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:I⊆R→R be a convex function and α,β∈I with α<β, then the following inequality holds,
ϕ(α+β2)≤1β−α∫βαϕ(x)dx≤ϕ(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 I⊂R∖{0} be a real interval. A function ϕ:I→R is said to be harmonically convex, if
ϕ(xytx+(1−t)y)≤tϕ(y)+(1−t)ϕ(x) | (1.2) |
for all x,y∈I 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 I⊂R∖{0} be a real interval. A function ϕ:I→Rϵ(0<ϵ≤1) is said to be generalized harmonically convex, if
ϕ(xytx+(1−t)y)≤tϵϕ(y)+(1−t)ϵϕ(x) | (1.3) |
for all x,y∈I 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:I⊂R∖{0}→Rϵ be a generalized harmonically convex function on fractal space and α,β∈I with α<β. If ϕ(x)∈I(ϵ)x[α,β], then
1Γ(1+ϵ)ϕ(2αβα+β)≤αϵβϵ(β−α)ϵαI(ϵ)βϕ(x)x2ϵ≤Γ(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<ϵ≤1) 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,γϵ∈Rϵ, then addition and multiplication operations satisfy
(a) αϵ+βϵ∈Rϵ, αϵβϵ∈Rϵ,
(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) (α−β)ϵ=αϵ−βϵ.
Definition 3. [18,19] If there exists the relation
|ϕ(x)−ϕ(x0)|<εϵ |
with |x−x0|<δ, for ε,δ>0 and ε,δ∈R.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=limx→x0Γ(ϵ+1)(ϕ(x)−ϕ(x0))(x−x0)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)times⏞Dϵx⋅⋅⋅Dϵxϕ(x) for any x∈I⊆R, then we denote ϕ∈D(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)limΔt→0N−1∑j=0f(tj)(Δtj)ϵ, |
where α=t0<t1<⋅⋅⋅<tN−1<tN=β,[tj,tj+1] is a partition of the interval [α,β], Δtj=tj+1−tj,Δt=max{Δt0,Δt1⋅⋅⋅ΔtN−1}.
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)]|βα−αI(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+(k−1)ϵ)x(k−1)ϵ; |
1Γ(ϵ+1)∫βαxkϵ(dx)ϵ=Γ(1+kϵ)Γ(1+(k+1)ϵ)(β(k+1)ϵ−α(k+1)ϵ),k>0. |
Lemma 4. [18,30] (Generalized Hölder's inequality) Let ϕ,φ∈Cϵ[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=(β−α)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(x−1)ϵ(1−t)(y−1)ϵ(dt)ϵ,x>0,y>0, |
(2) The generalized hypergeometric function is given by
2Fϵ1(α,β;γ;z)=1Bϵ(β,γ−β)1Γ(1+ϵ)∫10t(β−1)ϵ(1−t)(γ−β−1)ϵ(1−zt)−αϵ(dt)ϵ,γ>β>0,|z|<1. |
For convenience, we use the symbol At to denote tα+(1−t)β in the following sections.
