A hybrid analytical technique for solving nonlinear fractional order PDEs of power law kernel: Application to KdV and Fornberg-Witham equations

Shabir Ahmad; Aman Ullah; Ali Akgül; Fahd Jarad; Shabir Ahmad; Aman Ullah; Ali Akgül; Fahd Jarad

doi:10.3934/math.2022521

AIMS Mathematics

2022, Volume 7, Issue 5: 9389-9404. doi: 10.3934/math.2022521

Previous Article Next Article

Research article Special Issues

A hybrid analytical technique for solving nonlinear fractional order PDEs of power law kernel: Application to KdV and Fornberg-Witham equations

1.
Department of Mathematics, University of Malakand, Chakdara, Dir Lower, Khyber Pakhtunkhwa, Pakistan
2.
Art and Science Faculty, Department of Mathematics, Siirt University, TR-56100 Siirt, Turkey
3.
Department of Mathematics, Cankaya University, Etimesgut 06790, Ankara, Turkey
4.
King Abdulaziz University Jeddah, Saudi Arabia
5.
Department of Medical Research, China Medical University, Taichung 40402, Taiwan

Received: 08 December 2021 Revised: 21 February 2022 Accepted: 06 March 2022 Published: 11 March 2022
MSC : 35R11

It is important to deal with the exact solution of nonlinear PDEs of non-integer orders. Integral transforms play a vital role in solving differential equations of integer and fractional orders. To obtain analytical solutions to integer and fractional-order DEs, a few transforms, such as Laplace transforms, Sumudu transforms, and Elzaki transforms, have been widely used by researchers. We propose the Yang transform homotopy perturbation (YTHP) technique in this paper. We present the relation of Yang transform (YT) with the Laplace transform. We find a formula for the YT of fractional derivative in Caputo sense. We deduce a procedure for computing the solution of fractional-order nonlinear PDEs involving the power-law kernel. We show the convergence and error estimate of the suggested method. We give some examples to illustrate the novel method. We provide a comparison between the approximate solution and exact solution through tables and graphs.

Keywords:

Citation: Shabir Ahmad, Aman Ullah, Ali Akgül, Fahd Jarad. A hybrid analytical technique for solving nonlinear fractional order PDEs of power law kernel: Application to KdV and Fornberg-Witham equations[J]. AIMS Mathematics, 2022, 7(5): 9389-9404. doi: 10.3934/math.2022521

Related Papers:

[1]	Bandar Bin-Mohsin, Muhammad Uzair Awan, Muhammad Zakria Javed, Artion Kashuri, Muhammad Aslam Noor . Fractional integral estimations pertaining to generalized $ {\gamma} $-convex functions involving Raina's function and applications. AIMS Mathematics, 2022, 7(8): 13633-13663. doi: 10.3934/math.2022752
[2]	Thongchai Botmart, Soubhagya Kumar Sahoo, Bibhakar Kodamasingh, Muhammad Amer Latif, Fahd Jarad, Artion Kashuri . Certain midpoint-type Fejér and Hermite-Hadamard inclusions involving fractional integrals with an exponential function in kernel. AIMS Mathematics, 2023, 8(3): 5616-5638. doi: 10.3934/math.2023283
[3]	Shahid Mubeen, Rana Safdar Ali, Iqra Nayab, Gauhar Rahman, Kottakkaran Sooppy Nisar, Dumitru Baleanu . Some generalized fractional integral inequalities with nonsingular function as a kernel. AIMS Mathematics, 2021, 6(4): 3352-3377. doi: 10.3934/math.2021201
[4]	Yu-Pei Lv, Ghulam Farid, Hafsa Yasmeen, Waqas Nazeer, Chahn Yong Jung . Generalization of some fractional versions of Hadamard inequalities via exponentially $ (\alpha, h-m) $-convex functions. AIMS Mathematics, 2021, 6(8): 8978-8999. doi: 10.3934/math.2021521
[5]	Muhammad Imran Asjad, Waqas Ali Faridi, Mohammed M. Al-Shomrani, Abdullahi Yusuf . The generalization of Hermite-Hadamard type Inequality with exp-convexity involving non-singular fractional operator. AIMS Mathematics, 2022, 7(4): 7040-7055. doi: 10.3934/math.2022392
[6]	Shuhong Yu, Tingsong Du . Certain inequalities in frame of the left-sided fractional integral operators having exponential kernels. AIMS Mathematics, 2022, 7(3): 4094-4114. doi: 10.3934/math.2022226
[7]	Hu Ge-JiLe, Saima Rashid, Muhammad Aslam Noor, Arshiya Suhail, Yu-Ming Chu . Some unified bounds for exponentially $tgs$-convex functions governed by conformable fractional operators. AIMS Mathematics, 2020, 5(6): 6108-6123. doi: 10.3934/math.2020392
[8]	Miguel Vivas-Cortez, Muhammad Aamir Ali, Artion Kashuri, Hüseyin Budak . Generalizations of fractional Hermite-Hadamard-Mercer like inequalities for convex functions. AIMS Mathematics, 2021, 6(9): 9397-9421. doi: 10.3934/math.2021546
[9]	Muhammad Tariq, Asif Ali Shaikh, Sotiris K. Ntouyas, Jessada Tariboon . Some novel refinements of Hermite-Hadamard and Pachpatte type integral inequalities involving a generalized preinvex function pertaining to Caputo-Fabrizio fractional integral operator. AIMS Mathematics, 2023, 8(11): 25572-25610. doi: 10.3934/math.20231306
[10]	Artion Kashuri, Rozana Liko, Silvestru Sever Dragomir . New generalized integral inequalities with applications. AIMS Mathematics, 2019, 4(3): 984-996. doi: 10.3934/math.2019.3.984

Abstract

1. Introduction and preliminaries

Fractional calculus has emerged as one of the most important interdisciplinary subjects. In recent past it experienced rapid development and consequently several new generalizations of classical concepts of fractional calculus have been obtained in the literature, for example, see ^[9].

The classical Riemann-Liouville fractional integrals are defined as:

Definition 1.1 (^[9]). Let ${\mathcal{F}}\in L_{1}[a, b]$ . Then the Riemann-Liouville integrals $J_{a^+}^{\alpha}{\mathcal{F}}$ and $J_{b^-}^{\alpha}{\mathcal{F}}$ of order $\alpha > 0$ with $a\geq 0$ are defined by

$\begin{equation*} J_{a^+}^{\alpha}{\mathcal{F}}(x) = \frac{1}{\Gamma(\alpha)}\int\limits_a^x(x-v)^{\alpha-1}{\mathcal{F}}(v){\mathrm{d}}v,\quad x > a, \end{equation*}$

and

$\begin{equation*} J_{b^-}^{\alpha}{\mathcal{F}}(x) = \frac{1}{\Gamma(\alpha)}\int\limits_x^b(v-x)^{\alpha-1}{\mathcal{F}}(v){\mathrm{d}}v,\quad x < b, \end{equation*}$

where

$\begin{align*} \Gamma(x) = \int_0^\infty e^{-v}v^{x-1}{\mathrm{d}}v, \end{align*}$

is the well known Gamma function.

Diaz et al. ^[8] introduced the notion of generalized $k$ -gamma function. The integral form of $\Gamma_{k}$ is given by:

$\begin{equation*} \Gamma_{k}(x) = \int\limits_{0}^{\infty}v^{x-1}e^{-\frac{v^{k}}{k}}{\mathrm{d}}v,\quad \Re(x) > 0. \end{equation*}$

Note that

$\Gamma_{k}(x) = k^{\frac{x}{k}-1}\Gamma\left(\frac{x}{k}\right).$

$k$ -Beta function is defined as:

$\begin{equation*} \beta_{k}(x,y) = \frac{1}{k}\int\limits_{0}^{1}v^{\frac{x}{k}-1}(1-v)^{\frac{x}{k}-1}{\mathrm{d}}v. \end{equation*}$

Obviously

$\beta_{k}(x,y) = \frac{1}{k}\beta\left(\frac{x}{k},\frac{y}{k}\right).$

Sarikaya et al. ^[18] extended the notion of Riemann-Liouville fractional integrals to $k$ -Riemann-Liouville fractional integrals and discussed some of its interesting properties.

