\alpha -1}v|_0\le t^{\alpha -1}\|u\|,
$$
implies
$|u|_0\le \|u\|$.

Define the linear operators
\begin{equation}
Mu(t)=\int^1_0 G(t,s)p(s)u(s)ds
\end{equation}
and
\begin{equation}
Nu(t)=\int^1_0 G(t,s)q(s)u(s)\;ds.
\end{equation}

\begin{theorem}  
The operators $M,N:\mathcal{B}\to \mathcal{B}$ are compact linear operators.
\end{theorem}

\begin{proof} 
 We first show $M:\mathcal{B}\to \mathcal{B}$.  Let $u\in \mathcal{B}$.  So there is a $v\in C^1[0,1]$ 
such that $u=t^{\alpha-1}v$.  Since $v\in C^1[0,1]$ and
 $p\in C[0,1]$, let $L=|v|_0$ and let $P=|p|_0$.  Now
\begin{align*}
Mu(t) 
&=\int_0^{1}\frac{t^{\alpha -1}(1-s)^{\alpha -1}}{\Gamma (\alpha )}
p(s)u(s)ds-\int_0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha)}p(s)u(s)ds\\
&=t^{\alpha-1}\Big(\int_0^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}p(s)u(s)ds
-t^{1-\alpha}\int_0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha)}p(s)u(s)ds\Big).
\end{align*}
Define
\[
g(t)=\begin{cases}
0, & t=0,\\
t^{1-\alpha}\int_0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha)}p(s)u(s)ds,
& 0<t\le 1.
\end{cases}
\]
First, note $g\in C^1(0,1]$.  Now
\begin{align*}
|g(t)|
&=\big|t^{1-\alpha}\int_0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}p(s)u(s)ds
  \big|\\
&=\big|t^{1-\alpha}\int_0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}
 p(s)s^{\alpha-1}v(s)ds\big|\\
&\leq PLt^{1-\alpha}\int_0^t(t-s)^{\alpha -1}s^{\alpha -1}ds\\
&\leq PLt^{1-\alpha}t^{\alpha -1}\int_0^t(t-s)^{\alpha -1}ds \\
&=\frac{PLt^{\alpha}}{\alpha},
\end{align*}
where $\frac{PL}{\alpha}\ge0$.  So $\lim_{t\to 0^+}g(t)=g(0)=0$ and $g\in C[0,1]$.

Also, for $t>0$,
\begin{align*}
|g'(t)|
&=\Big|(1-\alpha )t^{-\alpha}\int_0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}
 p(s)u(s)ds \\
&\quad +(\alpha -1)t^{1-\alpha}\int_0^t
 \frac{(t-s)^{\alpha -2}}{\Gamma (\alpha)}p(s)u(s)ds\Big|\\
&\leq \Big|(1-\alpha )t^{-\alpha}\int_0^t
 \frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}p(s)s^{\alpha-1}v(s)ds\Big|\\
&\quad +\Big|t^{1-\alpha}\int_0^t\frac{(t-s)^{\alpha -2}}{\Gamma (\alpha -1)}
 p(s)s^{\alpha-1}v(s)ds\Big|\\
&\leq (\alpha -1)PLt^{-\alpha}t^{\alpha-1}\int_0^t(t-s)^{\alpha -1}ds
 +PLt^{1-\alpha}t^{\alpha-1}\int_0^t(t-s)^{\alpha -2}ds\\
&=  \big(\frac{\alpha-1}{\alpha}+\frac{1}{\alpha-1}\big)PLt^{\alpha-1}.
\end{align*}
So, $\lim_{t\to 0^{+}}g'(t)=0$.  Moreover, using the definition of derivative 
and L'Hospital's rule,
\[
g'(0)=\lim_{t\to 0^{+}} \frac{g(t)-g(0)}{t}=\lim_{t\to 0^{+}} \frac{g(t)}{t}
=\lim_{t\to 0^{+}} g'(t) =0,
\]
and so $g'\in C[0,1]$.

