Research article Special Issues

A collocation procedure for the numerical treatment of FitzHugh–Nagumo equation using a kind of Chebyshev polynomials

  • Received: 05 September 2024 Revised: 28 December 2024 Accepted: 09 January 2025 Published: 20 January 2025
  • MSC : 365M60, 11B39, 40A05, 34A08

  • This manuscript aims to provide numerical solutions for the FitzHugh–Nagumo (FH–N) problem. The suggested approximate solutions are spectral and may be achieved using the standard collocation technique. We introduce and utilize specific polynomials of the generalized Gegenbauer polynomials. These introduced polynomials have connections with Chebyshev polynomials. The polynomials' series representation, orthogonality property, and derivative expressions are among the new formulas developed for these polynomials. We transform these formulas to obtain their counterparts for the shifted polynomials, which serve as basis functions for the suggested approximate solutions. The convergence of the expansion is thoroughly examined. We provide several numerical tests and comparisons to confirm the applicability and accuracy of our proposed numerical algorithm.

    Citation: Waleed Mohamed Abd-Elhameed, Omar Mazen Alqubori, Ahmed Gamal Atta. A collocation procedure for the numerical treatment of FitzHugh–Nagumo equation using a kind of Chebyshev polynomials[J]. AIMS Mathematics, 2025, 10(1): 1201-1223. doi: 10.3934/math.2025057

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  • This manuscript aims to provide numerical solutions for the FitzHugh–Nagumo (FH–N) problem. The suggested approximate solutions are spectral and may be achieved using the standard collocation technique. We introduce and utilize specific polynomials of the generalized Gegenbauer polynomials. These introduced polynomials have connections with Chebyshev polynomials. The polynomials' series representation, orthogonality property, and derivative expressions are among the new formulas developed for these polynomials. We transform these formulas to obtain their counterparts for the shifted polynomials, which serve as basis functions for the suggested approximate solutions. The convergence of the expansion is thoroughly examined. We provide several numerical tests and comparisons to confirm the applicability and accuracy of our proposed numerical algorithm.



    Theory of inequalities play pivotal role in almost all branches of pure and applied mathematics. Theory of convex functions has played vital role in the development of theory of inequalities. In modern analysis many inequalities are direct consequences of the applications of convexity property of the functions. One of the most extensively as well as intensively studied inequality pertaining to convexity property of the functions is Hermite–Hadamard's inequality. This inequality provides necessary and sufficient condition for a function to be convex. It reads as: Let Φ:I=[1,2]RR be a convex function on closed interval [1,2], then

    Φ(1+22)12121Φ(τ)dτΦ(1)+Φ(2)2.

    In recent years several successful attempts have been made in obtaining novel improvements and generalizations of Hermite–Hadamard's inequality, see [1,2,3,4]. Dragomir and Pearce [5] have written a very informative monograph on Hermite–Hadamard's inequality and its applications. Interested readers can find very useful details pertaining to these inequalities. Another remarkable inequality which has played significant role in theory of inequalities is Jensen's inequality, see [6]. It reads as: Let Φ be a convex function on [1,2], then for all xi[1,2] and μi[0,1], where i=1,2,,n, we have

    Φ(ni=1μixi)ni=1μiΦ(xi).

    Following inequality is known as Jensen–Mercer's inequality in the literature:

    Φ(1+2ni=1μixi)Φ(1)+Φ(2)ni=1μiΦ(xi),

    for μi[0,1], where Φ is a convex function. For more details, see [7].

    Pavić [8] presented the generalized version of Jensen–Mercer's inequality as: Assume that Φ:[1,2]R be a convex function, where xi[1,2] are n–points. Let α,β,μi[0,1], γ[1,1] be coefficients of sums α+β+γ=ni=1μi=1, then

    Φ(α1+β2+γni=1μixi)αΦ(1)+βΦ(2)+γni=1μiΦ(xi). (1.1)

    Remark 1.1. Note that

    1) If we take α=1=β and γ=1 in (1.1), then we get Jensen–Mercer inequality.

    2) If we choose α=0=β and γ=1 in (1.1), then we obtain the well-known Jensen inequality.

    For some recent studies regarding Hermite-Hadamard-Mercer type inequalities, see [9,10].

    Fractional calculus is the branch of mathematics which deals with integrals and derivatives of any arbitrary real or complex order. The history of fractional calculus is old but in recent years it has received significant popularity and importance. This can be attributed mainly due to its great many applications in various fields of science and engineering. It provides many useful tools for solving differential equations, integral equations, and problems involving special functions of mathematical physics. Among several known forms of fractional integrals, the Riemann–Liouville fractional integral has been investigated extensively, which is defined as follows:

    Definition 1.1 ([11]). Let ΦL1[1,2] (the set of all integrable functions on [1,2]). The Riemann–Liouville integrals Jν1+Φ and Jν2Φ of order ν>0 are defined by

    Jν1+Φ(x1)=1Γ(ν)x11(x1τ)ν1Φ(τ)dτ,x1>1,

    and

    Jν2Φ(x1)=1Γ(ν)2x1(τx1)ν1Φ(τ)dτ,x1<2,

    Mubeen and Habibullah [12] introduced the notion of κ-Riemann–Liouville fractional integrals as: Let ΦL1[1,2], then

    Jν,κ1+Φ(x1)=1κΓκ(ν)x11(x1τ)νκ1Φ(τ)dτ,x1>1,Jν,κ2Φ(x1)=1κΓκ(ν)2x1(τx1)νκ1Φ(τ)dτ,x1<2,

    where Γκ(ν)=0τν1eτκκdτ,(ν)>0,κR+ is the κ–gamma function which was introduced and studied in [13].

