Lyapunov-type inequality for a fractional differential equation with fractional boundary conditions

Abstract

In this paper, a Lyapunov-type inequality is obtained for a fractional differential equation under fractional boundary conditions. We then use this inequality to obtain an interval where a certain Mittag-Leffler function has no real zeros.

1 Introduction

The Lyapunov inequality [1] has proved to be very useful in various problems related with differential equations; for example, see [2, 3] and the references therein. The Lyapunov inequality states that a necessary condition for the boundary value problem

$$ \left \{ \begin{array}{l} y''(t) + q(t)y(t) = 0,\quad a < t < b, \\ y(a)=0=y(b) \end{array} \right . $$

(1.1)

to have nontrivial solutions is that

$$ \int_{a}^{b} \bigl\vert q(s)\bigr\vert \, ds >\frac{4}{b-a}, $$

(1.2)

where $q:[ a , b ] \to R$ is a continuous function, and the zeros a and b of every solution $y(t)$ are consecutive. Since then, many generalizations of the Lyapunov inequality have appeared in the literature (see [4–9] and the references therein).

Recently, the research of Lyapunov-type inequalities for fractional boundary value problem has begun. In [10], Ferreira investigated a Lyapunov-type inequality for the Caputo fractional boundary value problem

$$ \left \{ \begin{array}{l} {}^{\mathrm{C}}_{a} D^{\alpha} y(t) + q(t)y(t) = 0, \quad a < t < b, \\ y(a)=0=y(b), \end{array} \right . $$

(1.3)

where ${}^{\mathrm{C}}_{a} D^{\alpha}$ is the Caputo fractional derivative of order α, $1 < \alpha\le2$, the zeros a and b are consecutive, and q is a real and continuous function. It was proved that if (1.3) has a nontrivial solution, then

$$ \int_{a}^{b}\bigl\vert q(t)\bigr\vert \, ds\geq\frac{\Gamma(\alpha)\alpha^{\alpha}}{ [(\alpha-1)(b-a)]^{\alpha-1}} . $$

(1.4)

Obviously, if we set $\alpha= 2$ in (1.4), one can obtain the Lyapunov classical inequality (1.2). In [11], the same author studied a differential equation that depends on the Riemann-Liouville fractional derivative and gave a Lyapunov-type inequality. In both works [10, 11], some interesting applications to the localization of real zeros of certain Mittag-Leffler functions were presented.

In [12], Jleli and Samet considered the fractional differential equation

$$ {}^{\mathrm{C}}_{a} D^{\alpha} y(t) + q(t)y(t) = 0,\quad a < t < b, 1 < \alpha\le2, $$

with the mixed boundary conditions

$$ y(a)=0=y'(b) $$

(1.5)

or

$$ y'(a)=0=y(b). $$

(1.6)

For boundary conditions (1.5) and (1.6), two Lyapunov-type inequalities were established respectively as follows:

$$ \int_{a}^{b} (b-s)^{\alpha-2} \bigl\vert q(s)\bigr\vert \, ds \geq\frac{\Gamma(\alpha)}{\max\{\alpha-1 , 2-\alpha\}(b-a)} $$

(1.7)

and

$$ \int_{a}^{b} (b-s)^{\alpha-1} \bigl\vert q(s)\bigr\vert \, ds \ge\Gamma(\alpha). $$

(1.8)

Motivated by the above works, we consider in this paper a Caputo fractional differential equation under boundary condition involving the Caputo fractional derivative. More precisely, we consider the boundary value problem

$$ \left \{ \begin{array}{l} {}^{\mathrm{C}}_{a} D^{\alpha} y(t) + q(t)y(t) = 0, \quad a < t < b, \\ y(a)=0, \qquad {}^{\mathrm{C}} _{a} D^{\beta} y(b)=0, \end{array} \right . $$

(1.9)

where $1 < \alpha\leq2$, $0 < \beta\le1$, and $q:[a , b]\rightarrow R$ is a continuous function. We write (1.9) as an equivalent integral equation and then, by using some properties of its Green function, we are able to get a corresponding Lyapunov-type inequality. After that, we show that this inequality can be used to obtain a real interval where a certain Mittag-Leffler function has no real zeros. Our results generalize the main results of Jleli and Samet [12].

2 Preliminaries

In this section, we introduce the definitions of the Riemann-Liouville fractional integral and the Caputo fractional derivative and give some lemmas which are used in this article.

