Skip to content

Advertisement

  • Research
  • Open Access

Minkowski’s inequality for the AB-fractional integral operator

  • 1, 2Email author,
  • 3,
  • 4,
  • 5 and
  • 6
Journal of Inequalities and Applications20192019:96

https://doi.org/10.1186/s13660-019-2045-3

  • Received: 4 February 2019
  • Accepted: 27 March 2019
  • Published:

Abstract

Recently, AB-fractional calculus has been introduced by Atangana and Baleanu and attracted a large number of scientists in different scientific fields for the exploration of diverse topics. An interesting aspect is the generalization of classical inequalities via AB-fractional integral operators. In this paper, we aim to generalize Minkowski inequality using the AB-fractional integral operator.

Keywords

  • AB-fractional integral operator
  • Minkowski inequality

1 Introduction

Nowadays the fractional calculus has an important role in diverse scientific fields due to its several applications in dynamical problems including signals, hydrodynamics, dynamics, fluid, viscoelastic theory, biology, control theory, image processing, computer networking, and many others [15]. A large number of scientists have worked on generalizations of existing results including theorems, definitions, models, and many more. A generalization of classical inequalities by means of fractional-order integral operators is considered as an interesting subject area. For instance, recently, Agarwal et al. [6] proved Hermite–Hadamard-type inequalities by using generalized k-fractional-integrals. Aldhaifallah et al. [7] used the \((k,s)\)-fractional integral operator to generalize the inequalities for a family/class of n positive functions. Set et al. [8] studied Hermite–Hadamard-type inequalities for a generalized fractional integral operator for functions with convex absolute values of derivatives. Khan et al. [9] produced the Minkowski inequality by using the Hahn integral operator. On the other hand, noninteger-order calculus, usually referred to as fractional calculus, is used to generalize integrals and derivatives, in particular, integrals involving inequalities. Recently, Dumitru and Arran [10] introduced a new formula for fractional derivatives and integrals by using the Mittag-Leffler kernel. More theoretical concepts regarding fractional operators with Mittag-Leffler kernels (Atangana–Baleanu operators) and the higher-order case have been discussed in [11, 12], whereas the generalization to the generalized Mittag-Leffler kernels to gain a semigroup property have been recently initiated in [13, 14]. Khan [15] studied inequalities for a class of n functions by means of Saigo fractional calculus. Jarad et al. [16] presented a Gronwall-type inequality for the analysis of the fractional-order Atangana–Baleanu differential equation and in [17] for generalized fractional derivatives.

Shuang and Qi [18] proved some Hermite–Hadamard-type inequalities for a class of s-convex functions and studied special means. Mehrez and Agarwal [19] produced new integral inequalities by means of classical Hermite–Hadamard inequalities and obtained particular cases of their results with applications to special means. Park et al. [20] investigated new generalized inequalities, which then were utilized for stability analysis. Sarikaya et al. [21] established fractional integral inequalities generalizing the classical results by using the local fractional approach.

The integral inequalities with Mittag-Leffler functions have been studied as a generalization of the classical inequalities. For instance, Farid et al. [22] generalized several classical inequalities using an extended Mittag-Leffler function and evaluated particular cases of their results. More related work can be found in [2325].

In this paper, we use the AB-fractional integral operator for generalization of classical Minkowski inequalities. Our results are more general and applicable than those in the classical case. There are many definitions of fractional integrals, for example, Riemann–Liouville, Hadamard, Liouville, Weyl, Erdelyi–Kober, and Katugampola [2629], which can be considered for getting the same results. Now we give some definitions and lemma related to the AB-fractional operator.

Definition 1.1

([30])

The fractional ABC-derivative in the Caputo sense of a function \(f \in H^{*}(a,b)\) is defined by
$$ {}^{ABC} {{}_{a}\mathcal{D}_{\tau }^{\nu }}f(\tau )=\frac{\mathbb{B}( \nu )}{1-\nu } \int _{a}^{\tau }f^{'}(s)E_{\nu } \biggl[\frac{-\nu (\tau -s)^{ \mu }}{1-\nu } \biggr]\,ds, $$
(1.1)
where \(b>a\) and \(\nu \in [0,1]\), and \(\mathbb{B}(\nu )>0\) satisfies the property \(\mathbb{B}(0)=\mathbb{B}(1)=1\).

Definition 1.2

The fractional ABC-derivative in the Riemann–Liouville sense of a function \(f \in H^{*}(a,b)\) is defined by
$$ {}^{ABR} {{}_{a}\mathcal{D}_{\tau }^{\nu }}f(\tau )= \frac{\mathbb{B}( \nu )}{1-\nu }\frac{d}{d\tau } \int _{a}^{\tau }f(s)E_{\nu } \biggl[ \frac{- \nu (\tau -s)^{\nu }}{1-\nu } \biggr]\,ds, $$
(1.2)
where \(b>a\) and \(\nu \in [0,1]\).

Definition 1.3

([31, 32])

The fractional AB-integral of the function \(f \in H^{*}(a,b)\) is given by
$$ ^{AB} {{}_{a}\mathcal{I}_{\tau }^{\nu }}f(\tau )= \frac{1-\nu }{ \mathbb{B}(\nu )}f(\tau )+\frac{\nu }{\mathbb{B}(\nu )\varGamma (\nu )} \int _{a}^{\tau }f(s) (\tau -s)^{\nu -1}\,ds, $$
(1.3)
where \(b>a\) and \(0<\nu <1 \).

