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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 88, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/88\hfil Lyapunov-type inequalities]
{Lyapunov-type inequalities for fractional boundary-value problems}

\author[M. Jleli, B. Samet \hfil EJDE-2015/88\hfilneg]
{Mohamed Jleli, Bessem Samet}

\address{Mohamed Jleli \newline
Department of Mathematics, College of Science,
King Saud University,
P.O. Box 2455, Riyadh 11451, Saudi Arabia}
\email{jleli@ksu.edu.sa}

\address{Bessem Samet \newline
Department of Mathematics, College of Science,
King Saud University,
P.O. Box 2455, Riyadh 11451, Saudi Arabia}
\email{bsamet@ksu.edu.sa}

\thanks{Submitted January 2, 2015. Published April 10, 2015.}
\subjclass[2000]{4A08, 34A40, 26D10, 33E12}
\keywords{Lyapunov's inequality; Caputo's fractional derivative; 
\hfill\break\indent Sturm-Liouville boundary condition}

\begin{abstract}
 In this article, we establish some Lyapunov-type inequalities for
 fractional boundary-value problems under Sturm-Liouville boundary conditions.
 As applications, we obtain intervals where linear combinations of  certain 
 Mittag-Leffler functions have no real zeros. We deduce also nonexistence
 results for some fractional boundary-value problems.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks


\section{Introduction}

The well-known Lyapunov result \cite{L} states that if a nontrivial solution
to the boundary-value problem
\begin{gather*}
u''(t)+q(t)u(t)=0,\quad a<t<b,\\
u(a)=u(b)=0,
\end{gather*}
exists, where $q: [a,b]\to \mathbb{R}$ is a continuous function, then
$$
\int_a^b |q(s)|\,ds>\frac{4}{b-a}\,.
$$
This result found many practical applications in differential and
difference equations (oscillation theory, disconjugacy, eigenvalue problems, etc.);
see \cite{C,CL,Pa,Y,Y2,Z} and references therein.

The search for Lyapunov-type inequalities in which the starting differential
equation is constructed via fractional differential operators has begun
very recently. The first work in this direction is due to Ferreira \cite{F0},
where he   derived  a Lyapunov-type inequality  for differential equations
depending on the Riemann-Liouville fractional derivative; that is,
for the boundary-value problem
\begin{gather*}
(_aD^\alpha u)(t)+q(t)u(t)=0,\quad a<t<b,\; 1<\alpha\leq 2,\\
u(a)=u(b)=0,
\end{gather*}
where $_aD^\alpha$ denotes the Riemann-Liouville fractional derivative of
order $\alpha$. Precisely, the author proved that if the above problem has
 a nontrivial solution, then
$$
\int_a^b |q(s)|\,ds > \Gamma(\alpha) \big(\frac{4}{b-a}\big)^{\alpha-1}\,.
$$
Clearly, if we let $\alpha=2$ in the above inequality, one obtains Lyapunov's
standard inequality. In \cite{F},  a Lyapunov-type inequality was
obtained by the same author for the Caputo fractional boundary-value problem
\begin{gather*}
(^C_aD^\alpha u)(t)+q(t)u(t)=0,\quad a<t<b,\; 1<\alpha\leq 2,\\
u(a)=u(b)=0,
\end{gather*}
where $^C_aD^\alpha$ denotes the Caputo  fractional derivative of order
$\alpha$. In this work, Ferreira proved that if the above problem has
a nontrivial solution, then
$$
\int_a^b |q(s)|\,ds
> \frac{\Gamma(\alpha) \alpha^\alpha}{[(\alpha-1)(b-a)]^{\alpha-1}}\,.
$$
For other works on Lyapunov-type inequalities for fractional boundary-value
 problems we refer the reader to \cite{J1,J2}.

Motivated by the above works, we consider  a
Caputo fractional differential equation with Sturm-Liouville boundary conditions.
More precisely, we consider the fractional boundary-value problem
\begin{equation}\label{SL}
(^C_aD^\alpha u)(t)+q(t)u(t)=0,\quad a<t<b,\; 1<\alpha<2,
\end{equation}
with the boundary conditions
\begin{equation}\label{BC}
pu(a)-ru'(a)=u(b)=0,
\end{equation}
where $p>0$, $r\geq 0$ and $q: [a,b]\to \mathbb{R}$ is a continuous function.
We distinguish two cases: the case
$\frac{r}{p}>\frac{b-a}{\alpha-1}$ and the case
$0\leq \frac{r}{p}\leq \frac{b-a}{\alpha-1}$. For each case, a Lyapunov-type
inequality is derived. The obtained results recover several existing inequalities
from the literature. As  applications, we obtain intervals where linear
combinations of certain Mittag-Leffler functions have no real zeros.
We deduce also nonexistence results for some fractional boundary-value problems.

