Lyapunov-type inequalities for sequential fractional boundary value problems using Hilfer’s fractional derivative

Wei Zhang; Wenbin Liu

doi:10.1186/s13660-019-2050-6

Research
Open Access

Lyapunov-type inequalities for sequential fractional boundary value problems using Hilfer’s fractional derivative

Wei Zhang¹ and
Wenbin Liu¹Email author

Journal of Inequalities and Applications20192019:98

https://doi.org/10.1186/s13660-019-2050-6

Received: 25 January 2019
Accepted: 1 April 2019
Published: 11 April 2019

Abstract

This paper is devoted to studying the Lyapunov-type inequality for sequential Hilfer fractional boundary value problems. We first provide some properties of Hilfer fractional derivative, and then establish Lyapunov-type inequalities for a sequential Hilfer fractional differential equation with two types of multi-point boundary conditions. Our results generalize and compliment the existing results in the literature.

Keywords

Lyapunov-type inequality
Sequential fractional differential equation
Hilfer fractional derivative
Multi-point boundary condition

MSC

34A08
34B15

1 Introduction

As is well known, Lyapunov inequality was first introduced by Lyapunov [1], who established a necessary condition for the existence of nontrivial solution of the boundary value problem (BVP for short):

$$ \textstyle\begin{cases} x''(t)+q(t)x(t) = 0,\quad t \in (a,b), \\ x(a) = x(b) = 0, \end{cases} $$

(1.1)

as the form

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

(1.2)

where $q \in C([a,b],\mathbb{R})$. Since then, Lyapunov inequality and Lyapunov-type inequality have been studied with great interest, and they have been proved to be an effective tool in the study of differential and difference equations, such as oscillation theory, disconjugacy, eigenvalue problems, etc. (see [2–5]).

In recent years, by the rise of theoretical research in fractional differential equations, there has been tremendous interest in the research of Lyapunov-type inequalities for fractional BVP, see [6–30] and the references cited therein.

In [11], Ferreira discussed the Lyapunov-type inequality for the following fractional BVP:

$$ \textstyle\begin{cases} (D_{a+}^{\alpha }x)(t)+q(t)x(t) = 0,\quad t \in (a,b), \\ x(a) = x(b) = 0, \end{cases} $$

(1.3)

where $D_{a+}^{\alpha }$ is the left Riemann–Liouville fractional derivative of order α, $\alpha \in (1,2]$ and $q \in C([a,b],\mathbb{R})$. The Lyapunov-type inequality for problem (1.3) was established as follows:

$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds} > \varGamma (\alpha ){ \biggl( {\frac{4}{{b-a}}} \biggr)^{\alpha - 1}}. $$

(1.4)

Furthermore, in 2016, Ferreira [12] considered the Lyapunov-type inequality for a sequential fractional BVP

$$ \textstyle\begin{cases} ({}_{a}{D^{\alpha }}{}_{a}{D^{\beta }}x)(t)+q(t)x(t) = 0, \quad t \in (a,b), \\ x(a) = x(b) = 0, \end{cases} $$

(1.5)

where ${}_{a}{D^{\gamma }}$, $\gamma = \alpha $, β stand for the left Riemann–Liouville fractional derivative or the left Caputo fractional derivative of order γ, $\gamma \in (0,1]$ and $1 < \alpha + \beta \leqslant 2$, $q \in C([a,b],\mathbb{R})$. Two interesting results have been obtained as follows:

(i)

Take ${}_{a}{D^{\gamma }}$, $\gamma = \alpha $, β, is the left Riemann–Liouville fractional derivative, then problem (1.5) has a nontrivial continuous solution provided that
$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds > \varGamma (\alpha + \beta )} { \biggl( {\frac{4}{ {b - a}}} \biggr)^{\alpha + \beta - 1}}. $$

(1.6)
(ii)

Take ${}_{a}{D^{\gamma }}$, $\gamma = \alpha $, β, is the left Caputo fractional derivative, then problem (1.5) has a nontrivial continuous solution provided that
$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert } \,ds>\frac{{\varGamma (\alpha + \beta )}}{{{{(b-a)} ^{\alpha + \beta - 1}}}}\frac{{{{(\alpha + 2\beta - 1)}^{\alpha + 2 \beta - 1}}}}{{{{(\alpha + \beta - 1)}^{\alpha + \beta - 1}}{\beta ^{\beta }}}}. $$

(1.7)

It is worth noting that the study of the Hilfer fractional differential equations has received a significant amount of attention in the last few years. Hilfer fractional derivative was proposed by Hilfer in 2000, which is a generalization of both Riemann–Liouville and Caputo fractional derivatives (see [31]). Meanwhile, the discussion of Lyapunov-type inequalities for fractional BVP with Hilfer fractional derivative can be found in papers [25, 29, 30].

In [25], Pathak investigated Lyapunov-type inequalities for the following Hilfer fractional differential equation:

$$ \bigl(D_{a+}^{\alpha ,\beta }x\bigr) (t)+q(t)x(t) = 0, \quad t \in (a,b), $$

(1.8)

with boundary conditions

$$ x(a) = x(b) = 0, $$

(1.9)

or

$$ x(a) = x'(b) = 0, $$

(1.10)

where $D_{a+}^{\alpha ,\beta }$ is the left Hilfer fractional derivative of order α and type β, $\alpha \in (1,2]$, $\beta \in [0,1]$, $q \in C([a,b],\mathbb{R})$. The author got two Lyapunov-type inequalities: for BVP (1.8), (1.9) as

$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds} \geqslant \frac{{\varGamma (\alpha ){{[\alpha - (2 - \alpha )(1 - \beta )]}^{\alpha - (2 - \alpha )(1 - \beta )}}}}{ {{{(b - a)}^{\alpha - 1}}{{(\alpha - 1)}^{\alpha - 1}}{{[\alpha - 1 + \beta (2 - \alpha )]}^{\alpha - 1 + \beta (2 - \alpha )}}}}, $$

(1.11)

for BVP (1.8), (1.10) as

$$ \int _{a}^{b} {{{(b - s)}^{\alpha - 2}} \bigl\vert q(s) \bigr\vert \,ds} \geqslant \frac{ {\varGamma (\alpha )[\alpha - 1 + \beta (2 - \alpha )]}}{{(b - a)\max \{ \alpha - 1,\beta (2 - \alpha )\} }}. $$

(1.12)

In [30], Wang obtained Lyapunov-type inequality for the fractional multi-point BVP involving Hilfer derivative:

$$ \textstyle\begin{cases} (D_{a+}^{\alpha ,\beta }x)(t)+q(t)x(t) = 0, \quad t \in (a,b), \\ x(a) = 0,\qquad x(b) = \sum_{i = 1}^{m - 2} {{\beta _{i}}x({\xi _{i}})} , \end{cases} $$

(1.13)

where $q \in C([a,b],\mathbb{R})$, $D_{a+}^{\alpha ,\beta }$ is the left Hilfer fractional derivative of order α and type β with $\alpha \in (1,2]$, $\beta \in [0,1]$, $a <{\xi _{1}} <{\xi _{2}} <\cdots <{\xi _{m-2}}< b$, ${\beta _{i}}\geqslant 0$ ($i=1,2,\ldots ,m-2$), ${(b-a)^{1-(2- \alpha )(1-\beta )}}> \sum_{i=1}^{m-2} {{\beta _{i}}} {( {\xi _{i}}-a)^{1-(2-\alpha )(1-\beta )}}$. The Lyapunov-type inequality for problem (1.13) is given as follows:

$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds} \geqslant \frac{{\varGamma (\alpha )}}{{{{(b-a)} ^{\alpha -1}}L}} \frac{1}{{1+\sum_{i=1}^{m-2} {{\beta _{i}}T(b)}}}, $$

where

$$\begin{aligned}& L = \frac{{{{(\alpha - 1)}^{\alpha - 1}}{{(\alpha - 1 + 2\beta - \alpha \beta )}^{\alpha - 1 + 2\beta - \alpha \beta }}}}{{{{(2\alpha - 2 + 2\beta - \alpha \beta )}^{2\alpha - 2 + 2\beta - \alpha \beta }}}}, \\& T(t) = \frac{{{{(t - a)}^{1 - (2 - \alpha )(1 - \beta )}}}}{{{{(b - a)}^{1 - (2 - \alpha )(1 - \beta )}} - \sum_{i = 1}^{m - 2} {{\beta _{i}}} {{({\xi _{i}} - a)}^{1 - (2 - \alpha )(1 - \beta )}}}}, \quad t \in [a,b]. \end{aligned}$$

Motivated by these results, in this paper we study Lyapunov-type inequalities for a sequential Hilfer fractional differential equation

$$ \bigl(D_{a+}^{{\alpha _{1}},{\beta _{1}}}D_{a+}^{{\alpha _{2}},{\beta _{2}}}x \bigr) (t) +q(t)x(t) = 0,\quad t \in (a,b), $$

(1.14)

with multi-point boundary conditions

$$ x(a) = 0,\qquad x(b) = \sum_{i = 1}^{m - 2} {{\sigma _{i}}x( {\xi _{i}})}, $$

(1.15)

or

$$ x(a) = 0, \qquad x'(b) = \sum _{i = 1}^{m - 2} {{\delta _{i}}} x( {\eta _{i}}), $$

