Driver identification and fatigue detection algorithm based on deep learning

Yuhua Ma; Ye Tao; Yuandan Gong; Wenhua Cui; Bo Wang; Yuhua Ma; Ye Tao; Yuandan Gong; Wenhua Cui; Bo Wang

doi:10.3934/mbe.2023355

Mathematical Biosciences and Engineering

2023, Volume 20, Issue 5: 8162-8189. doi: 10.3934/mbe.2023355

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Research article

Driver identification and fatigue detection algorithm based on deep learning

1.
School of Computer Science and Software Engineering, University of Science and Technology Liaoning, Anshan, China
2.
School of Electronic and Information Engineering, University of Science and Technology Liaoning, Anshan, China

Academic Editor: Yang Kuang

Received: 07 December 2022 Revised: 03 February 2023 Accepted: 12 February 2023 Published: 27 February 2023

In order to avoid traffic accidents caused by driver fatigue, smoking and talking on the phone, it is necessary to design an effective fatigue detection algorithm. Firstly, this paper studies the detection algorithms of driver fatigue at home and abroad, and analyzes the advantages and disadvantages of the existing algorithms. Secondly, a face recognition module is introduced to crop and align the acquired faces and input them into the Facenet network model for feature extraction, thus completing the identification of drivers. Thirdly, a new driver fatigue detection algorithm based on deep learning is designed based on Single Shot MultiBox Detector (SSD) algorithm, and the additional layer network structure of SSD is redesigned by using the idea of reverse residual. By adding the detection of drivers' smoking and making phone calls, adjusting the size and number of prior boxes of SSD algorithm, improving FPN network and SE network, the identification and verification of drivers can be realized. The experimental results showed that the number of parameters decreased from 96.62 MB to 18.24 MB. The average accuracy rate increased from 89.88% to 95.69%. The projected number of frames per second increased from 51.69 to 71.86. When the confidence threshold was set to 0.5, the recall rate of closed eyes increased from 46.69% to 65.87%, that of yawning increased from 59.72% to 82.72%, and that of smoking increased from 65.87% to 83.09%. These results show that the improved network model has better feature extraction ability for small targets.

Keywords:

Citation: Yuhua Ma, Ye Tao, Yuandan Gong, Wenhua Cui, Bo Wang. Driver identification and fatigue detection algorithm based on deep learning[J]. Mathematical Biosciences and Engineering, 2023, 20(5): 8162-8189. doi: 10.3934/mbe.2023355

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Abstract

1. Introduction

The well-known classical Lyapunov inequality ^[15] states that, if $u$ is a nontrivial solution of the Hill's equation

$\begin{equation} u^{\prime \prime }\left( t\right) +q\left( t\right) u\left( t\right) = 0, \text{ }a \lt t \lt b, \end{equation}$

(1.1)

subject to Dirichlet-type boundary conditions:

$\begin{equation} u\left( a\right) = u\left( b\right) = 0, \end{equation}$

(1.2)

then

$\begin{equation} \int_{a}^{b}\left\vert q\left( t\right) \right\vert dt \gt \frac{4}{b-a}, \end{equation}$

(1.3)

where $q:[a, b]\rightarrow \mathbb{R}$ is a real and continuous function.

Later, in 1951, Wintner ^[24], obtained the following inequality:

$\begin{equation} \int_{a}^{b}q^{+}\left( t\right) dt \gt \frac{4}{b-a}, \end{equation}$

(1.4)

where $q^{+}(t) = \max \left\{ q\left(t\right), 0\right\}.$

A more general inequality was given by Hartman and Wintner in ^[12], that is known as Hartman Wintner-type inequality:

$\begin{equation} \int_{a}^{b}\left( t-a\right) \left( b-t\right) q^{+}\left( t\right) dt \gt b-a, \end{equation}$

(1.5)

Since $\underset{t\in \lbrack a, b]}{\max }\left(t-a\right) \left(b-t\right) = \frac{\left(b-a\right) ^{2}}{4}$ , then, (1.5) implies (1.4).

