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Research article Special Issues

Nonlinear boundary value problems for a parabolic equation with an unknown source function

  • Received: 03 July 2019 Accepted: 20 August 2019 Published: 23 September 2019
  • MSC : 35K59, 35K61, 35Q35

  • We study nonlinear problems for a parabolic equation with unknown source functions. One of the problems is a system which contains the boundary value problem of the first kind and the equation for a time dependence of the sought source function. In the other problem the corresponding system is distinguished by boundary conditions. For these nonlinear systems, conditions of unique solvability in a class of smooth functions are obtained on the basis of the Rothe method. The proposed approach involves the proof of a priori estimates in the difference-continuous analogs of Holder ¨ spaces for the corresponding differential-difference nonlinear systems that approximate the original systems by the Rothe method. The considered nonlinear parabolic problems essentially differ from usual boundary value problems but have not only the theoretical interest. The present investigation is connected with the mathematical modeling of nonstationary filtration processes in porous media.

    Citation: Nataliya Gol’dman. Nonlinear boundary value problems for a parabolic equation with an unknown source function[J]. AIMS Mathematics, 2019, 4(5): 1508-1522. doi: 10.3934/math.2019.5.1508

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  • We study nonlinear problems for a parabolic equation with unknown source functions. One of the problems is a system which contains the boundary value problem of the first kind and the equation for a time dependence of the sought source function. In the other problem the corresponding system is distinguished by boundary conditions. For these nonlinear systems, conditions of unique solvability in a class of smooth functions are obtained on the basis of the Rothe method. The proposed approach involves the proof of a priori estimates in the difference-continuous analogs of Holder ¨ spaces for the corresponding differential-difference nonlinear systems that approximate the original systems by the Rothe method. The considered nonlinear parabolic problems essentially differ from usual boundary value problems but have not only the theoretical interest. The present investigation is connected with the mathematical modeling of nonstationary filtration processes in porous media.


    Let IR be an interval. Then a real-valued function f:IR is said to be convex (concave) if the inequality

    f[λx+(1λ)y]()λf(x)+(1λ)f(y)

    holds whenever x,yI and λ[0,1]. It is well-know that the convexity (concavity) has wild applications in pure and applied mathematics [1,2,3,4,5], and many inequalities [6,7,8,9,10,11,12,13,14,15,16,17,18,19] can be found in the literature via the convexity theory. Recently, the generalizations and variants for the convexity have attracted the attention of the researchers, for example, the GG- and GA-convexity [20], h-convexity [21], qausi-convexity [22], ρ-convexity [23], exponential convexity [24], harmonic convexity [25], s-convexity [26,27] and others.

    The classical Hermite-Hadamard inequality [28,29,30,31,32,33] is one of the most famous inequalities in convex function theory, which can be stated as follows:

    A real-valued function ψ:[b1,b2]R is convex if and only if

    ψ(b1+b22)1b2b1b2b1ψ(x)dxψ(b1)+ψ(b2)2. (1.1)

    If ψ is concave, then the inequalities in (1.1) remain valid in the reversed direction. The Hermite-Hadamard inequality (1.1) provides both upper and lower estimates of the integral mean of any convex function defined on a closed and bounded interval involving the endpoints and midpoints of the function's domain. Because of the excellent significance of the Hermite-Hadamard inequality, the literature is replete with ample amount of research articles dedicated to the generalizations, refinements, and extensions for the Hermite-Hadamard inequality for various families of convexity.

    Besides generalization via convexity, great effort has gone into extending (1.1) by means of fractional integral operators. Most popular of them is the Riemann-Liouville fractional integral operators given in the following definition.

    Definition 1. Let α>0, b1,b2R with b1<b2 and ψL[b1,b2]. Then the left and right Riemann-Liouville fractional integrals Jαb1+ψ and Jαb2ψ of order α are defined by

    Jαb1+ψ(x)=1Γ(α)xb1(xt)α1ψ(t)dt(x>b1)

    and

    Jαb2ψ(x)=1Γ(α)b2x(tx)α1ψ(t)dt(x<b2)

    respectively where Γ(α)=0ettα1dt is the gamma function.

    Sarikaya et al. [34] established the following fractional version of the Hermite-Hadamard inequality:

    Theorem 2. Let 0b1<b2 and ψ:[b1,b2]R be a positive convex function such that ψL[b1,b2]. Then the fractional integrals inequality

    ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]ψ(b1)+ψ(b2)2

    holds for α>0.