(1.16)

where $q \in C([a,b],\mathbb{R})$, $D_{a+}^{{\alpha _{i}},{\beta _{i}}}$, $i=1$, 2, are two left Hilfer fractional derivatives of order ${\alpha _{i}}$ and types ${\beta _{i}}$ with ${\alpha _{i}} \in (0,1]$, $1<{\alpha _{1}}+{\alpha _{2}} \leqslant 2$, ${\beta _{i}} \in [0,1]$. For the definition of Hilfer fractional derivative, see Sect. 2. A remarkable characteristic of this kind of fractional derivative is that the type ${\beta _{i}}$ allows $D_{a+}^{{\alpha _{i}},{\beta _{i}}}$ to interpolate continuously from the Riemann–Liouville case $D_{a+}^{{\alpha _{i}},0} \equiv D_{a+}^{\alpha _{i}}$ to the Caputo case $D_{a+}^{{\alpha _{i}},1} \equiv {}^{C}D_{a+}^{\alpha _{i}}$ (see [32]). To state our main results, we assume that the following conditions hold:

$(A_{1})$ :: $a < {\xi _{1}} < {\xi _{2}} < \cdots < {\xi _{m - 2}} < b, {\sigma _{i}} \geqslant 0$ ($i = 1,2, \ldots ,m - 2$) and
$$ {\Delta _{1}} := {(b-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} -\sum _{i = 1}^{m - 2} {{\sigma _{i}}} {({\xi _{i}} - a)^{ {\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} > 0. $$

$(A_{2})$ :: $a < {\eta _{1}} < {\eta _{2}} < \cdots < {\eta _{m - 2}} < b, {\delta _{i}} \geqslant 0$ ($i = 1,2, \ldots ,m - 2$) and
$$\begin{aligned} {\Delta _{2}} :=& \bigl[{\alpha _{2}} - (1 - {\alpha _{1}}) (1 - {\beta _{1}})\bigr]{(b - a)^{{\alpha _{2}} - 1 - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} \\ &{} - \sum_{i = 1}^{m - 2} {{\delta _{i}}} {({\eta _{i}} - a)^{{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} > 0. \end{aligned}$$

In the present work, we are focused on establishing the Lyapunov-type inequalities for a sequential Hilfer fractional differential equation with two types of multi-point boundary conditions. As far as we know, the Lyapunov-type inequality for fractional BVP with Hilfer derivative has seldom been considered up to now. The new insights of this paper can be presented as follows. On the one hand, we provide some properties of Hilfer fractional derivative, which are not introduced in the previous paper (see Sect. 2, Lemmas 2.6, 2.7, and 2.8). On the other hand, we extend the previous results. Many previous results are a special case of our work, which is embodied in Sect. 3. The main difficulties in this article are as follows. First, we have to construct Green’s function for BVPs (1.14), (1.15) and (1.14), (1.16). Second, it is difficult to estimate the maximum of Green’s function because Green’s functions do not satisfy the non-negativity.

The rest of this paper is organized as follows. In Sect. 2, we recall some definitions, lemmas of fractional calculus. In Sect. 3, by constructing Green’s functions and finding its corresponding maximum value, we prove our main results. Finally, a conclusion is given in Sect. 4.

2 Preliminaries

In this section, we recall some definitions and lemmas which are used throughout this paper.

Definition 2.1

([33–35])

Let $J = [a,b]$ ($-\infty < a < b < \infty $). The left Riemann–Liouville fractional integral of order $\alpha > 0$ of a function $x:(a,b) \to \mathbb{R}$ is defined by

$$ \bigl(I_{a+}^{\alpha }x\bigr) (t)=\frac{1}{{\varGamma (\alpha )}} \int _{a}^{t} {{{(t-s)}^{\alpha - 1}}x(s)\,ds} \quad (t>a), $$

provided that the right-hand side integral is pointwise defined on $[a,b]$.

Definition 2.2

([33–35])

The left-sided Riemann–Liouville fractional derivative of order $\alpha > 0$ ($n-1 < \alpha \leqslant n$, $n \in {\mathbb{N}^{+}}$) of a function $x:(a,b) \to \mathbb{R}$ is defined by

$$ \bigl(D_{a+}^{\alpha }x\bigr) (t) = \frac{{{d^{n}}}}{{d{t^{n}}}} \bigl(I_{a+}^{n- \alpha }x\bigr) (t)= \frac{1}{{\varGamma (n-\alpha )}} \frac{{{d^{n}}}}{{d {t^{n}}}} \int _{a}^{t} {{{(t-s)}^{n-\alpha - 1}}x(s)\,ds} \quad (t > a), $$

provided the right-hand side integral is pointwise defined on $[a,b]$.

Definition 2.3

([33–35])

The left-sided Caputo fractional derivative of order $\alpha > 0$ ($n - 1 < \alpha < n$, $n \in {\mathbb{N}^{+}}$) of a function $x:(a,b) \to \mathbb{R}$ is defined by

$$ \bigl({}^{C}D_{a+}^{\alpha }x\bigr) (t)= \bigl(I_{a+}^{n-\alpha }{x^{(n)}}\bigr) (t)= \frac{1}{ {\varGamma (n-\alpha )}} \int _{a}^{t} {{{(t-s)}^{n-\alpha - 1}} {x^{(n)}}(s)\,ds} \quad (t > a), $$

provided the right-hand side integral is pointwise defined on $[a,b]$.

Definition 2.4

([32])

The left-sided Hilfer fractional derivative of order $\alpha >0$ ($n-1 <\alpha \leqslant n$, $n \in {\mathbb{N}^{+}}$) and type $0\leq \beta \leq 1$ of a function $x:(a,b) \to \mathbb{R}$ is defined by

$$ \bigl(D_{a+}^{\alpha ,\beta }x\bigr) (t)= \bigl( {I_{a+}^{\beta (n-\alpha )} {D^{n}} \bigl( {I_{a+}^{(1-\beta )(n-\alpha )}x} \bigr)} \bigr) (t) \quad (t > a), $$

where ${D^{n}} = {{{d^{n}}} / {d{t^{n}}}}$.

Lemma 2.1

([33–35])

Let $\alpha > 0$, $n=[\alpha ] + 1$. If $x \in {L^{1}}(a,b)$, $I_{a+}^{n-\alpha }x \in A{C^{n}}[a,b]$, then

$$ I_{a+}^{\alpha }D_{a+}^{\alpha }x(t)=x(t)-\sum _{k=0}^{n-1} {\frac{{{{(t-a)}^{k-(n-\alpha )}}}}{{\varGamma (k-(n-\alpha )+1)}} \lim _{t \to a+}} \frac{{{d^{k}}}}{{d{t^{k}}}}\bigl(I_{a+} ^{n-\alpha }x\bigr) (t). $$

Lemma 2.2

([33–35])

If $\alpha > 0$, $\lambda > - 1$, then

$$ I_{a+}^{\alpha }{(t-a)^{\lambda }}=\frac{{\varGamma (\lambda +1)}}{ {\varGamma (\lambda + 1+\alpha )}}{(t-a)^{\alpha + \lambda }}, \qquad D_{a+} ^{\alpha }{(t-a)^{\lambda }} = \frac{{\varGamma (\lambda +1)}}{{\varGamma (\lambda + 1 - \alpha )}}{(t-a)^{\lambda - \alpha }}. $$

Lemma 2.3

([36])

Let $\alpha > 0$, $n = [\alpha ]+1$, $0 \leqslant \beta \leqslant 1$. If $x \in {L^{1}}(a,b)$, $I_{a+} ^{(n-\alpha )(1-\beta )}x \in A{C^{n}}[a,b]$, then

$$ \bigl(I_{a+}^{\alpha }D_{a+}^{\alpha ,\beta }x\bigr) (t)= x(t)- \sum_{k=0}^{n-1} { \frac{{{{(t-a)}^{k-(n-\alpha )(1-\beta )}}}}{{\varGamma (k-(n- \alpha )(1-\beta )+1)}}} \lim_{t \to a+} \frac{{{d ^{k}}}}{{d{t^{k}}}} \bigl(I_{a+}^{(n-\alpha )(1-\beta )}x\bigr) (t). $$

Lemma 2.4

([35])

Let $\alpha >0$, $n \in N$, and $D={d / {dx}}$. If the fractional derivatives $(D_{a+}^{\alpha }x)(t)$ and $(D_{a+}^{\alpha +n}x)(t)$ exist, then

$$ \bigl({D^{n}}D_{a+}^{\alpha }x\bigr) (t)= \bigl(D_{a+}^{\alpha + n}x\bigr) (t). $$

Lemma 2.5

([35])

Let $\alpha >0$, $n \in N$, and $D={d / {dx}}$. If the fractional derivatives $({D^{n}}x)(t)$ and $({}^{C}D _{a+}^{\alpha +n}x)(t)$ exist, then

$$ \bigl({}^{C}D_{a+}^{\alpha }{D^{n}}x\bigr) (t)=\bigl({}^{C}D_{a+}^{\alpha + n}x\bigr) (t). $$

Lemma 2.6

Let $\alpha > 0$, $n=[\alpha ]+1$, $0 \leqslant \beta \leqslant 1$, $m \in \mathbb{N}$, and $D={d / {dx}}$. If the fractional derivatives $({D^{m}}x)(t)$ and $(D_{a+}^{\alpha + m, \beta }x)(t)$ exist, then

$$ \bigl(D_{a+}^{\alpha ,\beta }{D^{m}}x\bigr) (t)= \bigl(D_{a+}^{\alpha + m,\beta }x\bigr) (t), $$

provided that

$$ {x^{(j)}}(a) = 0, \quad j = 0,1,2, \ldots ,m - 1. $$

Proof

Since ${x^{(j)}}(a)=0$, $j = 0,1,2,\ldots ,m - 1$, then we get

$$ \bigl(I_{a + }^{m}{D^{m}}x\bigr) (t) = x(t), $$

which yields the following equalities hold:

$$ \begin{aligned} \bigl(D_{a+}^{\alpha ,\beta }{D^{m}}x \bigr) (t)={} & \bigl( {I_{a+}^{\beta (n- \alpha )}{D^{n}}I_{a+}^{(1-\beta )(n-\alpha )}{D^{m}}x} \bigr) (t) \\ ={}& \bigl( {I_{a+}^{\beta (n-\alpha )}{D^{n+m}}I_{a+}^{(1-\beta )(n-\alpha )} \bigl(I_{a+}^{m}{D^{m}}x\bigr)} \bigr) (t) \\ ={}& \bigl( {I_{a+} ^{\beta (n-\alpha )}{D^{n+m}}I_{a+}^{(1-\beta )(n-\alpha )}x} \bigr) (t)=\bigl(D _{a+}^{\alpha + m,\beta }x\bigr) (t). \end{aligned} $$