The Lyapunov inequality and its generalizations have many applications in different fields such in oscillation theory, asymptotic theory, disconjugacy, eigenvalue problems.

Recently, many authors have extended the Lyapunov inequality (1.3) for fractional differential equations ^{[1,2,3,4,5,6,7,8,9,10,11,12,13,15,18,20,22,23,24]}. For this end, they substituted the ordinary second order derivative in (1.1) by a fractional derivative or a conformable derivative. The first result in which a fractional derivative is used instead of the ordinary derivative in equation (1.1), is the work of Ferreira ^[6]. He considered the following two-point Riemann-Liouville fractional boundary value problem

$\begin{equation*} D_{a^{+}}^{\alpha }u\left( t\right) +q(t)u\left( t\right) = 0, \text{ }a \lt t \lt b, \text{ }1 \lt \alpha \leq 2 \end{equation*}$

$\begin{equation*} u\left( a\right) = u\left( b\right) = 0. \end{equation*}$

And obtained the Lyapunov inequality:

$\begin{equation*} \int_{a}^{b}\left\vert q\left( t\right) \right\vert dt \gt \Gamma \left( \alpha \right) \left( \frac{4}{b-a}\right) ^{\alpha -1}. \end{equation*}$

Then, he studied in ^[7], the Caputo fractional differential equation

$\begin{equation*} ^{C}D_{a^{+}}^{\alpha }u\left( t\right) +q(t)u\left( t\right) = 0, \text{ } a \lt t \lt b, \text{ }1 \lt \alpha \leq 2 \end{equation*}$

under Dirichlet boundary conditions (1.2). In this case, the corresponding Lyapunov inequality has the form

$\begin{equation*} \int_{a}^{b}\left\vert q\left( t\right) \right\vert dt \gt \frac{\alpha ^{\alpha }\Gamma \left( \alpha \right) }{\left( \left( \alpha -1\right) \left( b-a\right) \right) ^{\alpha -1}}. \end{equation*}$

Later Agarwal and Özbekler in ^[1], complimented and improved the work of Ferreira ^[6]. More precisely, they proved that if $u$ is a nontrivial solution of the Riemann-Liouville fractional forced nonlinear differential equations of order $\alpha \in (0, 2]$ :

$\begin{equation*} D_{a^{+}}^{\alpha }u\left( t\right) +p\left( t\right) \left\vert u\left( t\right) \right\vert ^{\mu -1}u\left( t\right) +q\left( t\right) \left\vert u\left( t\right) \right\vert ^{\gamma -1}u\left( t\right) = f\left( t\right) , \text{ }a \lt t \lt b, \end{equation*}$

satisfying the Dirichlet boundary conditions (1.2), then the following Lyapunov type inequality

$\begin{equation*} \left( \int_{a}^{b}\left[ p^{+}\left( t\right) +q^{+}\left( t\right) \right] dt\right) \left( \int_{a}^{b}\left[ \mu _{0}p^{+}\left( t\right) +\gamma _{0}q^{+}\left( t\right) +\left\vert f\left( t\right) \right\vert \right] dt\right) \gt \frac{4^{2\alpha -3}\Gamma ^{2}\left( \alpha \right) }{\left( b-a\right) ^{2\alpha -2}}. \end{equation*}$

holds, where $p,$ $q,$ $f$ are real-valued functions, $0 < \gamma < 1 < \mu < 2$ , $\mu _{0} = \left(2-\mu \right) \mu ^{\mu /\left(2-\mu \right) }2^{2/\left(\mu -2\right) }$ and $\gamma _{0} = \left(2-\gamma \right) \gamma ^{\gamma /\left(2-\gamma \right) }2^{2/\left(\gamma -2\right) }$ .