Remark 1.4

Since the normalization function \(\mathbb{B}(\nu )>0\) is positive, it immediately follows that the AB-integral of a positive function is positive. We will rely on this fact throughout the proofs of the main results.

Lemma 1.5

([33])

The ABC-fractional derivative and AB-fractional integral of a function f satisfy the Newton–Leibnitz formula
$$ ^{AB}{{}_{a}\mathcal{I}_{\tau }^{\nu }} \bigl( ^{ABC}{{}_{a}\mathcal{D} _{\tau }^{\nu }}f( \tau ) \bigr)=f(\tau )-f(a). $$
(1.4)

Organization of the paper. This paper includes four sections. Introduction is given in Sect. 1, with a literature review, important definitions, and a lemma, which we will use in the proofs. In Sect. 2, we prove Minkowski’s inequality for the AB-fractional integral operator. Other AB-fractional integral inequalities are proved in Sect. 3. The summary is given in Sect. 4.

2 The AB-fractional Minkowski inequality

Theorem 2.1

Let \(\nu >0\) and \(p\geq 1\). Let \(u, v \in C_{\nu }[a,b]\) be two positive functions in \([0,\infty [\) such that \({}^{AB}{{}_{a}\mathcal{I} _{t}^{\nu }}u(t)<\infty \) and \({}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v(t)< \infty \) for all \(t>a\). If \(0<\alpha \leq \frac{u(t)}{v(t)}\leq \theta \) for some \(\alpha ,\theta \in \mathbb{R}_{+}^{*}\) and all \(t\in [a,b]\), then
$$ \bigl({}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u^{p}(t) \bigr)^{\frac{1}{p}}+ \bigl(^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v^{p}(t) \bigr)^{\frac{1}{p}} \leq \mathcal{A} \bigl[^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t) \bigr)^{p} \bigr]^{\frac{1}{p}}, $$
(2.1)
where
$$ \mathcal{A}=\frac{\theta (1+\alpha )+(\theta +1)}{(1+\alpha )(\theta +1)}. $$

Proof

From the condition \(\frac{u(t)}{v(t)}\leq \theta \) we obtain
$$ u(t)\leq \biggl(\frac{\theta }{\theta +1}\biggr) \bigl(u(t)+v(t)\bigr). $$
(2.2)
Taking the pth power of both sides of Eq. (2.2), we have
$$ u^{p}(t)\leq \biggl(\frac{\theta }{\theta +1} \biggr)^{p}\bigl(u(t)+v(t)\bigr)^{p}. $$
(2.3)
Multiplying both sides of (2.3) by \(\frac{1-\nu }{\mathbb{B}( \nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}u^{p}(t)\leq \biggl( \frac{\theta }{\theta +1}\biggr)^{p}\frac{1-\nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t) \bigr)^{p}. $$
(2.4)
Also, replacing t by s in Eq. (2.3) and multiplying both sides by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )} \), we get
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )} u^{p}(s) \leq \biggl( \frac{\theta }{\theta +1}\biggr)^{p} \frac{\nu (t-s)^{\nu -1}}{ \mathbb{B}(\nu )\varGamma (\nu )} \bigl(u(s)+v(s) \bigr)^{p}. $$
(2.5)
Integrating both sides of Eq. (2.4) with respect to s, we have
$$ \int _{a}^{t} \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u ^{p}(s)\,ds\leq \biggl(\frac{\theta }{\theta +1}\biggr)^{p} \int _{a}^{t} \frac{ \nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )} \bigl(u(s)+v(s) \bigr)^{p}\,ds. $$
(2.6)
Adding (2.4) and (2.6), we obtain
$$\begin{aligned} \frac{1-\nu }{\mathbb{B}(\nu )}u^{p}(t)+ \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )} u^{p}(s) \,ds \leq {}&\biggl(\frac{ \theta }{\theta +1}\biggr)^{p} \biggl[ \frac{1-\nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t)\bigr)^{p} \\ &{}+ \int _{a}^{t} \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )} \bigl(u(s)+v(s) \bigr)^{p}\,ds \biggr]. \end{aligned}$$
This implies
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u^{p}(s) \leq \biggl(\frac{\theta }{\theta +1}\biggr)^{p} {}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}\bigl(u(s)+v(s)\bigr)^{p}. $$
(2.7)
Taking the \(\frac{1}{p}\)th power of both sides of Eq. (2.7), we find
$$ \bigl(^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u^{p}(t) \bigr)^{\frac{1}{p}} \leq \frac{\theta }{\theta +1} \bigl[ ^{AB}{{}_{a} \mathcal{I}_{t}^{ \nu }}\bigl(u(t)+v(t)\bigr)^{p} \bigr]^{\frac{1}{p}}. $$
(2.8)
On the other hand, by using the condition \(0<\alpha \leq \frac{u(t)}{v(t)}\) we directly get
$$ v^{p}(t)\leq \frac{1}{(1+\alpha )^{p}}\bigl(u(t)+v(t) \bigr)^{p}. $$
(2.9)
Multiplying Eq. (2.9) by \(\frac{1-\nu }{\mathbb{B}(\nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}v^{p}(t)\leq \frac{1}{(1+\alpha )^{p}} \frac{1- \nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t)\bigr)^{p}. $$
(2.10)
Also, replacing t by s in Eq. (2.9) and multiplying both sides by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\), we get
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )} v^{p}(s) \leq \frac{1}{(1+\alpha )^{p}} \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}( \nu )\varGamma (\nu )} \bigl(u(s)+v(s)\bigr)^{p}. $$
(2.11)
Integrating both sides of Eq. (2.11) with respect to s, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )} v ^{p}(s)\,ds\leq \frac{1}{(1+\alpha )^{p}} \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{p}\,ds. $$
(2.12)
Adding (2.10) and (2.12), we obtain
$$\begin{aligned} \frac{1-\nu }{\mathbb{B}(\nu )}v^{p}(t) + \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )} v^{p}(s) \,ds\leq{}& \frac{1}{(1+ \alpha )^{p}}\frac{1-\nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t)\bigr)^{p} \\ &{}+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )} \bigl(u(s)+v(s) \bigr)^{p}\,ds ]. \end{aligned}$$
(2.13)
This leads to the AB-fractional integral inequality
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v^{p}(t) \leq \frac{1}{(1+\alpha )^{p}} {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{p}. $$
(2.14)
Taking the \(\frac{1}{p}\)th power of both sides of Eq. (2.14), we find
$$ \bigl(^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v^{p}(t) \bigr)^{\frac{1}{p}} \leq \frac{1}{1+\alpha } \bigl[ ^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}\bigl(u(t)+v(t)\bigr)^{p} \bigr]^{\frac{1}{p}}. $$
(2.15)
By Eqs. (2.8) and (2.15) we obtain
$$ \bigl(^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u^{p}(t) \bigr)^{\frac{1}{p}}+ \bigl(^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v^{p}(t) \bigr)^{\frac{1}{p}} \leq \mathcal{A} \bigl[ ^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}\bigl(u(t)+v(t)\bigr)^{p} \bigr]^{\frac{1}{p}}. $$
(2.16)
Thus, the proof of the AB-fractional integral inequality is completed. □