Now set
\[
\hat{v}(t) =\Big(\int_0^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}p(s)u(s)ds
-t^{1-\alpha}\int_0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha)}p(s)u(s)ds\Big).
\]
It is an easy calculation to verify that $\hat{v}(1)=0$.
Thus $Mu\in\mathcal{B}$.  So $M:\mathcal{B}\to\mathcal{B}$.  The proof that $N:\mathcal{B}\to \mathcal{B}$ is similar.

We now show that $M:\mathcal{B}\to\mathcal{B}$ is a compact operator.
Let $L>0$ and consider
\[
K=\{ u\in\mathcal{B}: \|u\|\le L\}
\]
or more appropriately consider
\[
\hat{K}=\{ v\in C^{1}[0,1]: v(1) =0, |v'|_0\le L\}.
\]
Define
\[
\hat{M}(v)(t)=\Big(\int_0^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}
 p(s)s^{\alpha -1}v(s)ds-t^{1-\alpha}\int_0^t
 \frac{(t-s)^{\alpha -1}}{\Gamma (\alpha)}p(s)s^{\alpha -1}v(s)ds\Big).
\]
To show that $M$ is compact on $\mathcal{B}$ it is sufficient to show that
$\{(\hat{M}(v))':v\in\hat{K}\}$ is uniformly bounded and equicontinuous
on $[0,1]$.  We provide the details for equicontinuity as the details
for uniform boundedness are straightforward.

Assume $|p|_0=P$ and assume $|v|_0\le L$. Let $\epsilon >0$.
As in the calculations above for $g'$, $ (\hat{M}(v))'(0)=0$ and 
\[
 |(\hat{M}(v))'(t)|\le \big(\frac{\alpha-1}{\alpha}+\frac{1}{\alpha-1}\big)
PLt^{\alpha-1}.
\]
Thus, there exists  $\delta_1>0$ such that if $|t|<\delta_1$ then
$|(\hat{M}(v))'(t)|<\frac{\epsilon}{2}$.

On $[\delta_1,1]$, $\{(\hat{M}(v))':v\in\hat{K}\}$ is shown to be equicontinuous by showing that $\{(\hat{M}(v))'':v\in\hat{K}\}$ is uniformly bounded.  Now
\begin{align*}
(\hat{M}(v))'(t)&=(1-\alpha )t^{-\alpha}\int_0^t
\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}p(s)u(s)ds\\
&\quad +(\alpha -1)t^{1-\alpha}\int_0^t\frac{(t-s)^{\alpha -2}}{\Gamma (\alpha)}
p(s)u(s)ds,
\end{align*}
and so
\begin{align*}
(\hat{M}(v))''(t)
&=-\alpha (1-\alpha )t^{-\alpha-1}\int_0^t
 \frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}p(s)s^{\alpha -1}v(s)ds\\
&\quad -(\alpha -1)^{2}t^{-\alpha}\int_0^t
 \frac{(t-s)^{\alpha -2}}{\Gamma (\alpha)}p(s)s^{\alpha -1}v(s)ds\\
&\quad -(\alpha -1)^{2}t^{-\alpha}\int_0^t
 \frac{(t-s)^{\alpha -2}}{\Gamma (\alpha )}p(s)s^{\alpha-1}v(s)ds\\
&\quad +(\alpha -1)(\alpha -2)t^{1-\alpha}\int_0^t
 \frac{(t-s)^{\alpha -3}}{\Gamma (\alpha -1)}p(s)s^{\alpha-1}v(s)ds.
\end{align*}
Each of the four terms can be bounded by a constant multiple of $t^{\alpha -2}$.

For the first term, notice
\[
\Big|t^{-\alpha-1}\int_0^t(t-s)^{\alpha -1}p(s)s^{\alpha -1}v(s)ds\Big|
\le PLt^{-\alpha-1}\Big|\int_0^t(t-s)^{\alpha -1}s^{\alpha -1}ds\Big|.
\]
Set $s=rt$.  So
\begin{align*}
t^{-\alpha-1}\Big|\int_0^t(t-s)^{\alpha -1}s^{\alpha -1}ds\Big|
&= t^{-\alpha -1}t^{\alpha -1}t^{\alpha -1}t
\Big|\int_0^{1}(1-r)^{\alpha -1}r^{\alpha -1}dr\Big|\\
&= t^{\alpha -2}|B(\alpha, \alpha )|,
\end{align*}
where $B$ denotes the beta function.

