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Research article

An effective homotopy analysis method to solve the cubic isothermal auto-catalytic chemical system

  • Received: 17 September 2017 Accepted: 31 January 2018 Published: 21 March 2018
  • We established an effective algorithm for the homotopy analysis method (HAM) to solve a cubic isothermal auto-catalytic chemical system (CIACS). Our solution comes in a rapidly convergent series where the intervals of convergence given by h-curves and to find the optimal values of h, we used the averaged residual errors. The HAM solutions are compared with the solutions obtained by Mathematica in-built numerical solver. We also show the behavior of the HAM solution.

    Citation: K. M. Saad, O. S. Iyiola, P. Agarwal. An effective homotopy analysis method to solve the cubic isothermal auto-catalytic chemical system[J]. AIMS Mathematics, 2018, 3(1): 183-194. doi: 10.3934/Math.2018.1.183

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  • We established an effective algorithm for the homotopy analysis method (HAM) to solve a cubic isothermal auto-catalytic chemical system (CIACS). Our solution comes in a rapidly convergent series where the intervals of convergence given by h-curves and to find the optimal values of h, we used the averaged residual errors. The HAM solutions are compared with the solutions obtained by Mathematica in-built numerical solver. We also show the behavior of the HAM solution.



    Let ϕ:IRR be a convex function and α,βI with α<β, then the following inequality holds,

    ϕ(α+β2)1βαβαϕ(x)dxϕ(α)+ϕ(β)2, (1.1)

    which is well known as Hermite-Hadamard's inequality [1] for convex functions. Both inequalities hold in the reversed direction if ϕ is concave.

    Convex function is an important function in mathematical analysis and has been applied in many aspects [2,3]. With the extension of the definition of convex function, Hermite-Hadamard's inequality has been deeply studied. Some improvement and generalizations for Hermite-Hadamard's inequality (1.1) can been found in the references [4,5,6,7,8,9,10,11,12].

    In [11], İşcan gave the definition of harmonically convexity as follows:

    Definition 1. Let IR{0} be a real interval. A function ϕ:IR is said to be harmonically convex, if

    ϕ(xytx+(1t)y)tϕ(y)+(1t)ϕ(x) (1.2)

    for all x,yI and t[0,1]. If the inequality in (1.2) is reversed, then ϕ is said to be harmonically concave.

    In recent years, many researchers presented many kinds of fractional calculus by different methods and explored their applications. For example, Riemann-Liouville fractional integrals and its applications in inequalities [13,14,15,16]. Recently, Yang stated the theory of local fractional calculus on Yang's fractal sets systematically in [17,18,19]. Local fractional calculus can explain the behavior of continuous but nowhere differentiable function. In view of the special advantages of local fractional calculus, more and more researchers extended their studies to Yang's fractal space, see [20,21,22,23,24,25,26,27,28,29].

    In [22], Sun introduced the definition of the generalized harmonically convex function on Yang's fractal sets as follows:

    Definition 2. Let IR{0} be a real interval. A function ϕ:IRϵ(0<ϵ1) is said to be generalized harmonically convex, if

    ϕ(xytx+(1t)y)tϵϕ(y)+(1t)ϵϕ(x) (1.3)

    for all x,yI and t[0,1]. If the inequality in (1.3) is reversed, then ϕ is said to be generalized harmonically concave. The sign ϵ represents the fractal dimension.

    Example 1. Let ϕ:(0,)Rϵ and ψ:(,0)Rϵ, then ϕ(x)=xϵ is a generalized harmonically convex function and ψ(x)=xϵ is a generalized harmonically concave function.

    The following result related to Hermite-Hadamard's inequalities holds.

    Theorem 1. [22] Let ϕ:IR{0}Rϵ be a generalized harmonically convex function on fractal space and α,βI with α<β. If ϕ(x)I(ϵ)x[α,β], then

    1Γ(1+ϵ)ϕ(2αβα+β)αϵβϵ(βα)ϵαI(ϵ)βϕ(x)x2ϵΓ(1+ϵ)Γ(1+2ϵ)[ϕ(α)+ϕ(β)]. (1.4)

    Based on the theory of local fractional calculus and the definition of the generalized harmonically convex function on Yang's fractal sets, the main aim of this paper is using a new integral identity and monotonicity of functions to establish some new Hermite-Hadamard type inequalities involving local fractional calculus.

    Let Rϵ(0<ϵ1) be ϵ-type set of the real line numbers on Yang's fractal sets, and give the following operation rules, see[17,18]. The sign ϵ represents the fractal dimension, not the exponential sign.

