# fractional-order boundary value problem with Sturm-Liouville boundary
conditions

Douglas R. Anderson  Department of Mathematics
Concordia College
Moorhead, MN 56562 USA<EMAIL_ADDRESS>and  Richard I. Avery  College of
Arts and Sciences
Dakota State University
Madison, SD 57042, USA<EMAIL_ADDRESS>
###### Abstract.

Using the new conformable fractional derivative, which differs from the
Riemann-Liouville and Caputo fractional derivatives, we reformulate the
second-order conjugate boundary value problem in this new setting. Utilizing
the corresponding positive fractional Green’s function, we apply a functional
compression-expansion fixed point theorem to prove the existence of a positive
solution. We then compare our results favorably to those based on the Riemann-
Liouville fractional derivative.

###### Key words and phrases:

conformable fractional derivative, boundary value problem, positivity, Green’s
function, conjugate conditions

###### 2010 Mathematics Subject Classification:

26A33

##  1\. Introduction

The search for the existence of positive solutions and multiple positive
solutions to nonlinear fractional boundary value problems has expanded greatly
over the past decade; for some recent examples please see
[3-9,11,14,15,17-20]. In all of these works and the references cited therein,
however, the definition of the fractional derivative used is either the Caputo
or the Riemann-Liouville fractional derivative, involving an integral
expression and the gamma function. Recently [10, 11, 14] a new definition has
been formulated and dubbed the conformable fractional derivative. In this
paper, we use this fractional derivative of order $\alpha$, given by

$D^{\alpha}f(t):=\lim_{\varepsilon\rightarrow 0}\frac{f(te^{\varepsilon
t^{-\alpha}})-f(t)}{\varepsilon},\quad D^{\alpha}f(0)=\lim_{t\rightarrow
0^{+}}D^{\alpha}f(t);$ (1.1)

note that if $f$ is differentiable, then

$D^{\alpha}f(t)=t^{1-\alpha}f^{\prime}(t),$ (1.2)

where $f^{\prime}(t)=\lim_{\varepsilon\rightarrow
0}[f(t+\varepsilon)-f(t)]/\varepsilon$. Using this new definition of the
fractional derivative, we investigate a conformable fractional boundary value
problem with Sturm-Liouville boundary conditions. With the fractional
differential equation and fractional boundary conditions established, we find
the corresponding Green’s function and prove its positivity under appropriate
assumptions. This work thus sets the stage for employing a functional
compression-expansion fixed point theorem to prove the existence of a positive
solution to the special case of conjugate boundary conditions. We then compare
our existence result to that of Bai and Lü [3].

##  2\. Two Iterated Fractional Derivatives

We begin by considering two iterated fractional derivatives in the
differential operator, together with two-point boundary conditions, as
illustrated in the nonlinear boundary value problem

$\displaystyle-D^{\beta}D^{\alpha}x(t)=f(t,x(t)),\qquad 0\leq t\leq 1,$ (2.1)
$\displaystyle\gamma x(0)-\delta D^{\alpha}x(0)=0=\eta x(1)+\zeta
D^{\alpha}x(1),$ (2.2)

where $\alpha,\beta\in(0,1]$ and the derivatives are conformable fractional
derivatives (1.1), with $\gamma,\delta,\eta,\zeta\geq 0$ and
$d:=\eta\delta+\gamma\zeta+\gamma\eta/\alpha>0$. Note that if $x$ is
$\alpha$-differentiable and $t^{1-\alpha}x^{\prime}$ is
$\beta$-differentiable, then using (1.2) we could rewrite (2.1) as

$-t^{1-\beta}\left(t^{1-\alpha}x^{\prime}(t)\right)^{\prime}=f(t,x(t)),\qquad
0\leq t\leq 1,$

where the prime indicates the derivative $d/dt$. Before we find Green’s
function for (2.1), (2.2), and prove that it is positive, we first define the
$\beta$-fractional integral.