In dealing with the second and third terms, first note
\[
\Big|t^{-\alpha}\int_0^t\frac{(t-s)^{\alpha -2}}{\Gamma (\alpha)}p(s)s^{\alpha -1}
v(s)ds\Big|
\le PL t^{-\alpha}\Big|\int_0^t(t-s)^{\alpha-2}s^{\alpha-1}ds\Big|.
\]
Set $s=rt$.  Then
\begin{align*}
t^{-\alpha}\Big|\int_0^t(t-s)^{\alpha-2}s^{\alpha-1}ds\Big|
&=t^{-\alpha}t^{\alpha-2}t^{\alpha-1}t
 \Big|\int_0^t(1-r)^{\alpha-2}r^{\alpha-1}dr\Big|\\
&=t^{\alpha-2}|B(\alpha,\alpha-1)|.
\end{align*}
Notice $B(\alpha,\alpha-1)$ is well-defined since $1<\alpha\le 2$.

Last, we obtain an analogous estimate for the fourth term. 
If $\alpha=2$, this term is zero.  If $1<\alpha<2$, first integrate 
by parts to obtain
\[
\int_0^t(t-s)^{\alpha -3}s^{\alpha -1}ds
=\frac{\alpha-1}{\alpha-2}\int^t_0(t-s)^{\alpha-2}s^{\alpha-2}ds.
\]
Thus,
\begin{align*}
\Big|t^{1-\alpha}\int_0^t\frac{(t-s)^{\alpha -3}}{\Gamma (\alpha -1)}
p(s)s^{\alpha-1}v(s)ds\Big|
&\le PL t^{1-\alpha}\Big|\int_0^t(t-s)^{\alpha -3}s^{\alpha -1}ds\Big|\\
&=PL t^{1-\alpha}\Big|\frac{\alpha-1}{\alpha-2}
\int^t_0(t-s)^{\alpha-2}s^{\alpha-2}ds\Big|.
\end{align*}
Again, by setting $s=rt$, we obtain
\begin{align*}
t^{1-\alpha}\Big|\int^t_0(t-s)^{\alpha-2}s^{\alpha-2}ds\Big|
&=t^{1-\alpha}t^{\alpha-2}t^{\alpha-2}t
\Big|\int^t_0 (1-r)^{\alpha-2}r^{\alpha-2}dr\Big|\\
&=t^{\alpha-2}|B(\alpha-1,\alpha-1)|.
\end{align*}
Again, $B(\alpha -1, \alpha -1)$ is well-defined since $1<\alpha < 2$.
Therefore, if $\alpha\ne 2$,
\begin{align*}
|(\hat{M}(v))''(t)|
&\le PL \Big[\frac{\alpha(\alpha-1)|B(\alpha,\alpha)|}{\Gamma(\alpha)}
 +\frac{2(\alpha-1)^2|B(\alpha,\alpha-1)|}{\Gamma(\alpha)}\\
&\quad +\frac{(\alpha-1)^2|B(\alpha-1,\alpha-1)|}{\Gamma(\alpha-1)}\Big]
t^{\alpha-2},
\end{align*}
and if $\alpha=2$,
\[
|(\hat{M}(v))''(t)|\le \frac{4PL}{3}.
\]
So $\{(\hat{M}(v))'':v\in\hat{K}\}$ is uniformly bounded on $[\delta_1, 1]$.

Since $\{(\hat{M}(v))'':v\in\hat{K}\}$ is uniformly bounded on $[\delta_1, 1]$,
there exists $\delta_2>0$ such that if $|t_1-t_2|<\delta_2$,
$t_1, t_2 \in [\delta_1, 1]$, then 
$|(\hat{M}(v))'(t_1)-(\hat{M}(v))'(t_2)|<\frac{\epsilon}{2}$.

