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Research article Special Issues

Theoretical analysis of induced MHD Sutterby fluid flow with variable thermal conductivity and thermal slip over a stretching cylinder

  • Received: 10 September 2022 Revised: 01 January 2023 Accepted: 18 January 2023 Published: 24 February 2023
  • MSC : 76A05, 76W99, 80M25, 93A30

  • In the current analysis, steady incompressible Sutterby fluid flows over a stretching cylinder are studied. The influence of variable thermal conductivity is considered in the presence of thermal slip, Darcy resistance, and sponginess. The impact of the induced magnetic field is considered to analyze the results at the cylindrical surface. The governing equations are established as partial differential equations using the boundary layer approximation. Appropriate transformations are used to convert partial differential equations into ordinary differential equations. The numerical technique, namely (bvp4c), is applied to ordinary differential equations to develop the results. The numerical results, such as heat transfer rate and skin friction, are revealed by tabular form to demonstrate the physical impact of governing factors. The physical impact of governing factors on induced magnetic hydrodynamic, velocity, and temperature profiles is presented through various graphs. The velocity function deteriorated due to the augmentation of the Sutterby fluid parameter.

    Citation: Nadeem Abbas, Wasfi Shatanawi, Taqi A. M. Shatnawi, Fady Hasan. Theoretical analysis of induced MHD Sutterby fluid flow with variable thermal conductivity and thermal slip over a stretching cylinder[J]. AIMS Mathematics, 2023, 8(5): 10146-10159. doi: 10.3934/math.2023513

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  • In the current analysis, steady incompressible Sutterby fluid flows over a stretching cylinder are studied. The influence of variable thermal conductivity is considered in the presence of thermal slip, Darcy resistance, and sponginess. The impact of the induced magnetic field is considered to analyze the results at the cylindrical surface. The governing equations are established as partial differential equations using the boundary layer approximation. Appropriate transformations are used to convert partial differential equations into ordinary differential equations. The numerical technique, namely (bvp4c), is applied to ordinary differential equations to develop the results. The numerical results, such as heat transfer rate and skin friction, are revealed by tabular form to demonstrate the physical impact of governing factors. The physical impact of governing factors on induced magnetic hydrodynamic, velocity, and temperature profiles is presented through various graphs. The velocity function deteriorated due to the augmentation of the Sutterby fluid parameter.



    For real numbers a,b,c with cN{0}, the Gaussian hypergeometric function is defined by

    F(a,b;c;x)=n=0(a,n)(b,n)(c,n)xnn!,|x|<1,

    where (a,0)=1 for a0 and (a,n) is the shifted factorial function given by

    (a,n)=a(a+1)(a+2)(a+n1)

    for nN. It is well known that the Gaussian hypergeometric function, F(a,b;c;x), has a broad range of applications, including in geometric function theory, the theory of mean values, and numerous other areas within mathematics and related disciplines. Many elementary and special functions in mathematical physics are either particular cases or limiting cases. Specifically, F(a,b;c;x) is said to be zero-balanced if c=a+b. For the case of c=a+b, as x1, Ramanujan's asymptotic formula satisfies

    F(a,b;a+b,x)=R(a,b)ln(1x)B(a,b)+O((1x)ln(1x)), (1.1)

    where

    B(a,b)=Γ(a)Γ(b)Γ(a+b)

    is the classical beta function [1] and

    R(a,b)=2γΨ(a)Ψ(b),

    here Ψ(z)=Γ(z)/Γ(z), Re(x)>0 is the psi function, and γ is the Euler–Mascheroni constant [1].

    Throughout this paper, let a[12,1), and we denote r=1r2 for r(0,1). The generalized elliptic integrals of the first and second kind are defined on (0,1) as follows [2]:

    Ka=Ka(r)=π2F(a,1a;1,r2),Ka(0)=π2,Ka(1)=, (1.2)

    and

    Ea=Ea(r)=π2F(a1,1a;1,r2),Ea(0)=π2,Ea(1)=sin(πa)2(1a). (1.3)

    Set Ka(r)=Ka(r),Ea(r)=Ea(r). Note that when a=12, Ka(r) and Ea(r) reduce to the classical complete elliptic integrals K(r) and E(r) of the first and second kind, respectively

    K(r)=π2F(12,12;1;r2),E(r)=π2F(12,12;1;r2).