The proof is completed. □

Lemma 2.7

Let $\alpha > 0$, $n=[\alpha ]+1$, $0 \leqslant \beta \leqslant 1$. If $x \in C[a,b]$, $I_{a+}^{1-\beta (n- \alpha )}x \in AC[a,b]$, then

$$ D_{a+}^{\alpha ,\beta }I_{a+}^{\alpha }x(t)=x(t). $$

(2.1)

Proof

On the one hand, if $\beta (n-\alpha ) =0$, i.e., $\beta =0$ or $\alpha =n$, then

$$ D_{a+}^{\alpha ,\beta }I_{a+}^{\alpha }x(t)=D_{a+}^{\alpha }I_{a+} ^{\alpha }x(t)=x(t) $$

or

$$ D_{a+}^{\alpha ,\beta }I_{a+}^{\alpha }x(t)=D^{n}I_{a+}^{n} x(t) =x(t). $$

On the other hand, if $\beta (n-\alpha )\neq 0$, by Definitions 2.2, 2.4 and Lemma 2.1, we have

$$ \begin{aligned} D_{a+}^{\alpha ,\beta }I_{a+}^{\alpha }x(t) ={} & \bigl( {I_{a+}^{ \beta (n-\alpha )}{D^{n}} \bigl( {I_{a+}^{n-\beta (n-\alpha )}x} \bigr)} \bigr) (t) \\ ={}& \bigl( {I_{a+}^{\beta (n-\alpha )}D _{a+}^{\beta (n-\alpha )}x} \bigr) (t)=x(t)-\frac{{{{(t-a)}^{ \beta (n-\alpha )-1}}}}{{\varGamma (\beta (n-\alpha ))}}\lim_{t \to a+} \bigl(I_{a+}^{1-\beta (n-\alpha )}x\bigr) (t). \end{aligned} $$

Since $x \in C[a,b]$, one has

$$ \lim_{t \to a+} \bigl(I_{a+}^{1-\beta (n-\alpha )}x\bigr) (t) =0. $$

Thus, (2.1) holds for $\beta (n-\alpha ) \neq 0$. The proof is completed. □

Lemma 2.8

Let $\alpha > 0$, $n=[\alpha ]+1$, $0 \leqslant \beta \leqslant 1$, $\lambda > - 1$, then

$$ D_{a+}^{\alpha ,\beta }{(t-a)^{\lambda + \beta (n-\alpha )}}=\frac{ {\varGamma (\lambda + 1 + \beta (n - \alpha ))}}{{\varGamma (\lambda + 1 - \alpha + \beta (n - \alpha ))}}{(t-a)^{\beta (n-\alpha )+\lambda - \alpha }}. $$

In particular,

$$ D_{a+}^{\alpha ,\beta }{(t-a)^{\alpha -j+\beta (n-\alpha )}}=0,\quad j = 1,2,\ldots ,n. $$

Proof

For $\lambda > - 1$, by Definition 2.4 and Lemma 2.2, we have

$$\begin{aligned} D_{a+}^{\alpha ,\beta }{(t-a)^{\lambda + \beta (n-\alpha )}} = & I_{a+} ^{\beta (n-\alpha )}{D^{n}}I_{a+}^{(1-\beta )(n-\alpha )}{(t-a)^{ \lambda + \beta (n-\alpha )}} \\ =& \frac{{\varGamma (\lambda + 1 + \beta (n -\alpha ))}}{{\varGamma (\lambda + 1 + n - \alpha )}}I_{a+}^{ \beta (n-\alpha )}{D^{n}} {(t-a)^{\lambda + n - \alpha }} \\ = & \frac{{\varGamma (\lambda + 1 + \beta (n - \alpha ))}}{{\varGamma ( \lambda - \alpha + 1)}}I_{a+}^{\beta (n - \alpha )}{(t-a)^{\lambda - \alpha }} \\ =& \frac{{\varGamma (\lambda + 1 + \beta (n - \alpha ))}}{ {\varGamma (\lambda - \alpha + \beta (n-\alpha )+1)}}{(t-a)^{\lambda - \alpha + \beta (n-\alpha )}}. \end{aligned}$$

In particular,

$$\begin{aligned} D_{a+}^{\alpha ,\beta }{(t-\alpha )^{\alpha - j + \beta (n - \alpha )}} =& \frac{{\varGamma (\alpha - j + 1 + \beta (n - \alpha ))}}{{\varGamma (n - j + 1)}}I_{a+ }^{\beta (n-\alpha )}{D^{n}} {(t-\alpha )^{n-j}} \\ =& 0,\quad \text{as } j = 1,2, \ldots ,n. \end{aligned}$$

Thus the proof of Lemma 2.8 is completed. □

3 Main result

Take the Banach space $(X,\| \cdot \|{_{\infty }})$,

$$ X = C [ {a,b} ]\quad \text{with the norm } \Vert x \Vert {_{\infty }} = \max_{t \in [a,b]} \bigl\vert x(t) \bigr\vert . $$

Lemma 3.1

Let $0 < {\alpha _{1}}$, ${\alpha _{2}} \leqslant 1$, $1 < {\alpha _{1}} + {\alpha _{2}} \leqslant 2$, $0 \leqslant {\beta _{1}}$, ${\beta _{2}} \leqslant 1$. Assume that $(A_{1})$ holds. Then, for $y\in X$, the function $x\in X$ is a solution of the following BVP:

$$\begin{aligned}& D_{a+}^{{\alpha _{1}},{\beta _{1}}}D_{a+}^{{\alpha _{2}},{\beta _{2}}}x(t) +y(t)=0, \quad t\in (a,b), \end{aligned}$$

(3.1)

$$\begin{aligned}& x(a)=0,\qquad x(b)=\sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}}),} \end{aligned}$$

(3.2)

if and only if x satisfies the integral equation

$$\begin{aligned} x(t) = \int _{a}^{b} {K(t,s)y(s)\,ds}+\frac{{{{(t - a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{1}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\sigma _{i}}K({\xi _{i}},s)y(s)\,ds} } ,\quad t \in [a,b], \end{aligned}$$

(3.3)

where

$$ K(t,s)=\frac{{{{(b-a)}^{(1-{\alpha _{1}})(1-{\beta _{1}})-{\alpha _{2}}}}}}{ {\varGamma ({\alpha _{1}}+{\alpha _{2}})}}\textstyle\begin{cases} {k_{1}}(t,s),\quad a \leqslant s \leqslant t \leqslant b, \\ {k_{2}}(t,s),\quad a \leqslant t \leqslant s \leqslant b, \end{cases} $$

and

$$\begin{aligned}& {k_{1}}(t,s) = {(t-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(b-s)^{{\alpha _{1}}+{\alpha _{2}}-1}}-{(b-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}} {(t- s)^{{\alpha _{1}}+{\alpha _{2}}-1}}, \\& {k_{2}}(t,s) = {(t-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(b-s)^{ {\alpha _{1}}+{\alpha _{2}}-1}}. \end{aligned}$$

Proof

By using Lemma 2.3 twice and combining Lemma 2.2, we get that x is a solution of (3.1) if and only if

$$ x(t)=-I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t)+{c_{1}} \frac{{{{(t-a)} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{{\varGamma ({\alpha _{2}}+1-(1-{\alpha _{1}})(1-{\beta _{1}}))}}+{c_{2}}\frac{{{{(t-a)}^{-(1- {\alpha _{2}})(1-{\beta _{2}})}}}}{{\varGamma (1-(1-{\alpha _{2}})(1-{\beta _{2}}))}}, $$

where ${c_{1}},{c_{2}} \in \mathbb{R}$. Considering the boundary conditions $x(a) = 0$, $x(b)=\sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}})}$, we get

$$ {c_{2}} = 0, \qquad {c_{1}}=\frac{{\varGamma [{\alpha _{2}}+1-(1-{\alpha _{1}})(1-{\beta _{1}})]}}{ {{{(b-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}} \Biggl[ {I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t){|_{t=b}}+\sum _{i=1} ^{m-2} {{\sigma _{i}}x({\xi _{i}})} } \Biggr]. $$

Thus,

$$\begin{aligned} x(t) = &-I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t)+\frac{{{{(t -a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} [ {I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t)| _{t = b} } ]}}{{{{(b-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}} \\ &{} +\frac{{{{(t-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{ {{{(b-a)}^{{\alpha _{2}} -(1-{\alpha _{1}})(1-{\beta _{1}})}}}}\sum_{i=1}^{m-2} {{ \sigma _{i}}x({\xi _{i}})} \\ =& \int _{a}^{b} {K(t,s)y(s)\,ds}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}}}}{{{{(b-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}} \sum_{i = 1}^{m-2} {{\sigma _{i}}x({\xi _{i}})}. \end{aligned}$$

(3.4)

Therefore,

$$ \sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}})}= \int _{a}^{b} {\sum_{i=1}^{m-2} {{\sigma _{i}}} K({\xi _{i}},s)y(s)\,ds}+ \frac{ { \sum_{i=1}^{m-2} {{\sigma _{i}}{{({\xi _{i}}-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}} }}{{{{(b-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}\sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}})}. $$

That is,

$$\begin{aligned} \sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}})}=\frac{{{{(b-a)} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{1}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\sigma _{i}}} K({\xi _{i}},s)y(s)\,ds}. \end{aligned}$$