In 2017, Guezane-Lakoud et al. ^[11], derived a new Lyapunov type inequality for a boundary value problem involving both left Riemann-Liouville and right Caputo fractional derivatives in presence of natural conditions

$\begin{equation*} -^{C}D_{b^{-}}^{\alpha }D_{a^{+}}^{\beta }u\left( t\right) +q(t)u\left( t\right) = 0, \text{ }a \lt t \lt b, \text{ }0 \lt \alpha , \beta \leq 1 \end{equation*}$

$\begin{equation*} u\left( a\right) = D_{a^{+}}^{\beta }u\left( b\right) = 0, \text{ } \end{equation*}$

then, they obtained the following Lyapunov inequality:

$\begin{equation*} \int_{a}^{b}\left\vert q\left( t\right) \right\vert dt \gt \frac{\left( \alpha +\beta -1\right) \Gamma \left( \alpha \right) \Gamma \left( \beta \right) }{ \left( b-a\right) ^{\alpha +\beta -1}}. \end{equation*}$

Recently, Ferreira in ^[9], derived a Lyapunov-type inequality for a sequential fractional right-focal boundary value problem

$\begin{equation*} ^{C}D_{a^{+}}^{\alpha }D_{a^{+}}^{\beta }u\left( t\right) +q(t)u\left( t\right) = 0, \text{ }a \lt t \lt b \end{equation*}$

$\begin{equation*} u\left( a\right) = D_{a^{+}}^{\gamma }u\left( b\right) = 0, \text{ } \end{equation*}$

where $0 < \alpha, \beta, \gamma \leq 1,$ $1 < \alpha +\beta \leq 2,$ then, they obtained the following Lyapunov inequality:

$\begin{equation*} \int_{a}^{b}\left( b-s\right) ^{\alpha +\beta -\gamma -1}\left\vert q\left( t\right) \right\vert dt \gt \frac{1}{C}, \end{equation*}$

where

$\begin{eqnarray*} C & = &\left( b-a\right) ^{\gamma }\max \left\{ \frac{\Gamma \left( \beta -\gamma +1\right) }{\Gamma \left( \alpha +\beta -\gamma \right) \Gamma \left( \beta +1\right) }, \right. \\ &&\left. \frac{1-\alpha }{\beta \Gamma \left( \alpha +\beta \right) }\left( \frac{\Gamma \left( \beta -\gamma +1\right) \Gamma \left( \alpha +\beta -1\right) }{\Gamma \left( \alpha +\beta -\gamma \right) \Gamma \left( \beta \right) }\right) ^{\frac{\alpha +\beta -1}{\alpha -1}}, \text{ with }\alpha \lt 1\right\} \end{eqnarray*}$

Note that more generalized Lyapunov type inequalities have been obtained for conformable derivative differential equations in ^[13]. For more results on Lyapunov-type inequalities for fractional differential equations, we refer to the recent survey of Ntouyas et al. ^[18].

In this work, we obtain Lyapunov type inequality for the following mixed fractional differential equation involving both right Caputo and left Riemann-Liouville fractional derivatives

$\begin{equation} -^{C}D_{b^{-}}^{\alpha }D_{a^{+}}^{\beta }u\left( t\right) +q(t)u\left( t\right) = 0, \text{ }a \lt t \lt b, \end{equation}$

(1.6)

satisfying the Dirichlet boundary conditions (1.2), here $0 < \beta \leq \alpha \leq 1,$ $1 < \alpha +\beta \leq 2,$ $^{C}D_{b^{-}}^{\alpha }$ denotes right Caputo derivative, $D_{a^{+}}^{\beta }$ denotes the left Riemann-Liouville and $q$ is a continuous function on $\left[a, b\right].$

So far, few authors have considered sequential fractional derivatives, and some Lyapunov type inequalities have been obtained. In this study, we place ourselves in a very general context, in that in each fractional operator, the order of the derivative can be different. Such problems, with both left and right fractional derivatives arise in the study of Euler-Lagrange equations for fractional problems of the calculus of variations ^[2,16,17]. However, the presence of a mixed left and right Caputo or Riemann-Liouville derivatives of order $0 < \alpha < 1$ leads to great difficulties in the study of the properties of the Green function since in this case it's given as a fractional integral operator.