    If αϵ,βϵ,γϵRϵ, then addition and multiplication operations satisfy

    (a) αϵ+βϵRϵ, αϵβϵRϵ,

    (b) αϵ+βϵ=βϵ+αϵ=(α+β)ϵ=(β+α)ϵ,

    (c) αϵ+(βϵ+γϵ)=(α+β)ϵ+γϵ,

    (d) αϵβϵ=βϵαϵ=(αβ)ϵ=(βα)ϵ,

    (e) αϵ(βϵγϵ)=(αϵβϵ)γϵ,

    (f) αϵ(βϵ+γϵ)=αϵβϵ+αϵγϵ,

    (g) αϵ+0ϵ=αϵ, αϵ+(α)ϵ=0ϵ and αϵ1ϵ=1ϵαϵ=αϵ,

    (h) (αβ)ϵ=αϵβϵ.

    Definition 3. [18,19] If there exists the relation

    |ϕ(x)ϕ(x0)|<εϵ

    with |xx0|<δ, for ε,δ>0 and ε,δR.Then the function ϕ(x) is called local fractional continuous at x0. If ϕ(x) is local fractional continuous on (α,β), we denote by ϕ(x)Cϵ(α,β).

    Definition 4. [17,19] Supposing that ϕ(x)Cϵ(α,β), the local fractional derivative of ϕ(x) of order ϵ at x=x0 is defined by

    ϕ(ϵ)(x0)=dϵϕ(x)dxϵ|x=x0=limxx0Γ(ϵ+1)(ϕ(x)ϕ(x0))(xx0)ϵ.

    For any x(α,β), there exists ϕ(ϵ)(x)=D(ϵ)x, denoted by ϕ(ϵ)(x)D(ϵ)x(α,β). Dϵ(α,β) is called ϵ-local fractional derivative set. If there exits ϕ((n+1)ϵ)(x)=(n+1)timesDϵxDϵxϕ(x) for any xIR, then we denote ϕD(n+1)ϵ(I), where n=0,1,2,

    Definition 5. [17,19] Let ϕ(x)Cϵ[α,β]. The local fractional integral of function ϕ(x) of order ϵ is defined by

    αI(ϵ)βϕ(x)=1Γ(ϵ+1)βαϕ(t)(dt)ϵ=1Γ(ϵ+1)limΔt0N1j=0f(tj)(Δtj)ϵ,

    where α=t0<t1<<tN1<tN=β,[tj,tj+1] is a partition of the interval [α,β], Δtj=tj+1tj,Δt=max{Δt0,Δt1ΔtN1}.

    Note that αI(ϵ)αϕ(x)=0, and αI(ϵ)βϕ(x)=βI(ϵ)αϕ(x) if α<β. We denote ϕ(x)I(ϵ)x[α,β] if there exits αI(ϵ)xϕ(x) for any x(α,β).

    Lemma 1. [17]

    (1) Suppose that ϕ(x)=φ(ϵ)(x)Cϵ[α,β], then

    αI(ϵ)βϕ(x)=φ(β)φ(α).

    (2) (Local fractional integration by parts)

    Suppose that ϕ(x),φ(x)Dϵ(α,β), and ϕ(ϵ)(x),φ(ϵ)(x)Cϵ[α,β], then

    αI(ϵ)βϕ(x)φ(ϵ)(x)=[ϕ(x)φ(x)]|βααI(ϵ)βϕ(ϵ)(x)φ(x).

    Lemma 2. [17] Suppose that ϕ(x)Cϵ[α,β] and α<γ<β, then

    αI(ϵ)βϕ(x)=αI(ϵ)γϕ(x)+γI(ϵ)βϕ(x).

    Lemma 3. [17]

    dϵxkϵdxϵ=Γ(1+kϵ)Γ(1+(k1)ϵ)x(k1)ϵ;
    1Γ(ϵ+1)βαxkϵ(dx)ϵ=Γ(1+kϵ)Γ(1+(k+1)ϵ)(β(k+1)ϵα(k+1)ϵ),k>0.

    Lemma 4. [18,30] (Generalized Hölder's inequality) Let ϕ,φCϵ[α,β],p,q>1, with 1p+1q=1, then

    1Γ(ϵ+1)βα|ϕ(x)φ(x)|(dx)ϵ(1Γ(ϵ+1)βα|ϕ(x)|p(dx)ϵ)1/p(1Γ(ϵ+1)βα|φ(x)|q(dx)ϵ)1/q.

    Lemma 5. [17]

    αI(ϵ)β1ϵ=(βα)ϵΓ(1+ϵ).