    Sarikaya et al. [14] were the first to derive fractional analogue of Hermite–Hadamard's inequality. Since then blend of techniques both from fractional calculus and convex analysis have been used in obtaining various fractional analogues of classical inequalities. For more details, see [15,16,17,18,19,20,21,22].

    Having inspiration from the ongoing research, we will establish some new Hermite–Hadamard–Mercer type of inequalities by using κ–Riemann–Liouville fractional integrals. Moreover, we will derive two new integral identities as auxiliary results. Applying two identities as auxiliary results, we will obtain some new variants of Hermite–Hadamard–Mercer type via κ–Riemann–Liouville fractional integrals. Several special cases will be deduce in details and some know results will be recaptured as well. In order to illustrate the efficiency of our main results, some applications regarding special means of positive real numbers and error estimations for trapezoidal quadrature formula will be provide as well.

    In this section, we discuss our main results.

    Theorem 2.1. Assume that Φ:[1,2]R be a convex function. Let α,β,[0,1], γ(0,1] be coefficients of sums α+β+γ=1 and ν,κ>0, then

    Φ(α1+β2+γx1+x22)Γκ(ν+κ)2γνκ(x2x1)νκ[(Jν,κ(α1+β2+γx2)Φ)(α1+β2+γx1)+(Jν,κ(α1+β2+γx1)+Φ)(α1+β2+γx2)]αΦ(1)+βΦ(2)+γΦ(x1)+Φ(x2)2,

    holds for all x1,x2[1,2] with x1<x2.

    Proof. Consider

    Φ(α1+β2+γx11+x212)=Φ(α1+β2+γx11+α1+β2+γx212).

    Using change of variable technique, for α1+β2+γx11=τ(α1+β2+γx1)+(1τ)(α1+β2+γx2) and α1+β2+γx21=(1τ)(α1+β2+γx1)+τ(α1+β2+γx2), we have

    Φ(α1+β2+γx1+x22)12[Φ(τ(α1+β2+γx1)+(1τ)(α1+β2+γx2))+Φ(τ(α1+β2+γx2)+(1τ)(α1+β2+γx1))].

    Multiplying both side of above inequality τνκ1 and integrating with respect to τ on [0,1], we get

    Φ(α1+β2+γx1+x22)ν2κ[10τνκ1Φ(τ(α1+β2+γx1)+(1τ)(α1+β2+γx2))dτ+10τνκ1Φ(τ(α1+β2+γx2)+(1τ)(α1+β2+γx1))dτ].

    After simplify, we obtain

    Φ(α1+β2+γx1+x22)Γκ(ν+κ)2κγνκ(x2x1)νκΓκ(ν)[α1+β2+γx2α1+β2+γx1(α1+β2+γx2u)νκ1Φ(u)du+α1+β2+γx2α1+β2+γx1(u(α1+β2+γx1))νκ1Φ(u)du].

    Consequently, we have

    Φ(α1+β2+γx1+x22)Γκ(ν+κ)2γνκ(x2x1)νκ[(Jν,κ(α1+β2+γx2)Φ)(α1+β2+γx1)+(Jν,κ(α1+β2+γx1)+Φ)(α1+β2+γx2)].

    To prove second inequality, from convexity of Φ, we have

    Φ(τ(α1+β2+γx1)+(1τ)(α1+β2+γx2))τΦ(α1+β2+γx1)+(1τ)Φ(α1+β2+γx2), (2.1)

    and

    Φ(τ(α1+β2+γx2)+(1τ)(α1+β2+γx1))τΦ(α1+β2+γx2)+(1τ)Φ(α1+β2+γx1). (2.2)

    Adding inequalities (2.1) and (2.2), and then multiplying both side of above inequality by τνκ1, and integrating with respect to τ on [0,1], we get

    10τνκΦ(τ(α1+β2+γx1)+(1τ)(α1+β2+γx2))dτ+10τνκ1Φ(τ(α1+β2+γx2)+(1τ)(α1+β2+γx1))dτ2νκ[αΦ(1)+βΦ(2)+γx1+x22].

    After simple calculation, we obtain second part of our result. This completes our proof.

    Corollary 2.1. If we choose α=0=β and γ=1 in Theorem 2.1, then

    Φ(x1+x22)Γκ(ν+κ)2(x2x1)νκ[(Jν,κx2Φ)(x1)+(Jν,κx1+Φ)(x2)]Φ(x1)+Φ(x2)2,

    holds for all x1,x2[1,2] with x1<x2, see [23].