We recall the concept of fractional integral and derivative of order $p > 0$ . For details, we refer the reader to ^[14,19,21]

The left and right Riemann-Liouville fractional integral of a function $g$ are defined respectively by

$\begin{eqnarray*} I_{a^{+}}^{p}g(t) & = &\frac{1}{\Gamma \left( p\right) }\int_{a}^{t}\frac{g(s) }{(t-s)^{1-p}}ds, \\ I_{b^{-}}^{p}g(t) & = &\frac{1}{\Gamma \left( p\right) }\int_{t}^{b}\frac{g(s) }{(s-t)^{1-p}}ds. \end{eqnarray*}$

The left and right Caputo derivatives of order $p > 0,$ of a function $g$ are respectively defined as follows:

$\begin{eqnarray*} ^{C}D_{a^{+}}^{p}g(t) & = &I_{a^{+}}^{n-p}g^{\left( n\right) }(t), \\ ^{C}D_{b^{-}}^{p}g(t) & = &\left( -1\right) ^{n}I_{b^{-}}^{n-p}g^{\left( n\right) }(t), \end{eqnarray*}$

and the left and right Riemann-Liouville fractional derivatives of order $p > 0,$ of a function $g$ \ are respectively defined as follows:

$\begin{eqnarray*} D_{a^{+}}^{p}g(t) & = &\frac{d^{n}}{dt^{n}}\left( I_{a^{+}}^{n-p}g\right) (t), \\ D_{b^{-}}^{p}g(t) & = &\left( -1\right) ^{n}\frac{d^{n}}{dt^{n}} I_{b^{-}}^{n-p}g(t), \end{eqnarray*}$

where $n$ is the smallest integer greater or equal than $p$ .

We also recall the following properties of fractional operators. Let $0 < p < 1$ , then:

1- $I_{a^{+}}^{pC}D_{a^{+}}^{p}f\left(t\right) = f\left(t\right) -f(a).$

2- $I_{b^{-}}^{pC}D_{b^{-}}^{p}f\left(t\right) = f\left(t\right) -f\left(b\right).$

3- $\left(I_{a^{+}}^{p}c\right) \left(t\right) = \frac{c\left(t-a\right) ^{p}}{\Gamma \left(p+1\right) }, c\in \mathbb{R}$

4- $D_{a^{+}}^{p}u(t) = ^{C}D_{a^{+}}^{p}u(t)$ , when $u\left(a\right) = 0.$

5- $D_{b^{-}}^{p}u(t) = ^{C}D_{b^{-}}^{p}u(t)$ , when $u\left(b\right) = 0.$

2. Lyapunov inequality

Next we transform the problem (1.6) with (1.2) to an equivalent integral equation.

Lemma 1. Assume that $0 < \alpha, \beta \leq 1.$ The function $u$ is a solution to the boundary value problem (1.6) with (1.2) if and only if $u$ satisfies the integral equation

$\begin{equation} u\left( t\right) = \int_{a}^{b}G(t, r)q(r)u(r)dr, \end{equation}$

(2.1)

where

$\begin{equation} G(t, r) = \frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left( \int_{a}^{\inf \left\{ r, t\right\} }\left( t-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right. \notag \end{equation}$

$\begin{equation} \left. -\frac{\left( t-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }} \int_{a}^{r}\left( b-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right) \end{equation}$

(2.2)

is the Green's function of problem (1.6) with (1.2).