    Let us introduce the special functions on Yang's fractal sets as follows:

    (1) The generalized Beta function is given by

    Bϵ(x,y)=1Γ(1+ϵ)10t(x1)ϵ(1t)(y1)ϵ(dt)ϵ,x>0,y>0,

    (2) The generalized hypergeometric function is given by

    2Fϵ1(α,β;γ;z)=1Bϵ(β,γβ)1Γ(1+ϵ)10t(β1)ϵ(1t)(γβ1)ϵ(1zt)αϵ(dt)ϵ,γ>β>0,|z|<1.

    For convenience, we use the symbol At to denote tα+(1t)β in the following sections.

    Lemma 6. Let I(0,) be an interval, ϕ:IRϵ (I is the interior of I) such that ϕDϵ(I) and ϕ(ϵ)Cϵ(α,β) for α,βI with α<β. Then the following equality holds

    ϕ(2αβα+β)Γ(1+ϵ)αϵβϵ(βα)ϵαI(ϵ)βϕ(x)x2ϵ=I1+I2+I3, (3.1)

    where

    I1=αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)1/20ϕ(ϵ)(αβAt)(dt)ϵ(At)2ϵ,I2=αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)11/2ϕ(ϵ)(αβAt)(dt)ϵ(At)2ϵ,I3=αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)10(12t)ϵϕ(ϵ)(αβAt)(dt)ϵ(At)2ϵ.

    Proof. Calculating I1,I2, from Lemma 1(1), we get

    I1=αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)1/20ϕ(ϵ)(αβAt)(dt)ϵ(At)2ϵ=1ϵ2ϵϕ(αβAt)|1/20=1ϵ2ϵ[ϕ(2αβα+β)ϕ(α)]

    and

    I2=αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)11/2ϕ(ϵ)(αβAt)(dt)ϵ(At)2ϵ=1ϵ2ϵϕ(αβAt)|11/2=1ϵ2ϵ[ϕ(2αβα+β)ϕ(β)].

    Calculating I3, by the local fractional integration by parts, we have

    I3=αϵβϵ(αβ)ϵ2ϵ1Γ(1+ϵ)10(12t)ϵ(At)2ϵϕ(ϵ)(αβAt)(dt)ϵ=(2t1)ϵ2ϵϕ(αβAt)|101Γ(1+ϵ)10Γ(1+ϵ)ϕ(αβAt)(dt)ϵ=ϕ(α)+ϕ(β)2ϵΓ(1+ϵ)Γ(1+ϵ)10ϕ(αβtα+(1t)β)(dt)ϵ.

    Using changing variable with x=αβAt, we have

    I3=ϕ(α)+ϕ(β)2ϵΓ(1+ϵ)(αββα)ϵ1Γ(1+ϵ)βαϕ(x)x2ϵ(dx)ϵ=ϕ(α)+ϕ(β)2ϵΓ(1+ϵ)αϵβϵ(βα)ϵαI(ϵ)βϕ(x)x2ϵ.

    Adding I1I3, the desired result is obtained. This completes the proof.

    Theorem 2. Let I(0,) be an interval, ϕ:IRϵ (I is the interior of I) is an increasing function on I such that ϕDϵ(I) and ϕ(ϵ)Cϵ[α,β] for α,βI with α<β. If |ϕ(ϵ)|q is generalized harmonically convex on [α,β] for some fixed q>1, then for all x[α,β], the following local fractional integrals inequality holds,

    |ϕ(2αβα+β)Γ(1+ϵ)αϵβϵ(βα)ϵαI(ϵ)βϕ(x)x2ϵ|ϕ(β)ϕ(α)2ϵ+αϵβϵ(βα)ϵ2ϵ(Kϵ1(α,β))11q(Kϵ2(α,β)|ϕ(ϵ)(α)|q+Kϵ3(α,β)|ϕ(ϵ)(β)|q)1q, (3.2)

    where

    Kϵ1(α,β)=β2ϵ[2Fϵ1(2,1;3;12(1αβ))Γ(1+ϵ)Γ(1+2ϵ)+2Fϵ1(2,2;3;1αβ)2ϵΓ(1+ϵ)Γ(1+2ϵ)2Fϵ1(2,1;2;1αβ)Γ(1+ϵ)],Kα2(α,β)=β2ϵ[12ϵ2Fϵ1(2,2;4;12(1αβ))(Γ(1+ϵ)Γ(1+2ϵ)Γ(1+2ϵ)Γ(1+3ϵ))+2ϵ2Fϵ1(2,3;4;1αβ)Γ(1+2ϵ)Γ(1+3ϵ)2Fϵ1(2,2;3;1αβ)Γ(1+ϵ)Γ(1+2ϵ)],Kϵ3(α,β)=β2ϵ[2Fϵ1(2,1;3;12(1αβ))Γ(1+ϵ)Γ(1+2ϵ)+2ϵ2Fϵ1(2,2;4;1αβ)(Γ(1+ϵ)Γ(1+2ϵ)Γ(1+2ϵ)Γ(1+3ϵ))2Fϵ1(2,1;3;1αβ)Γ(1+ϵ)Γ(1+2ϵ)].