    Theorem 2.2. Assume that Φ:[1,2]R be a convex function. Let α,β[0,1], γ(0,1] be coefficients of sums α+β+γ=1 and ν,κ>0, then

    Φ(α1+β2+γx1+x22)Γκ(ν+κ)(ω+1)νκ2γνκ(x2x1)νκ[(Jν,κ(α1+β2+γx1+ωx2ω+1)+Φ)(α1+β2+γx2)+(Jν,κ(α1+β2+γωx1+x2ω+1)Φ)(α1+β2+γx1)]αΦ(1)+βΦ(2)+γΦ(x1)+Φ(x2)2,

    holds for all x1,x2[1,2] with x1<x2, and ωN.

    Proof. Since Φ is convex function, then

    Φ(α1+β2+γx11+x122)12[Φ(α1+β2+γx11)+Φ(α1+β2+γx21)].

    Using change of variable technique for x11=τω+1x1+ω+1τω+1x2 and x21=ω+1τω+1x1+τω+1x2, we have

    Φ(α1+β2+γx1+x22)12[Φ(α1+β2+γ(τω+1x1+ω+1τω+1x2))+Φ(α1+β2+γ(ω+1τω+1x1+τω+1x2))].

    Multiplying both side of above inequality τνκ1 and integrating with respect to τ on [0,1], we get

    Φ(α1+β2+γx1+x22)ν2κ[10τνκ1Φ(α1+β2+γ(τω+1x1+ω+1τω+1x2))dτ+10τνκ1Φ(α1+β2+γ(ω+1τω+1x1+τω+1x2))dτ]=ν(ω+1)νκ2κγνκ(x2x1)νκ[α1+β2+γx2α1+β2+γx1+ωx2ω+1(α1+β2+γx2u)νκ1Φ(u)du+α1+β2+γωx1+x2ω+1α1+β2+γx1(u(α1+β2+γx1))νκ1Φ(u)du]=Γκ(ν+κ)(ω+1)νκ2γνκ(x2x1)νκ[(Jν,κ(α1+β2+γx1+ωx2ω+1)+Φ)(α1+β2+γx2)+(Jν,κ(α1+β2+γωx1+x2ω+1)Φ)(α1+β2+γx1)].

    The first inequality is proved. To prove second inequality, from convexity of property of Φ, we have

    Φ(α1+β2+γ(τω+1x1+ω+1τω+1x2))αΦ(1)+βΦ(2)+γ(τω+1Φ(x1)+ω+1τω+1Φ(x2)), (2.3)

    and

    Φ(α1+β2+γ(ω+1τω+1x1+τω+1x2))αΦ(1)+βΦ(2)+γ(τω+1Φ(x2)+ω+1τω+1Φ(x1)). (2.4)

    Adding inequalities (2.3) and (2.4), multiplying both side by τνκ1, and then integrating with respect to τ on [0,1], we obtain second inequality. This completes the proof.

    Corollary 2.2. If we choose α=0=β and γ=1 in Theorem 2.2, then

    Φ(x1+x22)Γκ(ν+κ)(ω+1)νκ2(x2x1)νκ[(Jν,κ(x1+ωx2ω+1)+Φ)(x2)+(Jν,κ(ωx1+x2ω+1)Φ)(x1)]Φ(x1)+Φ(x2)2,

    holds for all x1,x2[1,2] with x1<x2, and ωN.

    In this section, we derive two new auxiliary identities, which will be used in obtaining our further results.

    Lemma 3.1. Let Φ:[1,2]R be a differentiable function on (1,2) with 1<2. If ΦL1[1,2] and α,β[0,1], γ(0,1] be coefficients of sums α+β+γ=1 and ν,κ>0, then

    Φ(α1+β2+γx1)+Φ(α1+β2+γx2)2Γκ(ν+κ)2γνκ(x2x1)νκ×[(Jα,κ(α1+β2+γx2)Φ)(α1+β2+γx1)+(Jα,κ(α1+β2+γx1)+Φ)(α1+β2+γx2)]=γ(x2x1)2[10(1τ)νκΦ(α1+β2+γ(τx1+(1τ)x2))dτ10τνκΦ(α1+β2+γ(τx1+(1τ)x2))dτ],

    holds for all x1,x2[1,2] with x1<x2.

    Proof. Consider

    I:=γ(x2x1)2[10(1τ)νκΦ(α1+β2+γ(τx1+(1τ)x2))dτ10τνκΦ(α1+β2+γ(τx1+(1τ)x2))dτ]=γ(x2x1)2[I1I2],

    where

    I1:=10(1τ)νκΦ(α1+β2+γ(τx1+(1τ)x2))dτ=(1τ)νκΦ(α1+β2+γ(τx1+(1τ)x2))γ(x2x1)|10νκγ(x2x1)10(1τ)νκ1Φ(α1+β2+γ(τx1+(1τ)x2))dτ=Φ(α1+β2+γx2)γ(x2x1)Γκ(ν+κ)γνκ+1(x2x1)νκ+1(Jν,κ(α1+β2+γx2)Φ)(α1+β2+γx1),

    and

    I2:=10τνκΦ(α1+β2+γ(τx1+(1τ)x2))dτ=τνκΦ(α1+β2+γ(τx1+(1τ)x2))γ(x2x1)|10+νκγ(x2x1)10τνκ1Φ(α1+β2+γ(τx1+(1τ)x2))dτ=Φ(α1+β2+γx1)γ(x2x1)+Γκ(ν+κ)γνκ+1(x2x1)νκ+1(Jν,κ(α1+β2+γx1)+Φ)(α1+β2+γx2).