Proof. Firstly, we apply the right side fractional integral $I_{b^{-}}^{\alpha }$ to equation (1.6), then the left side fractional integral $I_{a^{+}}^{\beta }$ to the resulting equation and taking into account the properties of Caputo and\Riemann-Liouville fractional derivatives and the fact that $D_{a^{+}}^{\beta }u(t) = ^{C}D_{a^{+}}^{\beta }u(t)$ , we get

$\begin{equation} u\left( t\right) = I_{a^{+}}^{\beta }I_{b^{-}}^{\alpha }q(t)u(t)+\frac{ c\left( t-a\right) ^{\beta }}{\Gamma \left( \beta +1\right) }. \end{equation}$

(2.3)

In view of the boundary condition $u\left(b\right) = 0,$ we get

$\begin{equation*} c = \frac{-\Gamma \left( \beta +1\right) }{\left( b-a\right) ^{\beta }} I_{a^{+}}^{\beta }I_{b^{-}}^{\alpha }q(t)u(t)\mid _{t = b}. \end{equation*}$

Substituting $c$ in (2.3), it yields

$\begin{eqnarray*} u\left( t\right) & = &I_{a^{+}}^{\beta }I_{b^{-}}^{\alpha }q(t)u(t)-\frac{ \left( t-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }}I_{a^{+}}^{\beta }I_{b^{-}}^{\alpha }q(t)u(t)\mid _{t = b} \\ & = &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \int_{a}^{t}\left( t-s\right) ^{\beta -1}\left( \int_{s}^{b}\left( r-s\right) ^{\alpha -1}q(r)u(r)dr\right) ds \\ &&-\frac{\left( t-a\right) ^{\beta }}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) \left( b-a\right) ^{\beta }}\int_{a}^{b}\left( b-s\right) ^{\beta -1}\left( \int_{s}^{b}\left( r-s\right) ^{\alpha -1}q(r)u(r)dr\right) ds. \end{eqnarray*}$

Finally, by exchanging the order of integration, we get

$\begin{eqnarray*} u\left( t\right) & = &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) }\int_{a}^{t}\left( \int_{a}^{r}\left( t-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right) q(r)u(r)dr \\ &&+\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \int_{t}^{b}\left( \int_{a}^{t}\left( t-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right) q(r)u(r)dr \\ &&-\frac{\left( t-a\right) ^{\beta }}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) \left( b-a\right) ^{\beta }}\int_{a}^{b}\left( \int_{a}^{r}\left( b-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right) q(r)u(r)dr, \end{eqnarray*}$

thus

$\begin{equation*} u\left( t\right) = \int_{a}^{b}G(t, r)q(r)u(r)dr, \end{equation*}$

with

$\begin{equation*} G(t, r) = \frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left\{ \begin{array}{c} \int_{a}^{r}\left( t-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds \\ -\frac{\left( t-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }} \int_{a}^{r}\left( b-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds, a\leq r\leq t\leq b, \\ \int_{a}^{t}\left( t-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds \\ -\frac{\left( t-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }} \int_{a}^{r}\left( b-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds, a\leq t\leq r\leq b. \end{array} \right. \end{equation*}$

that can be written as

$\begin{eqnarray*} G(t, r) & = &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left( \int_{a}^{\inf \left\{ r, t\right\} }\left( t-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right. \\ &&\left. -\frac{\left( t-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }} \int_{a}^{r}\left( b-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right) . \end{eqnarray*}$

Conversely, we can verify that if $u$ satisfies the integral equation (2.1), then $u$ is a solution to the boundary value problem (1.6) with (1.2). The proof is completed.

In the next Lemma we give the property of the Green function $G$ that will be needed in the sequel.

Lemma 2. Assume that $0 < \beta \leq \alpha \leq 1, 1 < \alpha +\beta \leq 2,$ then the Green function $G\left(t, r\right)$ given in (2.2) of problem (1.6) with (1.2) satisfies the following property:

$\begin{equation*} \left\vert G(t, r)\right\vert \leq \frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) \left( \alpha +\beta -1\right) \left( \alpha +\beta \right) }\left( \frac{\alpha \left( b-a\right) }{\left( \beta +\alpha \right) }\right) ^{\alpha +\beta -1}, \end{equation*}$

for all $a\leq r\leq t\leq b$ .