Definition 2.1

Let $\alpha\geq0$ and f be a real function defined on $[a, b]$. The Riemann-Liouville fractional integral of order α is defined by ${}_{a} I^{0} f \equiv f$ and

$$ \bigl({}_{a} I^{\alpha}f\bigr) (t)=\frac{1}{\Gamma(\alpha)} \int _{a}^{t} (t-s)^{\alpha-1}f(s)\, ds,\quad \alpha>0, t \in[a, b]. $$

Definition 2.2

The Caputo derivative of fractional order $\alpha\ge0$ is defined by ${}_{a}^{\mathrm{C}} D^{0} f \equiv f$ and

$$ \bigl({}_{a}^{\mathrm{C}} D^{\alpha} f\bigr) (t) = \frac{1}{\Gamma(n-\alpha)}\int_{a}^{t} (t-s)^{n-\alpha-1} f^{(n)}(s)\, ds,\quad \alpha> 0, t \in[a, b], $$

where n is the smallest integer greater or equal to α.

The following results are standard within the fractional calculus theory involving the Caputo differential operator.

Lemma 2.1

([13], Chapter 2)

Let $\gamma> \alpha> 0$, $f\in C[a , b]$, then

$$ {}_{a}^{\mathrm{C}} D^{\alpha}\bigl({}_{a} I^{\gamma} f(t)\bigr) = {}_{a} I^{\gamma-\alpha}f(t),\quad t \in[a, b]. $$

Lemma 2.2

([14], Section 2)

Let $y \in C[a , b]$ and $1 < \alpha\le2$, then

$$ {}_{a} I^{\alpha}\bigl({}_{a}^{\mathrm{C}} D^{\alpha} y\bigr) (t) = y(t) + c_{0}+ c_{1}(t-a) $$

for some real constants $c_{0}$ and $c_{1}$.

3 Main results

We begin by writing problem (1.9) in its equivalent integral form.

Lemma 3.1

$y \in C[a, b]$ is a solution of the boundary value problem (1.9) if and only if y satisfies the integral equation

$$ y = \int_{a}^{b} G(t,s) q(s) y(s)\, ds, $$

where

$$ G(t,s) = H(t,s) (b-s)^{\alpha-\beta-1} $$

and

$$ H(t,s) = \left \{ \begin{array}{l@{\quad}l} \frac{\Gamma(2-\beta)(t-a)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}}-\frac {1}{\Gamma(\alpha)} (t-s)^{\alpha-1}(b-s)^{1+\beta-\alpha}, & a\leq s\leq t\leq b, \\ \frac{\Gamma(2-\beta)(t-a)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}}, & a \leq t\leq s\leq b. \end{array} \right . $$

(3.1)

Proof

From (1.9) and Lemma 2.2, we obtain

$$ y(t) = c_{0} + c_{1}(t-a) - \frac{1}{\Gamma(\alpha)} \int _{a}^{t} (t-s)^{\alpha-1} q(s) y(s)\, ds, $$

where $c_{0}$ and $c_{1}$ are some real constants. By the boundary condition $y(a)=0$, we can obtain that $c_{0}=0$. Thus, we have

$$ y(t) = c_{1}(t-a) - \bigl({}_{a} I^{\alpha} q y\bigr) (t). $$

(3.2)

By (3.2), we get

$$ {}_{a}^{\mathrm{C}} D^{\beta} y(t) = \frac{c_{1}}{\Gamma(2-\beta)}(t-a)^{1-\beta}-\frac{1}{ \Gamma(\alpha-\beta)} \int_{a}^{b} (t-s)^{\alpha-\beta-1} q(s) y(s)\, ds. $$

(3.3)

Since ${}_{a}^{\mathrm{C}} D^{\beta} y(b)=0$, we have from (3.3) that

$$ c_{1}=\frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}} \int_{a}^{b} (b-s)^{\alpha-\beta-1} q(s) y(s)\, ds. $$

(3.4)

Substitute (3.4) into (3.2), we obtain

$$ y(t)=\frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}}\int_{a}^{b}(b-s)^{\alpha-\beta-1}(t-a) q(s) y(s)\, ds -\frac{1}{\Gamma(\alpha)}\int_{a}^{t}(t-s)^{\alpha-1} q(s) y(s)\, ds, $$

which concludes the proof. □

Lemma 3.2

If $1 < \alpha< 2$ and $0 < \beta < 1$, then

$$ \Gamma(\alpha) < \frac{\Gamma(\alpha-\beta)}{\Gamma(2-\beta)} < \Gamma(\alpha-1). $$

(3.5)