    In view of Theorem 2, it is pertinent to note that the positivity of the function ψ and the numbers b1 and b2 is not necessary. From Definition 1, it is clear that b1 and b2 are any real numbers such that b1<b2.

    Our main contribution in this article is to use a similar technique used in [35] to obtain Theorem 2. This time, our main result contains both sided fractional integral operators in the Riemann-Liouville sense. In this method, we use a green function and in the process, we obtain some identities involving the left and right Riemann-Liouville fractional integral operators. These identities are subsequently employed to establish some new results for the class of convex, concave and monotone functions.

    Our main result will be anchored on the succeeding lemma.

    Lemma 3. (See [36,37]) Let G be the green function defined on [b1,b2]×[b1,b2] by

    G(λ,μ)={b1μ,b1μλ;b1λ,λμb2.

    Then any ψC2([b1,b2]) can be expressed as

    ψ(x)=ψ(b1)+(xb1)ψ(b2)+b2b1G(x,μ)ψ(μ)dμ. (2.1)

    We are now in a position to frame and prove our results.

    Theorem 4. Let ψC2([b1,b2]) be a convex function. Then, for any α>0, the following fractional integral inequalities hold:

    ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]ψ(b1)+ψ(b2)2.

    Proof. Setting x=b1+b22 in (2.1), we get

    ψ(b1+b22)=ψ(b1)+(b1+b22b1)ψ(b2)+b2b1G(b1+b22,μ)ψ(μ)dμ.

    Equivalently,

    ψ(b1+b22)=ψ(b1)+(b2b12)ψ(b2)+b2b1G(b1+b22,μ)ψ(μ)dμ. (2.2)

    Using (2.1), we do the following computations:

    Jαb1+ψ(b2)=1Γ(α)b2b1(b2x)α1ψ(x)dx=1Γ(α)b2b1(b2x)α1{ψ(b1)+(xb1)ψ(b2)+b2b1G(x,μ)ψ(μ)dμ}dx=1Γ(α){ψ(b1)b2b1(b2x)α1dx+ψ(b2)b2b1(b2x)α1(xb1)dx+b2b1b2b1(b2x)α1G(x,μ)ψ(μ)dμdx}=1Γ(α)[ψ(b1)(b2x)αα|b2b1+ψ(b2){(xb1)(b2x)αα|b2b1b2b1(b2x)ααdx}+b2b1b2b1(b2x)α1G(x,μ)ψ(μ)dμdx]=1Γ(α)[ψ(b1)(b2b1)αα+ψ(b2){0(b2x)α+1α(α+1)|b2b1}+b2b1b2b1(b2x)α1G(x,μ)ψ(μ)dμdx].

    Therefore,

    Jαb1+ψ(b2)=1Γ(α)[(b2b1)ααψ(b1)+ψ(b2)(b2b1)α+1α(α+1)+b2b1b2b1(b2x)α1G(x,μ)ψ(μ)dμdx]. (2.3)

    Similarly,

    Jαb2ψ(b1)=1Γ(α)b2b1(xb1)α1ψ(x)dx=1Γ(α)b2b1(xb1)α1{ψ(b1)+(xb1)ψ(b2)+b2b1G(x,μ)ψ(μ)dμ}dx=1Γ(α){ψ(b1)b2b1(xb1)α1dx+ψ(b2)b2b1(xb1)α1(xb1)dx+b2b1b2b1(xb1)α1G(x,μ)ψ(μ)dμdx}=1Γ(α)[ψ(b1)(xb1)αα|b2b1+ψ(b2)b2b1(xb1)αdx+b2b1b2b1(xb1)α1G(x,μ)ψ(μ)dμdx]=1Γ(α)[ψ(b1)(b2b1)αα+ψ(b2)(xb1)α+1α+1|b2b1+b2b1b2b1(xb1)α1G(x,μ)ψ(μ)dμdx].

    So,

    Jαb2ψ(b1)=1Γ(α)[(b2b1)ααψ(b1)+ψ(b2)(b2b1)α+1α+1+b2b1b2b1(xb1)α1G(x,μ)ψ(μ)dμdx]. (2.4)

    Now, adding (2.3) and (2.4) and then multiplying the resultant sum by Γ(α+1)2(b2b1)α to get:

    Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]=Γ(α+1)2(b2b1)α1Γ(α)[(b2b1)ααψ(b1)+ψ(b2)(b2b1)α+1α(α+1)+b2b1b2b1(b2x)α1G(x,μ)ψ(μ)dμdx+(b2b1)ααψ(b1)+ψ(b2)(b2b1)α+1α+1+b2b1b2b1(xb1)α1G(x,μ)ψ(μ)dμdx]=α2(b2b1)α[2(b2b1)ααψ(b1)+ψ(b2)(b2b1)α+1(α+1){1α+1}+b2b1b2b1(b2x)α1G(x,μ)ψ(μ)dμdx+b2b1b2b1(xb1)α1G(x,μ)ψ(μ)dμdx]=ψ(b1)+ψ(b2)(b2b1)2+α2(b2b1)α[b2b1b2b1(b2x)α1G(x,μ)ψ(μ)dμdx+b2b1b2b1(xb1)α1G(x,μ)ψ(μ)dμdx]. (2.5)

    Subtracting (2.5) from (2.2), we obtain

    ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]=ψ(b1)+(b2b12)ψ(b2)+b2b1G(b1+b22,μ)ψ(μ)dμψ(b1)ψ(b2)(b2b1)2α2(b2b1)α[b2b1b2b1(b2x)α1G(x,μ)ψ(μ)dμdx+b2b1b2b1(xb1)α1G(x,μ)ψ(μ)dμdx]=b2b1[G(b1+b22,μ)α2(b2b1)α{b2b1(b2x)α1G(x,μ)dx+b2b1(xb1)α1G(x,μ)dx}]ψ(μ)dμ.

    By the definition of the Green function,

    G(x,μ)={b1μ,if b1μxb1x,if xμb2,

    we obtain

    b2b1(b2x)α1G(x,μ)dx=1α(α+1)[(b2μ)α+1(b2b1)α+1] (2.6)

    and

    b2b1(xb1)α1G(x,μ)dx=1α(α+1){(α+1)(b1μ)(b2b1)α+(μb1)α+1}. (2.7)

    Substituting Eqs (2.6) and (2.7) into (2.6), then we obtain

    ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]=b2b1[G(b1+b22,μ)(b2μ)α+12(α+1)(b2b1)α+b2b12(α+1)b1μ2(μb1)α+12(α+1)(b2b1)α]ψ(μ)dμ. (2.8)

    Let

    f(μ)=G(b1+b22,μ)(b2μ)α+12(α+1)(b2b1)α+b2b12(α+1)b1μ2(μb1)α+12(α+1)(b2b1)α. (2.9)

    Here,

    G(b1+b22,μ)={b1μ,b1μb1+b22;b1b22,b1+b22μb2. (2.10)

    Now, if b1μb1+b22, then from (2.9) and (2.10), we have

    f(μ)=b1μ2(b2μ)α+12(α+1)(b2b1)α+b2b12(α+1)(μb1)α+12(α+1)(b2b1)α.
    f(μ)=12+(b2μ)α2(b2b1)α(μb1)α2(b2b1)α0.

    This shows that f is decreasing and f(b1)=0 then f(μ)0 for all μ[b1,b1+b22].

    If, on the other hand, b1+b22μb2, then

    f(μ)=b1b22(b2μ)α+12(α+1)(b2b1)α+b2b12(α+1)b1μ2(μb1)α+12(α+1)(b2b1)α=μb22(b2μ)α+12(α+1)(b2b1)α+b2b12(α+1)(μb1)α+12(α+1)(b2b1)α.

    Therefore,

    f(μ)=12+(b2μ)α2(b2b1)α(μb1)α2(b2b1)α,f(μ)=α(b2μ)α12(b2b1)αα(μb1)α12(b2b1)α0,

    which shows that f is decreasing and f(b2)=0 and thus f(μ)0. Therefore, f is increasing and f(b2)=0. Hence, f(μ)0 for all μ[b1+b22,b2]. Combining the two cases discussed above, we have that

    f(μ)0for allμ[b1,b2].

    Using (2.8), we deduce the first inequality:

    ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)].

    For the right hand side of the inequality, we recall:

    ψ(x)=ψ(b1)+(xb1)ψ(b2)+b2b1G(x,μ)ψ(μ)dμ,ψ(b2)=ψ(b1)+(b2b1)ψ(b2)+b2b1G(b2,μ)ψ(μ)dμ,ψ(b1)+ψ(b2)=2ψ(b1)+(b2b1)ψ(b2)+b2b1G(b2,μ)ψ(μ)dμ,ψ(b1)+ψ(b2)2=ψ(b1)+(b2b1)2ψ(b2)+12b2b1G(b2,μ)ψ(μ)dμ. (2.11)

    Subtracting Eq (2.5) from (2.11), we get:

    ψ(b1)+ψ(b2)2Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]=ψ(b1)+(b2b1)2ψ(b2)+12b2b1G(b2,μ)ψ(μ)dμψ(b1)ψ(b2)(b2b1)2α2(b2b1)α[b2b1b2b1(b2x)α1G(x,μ)ψ(μ)dμdx+b2b1b2b1(xb1)α1G(x,μ)ψ(μ)dμdx]=12b2b1[G(b2,μ)α2(b2b1)αb2b1(b2x)α1G(x,μ)dx+b2b1(xb1)α1G(x,μ)dx]ψ(μ)dμ=12b2b1[G(b2,μ)(b2μ)α+1(α+1)(b2b1)α+b2b1(α+1)b1+μ(μb1)α+1(α+1)(b2b1)α]ψ(μ)dμ. (2.12)

    If we set

    F(μ)=G(b2,μ)(b2μ)α+1(α+1)(b2b1)α+b2b1(α+1)b1+μ(μb1)α+1(α+1)(b2b1)α, (2.13)

    then for b1μb2,

    F(μ)=b1μ(b2μ)α+1(α+1)(b2b1)α+b2b1(α+1)b1+μ(μb1)α+1(α+1)(b2b1)α=(b2μ)α+1(α+1)(b2b1)α+b2b1(α+1)(μb1)α+1(α+1)(b2b1)α=(b2b1)α+1(b2μ)α+1(μb1)α+1(α+1)(b2b1)α.

    If b1μb1+b22, then

    F(μ)=(b2μ)α(μb1)α(b2b1)α0

    which proves that F is increasing and F(b1)=0, and hence F(μ)0.

    Suppose also b1+b22μb2. Then,

    F(μ)=(b2μ)α(μb1)α(b2b1)α0.

    This implies that F is a decreasing function and F(b2)=0, and thus F(μ)0. Also, ψ(μ)0 since ψ is convex. Hence, F(μ)0 for all μ[b1,b2] and using (2.12) amounts to:

    Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]ψ(b1)+ψ(b2)2.

    That completes the proof.

    Next, we present new Hermite-Hadamard type inequalities for the class of monotone and convex function.

    Theorem 5. Let ψC2([b1,b2]) and α>0, Then the following statements are true:

    (ⅰ) If |ψ| is an increasing function, then

    |ψ(b1)+ψ(b2)2Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]||ψ(b2)|α(b2b1)22(α+1)(α+2).

    (ⅱ) If |ψ| is a decreasing function, then

    |ψ(b1)+ψ(b2)2Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]||ψ(b1)|α(b2b1)22(α+1)(α+2).

    (ⅲ) If |ψ| is a convex function, then

    |ψ(b1)+ψ(b2)2Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]|max{|ψ(b1)|,|ψ(b2)|}α(b2b1)22(α+1)(α+2).

    Proof. To prove (ⅰ), we use the following identity obtained from (2.12):

    ψ(b1)+ψ(b2)2Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]=12b2b1[G(b2,μ)(b2μ)α+1(α+1)(b2b1)α+b2b1(α+1)b1+μ(μb1)α+1(α+1)(b2b1)α]ψ(μ)dμ. (2.14)

    Taking absolute values on both sides of (2.14) and using the fact that |ψ| is an increasing function, we have

    |ψ(b1)+ψ(b2)2Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]||ψ(b2)|12b2b1{(b2μ)α+1(α+1)(b2b1)α+b2b1α+1(μb1)α+1(α+1)(b2b1)α}dμ=|ψ(b2)|12{(b2μ)α+2(α+1)(α+2)(b2b1)α|b2b1+b2b1α+1μ|b2b1(μb1)α+2(α+1)(α+2)(b2b1)α|b2b1}=|ψ(b2)|12{(b2b1)α+2(α+1)(α+2)(b2b1)α+(b2b1)2α+1(b2b1)α+2(α+1)(α+2)(b2b1)α}=|ψ(b2)|12{2(b2b1)α+2(α+1)(α+2)(b2b1)α+(b2b1)2α+1}=|ψ(b2)|α(b2b1)22(α+1)(α+2).

    Therefore,

    |ψ(b1)+ψ(b2)2Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]||ψ(b2)|α(b2b1)22(α+1)(α+2).

    This establishes the inequality in (ⅰ).

    Part (ⅱ) can be proved in a similar way. For (ⅲ), we make use of (2.13) and the fact that every convex function ψ defined on the interval [b1,b2] is bounded above by max{|ψ(b1)|,|ψ(b2)|} to get:

    |ψ(b1)+ψ(b2)2Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]|max{|ψ(b1)|,|ψ(b2)|}2(α+1)(b2b1)α|b2b1[(b2μ)α+1+(b2b1)α+1(μb1)α+1]dμ|=max{|ψ(b1)|,|ψ(b2)|}α(b2b1)22(α+1)(α+2).