Before presenting our main results, let us start by recalling the concepts
of the Riemann-Liouville fractional integral and the Caputo fractional
derivative of order $\alpha\geq 0$. For more details, we refer to \cite{K}.

Let  $\alpha\geq 0$ and let $f$ be a real function defined on a certain
interval $[a,b]$. The Riemann-Liouville fractional integral of order $\alpha$
is defined by
$$
(_aI^0f)(t)=f(t)
$$
and
$$
(_aI^\alpha f)(t)=\frac{1}{\Gamma(\alpha)} \int_a^t (t-s)^{\alpha-1}f(s)\,ds,\quad \alpha>0,\, t\in [a,b].
$$
The Caputo fractional derivative of order $\alpha\geq 0$ is defined by
$$
(^C_aD^0f)(t)=f(t)
$$
and
$$
(^C_aD^\alpha f)(t)=(_aI^{m-\alpha}D^mf)(t),\quad \alpha>0,
$$
where $m$ is the smallest integer greater or equal to $\alpha$.

\section{Main results}

\subsection{Integral representation of the solution}

We start by writing \eqref{SL}-\eqref{BC} in its equivalent integral form.

\begin{lemma}\label{lem1}
$u\in C[a,b]$ is a solution to \eqref{SL}-\eqref{BC} if and only if $u$
is a solution to the  integral equation
$$
u(t)=\int_a^b G(t,s)q(s)u(s)\,ds, \quad t\in [a,b],
$$
where $G$, the Green function associated to \eqref{SL}-\eqref{BC}, is given by
$$
G(t,s)=\frac{1}{\Gamma(\alpha)} \begin{cases}
\frac{(\frac{r}{p}+t-a)(b-s)^{\alpha-1}}{\gamma}-(t-s)^{\alpha-1},
  & a\leq s\leq t\leq b,\\
\frac{(\frac{r}{p}+t-a)(b-s)^{\alpha-1}}{\gamma}, & a\leq t\leq s\leq b,
\end{cases}
$$
where $\gamma=\frac{r}{p}+b-a$.
\end{lemma}

\begin{proof}
The general solution to \eqref{SL} is
$$
u(t)=c_0+c_1(t-a)-\frac{1}{\Gamma(\alpha)}\int_a^t (t-s)^{\alpha-1}q(s)u(s)\,ds,
$$
where $c_0$ and $c_1$ are real constants. Taking the derivative of $u(t)$, we obtain
$$
u'(t)=c_1-\frac{(\alpha-1)}{\Gamma(\alpha)}\int_a^t  (t-s)^{\alpha-2}q(s)u(s)\,ds.
$$
Using the boundary condition $pu(a)-ru'(a)=0$, we obtain
\begin{equation}\label{c1}
pc_0-rc_1=0.
\end{equation}
The boundary condition $u(b)=0$ gives us
\begin{equation}\label{c2}
c_0+c_1(b-a)-\frac{1}{\Gamma(\alpha)}\int_a^b (b-s)^{\alpha-1}q(s)u(s)\,ds=0.
\end{equation}
Then \eqref{c1} and \eqref{c2} yield
$$
c_0=\frac{r}{p}c_1=\frac{r}{p\gamma\Gamma(\alpha)}\int_a^b (b-s)^{\alpha-1}
q(s)u(s)\,ds\,.
$$
Therefore,
\begin{align*}
u(t)&=\frac{r}{p\gamma\Gamma(\alpha)}\int_a^b (b-s)^{\alpha-1}q(s)u(s)\,ds
 +\frac{(t-a)}{\gamma\Gamma(\alpha)}\int_a^b (b-s)^{\alpha-1}q(s)u(s)\,ds\\
&\quad -\frac{1}{\Gamma(\alpha)}\int_a^t (t-s)^{\alpha-1}q(s)u(s)\,ds,
\end{align*}
which concludes the proof. 
\end{proof}

\subsection{Green function estimates}

Let
\begin{gather*}
g_1(t,s)=\frac{(\frac{r}{p}+t-a)(b-s)^{\alpha-1}}{\gamma}-(t-s)^{\alpha-1},\quad 
a\leq s\leq t\leq b,\\
g_2(t,s)=\frac{(\frac{r}{p}+t-a)(b-s)^{\alpha-1}}{\gamma},\quad
a\leq t\leq s\leq b.
\end{gather*}
We distinguish two cases.