Lemma 6. Let I⊂(0,∞) be an interval, ϕ:I∘→Rϵ (I∘ is the interior of I) such that ϕ∈Dϵ(I∘) and ϕ(ϵ)∈Cϵ(α,β) for α,β∈I∘ with α<β. Then the following equality holds
ϕ(2αβα+β)−Γ(1+ϵ)αϵβϵ(β−α)ϵαI(ϵ)βϕ(x)x2ϵ=I1+I2+I3, | (3.1) |
where
I1=αϵβϵ(β−α)ϵ2ϵ1Γ(1+ϵ)∫1/20ϕ(ϵ)(αβAt)(dt)ϵ(At)2ϵ,I2=−αϵβϵ(β−α)ϵ2ϵ1Γ(1+ϵ)∫11/2ϕ(ϵ)(αβAt)(dt)ϵ(At)2ϵ,I3=−αϵβϵ(β−α)ϵ2ϵ1Γ(1+ϵ)∫10(1−2t)ϵϕ(ϵ)(αβAt)(dt)ϵ(At)2ϵ. |
Proof. Calculating I1,I2, from Lemma 1(1), we get
I1=αϵβϵ(β−α)ϵ2ϵ1Γ(1+ϵ)∫1/20ϕ(ϵ)(αβAt)(dt)ϵ(At)2ϵ=1ϵ2ϵϕ(αβAt)|1/20=1ϵ2ϵ[ϕ(2αβα+β)−ϕ(α)] |
and
I2=−αϵβϵ(β−α)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(α−β)ϵ2ϵ1Γ(1+ϵ)∫10(1−2t)ϵ(At)2ϵϕ(ϵ)(αβAt)(dt)ϵ=(2t−1)ϵ2ϵϕ(αβAt)|10−1Γ(1+ϵ)∫10Γ(1+ϵ)ϕ(αβAt)(dt)ϵ=ϕ(α)+ϕ(β)2ϵ−Γ(1+ϵ)Γ(1+ϵ)∫10ϕ(αβtα+(1−t)β)(dt)ϵ. |
Using changing variable with x=αβAt, we have
I3=ϕ(α)+ϕ(β)2ϵ−Γ(1+ϵ)(αββ−α)ϵ1Γ(1+ϵ)∫βαϕ(x)x2ϵ(dx)ϵ=ϕ(α)+ϕ(β)2ϵ−Γ(1+ϵ)αϵβϵ(β−α)ϵαI(ϵ)βϕ(x)x2ϵ. |
Adding I1−I3, the desired result is obtained. This completes the proof.
Theorem 2. Let I⊂(0,∞) be an interval, ϕ:I∘→Rϵ (I∘ is the interior of I) is an increasing function on I∘ such that ϕ∈Dϵ(I∘) and ϕ(ϵ)∈Cϵ[α,β] for α,β∈I∘ 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(β−α)ϵαI(ϵ)βϕ(x)x2ϵ|≤ϕ(β)−ϕ(α)2ϵ+αϵβϵ(β−α)ϵ2ϵ(Kϵ1(α,β))1−1q(Kϵ2(α,β)|ϕ(ϵ)(α)|q+Kϵ3(α,β)|ϕ(ϵ)(β)|q)1q, | (3.2) |
where
Kϵ1(α,β)=β−2ϵ[2Fϵ1(2,1;3;12(1−αβ))Γ(1+ϵ)Γ(1+2ϵ)+2Fϵ1(2,2;3;1−αβ)2ϵΓ(1+ϵ)Γ(1+2ϵ)−2Fϵ1(2,1;2;1−αβ)Γ(1+ϵ)],Kα2(α,β)=β−2ϵ[12ϵ2Fϵ1(2,2;4;12(1−αβ))(Γ(1+ϵ)Γ(1+2ϵ)−Γ(1+2ϵ)Γ(1+3ϵ))+2ϵ2Fϵ1(2,3;4;1−αβ)Γ(1+2ϵ)Γ(1+3ϵ)−2Fϵ1(2,2;3;1−αβ)Γ(1+ϵ)Γ(1+2ϵ)],Kϵ3(α,β)=β−2ϵ[2Fϵ1(2,1;3;12(1−αβ))Γ(1+ϵ)Γ(1+2ϵ)+2ϵ2Fϵ1(2,2;4;1−αβ)(Γ(1+ϵ)Γ(1+