######  Definition 2.1.

Let $\beta\in(0,1]$ and $0\leq a<b$. A function
$f:[a,b]\rightarrow{\mathbb{R}}$ is $\beta$-fractional integrable on $[a,b]$
if the integral

$\int_{a}^{b}f(s)d_{\beta}s:=\int_{a}^{b}f(s)s^{\beta-1}ds$

exists and is finite. Thus the integral can be interpreted either as a
Riemann-Stieltjes integral or an improper Riemann integral.

######  Theorem 2.1.

Let $\alpha,\beta\in(0,1]$. The corresponding Green’s function for the
homogeneous problem

$-D^{\beta}D^{\alpha}x(t)=0$

satisfying the boundary conditions (2.2) is given by

$G(t,s)=\begin{cases}\frac{1}{d}\left[\delta+\frac{\gamma}{\alpha}s^{\alpha}\right]\left[\zeta+\frac{\eta}{\alpha}\left(1-t^{\alpha}\right)\right]&:s\leq
t,\\\
\frac{1}{d}\left[\delta+\frac{\gamma}{\alpha}t^{\alpha}\right]\left[\zeta+\frac{\eta}{\alpha}\left(1-s^{\alpha}\right)\right]&:t\leq
s,\end{cases}$ (2.3)

where we assume the parameters satisfy $\gamma,\delta,\eta,\zeta\geq 0$ and
$d=\eta\delta+\gamma\zeta+\gamma\eta/\alpha>0$.

###### Proof.

We will show that

$x(t)=\int_{0}^{1}G(t,s)h(s)d_{\beta}s,$

for $G$ given by (2.3), is a solution to the linear boundary value problem

$-D^{\beta}D^{\alpha}x(t)=h(t)$

with boundary conditions (2.2).

For any $t\in[0,1]$, using the branches of (2.3) we have

$\displaystyle x(t)$ $\displaystyle=$
$\displaystyle\frac{1}{d}\left[\zeta+\frac{\eta}{\alpha}\left(1-t^{\alpha}\right)\right]\int_{0}^{t}\left[\delta+\frac{\gamma}{\alpha}s^{\alpha}\right]h(s)d_{\beta}s$
$\displaystyle+\frac{1}{d}\left[\delta+\frac{\gamma}{\alpha}t^{\alpha}\right]\int_{t}^{1}\left[\zeta+\frac{\eta}{\alpha}\left(1-s^{\alpha}\right)\right]h(s)d_{\beta}s.$

Taking the $\alpha$-fractional derivative yields

$\displaystyle D^{\alpha}x(t)$ $\displaystyle=$
$\displaystyle-\frac{\eta}{d}\int_{0}^{t}\left[\delta+\frac{\gamma}{\alpha}s^{\alpha}\right]h(s)d_{\beta}s+\frac{\gamma}{d}\int_{t}^{1}\left[\zeta+\frac{\eta}{\alpha}\left(1-s^{\alpha}\right)\right]h(s)d_{\beta}s.$

Checking the first boundary condition, we see that

$\gamma x(0)-\delta D^{\alpha}x(0)=0.$

Moreover, in checking the second boundary condition we get

$\eta x(1)+\zeta D^{\alpha}x(1)=0.$

Taking the $\beta$-fractional derivative of the $\alpha$-fractional derivative
yields

$\displaystyle D^{\beta}D^{\alpha}x(t)$ $\displaystyle=$
$\displaystyle-\frac{\eta}{d}\left[\delta+\frac{\gamma}{\alpha}t^{\alpha}\right]h(t)t^{\beta-1}t^{1-\beta}-\frac{\gamma}{d}\left[\zeta+\frac{\eta}{\alpha}\left(1-t^{\alpha}\right)\right]h(t)t^{\beta-1}t^{1-\beta}$
$\displaystyle=$
$\displaystyle-\frac{1}{d}h(t)\left[\eta\delta+\gamma\zeta+\frac{\gamma\eta}{\alpha}\right]=-h(t),$

which is what we set out to prove. ∎

######  Corollary 2.1 (Fractional Conjugate and Right-Focal Problems).

Let $\alpha,\beta\in(0,1]$. The corresponding Green’s function for the
homogeneous problem

