Research article Special Issues

Fixed point approach to the Mittag-Leffler kernel-related fractional differential equations

  • Received: 07 November 2022 Revised: 28 January 2023 Accepted: 01 February 2023 Published: 06 February 2023
  • MSC : 34A08, 34A12, 47H10, 54H25

  • The goal of this paper is to present a new class of contraction mappings, so-called ηθ-contractions. Also, in the context of partially ordered metric spaces, some coupled fixed-point results for ηθ-contraction mappings are introduced. Furthermore, to support our results, two examples are provided. Finally, the theoretical results are applied to obtain the existence of solutions to coupled fractional differential equations with a Mittag-Leffler kernel.

    Citation: Hasanen A. Hammad, Hüseyin Işık, Hassen Aydi, Manuel De la Sen. Fixed point approach to the Mittag-Leffler kernel-related fractional differential equations[J]. AIMS Mathematics, 2023, 8(4): 8633-8649. doi: 10.3934/math.2023433

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  • The goal of this paper is to present a new class of contraction mappings, so-called ηθ-contractions. Also, in the context of partially ordered metric spaces, some coupled fixed-point results for ηθ-contraction mappings are introduced. Furthermore, to support our results, two examples are provided. Finally, the theoretical results are applied to obtain the existence of solutions to coupled fractional differential equations with a Mittag-Leffler kernel.



    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 FL1[a,b]. Then the Riemann-Liouville integrals Jαa+F and JαbF of order α>0 with a0 are defined by

    Jαa+F(x)=1Γ(α)xa(xv)α1F(v)dv,x>a,

    and

    JαbF(x)=1Γ(α)bx(vx)α1F(v)dv,x<b,

    where

    Γ(x)=0evvx1dv,

    is the well known Gamma function.

    Diaz et al. [8] introduced the notion of generalized k-gamma function. The integral form of Γk is given by:

    Γk(x)=0vx1evkkdv,(x)>0.

    Note that

    Γk(x)=kxk1Γ(xk).

    k-Beta function is defined as:

    βk(x,y)=1k10vxk1(1v)xk1dv.

    Obviously

    βk(x,y)=1kβ(xk,yk).

    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 F be piecewise continuous on I=(0,) and integrable on any finite subinterval of I=[0,]. Then for v>0, we consider k-Riemann-Liouville fractional integral of F of order α

    kJαaF(x)=1kΓk(α)xa(xv)αk1F(v)dv,x>a,k>0.

    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 F[a,b]. The fractional left side integral kIαa+F and right side integral kIαbF of order α(0,1) are defined as follows:

    Iαa+F(x)=1αxae1αα(xv)F(v)dv,    x>a,

    and

    IαbF(x)=1αbxe1αα(vx)F(v)dv,    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 FL[a,b]. The k-fractional left side integral kIαa+F and right side integral kIαbF of order α(0,1) for k>0 are defined as follows

    kIαa+F(x)=kαxaekαα(xv)F(v)dv,    x>a,

    and

    kIαbF(x)=kαbxekαα(vx)F(v)dv,    x<b.

    It is to be noted that by taking k1 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 θ(.,.), if

    x+vθ(y,x)K,x,yK,v[0,1].

    The preinvexity of the functions is defined as:

    Definition 1.5 ([21]). A function F:KR is said to be preinvex with respect to bifunction θ(.,.), if

    F(x+vθ(y,x))(1v)F(x)+vf(y),x,yK,v[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 θ:K×KRn. We say that the bifunction θ(.,.) satisfies the condition C, if for any x,yRn

    1. θ(x,x+vθ(y,x))=vθ(y,x),

    2. θ(y,x+vθ(y,x))=(1v)θ(y,x),

    for all v[0,1].

    Note that for any x,yRn and v1,v2[0,1] and from the condition C, we have, see [12]

    θ(x+v2θ(y,x),x+v1θ(y,x))=(v2v1)θ(y,x).