Proof. Firstly, for $a\leq r\leq t\leq b$ , we have $G(t, r)\geq 0$ . In fact, we have

$\begin{eqnarray*} G(t, r) & = &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left( \int_{a}^{r}\left( t-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right. \\ &&-\frac{\left( t-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }}\left. \int_{a}^{r}\left( b-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right) \\ &\geq &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left( \int_{a}^{r}\left( b-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right. \\ &&-\frac{\left( t-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }}\left. \int_{a}^{r}\left( b-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right) \end{eqnarray*}$

$\begin{equation} = \frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) }\left( 1- \frac{\left( t-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }}\right) \int_{a}^{r}\left( b-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\geq 0 \end{equation}$

(2.4)

in addition,

$\begin{eqnarray*} G(t, r) &\leq &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) }\left( \int_{a}^{r}\left( r-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right. \\ &&-\frac{\left( r-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }}\left. \int_{a}^{r}\left( b-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right) \\ &\leq &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left( \frac{(r-a)^{\alpha +\beta -1}}{\left( \alpha +\beta -1\right) } \right. \\ &&\left. -\frac{\left( r-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }} \int_{a}^{r}\left( b-a\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right) \end{eqnarray*}$

$\begin{equation} = \frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) }\left( \frac{(r-a)^{\alpha +\beta -1}}{\left( \alpha +\beta -1\right) }-\frac{ \left( r-a\right) ^{\beta +\alpha }}{\alpha \left( b-a\right) }\right) . \end{equation}$

(2.5)

Thus, from (2.4) and (2.5), we get

$\begin{equation} 0\leq G(t, r)\leq h\left( r\right) , \text{ }a\leq r\leq t\leq b, \end{equation}$

(2.6)

where

$\begin{equation*} h\left( s\right) : = \frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) }\left( \frac{(s-a)^{\alpha +\beta -1}}{\left( \alpha +\beta -1\right) }-\frac{\left( s-a\right) ^{\beta +\alpha }}{\alpha \left( b-a\right) }\right) , \end{equation*}$

it is clear that $h\left(s\right) \geq 0,$ for all $s\in \left[a, b\right]$ .

Now, for $a\leq t\leq r\leq b$ , we have

$\begin{eqnarray*} G(t, r) & = &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left( \int_{a}^{t}\left( t-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right. \\ &&\left. -\frac{\left( t-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }} \int_{a}^{r}\left( b-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right) \\ &\leq &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left( \int_{a}^{t}\left( t-s\right) ^{\beta -1}\left( t-s\right) ^{\alpha -1}ds\right. \\ &&\left. -\frac{\left( t-a\right) ^{\beta }}{\left( b-a\right) } \int_{a}^{r}\left( r-s\right) ^{\alpha -1}ds\right) \\ & = &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) }\left( \frac{(t-a)^{\alpha +\beta -1}}{\left( \alpha +\beta -1\right) }-\frac{ \left( t-a\right) ^{\beta }\left( r-a\right) ^{\alpha }}{\alpha \left( b-a\right) }\right) \end{eqnarray*}$

$\begin{equation} \leq \frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left( \frac{(t-a)^{\alpha +\beta -1}}{\left( \alpha +\beta -1\right) }- \frac{\left( t-a\right) ^{\beta +\alpha }}{\alpha \left( b-a\right) }\right) = h(t). \end{equation}$

(2.7)