To be more precise let ${\mathcal{F}}$ be piecewise continuous on $I^{*} = (0, \infty)$ and integrable on any finite subinterval of $I = [0, \infty]$ . Then for $v > 0$ , we consider $k$ -Riemann-Liouville fractional integral of ${\mathcal{F}}$ of order $\alpha$

$\begin{align*} _kJ_{a}^{\alpha}{\mathcal{F}}(x) = \frac{1}{k\Gamma_{k}(\alpha)}\int\limits_{a}^{x}(x-v)^{\frac{\alpha}{k}-1}{\mathcal{F}}(v){\mathrm{d}}v,\quad x > a,k > 0. \end{align*}$

It has been observed that $k$ -fractional integrals are significant generalizations of classical fractional integrals. For more details, see ^[18].

Ahmad et al. ^[1] defined fractional integral operators with an exponential kernel and obtained corresponding inequalities.

Definition 1.2. Let $\mathcal{F}\in [a, b]$ . The fractional left side integral $\, _{k}\mathcal{I}^{\alpha}_{a^{+}}\mathcal{F}$ and right side integral $\, _{k}\mathcal{I}^{\alpha}_{b^{-}}\mathcal{F}$ of order $\alpha\in(0, 1)$ are defined as follows:

$\mathcal{I}^{\alpha}_{a^{+}}\mathcal{F}(x) = \frac{1}{\alpha}\int\limits_{a}^{x}e^{-\frac{1-\alpha}{\alpha}(x-v)}\mathcal{F}(v){\mathrm{d}}v,\ \ \ \ x > a,$

and

$\mathcal{I}^{\alpha}_{b^{-}}\mathcal{F}(x) = \frac{1}{\alpha}\int\limits_{x}^{b}e^{-\frac{1-\alpha}{\alpha}(v-x)}\mathcal{F}(v){\mathrm{d}}v,\ \ \ \ x < b.$

Using the ideas of ^[1,18], we now introduce the notion of $k$ -fractional integral operators with an exponential kernel.

Definition 1.3. Let $\mathcal{F}\in L[a, b]$ . The $k$ -fractional left side integral $\, _{k}\mathcal{I}^{\alpha}_{a^{+}}\mathcal{F}$ and right side integral $\, _{k}\mathcal{I}^{\alpha}_{b^{-}}\mathcal{F}$ of order $\alpha\in(0, 1)$ for $k > 0$ are defined as follows

$_{k}\mathcal{I}^{\alpha}_{a^{+}}\mathcal{F}(x) = \frac{k}{\alpha}\int\limits_{a}^{x}e^{-\frac{k-\alpha}{\alpha}(x-v)}\mathcal{F}(v){\mathrm{d}}v,\ \ \ \ x > a,$

and

$_{k}\mathcal{I}^{\alpha}_{b^{-}}\mathcal{F}(x) = \frac{k}{\alpha}\int\limits_{x}^{b}e^{-\frac{k-\alpha}{\alpha}(v-x)}\mathcal{F}(v){\mathrm{d}}v,\ \ \ \ x < b.$

It is to be noted that by taking $k \to 1$ in Definition 1.3, we recapture Definition 1.2. Fractional analogues of integral inequalities have a great many applications in numerical quadrature, transform theory, probability, statistical problems etc. Therefore, a significant and rapid development in this field has been noticed, for details, see ^{[2,3,20,24,25]}. Sarikaya et al. ^[19] utilized the concepts of fractional integrals and obtained new fractional refinements of trapezium like inequalities. This article motivated many researchers and as a result several new fractional extensions of classical inequalities have been obtained in the literature, for example, see ^{[1,4,6,7,11,14,15,16,17,18,19,22,23]}. Recently Ahmad et al. ^[1] used fractional integral operators with an exponential kernel and obtained corresponding inequalities. Wu et al. ^[23] derived some new identities and bounds pertaining to fractional integrals with the exponential kernel.

The main motivation of this paper is to derive some new fractional refinements of trapezium like inequalities essentially using the new fractional integral operators with an exponential kernel to $k$ -fractional integral operators with an exponential kernel and the preinvexity property of the functions. In order to establish the significance of our main results, we offer some applications of our main results to means and $q$ -digamma functions. We hope that the ideas and techniques of this paper will inspire interested readers working in the field of inequalities.

Before we proceed further, we now recall some previously known concepts from convex analysis. We first, start with the definition of invex sets.

Definition 1.4 (^[10]). A set $K$ is said to be invex with respect to bifunction $\theta(., .)$ , if

$x+v\theta(y,x)\in K,\,\forall x,y\in K,\,v\in[0,1].$

The preinvexity of the functions is defined as:

Definition 1.5 (^[21]). A function $\mathcal{F}:K\to\mathbb{R}$ is said to be preinvex with respect to bifunction $\theta(., .)$ , if

${\mathcal{F}}(x+v\theta(y,x))\leq(1-v){\mathcal{F}}(x)+vf(y),\,\forall x,y\in K, v\in[0,1].$

In order to obtain some of the main results of the paper, we need the famous condition C, which was introduced by Mohan and Neogy ^[13]. This condition played a vital role in the development of several results involving preinvex functions.

Condition C. Let $\theta:K\times K\to\mathbb{R}^{n}$ . We say that the bifunction $\theta(., .)$ satisfies the condition C, if for any $x, y\in\mathbb{R}^{n}$

1. $\theta(x, x+v\theta(y, x)) = -v\theta(y, x)$ ,

2. $\theta(y, x+v\theta(y, x)) = (1-v)\theta(y, x)$ ,

for all $v\in[0, 1]$ .

Note that for any $x, y\in\mathbb{R}^{n}$ and $v_{1}, v_{2}\in[0, 1]$ and from the condition C, we have, see ^[12]

$\theta(x+v_{2}\theta(y,x),x+v_{1}\theta(y,x)) = (v_{2}-v_{1})\theta(y,x).$

2. Trapezium type inequalities

In this section, we derive some new fractional trapezium type inequalities involving the functions having preinvexity property. For the sake of simplicity, we set $\rho = \frac{k-\alpha}{\alpha}\theta(b, a)$ and $\rho_1 = \frac{1-\alpha}{\alpha}\theta(b, a)$ .

Theorem 2.1. Let $\mathcal{F}:[a, a+\theta(b, a)]\subseteq\mathbb{R}\to\mathbb{R}$ be a positive function with $\theta(b, a) > 0$ and $\mathcal{F}\in L[a, a+\theta(b, a)]$ . Suppose $\mathcal{F}$ is a preinvex function and $\theta(., .)$ satisfies condition C, then

$\begin{align} \mathcal{F}\left(\frac{2a+\theta(b,a)}{2}\right)\leq\frac{k-\alpha}{2k(1-e^{-\rho})}[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{F}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{F}(a)]\leq\frac{\mathcal{F}(a)+\mathcal{F}(b)}{2}. \end{align}$

(2.1)

Proof. By preinvexity of $\mathcal{F}$ , we have for every $x, y\in [a, a+\theta(b, a)]$ with $\lambda = \frac{1}{2}$

$\begin{align*} 2\mathcal{F}\left(x+\frac{\theta(y,x)}{2}\right)\leq[\mathcal{F}(x)+\mathcal{F}(y)], \end{align*}$

with $x = a+v\theta(b, a), y = a+(1-v)\theta(b, a)$ and using the condition C, we have

$\begin{align} &2\mathcal{F}\left(a+v\theta(b,a)+\frac{\theta(a+(1-v)\theta(b,a),a+v\theta(b,a))}{2}\right)\\ & = 2\mathcal{F}\left(a+v\theta(b,a)+\frac{(1-2v)\theta(b,a)}{2}\right)\\ & = 2\mathcal{F}\left(\frac{2a+\theta(b,a)}{2}\right)\\ &\leq\mathcal{F}(a+v\theta(b,a))+\mathcal{F}(a+(1-v)\theta(b,a)). \end{align}$

(2.2)

Multiplying both sides of above inequality by $e^{-\rho v}$ and integrating with respect to $v$ over $[0, 1]$ , we have

$\begin{align*} \frac{2(1-e^{-\rho})}{\rho}\mathcal{F}\left(\frac{2a+\theta(b,a)}{2}\right)&\leq\int\limits^{1}_{0}e^{-\rho v}f(a+v\theta(b,a)){\mathrm{d}}v+\int\limits^{1}_{0}e^{-\rho v}\mathcal{F}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ & = \frac{1}{\theta(b,a)}\left[\int\limits^{a+\theta(b,a)}_{a}e^{-\rho(\frac{s-a}{\theta(b,a)})}\mathcal{F}(s){\mathrm{d}}s+\int\limits^{a+\theta(b,a)}_{a}e^{-\rho(\frac{a+\theta(b,a)-s}{\theta(b,a)})}\mathcal{F}(s){\mathrm{d}}s\right]\\ & = \frac{\alpha}{k\theta(b,a)}\left[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{F}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{F}(a)\right]. \end{align*}$

As a result, we get

(2.3)

For the proof of second inequality, we note that $\mathcal{F}$ is a preinvex function, so we have

$\begin{align} \mathcal{F}(a+v\theta(b,a))+\mathcal{F}(a+(1-v)\theta(b,a))\leq\mathcal{F}(a)+\mathcal{F}(b). \end{align}$

(2.4)

Multiplying both sides by $e^{-\rho v}$ and integrating with respect to $v$ over $[0, 1]$ , we have

$\begin{align} \frac{\alpha}{k\theta(b,a)}\left[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{F}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{F}(a)\right]\leq\frac{1-e^{-\rho}}{\rho}[\mathcal{F}(a)+\mathcal{F}(b)], \end{align}$

(2.5)

Combining (2.3) and (2.5) completes the proof.