$-D^{\beta}D^{\alpha}x(t)=0$

satisfying the conjugate boundary conditions $x(0)=x(1)=0$ is given by

$G(t,s)=\begin{cases}\frac{1}{\alpha}s^{\alpha}\left(1-t^{\alpha}\right)&:s\leq
t,\\\ \frac{1}{\alpha}t^{\alpha}\left(1-s^{\alpha}\right)&:t\leq
s,\end{cases}$ (2.4)

and the corresponding Green’s function for the homogeneous problem

$-D^{\beta}D^{\alpha}x(t)=0$

satisfying the right-focal-type boundary conditions $x(0)=D^{\alpha}x(1)=0$ is
given by

$G(t,s)=\begin{cases}\frac{1}{\alpha}s^{\alpha}&:s\leq t,\\\
\frac{1}{\alpha}t^{\alpha}&:t\leq s.\end{cases}$ (2.5)

######  Remark 2.1.

Note that the conformable fractional Green’s function based on (1.1) and given
above for the conjugate boundary conditions in (2.4) differs from that found
for example in Bai and Lü [3], where the Riemann-Liouville fractional
derivative is used.

######  Theorem 2.2 (Bounds on Green’s Function).

For $G(t,s)$ given in (2.3), we have the following bounds. First,

$g_{1}(t)G(s,s)<G(t,s)\leq G(s,s)$ (2.6)

for $t,s\in[0,1]$, where

$g_{1}(t):=\min\left\\{\frac{\alpha\delta+\gamma
t^{\alpha}}{\alpha\delta+\gamma},\frac{\alpha\zeta+\eta\left(1-t^{\alpha}\right)}{\alpha\zeta+\eta}\right\\}.$
(2.7)

Next, for integers $n\geq 3$,

$\min_{\frac{1}{n}\leq t\leq 1-\frac{1}{n}}G(t,s)\geq g_{2}(s)G(s,s)$ (2.8)

for $g_{2}$ given by

$g_{2}(s):=\begin{cases}\displaystyle\frac{\alpha\zeta+\eta\left(1-\left(1-1/n\right)^{\alpha}\right)}{\alpha\zeta+\eta\left(1-s^{\alpha}\right)}&:s\in[0,r]\\\
\displaystyle\frac{\alpha\delta+\gamma(1/n)^{\alpha}}{\alpha\delta+\gamma
s^{\alpha}}&:s\in[r,1]\end{cases}$ (2.9)

for the constant

$r=\left(\frac{\alpha\gamma\zeta+\gamma\eta+\alpha\delta\eta(n-1)^{\alpha}}{\left(\alpha\delta\eta+\alpha\gamma\zeta+\gamma\eta\right)n^{\alpha}-\gamma\eta(n-1)^{\alpha}+\gamma\eta}\right)^{1/\alpha}\in\left(\frac{1}{n},1-\frac{1}{n}\right],$
(2.10)

where $r=1-1/n$ if $\gamma=0$. Finally, we also have

$\min_{\frac{1}{n}\leq t\leq 1-\frac{1}{n}}G(t,s)\geq g_{3}G(s,s)$ (2.11)

for the constant $g_{3}$ given by

$g_{3}\equiv\min\left\\{\frac{\alpha\zeta+\eta\left(1-\left(1-1/n\right)^{\alpha}\right)}{\alpha\zeta+\eta},\frac{\alpha\delta+\gamma(1/n)^{\alpha}}{\alpha\delta+\gamma}\right\\}$
(2.12)

for all $\alpha\in(0,1]$.

###### Proof.

It is straightforward to see that

$\frac{G(t,s)}{G(s,s)}=\begin{cases}\displaystyle\frac{\zeta+\frac{\eta}{\alpha}(1-t^{\alpha})}{\zeta+\frac{\eta}{\alpha}(1-s^{\alpha})}&:s\leq
t,\\\ &\\\
\displaystyle\frac{\delta+\frac{\gamma}{\alpha}t^{\alpha}}{\delta+\frac{\gamma}{\alpha}s^{\alpha}}&:t\leq
s;\end{cases}$ (2.13)

this expression yields both inequalities in (2.6) for $g_{1}$ as in (2.7).