Set $\delta =\min \{\delta_1, \delta_2\}$. If $|t_1-t_2|<\delta$, 
$t_1, t_2 \in [0, \delta_1]$, then
$$ 
|(\hat{M}(v))'(t_1)-(\hat{M}(v))'(t_2)|
\le |(\hat{M}(v))'(t_1)|+|(\hat{M}(v))'(t_2)|<\epsilon.
$$
If $|t_1-t_2|<\delta$, $t_1, t_2 \in [\delta_1, 1]$, then
$$ 
|(\hat{M}(v))'(t_1)-(\hat{M}(v))'(t_2)|\le \frac{\epsilon}{2}<\epsilon.
$$
If $|t_1-t_2|<\delta$, $0\le t_1<\delta_1\le t_2\le 1$, then
\begin{align*}
&|(\hat{M}(v))'(t_1)-(\hat{M}(v))'(t_2)|\\
&\le  |(\hat{M}(v))'(t_1)-(\hat{M}(v))'(\delta_1)|
 +|(\hat{M}(v))'(\delta_1)-(\hat{M}(v))'(t_2)|\\
&<  \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.
\end{align*}
Details for the operator $N$ are similar and the proof is complete.
\end{proof}

Define the cone
\[
\mathcal{P}=\{u\in \mathcal{B}: u(t)\geq 0\text{ for } t\in[0,1]\}.
\]

\begin{lemma} 
The cone $\mathcal{P}$ is solid in $B$ and hence reproducing.
\end{lemma}

\begin{proof} 
Define
\begin{equation}\label{cone}
\Omega:=\{u\in \mathcal{B}\;|\; u(t)>0\text{ for }t\in(0,1), v(0)>0, v'(1)<0,
 \text{ where }u=t^{\alpha-1}v\}.
\end{equation}
We will show $\Omega\subset \mathcal{P}^\circ$.  Let $u\in\Omega$.  
Since $v(0)>0$, there exists an $\epsilon_1>0$ such that $v(0)-\epsilon_1>0$.  
Since $v\in C^1[0,1]$, there exists an $a\in(0,1)$ such that $v(t)> \epsilon_1$ 
for all $t\in(0,a)$.  So $u(t)=t^{\alpha-1}v(t)> \epsilon_1t^{\alpha-1}$ 
for all $t\in(0,a)$.  Now, since $v'(1)<0$, there exists an $\epsilon_2>0$ 
such that $v'(1)+\epsilon_2<0$.  Then, since $v(1)=0$ and $-v'(1)>\epsilon_2$, 
there exists a $b\in(a,1)$ such that $v(t)>(1-t)\epsilon_2$ for all $t\in(b,1]$.  
Thus $u(t)> b^{\alpha-1}(1-t)\epsilon_2$ for all $t\in(b,1]$.  
Also, since $u(t)>0$ on $[a,b]$, there exists an $\epsilon_3>0$ such that 
$u(t)-\epsilon_3>0$ for all $t\in[a,b]$.

Let $\epsilon=\min\big\{\frac{\epsilon_1}{2},\frac{b^{\alpha-1}\epsilon_2}{2},
\frac{\epsilon_3}{2}\big\}$. Define 
$B_\epsilon(u)=\{\hat{u}\in\mathcal{B}:\|u-\hat{u}\|<\epsilon\}$. 
 Let $\hat{u}\in B_\epsilon(u)$.  So $\hat{u}=t^{\alpha-1}\hat{v}$, 
where $\hat{v}\in C^1[0,1]$ with $\hat{v}(1)=0$.  
Now 
$$
|\hat{u}(t)-u(t)|\leq t^{\alpha-1}\|\hat{u}-u\|<\epsilon t^{\alpha-1}. 
$$ 
So for $t\in(0,a)$, 
$\hat{u}(t)>u(t)-t^{\alpha-1}\epsilon>t^{\alpha-1}
\epsilon_1-t^{\alpha-1}\epsilon_1/2=t^{\alpha-1}\epsilon_1/2$. 
 So $\hat{u}(t)>0$ for $t\in(0,a)$.  By the Mean Value Theorem, 
for $t\in(b,1)$, $|\hat{u}(t)-u(t)|\leq(1-t)\|\hat{u}-u\|<(1-t)\epsilon$.  
So for $t\in(b,1)$, 
$$
\hat{u}(t)>u(t)-(1-t)\epsilon>b^{\alpha-1}(1-t)\epsilon_2-(1-t)b^{\alpha-1}
\epsilon_2/2=(1-t)b^{\alpha-1}\epsilon_2/2.
$$
So for $t\in(b,1)$, $\hat{u}(t)>0$.  
Also, $|\hat{u}(t)-u(t)|\le\|\hat{u}-u\|<\epsilon$.  So for $t\in[a,b]$, 
$\hat{u}(t)>u(t)-\epsilon>\epsilon_3-\epsilon_3/2>0$. 
So $\hat{u}(t)>0$ for all $t\in[a,b]$.  So $\hat{u}\in\mathcal{P}$ and thus 
$B_\epsilon(u)\subset\mathcal{P}$.  So $\Omega\subset\mathcal{P}^\circ$.
\end{proof}