Proof

Consider the logarithmic derivative of the gamma function

$$ \psi(x) = \frac{d}{dx} \ln\Gamma(x) = \frac{\Gamma^{\prime}(x)}{\Gamma(x)}. $$

(3.6)

We have by [15], p.264, that

$$ \psi(x) = - \gamma- \sum_{k=0}^{\infty} \biggl(\frac{1}{x+k} - \frac{1}{k+1} \biggr), $$

(3.7)

where γ is an Euler constant. From (3.7) we obtain

$$ \frac{d \psi(x)}{dx} = \sum_{k=0}^{\infty} \frac{1}{(x+k)^{2}} > 0. $$

(3.8)

Since $\alpha< 2$, we get by (3.6) and (3.8) that $\psi(\alpha- x) < \psi(2 - x)$, that is,

$$ \frac{\Gamma^{\prime}(\alpha-x)}{\Gamma(\alpha-x)} < \frac{\Gamma^{\prime}(2-x)}{\Gamma(2-x)}. $$

(3.9)

Let

$$ f(x) = \frac{\Gamma(\alpha-x)}{\Gamma(2-x)},\quad 0 < x < \alpha. $$

Then we have by (3.9) that

$$ f^{\prime}(x) = \frac{- \Gamma^{\prime}(\alpha-x) \Gamma(2-x) + \Gamma(\alpha-x) \Gamma^{\prime}(2-x)}{(\Gamma(2-x))^{2}} > 0. $$

Thus, $f(0) < f(\beta) < f(1)$ ($0 < \beta< 1$), which implies that (3.5) holds. □

Lemma 3.3

Assume that $0 < \beta\le1$ and $1 < \alpha\le1+\beta$ hold. Then

$$ \bigl\vert H(t,s)\bigr\vert \le \left \{ \begin{array}{l@{\quad}l} \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}(b-a)^{\beta}, & a \le t \le s \le b, \\ \max \{\frac{1}{\Gamma(\alpha)} - \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}, \frac{\Gamma(2-\beta )}{\Gamma(\alpha-\beta)}, \frac{2-\alpha}{\alpha-1} \cdot \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} \} (b-a)^{\beta}, & a \le s \le t \le b. \end{array} \right . $$

(3.10)

Proof

Throughout the proof we consider $\beta< 1$ since when $\beta= 1$ our study is reduced to the case in [12]. For $a \le t \le s \le b$, we easily know that

$$ \bigl\vert H(t, s)\bigr\vert \le \frac{\Gamma(2-\beta)(s-a)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}} \leq\frac{\Gamma(2-\beta)(b-a)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}} = \frac{\Gamma(2-\beta)(b-a)^{\beta}}{\Gamma(\alpha-\beta)}. $$

For convenience, let

$$ \psi (t,s)=\frac{\Gamma(2-\beta)(t-a)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta }}-\frac{1}{\Gamma(\alpha)} (t-s)^{\alpha-1}(b-s)^{1+\beta-\alpha}, \quad a \leq s\leq t\leq b. $$

Fixing arbitrary $s \in[a, b)$ and differentiating $\psi(t, s)$ with respect to t, we obtain

$$ \psi_{t}(t,s) = \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}}-\frac {1}{\Gamma(\alpha-1)} (t-s)^{\alpha-2}(b-s)^{1+\beta-\alpha},\quad s < t. $$

(3.11)

From (3.11) we easily know that $\psi_{t}(t^{*}, s) = 0$ if and only if

$$ t^{*} = s + \biggl[\frac{\Gamma(2-\beta) \Gamma(\alpha-1)}{\Gamma(\alpha-\beta)} \cdot\frac{(b-s)^{\alpha-\beta-1}}{(b-a)^{1-\beta}} \biggr]^{\frac {1}{\alpha-2}} $$

(3.12)

provided $t^{*} \le b$, i.e., as long as $s \le b - l$, where

$$ l = \biggl[\frac{\Gamma(\alpha-\beta)}{\Gamma(2-\beta) \Gamma(\alpha-1)} \biggr]^{\frac{1}{1-\beta}}(b-a) < b-a \quad ( \mbox{by (3.5)}). $$

Hence, if $s > b - l$, then

$$ \psi_{t}(t, s) < 0,\quad t \in (s, b). $$

(3.13)

On the other hand, we have

$$ \lim_{t \to s^{+}} \psi(t, s) = \frac{\Gamma(2-\beta)(s-a)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}}\quad \mbox{and} \quad \psi(b, s) = \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}(b-a)^{\beta}-\frac {1}{\Gamma(\alpha)} (b-s)^{\beta}. $$

