Energy consumption and surface roughness maps for low and moderate speed machining of Aluminum alloy 2014: An experimental study

Umer Shaukat; Scott Gohery; Tesfaye Molla; Umer Shaukat; Scott Gohery; Tesfaye Molla

doi:10.3934/matersci.2023032

AIMS Materials Science

2023, Volume 10, Issue 4: 575-588. doi: 10.3934/matersci.2023032

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Research article Topical Sections

Energy consumption and surface roughness maps for low and moderate speed machining of Aluminum alloy 2014: An experimental study

Umer Shaukat ^{1
,
,},
Scott Gohery ^{2
,
,},
Tesfaye Molla ^{2
,
,}

1.
Department of Mechanical Engineering, Pakistan Institute of Engineering and Applied Sciences, Nilore, Islamabad, Pakistan
2.
Department of Mechanical Engineering, the University of Melbourne, Parkville, VIC 3010, Australia

Received: 16 April 2023 Revised: 22 June 2023 Accepted: 04 July 2023 Published: 21 July 2023

The rising energy prices and soaring environmental concerns have put an immense pressure on the wide usage of machining processes. The total power consumption during machining includes the power consumed by the machine itself and the power used to remove the material from the workpiece. An accurate prediction of energy consumption during the machining process is the basis for energy reduction. In this study, the specific cutting energy and surface finish for low and moderate-speed orthogonal machining of the aluminum alloy 2014 are evaluated. The measured values for the specific cutting energy and surface roughness are presented as maps on a grid, which is based on the machining parameters including the following: (1) cutting speed and (2) undeformed chip thickness. The specific cutting energy map depicts low energy consumption values of 0.52 J/mm³ for the aluminum alloy 2014 at medium speed machining. The roughness maps depict high roughness values at high cutting speeds. Both maps help in optimizing the machining process to achieve a required surface roughness with minimal energy consumption. A review of a specific cutting energy map demonstrates that energy consumption decreases by increasing the cutting speeds. The decrease in energy consumption at moderate speeds corresponds to the low cutting forces. This potentially happens as a result of thermal softening of the material caused by adiabatic heating. This subsequently leads to an increase in the machinability of the aluminum alloy 2014 at moderate cutting speeds. Furthermore, the decreasing chip thickness and increasing shear angle as a result of increasing the cutting speed confirms the increased machinability of the workpiece at moderate speeds.

Keywords:

Citation: Umer Shaukat, Scott Gohery, Tesfaye Molla. Energy consumption and surface roughness maps for low and moderate speed machining of Aluminum alloy 2014: An experimental study[J]. AIMS Materials Science, 2023, 10(4): 575-588. doi: 10.3934/matersci.2023032

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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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