    It is well known that complete elliptic integrals play a crucial role in various areas of mathematics and physics. In particular, these integrals provide a foundation for investigating numerous special functions within conformal and quasiconformal mappings, including the Grötzsch ring function, Hübner's upper bound function, and the Hersch–Pfluger distortion function[3,4]. In 2000, Anderson, et al. [5] reintroduced the generalized elliptic integrals in geometric function theory. They discovered that the generalized elliptic integral of the first kind, denoted as Ka, originates from the Schwarz–Christoffel transformation [3] of the upper half–plane onto a parallelogram and established several monotonicity theorems for Ka and Ea. The generalized Grötzsch ring function in generalized modular equations and the generalized Hübner upper bound function can also be expressed in terms of generalized elliptic integrals[6]. Recently, generalized elliptic integrals have garnered significant attention from mathematicians. A wealth of properties and inequalities for these integrals can be found in the literature. Specifically, various properties of elliptic integrals and hypergeometric functions, including monotonicity, approximation, and discrete approximation, have been investigated in [7,8,9], with sharp inequalities derived for elliptic integrals. Additionally, studies in [10,11] primarily focus on inequalities between different means, such as the Toader mean, and Hölder mean, as well as their applications in elliptic integrals.

    For r(0,1), r=1r2, it is known that the arc-length of an ellipse with semiaxes 1 and r, denoted as L(1,r), is given by L(1,r)=4E(r). Muir indicated that L(1,r) can be approximated by 2π{(1+r322)23. Later, Vuorinen conjectured the following inequality for r(0,1):

    π2(1+r322)23<E(r),

    which was subsequently proven by Barnard et al.[12].

    The Hölder mean of positive numbers x,y>0 with order sR, is defined as

    Hs(x,y)={(xs+ys2)1s,s0,xy,s=0.

    It is easy to see Hs(x,y) is strictly increasing with respect to s. Alzer and Qiu [13] established the following inequalities:

    π2Hs1(1,r)<E(r)<π2Hs2(1,r) (1.4)

    with the best constants s1=3/2 and s2=log2/log(π/2)=1.5349, see [13,14] for details.

    The generalized weighted Hölder mean of positive numbers x,y, with weight ω and order s, is defined as [14]:

    Hs(x,y;ω)={[ωxs+(1ω)ys]1s,s0,xωy1ω,s=0. (1.5)

    Wang et al. [15] proved that for r(0,1), the following inequality holds:

    π2Hs1(1,r;α)<E(r)<π2Hs2(1,r;β), (1.6)

    and the best parameters α=α(s), β=β(s) satisfy

    α(s)={12,s(,32],1η,s(32,2),(2π)s,s[2,),β(s)={1,s(,0],(2π)s,s(0,s0),12,s[s0,),

    where s0=log2log(2/π)=1.5349, η=Fs(r0)>12, Fs=1[2E(r)/π]s1rs, and r0=r0(s)(0,1) is the value such that Fs(r) is strictly increasing on (0,r0) and strictly decreasing on (r0,1)) for s(32,2).

    The extension of the inequality (1.6) to the second kind of generalized elliptic integral Ea, where a[12,1), is a natural inquiry. This paper aims to address this question. One might wonder why the parameter a is restricted to the interval [12,1) rather than (0,1). For a(0,12), our analysis has revealed a lack of the expected monotonicity in the function Fa,p(x), as defined in Theorem 3.1. This monotonicity is crucial for establishing the desired inequalities.

    To achieve our purpose, we require some more properties of generalized elliptic integrals of the first and second kind. Therefore, Section 2 will introduce several lemmas that establish these properties. Section 3 will present our main results along with their corresponding proofs. In Section 4, we establish several functional inequalities involving Ea as applications. Finally, we give the conclusion of this article.

    In this section, several key formulas and lemmas are presented to support the proof of the main results. The derivative formulas of the generalized elliptic integrals are given.

    Lemma 2.1 ([5]). For a(0,1) and r(0,1), we have

    dKadr=2(1a)(Ear2Ka)rr2,dEadr=2(1a)(KaEa)r,ddr(KaEa)=2(1a)rEar2,ddr(Ear2Ka)=2arKa.

    The following lemma provides the monotonicity of some generalized elliptic integrals with respect to r, which can be found in [16].