    Substituting the values of I1 and I2 in I, we obtain our required result.

    Corollary 3.1. If we choose α=0=β and γ=1 in Lemma 3.1, then

    Φ(x1)+Φ(x2)2Γκ(ν+κ)2(x2x1)νκ[(Jν,κx2Φ)(x1)+(Jν,κx1+Φ)(x2)]=(x2x1)2[10(1τ)νκΦ(τx1+(1τ)x2)dτ10τνκΦ(τx1+(1τ)x2)dτ].

    Lemma 3.2. Let Φ:[1,2]R be a differentiable function on (1,2) with 1<2. If ΦL1[1,2] and α,β[0,1], γ(0,1] be coefficients of sums α+β+γ=1 and ν,κ>0, then

    Φ(α1+β2+γωx1+x2ω+1)+Φ(α1+β2+γx1+ωx2ω+1)ω+1Γκ(ν+κ)(ω+1)νκ1γνκ(x2x1)νκ[(Jν,κ(α1+β2+γωx1+x2ω+1)Φ)(α1+β2+γx1)+(Jν,κ(α1+β2+γx1+ωx2ω+1)+Φ)(α1+β2+γx2)]=γ(x2x1)(ω+1)2[10τνκΦ(α1+β2+γ(ω+1τω+1x1+τω+1x2))dτ10τνκΦ(α1+β2+γ(τω+1x1+ω+1τω+1x2))dτ],

    holds for all x1,x2[1,2] with x1<x2, and ωN.

    Proof. Consider

    J:=γ(x2x1)(ω+1)2[10τνκΦ(α1+β2+γ(ω+1τω+1x1+τω+1x2))dτ10τνκΦ(α1+β2+γ(τω+1x1+ω+1τω+1x2))dτ]=γ(x2x1)(ω+1)2[J1J2],

    where

    J1:=10τνκΦ(α1+β2+γ(ω+1τω+1x1+τω+1x2))dτ=(ω+1)τνκΦ(α1+β2+γ(ω+1τω+1x1+τω+1x2))γ(x2x1)|10(ω+1)νκγ(x2x1)10τνκ1Φ(α1+β2+γ(ω+1τω+1x1+τω+1x2))dτ=(ω+1)Φ(α1+β2+γωx1+x2ω+1)γ(x2x1)ν(ω+1)νκ+1κγνκ+1(x2x1)νκ+1α1+β2+γωx1+x2ω+1α1+β2+γx1(u(α1+β2+γx1))νκ1Φ(u)du=(ω+1)Φ(α1+β2+γωx1+x2ω+1)γ(x2x1)Γκ(ν+κ)(ω+1)νκ+1γνκ+1(x2x1)νκ+1(Jν,κ(α1+β2+γωx1+x2ω+1)Φ)(α1+β2+γx1),

    and

    J2:=10τνκΦ(α1+β2+γ(τω+1x1+ω+1τω+1x2))dτ=(ω+1)Φ(α1+β2+γx1+ωx2ω+1)γ(x2x1)+(ω+1)νκ+1Γκ(ν+κ)γνκ+1(x2x1)νκ+1(Jν,κ(α1+β2+γx1+ωx2ω+1)+Φ)(α1+β2+γx2).

    Substituting the values of J1 and J2 in J and multiplying both sides by γ(x2x1)(ω+1)2, we obtain our required result.

    Now we derive some new results related to Hermite-Hadamard-Mercer type inequality using Lemma 3.1 and Lemma 3.2.

    Theorem 3.1. Under the assumptions of Lemma 3.1, if |Φ| is a convex function, then

    |Φ(α1+β2+γx1)+Φ(α1+β2+γx2)2Γκ(ν+κ)2γνκ(x2x1)νκ×[(Jα,κ(α1+β2+γx2)Φ)(α1+β2+γx1)+(Jα,κ(α1+β2+γx1)+Φ)(α1+β2+γx2)]|γ(x2x1)2[(2κκ(12)νκ1ν+κ)(α|Φ(1)|+β|Φ(2)|)+κγν+κ(κ(12)νκ)(|Φ(x1)|+|Φ(x2)|)].