    In this section, we derive some new fractional trapezium type inequalities involving the functions having preinvexity property. For the sake of simplicity, we set ρ=kααθ(b,a) and ρ1=1ααθ(b,a).

    Theorem 2.1. Let F:[a,a+θ(b,a)]RR be a positive function with θ(b,a)>0 and FL[a,a+θ(b,a)]. Suppose F is a preinvex function and θ(.,.) satisfies condition C, then

    F(2a+θ(b,a)2)kα2k(1eρ)[kIα(a)+F(a+θ(b,a))+kIα(a+θ(b,a))F(a)]F(a)+F(b)2. (2.1)

    Proof. By preinvexity of F, we have for every x,y[a,a+θ(b,a)] with λ=12

    2F(x+θ(y,x)2)[F(x)+F(y)],

    with x=a+vθ(b,a),y=a+(1v)θ(b,a) and using the condition C, we have

    2F(a+vθ(b,a)+θ(a+(1v)θ(b,a),a+vθ(b,a))2)=2F(a+vθ(b,a)+(12v)θ(b,a)2)=2F(2a+θ(b,a)2)F(a+vθ(b,a))+F(a+(1v)θ(b,a)). (2.2)

    Multiplying both sides of above inequality by eρv and integrating with respect to v over [0,1], we have

    2(1eρ)ρF(2a+θ(b,a)2)10eρvf(a+vθ(b,a))dv+10eρvF(a+(1v)θ(b,a))dv=1θ(b,a)[a+θ(b,a)aeρ(saθ(b,a))F(s)ds+a+θ(b,a)aeρ(a+θ(b,a)sθ(b,a))F(s)ds]=αkθ(b,a)[kIα(a)+F(a+θ(b,a))+kIα(a+θ(b,a))F(a)].

    As a result, we get

    F(2a+θ(b,a)2)kα2k(1eρ)[kIα(a)+F(a+θ(b,a))+kIα(a+θ(b,a))F(a)]. (2.3)

    For the proof of second inequality, we note that F is a preinvex function, so we have

    F(a+vθ(b,a))+F(a+(1v)θ(b,a))F(a)+F(b). (2.4)

    Multiplying both sides by eρv and integrating with respect to v over [0,1], we have

    αkθ(b,a)[kIα(a)+F(a+θ(b,a))+kIα(a+θ(b,a))F(a)]1eρρ[F(a)+F(b)], (2.5)

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

    Theorem 2.2. Let F:[a,a+θ(b,a)]RR be a positive and preinvex function with θ(b,a)>0 and FL[a,a+θ(b,a)]. Let W be a non-negative, integrable and symmetric with respect to 2a+θ(b,a)2, then using the condition C, we have

    F(2a+θ(b,a)2)[kIα(a)+W(a+θ(b,a))+kIα(a+θ(b,a))W(a)]kIα(a)+(FW)(a+θ(b,a))+kIα(a+θ(b,a))(FW)(a)F(a)+F(b)2[kIα(a)+W(a+θ(b,a))+kIα(a+θ(b,a))W(a)]. (2.6)

    Proof. Since F is preinvex function on L[a,a+θ(b,a)], so multiplying inequality (2.2) by

    eρvW(a+vθ(b,a)), (2.7)

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

    2F(2a+θ(b,a)2)10eρvW(a+vθ(b,a))dv10eρvW(a+vθ(b,a))F(a+vθ(b,a))dv+10eρvW(a+vθ(b,a))F(a+(1v)θ(b,a))dv=10eρvW(a+vθ(b,a))F(a+vθ(b,a))dv+10eρvW(a+(1v)θ(b,a))F(a+(1v)θ(b,a))dv=1θ(b,a)[a+θ(b,a)aeρ(saθ(b,a))F(s)W(s)ds+a+θ(b,a)aeρ(a+θ(b,a)sθ(b,a))F(s)W(s)ds]=αkθ(b,a)[kIα(a)+(FW)(a+θ(b,a))+kIα(a+θ(b,a))(FW)(a)].