\subsection*{Case $\frac{r}{p}>\frac{b-a}{\alpha-1}$}

\begin{lemma}\label{G1}
Suppose that
$$
\frac{r}{p}>\frac{b-a}{\alpha-1}\,.
$$
Then
\begin{gather*}
0\leq G(t,s)\leq G(s,s), \quad (t,s)\in [a,b]\times [a,b],\\
\max_{a\leq s\leq b}G(s,s)=\frac{1}{\Gamma(\alpha)}
\frac{\frac{r}{p}(b-a)^{\alpha-1}}{(\frac{r}{p}+b-a)}\,.
\end{gather*}
\end{lemma}

\begin{proof} 
 Obviously, the function $g_2$ satisfies the following inequalities:
$$
0\leq g_2(t,s)\leq g_2(s,s), \quad a\leq t\leq s\leq b.
$$
Now, let us compute the derivative of $g_2(s,s)$ on $(a,b)$.
 After some simplifications, we obtain
$$
(g_2(s,s))'=\frac{(b-s)^{\alpha-2}}{\gamma} 
\Big(-\alpha s+(1-\alpha)(\frac{r}{p}-a)+b\Big).
$$
Then $(g_2(s,s))'$ has a unique zero, attained at the point
$$
s^*=\frac{b+(1-\alpha)(\frac{r}{p}-a)}{\alpha}\,.
$$
It is easy to see that $(g_2(s,s))'>0$ on $(-\infty,s^*)$ and 
$(g_2(s,s))'<0$ on $(s^*,b)$. On the other hand, from the condition 
$\frac{r}{p}>\frac{b-a}{\alpha-1}$, we obtain easily that $s^*<a$. 
By continuity of $g_2$, we deduce that
$$
\max_{a\leq s\leq b} g_2(s,s)=g_2(a,a)
=\frac{\frac{r}{p}(b-a)^{\alpha-1}}{(\frac{r}{p}+b-a)}\,.
$$
Thus
$$
0\leq g_2(t,s)\leq\frac{\frac{r}{p}(b-a)^{\alpha-1}}{(\frac{r}{p}+b-a)},
\quad a\leq t\leq s\leq b.
$$
Now, we turn our attention to the function $g_1(t,s)$. Let $s\in [a,b)$ be fixed. 
Differentiating $g_1(t,s)$ with respect to $t$, we obtain
$$
\partial_t g_1(t,s)=\frac{(b-s)^{\alpha-1}}{\gamma}-(\alpha-1)(t-s)^{\alpha-2},
\quad s<t.
$$
It follows from the above equality that $\partial_t g_1(t,s)=0$ if and only if
$$
t=t^*=s+\big[\frac{(b-s)^{\alpha-1}}{\gamma(\alpha-1)}\big]^{\frac{1}{\alpha-2}},
$$
provided $t^*\leq b$, i.e. as long as $a\leq s\leq b-(\alpha-1)\gamma$.
However, from the condition $\frac{r}{p}>\frac{b-a}{\alpha-1}$, 
we observe easily that $b-(\alpha-1)\gamma<a$. Then we deduce that 
$s>b-(\alpha-1)\gamma$, i.e. $t^*>b$. In this case,
 $\partial_t g_1(t,s)<0$, i.e. $g_1(\cdot,s)$ is strictly decreasing and, 
since $g_1(b,s)=0$, we conclude that
$$
0\leq g_1(t,s)\leq g_1(s,s)=g_2(s,s) \leq g_2(a,a)
\leq \frac{\frac{r}{p}(b-a)^{\alpha-1}}{(\frac{r}{p}+b-a)}\quad a\leq s\leq t\leq b,
$$
which concludes the proof. 
\end{proof}