    Proof. Since ϕ is an increasing function on I, and 0<α<2αβα+β<β, we can obtain

    ϕ(α)<ϕ(2αβα+β)<ϕ(β).

    From the proof of Lemma 6, we have

    |I1|+|I2|=ϕ(β)ϕ(α)2ϵ. (3.3)

    Taking modulus in equality (3.1), we obtain

    |ϕ(2αβα+β)Γ(1+ϵ)αϵβϵ(βα)ϵαI(ϵ)βϕ(x)x2ϵ||I1|+|I2|+|I3|=ϕ(β)ϕ(α)2ϵ+|I3|. (3.4)

    From Lemma 6, using the property of the modulus and the generalized Hölder's inequality, we have

    |I3|=|αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)10(12t)ϵϕ(ϵ)(αβAt)(dt)ϵ(At)2ϵ|αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)10|(12t)ϵA2ϵt||ϕ(ϵ)(αβAt)|(dt)ϵ=αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)10|(12t)ϵA2ϵt|(11q)+1q|ϕ(ϵ)(αβAt)|(dt)ϵαϵβϵ(βα)ϵ2ϵ[1Γ(1+ϵ)10|(12t)ϵA2ϵt|(dt)ϵ]11q[1Γ(1+ϵ)10|(12t)ϵA2ϵt||ϕ(ϵ)(αβAt)|q(dt)ϵ]1q. (3.5)

    Since |ϕ(ϵ)|q is generalized harmonically convex on [α,β], thus

    1Γ(1+ϵ)10|(12t)ϵA2ϵt||ϕ(ϵ)(αβAt)|q(dt)ϵ1Γ(1+ϵ)10|(12t)ϵA2ϵt|(tϵ|ϕ(ϵ)(β)|q+(1t)ϵ|ϕ(ϵ)(α)|q)(dt)ϵ=(1Γ(1+ϵ)10|12t|ϵtϵA2ϵt(dt)ϵ)|ϕ(ϵ)(α)|q+(1Γ(1+ϵ)10|12t|ϵ(1t)ϵA2ϵt(dt)ϵ)|ϕ(ϵ)(β)|q. (3.6)

    By calculating, we get

    1Γ(1+ϵ)10|12t|ϵA2ϵt(dt)ϵ=1Γ(1+ϵ)120(12t)ϵA2ϵt(dt)ϵ+1Γ(1+ϵ)112(2t1)ϵA2ϵt(dt)ϵ=2ϵΓ(1+ϵ)120(12t)ϵA2ϵt(dt)ϵ+1Γ(1+ϵ)10(2t)ϵA2ϵt(dt)ϵ1Γ(1+ϵ)101ϵA2ϵt(dt)ϵ=β2ϵ[1Γ(1+ϵ)10(1u)ϵ(1u2(1αβ))2ϵ(du)ϵ+2ϵΓ(1+ϵ)10tϵ(1(1αβ)t)2ϵ(dt)ϵ1Γ(1+ϵ)10(1(1αβ)t)2ϵ(dt)ϵ]=β2ϵ[2Fϵ1(2,1;3;12(1αβ))Bϵ(1,2)+2ϵ2Fϵ1(2,2;3;1αβ)Bϵ(2,1)2Fϵ1(2,1;2;1αβ)Bϵ(1,1)]=β2ϵ[2Fϵ1(2,1;3;12(1αβ))Γ(1+ϵ)Γ(1+2ϵ)+2Fϵ1(2,2;3;1αβ)2ϵΓ(1+ϵ)Γ(1+2ϵ)2Fϵ1(2,1;2;1αβ)Γ(1+ϵ)]=Kϵ1(α,β). (3.7)