Thus, we obtain

$$ \bigl\vert \psi(s, s)\bigr\vert \le \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} (b-a)^{\beta} $$

(3.14)

and

$$\begin{aligned} \bigl\vert \psi(b, s)\bigr\vert =& \biggl\vert \biggl( \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} - \frac{1}{\Gamma(\alpha)} \biggr) (b-s)^{\beta} + \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}\bigl((b-a)^{\beta}-(b-s)^{\beta }\bigr)\biggr\vert \\ \le& \max \biggl\{ \biggl(\frac{1}{\Gamma(\alpha)} -\frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} \biggr) (b-s)^{\beta}, \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} (s-a)^{\beta} \biggr\} \end{aligned}$$

(3.15)

by $(s-a)^{\beta}+(b-s)^{\beta} \ge(b-a)^{\beta}$ and (3.5). Thus, we have by (3.13)-(3.15) that

$$\begin{aligned} \bigl\vert \psi(t, s)\bigr\vert \le&\max\bigl\{ \bigl\vert \psi(s, s)\bigr\vert , \bigl\vert \psi(b, s)\bigr\vert \bigr\} \\ \le& \max \biggl\{ \biggl(\frac{1}{\Gamma(\alpha)} -\frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} \biggr), \frac{\Gamma (2-\beta)}{\Gamma(\alpha-\beta)} \biggr\} (b-a)^{\beta}, \\ & a < b-l < s < t \le b. \end{aligned}$$

(3.16)

It remains to verify the result when $s \le b-l$, i.e., when $t^{*} \le b$. It is easy to check that

$$ \psi_{t}(t, s) < 0\quad \mbox{for } t < t^{*}\quad \mbox{and}\quad \psi_{t}(t, s) \ge0 \quad \mbox{for } t \ge t^{*}. $$

Hence, we have

$$ \bigl\vert \psi(t, s)\bigr\vert \le\max\bigl\{ \bigl\vert \psi\bigl(t^{*}, s\bigr)\bigr\vert , \bigl\vert \psi(b, s)\bigr\vert , \bigl\vert \psi(s, s)\bigr\vert \bigr\} ,\quad a < s \le b-l, s \le t \le b. $$

(3.17)

By (3.12) we have

$$\begin{aligned} \bigl\vert \psi\bigl(t^{*}, s\bigr)\bigr\vert =& \biggl\vert \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}} \biggl[s + \biggl(\frac{\Gamma(2-\beta) \Gamma(\alpha-1)}{\Gamma(\alpha-\beta)} \cdot\frac{(b-s)^{\alpha-\beta-1}}{(b-a)^{1-\beta}} \biggr)^{\frac {1}{\alpha-2}}-a \biggr] \\ &{}-\frac{1}{\Gamma(\alpha)} \biggl(\frac{\Gamma(2-\beta) \Gamma(\alpha-1)}{\Gamma(\alpha-\beta)} \cdot\frac{(b-s)^{\alpha-\beta-1}}{(b-a)^{1-\beta}} \biggr)^{\frac {\alpha-1}{\alpha-2}} (b-s)^{1+\beta-\alpha}\biggr\vert \\ =& \biggl\vert \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}}(s-a) + \biggl(\frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}} \biggr)^{\frac{\alpha-1}{\alpha-2}} \\ &{}\cdot\bigl(\Gamma(\alpha-1)\bigr)^{\frac{1}{\alpha-2}} (b-s)^{\frac{\alpha-\beta-1}{\alpha-2}} \\ &{} -\frac{1}{\Gamma(\alpha)} \cdot \biggl(\frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}} \biggr)^{\frac{\alpha-1}{\alpha-2}} \cdot\bigl(\Gamma(\alpha-1)\bigr)^{\frac{\alpha-1}{\alpha-2}} (b-s)^{\frac{\alpha-\beta-1}{\alpha-2}} \biggr\vert \\ =& \biggl\vert \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}}(s-a) + \frac{\alpha-2}{\alpha-1} \biggl( \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}} \biggr)^{\frac{\alpha-1}{\alpha-2}} \\ &{}\cdot\bigl(\Gamma(\alpha-1) \bigr)^{\frac{1}{\alpha-2}} (b-s)^{\frac{\alpha-\beta-1}{\alpha-2}}\biggr\vert \\ \le&\max \biggl\{ \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}}(s-a), \frac{2-\alpha}{\alpha-1} \biggl( \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}} \biggr)^{\frac{\alpha-1}{\alpha-2}} \\ &{} \cdot\bigl(\Gamma(\alpha-1) \bigr)^{\frac{1}{\alpha-2}} (b-s)^{\frac{\alpha-\beta-1}{\alpha-2}} \biggr\} . \end{aligned}$$