Next, let

$u(t,s)=\frac{1}{d}\left[\delta+\frac{\gamma}{\alpha}t^{\alpha}\right]\left[\zeta+\frac{\eta}{\alpha}\left(1-s^{\alpha}\right)\right],$

so that

$G(t,s)=\begin{cases}u(s,t)&:s\leq t,\\\ u(t,s)&:t\leq s.\end{cases}$

Let $r$ be given by (2.10). Then we have

$\displaystyle\min_{\frac{1}{n}\leq t\leq 1-\frac{1}{n}}G(t,s)$
$\displaystyle=$ $\displaystyle\begin{cases}u(s,1-1/n)&:s\in[0,1/n],\\\
\min\\{u(s,1-1/n),u(1/n,s)\\}&:s\in[1/n,1-1/n],\\\
u(1/n,s)&:s\in[1-1/n,1],\end{cases}$ $\displaystyle=$
$\displaystyle\begin{cases}u(s,1-1/n)&:s\in[0,r],\\\
u(1/n,s)&:s\in[r,1],\end{cases}$ $\displaystyle=$
$\displaystyle\begin{cases}\frac{1}{d}\left[\delta+\frac{\gamma}{\alpha}s^{\alpha}\right]\left[\zeta+\frac{\eta}{\alpha}\left(1-\left(1-1/n\right)^{\alpha}\right)\right]&:s\in[0,r],\\\
\frac{1}{d}\left[\delta+\frac{\gamma}{\alpha}(1/n)^{\alpha}\right]\left[\zeta+\frac{\eta}{\alpha}\left(1-s^{\alpha}\right)\right]&:s\in[r,1],\end{cases}$

where $r$ is given in (2.10). By the monotonicity of $G(t,s)$, we have

$\max_{0\leq t\leq
1}G(t,s)=G(s,s)=\frac{1}{d}\left[\delta+\frac{\gamma}{\alpha}s^{\alpha}\right]\left[\zeta+\frac{\eta}{\alpha}\left(1-s^{\alpha}\right)\right],\quad
s\in[0,1].$

Therefore if we take $g_{2}$ as in (2.9), then $G(t,s)$ satisfies (2.8). Now
since

$\frac{\alpha\zeta+\eta\left(1-\left(1-1/n\right)^{\alpha}\right)}{\alpha\zeta+\eta\left(1-s^{\alpha}\right)}\geq\frac{\alpha\zeta+\eta\left(1-\left(1-1/n\right)^{\alpha}\right)}{\alpha\zeta+\eta},\quad
s\in[0,r]$

and

$\frac{\alpha\delta+\gamma(1/n)^{\alpha}}{\alpha\delta+\gamma
s^{\alpha}}\geq\frac{\alpha\delta+\gamma(1/n)^{\alpha}}{\alpha\delta+\gamma},\quad
s\in[r,1],$

we could in (2.8) use the constant $g_{3}$ given in (2.12) instead of (2.9).
This constant is well defined and strictly positive, since $d>0$ in (2.3)
implies neither $\delta$ and $\gamma$ nor $\eta$ and $\zeta$ can
simultaneously be zero. ∎

######  Remark 2.2.

For conjugate boundary conditions, $\gamma=\eta=1$ and $\delta=\zeta=0$; if
$\alpha=1$ and $n=4$, then $g_{3}\equiv 1/4$, the constant used in [13,
(3.4)]. For right-focal boundary conditions, $\gamma=\zeta=1$ and
$\delta=\eta=0$, so that clearly $g_{3}\equiv(1/n)^{\alpha}$ for all
$\alpha\in(0,1]$ and all integers $n\geq 3$. In the conjugate case
specifically, the constant bound (2.12) is new for fractional derivatives, as
the standard Riemann-Liouville fractional derivative does not allow one to
calculate a single constant bound; see [3, Remark 2.2].