2ϵ)−Γ(1+2ϵ)Γ(1+3ϵ))−2Fϵ1(2,1;3;1−αβ)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(β−α)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(β−α)ϵ2ϵ1Γ(1+ϵ)∫10(1−2t)ϵϕ(ϵ)(αβAt)(dt)ϵ(At)2ϵ|≤αϵβϵ(β−α)ϵ2ϵ1Γ(1+ϵ)∫10|(1−2t)ϵA2ϵt||ϕ(ϵ)(αβAt)|(dt)ϵ=αϵβϵ(β−α)ϵ2ϵ1Γ(1+ϵ)∫10|(1−2t)ϵA2ϵt|(1−1q)+1q|ϕ(ϵ)(αβAt)|(dt)ϵ≤αϵβϵ(β−α)ϵ2ϵ[1Γ(1+ϵ)∫10|(1−2t)ϵA2ϵt|(dt)ϵ]1−1q[1Γ(1+ϵ)∫10|(1−2t)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|(1−2t)ϵA2ϵt||ϕ(ϵ)(αβAt)|q(dt)ϵ≤1Γ(1+ϵ)∫10|(1−2t)ϵA2ϵt|(tϵ|ϕ(ϵ)(β)|q+(1−t)ϵ|ϕ(ϵ)(α)|q)(dt)ϵ=(1Γ(1+ϵ)∫10|1−2t|ϵtϵA2ϵt(dt)ϵ)|ϕ(ϵ)(α)|q+(1Γ(1+ϵ)∫10|1−2t|ϵ(1−t)ϵA2ϵt(dt)ϵ)|ϕ(ϵ)(β)|q. | (3.6) |
By calculating, we get
1Γ(1+ϵ)∫10|1−2t|ϵA2ϵt(dt)ϵ=1Γ(1+ϵ)∫120(1−2t)ϵA2ϵt(dt)ϵ+1Γ(1+ϵ)∫112(2t−1)ϵA2ϵt(dt)ϵ=2ϵΓ(1+ϵ)∫120(1−2t)ϵA2ϵt(dt)ϵ+1Γ(1+ϵ)∫10(2t)ϵA2ϵt(dt)ϵ−1Γ(1+ϵ)∫101ϵA2ϵt(dt)ϵ=β−2ϵ[1Γ(1+ϵ)∫10(1−u)ϵ(1−u2(1−αβ))−2ϵ(du)ϵ+2ϵΓ(1+ϵ)∫10tϵ(1−(1−αβ)t)−2ϵ(dt)ϵ−1Γ(1+ϵ)∫10(1−(1−αβ)t)−2ϵ(dt)ϵ]=β−2ϵ[2Fϵ1(2,1;3;12(1−αβ))Bϵ(1,2)+2ϵ2Fϵ1(2,2;3;1−αβ)Bϵ(2,1)−2Fϵ1(2,1;2;1−αβ)Bϵ(1,1)]=β−2ϵ[2Fϵ1(2,1;3;12(1−αβ))Γ(1+ϵ)Γ(1+2ϵ)+2Fϵ1(2,2;3;1−αβ)2ϵΓ(1+ϵ)Γ(1+2ϵ)−2Fϵ1(2,1;2;1−αβ)Γ(1+ϵ)]=Kϵ1(α,β). | (3.7) |
Similarly, we get
1Γ(1+ϵ)∫10|1−2t|ϵtϵA2ϵt(dt)ϵ=1Γ(1+ϵ)∫120(1−2t)ϵtϵA2ϵt(dt)ϵ+1Γ(1+ϵ)∫112(2t−1)ϵtϵA2ϵt(dt)ϵ=2ϵΓ(1+ϵ)∫120tϵ(1−2t)ϵA2ϵt(dt)ϵ+1Γ(1+ϵ)∫102ϵt2ϵA2ϵt(dt)ϵ−1Γ(1+ϵ)∫10tϵA2ϵt(dt)ϵ=β−2ϵ[12ϵ1Γ(1+ϵ)∫10uϵ(1−u)ϵ(1−u2(1−αβ))−2ϵ(du)ϵ+2ϵΓ(1+ϵ)∫10t2ϵ(1−(1−αβ)t)−2ϵ(dt)ϵ−1Γ(1+ϵ)∫10tϵ(1−(1−αβ)t)−2ϵ(dt)ϵ]=β−2ϵ[12ϵ2Fϵ1(2,2;4;12(1−αβ))(Γ(1+ϵ)Γ(1+2ϵ)−Γ(1+2ϵ)Γ(1+3ϵ))+2ϵ2Fϵ1(2,3;4;1−αβ)Γ(1+2ϵ)Γ(1+3ϵ)−2Fϵ1(2,2;3;1−αβ)Γ(1+ϵ)Γ(1+2ϵ)]=Kϵ2(α,β), | (3.8) |
and