(3.18)

Obviously, we have

$$ \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}}(s-a) \le\frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}(b-a)^{\beta}, $$

(3.19)

and we obtain from Lemma 3.2 and condition $\alpha - \beta- 1 \le0$ that

$$\begin{aligned}& \frac{2-\alpha}{\alpha-1} \biggl(\frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}} \biggr)^{\frac{\alpha-1}{\alpha-2}} \cdot\bigl(\Gamma(\alpha-1)\bigr)^{\frac{1}{\alpha-2}} (b-s)^{\frac{\alpha-\beta-1}{\alpha-2}} \\& \quad \le\frac{2-\alpha}{\alpha-1} \biggl(\frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)(b-a)^{1-\beta}} \biggr)^{\frac{\alpha-1}{\alpha-2}} \cdot\bigl(\Gamma(\alpha-1)\bigr)^{\frac{1}{\alpha-2}} (b-a)^{\frac{\alpha-\beta-1}{\alpha-2}} \\& \quad = \frac{2-\alpha}{\alpha-1} \cdot \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} \cdot \biggl( \frac{\Gamma(2-\beta) \Gamma(\alpha-1) }{\Gamma(\alpha-\beta)} \biggr)^{\frac{1}{\alpha-2}} \cdot (b-a)^{\beta} \\& \quad < \frac{2-\alpha}{\alpha-1} \cdot \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}(b-a)^{\beta}. \end{aligned}$$

(3.20)

Thus, from (3.18)-(3.20) we conclude that

$$ \bigl\vert \psi\bigl(t^{*}, s\bigr)\bigr\vert \le\max \biggl\{ 1, \frac{2-\alpha}{\alpha-2} \biggr\} \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} (b-a)^{\beta},\quad s \in[a, b-l] $$

(3.21)

holds. From (3.14), (3.15), (3.21) and (3.17), we know that inequality (3.10) is true. The proof is complete. □

Theorem 3.4

Let $0 < \beta\le1$ and $1 < \alpha\le1+\beta$. If a nontrivial continuous solution of the fractional boundary value problem (1.9) exists, then

$$ \int_{a}^{b} (b-s)^{\alpha-\beta-1} \bigl\vert q(s)\bigr\vert \, ds \geq \frac{(b-a)^{-\beta}}{\max \{\frac{1}{\Gamma(\alpha)} - \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}, \frac{\Gamma(2-\beta )}{\Gamma(\alpha-\beta)}, \frac{2-\alpha}{\alpha-1} \cdot \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} \}}. $$

(3.22)

Proof

Let $B = C[a, b]$ be the Banach space endowed with the norm $\|y\|_{\infty} = \sup_{t \in[a, b]} |y(t)|$. According to Lemma 3.1, the solution of (1.9) can be written as

$$ y(t) = \int_{a}^{b} (b-s)^{\alpha-\beta-1} H(t, s)q(s) y(s)\, ds. $$

Now, an application of Lemma 3.3 yields

$$\begin{aligned} \|y\|_{\infty} \leq&\max \biggl\{ \frac{1}{\Gamma(\alpha)} - \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}, \frac{\Gamma(2-\beta )}{\Gamma(\alpha-\beta)}, \frac{2-\alpha}{\alpha-1} \cdot \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} \biggr\} \\ &{}\cdot (b-a)^{\beta} \|y\|_{\infty} \int_{a}^{b} (b-s)^{\alpha-\beta-1} \bigl\vert q(s)\bigr\vert \, ds, \end{aligned}$$

which implies that (3.22) holds. □

Remark 3.1

If $\beta= 1$, then (3.22) reduces to the following Lyapunov-type inequality [12]:

$$ \int_{a}^{b} (b-s)^{\alpha-2} \bigl\vert q(s)\bigr\vert \, ds \geq\frac{\Gamma(\alpha)}{\max\{\alpha-1, 2-\alpha\} (b-a)}. $$

Remark 3.2

If $\alpha= 2$ and $0 < \beta< 1$, then we have by Lemma 3.1 that

$$ G(t,s) = \left \{ \begin{array}{l@{\quad}l} (t-a) (\frac{b-s}{b-a} )^{1-\beta} - (t-s), & a\leq s\leq t\leq b, \\ (t-a) (\frac{b-s}{b-a} )^{1-\beta}, & a \leq t\leq s\leq b. \end{array} \right . $$