    Similarly, we get

    1Γ(1+ϵ)10|12t|ϵtϵA2ϵt(dt)ϵ=1Γ(1+ϵ)120(12t)ϵtϵA2ϵt(dt)ϵ+1Γ(1+ϵ)112(2t1)ϵtϵA2ϵt(dt)ϵ=2ϵΓ(1+ϵ)120tϵ(12t)ϵA2ϵt(dt)ϵ+1Γ(1+ϵ)102ϵt2ϵA2ϵt(dt)ϵ1Γ(1+ϵ)10tϵA2ϵt(dt)ϵ=β2ϵ[12ϵ1Γ(1+ϵ)10uϵ(1u)ϵ(1u2(1αβ))2ϵ(du)ϵ+2ϵΓ(1+ϵ)10t2ϵ(1(1αβ)t)2ϵ(dt)ϵ1Γ(1+ϵ)10tϵ(1(1αβ)t)2ϵ(dt)ϵ]=β2ϵ[12ϵ2Fϵ1(2,2;4;12(1αβ))(Γ(1+ϵ)Γ(1+2ϵ)Γ(1+2ϵ)Γ(1+3ϵ))+2ϵ2Fϵ1(2,3;4;1αβ)Γ(1+2ϵ)Γ(1+3ϵ)2Fϵ1(2,2;3;1αβ)Γ(1+ϵ)Γ(1+2ϵ)]=Kϵ2(α,β), (3.8)

    and

    1Γ(1+ϵ)10|12t|ϵ(1t)ϵA2ϵt(dt)ϵ=2ϵΓ(1+ϵ)120(12t)ϵ(1t)ϵA2ϵt(dt)ϵ+1Γ(1+ϵ)10(2t1)ϵ(1t)ϵA2ϵt(dt)ϵ2ϵΓ(1+ϵ)120(12t)ϵA2ϵt(dt)ϵ+1Γ(1+ϵ)102ϵtϵ(1t)ϵA2ϵt(dt)ϵ1Γ(1+ϵ)10(1t)ϵA2ϵt(dt)ϵ=β2ϵ[1Γ(1+ϵ)10(1u)ϵ(1u2(1αβ))2ϵ(du)ϵ+2ϵΓ(1+ϵ)10tϵ(1t)ϵ(1(1αβ)t)2ϵ(dt)ϵ1Γ(1+ϵ)10(1t)ϵ(1(1αβ)t)2ϵ(dt)ϵ]=β2ϵ[2Fϵ1(2,1;3;12(1αβ))Γ(1+ϵ)Γ(1+2ϵ)+2ϵ2Fϵ1(2,2;4;1αβ)(Γ(1+ϵ)Γ(1+2ϵ)Γ(1+2ϵ)Γ(1+3ϵ))2Fϵ1(2,1;3;1αβ)Γ(1+ϵ)Γ(1+2ϵ)]=Kϵ3(α,β). (3.9)

    From (3.4)–(3.9), we get inequality (3.2). This completes the proof.

    Theorem 3. Let I(0,) be an interval, ϕ:IRϵ is an increasing function on I such that ϕDϵ(I) and ϕ(ϵ)Cϵ[α,β] for α,βI with α<β. If |ϕ(ϵ)|q is generalized harmonically convex on [α,β], q>1,1p+1q=1, then for all x[α,β], the following local fractional integrals inequality holds.

    |ϕ(2αβα+β)Γ(1+ϵ)αϵβϵ(βα)ϵαI(ϵ)βϕ(x)x2ϵ|ϕ(β)ϕ(α)2ϵ+αϵ(βα)ϵ2ϵβϵ[Γ(1+pϵ)Γ(1+(p+1)ϵ)]1p(Γ(1+ϵ)Γ(1+2ϵ))1q×[2Fϵ1(2q,2;3;1αβ)|ϕ(ϵ)(α)|q+2Fϵ1(2q,1;3;1αβ)|ϕ(ϵ)(β)|q]1q. (3.10)

    Proof. From inequality (3.4), we have

    |ϕ(2αβα+β)Γ(1+ϵ)αϵβϵ(βα)ϵαI(ϵ)βϕ(x)x2ϵ|ϕ(β)ϕ(α)2ϵ+|I3|. (3.11)

    From Lemma 6, using the property of the modulus and the generalized Hölder's inequality, we have

    |I3|αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)10|12t|ϵ|1A2ϵtϕ(ϵ)(αβAt)|(dt)ϵαϵβϵ(βα)ϵ2ϵ[1Γ(1+ϵ)10|12t|ϵp(dt)ϵ]1p[1Γ(1+ϵ)101A2ϵqt|ϕ(ϵ)(αβAt)|q(dt)ϵ]1q. (3.12)

    Since |ϕ(ϵ)|q is generalized harmonically convex on [α,β], we can get

    1Γ(1+ϵ)101A2ϵqt|ϕ(ϵ)(αβAt)|q(dt)ϵ1Γ(1+ϵ)101A2qϵt(tϵ|ϕ(ϵ)(β)|q+(1t)ϵ|ϕ(ϵ)(α)|q)(dt)ϵ=(1Γ(1+ϵ)10tϵA2qϵt(dt)ϵ)|ϕ(ϵ)(β)|q+(1Γ(1+ϵ)10(1t)ϵA2qϵt(dt)ϵ)|ϕ(ϵ)(α)|q. (3.13)