    Lemma 2.2. Let a(0,1). Then the following function:

    (1) rEar2Kar2 is increasing from (0,1) to (πa2,sin(πa)2(1a));

    (2) rEar2Kar2Ka is decreasing from (0,1) to (0,a);

    (3) rr2(KaEa)r2Ea is decreasing from (0,1) to (0,1a);

    (4) rKaEa)r2Ka is increasing from (0,1) to (1a,1);

    (5) rrc(KaEa)r2 is decreasing on (0,1) if and only if ca(2a).

    Lemmas 2.3 and 2.4 are important tools for proving the monotonicity of the related functions.

    Lemma 2.3 ([17]). Let α(x)=n=0anxn and β(x)=n=obnxn be real power series that converge on (r,r)(r>0), and bn>0 for all n. If the sequence {anbn}n0 is increasing (or decreasing) on (0,r), then so is α(x)β(x).

    Lemma 2.4 ([3]). For a,b(,) and a<b, let f,g:[a,b] be continuous on [a,b] and be differentiable on (a,b), and g(x)0 for all x(a,b). If f(x)g(x) is increasing (or decreasing) on (a,b), then so are

    f(x)f(a)g(x)g(a)andf(x)f(b)g(x)g(b).

    In particular, if f(a)=g(a)=0(orf(b)=g(b)=0), then the monotonicity of f(x)g(x) is the same as f(x)g(x).

    However, f(x)g(x) is not always monotonic; it is sometimes piecewise monotonic. An auxiliary function Hf,g [8] is defined as

    Hf,g:=fggf, (2.1)

    where f and g are differentiable on (a,b) and g0 on (a,b) for <a<b<. If f and g are twice differentiable on (a,b), the function Hf,g satisfies the following identities:

    (fg)=fgfgg2=gg2(fggf)=gg2Hf,g, (2.2)
    Hf,g=(fg)g. (2.3)

    Here, Hf,g establishes a connection between fg and fg.

    Lemma 2.5. Define the function f1(x) on [12,1) by

    f1(x)=2(1x)logxlog(sin(πx)/(π(1x))).

    Then 2x<f1(x)<2.

    Proof. To establish the right-hand side of the inequality, it suffices to prove that

    (1x)logxlogsin(πx)π(1x)>0.

    Denote

    g1(x)=(1x)logxlogsin(πx)π(1x).

    By differentiation, we obtain

    g1(x)=logx+1xxπcos(πx)sin(πx)11x,g1(x)=1x1x2+π2sin2(πx)1(1x)2.

    Observe that g1(12)=π210=0.130...<0, and lim. This implies that there exists x_0\in \lbrack\frac{1}{2}, 1) such that g'_1(x) is decreasing on \lbrack\frac{1}{2}, x_0) and increasing on (x_0, 1) . Since g'_1(\frac{1}{2}) = \log2-1 = -0.306\dots, and g'_1(1^-) = 0, it is clear that

    \begin{equation} g'_1(x)\leq { \max\left\{g'_1\left(\frac{1}{2}\right),g'_1(1^-)\right\}} = 0, \notag \end{equation}

    which implies that g_1(x) is decreasing on \lbrack \frac{1}{2}, 1) . Consequently,

    \begin{equation*} g_1(x) > g_1(1^-) = 0. \end{equation*}

    In order to establish the left-hand side of the inequality, we define

    \begin{equation} g_2(x) = 2(1-x)\log x-(2-x)\log \frac{\sin(\pi x)}{\pi(1-x)}. \notag \end{equation}

    Note

    \begin{equation} g_2\left(\frac{1}{2}\right) = \log \frac{1}{2}-\frac{3}{2}\log \frac{2}{\pi} = -0.015...,\; \; \; \; \; \; \; \; g_2(1^-) = 0. \end{equation} (2.4)

    Differentiating g_2(x) yields

    \begin{equation} g'_2(x) = -2\log x+\frac{2(1-x)}{x}-\frac{(2-x)\pi \cos(\pi x)}{\sin(\pi x)}-\frac{2-x}{1-x}+\log \frac{\sin(\pi x)}{\pi(1-x)}. \notag \end{equation}

    Observe that

    \begin{equation} g'_2\left(\frac{1}{2}\right) = \log \frac{8}{\pi}-1 = -0.065... < 0, \; g'_2\left(\frac{3}{4}\right) = \log \frac{32\sqrt{2}}{9\pi}+\frac{5\pi}{4}-\frac{13}{3} = 4.166... > 0. \notag \end{equation}

    Based on these observations and the intermediate value theorem, there exists x_2\in \lbrack \frac{1}{2}, 1) such that g'_2(x_2) = 0 and g_2(x) is decreasing on \lbrack\frac{1}{2}, x_2) and increasing on (x_2, 1) . Therefore, together with (2.4), we conclude that

    \begin{equation} g_2(x) < 0. \notag \end{equation}

    This completes the proof.