On the other hand,

$\begin{eqnarray*} G(t, r) &\geq &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) }\left( r-a\right) ^{\alpha -1}\int_{a}^{t}\left( t-s\right) ^{\beta -1}ds \\ &&\left. -\frac{\left( t-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }} \int_{a}^{r}\left( r-s\right) ^{\beta -1}\left( r-s\right) ^{\alpha -1}ds\right) \\ &\geq &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left( \frac{\left( t-a\right) ^{\alpha }\left( t-a\right) ^{\beta }}{\beta \left( b-a\right) }-\frac{\left( t-a\right) ^{\beta }}{\left( b-a\right) ^{\beta }}\frac{(r-a)^{\alpha +\beta -1}}{\left( \alpha +\beta -1\right) } \right) \\ &\geq &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left( \frac{\left( t-a\right) ^{\alpha +\beta }}{\beta \left( b-a\right) }- \frac{\left( t-a\right) ^{\beta }(r-a)^{\alpha -1}}{\left( \alpha +\beta -1\right) }\right) \\ &\geq &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) } \left( \frac{\left( t-a\right) ^{\alpha +\beta }}{\beta \left( b-a\right) }- \frac{(t-a)^{\alpha +\beta -1}}{\left( \alpha +\beta -1\right) }\right) , \end{eqnarray*}$

since $\beta \leq \alpha$ , we get

$\begin{equation} G(t, r)\geq -h\left( t\right) , \text{ }a\leq t\leq r\leq b. \end{equation}$

(2.8)

From (2.7) and (2.8) we obtain

$\begin{equation} \left\vert G(t, r)\right\vert \leq h\left( t\right) , \text{ }a\leq t\leq r\leq b. \end{equation}$

(2.9)

Finally, by differentiating the function $h,$ it yields

$\begin{equation*} h^{\prime }(s) = \frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) }(s-a)^{\alpha +\beta -2}\left( 1-\frac{\left( \beta +\alpha \right) \left( s-a\right) }{\alpha \left( b-a\right) }\right) . \end{equation*}$

We can see that $h^{\prime }(s) = 0$ for $s_{0} = a+\frac{\alpha \left(b-a\right) }{\left(\beta +\alpha \right) }\in \left(a, b\right),$ $h^{\prime }(s) < 0$ for $s > s_{0}$ and $h^{\prime }(s) > 0$ for $s < s_{0}.$ Hence, the function $h(s)$ has a unique maximum given by

$\begin{eqnarray*} \underset{s\in \left[ a, b\right] }{\max }h\left( s\right) & = &h\left( s_{0}\right) \\ & = &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) }\left( \frac{\left( \frac{\alpha \left( b-a\right) }{\left( \beta +\alpha \right) } \right) ^{\alpha +\beta -1}}{\left( \alpha +\beta -1\right) }-\frac{\left( \frac{\alpha \left( b-a\right) }{\left( \beta +\alpha \right) }\right) ^{\beta +\alpha }}{\alpha \left( b-a\right) }\right) \\ & = &\frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) \left( \alpha +\beta -1\right) \left( \alpha +\beta \right) }\left( \frac{\alpha \left( b-a\right) }{\left( \beta +\alpha \right) }\right) ^{\alpha +\beta -1}. \end{eqnarray*}$

From (2.6) and (2.9), we get $\left\vert G(t, r)\right\vert \leq h(s_{0})$ , from which the intended result follows.

Next, we state and prove the Lyapunov type inequality for problem (1.6) with (1.2).

Theorem 3. Assume that $0 < \beta \leq \alpha \leq 1$ and $1 < \alpha +\beta \leq 2.$ If the fractional boundary value problem (1.6) with (1.2) has a nontrivial continuous solution, then

$\begin{equation} \int_{a}^{b}\left\vert q\left( r\right) \right\vert dr\geq \frac{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) \left( \alpha +\beta -1\right) \left( \alpha +\beta \right) ^{\alpha +\beta }}{\left( \alpha \left( b-a\right) \right) ^{\alpha +\beta -1}}. \end{equation}$

(2.10)