Theorem 2.2. Let $\mathcal{F}:[a, a+\theta(b, a)]\subseteq\mathbb{R}\to\mathbb{R}$ be a positive and preinvex function with $\theta(b, a) > 0$ and $\mathcal{F}\in L[a, a+\theta(b, a)]$ . Let $W$ be a non-negative, integrable and symmetric with respect to $\frac{2a+\theta(b, a)}{2}$ , then using the condition C, we have

$\begin{align} &\mathcal{F}\left(\frac{2a+\theta(b,a)}{2}\right)[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{W}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{W}(a)]\\ &\leq\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{(FW)}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{(FW)}(a)\\ &\leq\frac{\mathcal{F}(a)+\mathcal{F}(b)}{2}[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{W}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{W}(a)]. \end{align}$

(2.6)

Proof. Since $\mathcal{F}$ is preinvex function on $L[a, a+\theta(b, a)]$ , so multiplying inequality (2.2) by

$\begin{align} e^{-\rho v}\mathcal{W}(a+v\theta(b,a)), \end{align}$

(2.7)

and then integrating with respect to $v$ over $[0, 1]$ , we get

$\begin{align*} &2\mathcal{F}\left(\frac{2a+\theta(b,a)}{2}\right)\int\limits_{0}^{1}e^{-\rho v}\mathcal{W}(a+v\theta(b,a)){\mathrm{d}}v\\ &\leq\int\limits^{1}_{0}e^{-\rho v}\mathcal{W}(a+v\theta(b,a))\mathcal{F}(a+v\theta(b,a)){\mathrm{d}}v +\int\limits^{1}_{0}e^{-\rho v}\mathcal{W}(a+v\theta(b,a))\mathcal{F}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ & = \int\limits^{1}_{0}e^{-\rho v}\mathcal{W}(a+v\theta(b,a))\mathcal{F}(a+v\theta(b,a)){\mathrm{d}}v+\int\limits^{1}_{0}e^{-\rho v}\mathcal{W}(a+(1-v)\theta(b,a))\mathcal{F}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ & = \frac{1}{\theta(b,a)}\left[\int\limits^{a+\theta(b,a)}_{a}e^{-\rho(\frac{s-a}{\theta(b,a)})}\mathcal{F}(s)\mathcal{W}(s){\mathrm{d}}s+\int\limits^{a+\theta(b,a)}_{a}e^{-\rho(\frac{a+\theta(b,a)-s}{\theta(b,a)})}\mathcal{F}(s)\mathcal{W}(s){\mathrm{d}}s\right]\\ & = \frac{\alpha}{k\theta{(b,a)}}[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{(FW)}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{(FW)}(a)]. \end{align*}$

Thus

$\begin{align*} &2\mathcal{F}\left(\frac{2a+\theta(b,a)}{2}\right)\int\limits_{0}^{1}e^{-\rho v}\mathcal{W}(a+v\theta(b,a)){\mathrm{d}}v\\ &\leq\frac{\alpha}{k\theta{(b,a)}}[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{FW}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{FW}(a)]. \end{align*}$

Since $\mathcal{W}$ is symmetric with respect to $\frac{2a+\theta(b, a)}{2},$ we have

$\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{W}(a+\theta(b,a)) = \,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{W}(a) = \frac{1}{2}[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{W}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{W}(a)].$

Thus we get the left side of inequality (2.6).

For the proof of right side of inequality (2.6), we multiply (2.7) and (2.4) and then integrating the resulting inequality with respect to $v$ over $[0, 1],$ we get the required result.

Theorem 2.3. Let $\mathcal{F, W}:[a, a+\theta(b, a)]\subseteq\mathbb{R}\to\mathbb{R}$ be nonnegative and preinvex function on $L[a, a+\theta(b, a)]$ with $\theta(b, a) > 0$ . If $\theta(., .)$ satisfies condition C, then the following inequalities for the $k$ -fractional integrals with exponential kernel holds:

$\begin{align} &\frac{\alpha}{2k\theta(b,a)}[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{(FW)}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{(FW)}(a)]\\ &\leq M(a,b)\frac{\rho^{2}-2\rho+4-e^{-\rho}(\rho^{2}+2\rho+4)}{2\rho^{3}}+N(a,b)\frac{\rho-2+e^{-\rho}(\rho+2)}{\rho^{3}}, \end{align}$

(2.8)

and

$\begin{align} &2\mathcal{F}\left(\frac{2a+\theta(b,a)}{2}\right)\mathcal{W}\left(\frac{2a+\theta(b,a)}{2}\right)\\ &\leq\frac{k-\alpha}{2k(1-e^{-\rho})}[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{FW}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{FW}(a)]\\ &\leq M(a,b)\frac{\rho-2+e^{-\rho}(\rho+2)}{\rho^{2}(1-e^{-\rho})}+N(a,b)\frac{\rho^{2}-2\rho+4-e^{-\rho}(\rho^{2}+2\rho+4)}{2\rho^{2}(1-e^{-\rho})}, \end{align}$

(2.9)

where

$M(a,b) = \mathcal{F}(a)\mathcal{W}(a)+\mathcal{F}(b)\mathcal{W}(b),$

and

$N(a,b) = \mathcal{F}(a)\mathcal{W}(b)+\mathcal{F}(b)\mathcal{W}(a).$

Proof. Since $\mathcal{F}, \mathcal{W}$ are preinvex functions on $[a, a+\theta(b, a)]$ , then we have

$\begin{align*} \mathcal{F}(a+v\theta(b,a))\mathcal{W}(a+v\theta(b,a))\leq (1-v)^{2}\mathcal{F}(a)\mathcal{W}(a)+v^{2}\mathcal{F}(b)\mathcal{W}(b)+v(1-v)N(a,b), \end{align*}$

and

$\begin{align*} \mathcal{F}(a+(1-v)\theta(b,a))\mathcal{W}(a+(1-v)\theta(b,a))\leq v^{2}\mathcal{F}(a)\mathcal{W}(a)+(1-v)^{2}\mathcal{F}(b)\mathcal{W}(b)+v(1-v)N(a,b). \end{align*}$

Adding above inequalities, we have

$\begin{align*} \mathcal{F}(a+v\theta(b,a))\mathcal{W}(a+v\theta(b,a))&+\mathcal{F}(a+(1-v)\theta(b,a))\mathcal{W}(a+(1-v)\theta(b,a))\\ &\leq(2v^{2}-2v+1)M(a,b)+2v(1-v)N(a,b). \end{align*}$

Multiplying both sides of above inequality by $e^{-\rho v}$ and integrating with respect to $v$ over $[0, 1]$ , we have

$\begin{align*} &\int\limits_{0}^{1}e^{-\rho v}\mathcal{F}(a+v\theta(b,a))\mathcal{W}(a+v\theta(b,a)){\mathrm{d}}v\\ &\quad+\int\limits_{0}^{1}e^{-\rho v}\mathcal{F}(a+(1-v)\theta(b,a))\mathcal{W}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ &\leq M(a,b)\int\limits_{0}^{1}e^{-\rho v}(2v^{2}-2v+1){\mathrm{d}}v+M(a,b)\int\limits_{0}^{1}e^{-\rho v}2v(1-v){\mathrm{d}}v\\ & = M(a,b)\frac{\rho^{2}-2\rho+4-e^{-\rho}(\rho^{2}+2\rho+4)}{2\rho^{3}}+N(a,b)\frac{\rho-2+e^{-\rho}(\rho+2)}{\rho^{3}}. \end{align*}$