    Thus

    2F(2a+θ(b,a)2)10eρvW(a+vθ(b,a))dvαkθ(b,a)[kIα(a)+FW(a+θ(b,a))+kIα(a+θ(b,a))FW(a)].

    Since W is symmetric with respect to 2a+θ(b,a)2, we have

    kIα(a)+W(a+θ(b,a))=kIα(a+θ(b,a))W(a)=12[kIα(a)+W(a+θ(b,a))+kIα(a+θ(b,a))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 F,W:[a,a+θ(b,a)]RR be nonnegative and preinvex function on L[a,a+θ(b,a)] with θ(b,a)>0. If θ(.,.) satisfies condition C, then the following inequalities for the k-fractional integrals with exponential kernel holds:

    α2kθ(b,a)[kIα(a)+(FW)(a+θ(b,a))+kIα(a+θ(b,a))(FW)(a)]M(a,b)ρ22ρ+4eρ(ρ2+2ρ+4)2ρ3+N(a,b)ρ2+eρ(ρ+2)ρ3, (2.8)

    and

    2F(2a+θ(b,a)2)W(2a+θ(b,a)2)kα2k(1eρ)[kIα(a)+FW(a+θ(b,a))+kIα(a+θ(b,a))FW(a)]M(a,b)ρ2+eρ(ρ+2)ρ2(1eρ)+N(a,b)ρ22ρ+4eρ(ρ2+2ρ+4)2ρ2(1eρ), (2.9)

    where

    M(a,b)=F(a)W(a)+F(b)W(b),

    and

    N(a,b)=F(a)W(b)+F(b)W(a).

    Proof. Since F,W are preinvex functions on [a,a+θ(b,a)], then we have

    F(a+vθ(b,a))W(a+vθ(b,a))(1v)2F(a)W(a)+v2F(b)W(b)+v(1v)N(a,b),

    and

    F(a+(1v)θ(b,a))W(a+(1v)θ(b,a))v2F(a)W(a)+(1v)2F(b)W(b)+v(1v)N(a,b).

    Adding above inequalities, we have

    F(a+vθ(b,a))W(a+vθ(b,a))+F(a+(1v)θ(b,a))W(a+(1v)θ(b,a))(2v22v+1)M(a,b)+2v(1v)N(a,b).

    Multiplying both sides of above inequality by eρv and integrating with respect to v over [0,1], we have

    10eρvF(a+vθ(b,a))W(a+vθ(b,a))dv+10eρvF(a+(1v)θ(b,a))W(a+(1v)θ(b,a))dvM(a,b)10eρv(2v22v+1)dv+M(a,b)10eρv2v(1v)dv=M(a,b)ρ22ρ+4eρ(ρ2+2ρ+4)2ρ3+N(a,b)ρ2+eρ(ρ+2)ρ3.

    So,

    α2kθ(b,a)[kIα(a)+(FW)(a+θ(b,a))+kIα(a+θ(b,a))(FW)(a)]M(a,b)ρ22ρ+4eρ(ρ2+2ρ+4)2ρ3+N(a,b)ρ2+eρ(ρ+2)ρ3.

    For the proof of inequality (2.9), using the preinvexity of F,W and condition C, we have

    F(2a+θ(b,a)2)W(2a+θ(b,a)2)=F(a+(1v)θ(b,a)+12θ(a+vθ(b,a),a+(1v)θ(b,a))×W(a+(1v)θ(b,a)+12θ(a+vθ(b,a),a+(1v)θ(b,a))(F(a+vθ(b,a))+F(a+(1v)θ(b,a))2)(W(a+vθ(b,a))+W(a+(1v)θ(b,a))2)F(a+vθ(b,a))W(a+vθ(b,a))4+F(a+(1v)θ(b,a))W(a+(1v)θ(b,a))4+v(1v)2M(a,b)+2v22v+14N(a,b). (2.10)

    Multiplying both sides of inequality (2.10) by eρv and integrating with respect to v over [0,1], we get

    1eρρF(2a+θ(b,a)2)W(2a+θ(b,a)2)10eρvF(a+vθ(b,a))W(a+vθ(b,a))4dv+10eρvF(a+(1v)θ(b,a))W(a+(1v)θ(b,a))4dv+M(a,b)10eρvv(1v)2dv+N(a,b)10eρv2v22v+14dv,

    which completes the proof.