    Proof. Using Lemma 3.1, property of modulus, and convexity property of |Φ|, we have

    |Φ(α1+β2+γx1)+Φ(α1+β2+γx2)2Γκ(ν+κ)2γνκ(x2x1)νκ×[(Jα,κ(α1+β2+γx2)Φ)(α1+β2+γx1)+(Jα,κ(α1+β2+γx1)+Φ)(α1+β2+γx2)]|γ(x2x1)2[10|(1τ)νκτνκ||Φ(α1+β2+γ(τx1+(1τ)x2))|dτ]γ(x2x1)2[120[(1τ)νκτνκ]|Φ(α1+β2+γ(τx1+(1τ)x2))|dτ+112[τνκ(1τ)νκ]|Φ(α1+β2+γ(τx1+(1τ)x2))|dτ]γ(x2x1)2[120[(1τ)νκτνκ][α|Φ(1)|+β|Φ(2)|+γ(τ|Φ(x1)|+(1τ)|Φ(x2)|)]dτ+112[τνκ(1τ)νκ][α|Φ(1)|+β|Φ(2)|+γ(τ|Φ(x1)|+(1τ)|Φ(x2)|)]dτ].

    After simple calculations, we obtain the required result.

    Corollary 3.2. If we take α=0=β and γ=1 in Theorem 3.1, then

    |Φ(x1)+Φ(x2)2Γκ(ν+κ)2(x2x1)νκ[(Jν,κx2Φ)(x1)+(Jα,κx1+Φ)(x2)]|(x2x1)2[κν+κ(κ(12)νκ)(|Φ(x1)|+|Φ(x2)|)].

    Corollary 3.3. If we choose ν=1=κ in Theorem 3.1, then

    |Φ(α1+β2+γx1)+Φ(α1+β2+γx2)212γ(x2x1)α1+β2+γx2α1+β2+γx1Φ(u)du|γ(x2x1)2[α|Φ(1)|+β|Φ(2)|2+γ(|Φ(x1)|+|Φ(x2)|)4].

    Remark 3.1. Using Lemma 3.1, Hölder's inequality or power mean inequality, interested reader can obtain new interesting integral inequalities. We omit here their proofs.

    Theorem 3.2. Under the assumptions of Lemma 3.2, if |Φ| is a convex function, then

    |Φ(α1+β2+γωx1+x2ω+1)+Φ(α1+β2+γx1+ωx2ω+1)ω+1Γκ(ν+κ)(ω+1)νκ1γνκ(x2x1)νκ×[(Jν,κ(α1+β2+γωx1+x2ω+1)Φ)(α1+β2+γx1)+(Jν,κ(α1+β2+γx1+ωx2ω+1)+Φ)(α1+β2+γx2)]|2γ(x2x1)(ω+1)2(κν+κ)[α|Φ(1)|+β|Φ(2)|+γ|Φ(x1)|+|Φ(x2)|2].

    Proof. Using Lemma 3.2, property of modulus, and convexity property of |Φ|, we have

    |Φ(α1+β2+γωx1+x2ω+1)+Φ(α1+β2+γx1+ωx2ω+1)ω+1Γκ(ν+κ)(ω+1)νκ1γνκ(x2x1)νκ×[(Jν,κ(α1+β2+γωx1+x2ω+1)Φ)(α1+β2+γx1)+(Jν,κ(α1+β2+γx1+ωx2ω+1)+Φ)(α1+β2+γx2)]|γ(x2x1)(ω+1)210τνκ[|Φ(α1+β2+γ(ω+1τω+1x1+τω+1x2))|+|Φ(α1+β2+γ(τω+1x1+ω+1τω+1x2))|]dτγ(x2x1)(ω+1)210τνκ[α|Φ(1)|+β|Φ(2)|+γ(ω+1τω+1|Φ(x1)|+τω+1|Φ(x2)|)+α|Φ(1)|+β|Φ(2)|+γ(τω+1|Φ(x1)|+ω+1τω+1|Φ(x2)|)]dτ=2γ(x2x1)(ω+1)2(κν+κ)[α|Φ(1)|+β|Φ(2)|+γ|Φ(x1)|+|Φ(x2)|2].

    This completes the proof.

    Corollary 3.4. If we take ν=ω=κ=1 in Theorem 3.2, then

    |Φ(α1+β2+γx1+x22)1γ(x2x1)α1+β2+γx2α1+β2+γx1Φ(u)du|γ(x2x1)4[α|Φ(1)|+β|Φ(2)|+γ|Φ(x1)|+|Φ(x2)|2].

    Corollary 3.5. If we choose α=0=β and γ=1 in Theorem 3.2, then

    |Φ(ωx1+x2ω+1)+Φ(x1+ωx2ω+1)ω+1Γκ(ν+κ)(ω+1)νκ1(x2x1)νκ[(Jν,κ(ωx1+x2ω+1)Φ)(x1)+(Jν,κ(x1+ωx2ω+1)+Φ)(x2)]|(x2x1)(ω+1)2(κν+κ)[|Φ(x1)|+|Φ(x2)|].