    Remark 6. By setting α=1 in Theorem 5, we get the following inequalities:

    |ψ(b1)+ψ(b2)21b2b1b2b1ψ(t)dt||ψ(b2)|(b2b1)212,|ψ(b1)+ψ(b2)21b2b1b2b1ψ(t)dt||ψ(b1)|(b2b1)212,|ψ(b1)+ψ(b2)21b2b1b2b1ψ(t)dt|max{|ψ(b1)|,|ψ(b2)|}(b2b1)212.

    Theorem 7. Let ψC2([b1,b2]) and α>0. Then the following statements are true:

    (i) If |ψ| is an increasing function, then

    |ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]|(b2b1)2(α2α+2)16(α+1)(α+2)[|ψ(b1+b22)|+|ψ(b2)|].

    (ii) If |ψ| is a decreasing function, then

    |ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]|(b2b1)2(α2α+2)16(α+1)(α+2)[|ψ(b1)|+|ψ(b1+b22)|].

    (iii) If |ψ| is a convex function, then

    |ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]|(b2b1)2(α2α+2)16(α+1)(α+2)[max{|ψ(b1)|,|ψ(b1+b22)|}+{max|ψ(b1+b22)|,|ψ(b2)|}].

    Proof. The inequality in (ⅰ) is obtained by using (2.8):

    ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]=b1+b22b1[G(b1+b22,μ)(b2μ)α+12(α+1)(b2b1)α+b2b12(α+1)b1μ2(μb1)α+12(α+1)(b2b1)α]ψ(μ)dμ+b2b1+b22[G(b1+b22,μ)(b2μ)α+12(α+1)(b2b1)α+b2b12(α+1)b1μ2(μb1)α+12(α+1)(b2b1)α]ψ(μ)dμ.

    Taking absolute values and using triangle inequality, we get

    |ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]||ψ(b1+b22)||b1+b22b1{b1μ2(b2μ)α+12(α+1)(b2b1)α+b2b12(α+1)(μb1)α+12(α+1)(b2b1)α}dμ|+|ψ(b2)||b2b1+b22{μb22(b2μ)α+12(α+1)(b2b1)α+b2b12(α+1)(μb1)α+12(α+1)(b2b1)α}dμ|=|ψ(b1+b22)||{2b1μμ24|b1+b22b1+(b2μ)α+22(α+1)(α+2)(b2b1)α|b1+b22b1+(b2b1)2(α+1)μ|b1+b22b1(μb1)α+22(α+1)(α+2)(b2b1)α|b1+b22b1}|+|ψ(b2)||{μ22b2μ4|b2b1+b22+(b2μ)α+22(α+1)(α+2)(b2b1)α|b2b1+b22+b2b12(α+1)μ|b2b1+b22(μb1)α+22(α+1)(α+2)(b2b1)α|b2b1+b22}|=|ψ(b1+b22)||2b1(b1+b22)(b1+b22)242b1(b1)(b1)24+(b2b1+b22)α+22(α+1)(α+2)(b2b1)α(b2b1)α+22(α+1)(α+2)(b2b1)α+(b2b1)2(α+1)(b1+b22b1)(b1+b22b1)α+22(α+1)(α+2)(b2b1)α+(b1b1)α+22(α+1)(α+2)(b2b1)α|+|ψ(b2)||(b2)22(b2)24(b1+b22)22b2(b1+b22)4(b2b1+b22))α+22(α+1)(α+2)(b2b1)α+b2b12(α+1)(b2b1+b22)(b2b1)α+22(α+1)(α+2)(b2b1)α|
    =|ψ(b1+b22)||(b2b1)216+(b2b1)22α+3(α+1)(α+2)(b2b1)22(α+1)(α+2)+(b2b1)24(α+1)(b2b1)22α+3(α+1)(α+2)|+|ψ(b2)||(b2b1)216(b2b1)22α+3(α+1)(α+2)+(b2b1)24(α+1)(b2b1)22(α+1)(α+2)+(b2b1)22α+3(α+1)(α+2)|=|ψ(b1+b22)||(b2b1)216(b2b1)22(α+1)(α+2)+(b2b1)24(α+1)|+|ψ(b2)||(b2b1)216+(b2b1)24(α+1)(b2b1)22(α+1)(α+2)|=|ψ(b1+b22)|(b2b1)2|α2+α216(α+1)(α+2)|+|ψ(b2)|(b2b1)2|α2+α216(α+1)(α+2)|={|ψ(b1+b22)|+|ψ(b2)|}(b2b1)2[α2α+216(α+1)(α+2)]={|ψ(b1+b22)|+|ψ(b2)|}(b2b1)2(α2α+2)16(α+1)(α+2).