So,

$\begin{align*} &\frac{\alpha}{2k\theta(b,a)}[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{(FW)}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{(FW)}(a)]\\ &\leq M(a,b)\frac{\rho^{2}-2\rho+4-e^{-\rho}(\rho^{2}+2\rho+4)}{2\rho^{3}}+N(a,b)\frac{\rho-2+e^{-\rho}(\rho+2)}{\rho^{3}}. \end{align*}$

For the proof of inequality (2.9), using the preinvexity of $\mathcal{F}, \mathcal{W}$ and condition C, we have

$\begin{align} &\mathcal{F}\left(\frac{2a+\theta(b,a)}{2}\right)\mathcal{W}\left(\frac{2a+\theta(b,a)}{2}\right)\\ & = \mathcal{F}\left(a+(1-v)\theta(b,a)+\frac{1}{2}\theta(a+v\theta(b,a),a+(1-v)\theta(b,a)\right)\\ &\times\mathcal{W}\left(a+(1-v)\theta(b,a)+\frac{1}{2}\theta(a+v\theta(b,a),a+(1-v)\theta(b,a)\right)\\ &\leq\left(\frac{\mathcal{F}(a+v\theta(b,a))+\mathcal{F}(a+(1-v)\theta(b,a))}{2}\right)\left(\frac{\mathcal{W}(a+v\theta(b,a))+\mathcal{W}(a+(1-v)\theta(b,a))}{2}\right)\\ &\leq\frac{\mathcal{F}(a+v\theta(b,a))\mathcal{W}(a+v\theta(b,a))}{4}+\frac{\mathcal{F}(a+(1-v)\theta(b,a))\mathcal{W}(a+(1-v)\theta(b,a))}{4}\\ &+\frac{v(1-v)}{2}M(a,b)+\frac{2v^{2}-2v+1}{4}N(a,b). \end{align}$

(2.10)

Multiplying both sides of inequality (2.10) by $e^{-\rho v}$ and integrating with respect to $v$ over $[0, 1]$ , we get

$\begin{align*} &\frac{1-e^{-\rho}}{\rho}\mathcal{F}\left(\frac{2a+\theta(b,a)}{2}\right)\mathcal{W}\left(\frac{2a+\theta(b,a)}{2}\right)\\ &\leq\int\limits_{0}^{1}e^{-\rho v}\frac{\mathcal{F}(a+v\theta(b,a))\mathcal{W}(a+v\theta(b,a))}{4}{\mathrm{d}}v\\ &+\int\limits_{0}^{1}e^{-\rho v}\frac{\mathcal{F}(a+(1-v)\theta(b,a))\mathcal{W}(a+(1-v)\theta(b,a))}{4}{\mathrm{d}}v\\ &+M(a,b)\int\limits_{0}^{1}e^{-\rho v}\frac{v(1-v)}{2}{\mathrm{d}}v+N(a,b)\int\limits_{0}^{1}e^{-\rho v}\frac{2v^{2}-2v+1}{4}{\mathrm{d}}v, \end{align*}$

which completes the proof.

3. Inequalities involving first and second order differentiable preinvex functions

Lemma 3.1. Assume that $\mathcal{F}:[a, a+\theta(b, a)]\subseteq\mathbb{R}\to\mathbb{R}$ is a differentiable function and $\mathcal{F^{\prime}}\in L[a, a+\theta(b, a)]$ . Then the following equality for the $k$ -fractional integrals with exponential kernel holds:

$\begin{align} R_{ab}& = \frac{\theta(b,a)}{2}\int\limits_{0}^{1}u\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ &-\frac{\theta(b,a)}{2(1-e^{-\rho})}\left[\int\limits_{0}^{1}e^{-\rho v}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v-\int\limits_{0}^{1}e^{-\rho(1-v)}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right], \end{align}$

(3.1)

where

$\begin{align*} R_{ab} = \frac{k-\alpha}{2k(1-e^{-\rho})}[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{F}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{F}(a)]-\mathcal{F}\left(\frac{2a+\theta(b,a)}{2}\right), \end{align*}$

and

$u = \left\{ \begin{array}{rcl} 1, \ & \mathit{{for}} & 0\leq v\leq\frac{1}{2}, \\\\ -1, & \mathit{{for}} & \frac{1}{2}\leq v\leq1. \end{array}\right.$

Proof. By simple calculations, we have

$\begin{align} &\int\limits_{0}^{1}e^{-\rho v}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ & = -\frac{1}{\theta(b,a)}\left[e^{-\rho v}\mathcal{F}(a+(1-v)\theta(b,a))\bigg|_{0}^{1}+\rho\int\limits_{0}^{1}e^{-\rho v}\mathcal{F}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ & = -\frac{1}{\theta(b,a)}\left[\mathcal{F}(a+\theta(b,a))-e^{-\rho}\mathcal{F}(a)+\rho\int\limits_{0}^{1}e^{-\rho v}\mathcal{F}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ & = \frac{1}{\theta(b,a)}\left[\mathcal{F}(a+\theta(b,a))-e^{-\rho}\mathcal{F}(a)-\frac{\rho}{\theta(b,a)}\int\limits_{a}^{a+\theta(b,a)}e^{-\rho \frac{a+\theta(b,a)-s}{\theta(b,a)}}\mathcal{F}(s){\mathrm{d}}s\right]\\ & = \frac{1}{\theta(b,a)}\left[\mathcal{F}(a+\theta(b,a))-e^{-\rho}\mathcal{F}(a)-\frac{k-\alpha}{\alpha}\int\limits_{a}^{a+\theta(b,a)}e^{-\frac{(k-\alpha)}{\alpha} (a+\theta(b,a)-s)}\mathcal{F}(s){\mathrm{d}}s\right]\\ & = \frac{1}{\theta(b,a)}\left[\mathcal{F}(a+\theta(b,a))-e^{-\rho}\mathcal{F}(a)-\frac{k-\alpha}{k}\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{F}(a+\theta(b,a))\right], \end{align}$

(3.2)

similarly

$\begin{align} &\int\limits_{0}^{1}e^{-\rho (1-v)}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ & = -\frac{1}{\theta(b,a)}\left[e^{-\rho (1-v)}\mathcal{F}(a+(1-v)\theta(b,a))\bigg|_{0}^{1}-\rho\int\limits_{0}^{1}e^{-\rho (1-v)}\mathcal{F}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ & = -\frac{1}{\theta(b,a)}\left[\mathcal{F}(a)-e^{-\rho}\mathcal{F}(a+\theta(b,a))-\rho\int\limits_{0}^{1}e^{-\rho v}\mathcal{F}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ & = \frac{1}{\theta(b,a)}\left[e^{-\rho}\mathcal{F}(a+\theta(b,a))-\mathcal{F}(a)+\frac{\rho}{\theta(b,a)}\int\limits_{a}^{a+\theta(b,a)}e^{-\rho \frac{s-a}{\theta(b,a)}}\mathcal{F}(s){\mathrm{d}}s\right]\\ & = \frac{1}{\theta(b,a)}\left[e^{-\rho}\mathcal{F}(a+\theta(b,a))-\mathcal{F}(a)+\frac{k-\alpha}{\alpha}\int\limits_{a}^{a+\theta(b,a)}e^{-\frac{(k-\alpha)}{\alpha} (s-a)}\mathcal{F}(s){\mathrm{d}}s\right]\\ & = \frac{1}{\theta(b,a)}\left[e^{-\rho}\mathcal{F}(a+\theta(b,a))-\mathcal{F}(a)+\frac{k-\alpha}{k}\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{F}(a)\right]. \end{align}$

(3.3)

Also note that

(3.4)

Substituting (3.2), (3.3) and (3.4) in (3.1) completes the proof.