3 Other types of inequalities

Theorem 3.1

Let \(\nu >0\) and \(p>1,q>1,\frac{1}{p}+\frac{1}{q}=1\). Let \(u, v \in C_{\nu }[a,b]\) be two positive functions in \([0,\infty [\) such that \({}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t)<\infty \) and \({}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}v(t)<\infty \) for all \(t>a\). If \(0<\alpha \leq \frac{u(t)}{v(t)}\leq \theta \) for some \(\alpha ,\theta \in \mathbb{R}_{+}^{*}\) and all \(t\in [a,b]\), then
$$ \bigl(^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t) \bigr)^{\frac{1}{p}} \bigl(^{AB} {{}_{a}\mathcal{I}_{t}^{\nu }}v(t) \bigr)^{\frac{1}{q}}\leq \biggl(\frac{ \theta }{\alpha } \biggr)^{\frac{1}{pq}} \bigl[^{AB}{{}_{a}\mathcal{I}_{t} ^{\nu }} \bigl(u^{\frac{1}{p}}(t)v^{\frac{1}{q}}(t) \bigr) \bigr]. $$
(3.1)

Proof

Using the condition \(\frac{u(t)}{v(t)}\leq \theta \), we get
$$ u^{\frac{1}{q}}\leq \theta ^{\frac{1}{q}} v^{\frac{1}{q}}. $$
(3.2)
Multiplying (3.2) by \(u^{\frac{1}{p}}\) and using the condition \(\frac{1}{p}+\frac{1}{q}=1\), we have
$$ u\leq \theta ^{\frac{1}{q}} u^{\frac{1}{p}} v^{\frac{1}{q}}. $$
(3.3)
Now let us use (3.3) twice. First, multiplying by \(\frac{1- \nu }{\mathbb{B}(\nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}u\leq \theta ^{\frac{1}{q}} \frac{1- \nu }{\mathbb{B}(\nu )}u^{\frac{1}{p}} v^{\frac{1}{q}}. $$
(3.4)
Second, multiplying by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu ) \varGamma (\nu )}\), we obtain
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u\leq \theta ^{\frac{1}{q}} \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u^{\frac{1}{p}} v^{\frac{1}{q}}. $$
(3.5)
Integrating both sides of Eq. (3.5) from 0 to t, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(s)\,ds \leq \theta ^{\frac{1}{q}} \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{ \mathbb{B}(\nu )\varGamma (\nu )}u(s)^{\frac{1}{p}} v(s)^{\frac{1}{q}}\,ds. $$
(3.6)
Now, by adding Eq. (3.4) and Eq. (3.6) we find
$$\begin{aligned} \frac{1-\nu }{\mathbb{B}(\nu )}u(t)+ \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(s)\,ds \leq{} & \theta ^{ \frac{1}{q}} \biggl[\frac{1-\nu }{\mathbb{B}(\nu )}u(t)^{\frac{1}{p}} v(t)^{ \frac{1}{q}} \\ &{}+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )}u(s)^{\frac{1}{p}} v(s)^{\frac{1}{q}}\,ds \biggr]. \end{aligned}$$
This implies
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t) \leq \theta ^{\frac{1}{q}} \bigl[^{AB} {{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u^{\frac{1}{p}}(t)v^{\frac{1}{q}}(t) \bigr) \bigr]. $$
(3.7)
Taking the \(\frac{1}{p}\)th power of both sides of (3.7), we have
$$ \bigl[^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t) \bigr]^{\frac{1}{p}}\leq \theta ^{\frac{1}{pq}} \bigl[^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u ^{\frac{1}{p}}(t)v^{\frac{1}{q}}(t) \bigr) \bigr]^{\frac{1}{p}}. $$
(3.8)
Now, by the condition \(\alpha \leq \frac{u(t)}{v(t)}\) we have
$$ v^{\frac{1}{p}}\leq \alpha ^{\frac{-1}{p}} u^{\frac{1}{p}}. $$
(3.9)
Multiplying Eq. (3.9) by \(v^{\frac{1}{q}}\), we get
$$ v\leq \alpha ^{\frac{-1}{p}} u^{\frac{1}{p}}v^{\frac{1}{q}}. $$
(3.10)
Now let us use (3.10) twice. First, multiplying by \(\frac{1- \nu }{\mathbb{B}(\nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}v\leq \alpha ^{\frac{-1}{p}} \frac{1- \nu }{\mathbb{B}(\nu )}u^{\frac{1}{p}} v^{\frac{1}{q}}. $$
(3.11)
Second, multiplying by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu ) \varGamma (\nu )}\), we obtain
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v\leq \alpha ^{\frac{-1}{p}} \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu ) \varGamma (\nu )}u^{\frac{1}{p}} v^{\frac{1}{q}}. $$
(3.12)
Integrating both sides of Eq. (3.12) from 0 to t, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v(s)\,ds \leq \alpha ^{\frac{-1}{p}} \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{ \mathbb{B}(\nu )\varGamma (\nu )}u(s)^{\frac{1}{p}} v(s)^{\frac{1}{q}}\,ds. $$
(3.13)
Now, by adding Eq. (3.11) and Eq. (3.13) we find
$$\begin{aligned} \frac{1-\nu }{\mathbb{B}(\nu )}v(t)+ \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v(s)\,ds \leq {}& \alpha ^{ \frac{-1}{p}} \biggl[\frac{1-\nu }{\mathbb{B}(\nu )}u(t)^{\frac{1}{p}} v(t)^{ \frac{1}{q}} \\ &{}+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )}u(s)^{\frac{1}{p}} v(s)^{\frac{1}{q}}\,ds \biggr]. \end{aligned}$$
(3.14)
This implies
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v(t) \leq \alpha ^{\frac{-1}{p}} \bigl[^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u^{\frac{1}{p}}(t)v^{ \frac{1}{q}}(t) \bigr) \bigr]. $$
(3.15)
Taking the \(\frac{1}{q}\)th power of both sides of (3.15), we have
$$ \bigl[^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v(s) \bigr]^{\frac{1}{q}}\leq \alpha ^{\frac{-1}{pq}} \bigl[^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u ^{\frac{1}{p}}(t)v^{\frac{1}{q}}(t) \bigr) \bigr]^{\frac{1}{q}}. $$
(3.16)
Finally, multiplying Eq. (3.8) and Eq. (3.16), we obtain
$$ \bigl(^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t) \bigr)^{\frac{1}{p}} \bigl(^{AB} {{}_{a}\mathcal{I}_{t}^{\nu }}v(t) \bigr)^{\frac{1}{q}}\leq \biggl(\frac{ \theta }{\alpha } \biggr)^{\frac{1}{pq}} \bigl[^{AB}{{}_{a}\mathcal{I}_{t} ^{\nu }} \bigl(u^{\frac{1}{p}}(t)v^{\frac{1}{q}}(t) \bigr) \bigr]. $$
(3.17)
 □

Theorem 3.2

Let \(\nu >0\) and \(p>1,q>1,\frac{1}{p}+\frac{1}{q}=1\). Let \(u, v \in C_{\nu }[a,b]\) be two positive functions in \([0,\infty [\) such that \({}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u^{p}(t)<\infty\), \({}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}u^{q}(t)<\infty \), \({}^{AB}{{}_{a}\mathcal{I}_{t} ^{\nu }}v^{p}(t)<\infty \), and \({}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v ^{q}(t)<\infty \) for all \(t>a\). If \(0<\alpha \leq \frac{u(t)}{v(t)} \leq \theta \) for some \(\alpha ,\theta \in \mathbb{R}_{+}^{*}\) and all \(t\in [a,b]\), then
$$ {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t) \bigr)\leq \mathcal{A}^{*} {}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u^{p}(t)+v^{p}(t) \bigr)+ \mathcal{B}^{*}_{m} {}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u^{q}(t)+v ^{q}(t) \bigr), $$
(3.18)
where
$$ \mathcal{A}^{*}=\frac{2^{p-1}\theta ^{p}}{p(\theta +1)^{p}},\qquad \mathcal{B}^{*}_{m}= \frac{2^{q-1}}{q(1+\alpha )^{q}.} $$