    Theorem 3.3. Under the assumptions of Lemma 3.2, if |Φ|q is a convex function, then

    |Φ(α1+β2+γωx1+x2ω+1)+Φ(α1+β2+γx1+ωx2ω+1)ω+1Γκ(ν+κ)(ω+1)νκ1γνκ(x2x1)νκ×[(Jν,κ(α1+β2+γωx1+x2ω+1)Φ)(α1+β2+γx1)+(Jν,κ(α1+β2+γx1+ωx2ω+1)+Φ)(α1+β2+γx2)]|γ(x2x1)(ω+1)2(κνp+κ)1p[(α|Φ(1)|q+β|Φ(2)|q+γ(12(ω+1)|Φ(x1)|q+2(ω+1)12(ω+1)|Φ(x2)|q))1q+(α|Φ(1)|q+β|Φ(2)|q+γ(12(ω+1)|Φ(x2)|q+2(ω+1)12(ω+1)|Φ(x1)|q))1q],

    where 1p+1q=1 and q>1.

    Proof. Using Lemma 3.2, property of modulus, Hölder's inequality and the convexity property of |Φ|q, we have

    |Φ(α1+β2+γωx1+x2ω+1)+Φ(α1+β2+γx1+ωx2ω+1)ω+1Γκ(ν+κ)(ω+1)νκ1γνκ(x2x1)νκ×[(Jν,κ(α1+β2+γωx1+x2ω+1)Φ)(α1+β2+γx1)+(Jν,κ(α1+β2+γx1+ωx2ω+1)+Φ)(α1+β2+γx2)]|γ(x2x1)(ω+1)2(10τνpκdτ)1p[(10|Φ(α1+β2+γ(ω+1τω+1x1+τω+1x2))|qdτ)1q+(10|Φ(α1+β2+γ(τω+1x1+ω+1τω+1x2))|qdτ)1q]γ(x2x1)(ω+1)2(κνp+κ)1p[(10[α|Φ(1)|q+β|Φ(2)|q+γ(ω+1τω+1|Φ(x1)|q+τω+1|Φ(x2)|q)]dτ)1q+(10[α|Φ(1)|q+β|Φ(2)|q+γ(τω+1|Φ(x1)|q+ω+1τω+1|Φ(x2)|q)]dτ)1q]=γ(x2x1)(ω+1)2(κνp+κ)1p[(α|Φ(1)|q+β|Φ(2)|q+γ(12(ω+1)|Φ(x1)|q+2(ω+1)12(ω+1)|Φ(x2)|q))1q+(α|Φ(1)|q+β|Φ(2)|q+γ(12(ω+1)|Φ(x2)|q+2(ω+1)12(ω+1)|Φ(x1)|q))1q].

    This completes the proof.

    Corollary 3.6. If we take ν=ω=κ=1 in Theorem 3.3, then

    |Φ(α1+β2+γx1+x22)1γ(x2x1)α1+β2+γx2α1+β2+γx1Φ(u)du|γ(x2x1)4(1p+1)1p[(α|Φ(1)|q+β|Φ(2)|q+γ(14|Φ(x1)|q+34|Φ(x2)|q))1q+(α|Φ(1)|q+β|Φ(2)|q+γ(14|Φ(x2)|q+34|Φ(x1)|q))1q].

    Corollary 3.7. If we choose α=0=β and γ=1 in Theorem 3.3, then

    |Φ(ωx1+x2ω+1)+Φ(x1+ωx2ω+1)ω+1Γκ(ν+κ)(ω+1)νκ1(x2x1)νκ[(Jν,κ(ωx1+x2ω+1)Φ)(x1)+(Jν,κ(x1+ωx2ω+1)+Φ)(x2)]|(x2x1)(ω+1)2(κνp+κ)1p[(12(ω+1)|Φ(x1)|q+2(ω+1)12(ω+1)|Φ(x2)|q)1q+(12(ω+1)|Φ(x2)|q+2(ω+1)12(ω+1)|Φ(x1)|q)1q].

    Theorem 3.4. Under the assumptions of Lemma 3.2, if |Φ|q is a convex function for q1, then

    |Φ(α1+β2+γωx1+x2ω+1)+Φ(α1+β2+γx1+ωx2ω+1)ω+1Γκ(ν+κ)(ω+1)νκ1γνκ(x2x1)νκ×[(Jν,κ(α1+β2+γωx1+x2ω+1)Φ)(α1+β2+γx1)+(Jν,κ(α1+β2+γx1+ωx2ω+1)+Φ)(α1+β2+γx2)]|γ(x2x1)(ω+1)2(κν+κ)11q[(καν+κ|Φ(1)|q+κβν+κ|Φ(2)|q+γ(kω(ω+2κ)+κ2(ω+1)(ν+κ)(ν+2κ)|Φ(x1)|q+κ(ω+1)(ν+2κ)|Φ(x2)|q))1q+(καν+κ|Φ(1)|q+κβν+κ|Φ(2)|q+γ(κ(ω+1)(ν+2κ)|Φ(x1)|q+kω(ν+2κ)+κ2(ω+1)(ν+κ)(ν+2κ)|Φ(x2)|q))1q].