The following corollary is needed in Section 4 for the main existence theorem
and example found there.

######  Corollary 2.2.

Let $\alpha,\beta\in(0,1]$. For every $s\in[0,1]$ we have

$\max_{t\in[0,1]}G(t,s)\leq\left(\frac{1}{1-\left(\frac{3}{4}\right)^{\alpha}}\right)\min_{t\in\left[\frac{1}{4},\frac{3}{4}\right]}G(t,s),$

where $G(t,s)$ is the Green’s function (2.4) for the homogeneous problem

$-D^{\beta}D^{\alpha}x(t)=0$

satisfying the conjugate boundary conditions $x(0)=x(1)=0$.

###### Proof.

By (2.4),

$\max_{t\in[0,1]}G(t,s)=G(s,s).$

Then

$\displaystyle\min_{t\in\left[\frac{1}{4},\frac{3}{4}\right]}\frac{G(t,s)}{G(s,s)}$
$\displaystyle=$
$\displaystyle\begin{cases}\frac{1-t^{\alpha}}{1-s^{\alpha}}&:0\leq s\leq
t\leq 3/4\\\ \frac{t^{\alpha}}{s^{\alpha}}&:1/4\leq t\leq s\leq 1\end{cases}$
$\displaystyle\geq$
$\displaystyle\begin{cases}\frac{1-(3/4)^{\alpha}}{1-(0)^{\alpha}}&:0\leq
s\leq 3/4\\\ \frac{(1/4)^{\alpha}}{(1)^{\alpha}}&:1/4\leq s\leq 1\end{cases}$
$\displaystyle=$ $\displaystyle 1-(3/4)^{\alpha},$

since $(1/4)^{\alpha}\geq 1-(3/4)^{\alpha}$ for all $\alpha\in(0,1]$. One
could also use (2.11) and (2.12). ∎

##  3\. Fixed Point Preliminaries

In this section we will state the fixed point theorem and the definitions that
are used in the fixed point theorem which will be used to verify the existence
of a positive solution to the fractional-order boundary value problem with
conjugate boundary conditions.

######  Definition 3.1.

Let $E$ be a real Banach space. A nonempty closed convex set $P\subset E$ is
called a _cone_ if it satisfies the following two conditions:

  1. $(i)$

$x\in P,\lambda\geq 0$ implies $\lambda x\in P$;

  2. $(ii)$

$x\in P,-x\in P$ implies $x=0$.

Every cone $P\subset E$ induces an ordering in $E$ given by

$x\leq y\;\;\mbox{if and only if}\;\;y-x\in P.$

######  Definition 3.2.

An operator is called completely continuous if it is continuous and maps
bounded sets into precompact sets.

######  Definition 3.3.

A map $\xi$ is said to be a nonnegative continuous concave functional on a
cone $P$ of a real Banach space $E$ if

$\xi:P\rightarrow[0,\infty)$

is continuous and

$\xi(tx+(1-t)y)\geq t\xi(x)+(1-t)\xi(y)$

for all $x,y\in P$ and $t\in[0,1]$. Similarly we say the map $\phi$ is a
nonnegative continuous convex functional on a cone $P$ of a real Banach space
$E$ if

$\phi:P\rightarrow[0,\infty)$

is continuous and

$\phi(tx+(1-t)y)\leq t\phi(x)+(1-t)\phi(y)$

for all $x,y\in P$ and $t\in[0,1]$. We say the map $\psi$ is a sub-linear
functional if

$\psi(tx)\leq t\psi(x)\;\;\hbox{for all}\;\;x\in P,t\in[0,1].$

######  Definition 3.4.

Let $P$ be a cone in a real Banach space $E$ and $\Omega$ be a bounded open
subset of $E$ with $0\in\Omega$. Then a continuous functional
$\phi:P\rightarrow[0,\infty)$ is said to satisfy property $(A1)$ if one of the
following conditions hold:

  1. $(i)$

$\phi$ is convex, $\phi(0)=0$, $\phi(x)\neq 0$ if $x\neq 0$, and
$\displaystyle{\inf_{x\in P\cap\partial\Omega}\phi(x)>0}$,

  2. $(ii)$

$\phi$ is sublinear, $\phi(0)=0$, $\phi(x)\neq 0$ if $x\neq 0$, and
$\displaystyle{\inf_{x\in P\cap\partial\Omega}\phi(x)>0}$,

  3. $(iii)$

$\phi$ is concave and unbounded.