Proof

Using the condition \(\frac{u(t)}{v(t)}\leq \theta \), we obtain
$$ u(t)\leq \biggl(\frac{\theta (u(t)+v(t))}{1+\theta }\biggr). $$
(3.19)
Taking the pth power of both sides of Eq. (2.2), we have
$$ u^{p}(t)\leq \biggl(\frac{\theta }{\theta +1} \biggr)^{p}\bigl(u(t)+v(t)\bigr)^{p}. $$
(3.20)
Multiplying both sides of (3.20) by \(\frac{1-\nu }{\mathbb{B}( \nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}u^{p}(t)\leq \biggl( \frac{\theta }{\theta +1}\biggr)^{p}\frac{1-\nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t) \bigr)^{p}. $$
(3.21)
Also, replacing t by s in Eq. (3.20) and multiplying both sides by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\), we get
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u^{p}(s) \leq \biggl( \frac{\theta }{\theta +1}\biggr)^{p}\frac{\nu (t-s)^{\nu -1}}{ \mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{p}. $$
(3.22)
Integrating both sides of Eq. (3.21) with respect to s, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u ^{p}(s) \,ds\leq \biggl(\frac{\theta }{\theta +1}\biggr)^{p} \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{p}\,ds. $$
(3.23)
Adding (3.21) and (3.23), we obtain
$$\begin{aligned} &\frac{1-\nu }{\mathbb{B}(\nu )}u^{p}(t)+ \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u^{p}(s) \,ds\\ &\quad\leq \biggl(\frac{\theta }{ \theta +1}\biggr)^{p} \biggl[\frac{1-\nu }{\mathbb{B}(\nu )} \bigl(u(t)+v(t)\bigr)^{p}+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{p}\,ds \biggr]. \end{aligned}$$
This implies
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u^{p}(t) \leq \biggl(\frac{\theta }{\theta +1}\biggr)^{p} {}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}\bigl(u(t)+v(t)\bigr)^{p}. $$
(3.24)
Multiplying (2.7) by the constant \(\frac{1}{p}\), we find
$$ \frac{1}{p} \bigl(^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}u^{p}(t) \bigr)\leq \frac{1}{p} \biggl( \frac{\theta }{\theta +1} \biggr)^{p} \bigl[ ^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{p} \bigr]. $$
(3.25)
On the other hand, by using the condition \(0<\alpha \leq \frac{u(t)}{v(t)}\) we directly get
$$ v^{q}(t)\leq \frac{1}{(1+\alpha )^{q}}\bigl(u(t)+v(t) \bigr)^{q}. $$
(3.26)
Multiplying (3.26) by \(\frac{1-\nu }{\mathbb{B}(\nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}v^{q}(t)\leq \frac{1}{(1+\alpha )^{q}} \frac{1- \nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t)\bigr)^{q}. $$
(3.27)
Also, replacing t by s in Eq. (3.26) and multiplying both sides by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\), we get
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v^{q}(s) \leq \frac{1}{(1+\alpha )^{q}} \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}( \nu )\varGamma (\nu )}\bigl(u(s)+v(s)\bigr)^{q}. $$
(3.28)
Integrating both sides of Eq. (3.28) with respect to s, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v ^{q}(s) \,ds\leq \frac{1}{(1+\alpha )^{q}} \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{q}\,ds. $$
(3.29)
Adding (3.27) and (3.29), we obtain
$$\begin{aligned} \frac{1-\nu }{\mathbb{B}(\nu )}v^{q}(t)+ \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v^{q}(s) \,ds \leq{} & \frac{1}{(1+ \alpha )^{q}} \biggl[\frac{1-\nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t) \bigr)^{q} \\ &{}+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )}\bigl(u(s)+v(s) \bigr)^{q}\,ds \biggr]. \end{aligned}$$
This implies
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v^{q}(t) \leq \frac{1}{(1+\alpha )^{q}} {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{q}. $$
(3.30)
Multiplying (2.14) by\(\frac{1}{q}\), we have
$$ \frac{1}{q} \bigl(^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}v^{q}(t) \bigr)\leq \frac{1}{q} \frac{1}{(1+\alpha )^{q}} \bigl[ ^{AB}{{}_{a} \mathcal{I}_{t} ^{\nu }}\bigl(u(t)+v(t)\bigr)^{q} \bigr]. $$
(3.31)
By means of Eqs. (3.25) and (3.31) we get
$$\begin{aligned} &\frac{1}{p} \bigl(^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}u^{p}(t) \bigr)+ \frac{1}{q} \bigl(^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v^{q}(t) \bigr) \\ &\quad\leq \frac{1}{p} \biggl(\frac{\theta }{\theta +1} \biggr)^{p} \bigl[ ^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{p} \bigr]+\frac{1}{q} \frac{1}{(1+ \alpha )^{q}} \bigl[ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{q} \bigr]. \end{aligned}$$