    By calculating, we have

    1Γ(1+ϵ)10tϵA2qϵt(dt)ϵ=β2qϵ1Γ(1+ϵ)10tϵ(1(1αβ)t)2qϵ(dt)ϵ=β2qϵ2Fϵ1(2q,2;3;1αβ)Bϵ(2,1),=Γ(1+ϵ)β2qϵΓ(1+2ϵ)2Fϵ1(2q,2;3;1αβ), (3.14)
    1Γ(1+ϵ)10(1t)ϵA2qϵt(dt)ϵ=β2qϵ1Γ(1+ϵ)10(1t)ϵ(1(1αβ)t)2qϵ(dt)ϵ=β2qϵ2Fϵ1(2q,1;3;1αβ)Bϵ(1,2),=Γ(1+ϵ)β2qϵΓ(1+2ϵ)2Fϵ1(2q,1;3;1αβ), (3.15)

    and

    1Γ(1+ϵ)10|12t|ϵp(dt)ϵ=Γ(1+pϵ)Γ(1+(p+1)ϵ). (3.16)

    Thus, combining (3.11)–(3.16), we obtain the required inequality. The proof is completed.

    Theorem 4. Let I(0,) be an interval, ϕ:IRϵ is an increasing function on I such that ϕDϵ(I) and ϕ(ϵ)Cϵ[α,β] for α,βI with α<β. If |ϕ(ϵ)|q is generalized harmonically convex on [α,β], q>1,1p+1q=1, then for all x[α,β], the following local fractional integrals inequality holds.

    |ϕ(2αβα+β)Γ(1+ϵ)αϵβϵ(βα)ϵαI(ϵ)βϕ(x)x2ϵ|ϕ(β)ϕ(α)2ϵ+αϵ(βα)ϵ2ϵβϵ[2Fϵ1(2p,1;2;1αβ)Γ(1+ϵ)]1p(Γ(1+qϵ)2ϵΓ(1+(q+1)ϵ))1q×[|ϕ(ϵ)(α)|q+|ϕ(ϵ)(β)|q]1q. (3.17)

    Proof. From Lemma 6, using the generalized Hölder's inequality and the generalized harmonically convexity of |ϕ(ϵ)|q, we have

    |I3|αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)10|12t|ϵ1A2ϵt|ϕ(ϵ)(αβAt)|(dt)ϵαϵβϵ(βα)ϵ2ϵ[1Γ(1+ϵ)101A2ϵpt(dt)ϵ]1p[1Γ(1+ϵ)10|12t|ϵq|ϕ(ϵ)(αβAt)|q(dt)ϵ]1qαϵβϵ(βα)ϵ2ϵ[1Γ(1+ϵ)101A2ϵpt(dt)ϵ]1p×[1Γ(1+ϵ)10|12t|ϵq(tϵ|ϕ(ϵ)(β)|q+(1t)ϵ|ϕ(ϵ)(α)|q)(dt)ϵ]1q. (3.18)

    By calculating, we have

    1Γ(1+ϵ)101A2pϵt(dt)ϵ=β2pϵ1Γ(1+ϵ)10(1(1αβ)t)2pϵ(dt)ϵ=β2pϵ2Fϵ1(2p,1;2;1αβ)Bϵ(1,1),=2Fϵ1(2p,1;2;1αβ)β2pϵΓ(1+ϵ), (3.19)
    1Γ(1+ϵ)10|12t|ϵqtϵ(dt)ϵ=1Γ(1+ϵ)1/20(12t)ϵqtϵ(dt)ϵ+1Γ(1+ϵ)11/2(2t1)ϵqtϵ(dt)ϵ=Γ(1+qϵ)2ϵΓ(1+(q+1)ϵ), (3.20)

    and

    1Γ(1+ϵ)10|12t|ϵq(1t)ϵ(dt)ϵ=1Γ(1+ϵ)1/20(12t)ϵq(1t)ϵ(dt)ϵ+1Γ(1+ϵ)11/2(2t1)ϵq(1t)ϵ(dt)ϵ=Γ(1+qϵ)2ϵΓ(1+(q+1)ϵ). (3.21)

    From (3.11) in Theorem 3, combining (3.18)–(3.21), we obtain the required inequality. The proof is completed.