    The second part can be deduced by using the same procedure. For Part (ⅲ), we also employ the fact that the convex function ψ is bounded above by max{|ψ(b1)|,|ψ(b2)|} since it is defined on the interval [b1,b2]. That is, we obtain from (2.8):

    |ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+(b2)+Jαb2(b1)]|max{|ψ(b1)|+|ψ(b1+b22)|,|ψ(b1+b22)|+|ψ(b2)|}×(b2b1)2(α2α+2)16(α+1)(α+2)=[max{|ψ(b1)|,|ψ(b1+b22)|}+max{|ψ(b1+b22)|,|ψ(b2)|}]×(b2b1)2(α2α+2)16(α+1)(α+2)

    which is the desired inequality.

    Remark 8. In Theorem 7, if we take α = 1, then we obtain

    |ψ(b1+b22)1b2b1b2b1ψ(t)dt|(b2b1)248[|ψ(b1+b22)|+|ψ(b2)|],|ψ(b1+b22)1b2b1b2b1ψ(t)dt|(b2b1)248[|ψ(b1)|+|ψ(b1+b22)|],|ψ(b1+b22)1b2b1b2b1ψ(t)dt|(b2b1)248[max{|ψ(b1)|,|ψ(b1+b22)|}+max{|ψ(b1+b22)|,|ψ(b2)|}].

    Theorem 9. Assume that ψC2([b1,b2]) and |ψ| is a convex function. Then for any α>0 the following inequality holds

    |Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]ψ(b1+b22)|(b2b1)2(α2α+2)16(α+1)(α+2)[|ψ(b1)|+|ψ(b2)|].

    Proof. The inequality in (ⅰ) is obtained by using (2.8):

    Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]ψ(b1+b22)=b1+b22b1[(b2μ)α+12(α+1)(b2b1)αb2b12(α+1)+μb12+(μb1)α+12(α+1)(b2b1)α]ψ(μ)dμ+b2b1+b22[(b2μ)α+12(α+1)(b2b1)αb2b12(α+1)+b2μ2+(μb1)α+12(α+1)(b2b1)α]ψ(μ)dμ.

    Suppose μ=(1t)b1+tb2 where dμ=(b2b1)dt,

    =120[(b2(1t)b1tb2)α+12(α+1)(b2b1)αb2b12(α+1)+(1t)b1+tb2b12+((1t)b1+tb2b1)α+12(α+1)(b2b1)α]ψ((1t)b1+tb2)(b2b1)dt+112[(b2(1t)b1tb2)α+12(α+1)(b2b1)αb2b12(α+1)+b2(1t)b1tb22+((1t)b1+tb2b1)α+12(α+1)(b2b1)α]ψ((1t)b1+tb2)(b2b1)dt=(b2b1)22(α+1)[120[(1t)α+11+(α+1)t+tα+1]ψ((1t)b1+tb2)dt+112[(1t)α+1+α(1t)t+tα+1]ψ((1t)b1+tb2)dt]. (2.15)

    Taking absolute and triangular inequality,

    (b2b1)22(α+1)[120[(1t)α+11+(α+1)t+tα+1](1t)|ψ(b1)|dt+120[(1t)α+11+(α+1)t+tα+1]t|ψ(b2)|dt+112[(1t)α+1+α(1t)t+tα+1](1t)|ψ(b1)|dt+112[(1t)α+1+α(1t)t+tα+1]t|ψ(b2)|]dt]=(b2b1)22(α+1)[|ψ(b1)|{120((1t)α+2(1t)+(α+1)(tt2)+tα+1tα+2)dt+112((1t)α+2+α(1t)2(tt2)+tα+1tα+2)dt}+|ψ(b2)|{120(t(1t)α+1t+(α+1)t2+tα+2)dt+112(t(1t)α+1+α(tt2)t2+tα+2)dt}]=(b2b1)22(α+1)[|ψ(b1)|{I1}+|ψ(b2)|{I2}]. (2.16)

    Putting the values of I1 and I2 in above (2.16), we get:

    =(b2b1)2(α2α+2)16(α+1)(α+2)[|ψ(b1)|+|ψ(b2)|].