    Lemma 3.1. Assume that F:[a,a+θ(b,a)]RR is a differentiable function and FL[a,a+θ(b,a)]. Then the following equality for the k-fractional integrals with exponential kernel holds:

    Rab=θ(b,a)210uF(a+(1v)θ(b,a))dvθ(b,a)2(1eρ)[10eρvF(a+(1v)θ(b,a))dv10eρ(1v)F(a+(1v)θ(b,a))dv], (3.1)

    where

    Rab=kα2k(1eρ)[kIα(a)+F(a+θ(b,a))+kIα(a+θ(b,a))F(a)]F(2a+θ(b,a)2),

    and

    u={1, for0v12,1,for12v1.

    Proof. By simple calculations, we have

    10eρvF(a+(1v)θ(b,a))dv=1θ(b,a)[eρvF(a+(1v)θ(b,a))|10+ρ10eρvF(a+(1v)θ(b,a))dv]=1θ(b,a)[F(a+θ(b,a))eρF(a)+ρ10eρvF(a+(1v)θ(b,a))dv]=1θ(b,a)[F(a+θ(b,a))eρF(a)ρθ(b,a)a+θ(b,a)aeρa+θ(b,a)sθ(b,a)F(s)ds]=1θ(b,a)[F(a+θ(b,a))eρF(a)kααa+θ(b,a)ae(kα)α(a+θ(b,a)s)F(s)ds]=1θ(b,a)[F(a+θ(b,a))eρF(a)kαkkIα(a)+F(a+θ(b,a))], (3.2)

    similarly

    10eρ(1v)F(a+(1v)θ(b,a))dv=1θ(b,a)[eρ(1v)F(a+(1v)θ(b,a))|10ρ10eρ(1v)F(a+(1v)θ(b,a))dv]=1θ(b,a)[F(a)eρF(a+θ(b,a))ρ10eρvF(a+(1v)θ(b,a))dv]=1θ(b,a)[eρF(a+θ(b,a))F(a)+ρθ(b,a)a+θ(b,a)aeρsaθ(b,a)F(s)ds]=1θ(b,a)[eρF(a+θ(b,a))F(a)+kααa+θ(b,a)ae(kα)α(sa)F(s)ds]=1θ(b,a)[eρF(a+θ(b,a))F(a)+kαkkIα(a+θ(b,a))F(a)]. (3.3)

    Also note that

    θ(b,a)210uF(a+(1v)θ(b,a))dv=θ(b,a)2[120F(a+(1v)θ(b,a))dv112F(a+(1v)θ(b,a))dv]=12[F(a+(1v)θ(b,a))|120F(a+(1v)θ(b,a))|112]=F(a)F(2a+θ(b,a)2)2F(2a+θ(b,a)2)F(a+θ(b,a))2. (3.4)

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

    Theorem 3.1. Assume that F:[a,a+θ(b,a)]RR is a differentiable function and FL[a,a+θ(b,a)] and |F| is preinvex on [a,a+θ(b,a)]. Then the following inequality for the k-fractional integrals with exponential kernel holds:

    |Rab|θ(b,a)2(12tanh(ρ4)ρ)(|F(a)|+|F(b)|).