    Theorem 5. Let I(0,) be an interval, ϕ:IRϵ is an increasing function on I such that ϕDϵ(I) and ϕ(ϵ)Cϵ[α,β] for α,βI with α<β. If |ϕ(ϵ)|q is generalized harmonically convex on [α,β], q>1,1p+1q=1, then for all x[α,β], the following local fractional integrals inequality holds.

    |ϕ(2αβα+β)Γ(1+ϵ)αϵβϵ(βα)ϵαI(ϵ)βϕ(x)x2ϵ|ϕ(β)ϕ(α)2ϵ+αϵ(βα)ϵ2ϵβϵ[Γ(1+pϵ)Γ(1+(p+1)ϵ)]1p[(2Fϵ1(2p,1;p+2;1αβ))1p+(2Fϵ1(2p,p+1;p+2;1αβ))1p](Γ(1+ϵ)Γ(1+2ϵ))1q[|ϕ(ϵ)(α)|q+|ϕ(ϵ)(β)|q]1q. (3.22)

    Proof. Note that (αβ)ϵ=αϵβϵ. From Lemma 6, using the generalized Hölder's inequality and the generalized harmonically convexity of |ϕ(ϵ)|q, we have

    |I3|αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)10|12t|ϵA2ϵt|ϕ(ϵ)(αβAt)|(dt)ϵ=αϵβϵ(βα)ϵ2ϵ1Γ(1+ϵ)10|(1t)ϵtϵ|A2ϵt|ϕ(ϵ)(αβAt)|(dt)ϵαϵβϵ(βα)ϵ2ϵ[1Γ(1+ϵ)10(1t)ϵA2ϵt|ϕ(ϵ)(αβAt)|(dt)ϵ+1Γ(1+ϵ)10tϵA2ϵt|ϕ(ϵ)(αβAt)|(dt)ϵ]αϵβϵ(βα)ϵ2ϵ[(1Γ(1+ϵ)10(1t)ϵpA2ϵpt(dt)ϵ)1/p(1Γ(1+ϵ)10|ϕ(ϵ)(αβAt)|q(dt)ϵ)1/q+(1Γ(1+ϵ)10tϵpA2ϵpt(dt)ϵ)1/p(1Γ(1+ϵ)10|ϕ(ϵ)(αβAt)|q(dt)ϵ)1/q]=αϵβϵ(βα)ϵ2ϵ[(1Γ(1+ϵ)10(1t)ϵpA2ϵpt(dt)ϵ)1/p+(1Γ(1+ϵ)10tϵpA2ϵpt(dt)ϵ)1/p]×(1Γ(1+ϵ)10|ϕ(ϵ)(αβAt)|q(dt)ϵ)1/qαϵβϵ(βα)ϵ2ϵ[(1Γ(1+ϵ)10(1t)ϵpA2ϵpt(dt)ϵ)1/p+(1Γ(1+ϵ)10tϵpA2ϵpt(dt)ϵ)1/p]×(1Γ(1+ϵ)10[tϵ|ϕ(ϵ)(β)|q+(1t)ϵ|ϕ(ϵ)(α)|q](dt)ϵ)1/q. (3.23)

    By calculating, we have

    1Γ(1+ϵ)10(1t)ϵpA2ϵpt(dt)ϵ=β2pϵ1Γ(1+ϵ)10(1t)ϵp[1(1αβ)t]2pϵ(dt)ϵ=β2pϵ2Fϵ1(2p,1;p+2;1αβ)Bϵ(1,p+1)=β2pϵΓ(1+pϵ)Γ(1+(p+1)ϵ)2Fϵ1(2p,1;p+2;1αβ). (3.24)

    Similarly,

    1Γ(1+ϵ)10tϵpA2ϵpt(dt)ϵ=β2pϵ2Fϵ1(2p,p+1;p+2;1αβ)Bϵ(p+1,1)=β2pϵΓ(1+pϵ)Γ(1+(p+1)ϵ)2Fϵ1(2p,p+1;p+2;1αβ). (3.25)

    And

    1Γ(1+ϵ)10tϵ(dt)ϵ=1Γ(1+ϵ)10(1t)ϵ(dt)ϵ=Γ(1+ϵ)Γ(1+2ϵ). (3.26)

    From (3.11) in Theorem 3, combining (3.23)–(3.26), we obtain the required inequality. The proof is completed.

    We consider the following ϵ-type generalized special means of the real line numbers αϵ,βϵ with α<β on Yang's fractal sets.