\subsection*{Case $0\leq \frac{r}{p}\leq \frac{b-a}{\alpha-1}$}

\begin{lemma}\label{G2}
Suppose that
$$
0\leq \frac{r}{p}\leq \frac{b-a}{\alpha-1}\,.
$$
Then
$$
\Gamma(\alpha)|G(t,s)|\leq \max\{\mathcal{A}(\alpha,r/p),
\mathcal{B}(\alpha,r/p)\}, \quad (t,s)\in [a,b]\times [a,b],
$$
where
\begin{gather*}
\mathcal{A}(\alpha,r/p)=
\frac{(b-a)^{\alpha-1}}{(\frac{r}{p}+b-a)}
\bigg(\Big(\frac{(b-a)^{\alpha-1}}{(\frac{r}{p}+b-a)
(\alpha-1)^{\alpha-1}}\Big)^{\frac{1}{\alpha-2}}(2-\alpha)-\frac{r}{p}\bigg),\\
\mathcal{B}(\alpha,r/p)=
(\frac{r}{p}+b-a)^{\alpha-1} \frac{(\alpha-1)^{\alpha-1}}{\alpha^\alpha}\,.
\end{gather*}
\end{lemma}

\begin{proof} 
Following the proof of Lemma \ref{G1}, we have
$$
0\leq g_2(t,s)\leq g_2(s,s), \quad a\leq t\leq s\leq b
$$
and  $(g_2(s,s))'$ has a unique zero, attained at the point
$$
s^*=\frac{b+(1-\alpha)(\frac{r}{p}-a)}{\alpha}\,.
$$
Under the condition $0\leq \frac{r}{p}\leq \frac{b-a}{\alpha-1}$, 
it is easy to observe that $s^*\in [a,b]$. 
Moreover, $(g_2(s,s))'>0$ on $(-\infty,s^*)$ and $(g_2(s,s))'<0$ on $(s^*,b)$. 
Then
$$
\max_{a\leq s\leq b} g_2(s,s)=g_2(s^*,s^*)=\mathcal{B}(\alpha,r/p).
$$
Thus we have
$$
0\leq g_2(t,s)\leq \mathcal{B}(\alpha,r/p),\quad a\leq t\leq s\leq b.
$$
Following the proof of Lemma \ref{G1}, for a fixed $s\in [a,b)$,
$\partial_t g_1(t,s)=0$ if and only if
$$
t=t^*=s+\big[\frac{(b-s)^{\alpha-1}}{\gamma(\alpha-1)}\big]^{\frac{1}{\alpha-2}},
$$
provided $t^*\leq b$, i.e. as long as $a\leq s\leq b-(\alpha-1)\gamma$.
So, if $s>b-(\alpha-1)\gamma$ (i.e. $\partial_t g_1(t,s)$ has no zeros), 
then $\partial_t g_1(t,s)<0$, i.e. $g_1(\cdot,s)$ is strictly decreasing and, 
since $g_1(b,s)=0$, we obtain
$$
\max_{s\leq t\leq b}g_1(t,s)=g_1(s,s)=g_2(s,s),\quad s\in (b-(\alpha-1)\gamma,b).
$$
It is easy to check that
$$
s^*\in (b-(\alpha-1)\gamma,b).
$$
Thus we have
$$
0\leq g_1(t,s)\leq g_2(s^*,s^*)=\mathcal{B}(\alpha,r/p),\quad 
b-(\alpha-1)\gamma<s\leq t\leq b.
$$
Now, we have to check the case when $a\leq s\leq b-(\alpha-1)\gamma$;
 i.e., $t^*\leq b$. It is easy to see that $\partial_t g_1(t,s)<0$ for $t<t^*$ 
and that $\partial_t g_1(t,s)\geq 0$ for $t\geq t^*$. This together with the 
fact that $g_1(b,s)=0$ implies that $g_1(t^*,s)\leq 0$ and, therefore,
 we only have to show that
$$
|g_1(t^*,s)|\leq \max\left\{\mathcal{A}(\alpha,r/p),\mathcal{B}(\alpha,r/p)\right\}, \quad s\in [a,b-(\alpha-1)\gamma].
$$
After some simplifications, we obtain
$$
|g_1(t^*,s)|=\frac{(b-s)^{\frac{(\alpha-1)^2}{\alpha-2}}(2-\alpha)}
{\gamma^{\frac{\alpha-1}{\alpha-2}}(\alpha-1)^{\frac{\alpha-1}{\alpha-2}}}
-\frac{(b-s)^{\alpha-1}}{\gamma}(s-a+\frac{r}{p}).
$$
Let us define the function
$$
h(s)=\frac{(b-s)^{\frac{(\alpha-1)^2}{\alpha-2}}(2-\alpha)}
{\gamma^{\frac{\alpha-1}{\alpha-2}}(\alpha-1)^{\frac{\alpha-1}{\alpha-2}}}
-\frac{(b-s)^{\alpha-1}}{\gamma}(s-a+\frac{r}{p}),\quad
s\in [a,b-(\alpha-1)\gamma].
$$
Now, we differentiate $h$ in the interior of $[a,b-(\alpha-1)\gamma]$. We obtain
$$
h'(s)=
\frac{(b-s)^{\frac{(\alpha-1)^2}{\alpha-2}-1}}{(\alpha-1)
^{\frac{3-\alpha}{\alpha-2}}\gamma^{\frac{\alpha-1}{\alpha-2}}}
+\frac{(\alpha-1)(s-a+\frac{r}{p})(b-s)^{\alpha-2}}{\gamma}
-\frac{(b-s)^{\alpha-1}}{\gamma}\,.
$$
It is clear that $h'$ is an increasing function in $[a,b-(\alpha-1)\gamma]$.
Then we have
$$
h'(s)\leq h'(b-(\alpha-1)\gamma).
$$
On the other hand, after some simplifications, we obtain
$$
h'(b-(\alpha-1)\gamma)=0,
$$
which yields $h'(s)\leq 0$. Therefore,
$$
\max_{a\leq s\leq b-(\alpha-1)\gamma} h(s)=h(a)=\mathcal{A}(\alpha,r/p),
$$
which concludes the proof. 
\end{proof}