(3.5)

By substituting (3.5) into (3.4), we obtain

$$ x(t)= \int _{a}^{b} {K(t,s)y(s)\,ds}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{1}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\sigma _{i}}} K({\xi _{i}},s)y(s)\,ds} . $$

The proof is completed. □

Lemma 3.2

Let $0 < {\alpha _{1}}, {\alpha _{2}} \leqslant 1$, $1 < {\alpha _{1}} + {\alpha _{2}} \leqslant 2$, $0 \leqslant {\beta _{1}}, {\beta _{2}} \leqslant 1$. Assume that $(A_{2})$ holds. Then, for $y \in X$, the function $x\in X$ is a solution of the following BVP:

$$\begin{aligned}& D_{a+}^{{\alpha _{1}},{\beta _{1}}}D_{a+}^{{\alpha _{2}},{\beta _{2}}}x(t) +y(t)=0,\quad t\in (a,b), \\ \end{aligned}$$

(3.6)

$$\begin{aligned}& x(a)=0,\qquad x'(b)=\sum _{i=1}^{m-2} {{\delta _{i}}} x({\eta _{i}}), \end{aligned}$$

(3.7)

if and only if x satisfies the integral equation

$$\begin{aligned} x(t)= \int _{a}^{b} {H(t,s)y(s)\,ds}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{2}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}} H({\eta _{i}},s)y(s)\,ds},\quad t \in [a,b], \end{aligned}$$

(3.8)

where

$$ H(t,s) = \frac{{{{(b - s)}^{{\alpha _{1}} + {\alpha _{2}} - 2}}}}{ {\varGamma ({\alpha _{1}} + {\alpha _{2}})[{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})]}}{H_{1}}(t,s), $$

and

$$\begin{aligned}& {H_{1}}(t,s) = \textstyle\begin{cases} {h_{1}}(t,s),\quad a \leqslant s \leqslant t \leqslant b, \\ {h_{2}}(t,s),\quad a \leqslant t \leqslant s \leqslant b, \end{cases}\displaystyle \\& {h_{1}}(t,s) = ({\alpha _{1}} + {\alpha _{2}} - 1){(b - a)^{2 - ({\alpha _{1}} + {\alpha _{2}}) - {\beta _{1}}(1 - {\alpha _{1}})}} {(t - a)^{{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} \\& \hphantom{{h_{1}}(t,s) =}{}- \bigl[{\alpha _{2}} - (1 - {\alpha _{1}}) (1 - {\beta _{1}})\bigr]\frac{{{{(t - s)}^{{\alpha _{1}} + {\alpha _{2}} - 1}}}}{{{{(b - s)}^{{\alpha _{1}} + {\alpha _{2}} - 2}}}}, \\& {h_{2}}(t,s) = ({\alpha _{1}} + {\alpha _{2}} - 1){(b - a)^{2 - ( {\alpha _{1}} + {\alpha _{2}}) - {\beta _{1}}(1 - {\alpha _{1}})}} {(t - a)^{ {\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}}. \end{aligned}$$

Proof

By a similar method used in Lemma 3.1, we obtain

$$ x(t)=-I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t)+{c_{1}}\frac{{{{(t-a)} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{{\varGamma [{\alpha _{1}}+{\alpha _{2}}+{\beta _{1}}(1-{\alpha _{1}})]}}, $$

where ${c_{1}} \in \mathbb{R}$. Then, taking derivative to the both sides of the above equality, we have

$$\begin{aligned} x'(t) = &- \frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{\varGamma ({\alpha _{1}}+ {\alpha _{2}})}} \int _{a}^{t} {{{(t-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}}} y(s)\,ds \\ &{}+{c_{1}}\frac{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]{{(t-a)}^{{\alpha _{2}}-1-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{ {\varGamma [{\alpha _{1}}+{\alpha _{2}}+{\beta _{1}}(1 - {\alpha _{1}})]}}. \end{aligned}$$

Using the boundary condition $x'(b) = \sum_{i = 1}^{m - 2} {{\delta _{i}}x({\eta _{i}})}$, we get

$$\begin{aligned} {c_{1}} = & \frac{{\varGamma [{\alpha _{1}}+{\alpha _{2}}+{\beta _{1}}(1- {\alpha _{1}})]}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]{{(b-a)} ^{{\alpha _{2}}-1-(1-{\alpha _{1}})(1- {\beta _{1}})}}}} \\ &{} \times \Biggl[ {\frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{\varGamma ({\alpha _{1}}+ {\alpha _{2}})}} \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}}} y(s)\,ds+ \sum_{i = 1}^{m-2} {{\delta _{i}}x({\eta _{i}})} } \Biggr]. \end{aligned}$$

Hence,

$$\begin{aligned} x(t) = &-I_{a+}^{{\alpha _{1}}+{\alpha _{2}}}y(t)+ \frac{{{{(t-a)}^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}({\alpha _{1}}+{\alpha _{2}}-1) \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+ {\alpha _{2}}-2}}} y(s)\,ds}}{ {[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]{{(b-a)}^{{\alpha _{2}}-1-(1- {\alpha _{1}})(1-{\beta _{1}})}}\varGamma ({\alpha _{1}}+{\alpha _{2}})}} \\ &{}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{ {[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]{{(b-a)}^{{\alpha _{2}}-1-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}\sum_{i=1}^{m-2} {{ \delta _{i}}x( {\eta _{i}})} \\ =& \int _{a}^{b} {H(t,s)y(s)\,ds}+\frac{{{{(t-a)}^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{{[{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})]{{(b-a)}^{{\alpha _{2}}-1-(1- {\alpha _{1}})(1- {\beta _{1}})}}}} \\ &{}\times \sum_{i=1}^{m-2} {{\delta _{i}}x({\eta _{i}})}. \end{aligned}$$

(3.9)

Therefore,

$$ \sum_{i=1}^{m-2} {{\delta _{i}}x({\eta _{i}})}= \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}H({\eta _{i}},s)} y(s)\,ds}+ \frac{ { \sum_{i=1}^{m-2} {{\delta _{i}}{{({\eta _{i}}-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}} \sum_{i=1}^{m-2} {{\delta _{i}}x({\eta _{i}})} }}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})] {{(b -a)}^{{\alpha _{2}}-1-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}. $$

It follows that

$$\begin{aligned} \sum_{i=1}^{m-2} {{\delta _{i}}x({\eta _{i}})}=\frac{{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}{{{\Delta _{2}}{{(b-a)}^{1- {\alpha _{2}}+(1-{\alpha _{1}})(1-{\beta _{1}})}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}H({\eta _{i}},s)} y(s)\,ds}. \end{aligned}$$

(3.10)

If we plug (3.10) back into (3.9), we obtain

$$ x(t)= \int _{a}^{b} {H(t,s)y(s)\,ds}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{2}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}H({\eta _{i}},s)} y(s)\,ds} , $$

which completes the proof. □

Lemma 3.3

(See [14])

If $1<\omega <2$, then

$$ \frac{{2-\omega }}{{{{(\omega -1)}^{{{(\omega -1)} / {(\omega -2)}}}}}} \leqslant \frac{ {{{(\omega -1)}^{\omega -1}}}}{{{\omega ^{\omega }}}}. $$

Lemma 3.4

The functions $K(t,s)$ and ${H_{1}}(t,s)$ defined in (3.3) and (3.8) satisfy the following properties:

(i)

$K(t,s)$ and ${H_{1}}(t,s)$ are two continuous functions for any $(t,s) \in [a,b] \times [a,b]$;
(ii)

$|K(t,s)| \leqslant \frac{{{{[(b-a)({\alpha _{1}}+{\alpha _{2}}-1)]}^{{\alpha _{1}}+{\alpha _{2}}-1}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{{\varGamma ({\alpha _{1}}+{\alpha _{2}}){{[2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})]} ^{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}}}$ for all $(t,s) \in [a,b] \times [a,b]$;
(iii)

$|{H_{1}}(t,s)| \leqslant (b-a)\max \{ {\alpha _{1}}+ {\alpha _{2}}-1,{\beta _{1}}(1-{\alpha _{1}})\}$ for every $(t,s) \in [a,b] \times [a,b]$.

Proof

Obviously, (i) is true. To prove (ii), for $(t,s) \in [a,b] \times [a,b]$, it is straightforward to show that

$$ 0 \leqslant {k_{2}}(t,s) \leqslant {k_{2}}(s,s). $$

Differentiating ${k_{1}}(t,s)$ with respect to s, we have

$$\begin{aligned} \frac{{\partial {k_{1}}(t,s)}}{{\partial s}} = &-({\alpha _{1}}+{\alpha _{2}}-1){(t-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(b-s)^{ {\alpha _{1}}+{\alpha _{2}}-2}} \\ &{}+({\alpha _{1}}+{\alpha _{2}}-1) {(b-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(t-s)^{{\alpha _{1}}+{\alpha _{2}}-2}} \\ =&({\alpha _{1}}+{\alpha _{2}}-1){(b-a)^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(t-s)^{{\alpha _{1}}+ {\alpha _{2}}-2}} \\ &{}\times \biggl[ {1-{{ \biggl( {\frac{{t-s}}{ {b-s}}} \biggr)}^{2-({\alpha _{1}} + {\alpha _{2}})}} {{ \biggl( {\frac{ {t-a}}{{b-a}}} \biggr)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}} \biggr] \\ \geqslant & 0, \end{aligned}$$

which shows ${k_{1}}(t,s)$ is increasing with respect to $s \in [a,t]$. Thus,

$$ {k_{1}}(t,a) \leqslant {k_{1}}(t,s) \leqslant {k_{1}}(t,t). $$

Since

$$\begin{aligned} {k_{1}}(t,a) = &{(t-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(b-a)^{{\alpha _{1}}+{\alpha _{2}}-1}}-{(b-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}} {(t-a)^{{\alpha _{1}}+{\alpha _{2}}-1}} \\ =&{(t-a)^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {(b-a)^{{\alpha _{1}}+ {\alpha _{2}}-1}} \biggl[ {1-{{ \biggl( { \frac{{b-a}}{{t-a}}} \biggr)} ^{{\beta _{1}}(1-{\alpha _{1}})}}} \biggr] \leqslant 0. \end{aligned}$$