    Proof. Using Lemma 3.1, preinvexity of |F| and increasing property of exponential function, we have

    |Rab|=|θ(b,a)210uF(a+(1v)θ(b,a))dvθ(b,a)2(1eρ)[10eρvF(a+(1v)θ(b,a))dv10eρ(1v)F(a+(1v)θ(b,a))dv]|θ(b,a)2(1eρ)[120(1eρeρv+eρ(1v))|F(a+(1v)θ(b,a))|dv112(1eρeρ(1v)+eρv)|F(a+(1v)θ(b,a))|dv]θ(b,a)2(1eρ)[120(1eρeρv+eρ(1v))(v|F(a)|+(1v)|F(b)|)dv112(1eρeρ(1v)+eρv)(v|F(a)|+(1v)|F(b)|)dv]=θ(b,a)2(1eρ)[120(1eρeρv+eρ(1v))(v|F(a)|+(1v)|F(b)|)dv120(1eρeρv+eρ(1v))((1v)|F(a)|+v|F(b)|)dv]=θ(b,a)2(1eρ)120(1eρeρv+eρ(1v))(|F(a)|+|F(b)|)dv=θ(b,a)2(1eρ)[1eρ21ρ(1eρ2)2](|F(a)|+|F(b)|)=θ(b,a)2(12tanh(ρ4)ρ)(|F(a)|+|F(b)|),

    which completes the proof.

    Lemma 3.2. Assume that F:[a,a+θ(b,a)]RR is a differentiable function and FL[a,a+θ(b,a)]. Then the following equality for the fractional integrals with exponential kernel holds:

    Lab=θ(b,a)2(1eρ)[10eρvF(a+(1v)θ(b,a))dv10eρ(1v)F(a+(1v)θ(b,a))dv], (3.5)

    where

    Lab=F(a)+F(a+θ(b,a)2kα2k(1eρ)[kIα(a)+F(a+θ(b,a))+kIα(a+θ(b,a))F(a)].

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

    Theorem 3.2. Assume that F:[a,a+θ(b,a)]RR is a differentiable function and FL[a,a+θ(b,a)]. If |F| is preinvex function on [a,a+θ(b,a)], then the following inequality for the k-fractional integrals with exponential kernel holds:

    |Lab|θ(b,a)2ρtanh(ρ4)(|F(a)|+|F(b)|).

    Proof. Using Lemma 3.2, preinvexity of |F| and increasing property of exponential function, we have

    |Lab|θ(b,a)210|eρveρ(1v)|1eρ|F(a+vθ(b,a))|dvθ(b,a)2[10|eρveρ(1v)|1eρv|F(a)|dv+10|eρveρ(1v)|1eρ(1v)|F(b)|dv]=θ(b,a)2|F(a)|[120eρveρ(1v)1eρvdv+112eρ(1v)eρv1eρvdv]+θ(b,a)2|F(b)|[120eρveρ(1v)1eρ(1v)dv+112eρ(1v)eρv1eρ(1v)dv]=θ(b,a)2ρtanh(ρ4)(|F(a)|+|F(b)|),

    which completes the proof.

    Lemma 3.3. Assume that F:[a,a+θ(b,a)]RR is twice differentiable function and FL[a,a+θ(b,a)]. Then the following equality for the k-fractional integrals with exponential kernel holds:

    Lab=θ2(b,a)2ρ(1eρ)10(1+eρeρveρ(1v))F(a+(1v)θ(b,a))dv. (3.6)

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

    10eρvF(a+(1v)θ(b,a))dv=1ρ[eρvF(a+(1v)θ(b,a))|10+θ(b,a)10eρvF(a+(1v)θ(b,a))dv]=1ρ[eρF(a)F(a+θ(b,a))+θ(b,a)10eρvF(a+(1v)θ(b,a))dv], (3.7)

    similarly

    10eρ(1v)F(a+(1v)θ(b,a))dv=1ρ[eρ(1v)F(a+(1v)θ(b,a))|10+θ(b,a)10eρ(1v)F(a+(1v)θ(b,a))dv]=1ρ[F(a)eρF(a+θ(b,a))+θ(b,a)10eρ(1v)F(a+(1v)θ(b,a))dv]. (3.8)

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

    Lab=θ(b,a)2ρ(1eρ)[(1+eρ)(F(a+θ(b,a))F(a))θ(b,a)10(eρv+eρ(1v))F(a+(1v)θ(b,a))dv]=θ2(b,a)2ρ(1eρ)10(1+eρeρveρ(1v))F(a+(1v)θ(b,a))dv,

    which completes the proof.