Theorem 3.1. Assume that $\mathcal{F}:[a, a+\theta(b, a)]\subseteq\mathbb{R}\to\mathbb{R}$ is a differentiable function and $\mathcal{F^{\prime}}\in L[a, a+\theta(b, a)]$ and $|\mathcal{F^{\prime}}|$ is preinvex on $[a, a+\theta(b, a)]$ . Then the following inequality for the $k$ -fractional integrals with exponential kernel holds:

$\begin{align*} |R_{ab}|\leq\frac{\theta(b,a)}{2}\left(\frac{1}{2}-\frac{\tanh\left(\frac{\rho}{4}\right)}{\rho}\right)(|\mathcal{F}^{\prime}(a)|+|\mathcal{F}^{\prime}(b)|). \end{align*}$

Proof. Using Lemma 3.1, preinvexity of $|\mathcal{F^{\prime}}|$ and increasing property of exponential function, we have

$\begin{align*} |R_{ab}|& = \left|\frac{\theta(b,a)}{2}\int\limits_{0}^{1}u\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right.\\ &-\left.\frac{\theta(b,a)}{2(1-e^{-\rho})}\left[\int\limits_{0}^{1}e^{-\rho v}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v-\int\limits_{0}^{1}e^{-\rho(1-v)}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\right|\\ &\leq\frac{\theta(b,a)}{2(1-e^{-\rho})}\left[\int\limits_{0}^{\frac{1}{2}}\left(1-e^{-\rho}-e^{-\rho v}+e^{-\rho(1-v)}\right)|\mathcal{F^{\prime}}(a+(1-v)\theta(b,a))|{\mathrm{d}}v\right.\\ &\left.-\int\limits_{\frac{1}{2}}^{1}\left(1-e^{-\rho}-e^{-\rho (1-v)}+e^{-\rho v}\right)|\mathcal{F^{\prime}}(a+(1-v)\theta(b,a))|{\mathrm{d}}v\right]\\ &\leq\frac{\theta(b,a)}{2(1-e^{-\rho})}\left[\int\limits_{0}^{\frac{1}{2}}\left(1-e^{-\rho}-e^{-\rho v}+e^{-\rho(1-v)}\right)(v|\mathcal{F^{\prime}}(a)|+(1-v)|\mathcal{F^{\prime}}(b)|){\mathrm{d}}v\right.\\ &\left.-\int\limits_{\frac{1}{2}}^{1}\left(1-e^{-\rho}-e^{-\rho (1-v)}+e^{-\rho v}\right)(v|\mathcal{F^{\prime}}(a)|+(1-v)|\mathcal{F}^{\prime}(b)|){\mathrm{d}}v\right] \\ & = \frac{\theta(b,a)}{2(1-e^{-\rho})}\left[\int\limits_{0}^{\frac{1}{2}}\left(1-e^{-\rho}-e^{-\rho v}+e^{-\rho(1-v)}\right)(v|\mathcal{F^{\prime}}(a)|+(1-v)|\mathcal{F^{\prime}}(b)|){\mathrm{d}}v\right.\\ &\left.-\int\limits_{0}^{\frac{1}{2}}\left(1-e^{-\rho}-e^{-\rho v}+e^{-\rho (1-v)}\right)((1-v)|\mathcal{F^{\prime}}(a)|+v|\mathcal{F}^{\prime}(b)|){\mathrm{d}}v\right]\\ & = \frac{\theta(b,a)}{2(1-e^{-\rho})}\int\limits_{0}^{\frac{1}{2}}\left(1-e^{-\rho}-e^{-\rho v}+e^{-\rho(1-v)}\right)(|\mathcal{F^{\prime}}(a)|+|\mathcal{F^{\prime}}(b)|){\mathrm{d}}v\\ & = \frac{\theta(b,a)}{2(1-e^{-\rho})}\left[\frac{1-e^{-\rho}}{2}-\frac{1}{\rho}(1-e^{\frac{-\rho}{2}})^{2}\right](|\mathcal{F^{\prime}}(a)|+|\mathcal{F^{\prime}}(b)|)\\ & = \frac{\theta(b,a)}{2}\left(\frac{1}{2}-\frac{\tanh\left(\frac{\rho}{4}\right)}{\rho}\right)(|\mathcal{F}^{\prime}(a)|+|\mathcal{F}^{\prime}(b)|), \end{align*}$

which completes the proof.

Lemma 3.2. Assume that $\mathcal{F}:[a, a+\theta(b, a)]\subseteq\mathbb{R}\to\mathbb{R}$ is a differentiable function and $\mathcal{F^{\prime}}\in L[a, a+\theta(b, a)]$ . Then the following equality for the fractional integrals with exponential kernel holds:

$\begin{align} L_{ab} = \frac{\theta(b,a)}{2(1-e^{-\rho})}\left[\int\limits_{0}^{1}e^{-\rho v}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v-\int\limits_{0}^{1}e^{-\rho(1-v)}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right], \end{align}$

(3.5)

where

$L_{ab} = \frac{\mathcal{F}(a)+\mathcal{F}(a+\theta(b,a)}{2}-\frac{k-\alpha}{2k(1-e^{-\rho})}[\,_{k}\mathcal{I}^{\alpha}_{(a)^{+}}\mathcal{F}(a+\theta(b,a))+\,_{k}\mathcal{I}^{\alpha}_{(a+\theta(b,a))^{-}}\mathcal{F}(a)].$

Proof. Using 3.2 and 3.3, we get the required result.

Theorem 3.2. Assume that $\mathcal{F}:[a, a+\theta(b, a)]\subseteq\mathbb{R}\to\mathbb{R}$ is a differentiable function and $\mathcal{F^{\prime}}\in L[a, a+\theta(b, a)]$ . If $|\mathcal{F^{\prime}}|$ is preinvex function on $[a, a+\theta(b, a)]$ , then the following inequality for the $k$ -fractional integrals with exponential kernel holds:

$\begin{align*} |L_{ab}|\leq\frac{\theta(b,a)}{2\rho}\tanh\left(\frac{\rho}{4}\right)(|\mathcal{F}^{\prime}(a)|+|\mathcal{F}^{\prime}(b)|). \end{align*}$

Proof. Using Lemma 3.2, preinvexity of $|\mathcal{F^{\prime}}|$ and increasing property of exponential function, we have

$\begin{align*} |L_{ab}|&\leq\frac{\theta(b,a)}{2}\int\limits_{0}^{1}\frac{|e^{-\rho v}-e^{-\rho(1-v)}|}{1-e^{-\rho}}|\mathcal{F^{\prime}}(a+v\theta(b,a))|{\mathrm{d}}v\\ &\leq\frac{\theta(b,a)}{2}\left[\int\limits_{0}^{1}\frac{|e^{-\rho v}-e^{-\rho(1-v)}|}{1-e^{-\rho}}v|\mathcal{F^{\prime}}(a)|{\mathrm{d}}v+\int\limits_{0}^{1}\frac{|e^{-\rho v}-e^{-\rho(1-v)}|}{1-e^{-\rho}}(1-v)|\mathcal{F^{\prime}}(b)|{\mathrm{d}}v\right] \\& = \frac{\theta(b,a)}{2}|\mathcal{F^{\prime}}(a)|\left[\int\limits_{0}^{\frac{1}{2}}\frac{e^{-\rho v}-e^{-\rho(1-v)}}{1-e^{-\rho}}v{\mathrm{d}}v+\int\limits^{1}_{\frac{1}{2}}\frac{e^{-\rho (1-v)}-e^{-\rho v}}{1-e^{-\rho}}v{\mathrm{d}}v\right]\\ &+\frac{\theta(b,a)}{2}|\mathcal{F^{\prime}}(b)|\left[\int\limits_{0}^{\frac{1}{2}}\frac{e^{-\rho v}-e^{-\rho(1-v)}}{1-e^{-\rho}}(1-v){\mathrm{d}}v+\int\limits^{1}_{\frac{1}{2}}\frac{e^{-\rho (1-v)}-e^{-\rho v}}{1-e^{-\rho}}(1-v){\mathrm{d}}v\right]\\ & = \frac{\theta(b,a)}{2\rho}\tanh\left(\frac{\rho}{4}\right)(|\mathcal{F}^{\prime}(a)|+|\mathcal{F}^{\prime}(b)|), \end{align*}$

which completes the proof.