\begin{lemma}\label{u0}
The bounded linear operators $M$ and $N$ are $u_0$-positive with respect to $\mathcal{P}$.
\end{lemma}

\begin{proof}
First, we show $M:\mathcal{P}\backslash \{0\}\to\Omega\subset\mathcal{P}^{\circ}$.  
Let $u\in\mathcal{P}$.  So $u(t)\geq0$.  Then since $G(t,s)\geq0$ on $[0,1]\times[0,1]$ 
and $p(t)\geq0$ on $[0,1]$,
\[
Mu(x)=\int^{1}_0G(t,s)p(s)u(s)ds\geq0,
\]
for $0\leq t\leq 1$.  So $M:\mathcal{P}\to\mathcal{P}$.

Now let $u\in\mathcal{P}\backslash \{0\}$.  So there exists a compact interval 
$[\alpha,\beta]\subset[0,1]$ such that $u(t)>0$ and $p(t)>0$ for all 
$t\in[\alpha,\beta]$.  Then, since $G(t,s)>0$ on $(0,1)\times(0,1)$,
\begin{align*}
Mu(t)&=\int^{1}_0G(t,s)p(s)u(s)ds\\
&\geq\int^{\beta}_{\alpha}G(t,s)p(s)u(s)ds
>0,
\end{align*}
for $0<t<1$.
Now
\[
Mu(t)=t^{\alpha-1}\Big(\int_0^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}
p(s)u(s)ds-t^{1-\alpha}\int_0^t
\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha)}p(s)u(s)ds\Big).
\]
Let
\[
v(t)=\int_0^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}p(s)u(s)ds
-t^{1-\alpha}\int_0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha)}p(s)u(s)ds.
\]
So $v(0)=\int_0^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}p(s)u(s)ds>0$ and
\[
v'(1)
=-(1-\alpha )\int_0^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}p(s)u(s)ds
 -(\alpha -1)\int_0^{1}\frac{(1-s)^{\alpha -2}}{\Gamma (\alpha)}p(s)u(s)ds<0.
\]
So $M:\mathcal{P}\backslash\{0\}\to\Omega\subset\mathcal{P}^\circ$.

Now choose any $u_0\in\mathcal{P}\backslash \{0\}$, and let $u\in\mathcal{P}\backslash \{0\}$.  
So $Mu\in\Omega\subset\mathcal{P}^{\circ}$.  Choose $k_1>0$ sufficiently small and $k_2$ 
sufficiently large so that $Mu-k_1u_0\in\mathcal{P}^{\circ}$ and 
$u_0-\frac{1}{k_2}Mu\in\mathcal{P}^{\circ}$.  So $k_1u_0\leq Mu$ with respect to $\mathcal{P}$ 
and $Mu\leq k_2u_0$ with respect to $\mathcal{P}$.  
Thus $k_1u_0\leq Mu\leq k_2u_0$ with respect to $\mathcal{P}$ and so $M$ is $u_0$-positive 
with respect to $P$.  Similarly, $N$ is $u_0$-positive.
\end{proof}

\begin{remark} \label{remark} \rm
Notice that
\[
\Lambda u=Mu=\int^{1}_0G(t,s)p(s)u(s)ds,
\]
if and only if
\[
u(t)=\frac{1}{\Lambda}\int^{1}_0G(t,s)p(s)u(s)ds,
\]
if and only if
\[
D_{0+}^{\alpha} u(t)+\frac{1}{\Lambda}p(t)u(t)=0, \;\;0<t<1,
\]
with
$u(0)=u(1)=0$.