\subsection{Lyapunov-type inequalities}

We are ready to state and prove our main results.

\begin{theorem}\label{thm1}
If there exists a nontrivial continuous solution of the fractional 
boundary-value problem
\begin{gather*}
(^C_aD^\alpha u)(t)+q(t)u(t)=0,\quad a<t<b,\, 1<\alpha<2,\\
pu(a)-ru'(a)=u(b)=0,
\end{gather*}
 where $p>0$,  $\frac{r}{p}>\frac{b-a}{\alpha-1}$ and 
$q: [a,b]\to \mathbb{R}$ is a continuous function, then
\begin{equation}\label{E1}
\int_a^b |q(s)|\,ds \geq \big(1+\frac{p}{r} (b-a)\big)\frac{\Gamma(\alpha)}
{(b-a)^{\alpha-1}}\,.
\end{equation}
\end{theorem}

\begin{proof} 
Let $X=C[a,b]$ be the Banach space endowed with the norm
$$
\|y\|_\infty=\max\{|y(t)|: a\leq t\leq b\}.
$$
It follows from Lemma \ref{lem1} that
$$
u(t)=\int_a^b G(t,s)q(s)u(s)\,ds, \quad t\in [a,b].
$$
We obtain
$$
|u(t)|\leq  \|u\|_\infty \max|G(t,s)|_{a\leq t,s\leq b} \int_a^b |q(s)|\,ds.
$$
Now, Lemma \ref{G1} yields
$$
\|u\|_\infty\leq \|u\|_\infty \frac{1}{\Gamma(\alpha)}\frac{\frac{r}{p}(b-a)^{\alpha-1}}{(\frac{r}{p}+b-a)}\int_a^b |q(s)|\,ds,
$$
from which the  inequality \eqref{E1} follows.
\end{proof}

Similarly, using Lemma \ref{lem1} and Lemma \ref{G2}, we obtain the following result.


\begin{theorem}\label{thm2}
If there exists a nontrivial continuous solution of the fractional boundary-value 
problem
\begin{gather*}
(^C_aD^\alpha u)(t)+q(t)u(t)=0,\quad a<t<b,\; 1<\alpha<2,\\
pu(a)-ru'(a)=u(b)=0,
\end{gather*}
where $p>0$,  $0\leq \frac{r}{p}\leq \frac{b-a}{\alpha-1}$ and 
$q: [a,b]\to \mathbb{R}$ is a continuous function, then
\begin{equation}\label{E2}
\int_a^b |q(s)|\,ds 
\geq \frac{\Gamma(\alpha)}{\max\{\mathcal{A}(\alpha,r/p),\mathcal{B}(\alpha,r/p)\}}\,.
\end{equation}
\end{theorem}