Thus,

$$ \bigl\vert {k_{1}}(t,s) \bigr\vert \leqslant {\mathrm{{max}}} \Bigl\{ {\max_{t \in [a,b]} {k_{1}}(t,t),\max _{t \in [a,b]} \bigl(-{k_{1}}(t,a)\bigr)} \Bigr\} . $$

We consider the functions

$$\begin{aligned}& f(t) = {k_{1}}(t,t) = {(t-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}} {(b-t)^{{\alpha _{1}}+{\alpha _{2}}-1}},\quad t \in [a,b], \\& \tilde{f}(t) = -{k_{1}}(t,a) \\& \hphantom{\tilde{f}(t)}={(b-a)^{{\alpha _{1}}+{\alpha _{2}}-1}} {(t-a)^{{\alpha _{1}}+{\alpha _{2}}-1}} \bigl[ {{{(b-a)}^{ {\beta _{1}}(1-{\alpha _{1}})}}-{{(t-a)}^{{\beta _{1}}(1- {\alpha _{1}})}}} \bigr],\quad t \in [a,b]. \end{aligned}$$

Differentiating $f(t)$ on $(a,b)$, we have

$$\begin{aligned} f'(t) = & \bigl[{\alpha _{2}} - (1 - { \alpha _{1}}) (1 - {\beta _{1}})\bigr]{(t - a)^{ {\alpha _{2}} - 1 - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{ {\alpha _{1}} + {\alpha _{2}} - 1}} \\ &{}- ({\alpha _{1}} + {\alpha _{2}} - 1){(t - a)^{{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{{\alpha _{1}} + {\alpha _{2}} - 2}} \\ =& {(t - a)^{{\alpha _{2}} - 1 - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{{\alpha _{1}} + {\alpha _{2}} - 2}} \\ &{} \times \bigl\{ {\bigl[{\alpha _{2}} - (1 - {\alpha _{1}}) (1 - {\beta _{1}})\bigr](b - t) - ({\alpha _{1}} + {\alpha _{2}} - 1) (t - a)} \bigr\} , \end{aligned}$$

and

$$\begin{aligned} f''(t) = & \bigl[{\alpha _{2}} - (1 - {\alpha _{1}}) (1 - {\beta _{1}})\bigr] \bigl[{ \alpha _{2}} - 1 - (1 - {\alpha _{1}}) (1 - {\beta _{1}})\bigr] \\ &{} \times {(t - a)^{{\alpha _{2}} - 2 - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{{\alpha _{1}} + {\alpha _{2}} - 1}} \\ &{} - 2({\alpha _{1}} + {\alpha _{2}} - 1)\bigl[{\alpha _{2}} - (1 - {\alpha _{1}}) (1 - {\beta _{1}}) \bigr] \\ &{} \times {(t - a)^{{\alpha _{2}} - 1 - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{{\alpha _{1}} + {\alpha _{2}} - 2}} \\ &{} + ({\alpha _{1}} + {\alpha _{2}} - 1) ({\alpha _{1}} + {\alpha _{2}} - 2){(t - a)^{{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}} {(b - t)^{{\alpha _{1}} + {\alpha _{2}} - 3}}. \end{aligned}$$

By calculating, we get $f'(t)=0$ has a unique zero in $(a,b)$ as follows:

$$\begin{aligned} t ={t^{*}} = & a + \frac{{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}{{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}(b-a) \\ =& b-\frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}(b-a) \in (a,b). \end{aligned}$$

(3.11)

Because

$$\begin{aligned}& {\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}}) \geqslant 0,\qquad {\alpha _{2}}-1-(1-{\alpha _{1}}) (1-{\beta _{1}}) \leqslant 0, \\& {\alpha _{1}}+{\alpha _{2}}-1>0, \qquad {\alpha _{1}}+{\alpha _{2}}-2 \leqslant 0, \end{aligned}$$

it is easy to verify that

$$ f''\bigl({t^{*}}\bigr) \leqslant 0. $$

Hence, we obtain

$$\begin{aligned} \max_{t \in [a,b]} f(t) =&f\bigl({t^{*}}\bigr) \\ =& \biggl[ {\frac{{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}{ {2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}(b-a)} \biggr]^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})} \\ &{}\times \biggl[ {\frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}(b-a)} \biggr]^{{\alpha _{1}}+{\alpha _{2}}-1} \\ =&\frac{{}}{{}}{ \bigl[ {{\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})} \bigr]^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {({\alpha _{1}}+ {\alpha _{2}}-1)^{{\alpha _{1}}+ {\alpha _{2}}-1}} \\ &{}\times { \biggl[ {\frac{{b-a}}{{2{\alpha _{2}}-(1-{\alpha _{1}})(2- {\beta _{1}})}}} \biggr]^{{\alpha _{1}}+2{\alpha _{2}}-1-(1-{\alpha _{1}})(1- {\beta _{1}})}}. \end{aligned}$$

We now prove that $\max_{t \in [a,b]} \tilde{f}(t) \leqslant \max_{t \in [a,b]} f(t)$. In fact, if ${\beta _{1}}(1-{\alpha _{1}})=0$, then $\tilde{f}(t) \equiv 0$, and the conclusion is obvious. If ${\beta _{1}}(1-{\alpha _{1}}) \ne 0$, differentiating $\tilde{f}(t)$ on $(a,b)$, we have

$$\begin{aligned} \tilde{f}'(t) = & ({\alpha _{1}}+{\alpha _{2}}-1){(b-a)^{{\alpha _{2}}-(1- {\beta _{1}})(1-{\alpha _{1}})}} {(t-a)^{{\alpha _{1}}+{\alpha _{2}}-2}} \\ &{}-\bigl[{\alpha _{2}}-(1-{\beta _{1}}) (1-{\alpha _{1}})\bigr]{(b-a)^{{\alpha _{1}}+{\alpha _{2}}-1}} {(t-a)^{{\alpha _{2}}-1-(1-{\beta _{1}})(1-{\alpha _{1}})}}, \end{aligned}$$

and

$$\begin{aligned} \tilde{f}''(t) = & ({\alpha _{1}}+{\alpha _{2}}-1) ({\alpha _{1}}+{\alpha _{2}}-2){(b-a)^{{\alpha _{2}}-(1-{\beta _{1}})(1-{\alpha _{1}})}} {(t-a)^{ {\alpha _{1}}+{\alpha _{2}}-3}} \\ &{} -\bigl[{\alpha _{2}}-(1-{\beta _{1}}) (1- {\alpha _{1}})\bigr] \bigl[{\alpha _{2}}-1-(1-{\beta _{1}}) (1-{\alpha _{1}})\bigr] \\ & {} \times {(b-a)^{{\alpha _{1}}+{\alpha _{2}}-1}} {(t-a)^{{\alpha _{2}}-2-(1- {\beta _{1}})(1-{\alpha _{1}})}}. \end{aligned}$$

(3.12)

By calculating, we get $\tilde{f}'(t) = 0$ has a unique zero in $(a,b)$ as follows:

$$ t={t_{*}}=a+{ \biggl[ {\frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}} \biggr]^{{1 / {{\beta _{1}}(1-{\alpha _{1}})}}}}(b-a) \in (a,b). $$

(3.13)

Submitting (3.13) into (3.12), we have

$$\begin{aligned} \tilde{f}''({t_{*}}) = & - \bigl[ {{\alpha _{2}}-(1-{\alpha _{1}}) (1- {\beta _{1}})} \bigr]{\beta _{1}}(1-{\alpha _{1}}){(b-a)^{-2(1-{\alpha _{2}})-(2-{\beta _{1}})(1-{\alpha _{1}})}} \\ &{}\times { \biggl[ {\frac{ {{\alpha _{1}}+{\alpha _{2}}-1}}{{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}} \biggr]^{{{[{\alpha _{2}}-2-(1-{\beta _{1}})(1-{\alpha _{1}})]} / {{\beta _{1}}(1-{\alpha _{1}})}}}} \\ \leqslant & 0. \end{aligned}$$

Thus,

$$\begin{aligned} \max_{t \in [a,b]} \tilde{f}(t) =&\max _{t \in [a,b]} \tilde{f}({t_{*}}) \\ =& \frac{{{\beta _{1}}(1-{\alpha _{1}}){{(b-a)}^{2({\alpha _{1}}+{\alpha _{2}}-1)+{\beta _{1}}(1-{\alpha _{1}})}}}}{{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}} \\ &{}\times { \biggl[ {\frac{{{\alpha _{1}}+{\alpha _{2}}-1}}{{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}} \biggr]^{{{({\alpha _{1}}+ {\alpha _{2}}-1)} / {{\beta _{1}}(1-{\alpha _{1}})}}}}. \end{aligned}$$