Lemma 3.3. Assume that $\mathcal{F}:[a, a+\theta(b, a)]\subseteq\mathbb{R}\to\mathbb{R}$ is twice differentiable function and $\mathcal{F^{\prime\prime}}\in L[a, a+\theta(b, a)]$ . Then the following equality for the $k$ -fractional integrals with exponential kernel holds:

$\begin{align} L_{ab} = \frac{\theta^{2}(b,a)}{2\rho(1-e^{-\rho})}\int\limits_{0}^{1}\left(1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}\right)\mathcal{F^{\prime\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v. \end{align}$

(3.6)

Proof. Using (3.5) and integration by parts, we have

$\begin{align} &\int\limits_{0}^{1}e^{-\rho v}\mathcal{F}^{\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ & = -\frac{1}{\rho}\left[e^{-\rho v}\mathcal{F}^{\prime}(a+(1-v)\theta(b,a))\bigg|_{0}^{1}+\theta(b,a)\int\limits_{0}^{1}e^{-\rho v}\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ & = -\frac{1}{\rho}\left[e^{-\rho}\mathcal{F}^{\prime}(a)-\mathcal{F}^{\prime}(a+\theta(b,a))+\theta(b,a)\int\limits_{0}^{1}e^{-\rho v}\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right], \end{align}$

(3.7)

similarly

$\begin{align} &\int\limits_{0}^{1}e^{-\rho (1-v)}\mathcal{F}^{\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ & = \frac{1}{\rho}\left[e^{-\rho (1-v)}\mathcal{F}^{\prime}(a+(1-v)\theta(b,a))\bigg|_{0}^{1}+\theta(b,a)\int\limits_{0}^{1}e^{-\rho (1-v)}\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ & = \frac{1}{\rho}\left[\mathcal{F}^{\prime}(a)-e^{-\rho}\mathcal{F}^{\prime}(a+\theta(b,a))+\theta(b,a)\int\limits_{0}^{1}e^{-\rho (1-v)}\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]. \end{align}$

(3.8)

Substituting (3.7) and (3.8) in (3.5), we have

$\begin{align*} L_{ab}& = \frac{\theta(b,a)}{2\rho(1-e^{-\rho})}\left[(1+e^{-\rho})(\mathcal{F}^{\prime}(a+\theta(b,a))-\mathcal{F}^{\prime}(a))\right.\\ &\left.-\theta(b,a)\int\limits_{0}^{1}(e^{-\rho v}+e^{-\rho (1-v)})\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ & = \frac{\theta^{2}(b,a)}{2\rho(1-e^{-\rho})}\int\limits_{0}^{1}\left(1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}\right)\mathcal{F^{\prime\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v, \end{align*}$

which completes the proof.

Theorem 3.3. Assume that $\mathcal{F}:[a, a+\theta(b, a)]\subseteq\mathbb{R}\to\mathbb{R}$ is twice differentiable function. If $\mathcal{F^{\prime\prime}}\in L[a, a+\theta(b, a)]$ and $|\mathcal{F^{\prime\prime}}|$ is preinvex on $[a, a+\theta(b, a)]$ , then the following inequality for the $k$ -fractional integrals with exponential kernel holds:

$\begin{align*} |L_{ab}|\leq\frac{\theta^{2}(b,a)}{2\rho(1-e^{-\rho})}\left(\frac{1+e^{-\rho}}{2}-\frac{1-e^{-\rho}}{\rho}\right)(|\mathcal{F}^{\prime\prime}(a)|+|\mathcal{F}^{\prime\prime}(b)|). \end{align*}$

Proof. It is to be noted that

$\begin{align} \int\limits_{0}^{1}\left(1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}\right)v{\mathrm{d}}v = \frac{1+e^{-\rho}}{2}-\frac{1-e^{-\rho}}{\rho}, \end{align}$

(3.9)

and

$\begin{align} \int\limits_{0}^{1}\left(1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}\right)(1-v){\mathrm{d}}v = \frac{1+e^{-\rho}}{2}-\frac{1-e^{-\rho}}{\rho}. \end{align}$

(3.10)

Using (3.6), (3.9), (3.10) and the preinvexity of $|\mathcal{F}^{\prime\prime}|,$ we have

$\begin{align*} L_{ab}& = \bigg|\frac{\theta^{2}(b,a)}{2\rho(1-e^{-\rho})}\int\limits_{0}^{1}\left(1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}\right)\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\bigg|\\ &\leq\frac{\theta^{2}(b,a)}{2\rho(1-e^{-\rho})}\int\limits_{0}^{1}\left(1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}\right)|\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a))|{\mathrm{d}}v\\ &\leq\frac{\theta^{2}(b,a)}{2\rho(1-e^{-\rho})}\int\limits_{0}^{1}\left(1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}\right)(v|\mathcal{F}^{\prime\prime}(a)|+(1-v)|\mathcal{F}^{\prime\prime}(b)|){\mathrm{d}}v\\ & = \frac{\theta^{2}(b,a)}{2\rho(1-e^{-\rho})}\left(\frac{1+e^{-\rho}}{2}-\frac{1-e^{-\rho}}{\rho}\right)(|\mathcal{F}^{\prime\prime}(a)|+|\mathcal{F}^{\prime\prime}(b)|), \end{align*}$

the proof is complete.

Lemma 3.4. Assume that $\mathcal{F}:[a, a+\theta(b, a)]\subseteq\mathbb{R}\to\mathbb{R}$ is twice differentiable function and $\mathcal{F^{\prime\prime}}\in L[a, a+\theta(b, a)]$ . Then the following equality for the $k$ -fractional integrals with exponential kernel holds:

$\begin{align} R_{ab} = \frac{\theta^{2}(b,a)}{2}\int\limits_{0}^{1}h(v)\mathcal{F^{\prime\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v, \end{align}$

(3.11)

where

$h(v) = \left\{ \begin{array}{rcl} v-\frac{1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}}{\rho(1-e^{-\rho})}, \ & \mathit{{for}} & 0\leq v\leq\frac{1}{2}, \\\\ (1-v)-\frac{1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}}{\rho(1-e^{-\rho})}, & \mathit{{for}} & \frac{1}{2}\leq v\leq1. \end{array}\right.$

Proof. Using (3.1), we have

$\begin{align*} R_{ab}& = \frac{\theta(b,a)}{2}\int\limits_{0}^{1}u\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ &-\frac{\theta(b,a)}{2(1-e^{-\rho})}\left[\int\limits_{0}^{1}e^{-\rho v}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v-\int\limits_{0}^{1}e^{-\rho(1-v)}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right], \end{align*}$

Thus

$\begin{align} &\frac{\theta(b,a)}{2}\int\limits_{0}^{1}u\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ & = \frac{\theta(b,a)}{2}\left[\int\limits_{0}^{\frac{1}{2}}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v-\int\limits_{\frac{1}{2}}^{1}\mathcal{F^{\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ & = \frac{\theta(b,a)}{2}\left[v\mathcal{F}^{\prime}(a+(1-v)\theta(b,a))\bigg|_{0}^{\frac{1}{2}}+\theta(b,a)\int\limits_{0}^{\frac{1}{2}}v\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ &-\frac{\theta(b,a)}{2}\left[v\mathcal{F}^{\prime}(a+(1-v)\theta(b,a))\bigg|_{\frac{1}{2}}^{1}+\theta(b,a)\int\limits_{\frac{1}{2}}^{1}v\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ & = \frac{\theta(b,a)}{2}\left[\frac{1}{2}\mathcal{F}^{\prime}\left(\frac{2a+\theta(b,a)}{2}\right)+\theta(b,a)\int\limits_{0}^{\frac{1}{2}}v\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ &-\frac{\theta(b,a)}{2}\left[\mathcal{F}^{\prime}(a)-\frac{1}{2}\mathcal{F}^{\prime}\left(\frac{2a+\theta(b,a)}{2}\right)+\theta(b,a)\int\limits_{\frac{1}{2}}^{1}v\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\right]\\ & = \frac{\theta(b,a)}{2}\left[\mathcal{F}^{\prime}\left(\frac{2a+\theta(b,a)}{2}\right)-\mathcal{F}^{\prime}(a)\right]\\ &+\frac{\theta^{2}(b,a)}{2}\int\limits_{0}^{\frac{1}{2}}v\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v-\frac{\theta^{2}(b,a)}{2}\int\limits_{\frac{1}{2}}^{1}v\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v \\ & = \frac{\theta^{2}(b,a)}{2}\int\limits_{\frac{1}{2}}^{1}\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ &+\frac{\theta^{2}(b,a)}{2}\int\limits_{0}^{\frac{1}{2}}v\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v-\frac{\theta^{2}(b,a)}{2}\int\limits_{\frac{1}{2}}^{1}v\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v\\ & = \frac{\theta^{2}(b,a)}{2}\int\limits_{0}^{\frac{1}{2}}v\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v+\frac{\theta^{2}(b,a)}{2}\int\limits_{\frac{1}{2}}^{1}(1-v)\mathcal{F}^{\prime\prime}(a+(1-v)\theta(b,a)){\mathrm{d}}v. \end{align}$

(3.12)

Substituting (3.7), (3.8) and (3.12) in (3.1), we get the required result.