######  Definition 3.5.

Let $P$ be a cone in a real Banach space $E$ and $\Omega$ be a bounded open
subset of $E$ with $0\in\Omega$. Then a continuous functional
$\phi:P\rightarrow[0,\infty)$ is said to satisfy property $(A2)$ if one of the
following conditions hold:

  1. $(i)$

$\phi$ is convex, $\phi(0)=0$ and $\phi(x)\neq 0$ if $x\neq 0$,

  2. $(ii)$

$\phi$ is sublinear, $\phi(0)=0$ and $\phi(x)\neq 0$ if $x\neq 0$,

  3. $(iii)$

$\phi(x+y)\geq\phi(x)+\phi(y)$ for all $x,y\in P$, $\phi(0)=0$, $\phi(x)\neq
0$ if $x\neq 0$.

The following theorem is Avery, Henderson, and O’Regan’s functional
compression-expansion fixed point theorem [2], which generalized the
functional compression fixed point theorems of Anderson-Avery [1] and Sun-
Zhang [17].

######  Theorem 3.1.

Let $\Omega_{1}$ and $\Omega_{2}$ be two bounded open sets in a Banach Space
$E$ such that $0\in\Omega_{1}$ and $\overline{\Omega_{1}}\subseteq\Omega_{2}$
and $P$ is a cone in $E$. Suppose
$A:P\cap(\overline{\Omega_{2}}-\Omega_{1})\rightarrow P$ is completely
continuous, $\xi$ and $\psi$ are nonnegative continuous functionals on $P$,
and one of the two conditions:

  1. $(K1)$

$\xi$ satisfies property $(A1)$ with $\xi(Ax)\geq\xi(x)$, for all $x\in
P\cap\partial\Omega_{1}$, and $\psi$ satisfies property $(A2)$ with
$\psi(Ax)\leq\psi(x)$, for all $x\in P\cap\partial\Omega_{2}$ , or

  2. $(K2)$

$\xi$ satisfies property $(A2)$ with $\xi(Ax)\leq\xi(x)$, for all $x\in
P\cap\partial\Omega_{1}$, and $\psi$ satisfies property $(A1)$ with
$\psi(Ax)\geq\psi(x)$, for all $x\in P\cap\partial\Omega_{2}$,

is satisfied. Then $A$ has at least one fixed point in
$P\cap(\overline{\Omega_{2}}-\Omega_{1})$.

##  4\. Existence of a Positive Solution

Let the Banach space

$E=C[0,1]$

be endowed with the maximum norm,

$\left\|x\right\|=\displaystyle{\max_{0\leq t\leq 1}}\left|x(t)\right|,$

and define the cone $P\subset E$ by

$P=\left\\{x\in E\left|\begin{array}[]{l}\mbox{nonnegative valued on
[0,1]},\;\mbox{and}\\\
\|x\|\leq\left(\frac{1}{1-\left(\frac{3}{4}\right)^{\alpha}}\right)\displaystyle{\min_{t\in\left[\frac{1}{4},\frac{3}{4}\right]}x(t)}\end{array}\right.\right\\}.$

Let the nonnegative continuous functionals $\phi$ and $\psi$ be defined on the
cone $P$ by

$\displaystyle\psi(x)=\min_{t\in\left[\frac{1}{4},\frac{3}{4}\right]}x(t)\;\;\;\mbox{and}$
(4.1) $\displaystyle\phi(x)=\max_{t\in[0,1]}x(t)=\|x\|.$ (4.2)

The following theorem is our main result.