1Γ(1+ϵ)∫10|1−2t|ϵ(1−t)ϵA2ϵt(dt)ϵ=2ϵΓ(1+ϵ)∫120(1−2t)ϵ(1−t)ϵA2ϵt(dt)ϵ+1Γ(1+ϵ)∫10(2t−1)ϵ(1−t)ϵA2ϵt(dt)ϵ≤2ϵΓ(1+ϵ)∫120(1−2t)ϵA2ϵt(dt)ϵ+1Γ(1+ϵ)∫102ϵtϵ(1−t)ϵA2ϵt(dt)ϵ−1Γ(1+ϵ)∫10(1−t)ϵA2ϵt(dt)ϵ=β−2ϵ[1Γ(1+ϵ)∫10(1−u)ϵ(1−u2(1−αβ))−2ϵ(du)ϵ+2ϵΓ(1+ϵ)∫10tϵ(1−t)ϵ(1−(1−αβ)t)−2ϵ(dt)ϵ−1Γ(1+ϵ)∫10(1−t)ϵ(1−(1−αβ)t)−2ϵ(dt)ϵ]=β−2ϵ[2Fϵ1(2,1;3;12(1−αβ))Γ(1+ϵ)Γ(1+2ϵ)+2ϵ2Fϵ1(2,2;4;1−αβ)(Γ(1+ϵ)Γ(1+2ϵ)−Γ(1+2ϵ)Γ(1+3ϵ))−2Fϵ1(2,1;3;1−αβ)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:I∘→Rϵ is an increasing function on I∘ such that ϕ∈Dϵ(I∘) and ϕ(ϵ)∈Cϵ[α,β] for α,β∈I∘ 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(β−α)ϵαI(ϵ)βϕ(x)x2ϵ|≤ϕ(β)−ϕ(α)2ϵ+αϵ(β−α)ϵ2ϵβϵ[Γ(1+pϵ)Γ(1+(p+1)ϵ)]1p(Γ(1+ϵ)Γ(1+2ϵ))1q×[2Fϵ1(2q,2;3;1−αβ)|ϕ(ϵ)(α)|q+2Fϵ1(2q,1;3;1−αβ)|ϕ(ϵ)(β)|q]1q. | (3.10) |
Proof. From inequality (3.4), we have
|ϕ(2αβα+β)−Γ(1+ϵ)αϵβϵ(β−α)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(β−α)ϵ2ϵ1Γ(1+ϵ)∫10|1−2t|ϵ|1A2ϵtϕ(ϵ)(αβAt)|(dt)ϵ≤αϵβϵ(β−α)ϵ2ϵ[1Γ(1+ϵ)∫10|1−2t|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)ϵ≤1Γ(1+ϵ)∫101A2qϵt(tϵ|ϕ(ϵ)(β)|q+(1−t)ϵ|ϕ(ϵ)(α)|q)(dt)ϵ=(1Γ(1+ϵ)∫10tϵA2qϵt(dt)ϵ)|ϕ(ϵ)(β)|q+(1Γ(1+ϵ)∫10(1−t)ϵA2qϵt(dt)ϵ)|ϕ(ϵ)(α)|q. | (3.13) |
By calculating, we have
1Γ(1+ϵ)∫10tϵA2qϵt(dt)ϵ=β−2qϵ1Γ(1+ϵ)∫10tϵ(1−(1−αβ)t)−2qϵ(dt)ϵ=β−2qϵ2Fϵ1(2q,2;3;1−αβ)Bϵ(2,1),=Γ(1+ϵ)β2qϵΓ(1+2ϵ)2Fϵ1(2q,2;3;1−αβ), | (3.14) |
1Γ(1+ϵ)∫10(1−t)ϵA2qϵt(dt)ϵ=β−2qϵ1Γ(1+ϵ)∫10(1−t)ϵ(1−(1−αβ)t)−2qϵ(dt)ϵ=β−2qϵ2Fϵ1(2q,1;3;1−αβ)Bϵ(1,2),=Γ(1+ϵ)β2qϵΓ(1+2ϵ)2Fϵ1(2q,1;3;1−αβ), | (3.15) |
and