(3.32)
To complete our proof, we have to use Young’s inequality
$$ u(t)v(t)\leq \frac{u^{p}(t)}{p}+\frac{v^{q}(t)}{q}. $$
(3.33)
Multiplying (3.33) by \(\frac{1-\nu }{\mathbb{B}(\nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}u(t)v(t)\leq \frac{1-\nu }{\mathbb{B}( \nu )} \biggl( \frac{u^{p}(t)}{p}+\frac{v^{q}(t)}{q} \biggr). $$
(3.34)
Also, replacing t by s in Eq. (3.33) and multiplying both sides by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\), we get
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(s)v(s) \leq \frac{\nu (t-s)^{\nu -1}}{p \mathbb{B}(\nu )\varGamma (\nu )}u^{p}(s)+ \frac{ \nu (t-s)^{\nu -1}}{q \mathbb{B}(\nu )\varGamma (\nu )}v^{q}(s). $$
(3.35)
Integrating both sides of Eq. (3.35) with respect to s, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(s)v(s)\,ds \leq \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{p \mathbb{B}(\nu )\varGamma ( \nu )}u^{p}(s) \,ds+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{q \mathbb{B}( \nu )\varGamma (\nu )}v^{q}(s) \,ds. $$
(3.36)
Adding (3.34) and (3.36), we obtain
$$\begin{aligned} & \frac{1-\nu }{\mathbb{B}(\nu )}u(t)v(t) + \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(s)v(s)\,ds \\ &\quad\leq \frac{1-\nu }{ \mathbb{B}(\nu )} \biggl(\frac{u^{p}(t)}{p}+\frac{v^{q}(t)}{q} \biggr) \\ &\qquad{}+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{p \mathbb{B}(\nu )\varGamma ( \nu )}u^{p}(s) \,ds+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{q \mathbb{B}( \nu )\varGamma (\nu )}v^{q}(s) \,ds. \end{aligned}$$
(3.37)
This implies
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t)v(t) \leq \frac{1}{p} {}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}u^{p}(t)+ \frac{1}{q} {}^{AB}{{}_{a}\mathcal{I} _{t}^{\nu }}v^{q}(t). $$
(3.38)
Using (3.32) and (3.38), we have
$$\begin{aligned} &{} ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t)v(t) \\ &\quad \leq \frac{1}{p} \biggl(\frac{ \theta }{\theta +1} \biggr)^{p} \bigl[ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{p} \bigr]+\frac{1}{q} \frac{1}{(1+\alpha )^{q}} \bigl[ ^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{q} \bigr]. \end{aligned}$$
(3.39)
Using the inequality
$$ (u+v)^{r}\leq 2^{r-1}\bigl(u^{r}+v^{r} \bigr),\quad u,v\geq 0,r>1, $$
(3.40)
with \(r=p\) and multiplying (3.40) by the constant \(\frac{1-\nu }{ \mathbb{B}(\nu )}\), we find
$$ \frac{1-\nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t)\bigr)^{p}\leq 2^{p-1}\frac{1- \nu }{\mathbb{B}(\nu )}\bigl(u(t)^{p}+v(t)^{p} \bigr). $$
(3.41)
Then multiplying Eq. (3.40) with \(r=p \) by \(\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\), we get
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}(u+v)^{p} \leq 2^{p-1} \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u ^{p}+v^{p}\bigr). $$
(3.42)
Integrating Eq. (3.42) from a to t, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{p}\,ds \leq 2^{p-1} \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu ) \varGamma (\nu )} \bigl(u^{p}(s)+v^{p}(s)\bigr)\,ds. $$
(3.43)
Adding Eq. (3.41) and Eq. (3.43), we obtain
$$\begin{aligned} &\frac{1-\nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t)\bigr)^{p}+ \int _{a}^{t}\frac{ \nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{p}\,ds \\ &\quad \leq2^{p-1} \biggl(\frac{1-\nu }{\mathbb{B}(\nu )} \bigl(u(t)^{p}+v(t)^{p}\bigr)+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )} \bigl(u^{p}(s)+v^{p}(s)\bigr)\,ds \biggr). \end{aligned}$$
(3.44)
This implies
$$ {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{p}\leq 2^{p-1} {}^{AB} {{}_{a}\mathcal{I}_{t}^{\nu }}\bigl(u^{p}(t)+v^{p}(t) \bigr). $$
(3.45)
Repeating the same process with \(r=q\), we get
$$ {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{q}\leq 2^{q-1} {}^{AB} {{}_{a}\mathcal{I}_{t}^{\nu }}\bigl(u^{q}(t)+v^{q}(t) \bigr). $$
(3.46)
Substituting by (3.45) and (3.46) into Eq. (3.39), the proof completed. □