Take $\omega = \frac{{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}{{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}$, then $1< \omega <2$. It follows from Lemma 3.3 that

$$\begin{aligned} \max_{t \in [a,b]} \tilde{f}(t) = & {(b-a)^{2({\alpha _{1}}+{\alpha _{2}}-1)+{\beta _{1}}(1-{\alpha _{1}})}}(2- \omega ){( \omega -1)^{{{(\omega -1)} / {(2-\omega )}}}} \\ \leqslant & {(b-a)^{2( {\alpha _{1}}+{\alpha _{2}}-1)+{\beta _{1}}(1-{\alpha _{1}})}}\frac{{{{( \omega -1)}^{(\omega -1)}}}}{{{\omega ^{\omega }}}} \\ = & { \biggl\{ {\frac{{{{({\alpha _{1}}+{\alpha _{2}}-1)}^{({\alpha _{1}}+ {\alpha _{2}}-1)}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}}}}{{{{[2{\alpha _{2}}-(1- {\alpha _{1}})(2-{\beta _{1}})]}^{2{\alpha _{2}}-(1-{\alpha _{1}})(2- {\beta _{1}})}}}}} \biggr\} ^{\frac{1}{{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}}}} \\ &{} \times {(b-a)^{2({\alpha _{1}}+{\alpha _{2}}-1)+ {\beta _{1}}(1-{\alpha _{1}})}} \\ \leqslant & {(b-a)^{2({\alpha _{1}}+ {\alpha _{2}}-1)+{\beta _{1}}(1-{\alpha _{1}})}} {{({\alpha _{1}}+{\alpha _{2}}-1)}^{({\alpha _{1}}+{\alpha _{2}}-1)}} \\ &{} \times \frac{{{{[ {\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})]}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{{[2{\alpha _{2}}-(1-{\alpha _{1}})(2- {\beta _{1}})]}^{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}}} \\ =& \max_{t \in [a,b]} f(t). \end{aligned}$$

From the above we get

$$\begin{aligned} \bigl\vert {k_{1}}(t,s) \bigr\vert \leqslant & \max _{t \in [a,b]} {k_{1}}(t,t)=\max_{s \in [a,b]} {k_{2}}(s,s) \\ =& { \bigl[ {{\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})} \bigr]^{ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {({\alpha _{1}}+{\alpha _{2}}-1)^{{\alpha _{1}}+{\alpha _{2}} -1}} \\ &{} \times { \biggl[ {\frac{ {b-a}}{{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}} \biggr]^{2 {\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}. \end{aligned}$$

Therefore,

$$\begin{aligned} \bigl\vert K(t,s) \bigr\vert \leqslant & \frac{{{\max_{t \in [a,b]}}{k_{1}}(t,t)}}{ {\varGamma ({\alpha _{1}}+{\alpha _{2}}){{(b-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}} \\ =& \frac{{{{[(b-a)({\alpha _{1}}+{\alpha _{2}}-1)]}^{{\alpha _{1}}+{\alpha _{2}}-1}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{ {\varGamma ({\alpha _{1}}+{\alpha _{2}}){{[2{\alpha _{2}}-(1-{\alpha _{1}})(2- {\beta _{1}})]}^{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}}}. \end{aligned}$$

To prove (iii), for $(t,s) \in [a,b] \times [a,b]$, obviously, we have that the following inequalities hold:

$$ 0 \leqslant {h_{2}}(t,s) \leqslant {h_{2}}(s,s)={h_{1}}(s,s). $$

Differentiating ${h_{1}}(t,s)$ with respect to t, we have

$$\begin{aligned} \frac{{\partial {h_{1}}(t,s)}}{{\partial t}} = & ({\alpha _{1}}+{\alpha _{2}}-1)\bigl[{\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})\bigr] \\ &{} \times {(b-a)^{2-({\alpha _{1}}+{\alpha _{2}})-{\beta _{1}}(1-{\alpha _{1}})}} {(t-a)^{{\alpha _{2}}-1-(1-{\alpha _{1}})(1-{\beta _{1}})}} \\ &{}- \bigl[ {\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})\bigr]({\alpha _{1}}+{\alpha _{2}}-1) \frac{{{{(t-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}}}}{{{{(b-s)} ^{{\alpha _{1}}+{\alpha _{2}}-2}}}} \\ =& ({\alpha _{1}}+{\alpha _{2}}-1)\bigl[ {\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})\bigr] \\ &{}\times \biggl[ {-{{ \biggl( {\frac{{b-s}}{{t-s}}} \biggr)}^{2-({\alpha _{1}}+{\alpha _{2}})}}+{{ \biggl( {\frac{{b-a}}{{t-a}}} \biggr)}^{1-{\alpha _{2}}+(1- {\beta _{1}})(1- {\alpha _{1}})}}} \biggr] \\ \leqslant & 0. \end{aligned}$$

Hence, ${h_{1}}(t,s)$ is a decreasing function of $t \in [s,b]$, which implies that

$$\begin{aligned}& \begin{aligned} &{h_{1}}(b,s)\leqslant {h_{1}}(t,s) \leqslant {h_{1}}(s,s)= {h_{2}}(s,s), \\ &{h_{1}}(s,s)=({\alpha _{1}}+{\alpha _{2}}-1){(b-a)^{2-( {\alpha _{1}}+{\alpha _{2}})-{\beta _{1}}(1-{\alpha _{1}})}} {(s-a)^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}} \\ &\hphantom{{h_{1}}(s,s)}\leqslant ({\alpha _{1}}+{\alpha _{2}}-1) (b-a), \\ &{h_{1}}(b,s)= ({\alpha _{1}}+{\alpha _{2}}-1) (b-a)-\bigl[ {\alpha _{2}}-(1-{\alpha _{1}}) (1-{\beta _{1}})\bigr](b-s). \end{aligned} \end{aligned}$$

(3.14)

It is trivial to show that ${h_{1}}(t,s)$ is an increasing function with respect to $s \in [a,b]$. Thus,

$$ {h_{1}}(b,a) \leqslant {h_{1}}(b,s) \leqslant {h_{1}}(b,b). $$

Note that

$$\begin{aligned}& {h_{1}}(b,b)= ({\alpha _{1}}+{\alpha _{2}}-1) (b-a)> 0, \\& {h_{1}}(b,a) =-{\beta _{1}}(1-{\alpha _{1}}) (b-a)< 0. \end{aligned}$$

We get

$$ \bigl\vert {{h_{1}}(b,s)} \bigr\vert \leqslant (b-a)\max \bigl\{ {{\alpha _{1}}+{\alpha _{2}}-1,{\beta _{1}}(1-{\alpha _{1}})} \bigr\} . $$

(3.15)

Combining (3.14) and (3.15), we obtain

$$ \bigl\vert {{h_{1}}(t,s)} \bigr\vert \leqslant (b-a)\max \bigl\{ {{\alpha _{1}}+{\alpha _{2}}-1,{\beta _{1}}(1-{ \alpha _{1}})} \bigr\} . $$

Therefore,

$$ \bigl\vert {H_{1}}(t,s) \bigr\vert \leqslant (b-a)\max \bigl\{ { \alpha _{1}}+{\alpha _{2}}-1, {\beta _{1}}(1-{ \alpha _{1}})\bigr\} . $$

Then we complete the proof of Lemma 3.4. □

Remark 3.1

From the proof of Lemma 3.4, we have the following conclusions:

(i)

$K(t,s)$ has a unique maximum, given by
$$\begin{aligned}& \max_{(t,s) \in {{[a,b]}^{2}}} \bigl\vert K(t,s) \bigr\vert \\& \quad = K\bigl({t ^{*}},{t^{*}}\bigr) \\& \quad = \frac{{{{[(b-a)({\alpha _{1}}+{\alpha _{2}}-1)]} ^{{\alpha _{1}}+{\alpha _{2}}-1}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}}}{ {\varGamma ({\alpha _{1}}+{\alpha _{2}}){{[2{\alpha _{2}}-(1-{\alpha _{1}})(2- {\beta _{1}})]}^{2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})}}}}, \end{aligned}$$
where $t^{*}$ is defined by (3.11);
(ii)

$\max_{(t,s) \in {{[a,b]}^{2}}} | {H_{1}}(t,s)| = (b-a)\max \{ {\alpha _{1}}+{\alpha _{2}}-1,{\beta _{1}}(1- {\alpha _{1}})\}$ and
$$ \bigl\vert {H_{1}}(t,s) \bigr\vert = \textstyle\begin{cases} (b-a)({\alpha _{1}}+{\alpha _{2}}-1),\quad \text{if and only if } t=s=b, \\ (b-a){\beta _{1}}(1-{\alpha _{1}}),\quad \text{if and only if } t=b, s=a. \end{cases} $$

Theorem 3.1

Assume that $(\mathrm{{A}_{1})}$ holds. If the fractional BVP (1.14), (1.15) has a nontrivial continuous solution for a real-valued continuous function q, then

$$\begin{aligned}& \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds} \\& \quad > \frac{{\varGamma ({\alpha _{1}}+{\alpha _{2}}) {{[2{\alpha _{2}}-(1-{\alpha _{1}})(2-{\beta _{1}})]}^{2{\alpha _{2}}-(1- {\alpha _{1}})(2-{\beta _{1}})}}{\Delta _{1}}}}{{{{(b-a)}^{{\alpha _{1}} +{\alpha _{2}}-1}}{{({\alpha _{1}}+{\alpha _{2}}-1)}^{{\alpha _{1}}+{\alpha _{2}}-1}}{{[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]}^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} {{\tilde{\Delta }}_{1}}}}, \end{aligned}$$

(3.16)

where

$$ {\tilde{\Delta }_{1}}: = {\Delta _{1}}+\sum _{i=1}^{m-2} {{\sigma _{i}}} {(b-a)^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}}. $$

Proof

Assume $x(t)$ is a nontrivial solution of BVP (1.14), (1.15), then

$$ x(t)= \int _{a}^{b} {K(t,s)q(s)x(s)\,ds}+ \frac{{{{(t-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{1}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\sigma _{i}}} K({\xi _{i}},s)q(s)x(s)\,ds}, $$

and

$$\begin{aligned} \bigl\vert x(t) \bigr\vert \leqslant & \int _{a}^{b} { \bigl\vert K(t,s) \bigr\vert \cdot \bigl\vert q(s) \bigr\vert \cdot \bigl\vert x(s) \bigr\vert \,ds} \\ &{} + \frac{{{{(t - a)}^{{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}}}}{{{\Delta _{1}}}} \int _{a}^{b} {\sum_{i = 1}^{m - 2} {{\sigma _{i}}} \bigl\vert K({\xi _{i}},s) \bigr\vert \cdot \bigl\vert q(s) \bigr\vert \cdot \bigl\vert x(s) \bigr\vert \,ds}, \quad t \in [a,b]. \end{aligned}$$