1Γ(1+ϵ)∫10|1−2t|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:I∘→Rϵ is an increasing function on I∘ such that ϕ∈Dϵ(I∘) and ϕ(ϵ)∈Cϵ[α,β] for α,β∈I∘ 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(β−α)ϵαI(ϵ)βϕ(x)x2ϵ|≤ϕ(β)−ϕ(α)2ϵ+αϵ(β−α)ϵ2ϵβϵ[2Fϵ1(2p,1;2;1−αβ)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(β−α)ϵ2ϵ1Γ(1+ϵ)∫10|1−2t|ϵ1A2ϵt|ϕ(ϵ)(αβAt)|(dt)ϵ≤αϵβϵ(β−α)ϵ2ϵ[1Γ(1+ϵ)∫101A2ϵpt(dt)ϵ]1p[1Γ(1+ϵ)∫10|1−2t|ϵq|ϕ(ϵ)(αβAt)|q(dt)ϵ]1q≤αϵβϵ(β−α)ϵ2ϵ[1Γ(1+ϵ)∫101A2ϵpt(dt)ϵ]1p×[1Γ(1+ϵ)∫10|1−2t|ϵq(tϵ|ϕ(ϵ)(β)|q+(1−t)ϵ|ϕ(ϵ)(α)|q)(dt)ϵ]1q. | (3.18) |
By calculating, we have
1Γ(1+ϵ)∫101A2pϵt(dt)ϵ=β−2pϵ1Γ(1+ϵ)∫10(1−(1−αβ)t)−2pϵ(dt)ϵ=β−2pϵ2Fϵ1(2p,1;2;1−αβ)Bϵ(1,1),=2Fϵ1(2p,1;2;1−αβ)β2pϵΓ(1+ϵ), | (3.19) |
1Γ(1+ϵ)∫10|1−2t|ϵqtϵ(dt)ϵ=1Γ(1+ϵ)∫1/20(1−2t)ϵqtϵ(dt)ϵ+1Γ(1+ϵ)∫11/2(2t−1)ϵqtϵ(dt)ϵ=Γ(1+qϵ)2ϵΓ(1+(q+1)ϵ), | (3.20) |
and
1Γ(1+ϵ)∫10|1−2t|ϵq(1−t)ϵ(dt)ϵ=1Γ(1+ϵ)∫1/20(1−2t)ϵq(1−t)ϵ(dt)ϵ+1Γ(1+ϵ)∫11/2(2t−1)ϵq(1−t)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:I∘→Rϵ is an increasing function on I∘ such that ϕ∈Dϵ(I∘) and ϕ(ϵ)∈Cϵ[α,β] for α,β∈I∘ 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(β−α)ϵαI(ϵ)βϕ(x)x2ϵ|≤ϕ(β)−ϕ(α)2ϵ+αϵ(β−α)ϵ2ϵβϵ[Γ(1+pϵ)Γ(1+(p+1)ϵ)]1p[(2Fϵ1(2p,1;p+2;1−αβ))1p+(2Fϵ1(2p,p+1;p+2;1−αβ))1p](Γ(1+ϵ)Γ(1+2ϵ))1q[|ϕ(ϵ)(α)|q+|ϕ(ϵ)(β)|q]1q. | (3.22) |
Proof. Note that (α−β)ϵ=αϵ−βϵ. From Lemma 6, using the generalized Hölder's inequality and the generalized harmonically convexity of |ϕ(ϵ)|q, we have
|I3|≤αϵβϵ(β−α)ϵ2ϵ1Γ(1+ϵ)∫10|1−2t|ϵA2ϵt|ϕ(ϵ)(αβAt)|(dt)ϵ=αϵβϵ(β−α)ϵ2ϵ1Γ(1+ϵ)∫10|(1−t)ϵ−tϵ|A2ϵt|ϕ(ϵ)(αβAt)|(dt)ϵ≤αϵβϵ(β−α)ϵ2ϵ[1Γ(1+ϵ)∫10(1−t)ϵA2ϵt|ϕ(ϵ)(αβAt)|(dt)ϵ+1Γ(1+ϵ)∫10tϵA2ϵt|ϕ(ϵ)(αβAt)|(dt)ϵ]≤αϵβϵ(β−α)ϵ2ϵ[(1Γ(1+ϵ)∫10(1−t)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(β−α)ϵ2ϵ[(1Γ(1+ϵ)∫10(1−t)ϵpA2ϵpt(dt)ϵ)1/p+(1Γ(1+ϵ)∫10tϵpA2ϵpt(dt)ϵ)1/p]×(1Γ(1+ϵ)∫10|ϕ(ϵ)(αβAt)|q(dt)ϵ)1/q≤αϵβϵ(β−α)ϵ2ϵ[(1Γ(1+ϵ)∫10(1−t)ϵpA2ϵpt(dt)ϵ)1/p+(1Γ(1+ϵ)∫10tϵpA2ϵpt(dt)ϵ)1/p]×(1Γ(1+ϵ)∫10[tϵ|ϕ(ϵ)(β)|q+(1−t)ϵ|ϕ(ϵ)(α)|q](dt)ϵ)1/q. | (3.23) |