    Remark 10. In Theorem 9, if we take α=1, then we obtain

    |ψ(b1+b22)1b2b1b2b1ψ(t)dt|(b2b1)248[|ψ(b1)|+|ψ(b2)|].

    Theorem 11. Assume that ψC2([b1,b2]) and |ψ| is a convex function. Then for any α>0 the following inequality holds

    |ψ(b1)+ψ(b2)2Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]|α(b2b1)24(α+1)(α+2)[|ψ(b1)|+|ψ(b2)|].

    Proof. We start by recalling the following identity from (2.12):

    ψ(b1)+ψ(b2)2Γ(α+1)(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]=12b2b1[G(b2,μ)(b2μ)α+1(α+1)(b2b1)α+b2b1(α+1)b1+μ(μb1)α+1(α+1)(b2b1)α]ψ(μ)dμ=12b2b1[b1μ(b2μ)α+1(α+1)(b2b1)α+b2b1(α+1)b1+μ(μb1)α+1(α+1)(b2b1)α]ψ(μ)dμ=12b2b1[(b2μ)α+1(α+1)(b2b1)α+b2b1(α+1)(μb1)α+1(α+1)(b2b1)α]ψ(μ)dμ.

    If μ=(1t)b1+tb2 with t[0,1], then we obtain

    ψ(b1)+ψ(b2)2Γ(α+1)(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]=1210[(b2(1t)b1tb2)α+1(α+1)(b2b1)α+b2b1(α+1)((1t)b1+tb2b1)α+1(α+1)(b2b1)α]×ψ((1t)b1+tb2)(b2b1)dt=1210[(b2b1)α+1(1t)α+1(α+1)(b2b1)α+b2b1(α+1)tα+1(b2b1)α+1(α+1)(b2b1)α]×ψ((1t)b1+tb2)(b2b1)dt(b2b1)22(α+1)10[(1t)α+1+1tα+1][(1t)|ψ(b1)|+t|ψb2)|]dt=(b2b1)22(α+1)[|ψ(b1)|10{(1t)α+2+1ttα+1+tα+2}dt+|ψb2)|10{t(1t)α+1+ttα+2}dt].

    Taking absolute values and applying the triangle inequality amounts to the intended result.

    Remark 12. Let α=1. Then the inequality in Theorem 11 becomes:

    |ψ(b1)+ψ(b2)21b2b1b2b1ψ(t)dt|(b2b1)224[|ψ(b1)|+|ψ(b2)|].

    Next, we present results associated with the concave functions.

    Theorem 13. Let ψC2([b1,b2]) and |ψ| be a concave function. Then, for any α>0,

    |ψ(b1+b22)Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]|(b2b1)2(α2α+2)16(α+1)(α+2)[×|ψ(b1(2+2α+3(α+2)2α+3(α+2)(α+3)+2α724)+b2(2α+322α+3(α+2)(α+3)+α224)α2α+28(α+2))|+|ψ(b1(α224+2α+322α+3(α+2)(α+3))+b2(2α724+2α+3(α+2)+22α+3(α+2)(α+3))α2α+28(α+2))|].

    Proof. From (2.15), we have:

    Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]ψ(b1+b22)=(b2b1)22(α+1)[120[(1t)α+11+(α+1)t+tα+1]ψ((1t)b1+tb2)dt+112[(1t)α+1+α(1t)t+tα+1]ψ((1t)b1+tb2)dt].

    Now, we use Jensen's integral inequality to get:

    (b2b1)22(α+1)[120[(1t)α+11+(α+1)t+tα+1]dt×|ψ(120((1t)α+11+(α+1)t+tα+1)((1t)b1+tb2)dt120(1t)α+11+(α+1)t+tα+1)dt)|+112((1t)α+1+α(1t)t+tα+1)dt×|ψ(112((1t)α+1+α(1t)t+tα+1)((1t)b1+tb2)dt112((1t)α+1+α(1t)t+tα+1)dt)|].=(b2b1)22(α+1)[120[I1]dt|ψ(120(I2)dt120(I1)dt)|+112(A1)dt|ψ(112(A2)dt112(A1)dt)|]. (2.17)

    For this we calculate:

    I1=120[(1t)α+11+(α+1)t+tα+1]dt=(1t)α+2α+2|120t|120+(α+1)t22|120+tα+2α+2|120=12α+2(α+2)+1α+212+α+18+12α+2(α+2)=α2α+28(α+2),
    I2=120((1t)α+11+(α+1)t+tα+1)((1t)b1+tb2)dt=b1120((1t)α+2(1t)+(α+1)(tt2)+tα+1tα+2)dt+b2120(t(1t)α+1t+(α+1)t2+tα+2)dt=b1((1t)α+3α+3t+t22+(α+1)(t22t33)+tα+2α+2tα+3α+3)|120+b2(t(1t)α+2α+2120(1t)α+2α+2dtt22+(α+1)t33+tα+3α+3)|120=b1(2+2α+3(α+2)2α+3(α+2)(α+3)+2α724)+b2(2α+322α+3(α+2)(α+3)+α224),
    A1=112((1t)α+1+α(1t)t+tα+1)dt=(1t)α+2α+2|112+α(tt22)|112t22|112+tα+2α+2|112=α2α+28(α+2)

    and

    A2=112((1t)α+1+α(1t)t+tα+1)((1t)b1+tb2)dt=b1112((1t)α+2+α(1t)2t+t2+tα+1tα+2)dt+b2112(t(1t)α+1+α(tt2)t2+tα+2)dt=b1[(1t)α+3α+3|112α(1t)33|112t22|112+t33|112+tα+2α+2|112tα+3α+3|112]+b2[t(1t)α+2α+2112(1t)α+2(α+2)dt+α(t22t33)t33+tα+3α+3]112=b1[α224+2α+322α+3(α+2)(α+3)]+b2[2α724+2α+3(α+2)+22α+3(α+2)(α+3)].

    Putting the values of I1, I2, A1 and A2 in (2.17), we obtain:

    =(b2b1)22(α+1)[α2α+28(α+2)|ψ(b1(2+2α+3(α+2)2α+3(α+2)(α+3)+2α724)+b2(2α+322α+3(α+2)(α+3)+α224)α2α+28(α+2))|+α2α+28(α+2)|ψ(b1(α224+2α+322α+3(α+2)(α+3))+b2(2α724+2α+3(α+2)+22α+3(α+2)(α+3))α2α+28(α+2))|]=(b2b1)2(α2α+2)16(α+1)(α+2)×[|ψ(b1(2+2α+3(α+2)2α+3(α+2)(α+3)+2α724)+b2(2α+322α+3(α+2)(α+3)+α224)α2α+28(α+2))|+|ψ(b1(α224+2α+322α+3(α+2)(α+3))+b2(2α724+2α+3(α+2)+22α+3(α+2)(α+3))α2α+28(α+2))|].

    Remark 14. If we take α=1 in Theorem 13, then we get:

    |ψ(b1+b22)1b2b1b2b1ψ(t)dt|(b2b1)248[|ψ(5b1+3b28)|+|ψ(3b1+5b28)|].

    Theorem 15. Let ψC2([b1,b2]) and |ψ| be a concave function. Then, for any α>0,

    |ψ(b1)+ψ(b2)2Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]|α(b2b1)22(α+1)(α+2)|ψ(b1+b22)|.

    Proof. For this, we use (2.14) to obtain the following inequality:

    |ψ(b1)+ψ(b2)2Γ(α+1)2(b2b1)α[Jαb1+ψ(b2)+Jαb2ψ(b1)]|=(b2b1)22(α+1)10[(1t)α+1+1tα+1][ψ((1t)b1+tb2)]dt.

    We use Jensen's integral inequality to get:

    (b2b1)22(α+1){10((1t)α+1+1tα+1)dt×|ψ(10((1t)α+1+1tα+1)((1t)b1+tb2)dt10((1t)α+1+1tα+1)dt)|}
    =(b2b1)22(α+1){αα+2|ψ(b1(α2(α+2))+b2(α2(α+2))αα+2)|}=(b2b1)22(α+1){αα+2|ψ(b1+b22)|}.

    Remark 16. If we set α=1, then Theorem 15 amounts to:

    |ψ(b1)+ψ(b2)21b2b1b2b1ψ(t)dt|(b2b1)212|ψ(b1+b22)|.

    By means of a green function, we outlined a new method of proving the Hermite-Hadamard inequality involving the Riemann-Liouville fractional integral operators. In the process of doing this, some new identities were obtained. We applied these identities to prove more results in this direction. Results established in this paper can be recast by using the green function G2(λ,μ) defined on [b1,b2]×[b1,b2] by

    G2(λ,μ)={λb1,b1μλ;μb1,λμb2.

    The authors would like to express their sincere thanks to the editor and the anonymous reviewers for their helpful comments and suggestions. The work was supported by the Natural Science Foundation of China (Grant Nos. 61673169, 11301127, 11701176, 11626101, 11601485).

    There is no conflict of interest to report.



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