    Proof. Using Lemma 3.2, property of modulus, power mean inequality and the convexity property of |Φ|q, we have

    |Φ(α1+β2+γωx1+x2ω+1)+Φ(α1+β2+γx1+ωx2ω+1)ω+1Γκ(ν+κ)(ω+1)νκ1γνκ(x2x1)νκ×[(Jν,κ(α1+β2+γωx1+x2ω+1)Φ)(α1+β2+γx1)+(Jν,κ(α1+β2+γx1+ωx2ω+1)+Φ)(α1+β2+γx2)]|γ(x2x1)(ω+1)2(10τνκdτ)11q[(10τνκ|Φ(α1+β2+γ(ω+1τω+1x1+τω+1x2))|qdτ)1q+(10τνκ|Φ(α1+β2+γ(τω+1x1+ω+1τω+1x2))|qdτ)1q]γ(x2x1)(ω+1)2(κν+κ)11q[(10τνκ[α|Φ(1)|q+β|Φ(2)|q+γ(ω+1τω+1|Φ(x1)|q+τω+1|Φ(x2)|q)]dτ)1q+(10τνκ[α|Φ(1)|q+β|Φ(2)|q+γ(τω+1|Φ(x1)|q+ω+1τω+1|Φ(x2)|q)]dτ)1q]=γ(x2x1)(ω+1)2(κν+κ)11q[(καν+κ|Φ(1)|q+κβν+κ|Φ(2)|q+γ(kω(ω+2κ)+κ2(ω+1)(ν+κ)(ν+2κ)|Φ(x1)|q+κ(ω+1)(ν+2κ)|Φ(x2)|q))1q+(καν+κ|Φ(1)|q+κβν+κ|Φ(2)|q+γ(κ(ω+1)(ν+2κ)|Φ(x1)|q+kω(ν+2κ)+κ2(ω+1)(ν+κ)(ν+2κ)|Φ(x2)|q))1q].

    This completes the proof.

    Corollary 3.8. If we take ν=ω=κ=1 in Theorem 3.4, then

    |Φ(α1+β2+γx1+x22)1γ(x2x1)α1+β2+γx2α1+β2+γx1Φ(u)du|γ(x2x1)4(12)11q[(α2|Φ(1)|q+β2|Φ(2)|q+γ(13|Φ(x1)|q+16|Φ(x2)|q))1q+(α2|Φ(1)|q+β2|Φ(2)|q+γ(16|Φ(x1)|q+13|Φ(x2)|q))1q].

    Corollary 3.9. If we choose α=0=β and γ=1 in Theorem 3.4, then

    |Φ(ωx1+x2ω+1)+Φ(x1+ωx2ω+1)ω+1Γκ(ν+κ)(ω+1)νκ1(x2x1)νκ[(Jν,κ(ωx1+x2ω+1)Φ)(x1)+(Jν,κ(x1+ωx2ω+1)+Φ)(x2)]|(x2x1)(ω+1)2(κν+κ)11q[(kω(ω+2κ)+κ2(ω+1)(ν+κ)(ν+2κ)|Φ(x1)|q+κ(ω+1)(ν+2κ)|Φ(x2)|q)1q+(κ(ω+1)(ν+2κ)|Φ(x1)|q+kω(ν+2κ)+κ2(ω+1)(ν+κ)(ν+2κ)|Φ(x2)|q)1q].

    In this section, we will discuss some applications regarding our results for special means and error estimations.

    Let recall the following two special means:

    ● The arithmetic mean is defined as

    A(x1,x2):=x1+x22.

    ● The generalized log–mean is given by

    Ln(x1,x2):=[xn+12xn+11(n+1)(x2x1)]1n,nZ{1,0},

    where 0<x1<x2 are real numbers.

    Using above special means we can establish some new inequalities as follows:

    Proposition 4.1. Let x1,x2[1,2] with 0<1<2 and α,β,[0,1], γ(0,1] be coefficients of sums α+β+γ=1, then for n>1, we have

    |A((α1+β2+γx1)n+2,(α1+β2+γx2)n+2)12Ln+2n+2(α1+β2+γx1,α1+β2+γx2)|γ(n+2)(x2x1)2[A(α1n+1,β2n+1)+γ2A(xn+11,xn+12)]. (4.1)

    Proof. The proof directly follows from Theorem 3.1 applying for Φ(x)=xn+2 and ν=1=κ.

    Proposition 4.2. Let x1,x2[1,2] with 0<1<2 and α,β,[0,1], γ(0,1] be coefficients of sums α+β+γ=1, then for n>1, we have

    |(2A(α1,β2)+γA(x1,x2))n+2Ln+2n+2(α1+β2+γx1,α1+β2+γx2)|γ(n+2)(x2x1)2[A(α1n+1,β2n+1)+γ2A(xn+11,xn+12)]. (4.2)

    Proof. The proof directly follows from Theorem 3.2 applying for Φ(x)=xn+2 and ν=ω=κ=1.

    Proposition 4.3. Let x1,x2[1,2] with 0<1<2 and α,β,[0,1], γ(0,1] be coefficients of sums α+β+γ=1, then for n>1 with 1p+1q=1 and q>1, we have

    |(2A(α1,β2)+γA(x1,x2))n+2Ln+2n+2(α1+β2+γx1,α1+β2+γx2)|γ(n+2)(x2x1)4(1p+1)1p[(2A(α1q(n+1),β2q(n+1))+γ2A(xq(n+1)1,3xq(n+1)2))1q+(2A(α1q(n+1),β2q(n+1))+γ2A(3xq(n+1)1,xq(n+1)2))1q]. (4.3)

    Proof. The proof directly follows from Theorem 3.3 applying for Φ(x)=xn+2 and ν=ω=κ=1.