Theorem 3.4. Assume that $\mathcal{F}:[a, a+\theta(b, a)]\subseteq\mathbb{R}\to\mathbb{R}$ is a twice differentiable function. If $\mathcal{F^{\prime\prime}}\in L[a, a+\theta(b, a)]$ and $|\mathcal{F^{\prime\prime}}|$ is preinvex on $[a, a+\theta(b, a)]$ , then the following inequality for the $k$ -fractional integrals with exponential kernel holds:

$\begin{align*} |R_{ab}|\leq\frac{\theta^{2}(b,a)}{2}\left(\frac{1}{8}+\frac{1+e^{-\rho}}{2\rho(1-e^{-\rho})}-\frac{1}{\rho^{2}}\right)(|\mathcal{F}^{\prime\prime}(a)|+|\mathcal{F}^{\prime\prime}(b)|). \end{align*}$

Proof. Using (3.11) and preinvexity of $|\mathcal{F^{\prime\prime}}|$ , we have

$\begin{align*} |R_{ab}|& = \bigg|\frac{\theta^{2}(b,a)}{2}\int\limits_{0}^{1}h(v)\mathcal{F^{\prime\prime}}(a+(1-v)\theta(b,a)){\mathrm{d}}v\bigg|\\ &\leq\frac{\theta^{2}(b,a)}{2}\int\limits_{0}^{1}h(v)|\mathcal{F^{\prime\prime}}(a+(1-v)\theta(b,a))|{\mathrm{d}}v\\ &\leq\frac{\theta^{2}(b,a)}{2}\int\limits_{0}^{\frac{1}{2}}\left(v-\frac{1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}}{\rho(1-e^{-\rho})}\right)(v|\mathcal{F}^{\prime\prime}(a)|+(1-v)|\mathcal{F}^{\prime\prime}(b)|){\mathrm{d}}v\\ &+\frac{\theta^{2}(b,a)}{2}\int\limits_{\frac{1}{2}}^{1}\left(1-v-\frac{1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}}{\rho(1-e^{-\rho})}\right)(v|\mathcal{F}^{\prime\prime}(a)|+(1-v)|\mathcal{F}^{\prime\prime}(b)|){\mathrm{d}}v\\ & = \frac{\theta^{2}(b,a)}{2}\left[\int\limits_{0}^{\frac{1}{2}}(v^{2}|\mathcal{F}^{\prime\prime}(a)|+v(1-v)|\mathcal{F}^{\prime\prime}(b)|){\mathrm{d}}v+\int\limits_{\frac{1}{2}}^{1}(v(1-v)|\mathcal{F}^{\prime\prime}(a)|+(1-v)^{2}|\mathcal{F}^{\prime\prime}(b)|){\mathrm{d}}v\right]\\ &+\frac{1}{\rho(1-e^{-\rho})}\int\limits_{0}^{1}\left(1+e^{-\rho}-e^{-\rho v}-e^{-\rho(1-v)}\right)(v|\mathcal{F}^{\prime\prime}(a)|+(1-v)|\mathcal{F}^{\prime\prime}(b)|){\mathrm{d}}v\\ & = \frac{\theta^{2}(b,a)}{2}\left(\frac{1}{8}+\frac{1+e^{-\rho}}{2\rho(1-e^{-\rho})}-\frac{1}{\rho^{2}}\right)(|\mathcal{F}^{\prime\prime}(a)|+|\mathcal{F}^{\prime\prime}(b)|). \end{align*}$

The proof is complete.

Remark 3.1. We would like to remark here that by taking $k\to1$ , new results can be obtained from our results.

Applications

We now discuss some applications of the results obtained in previous section. Before we proceed further let us recall the definition of arithmetic mean.

The arithmetic mean is defined as

$\mathcal{A}(a,b): = \frac{a+b}{2}, \quad a\neq b.$

Proposition 3.1. Suppose all the assumptions of Theorem 3.1 are satisfied, then

$\begin{align*} &\left|\alpha {\mathcal{A}}(a^2,b^2)+\frac{\alpha^2}{(1-\alpha)^2}[(a-b)(1-\alpha)+2\alpha]-{\mathcal{A}}^2(a,b)\right|\\ &\leq(b-a){\mathcal{A}}(a,b)\left(\frac{1}{2}-\frac{tanh\left(\frac{\rho_1}{4}\right)}{\rho_1}\right). \end{align*}$

Proof. The proof directly follows from Theorem 3.1 by setting $\theta(b, a) = b-a, k = 1$ and $\mathcal{F}(x) = x^2$ .

Proposition 3.2. Suppose all the assumptions of Theorem 3.2 are satisfied, then

$\begin{align*} &\left|(1-\alpha) {\mathcal{A}}(a^2,b^2)-\frac{\alpha^2}{(1-\alpha)^2}[(a-b)(1-\alpha)+2\alpha]\right|\\ &\leq(b-a){\mathcal{A}}(a,b)tanh\left(\frac{\rho_1}{4}\right). \end{align*}$

Proof. The proof directly follows from Theorem 3.2 by setting $\theta(b, a) = b-a, k = 1$ and $\mathcal{F}(x) = x^2$ .

Proposition 3.3. Suppose all the assumptions of Theorem 3.3 are satisfied, then

$\begin{align*} &\left|(1-\alpha) {\mathcal{A}}(a^2,b^2)-\frac{\alpha^2}{(1-\alpha)^2}[(a-b)(1-\alpha)+2\alpha]\right|\\ &\leq \frac{2(b-a)^2{\mathcal{A}}(a,b)}{\rho_1(1-e^{-\rho_1})}\left(\frac{1+e^{-\rho_1}}{2}-\frac{1-e^{-\rho_1}}{\rho_1}\right). \end{align*}$

Proof. The proof directly follows from Theorem 3.3 by setting $\theta(b, a) = b-a, k = 1$ and $\mathcal{F}(x) = x^2$ .

Proposition 3.4. Suppose all the assumptions of Theorem 3.4 are satisfied, then

$\begin{align*} &\left|\alpha {\mathcal{A}}(a^2,b^2)+\frac{\alpha^2}{(1-\alpha)^2}[(a-b)(1-\alpha)+2\alpha]-{\mathcal{A}}^2(a,b)\right|\\ &\leq{2(b-a)^2{\mathcal{A}}(a,b)}\left(\frac{1}{8}+\frac{1+e^{-\rho_1}}{2\rho_1(1-e^{-\rho_1})}-\frac{1}{\rho_1^{2}}\right). \end{align*}$

Proof. The proof directly follows from Theorem 3.4 by setting $\theta(b, a) = b-a, k = 1$ and $\mathcal{F}(x) = x^2$ . We now discuss applications to $q$ -digamma functions, which is defined as:

Suppose $0 < q < 1$ , the $q$ -digamma function $\chi_q(u)$ is given as

$\begin{align*} \chi_q(u)& = -\ln(1-q)+\ln(q)\sum\limits_{i = 0}^{\infty}\frac{q^{i+u}}{1-q^{i+u}}.\\ & = -\ln(1-q)+\ln(q)\sum\limits_{i = 0}^{\infty}\frac{q^{iu}}{1-q^{iu}}. \end{align*}$

For $q > 1, t > 0$ , then $q$ -digamma function $\chi_q$ can be given as

$\begin{align*} \chi_q(u)& = -\ln(q-1)+\ln(q)\left[u-\frac{1}{2}-\sum\limits_{i = 0}^{\infty}\frac{q^{-(i+u)}}{1-q^{-(i+u)}}\right].\\ & = -\ln(q-1)+\ln(q)\left[u-\frac{1}{2}-\sum\limits_{i = 0}^{\infty}\frac{q^{-iu}}{1-q^{-iu}}\right]. \end{align*}$

From the above definition, it is clear that $\chi'_q$ is completely monotone on $(0, \infty)$ for $q > 0$ . This implies that $\chi'_q$ is convex. For more details, see ^[5].