######  Theorem 4.1.

Let $\alpha,\beta\in(0,1]$ and suppose there exists positive numbers $r$ and
$R$ such that
$0<\left(\frac{1}{1-\left(\frac{3}{4}\right)^{\alpha}}\right)r<R,$ and suppose
$f$ satisfies the following conditions:

  1. $(i)$

$f(s,x)\leq R(\alpha+\beta)(2\alpha+\beta)$ for all $s\in[0,1]$ and all
$x\in[0,R]$,

  2. $(ii)$

$f(s,x)\geq rN$ for all $s\in[1/4,3/4]$ and for all
$x\in\left[r,\frac{r}{1-\left(\frac{3}{4}\right)^{\alpha}}\right]$,

where

$N=\left(\left(1-\left(\frac{3}{4}\right)^{\alpha}\right)\int_{1/4}^{3/4}G(s,s)d_{\beta}s\right)^{-1}$

and

$\displaystyle\left(1-\left(\frac{3}{4}\right)^{\alpha}\right)\int_{1/4}^{3/4}G(s,s)d_{\beta}s$
$\displaystyle=$
$\displaystyle\left(1-\left(\frac{3}{4}\right)^{\alpha}\right)\left\\{\left(\frac{3}{4}\right)^{\alpha+\beta}\left[\frac{1}{\alpha+\beta}-\frac{(3/4)^{\alpha}}{2\alpha+\beta}\right]\right.$
$\displaystyle\left.-\left(\frac{1}{4}\right)^{\alpha+\beta}\left[\frac{1}{\alpha+\beta}-\frac{(1/4)^{\alpha}}{2\alpha+\beta}\right]\right\\}$

Then, the second order conjugate boundary value problem has at least one
positive solution $x^{*}$ such that

$r\leq\min_{t\in\left[\frac{1}{4},\frac{3}{4}\right]}x^{*}(t)\;\;\mbox{and}\;\;\max_{t\in\left[0,1\right]}x^{*}(t)\leq
R.$

###### Proof.

Define the completely continuous operator $A$ by

$Ax(t)=\int_{0}^{1}G(t,s)f(s,x(s))d_{\beta}s$

then if we can show that $A$ has a fixed point in $P$ then we have verified
the existence of a positive solution. Let $x\in P$, then from properties of
$G(t,s)$ we have that $Ax(t)\geq 0$ and

$\displaystyle\phi(Ax)$ $\displaystyle=$
$\displaystyle\max_{t\in[0,1]}\int_{0}^{1}G(t,s)f(s,x(s))d_{\beta}s$
$\displaystyle\leq$
$\displaystyle\left(\frac{1}{1-\left(\frac{3}{4}\right)^{\alpha}}\right)\min_{t\in\left[\frac{1}{4},\frac{3}{4}\right]}\int_{0}^{1}G(t,s)f(s,x(s))d_{\beta}s$
$\displaystyle=$
$\displaystyle\left(\frac{1}{1-\left(\frac{3}{4}\right)^{\alpha}}\right)\psi(Ax)$

thus, $Ax\in P$ and we have verified that $A:P\rightarrow P$.

For all $x\in P$ we have

$\psi(x)\leq\phi(x),$

thus if we let

$\Omega_{1}=\\{x\;:\;\psi(x)<r\\}\;\;\mbox{and}\;\;\Omega_{2}=\\{x\;:\;\phi(x)<R\\}$

we have that

$0\in\Omega_{1}\;\;\mbox{and}\;\;\overline{\Omega_{1}}\subseteq\Omega_{2}$

since if $x\in\overline{\Omega_{1}}$ then

$\min_{t\in\left[\frac{1}{4},\frac{3}{4}\right]}x(t)\leq r$

hence since $x\in P$ we have

$\max_{t\in\left[0,1\right]}x(t)\leq\left(\frac{1}{1-\left(\frac{3}{4}\right)^{\alpha}}\right)\min_{t\in\left[\frac{1}{4},\frac{3}{4}\right]}x(t)\leq\left(\frac{1}{1-\left(\frac{3}{4}\right)^{\alpha}}\right)r<R.$

Clearly $\Omega_{1}$ and $\Omega_{2}$ being bounded open subsets of $P$.