Proof. Let $X = C[a, b]$ be the Banach space endowed with norm $\left\vert \left\vert u\right\vert \right\vert = \underset{t\in \left[a, b\right] }{\max } \left\vert u\left(t\right) \right\vert$ . It follows from Lemma 1 that a solution $u\in X$ to the boundary value problem (1.6) with (1.2) satisfies

$\begin{eqnarray*} \left\vert u\left( t\right) \right\vert &\leq &\int_{a}^{b}\left\vert G(t, r)\right\vert \left\vert q\left( r\right) \right\vert \left\vert u\left( r\right) \right\vert dr \\ &\leq &\left\Vert u\right\Vert \int_{a}^{b}\left\vert G(t, r)\right\vert q\left( r\right) dr, \end{eqnarray*}$

Now, applying Lemma 2 to equation (2.1), it yields

$\begin{equation*} \left\vert u\left( t\right) \right\vert \leq \frac{1}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) \left( \alpha +\beta -1\right) \left( \alpha +\beta \right) }\left( \frac{\alpha \left( b-a\right) }{\left( \beta +\alpha \right) }\right) ^{\alpha +\beta -1}\left\Vert u\right\Vert \int_{a}^{b}\left\vert q\left( r\right) \right\vert dr \end{equation*}$

Hence,

$\begin{equation*} \left\Vert u\right\Vert \leq \frac{\left( \alpha \left( b-a\right) \right) ^{\alpha +\beta -1}}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) \left( \alpha +\beta -1\right) \left( \alpha +\beta \right) ^{\alpha +\beta }}\left\Vert u\right\Vert \int_{a}^{b}\left\vert q\left( r\right) \right\vert dr, \end{equation*}$

from which the inequality (2.10) follows. Note that the constant in (2.10) is not sharp. The proof is completed.

Remark 4. Note that, according to boundary conditions (1.2), the Caputo derivatives $^{C}D_{b^{-}}^{\alpha }$ and $\ ^{C}D_{a^{+}}^{\beta }$ coincide respectively with the Riemann-Liouville derivatives $D_{b^{-}}^{\alpha }$ and $D_{a^{+}}^{\beta }$ . So, equation (1.6) is reduced to the one containing only Caputo derivatives or only Riemann-Liouville derivatives, i.e.,

$\begin{equation*} -^{C}D_{b^{-}}^{\alpha C}D_{a^{+}}^{\beta }u\left( t\right) +q(t)u\left( t\right) = 0, \text{ }a \lt t \lt b \end{equation*}$

$\begin{equation*} -D_{b^{-}}^{\alpha }D_{a^{+}}^{\beta }u\left( t\right) +q(t)u\left( t\right) = 0, \text{ }a \lt t \lt b \end{equation*}$

Furthermore, by applying the reflection operator $(Qf)(t) = f(a+b-t)$ and taking into account that $Q^{C}D_{a+}^{\alpha } = ^{C}D_{b^{-}}^{\alpha }Q$ and $Q^{C}D_{b^{-}}^{\beta } = ^{C}D_{a^{+}}^{\beta }Q$ (see ^[21]), we can see that, the boundary value problem (1.6) with (1.2) is equivalent to the following problem

$\begin{equation*} -^{C}D_{a^{+}}^{\alpha }D_{b^{-}}^{\beta }u\left( t\right) +q(t)u\left( t\right) = 0, \text{ }a \lt t \lt b, \end{equation*}$

$\begin{equation*} u\left( a\right) = u\left( b\right) = 0. \end{equation*}$

Remark 5. If we take $\alpha = \beta = 1,$ then the Lyapunov type inequality (2.3) is reduced to

$\begin{equation*} \int_{a}^{b}\left\vert q\left( t\right) \right\vert dt\geq \frac{4}{b-a}. \end{equation*}$

Acknowledgements

The authors thank the anonymous referees for their valuable comments and suggestions that improved this paper.

Conflict of interest

All authors declare no conflicts of interest in this paper.

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