Since x is a nontrivial solution, which will require $q(s)\not \equiv 0$ on $[a,b]$. Moreover, from $q(s) \in C[a,b]$, we obtain that there exists an interval $[{a_{1}},{b_{1}}] \subset [a,b]$ such that $|q(s)| > 0$ on $[{a_{1}},{b_{1}}]$. Then, by Lemma 3.4 and Remark 3.1, we have

$$\begin{aligned} \Vert x \Vert {_{\infty }} < & \frac{{{{[(b - a)({\alpha _{1}} + {\alpha _{2}} - 1)]}^{{\alpha _{1}} + {\alpha _{2}} - 1}}{{[{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})]}^{{\alpha _{2}} - (1 - {\alpha _{1}})(1 - {\beta _{1}})}}{{\tilde{\Delta }}_{1}}}}{{\varGamma ( {\alpha _{1}} + {\alpha _{2}}){{[2{\alpha _{2}} - (1 - {\alpha _{1}})(2 - {\beta _{1}})]}^{2{\alpha _{2}} - (1 - {\alpha _{1}})(2 - {\beta _{1}})}}{\Delta _{1}}}} \\ &{}\times \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds} \Vert x \Vert {_{\infty }}. \end{aligned}$$

Thus, inequality (3.16) holds. This completes the proof of Theorem 3.1. □

Theorem 3.2

Assume that $(\mathrm{{A}_{2})}$ holds. If the fractional BVP (1.14), (1.16) has a nontrivial continuous solution for a real-valued continuous function q, then

$$ \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}} \bigl\vert q(s) \bigr\vert \,ds} > \frac{ {\varGamma ({\alpha _{1}}+{\alpha _{2}})[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]{\Delta _{2}}}}{{\max \{ {\alpha _{1}}+{\alpha _{2}}-1, {\beta _{1}}(1-{\alpha _{1}})\} {{\tilde{\Delta }}_{2}}}}, $$

(3.17)

where

$$ {\tilde{\Delta }_{2}}:={\Delta _{2}}(b-a)+\sum _{i=1}^{m-2} {{\delta _{i}}} {(b-a)^{{\alpha _{1}}+{\alpha _{2}}+{\beta _{1}}(1-{\alpha _{1}})}}. $$

Proof

Assume that $x(t)$ is a nontrivial solution of BVP (1.14), (1.16), then

$$\begin{aligned} x(t) = & \int _{a}^{b}{H(t,s)q(s)x(s)\,ds}+ \frac{{{{(t-a)}^{{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})}}}}{{{\Delta _{2}}}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}H({\eta _{i}},s)} q(s)x(s)\,ds} \\ =& \frac{ { \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}}{H_{1}}(t,s)q(s)x(s)\,ds}}}{ {\varGamma ({\alpha _{1}}+{\alpha _{2}})[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]}} \\ &{}+\frac{{{{(t-a)}^{{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})}} \int _{a}^{b}{\sum_{i=1}^{m-2} {{\delta _{i}}{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}- 2}}{H_{1}}({\eta _{i}},s)q(s)x(s)\,ds}}}}{{{\Delta _{2}}\varGamma ({\alpha _{1}}+{\alpha _{2}})[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]}}, \end{aligned}$$

and

$$\begin{aligned} \bigl\vert x(t) \bigr\vert \leqslant & \frac{{ \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}} \vert {H_{1}}(t,s) \vert \cdot \vert q(s) \vert \cdot \vert x(s) \vert \,ds}}}{{\varGamma ({\alpha _{1}}+{\alpha _{2}})[ {\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})]}} \\ &{}+ \frac{{{{(t-a)} ^{{\alpha _{2}}-(1-{\alpha _{1}})(1-{\beta _{1}})}} \int _{a}^{b} {\sum_{i=1}^{m-2} {{\delta _{i}}{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}}} \vert {H_{1}}({\eta _{i}},s) \vert \cdot \vert q(s) \vert \cdot \vert x(s) \vert \,ds}}}{ {{\Delta _{2}}\varGamma ({\alpha _{1}}+{\alpha _{2}})[{\alpha _{2}}-(1-{\alpha _{1}})(1- {\beta _{1}})]}}. \end{aligned}$$

An argument similar to the one used in Theorem 3.1 shows that there exists an interval $[{a_{2}},{b_{2}}] \subset [a,b]$ such that $|q(s)| > 0$ on $[{a_{2}},{b_{2}}]$. Now, applying Lemma 3.4 and Remark 3.1, we have

$$ \Vert x \Vert {_{\infty }} < \frac{{\max \{{\alpha _{1}}+{\alpha _{2}}-1,{\beta _{1}}(1-{\alpha _{1}})\} {{\tilde{\Delta }}_{2}}}}{{\varGamma ({\alpha _{1}}+{\alpha _{2}})[{\alpha _{2}}-(1- {\alpha _{1}})(1-{\beta _{1}})] {\Delta _{2}}}} \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}} \bigl\vert q(s) \bigr\vert \,ds} \Vert x \Vert {_{\infty }}, $$

from which inequality in (3.17) follows. The proof is completed. □

Theorem 3.1 gives the following corollaries.

Corollary 3.1

The necessary condition for the existence of a nontrivial solution for BVP (1.1) is

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

(3.18)

Proof

Apply Theorem 3.1 for ${\alpha _{1}}= {\alpha _{2}}=1$, ${\sigma _{i}}=0$ ($i = 1,2, \ldots ,m - 2$), then (3.18) holds. Obviously, (3.18) coincides with the classical Lyapunov inequality, i.e., inequality (1.2). □

Corollary 3.2

The necessary condition for the existence of a nontrivial solution for BVP (1.5) of case (i) is

$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds} > \frac{{\varGamma ({\alpha }+{\beta }){2^{2( {\alpha }+{\beta }-1)}}}}{{{{(b-a)}^{{\alpha }+{\beta }-1}}}}. $$

(3.19)

Proof

Apply Theorem 3.1 for ${\alpha _{1}}= \alpha $, ${\alpha _{2}}=\beta $, ${\beta _{1}}={\beta _{2}}=0$, ${\sigma _{i}}=0$ ($i=1,2,\ldots ,m-2$), then (3.19) holds. Obviously, (3.19) coincides with inequality (1.6). □

Corollary 3.3

The necessary condition for the existence of a nontrivial solution for BVP (1.3) is

$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds}>\varGamma (\alpha ){ \biggl({\frac{4}{{b-a}}} \biggr) ^{\alpha -1}}. $$

(3.20)

Proof

Apply Theorem 3.1 for ${\alpha _{1}}=1$, ${\beta _{2}}=0$, $\alpha =1+{\alpha _{2}}$, ${\sigma _{i}}=0$ ($i=1,2, \ldots ,m-2$), then (3.20) holds. Obviously, (3.20) coincides with inequality (1.4). □

Corollary 3.4

The necessary condition for the existence of a nontrivial solution for BVP (1.5) of case (ii) is

$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma ({\alpha }+{\beta }){{({\alpha }+2 {\beta }-1)}^{{\alpha }+2{\beta }-1}}}}{{{{(b-a)}^{{\alpha }+{\beta }- 1}}{{({\alpha }+{\beta }-1)}^{{\alpha }+{\beta }-1}}\beta ^{{\beta }}}}. $$

(3.21)

Proof

Apply Theorem 3.1 for ${\alpha _{1}} =\alpha , {\alpha _{2}}=\beta , {\beta _{1}}={\beta _{2}}=1, {\sigma _{i}}=0$ ($i=1,2,\ldots ,m-2$), then (3.21) holds, which coincides with inequality (1.7). □

Corollary 3.5

Consider the following fractional BVP:

$$ \textstyle\begin{cases} {}^{C}D_{a+}^{\alpha }x(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=x(b)=0, \end{cases} $$

(3.22)

where $q\in C([a,b], \mathbb{R})$, ${}^{C}D_{a+}^{\alpha }$ is the left Caputo fractional derivative. If (3.22) has a nontrivial continuous solution in $[a,b]$, then

$$ \int _{a}^{b}{ \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha ){\alpha ^{\alpha }}}}{ {{{[(b-a)(\alpha -1)]}^{\alpha -1}}}}. $$

(3.23)

Proof

Apply Theorem 3.1 for ${\alpha _{2}}=1$, ${\beta _{1}}=1$, $\alpha ={\alpha _{1}}+1$, ${\sigma _{i}}=0$ ($i=1,2, \ldots ,m-2T$, then (3.23) holds. Corollary 3.5 coincides with [14, Theorem 1]. □

Corollary 3.6

The necessary condition for the existence of a nontrivial solution for BVP (1.8), (1.9) is

$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha ){{[\alpha -(2-\alpha )(1-\beta )]}^{\alpha -(2-\alpha )(1-\beta )}}}}{{{{(b-a)}^{\alpha -1}} {{(\alpha -1)}^{\alpha - 1}}{{[\alpha -1+\beta (2-\alpha )]}^{\alpha -1+\beta (2-\alpha )}}}}. $$

(3.24)

Proof

Apply Theorem 3.1 for ${\alpha _{2}} =1$, $\alpha ={\alpha _{1}}+1$, $\beta ={\beta _{1}}$, ${\sigma _{i}} =0$ ($i=1,2,\ldots ,m-2$), then (3.24) holds, which coincides with inequality (1.11). By Remark 3.1, we show that the non-strict inequality (1.11) can be replaced by strict inequality (3.24). □