Theorem 3.3

Let \(\nu >0\), and let \(u, v \in C_{\nu }[a,b]\) be two positive functions in \([0,\infty [\) such that \({}^{AB}{{}_{a}\mathcal{I}_{t}^{ \nu }}u(t)<\infty \) and \({}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v(t)<\infty \) for all \(t>a\), If \(0<\alpha \leq \frac{u(t)}{v(t)}\leq \theta \) for some \(\alpha ,\theta \in \mathbb{R}_{+}^{*}\) and all \(t\in [a,b]\), then
$$ \frac{1}{\theta } {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)v(t) \bigr)\leq {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t) \bigr)^{2} \leq \frac{1}{\alpha } {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)v(t) \bigr), $$
(3.47)

Proof

Using the condition
$$ 0< \alpha \leq \frac{u(t)}{v(t)}\leq \theta , $$
(3.48)
we conclude that
$$\begin{aligned} & (1+\alpha )v(t)\leq \bigl(u(t)+v(t)\bigr)\leq (\theta +1)v(t), \end{aligned}$$
(3.49)
$$\begin{aligned} & \frac{\theta +1}{\theta }u(t)\leq \bigl(u(t)+v(t)\bigr)\leq \frac{1+\alpha }{ \alpha }u(t). \end{aligned}$$
(3.50)
By (3.49) and (3.50) we obtain
$$\begin{aligned} \frac{1}{\theta }u(t)v(t)\leq \frac{(u(t)+v(t))^{2}}{(1+\alpha )( \theta +1)} \leq \frac{1}{\alpha }u(t)v(t). \end{aligned}$$
(3.51)
Multiplying (3.51) by \(\frac{1-\nu }{\mathbb{B}(\nu )} \) and then by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\), we get
$$\begin{aligned} & \frac{1}{\theta }\frac{1-\nu }{\mathbb{B}(\nu )}u(t)v(t)\leq \frac{1- \nu }{\mathbb{B}(\nu )}\frac{(u(t)+v(t))^{2}}{(1+\alpha )(\theta +1)} \leq \frac{1}{\alpha }\frac{1-\nu }{\mathbb{B}(\nu )}u(t)v(t), \end{aligned}$$
(3.52)
$$\begin{aligned} & \frac{1}{\theta }\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )}u(t)v(t)\leq \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )}\frac{(u(t)+v(t))^{2}}{(1+\alpha )(\theta +1)} \leq \frac{1}{ \alpha }\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(t)v(t). \end{aligned}$$
(3.53)
Integrating Eq. (3.53) from 0 to t with respect to s, we have
$$\begin{aligned} \frac{1}{\theta } \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}( \nu )\varGamma (\nu )}u(s)v(s)\,ds &\leq \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\frac{(u(s)+v(s))^{2}}{(1+ \alpha )(\theta +1)} \,ds \\ &\leq \frac{1}{\alpha } \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{ \mathbb{B}(\nu )\varGamma (\nu )}u(s)v(s)\,ds. \end{aligned}$$
(3.54)
Adding Eqs. (3.52) and Eq. (3.54), we obtain the required inequality. □