Claim 1: If $x\in P\cap\partial\Omega_{2}$, then $\phi(Ax)\leq\phi(x)$. Let
$x\in\partial\Omega_{2}$, thus $\phi(x)=R$ hence by condition $(i)$ we have

$\displaystyle\phi(Ax)$ $\displaystyle=$
$\displaystyle\max_{t\in\left[0,1\right]}\int_{0}^{1}G(t,s)f(s,x(s))d_{\beta}s$
$\displaystyle\leq$ $\displaystyle
R(\alpha+\beta)(2\alpha+\beta)\int_{0}^{1}G(s,s)d_{\beta}s$ $\displaystyle=$
$\displaystyle R=\phi(x).$

Claim 2: If $x\in P\cap\partial\Omega_{1}$, then $\psi(Ax)\geq\psi(x)$. Let
$x\in\partial\Omega_{1}$, thus $\psi(x)=r$ and
$\|x\|\leq\frac{r}{1-\left(\frac{3}{4}\right)^{\alpha}}$, hence by condition
$(ii)$ we have

$\displaystyle\psi(Ax)$ $\displaystyle=$
$\displaystyle\min_{t\in\left[\frac{1}{4},\frac{3}{4}\right]}\int_{0}^{1}G(t,s)f(s,x(s))d_{\beta}s$
$\displaystyle\geq$ $\displaystyle
rN\left(1-\left(\frac{3}{4}\right)^{\alpha}\right)\int_{1/4}^{3/4}G(s,s)d_{\beta}s$
$\displaystyle=$ $\displaystyle r=\psi(x).$

Clearly $\phi$ satisfies property $(A1)(i)$ and $\psi$ satisfies property
$(A2)(iii)$ thus the hypothesis $(K1)$ of Theorem 3.1 is satisfied, and
therefore $A$ has a fixed point in $\overline{\Omega_{2}}-\Omega_{1}$.

∎

######  Example 1.

To compare our results with those in [3, Example 3.1], where the authors use
the Riemann-Liouville fractional derivative of order $\frac{3}{2}\in(1,2]$, we
take $\alpha=1$, $\beta=\frac{1}{2}$, $r=\frac{11}{1000}$, $R=\frac{9}{25}$,
and $f(s,x)=1+\frac{1}{4}\sin s+x^{2}$ in Theorem 4.1 to get the following.
One can check that $0<4r<R$, and $f$ satisfies the following conditions:

  1. $(i)$

$f(s,x)\leq\frac{15}{4}R=\frac{27}{20}$ for all $s\in[0,1]$ and all
$x\in\left[0,\frac{9}{25}\right]$,

  2. $(ii)$

$f(s,x)\geq\frac{960r}{33\sqrt{3}-17}=\frac{264}{25(33\sqrt{3}-17)}$ for all
$s\in\left[\frac{1}{4},\frac{3}{4}\right]$ and for all
$x\in\left[r,4r\right]$.

Thus by Theorem 4.1 the $\frac{3}{2}$-order conjugate boundary value problem

$-D^{0.5}x^{\prime}(t)=1+\frac{1}{4}\sin t+x(t)^{2},\quad x(0)=x(1)=0$

has at least one positive solution $x^{*}$ such that

$\frac{11}{1000}\leq\min_{t\in\left[\frac{1}{4},\frac{3}{4}\right]}x^{*}(t)\;\;\mbox{and}\;\;\max_{t\in\left[0,1\right]}x^{*}(t)\leq\frac{9}{25}.$

In [3, Example 3.1], the result is the existence of a positive solution
$x^{\dagger}$ such that

$\frac{1}{14}\leq\max_{t\in\left[0,1\right]}x^{\dagger}(t)\leq 1,$

with no information on the minimum value of the function.

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