Corollary 3.7

Assume that the following boundary value problem

$$ \textstyle\begin{cases} D_{a+}^{\alpha ,\beta }x(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=0, \qquad x(b)=\sum_{i=1}^{m-2} {{\sigma _{i}}x({\xi _{i}}),} \end{cases} $$

where $q\in C([a,b], \mathbb{R})$, $D_{a+}^{\alpha ,\beta }$ is the Hilfer fractional derivative of order α and type $\beta \in [0,1]$, has a nontrivial continuous solution in $[a,b]$, then

$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha ){{[2-(2-\alpha )(2- \beta )]}^{2-(2-\alpha )(2-\beta )}}{\Delta _{11}}}}{{{{(b-a)}^{\alpha -1}}{{(\alpha -1)}^{\alpha -1}}{{[1- (2-\alpha )(1-\beta )]}^{1-(2- \alpha )(1-\beta )}}{{\tilde{\Delta }}_{11}}}}, $$

(3.25)

where

$$\begin{aligned}& {\Delta _{11}}:= {(b-a)^{1-(2-\alpha )(1-\beta )}}-\sum _{i=1} ^{m-2} {{\sigma _{i}}} {({\xi _{i}}-a)^{1-(2-\alpha )(1-\beta )}}, \\& {\tilde{\Delta }_{11}}:= {\Delta _{11}}+\sum _{i=1}^{m-2} {{\sigma _{i}}} {(b-a)^{1-(2-\alpha )(1-\beta )}}. \end{aligned}$$

Proof

Apply Theorem 3.1 for ${\alpha _{2}} =1$, $\alpha ={\alpha _{1}}+1$, $\beta ={\beta _{1}}$, then (3.25) holds, which coincides with [30, Theorem 3.1]. □

Theorem 3.2 gives the following corollaries.

Corollary 3.8

If a nontrivial continuous solution of the following boundary value problem

$$ \textstyle\begin{cases} x''(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=x'(b)=0, \end{cases} $$

exists, where $q\in C([a,b], \mathbb{R})$, then

$$ \int _{a}^{b} { \bigl\vert q(s) \bigr\vert \,ds} > \frac{1}{{b-a}}. $$

(3.26)

Proof

Apply Theorem 3.2 for ${\alpha _{1}} ={\alpha _{2}}=1$, ${\delta _{i}}=0$ ($i=1,2,\ldots ,m-2$), then (3.26) holds, which coincides with [16, Corollary 5]. □

Corollary 3.9

Suppose that the following boundary value problem

$$ \textstyle\begin{cases} D_{a+}^{{\alpha _{1}}}D_{a+}^{{\alpha _{2}}}x(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=x'(b)=0, \end{cases} $$

where $q\in C([a,b], \mathbb{R})$, $D_{a+}^{(\cdot )}$ is the left Riemann–Liouville fractional derivative, has a nontrivial continuous solution in $[a,b]$, then

$$ \int _{a}^{b}{{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{ {\varGamma ({\alpha _{1}}+{\alpha _{2}})}}{{b-a}}. $$

(3.27)

Proof

Apply Theorem 3.2 for ${\beta _{1}}={\beta _{2}}=0$, ${\delta _{i}}=0$ ($i=1,2,\ldots ,m-2$), then (3.27) holds. □

Corollary 3.10

Consider the following fractional BVP:

$$ \textstyle\begin{cases} D_{a+}^{\alpha }x(t)+q(t)x(t) = 0,\quad t \in (a,b), \\ x(a) = x'(b) = 0, \end{cases} $$

(3.28)

where $q\in C([a,b], \mathbb{R})$, $D_{a+}^{\alpha }$ is the Riemann–Liouville fractional derivative of fractional order α. If (3.28) has a nontrivial continuous solution in $[a,b]$, then

$$ \int _{a}^{b} {{{(b-s)}^{\alpha -2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha )}}{{b-a}}. $$

(3.29)

Proof

Apply Theorem 3.2 for ${\alpha _{1}}=1$, ${\beta _{2}} =0$, $\alpha =1+{\alpha _{2}}$, ${\delta _{i}}=0$ ($i=1,2,\ldots ,m -2$), then (3.29) holds. □

Corollary 3.11

Consider the following fractional BVP:

$$ \textstyle\begin{cases} {}^{C}D_{a+}^{{\alpha _{1}}}{}^{C}D_{a+}^{{\alpha _{2}}}x(t)+q(t)x(t) =0,\quad t \in (a,b), \\ x(a)=x'(b)=0, \end{cases} $$

(3.30)

where $q\in C([a,b], \mathbb{R})$, ${}^{C}D_{a+}^{(\cdot )}$ is the left Caputo fractional derivative. If (3.30) has a nontrivial continuous solution in $[a,b]$, then

$$ \int _{a}^{b} {{{(b-s)}^{{\alpha _{1}}+{\alpha _{2}}-2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{ {\varGamma ({\alpha _{1}}+{\alpha _{2}}){\alpha _{2}}}}{{(b-a)\max \{ {\alpha _{1}}+ {\alpha _{2}}-1,1-{\alpha _{1}}\} }}. $$

(3.31)

Proof

Apply Theorem 3.2 for ${\beta _{1}}={\beta _{2}}=1, {\delta _{i}}=0$ ($i = 1,2,\ldots ,m-2$), then (3.31) holds. □

Corollary 3.12

Consider the following fractional BVP:

$$ \textstyle\begin{cases} {}^{C}D_{a+}^{\alpha }x(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=x'(b)=0, \end{cases} $$

(3.32)

where $q\in C([a,b], \mathbb{R})$, ${}^{C}D_{a+}^{\alpha }$ is the left Caputo fractional derivative of order α. If (3.32) has a nontrivial continuous solution in $[a,b]$, then

$$ \int _{a}^{b}{{{(b-s)}^{\alpha -2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha )}}{ {(b-a)\max \{\alpha -1,2-\alpha \}}}. $$

(3.33)

Proof

Apply Theorem 3.2 for ${\alpha _{2}} =1,{\beta _{1}}=1,\alpha ={\alpha _{1}}+1,{\delta _{i}}=0$ ($i =1,2,\ldots ,m-2$), then (3.33) holds, which coincides with [17] and [23, Theorem 3]. □

Corollary 3.13

The necessary condition for the existence of a nontrivial solution for BVP (1.8), (1.10) is

$$ \int _{a}^{b} {{{(b-s)}^{\alpha -2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha )[\alpha -1+\beta (2-\alpha )]}}{{(b-a)\max \{\alpha -1,\beta (2- \alpha )\}}}, $$

(3.34)

Proof

Apply Theorem 3.2 for ${\alpha _{2}} =1$, $\alpha ={\alpha _{1}}+1$, $\beta ={\beta _{1}}$, ${\delta _{i}} =0$ ($i=1,2,\ldots ,m-2$), then (3.34) holds, which coincides with inequality (1.12). By Remark 3.1, we show that the non-strict inequality (1.12) can be replaced by strict inequality (3.34). □

Corollary 3.14

If a nontrivial continuous solution of the following fractional boundary value problem

$$ \textstyle\begin{cases} D_{a+}^{\alpha ,\beta }x(t)+q(t)x(t)=0,\quad t \in (a,b), \\ x(a)=0,\qquad x'(b)=\sum_{i=1}^{m-2} {{\delta _{i}}x({\eta _{i}}),} \end{cases} $$

exists, $q\in C([a,b], \mathbb{R})$, $D_{a+}^{\alpha ,\beta }$ is the Hilfer fractional derivative of order α and type $\beta \in [0,1]$, then

$$ \int _{a}^{b} {{{(b-s)}^{\alpha -2}} \bigl\vert q(s) \bigr\vert \,ds}>\frac{{\varGamma (\alpha )[\alpha -1+\beta (2-\alpha )]{\Delta _{22}}}}{{\max \{\alpha -1, \beta (2-\alpha )\} {{\tilde{\Delta }}_{22}}}}, $$

(3.35)

where

$$\begin{aligned}& {\Delta _{22}}:= \bigl[\alpha -1+\beta (2-\alpha ) \bigr]{(b-a)^{-(2-\alpha )(1- \beta )}}-\sum_{i=1}^{m-2}{{ \delta _{i}}} {({\eta _{i}}-a)^{ \alpha -1+\beta (2-\alpha )}}, \\& {{\tilde{\Delta }}_{22}}:={\Delta _{2}}(b-a)+\sum _{i=1}^{m-2} {{\delta _{i}}} {(b-a)^{\alpha + \beta (2-\alpha )}}. \end{aligned}$$

Proof

Apply Theorem 3.2 for ${\alpha _{2}}=1$, $\alpha = {\alpha _{1}}+1$, $\beta = {\beta _{1}}$, then (3.35) holds. □

4 Conclusion

In this paper, the Lyapunov-type inequalities of sequential Hilfer fractional BVPs were investigated for the first time. Since the Hilfer fractional derivative is a generalization of both Riemann–Liouville and Caputo types fractional derivatives, this requires that our results can be reduced to the corresponding classical results, and we do it. So our work is meaningful and the results we obtained are more general. There is some work to be done in the future such as: finding Lyapunov-type inequalities for higher order Hilfer fractional BVPs; studying Lyapunov-type inequalities for Hilfer fractional p-Laplacian equation, and so on.

Declarations

Acknowledgements

The authors wish to express their sincere appreciation to the editor and two anonymous referees for valuable comments and suggestions.

Availability of data and materials

Not applicable.

Funding

This research is supported by the National Natural Science Foundation of China (No. 11271364).

Authors’ contributions

The authors have made equal contributions to each part of this paper. All the authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)

School of Mathematics, China University of Mining and Technology, Xuzhou, P.R. China

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