Similar to the proof of Theorem 3.4, it is easy to obtain that the following Lyapunov-type inequality holds:

$$ \int_{a}^{b} \bigl\vert q(s)\bigr\vert \, ds \geq\frac{1}{ (b-a)^{\beta}}. $$

In the following, we will use Lyapunov-type inequalities (3.22) to obtain intervals where certain Mittag-Leffler functions have no real zeros. Let $z \in\mathbb{R}$ and consider the real zeros of the Mittag-Leffler functions $E_{\alpha, \gamma}(z)$, where

$$ E_{\alpha, \gamma}(z) = \sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma(k\alpha+\gamma)},\quad \alpha>0, \gamma> 0 \mbox{ and } z \in\mathbb{C}. $$

Obviously, $E_{\alpha, \gamma}(z)>0$ for all $z \ge 0$. Hence, the real zeros of $E_{\alpha, \gamma}(z)$, if they exist, must be negative real numbers.

Theorem 3.5

Assume that $0 < \beta\le1$ and $1 < \alpha\le1+\beta$ hold. Then the Mittag-Leffler function $E_{\alpha, 2-\beta}(x)$ has no real zeros for

$$ x \in \biggl(- \frac{\alpha-\beta}{\max \{\frac{1}{\Gamma(\alpha)} - \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}, \frac{\Gamma(2-\beta )}{\Gamma(\alpha-\beta)}, \frac{2-\alpha}{\alpha-1} \cdot \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} \}}, 0 \biggr]. $$

Proof

Let $a=0$ and $b=1$. Consider the following fractional Sturm-Liouville eigenvalue problem:

$$\begin{aligned}& {}^{\mathrm{C}}_{0} D^{\alpha}y(t)+ \lambda y(t) = 0,\quad t \in(0, 1), \end{aligned}$$

(3.23)

$$\begin{aligned}& y(0)= {}^{\mathrm{C}}_{0} D^{\beta} y(1)=0. \end{aligned}$$

(3.24)

By the Laplace transform method as in [13, 16, 17], the general solution of the fractional differential equation (3.23) can be given as follows:

$$ y(t) = A E_{\alpha, 1}\bigl(-\lambda t^{\alpha}\bigr) + B t E_{\alpha, 2}\bigl(-\lambda t^{\alpha}\bigr). $$

(3.25)

In the following discussion we will use the general solution (3.25) and its fractional Caputo derivative

$$ {}^{\mathrm{C}}_{0} D^{\beta} y(t) = A t^{-\beta} E_{\alpha, 1-\beta}\bigl(-\lambda t^{\alpha}\bigr) + B t^{1-\beta} E_{\alpha, 2-\beta }\bigl(-\lambda t^{\alpha}\bigr). $$

(3.26)

By (3.25), (3.26) and the boundary conditions (3.24), we obtain that

$$ A = 0,\qquad B E_{\alpha, 2-\beta}(-\lambda) = 0. $$

Thus, the eigenvalues $\lambda\in\mathbb{R}$ of (3.23) and (3.24) are the solutions of

$$ E_{\alpha, 2-\beta}(-\lambda) = 0, $$

(3.27)

and the corresponding eigenfunctions are given by

$$ y(t) = t E_{\alpha, 2-\beta}\bigl(-\lambda t^{\alpha}\bigr), \quad t \in[0, 1]. $$

By Theorem 3.4, if a real eigenvalue λ of (3.23) and (3.24) exists, i.e., −λ is a zero of (3.27), then

$$ \lambda\int_{0}^{1} (1-s)^{\alpha-\beta-1}\, ds \geq \frac{1}{\max \{\frac{1}{\Gamma(\alpha)} - \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}, \frac{\Gamma(2-\beta )}{\Gamma(\alpha-\beta)}, \frac{2-\alpha}{\alpha-1} \cdot \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} \}}, $$

that is,

$$ \lambda\geq\frac{\alpha-\beta}{\max \{\frac{1}{\Gamma(\alpha)} - \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}, \frac{\Gamma(2-\beta )}{\Gamma(\alpha-\beta)}, \frac{2-\alpha}{\alpha-1} \cdot \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)} \}}, $$

which concludes the proof. □

Remark 3.3

If $\beta= 1$, then Theorem 3.5 reduces to Theorem 3 in [12].