4 Conclusion

In this paper, we have considered Minkowski’s inequality for the AB-fractional integral operator. We have also obtained some other types of integral inequalities for the AB-fractional integral operator. By the help of this work we obtained more general inequalities than in the classical cases. For possible further work, we suggest to apply the obtained inequalities to prove the existence of solutions of fractional differential equations.

Declarations

Acknowledgements

The author Thabet Abdeljawad would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM), group number RG-DES-2017-01-17. All the authors are very grateful to the editorial board and the reviewers, whose comments improved the quality of the paper.

Availability of data and materials

Data sharing not applicable to this paper as no datasets were generated during the current study.

Funding

Not applicable.

Authors’ contributions

All the authors have equal contributions in this paper. All authors read and approved the final manuscript.

Competing interests

The authors have no conflict of interests regarding the publication of this paper.

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

Authors’ Affiliations

(1)
College of Engineering Mechanics and Materials, Hohai University, Nanjing, P.R. China
(2)
Department of Mathematics, Shaheed BB University, Khybar Pakhtunkhwa, Pakistan
(3)
Department of Mathematics and General Sciences, Prince Sultan University, Riyadh, Saudi Arabia
(4)
Department of Mathematics, Faculty of Sciences, Van Yuzuncu Yil University, Van, Turkey
(5)
Department of Mathematics, College of Science, Hohai University, Nanjing, P.R. China
(6)
Department of Mathematics, University of Peshawar, Khybar Pakhtunkhwa, Pakistan