\subsection{Particular cases}

\subsection*{Case $r=0$}

In the case $r=0$, from Theorem \ref{thm2}, taking $r=0$ in \eqref{E2}, we obtain
$$
\int_a^b |q(s)|\,ds \geq \frac{\Gamma(\alpha)}
{\max\{\mathcal{A}(\alpha,0),\mathcal{B}(\alpha,0)\}}\,.
$$
On the other hand, we have
\begin{gather*}
\mathcal{A}(\alpha,0)
=\frac{2-\alpha}{(\alpha-1)^{\frac{\alpha-1}{\alpha-2}}}(b-a)^{\alpha-1},\\
\mathcal{B}(\alpha,0)=
 \frac{(\alpha-1)^{\alpha-1}}{\alpha^\alpha}(b-a)^{\alpha-1}\,.
\end{gather*}
Using the inequality (see \cite{F})
$$
\frac{2-\alpha}{(\alpha-1)^{\frac{\alpha-1}{\alpha-2}}}\leq
\frac{(\alpha-1)^{\alpha-1}}{\alpha^\alpha},\quad 1<\alpha<2,
$$
we deuce that
$$
\max\{\mathcal{A}(\alpha,0),\mathcal{B}(\alpha,0)\}=\mathcal{B}(\alpha,0).
$$
Thus we obtain the following result (see \cite[Theorem 1]{F}).

\begin{corollary} \label{cor2.6}
If there exists a nontrivial continuous solution of the 
fractional boundary-value problem
\begin{gather*}
(^C_aD^\alpha u)(t)+q(t)u(t)=0,\quad a<t<b,\; 1<\alpha<2,\\
u(a)=u(b)=0,
\end{gather*}
 where  $q: [a,b]\to \mathbb{R}$ is a continuous function, then
$$
\int_a^b |q(s)|\,ds 
\geq \frac{\Gamma(\alpha)\alpha^\alpha}{[(\alpha-1)(b-a)]^{\alpha-1}}\,.
$$
\end{corollary}

\subsection*{Case $\frac{r}{p}=\frac{b-a}{\alpha-1}$ with $\alpha\simeq 2$}

In the case $\frac{r}{p}=\frac{b-a}{\alpha-1}$, from Theorem \ref{thm2}, 
taking $\frac{r}{p}=\frac{b-a}{\alpha-1}$ in \eqref{E2}, we obtain
$$
\int_a^b |q(s)|\,ds \geq \frac{\Gamma(\alpha)}
{\max\{\mathcal{A}\big(\alpha,\frac{b-a}{\alpha-1}\big),
\mathcal{B}\big(\alpha,\frac{b-a}{\alpha-1}\big)\}}\,.
$$
An easy computation gives us
\begin{gather*}
\mathcal{A}\big(\alpha,\frac{b-a}{\alpha-1}\big)
=\frac{(b-a)^{\alpha-1}}{\alpha} 
\big(\frac{2-\alpha}{\alpha^{\frac{1}{\alpha-2}}}-1\big),
\\
\mathcal{B}\big(\alpha,\frac{b-a}{\alpha-1}\big)
=\frac{(b-a)^{\alpha-1}}{\alpha}\,.
\end{gather*}
Thus we have
$$
\mathcal{A}\big(\alpha,\frac{b-a}{\alpha-1}\big)
-\mathcal{B}\big(\alpha,\frac{b-a}{\alpha-1}\big)=
\frac{(b-a)^{\alpha-1}}{\alpha} 
\big(\frac{2-\alpha}{\alpha^{\frac{1}{\alpha-2}}}-2\big)\,.
$$
On the other hand, 
$$
\lim_{\alpha\to 2^-} \frac{2-\alpha}{\alpha^{\frac{1}{\alpha-2}}}=+\infty.
$$
Then there exists $\delta>0$ such that
$$
2-\delta< \alpha<2 \Rightarrow  \frac{2-\alpha}{\alpha^{\frac{1}{\alpha-2}}}>2.
$$
Thus for $2-\delta< \alpha<2$, we have
$$
\max\big\{\mathcal{A}\big(\alpha,\frac{b-a}{\alpha-1}\big),
\mathcal{B}\big(\alpha,\frac{b-a}{\alpha-1}\big)\big\}
=\mathcal{A}\big(\alpha,\frac{b-a}{\alpha-1}\big).
$$
Hence we have the following result.