    Theorem 3.3. Assume that F:[a,a+θ(b,a)]RR is twice differentiable function. If FL[a,a+θ(b,a)] and |F| is preinvex on [a,a+θ(b,a)], then the following inequality for the k-fractional integrals with exponential kernel holds:

    |Lab|θ2(b,a)2ρ(1eρ)(1+eρ21eρρ)(|F(a)|+|F(b)|).

    Proof. It is to be noted that

    10(1+eρeρveρ(1v))vdv=1+eρ21eρρ, (3.9)

    and

    10(1+eρeρveρ(1v))(1v)dv=1+eρ21eρρ. (3.10)

    Using (3.6), (3.9), (3.10) and the preinvexity of |F|, we have

    Lab=|θ2(b,a)2ρ(1eρ)10(1+eρeρveρ(1v))F(a+(1v)θ(b,a))dv|θ2(b,a)2ρ(1eρ)10(1+eρeρveρ(1v))|F(a+(1v)θ(b,a))|dvθ2(b,a)2ρ(1eρ)10(1+eρeρveρ(1v))(v|F(a)|+(1v)|F(b)|)dv=θ2(b,a)2ρ(1eρ)(1+eρ21eρρ)(|F(a)|+|F(b)|),

    the proof is complete.

    Lemma 3.4. Assume that F:[a,a+θ(b,a)]RR is twice differentiable function and FL[a,a+θ(b,a)]. Then the following equality for the k-fractional integrals with exponential kernel holds:

    Rab=θ2(b,a)210h(v)F(a+(1v)θ(b,a))dv, (3.11)

    where

    h(v)={v1+eρeρveρ(1v)ρ(1eρ), for0v12,(1v)1+eρeρveρ(1v)ρ(1eρ),for12v1.

    Proof. Using (3.1), we have

    Rab=θ(b,a)210uF(a+(1v)θ(b,a))dvθ(b,a)2(1eρ)[10eρvF(a+(1v)θ(b,a))dv10eρ(1v)F(a+(1v)θ(b,a))dv],

    Thus

    θ(b,a)210uF(a+(1v)θ(b,a))dv=θ(b,a)2[120F(a+(1v)θ(b,a))dv112F(a+(1v)θ(b,a))dv]=θ(b,a)2[vF(a+(1v)θ(b,a))|120+θ(b,a)120vF(a+(1v)θ(b,a))dv]θ(b,a)2[vF(a+(1v)θ(b,a))|112+θ(b,a)112vF(a+(1v)θ(b,a))dv]=θ(b,a)2[12F(2a+θ(b,a)2)+θ(b,a)120vF(a+(1v)θ(b,a))dv]θ(b,a)2[F(a)12F(2a+θ(b,a)2)+θ(b,a)112vF(a+(1v)θ(b,a))dv]=θ(b,a)2[F(2a+θ(b,a)2)F(a)]+θ2(b,a)2120vF(a+(1v)θ(b,a))dvθ2(b,a)2112vF(a+(1v)θ(b,a))dv=θ2(b,a)2112F(a+(1v)θ(b,a))dv+θ2(b,a)2120vF(a+(1v)θ(b,a))dvθ2(b,a)2112vF(a+(1v)θ(b,a))dv=θ2(b,a)2120vF(a+(1v)θ(b,a))dv+θ2(b,a)2112(1v)F(a+(1v)θ(b,a))dv. (3.12)

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

    Theorem 3.4. Assume that F:[a,a+θ(b,a)]RR is a twice differentiable function. If FL[a,a+θ(b,a)] and |F| is preinvex on [a,a+θ(b,a)], then the following inequality for the k-fractional integrals with exponential kernel holds:

    |Rab|θ2(b,a)2(18+1+eρ2ρ(1eρ)1ρ2)(|F(a)|+|F(b)|).