References

  1. Baleanu, D., Khan, H., Jafari, H., Khan, R.A., Alipour, M.: On existence results for solutions of a coupled system of hybrid boundary value problems with hybrid conditions. Adv. Differ. Equ. 2015(1), 318 (2015) MathSciNetView ArticleGoogle Scholar
  2. Baleanu, D., Mustafa, O.G., Agarwal, P.R.: An existence result for a superlinear fractional differential equation. Appl. Math. Lett. 23(9), 1129–1132 (2010) MathSciNetView ArticleGoogle Scholar
  3. Baleanu, D., Mustafa, O.G., Agarwal, R.P.: On the solution set for a class of sequential fractional differential equations. J. Phys. A, Math. Theor. 43(38), 385209 (2010) MathSciNetView ArticleGoogle Scholar
  4. Baleanu, D., Agarwal, R.P., Khan, H., Khan, R.A., Jafari, H.: On the existence of solution for fractional differential equations of order \(3 < \delta \leq 4\). Adv. Differ. Equ. 2015, 362 (2015) View ArticleGoogle Scholar
  5. Baleanu, D., Agarwal, R.P., Mohammadi, H., Rezapour, S.: Some existence results for a nonlinear fractional differential equation on partially ordered Banach spaces. Bound. Value Probl. 2013, 112 (2013) MathSciNetView ArticleGoogle Scholar
  6. Agarwal, P., Jleli, M., Tomar, M.: Certain Hermite–Hadamard type inequalities via generalized k-fractional integrals. J. Inequal. Appl. 2017(1), 55 (2017) MathSciNetView ArticleGoogle Scholar
  7. Aldhaifallah, M., Tomar, M., Nisar, K.S., Purohit, S.D.: Some new inequalities for \((k, s)\)-fractional integrals. J. Nonlinear Sci. Appl. 9, 5374–5381 (2016) MathSciNetView ArticleGoogle Scholar
  8. Set, E., Noor, M.A., Awan, M.U., Gzpinar, A.: Generalized Hermite–Hadamard type inequalities involving fractional integral operators. J. Inequal. Appl. 2017, 169 (2017) MathSciNetView ArticleGoogle Scholar
  9. Khan, H., Tunç, C., Alkhazan, A., Ameen, B., Khan, A.: A generalization of Minkowski’s inequality by Hahn integral operator. J. Taibah Univ. Sci. 12(5), 506–513 (2018) View ArticleGoogle Scholar
  10. Baleanu, D., Fernandez, A.: On some new properties of fractional derivatives with Mittag-Leffler kernel. Commun. Nonlinear Sci. Numer. Simul. 59, 444–462 (2018) MathSciNetView ArticleGoogle Scholar
  11. Abdeljawad, T., Baleanu, D.: Integration by parts and its applications of a new nonlocal fractional derivative with Mittag-Leffler nonsingular kernel. J. Nonlinear Sci. Appl. 10(3), 1098–1107 (2017) MathSciNetView ArticleGoogle Scholar
  12. Abdeljawad, T.: A Lyapunov type inequality for fractional operators with nonsingular Mittag-Leffler kernel. J. Inequal. Appl. 2017, 130 (2017). https://doi.org/10.1186/s13660-017-1400-5 MathSciNetView ArticleMATHGoogle Scholar
  13. Abdeljawad, T., Baleanu, D.: On fractional derivatives with generalized Mittag-Leffler kernels. Adv. Differ. Equ. 2018, 468 (2018) MathSciNetView ArticleGoogle Scholar
  14. Abdeljawad, T.: Fractional operators with generalized Mittag-Leffler kernels and their iterated differintegrals. Chaos 29, 023102 (2019). https://doi.org/10.1063/1.5085726 MathSciNetView ArticleMATHGoogle Scholar
  15. Khan, H., Tunç, C., Baleanu, D., Khan, A., Alkhazzan, A.: Inequalities for n-class of functions using the Saigo fractional integral operator. Rev. R. Acad. Cienc. Exactas Fís. Nat., Ser. A Mat. 1–4 (2019, to appear). https://doi.org/10.1007/s13398-019-00624-5
  16. Jarad, F., Abdeljawad, T., Hammouch, Z.: On a class of ordinary differential equations in the frame of Atangana–Baleanu fractional derivative. Chaos Solitons Fractals 117, 16–20 (2021) MathSciNetView ArticleGoogle Scholar
  17. Adjabi, Y., Jarad, F., Abdeljawad, T.: On generalized fractional operators and a Gronwall type inequality with applications. Filomat 31(17), 5457–5473 (2017) MathSciNetView ArticleGoogle Scholar
  18. Shuang, Y., Qi, F.: Integral inequalities of Hermite–Hadamard type for extended s-convex functions and applications. Mathematics 6(11), 223 (2018) View ArticleGoogle Scholar
  19. Mehrez, K., Agarwal, P.: New Hermite–Hadamard type integral inequalities for convex functions and their applications. J. Comput. Appl. Math. 350, 274–285 (2019) MathSciNetView ArticleGoogle Scholar
  20. Park, M.J., Kwon, O.M., Ryu, J.H.: Generalized integral inequality: application to time-delay systems. Appl. Math. Lett. 77, 6–12 (2018) MathSciNetView ArticleGoogle Scholar
  21. Sarikaya, M.Z., Tunc, T., Budak, H.: On generalized some integral inequalities for local fractional integrals. Appl. Math. Comput. 276, 316–323 (2016) MathSciNetMATHGoogle Scholar
  22. Farid, G., Khan, K.A., Latif, N., Rehman, A.U., Mehmood, S.: General fractional integral inequalities for convex and m-convex functions via an extended generalized Mittag-Leffler function. J. Inequal. Appl. 2018, 243 (2018) MathSciNetView ArticleGoogle Scholar
  23. Sarikaya, M.Z., Set, E., Yaldiz, H., Basak, N.: Hermite–Hadamard’s inequality for fractional integrals and related fractional inequalities. Math. Comput. Model. 57, 2403–2407 (2013) View ArticleGoogle Scholar
  24. Mohammed, P.O.: On new trapezoid type inequalities for h-convex functions via generalized fractional integrals. Turk. J. Anal. Number Theory 6, 125–128 (2018) View ArticleGoogle Scholar
  25. Mohammed, P.O., Sarikaya, M.Z.: Hermite–Hadamard type inequalities for F-convex function involving fractional integrals. J. Inequal. Appl. 2018, 359 (2018) MathSciNetView ArticleGoogle Scholar
  26. Jafari, H., Jassim, H.K., Moshokoa, S.P., Ariyan, V.M., Tchier, F.: Reduced differential transform method for partial differential equations within local fractional derivative operators. Adv. Mech. Eng. 8(4), 1–6 (2016) View ArticleGoogle Scholar
  27. Richard, H.: Fractional Calculus: An Introduction for Physicists. World Scientific, Singapore (2014) MATHGoogle Scholar
  28. Podlubny, I.: Fractional Differential Equations: An Introduction to Fractional Derivatives, Fractional Differential Equations, to Methods of Their Solution and Some of Their Applications. Elsevier, Amsterdam (1998) MATHGoogle Scholar
  29. Sun, H., Zhang, Y., Chen, W., Reeves, D.M.: Use of a variable-index fractional-derivative model to capture transient dispersion in heterogeneous media. J. Contam. Hydrol. 157, 47–58 (2014) View ArticleGoogle Scholar
  30. Owolabi, K.M.: Modelling and simulation of a dynamical system with the Atangana–Baleanu fractional derivative. Eur. Phys. J. Plus 133(1), 15 (2018) View ArticleGoogle Scholar
  31. Kumar, D., Singh, J., Baleanu, D.: Analysis of regularized long-wave equation associated with a new fractional operator with Mittag-Leffler type kernel. Phys. A, Stat. Mech. Appl. 492, 155–167 (2018) MathSciNetView ArticleGoogle Scholar
  32. Jianke, Z., Gaofeng, W., Xiaobin, Z., Chang, Z.: Generalized Euler–Lagrange equations for fuzzy fractional variational problems under gH-Atangana–Baleanu differentiability. Hindawi 2018, Article ID 2740678 (2018). https://doi.org/10.1155/2018/2740678 MathSciNetView ArticleMATHGoogle Scholar
  33. Abdeljawad, T., Baleanu, D.: Discrete fractional differences with nonsingular discrete Mittag-Leffler kernels. Adv. Differ. Equ. 2016, 232 (2016). https://doi.org/10.1186/s13662-016-0949-5 MathSciNetView ArticleMATHGoogle Scholar

Copyright

© The Author(s) 2019

Advertisement