\begin{corollary}
There exists $\delta>0$ such that  if there exists
a nontrivial continuous solution of
 the fractional boundary-value problem
\begin{gather*}
(^C_aD^\alpha u)(t)+q(t)u(t)=0,\quad a<t<b,\, 2-\delta<\alpha<2,\\
pu(a)-ru'(a)=u(b)=0,
\end{gather*}
 where  $\frac{r}{p}=\frac{b-a}{\alpha-1}$ and $q: [a,b]\to \mathbb{R}$ 
is a continuous function, then
$$
\int_a^b |q(s)|\,ds \geq \frac{\Gamma(\alpha)
\alpha^{\frac{\alpha-1}{\alpha-2}}}{(b-a)^{\alpha-1}
(2-\alpha-\alpha^{\frac{1}{\alpha-2}})}\,.
$$
\end{corollary}

\subsection*{Case $p\simeq 0$}

Letting $p\to 0^+$ in the inequality \eqref{E1},  from 
Theorem \ref{thm1} we obtain the following result.

\begin{corollary}\label{u}
If there exists a nontrivial continuous solution of the fractional 
boundary-value problem
\begin{align*}
&(^C_aD^\alpha u)(t)+q(t)u(t)=0,\quad a<t<b,\, 1<\alpha<2,\\
&u'(a)=u(b)=0,
\end{align*}
 where $q: [a,b]\to \mathbb{R}$ is a continuous function, then
\begin{equation}\label{PE1}
\int_a^b |q(s)|\,ds \geq \frac{\Gamma(\alpha)}{(b-a)^{\alpha-1}}\,.
\end{equation}
\end{corollary}

Taking $\alpha=2$ in the inequality \eqref{PE1}, we obtain the following result.

\begin{corollary}
If there exists a nontrivial continuous solution of the  boundary-value problem
\begin{align*}
&u''(t)+q(t)u(t)=0,\quad a<t<b,\\
&u'(a)=u(b)=0,
\end{align*}
 where $q: [a,b]\to \mathbb{R}$ is a continuous function, then
$$
\int_a^b |q(s)|\,ds \geq \frac{1}{b-a}\,.
$$
\end{corollary}


\section{Applications}

In this section, we present some applications of our main results.

\subsection{Real zeros of certain Mittag-Leffler functions}

Let $\alpha,\beta> 0$  be fixed. The complex function
$$
E_{\alpha,\beta}(z)=\sum_{k=0}^\infty \frac{z^k}{\Gamma(k\alpha+\beta)},\quad 
\alpha>0, \beta>0, z\in \mathbb{C}
$$
is analytic in the whole complex plane; it will be referred to
\cite{M,P} as the Mittag-Leffler function with parameters $(\alpha,\beta)$.

Next, using the above Lyapunov-type inequalities, we give  
intervals where  linear combinations of some Mittag-Leffler
functions have no real zeros.

\begin{theorem}
Let $1<\alpha<2$. The Mittag-Leffler function $E_{\alpha,1}(x)$ 
 has no real zeros for
$$
x\in (-\Gamma(\alpha),0].
$$
\end{theorem}

\begin{proof} 
Let $(a,b)=(0,1)$, and consider the  fractional 
Sturm-Liouville eigenvalue problem
\begin{gather*}
(^C_0D^\alpha u)(t)+\lambda u(t)=0,\quad 0<t<1,\\
u'(0)=u(1)=0.
\end{gather*}
By \cite{D}, we know that the eigenvalues $\lambda\in \mathbb{R}$ 
of the above problem satisfy
$$
\lambda>0\quad\text{and}\quad E_{\alpha,1}(-\lambda)=0.
$$
The corresponding eigenfunctions are 
$$
u(t)=AE_{\alpha,1}(-\lambda t^\alpha),\quad t\in [0,1].
$$
By Corollary \ref{u}, if a real eigenvalue $\lambda$ exists; i.e.,
 $E_{\alpha,1}(-\lambda)=0$, then $\lambda \geq \Gamma(\alpha)$, 
which concludes the proof. 
\end{proof}