    Proof. Using (3.11) and preinvexity of |F|, we have

    |Rab|=|θ2(b,a)210h(v)F(a+(1v)θ(b,a))dv|θ2(b,a)210h(v)|F(a+(1v)θ(b,a))|dvθ2(b,a)2120(v1+eρeρveρ(1v)ρ(1eρ))(v|F(a)|+(1v)|F(b)|)dv+θ2(b,a)2112(1v1+eρeρveρ(1v)ρ(1eρ))(v|F(a)|+(1v)|F(b)|)dv=θ2(b,a)2[120(v2|F(a)|+v(1v)|F(b)|)dv+112(v(1v)|F(a)|+(1v)2|F(b)|)dv]+1ρ(1eρ)10(1+eρeρveρ(1v))(v|F(a)|+(1v)|F(b)|)dv=θ2(b,a)2(18+1+eρ2ρ(1eρ)1ρ2)(|F(a)|+|F(b)|).

    The proof is complete.

    Remark 3.1. We would like to remark here that by taking k1, 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

    A(a,b):=a+b2,ab.

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

    |αA(a2,b2)+α2(1α)2[(ab)(1α)+2α]A2(a,b)|(ba)A(a,b)(12tanh(ρ14)ρ1).

    Proof. The proof directly follows from Theorem 3.1 by setting θ(b,a)=ba,k=1 and F(x)=x2.

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

    |(1α)A(a2,b2)α2(1α)2[(ab)(1α)+2α]|(ba)A(a,b)tanh(ρ14).

    Proof. The proof directly follows from Theorem 3.2 by setting θ(b,a)=ba,k=1 and F(x)=x2.

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

    |(1α)A(a2,b2)α2(1α)2[(ab)(1α)+2α]|2(ba)2A(a,b)ρ1(1eρ1)(1+eρ121eρ1ρ1).

    Proof. The proof directly follows from Theorem 3.3 by setting θ(b,a)=ba,k=1 and F(x)=x2.

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

    |αA(a2,b2)+α2(1α)2[(ab)(1α)+2α]A2(a,b)|2(ba)2A(a,b)(18+1+eρ12ρ1(1eρ1)1ρ21).

    Proof. The proof directly follows from Theorem 3.4 by setting θ(b,a)=ba,k=1 and F(x)=x2. We now discuss applications to q-digamma functions, which is defined as:

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

    χq(u)=ln(1q)+ln(q)i=0qi+u1qi+u.=ln(1q)+ln(q)i=0qiu1qiu.

    For q>1,t>0, then q-digamma function χq can be given as

    χq(u)=ln(q1)+ln(q)[u12i=0q(i+u)1q(i+u)].=ln(q1)+ln(q)[u12i=0qiu1qiu].

    From the above definition, it is clear that χq is completely monotone on (0,) for q>0. This implies that χq is convex. For more details, see [5].

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

    χq(a+b2)1α2(1eρ1)[bae1αα(bv)χq(v)dv+bae1αα(va)χq(v)dv]χq(a)+χq(b)2.

    Proof. The proof is direct consequence of Theorem 2.1, by choosing θ(b,a)=ba,k=1 and F(v)χq(v).

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

    |1α2(1eρ1)[bae1αα(bv)χq(v)dv+bae1αα(va)χq(v)dv]χq(a+b2)|ba2(12tanh(ρ14)ρ1)(|χq(a)|+|χq(b)|).

    Proof. The proof is direct consequence of Theorem 3.1, by choosing θ(b,a)=ba,k=1 and F(v)χq(v).

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

    |χq(a)+χq(b)21α2(1eρ1)[bae1αα(bv)χq(v)dv+bae1αα(va)χq(v)dv]|ba2tanh(ρ14)(|χq(a)|+|χq(b)|).

    Proof. The proof is direct consequence of Theorem 3.1, by choosing θ(b,a)=ba,k=1 and F(v)χq(v).

    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.

    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).



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