Proposition 3.5. Under the assumption of Theorem 2.1, the following inequality holds:

$\begin{align*} \chi_q\left(\frac{{a}+{b}}{2}\right)\leq\frac{1-\alpha}{2(1-e^{-\rho_1})}\left[\int_a^be^{-\frac{1-\alpha}{\alpha}(b-v)}\chi_q(v){\mathrm{d}}v +\int_a^be^{-\frac{1-\alpha}{\alpha}(v-a)}\chi_q(v){\mathrm{d}}v\right]\leq\frac{\chi_q(a)+\chi_q(b)}{2}. \end{align*}$

Proof. The proof is direct consequence of Theorem 2.1, by choosing $\theta(b, a) = b-a, k = 1$ and $\mathcal{F}(v)\rightarrow \chi_q(v)$ .

Proposition 3.6. Under the assumption of Theorem 3.1, the following inequality holds:

$\begin{align*} &\left|\frac{1-\alpha}{2(1-e^{-\rho_1})}\left[\int_a^be^{-\frac{1-\alpha}{\alpha}(b-v)}\chi_q(v){\mathrm{d}}v +\int_a^be^{-\frac{1-\alpha}{\alpha}(v-a)}\chi_q(v){\mathrm{d}}v\right]-\chi_q\left(\frac{a+b}{2}\right)\right|\\ &\leq\frac{b-a}{2}\left(\frac{1}{2}-\frac{\tanh\left(\frac{\rho_1}{4}\right)}{\rho_1}\right)(|\chi_q^{\prime}(a)|+|\chi_q^{\prime}(b)|). \end{align*}$

Proof. The proof is direct consequence of Theorem 3.1, by choosing $\theta(b, a) = b-a, k = 1$ and $\mathcal{F}(v)\rightarrow \chi_q(v)$ .

Proposition 3.7. Under the assumption of Theorem 3.1, the following inequality holds:

$\begin{align*} &\left|\frac{\chi_q(a)+\chi_q(b)}{2}-\frac{1-\alpha}{2(1-e^{-\rho_1})}\left[\int_a^be^{-\frac{1-\alpha}{\alpha}(b-v)}\chi_q(v){\mathrm{d}}v +\int_a^be^{-\frac{1-\alpha}{\alpha}(v-a)}\chi_q(v){\mathrm{d}}v\right]\right|\\ &\leq\frac{b-a}{2}\tanh\left(\frac{\rho_1}{4}\right)(|\chi_q^{\prime}(a)|+|\chi_q^{\prime}(b)|). \end{align*}$

Proof. The proof is direct consequence of Theorem 3.1, by choosing $\theta(b, a) = b-a, k = 1$ and $\mathcal{F}(v)\rightarrow \chi_q(v)$ .

4. Conclusions

In the article, we have extended the fractional integral operators with an exponential kernel to $k$ -fractional integral operators with an exponential kernel and derived several new trapezium type integral inequalities involving the new fractional integral operator essentially using the functions having preinvexity property. We have also discussed some interesting applications of our obtained results, which show the significance of our main results. It is also worth mentioning here that our obtained results are the generalizations of some previously known results and our ideas may lead to a lot of follow-up research.

Acknowledgment

The authors are thankful to the editor and anonymous reviewers for their valuable comments and suggestions. This research was funded by Dirección de Investigación from Pontificia Universidad Católica del Ecuador in the research project entitled, "Some integrals inequalities and generalized convexity" (Algunas desigualdades integrales para funciones con algún tipo de convexidad generalizada y aplicaciones).

References

[1]	D. Baleanu, K. Diethelm, E. Scalas, J. J. Trujillo, Fractional calculus: Models and numerical methods, World Scientific, 2012.
[2]	R. T. Alqahtani, S. Ahmad, A. Akgül, Dynamical analysis of Bio-Ethanol production model under generalized nonlocal operator in Caputo Sense, Mathematics, 9 (2021), 2370. https://doi.org/10.3390/math9192370 doi: 10.3390/math9192370
[3]	R. Ozarslan, E. Bas, D. Baleanu, B. Acay, Fractional physical problems including wind-influenced projectile motion with Mittag-Leffler kernel, AIMS Mathematics, 5 (2020), 467–481. https://doi.org/10.3934/math.2020031 doi: 10.3934/math.2020031
[4]	S. Ahmad, A. Ullah, A. Akgül, M. De la Sen, A study of fractional order Ambartsumian equation involving exponential decay kernel, AIMS Mathematics, 6 (2021), 9981–9997. https://doi.org/10.3934/math.2021580 doi: 10.3934/math.2021580
[5]	B. Acay, E. Bas, T. Abdeljawad, Fractional economic models based on market equilibrium in the frame of different type kernels, Chaos Soliton. Fract., 130 (2020), 109438. https://doi.org/10.1016/j.chaos.2019.109438 doi: 10.1016/j.chaos.2019.109438
[6]	S. Ahmad, A. Ullah, K. Shah, A. Akgül, Computational analysis of the third order dispersive fractional PDE under exponential-decay and Mittag-Leffler type kernels, Numer. Meth. Part. Differ. Equ., 2020. https://doi.org/10.1002/num.22627
[7]	S. Ahmad, A. Ullah, A. Akgül, D. Baleanu, Analysis of the fractional tumour-immune-vitamins model with Mittag–Leffler kernel, Res. Phys., 19 (2020), 103559. https://doi.org/10.1016/j.rinp.2020.103559 doi: 10.1016/j.rinp.2020.103559
[8]	Gulalai, S. Ahmad, F. A. Rihan, A. Ullah, Q. M. Al-Mdallal, A. Akgül, Nonlinear analysis of a nonlinear modified KdV equation under Atangana Baleanu Caputo derivative, AIMS Mathematics, 7 (2022), 7847–7865. https://doi.org/10.3934/math.2022439 doi: 10.3934/math.2022439
[9]	S. Saifullah, A. Ali, M. Irfan, K. Shah, Time-fractional Klein–Gordon equation with solitary/shock waves solutions, Math. Probl. Eng., 2021 (2021), 6858592. https://doi.org/10.1155/2021/6858592 doi: 10.1155/2021/6858592
[10]	F. Rahman, A. Ali, S. Saifullah, Analysis of time-fractional $\varPhi^{4}$ equation with singular and non-singular Kernels, Int. J. Appl. Comput. Math., 7 (2021), 192. https://doi.org/10.1007/s40819-021-01128-w doi: 10.1007/s40819-021-01128-w
[11]	S. Saifullah, A. Ali, Z. A. Khan, Analysis of nonlinear time-fractional Klein-Gordon equation with power law kernel, AIMS Mathematics, 7 (2022), 5275–5290. https://doi.org/10.3934/math.2022293 doi: 10.3934/math.2022293
[12]	G. K. Watugala, Sumudu transform-a new integral transform to solve differential equations and control engineering problems, Math. Eng. Ind., 6 (1998), 319–329.
[13]	T. M. Elzaki, S. M. Elzaki, Application of new integral transform Elzaki transform to partial differential equations, Glob. J. Pure Appl. Math., 7 (2011), 65–70.
[14]	K. S. Aboodh, Application of new integral transform "Aboodh Transform" to partial differential equations, Glob. J. Pure Appl. Math., 10 (2014), 249–254.
[15]	J. L. Schiff, Laplace transform: Theory and applications, New York: Springer, 1999. https://doi.org/10.1007/978-0-387-22757-3
[16]	J. H. He, Homotopy perturbation technique, Comput. Meth. Appl. Mech. Eng., 178 (1999), 257–262.
[17]	J. H. He, Application of homotopy perturbation method to nonlinear wave equations, Chaos Soliton. Fract., 26 (2005), 695–700. https://doi.org/10.1016/j.chaos.2005.03.006 doi: 10.1016/j.chaos.2005.03.006
[18]	S. Das, P. K. Gupta, An approximate analytical solution of the fractional diusion equation with absorbent term and external force by homotopy perturbation method, Zeitschrift für Naturforschung A, 65 (2014), 182–190. https://doi.org/10.1515/zna-2010-0305 doi: 10.1515/zna-2010-0305
[19]	S. Ahmad, A. Ullah, A. Akgül, M. De la Sen, A novel homotopy perturbation method with applications to nonlinear fractional order KdV and Burger equation with exponential-decay kernel, J. Funct. Spaces, 2021 (2021), 8770488. https://doi.org/10.1155/2021/8770488 doi: 10.1155/2021/8770488
[20]	X. J. Yang, A new integral transform method for solving steady heat-transfer problem, Therm. Sci., 20 (2016), S639–S642.

Reader Comments

Your name:*

Email:*
© 2022 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)