So the eigenvalues of \eqref{1},\eqref{3} are reciprocals of eigenvalues 
of $M$, and conversely.  Similarly, eigenvalues of \eqref{2},\eqref{3} 
are reciprocals of eigenvalues of $N$, and conversely.
\end{remark}

\begin{theorem} \label{eigenvalue}
Let $\mathcal{B}$, $\mathcal{P}$, $M$, and $N$ be defined as earlier.  
Then $M$ (and $N$) has an eigenvalue that is simple, positive, 
and larger than the absolute value of any other eigenvalue, with an 
essentially unique eigenvector that can be chosen to be in $\mathcal{P}^{\circ}$.
\end{theorem}

\begin{proof}
Since $M$ is a compact linear operator that is $u_0$-positive with respect 
to $\mathcal{P}$, by Theorem \ref{TheoremOne}, $M$ has an essentially unique eigenvector, 
say $u\in\mathcal{P}$, and eigenvalue $\Lambda$ with the above properties. 
Since $u\neq0$, $Mu\in\Omega\subset\mathcal{P}^{\circ}$ and 
$u=M\left(\frac{1}{\Lambda}u\right)\in\mathcal{P}^{\circ}$.
\end{proof}


\begin{theorem}\label{comparison}
Let $\mathcal{B}$, $\mathcal{P}$, $M$, and $N$ be defined as earlier.  
Let $p(t)\leq q(t)$ on $[0,1]$.  Let $\Lambda_1$ and $\Lambda_2$ be the 
eigenvalues defined in Theorem \ref{eigenvalue} associated with $M$ and $N$, 
respectively, with the essentially unique eigenvectors $u_1$ and $u_2\in\mathcal{P}^{\circ}$.  Then $\Lambda_1\leq\Lambda_2$, and $\Lambda_1=\Lambda_2$ if and only if $p(t)=q(t)$ on $[0,1]$.
\end{theorem}

\begin{proof}
Let $p(t)\leq q(t)$ on $[0,1]$.  So for any $u\in\mathcal{P}$ and $t\in [0,1]$,
\[
(Nu-Mu)(t)=\int^{1}_0G(t,s)(q(s)-p(s))u(s)ds\geq0.
\]
So $Nu-Mu\in\mathcal{P}$ for all $u\in\mathcal{P}$, or $M\leq N$ with respect to $\mathcal{P}$.
Then by Theorem \ref{TheoremTwo}, $\Lambda_1\leq\Lambda_2$.

If $p(t)=q(t)$, then $\Lambda_1=\Lambda_2$.  Now suppose $p(t)\neq q(t)$.  
So $p(t)<q(t)$ on some subinterval $[\alpha,\beta]\subset[0,1]$.  
Then $(N-M)u_1\in\Omega\subset\mathcal{P}^{\circ}$ and so there exists $\epsilon>0$ 
such that $(N-M)u_1-\epsilon u_1\in\mathcal{P}$. So  
$\Lambda_1u_1+\epsilon u_1=Mu_1+\epsilon u_1\leq Nu_1$, implying 
$Nu_1\geq (\Lambda_1+\epsilon)u_1$.  Since $N\leq N$ and $Nu_2=\Lambda_2u_2$,
 by Theorem \ref{TheoremTwo}, $\Lambda_1+\epsilon \leq \Lambda_2$, 
or $\Lambda_1<\Lambda_2$.
\end{proof}

By Remark \ref{remark}, the following theorem is an immediate consequence 
of Theorems \ref{eigenvalue} and \ref{comparison}.

\begin{theorem}
Assume the hypotheses of Theorem \ref{comparison}.  
Then there exists smallest positive eigenvalues $\lambda_1$ and
 $\lambda_2$ of \eqref{1},\eqref{3} and \eqref{2},\eqref{3}, respectively, 
each of which is simple, positive, and less than the absolute value of 
any other eigenvalue of the corresponding problems.  Also, eigenfunctions 
corresponding to $\lambda_1$ and $\lambda_2$ may be chosen to belong 
to $\mathcal{P}^{\circ}$.  Finally, $\lambda_1\geq\lambda_2$, and $\lambda_1=\lambda_2$ 
if and only if $p(t)=q(t)$ for all $t\in[0,1]$.
\end{theorem}


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\end{document}