By calculating, we have
1Γ(1+ϵ)∫10(1−t)ϵpA2ϵpt(dt)ϵ=β−2pϵ1Γ(1+ϵ)∫10(1−t)ϵp[1−(1−αβ)t]−2pϵ(dt)ϵ=β−2pϵ2Fϵ1(2p,1;p+2;1−αβ)Bϵ(1,p+1)=β−2pϵΓ(1+pϵ)Γ(1+(p+1)ϵ)2Fϵ1(2p,1;p+2;1−αβ). | (3.24) |
Similarly,
1Γ(1+ϵ)∫10tϵpA2ϵpt(dt)ϵ=β−2pϵ2Fϵ1(2p,p+1;p+2;1−αβ)Bϵ(p+1,1)=β−2pϵΓ(1+pϵ)Γ(1+(p+1)ϵ)2Fϵ1(2p,p+1;p+2;1−αβ). | (3.25) |
And
1Γ(1+ϵ)∫10tϵ(dt)ϵ=1Γ(1+ϵ)∫10(1−t)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)ϵ−α(p+1)ϵ(β−α)ϵ]1/p,p∈R∖{−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,k≥1 and q≥1. 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(β−α)ϵαI(ϵ)βϕ(x)x2ϵ=Γ(1+kϵ)Γ(1+(k+1)ϵ)Lk−1(k−1)ϵ(α,β)G2ϵ(α,β), |
ϕ(β)−ϕ(α)2ϵ=(β−α)ϵ2ϵLkkϵ(α,β). |
Proposition 1. From Theorem 2, we obtain the following inequality
|Hk+1ϵ(α,β)−Γ(1+ϵ)Lk−1(k−1)ϵ(α,β)G2ϵ(α,β)|≤(β−α)ϵΓ(1+(k+1)ϵ)2ϵΓ(1+kϵ)[Lkkϵ(α,β)+αϵβϵ(Kϵ1(α,β))1−1q(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)Lk−1(k−1)ϵ(α,β)G2ϵ(α,β)|≤(β−α)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;1−αβ)αkqϵ+2Fϵ1(2q,1;3;1−αβ)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)Lk−1(k−1)ϵ(α,β)G2ϵ(α,β)|≤(β−α)ϵΓ(1+(k+1)ϵ)2ϵΓ(1+kϵ)[Lkkϵ(α,β)+αϵβϵ(2Fϵ1(2p,1;2;1−αβ)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)Lk−1(k−1)ϵ(α,β)G2ϵ(α,β)|≤(β−α)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;1−αβ))1p+(2Fϵ1(2p,p+1;p+2;1−αβ))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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