    (1) The generalized arithmetic mean

    Aϵ(α,β)=αϵ+βϵ2ϵ;

    (2) The generalized p-logarithmic mean

    Lpϵ(α,β)=[Γ(1+pϵ)Γ(1+(p+1)ϵ)β(p+1)ϵα(p+1)ϵ(βα)ϵ]1/p,pR{1,0};

    (3) The generalized geometric mean

    Gϵ(α,β)=(αϵβϵ)12;

    (4) The generalized harmonic mean

    Hϵ(α,β)=(2αβ)ϵαϵ+βϵ.

    Consider the function ϕ:(0,)Rϵ, ϕ(x)=Γ(1+kϵ)Γ(1+(k+1)ϵ)x(k+1)ϵ, x>0,k1 and q1. Because the function φ(x)=|ϕ(ϵ)(x)|q=xkqϵ is generalized convex and nondecreasing on (0,), by Proposition 3.3 in [22], the function φ(x) is generalized harmonically convex on (0,).

    Let ϕ(x)=Γ(1+kϵ)Γ(1+(k+1)ϵ)x(k+1)ϵ, x>0,k>1 and q>1. Then

    ϕ(2αβα+β)=Γ(1+kϵ)Γ(1+(k+1)ϵ)Hk+1ϵ(α,β),
    αϵβϵ(βα)ϵαI(ϵ)βϕ(x)x2ϵ=Γ(1+kϵ)Γ(1+(k+1)ϵ)Lk1(k1)ϵ(α,β)G2ϵ(α,β),
    ϕ(β)ϕ(α)2ϵ=(βα)ϵ2ϵLkkϵ(α,β).

    Proposition 1. From Theorem 2, we obtain the following inequality

    |Hk+1ϵ(α,β)Γ(1+ϵ)Lk1(k1)ϵ(α,β)G2ϵ(α,β)|(βα)ϵΓ(1+(k+1)ϵ)2ϵΓ(1+kϵ)[Lkkϵ(α,β)+αϵβϵ(Kϵ1(α,β))11q(Kϵ2(α,β)αkqϵ+Kϵ3(α,β)βkqϵ)1q],

    where Kϵ1(α,β),Kϵ2(α,β) and Kϵ3(α,β) as in Theorem 2.

    Proposition 2. From Theorem 3, we obtain the following inequality

    |Hk+1ϵ(α,β)Γ(1+ϵ)Lk1(k1)ϵ(α,β)G2ϵ(α,β)|(βα)ϵΓ(1+(k+1)ϵ)2ϵΓ(1+kϵ)[Lkkϵ(α,β)+αϵβϵ(Γ(1+pϵ)Γ(1+(p+1)ϵ))1p(Γ(1+ϵ)Γ(1+2ϵ))1q×(2Fϵ1(2q,2;3;1αβ)αkqϵ+2Fϵ1(2q,1;3;1αβ)βkqϵ)1q],

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

    Proposition 3. From Theorem 4, we obtain the following inequality

    |Hk+1ϵ(α,β)Γ(1+ϵ)Lk1(k1)ϵ(α,β)G2ϵ(α,β)|(βα)ϵΓ(1+(k+1)ϵ)2ϵΓ(1+kϵ)[Lkkϵ(α,β)+αϵβϵ(2Fϵ1(2p,1;2;1αβ)Γ(1+ϵ))1p(Γ(1+qϵ)2ϵΓ(1+(q+1)ϵ))1q×(αkqϵ+βkqϵ)1q],

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

    Proposition 4. From Theorem 5, we obtain the following inequality

    |Hk+1α(α,β)Γ(1+ϵ)Lk1(k1)ϵ(α,β)G2ϵ(α,β)|(βα)ϵΓ(1+(k+1)ϵ)2ϵΓ(1+kϵ){Lkkϵ(α,β)+αϵβϵ(Γ(1+pϵ)Γ(1+(p+1)ϵ))1p[(2Fϵ1(2p,1;p+2;1αβ))1p+(2Fϵ1(2p,p+1;p+2;1αβ))1p](Γ(1+ϵ)Γ(1+2ϵ))1q(αkqϵ+βkqϵ)1q},

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

    In this paper, the research on Hermite-Hadamard type inequalities is extended to Yang's fractal space. By using the definitions of generalized harmonically convex function and the theory of local fractional calculus, we construct some new Hermite-Hadamard type integral inequalities for monotonically increasing functions with generalized harmonically convexity. Some applications related to the special mean are established by using the obtained inequalities, which shows that our results have certain application significance. Our research may inspire more scholars to further explore Hermite-Hadamard type integral inequalities on Yang's fractal sets.

    This work is supported by the Natural Science Foundation of Hunan Province (No. 2020JJ4554) and Scientific Research Project of Hunan Provincial Education Department (No. 18B433).

    This work does not have any conflicts of interest.

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