\begin{theorem} \label{thm3.2}
Let $1<\alpha<2$, $p>0$, $\frac{r}{p}>\frac{1}{\alpha-1}$. 
The linear combination of Mittag-Leffler functions given by
$$
pE_{\alpha,2}(x)+qrE_{\alpha,1}(x)
$$
has no real zeros for
$$
x\in (-(1+\frac{p}{r})\Gamma(\alpha),0].
$$
\end{theorem}

\begin{proof} 
Let $(a,b)=(0,1)$, and consider the following fractional Sturm-Liouville
eigenvalue problem
\begin{gather*}
(^C_0D^\alpha u)(t)+\lambda u(t)=0,\quad 0<t<1,\\
pu(0)-ru'(0)=u(1)=0.
\end{gather*}
By \cite{D}, we know that the eigenvalues $\lambda\in \mathbb{R}$ 
of the above problem satisfies
$$
\lambda>0\quad\text{and}\quad pE_{\alpha,2}(-\lambda)+qrE_{\alpha,1}(-\lambda)=0.
$$
The corresponding eigenfunctions are
$$
u(t)=A\big(E_{\alpha,1}(-\lambda t^\alpha)
+\frac{p}{r}tE_{\alpha,2}(-\lambda t^\alpha)\big), \quad t\in [0,1].
$$
By Theorem \ref{thm1}, if a real eigenvalue $\lambda$ exists,  
then $\lambda \geq (1+\frac{p}{r})\Gamma(\alpha)$, which concludes the proof. 
\end{proof}

\subsection{Applications to fractional boundary-value problems}

In this section, we apply the results on the Liapunov-type inequalities 
obtained previoulsy to study the nonexistence of solutions for certain 
fractional boundary-value problems.
Consider the fractional boundary-value problem
\begin{equation}\label{ex1}
(^C_0D^\alpha u)(t)+q(t)u(t)=0,\quad 0<t<1,\; 3/2<\alpha<2,
\end{equation}
with the boundary conditions
\begin{equation}\label{BC1}
u(0)-2u'(0)=u(1)=0,
\end{equation}
where $q: [a,b]\to \mathbb{R}$ is a continuous function.
We have the following result.

\begin{theorem}\label{TBVP1}
Assume that
\begin{equation}\label{as1}
\int_0^1 |q(s)|\,ds < \frac{3}{2} \Gamma(\alpha).
\end{equation}
Then\eqref{ex1}-\eqref{BC1} has no nontrivial solution.
\end{theorem}

\begin{proof} 
Assume the contrary, i.e. \eqref{ex1}-\eqref{BC1}  has a nontrivial 
solution $u(t)$. By Theorem \ref{thm1} with $(p,r)=(1,2)$, we obtain
$$
\int_0^1 |q(s)|\,ds \geq \frac{3}{2}\Gamma(\alpha),
$$
which contradicts assumption \eqref{as1}. 
\end{proof}

Consider now the fractional boundary-value problem
\begin{equation}\label{ex2}
(^C_0D^\alpha u)(t)+q(t)u(t)=0,\quad 0<t<1,\; 1<\alpha<2,
\end{equation}
with the boundary conditions
\begin{equation}\label{BC2}
2u(0)-u'(0)=u(1)=0,
\end{equation}
where $q: [a,b]\to \mathbb{R}$ is a continuous function.
We have the following result.

\begin{theorem}\label{TBVP2}
Assume that
\begin{equation}\label{as2}
\int_0^1 |q(s)|\,ds 
< \frac{\Gamma(\alpha)}{\max\{\mathcal{A}(\alpha,1/2),\mathcal{B}(\alpha,1/2)\}}\,.
\end{equation}
Then \eqref{ex2}-\eqref{BC2} has no nontrivial solution.
\end{theorem}

\begin{proof} 
Assume the contrary; i.e., \eqref{ex2}-\eqref{BC2}  has a nontrivial solution
 $u(t)$. By Theorem \ref{thm2} with $(p,r)=(2,1)$, we obtain
$$
\int_0^1 |q(s)|\,ds \geq \frac{\Gamma(\alpha)}
{\max\{\mathcal{A}(\alpha,1/2),\mathcal{B}(\alpha,1/2)\}},
$$
which contradicts assumption \eqref{as2}. 
\end{proof}

\subsection*{Acknowledgements}
The authors would like to extend their sincere appreciation to 
the Deanship of Scientific Research at King Saud University for 
the funding of this research through the Research Group Project No. RGP-VPP-237.

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\end{document}