    Proposition 4.4. Let x1,x2[1,2] with 0<1<2 and α,β,[0,1], γ(0,1] be coefficients of sums α+β+γ=1, then for n>1 and q1, we have

    |(2A(α1,β2)+γA(x1,x2))n+2Ln+2n+2(α1+β2+γx1,α1+β2+γx2)|γ(n+2)(x2x1)4(12)11q[(A(α1q(n+1),β2q(n+1))+γ3A(2xq(n+1)1,xq(n+1)2))1q+(A(α1q(n+1),β2q(n+1))+γ3A(xq(n+1)1,2xq(n+1)2))1q]. (4.4)

    Proof. The proof directly follows from Theorem 3.4 applying for Φ(x)=xn+2 and ν=ω=κ=1.

    Remark 4.1. For suitable choices of function Φ, many other interesting inequalities regarding new special means can be derived. We omit here their proofs and the details are left to the interested reader.

    Let consider some applications of the integral inequalities obtained above, to find new error bounds for the trapezoidal quadrature formula. First, we fix three parameters α,β,[0,1], γ(0,1] such that α+β+γ=1.

    For 2>1>0, let U:1=χ0<χ1<<χn1<χn=2 be a partition of [1,2] and xi,1,xi,2[χi,χi+1] for all i=0,1,2,,n1.

    We denote, respectively,

    S(U,Φ):=γn1i=0Φ(αχi+βχi+1+γxi,1+xi,22)i,

    and

    α1+β2+γx2α1+β2+γx1Φ(u)du:=S(U,Φ)+R(U,Φ),

    where R(U,Φ) is the remainder term and i=χi+1χi.

    Using above notations, we are in position to prove the following error estimations.

    Proposition 4.5. Under the assumptions of Theorem 3.2, if we take ν=ω=κ=1, then the following inequality holds:

    |R(U,Φ)|γ4n1i=02i[α|Φ(χi)|+β|Φ(χi+1)|+γ|Φ(xi,1)|+|Φ(xi,2)|2].

    Proof. Using the Theorem 3.2 on subinterval [χi,χi+1] of closed interval [1,2] and choosing ν=ω=κ=1, for all i=0,1,2,,n1, we have

    |γΦ(αχi+βχi+1+γxi,1+xi,22)iαχi+βχi+1+γxi,2αχi+βχi+1+γxi,1Φ(u)du| (4.5)
    γ42i[α|Φ(χi)|+β|Φ(χi+1)|+γ|Φ(xi,1)|+|Φ(xi,2)|2].

    Summing inequality (4.5) over i from 0 to n1 and using the properties of the modulus, we obtain the desired inequality.

    Proposition 4.6. Under the assumptions of Theorem 3.3, if we take ν=ω=κ=1, then the following inequality holds:

    |R(U,Φ)|γ4(1p+1)1pn1i=02i[(α|Φ(χi)|q+β|Φ(χi+1)|q+γ(14|Φ(xi,1)|q+34|Φ(xi,2)|q))1q+(α|Φ(χi)|q+β|Φ(χi+1)|q+γ(14|Φ(xi,2)|q+34|Φ(xi,1)|q))1q].

    Proof. Applying the same technique as in Proposition 4.5 but using Theorem 3.3 and choosing ν=ω=κ=1.

    Proposition 4.7. Under the assumptions of Theorem 3.4, if we take ν=ω=κ=1, then the following inequality holds:

    |R(U,Φ)|γ4(12)11qn1i=02i[(α2|Φ(χi)|q+β2|Φ(χi+1)|q+γ(13|Φ(xi,1)|q+16|Φ(xi,2)|q))1q+(α2|Φ(χi)|q+β2|Φ(χi+1)|q+γ(16|Φ(xi,1)|q+13|Φ(xi,2)|q))1q].

    Proof. Applying the same technique as in Proposition 4.5 but using Theorem 3.4 and choosing ν=ω=κ=1.

    In this paper, we have established some new Hermite–Hadamard–Mercer type of inequalities by using κ–Riemann–Liouville fractional integrals. Moreover, we have derived two new integral identities as auxiliary results. From the applied identities as auxiliary results, we have obtained some new variants of Hermite–Hadamard–Mercer type via κ–Riemann–Liouville fractional integrals. Several special cases are deduced in details and some know results are recaptured as well. In order to illustrate the efficiency of our main results, some applications regarding special means of positive real numbers and error estimations for trapezoidal quadrature formula are provided as well. To the best of our knowledge these results are new in the literature. Since the class of convex functions have large applications in many mathematical areas, they can be applied to obtain several results in convex analysis, special functions, quantum mechanics, related optimization theory, mathematical inequalities and may stimulate further research in different areas of pure and applied sciences.

    Authors are thankful to the editor and the reviewer 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).

    The authors declare that they have no competing interests.



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