    Lemma 2.6. For each a\in \lbrack\frac{1}{2}, 1) , the function

    \begin{equation} f_2(r) = \frac{r'^{2-a(2-a)}[a(\mathcal{K}_a-\mathcal{E}_a)-(1-a)(\mathcal{E}_a-r'^2\mathcal{K}_a)]}{\mathcal{E}_a-r'^2\mathcal{K}_a-ar^2\mathcal{E}_a} \notag \end{equation}

    is decreasing from (0, 1) to \left(0, \frac{a}{2-a}\right) .

    Proof. Following from (1.2) and (1.3), we deduce that

    \begin{equation*} \begin{aligned} a(\mathcal{K}_a-\mathcal{E}_a)-(1-a)(\mathcal{E}_a-r'^2\mathcal{K}_a)& = \frac{\pi}{4}a^2(1-a)r^4F(a+1,2-a;3;r^2){,}\\ \mathcal{E}_a-r'^2\mathcal{K}_a-ar^2\mathcal{E}_a& = \frac{a(1-a)(2-a)\pi}{4}r^4F(a,2-a;3;r^2). \end{aligned} \end{equation*}

    To establish the desired monotonicity of f_2(r) , it suffices to prove that the function f_3(x) , defined on (0, 1) by

    \begin{equation} f_3(x) = \frac{(1-x)^{p(a)}F(a+1,2-a;3;x)}{F(a,2-a;3;x)}, \notag \end{equation}

    is decreasing on (0, 1) , where p(a) = \frac{2-a(2-a)}{2} . Using the power series expansion, the function can be expressed as

    \begin{equation} x \mapsto \frac{\sum_{n = 1}^{\infty}U_nx^n}{\sum_{n = 1}^{\infty}V_nx^n}, \notag \end{equation}

    where the coefficients U_n and V_n satisfy the recursive relations, as detailed in [18]:

    \begin{align} U_0 = &1,\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; V_0 = 1,\\ U_{n+1} = &a_nU_n-b_nU_{n-1},\; \; \; \; \; \; V_n = \frac{(a)_n(2-a)_n}{(3)_nn!}{,} \end{align} (2.5)

    with

    \begin{equation*} \begin{aligned} a_n = &\frac{4n^2+2(3-a^2+2a)n+(-5a^2+8a-2)}{2(n+1)(n+3)},\\ b_n = &\frac{(2n+4a-2-a^2)(2n-a^2)}{4(n+1)(n+3)}. \end{aligned} \end{equation*}

    By Lemma 2.3, we aim to prove that the sequence \left\{\frac{{U_n}}{{V_n}}\right\}_{n\geq0} is decreasing. Note that

    \begin{equation*} U_n > 0,\; \; \; \; V_n > 0, \end{equation*}

    and

    \begin{equation*} \frac{U_0}{V_0} = 1,\; \; \frac{U_1}{V_1} = \frac{-5a^2+8a+2}{2a(2-a)},\; \; \frac{U_2}{V_2} = \frac{-8a^3+10a^2+a-2}{a(3-a)(1+a)}. \end{equation*}

    Observe that

    \begin{equation*} \frac{U_0}{V_0} > \frac{U_1}{V_1} > \frac{U_2}{V_2}, \notag \end{equation*}

    which implies

    \begin{align*} U_1-\frac{V_1}{V_0}U_0 < 0, \quad U_2-\frac{V_2}{V_1}U_1 < 0. \end{align*}

    Assuming that U_{k}-\frac{V_{k}}{V_{k-1}}{U_{k-1}} < 0 for all 1 \leq k\leq n , we prove by induction that U_{n+1}-\frac{V_{n+1}}{V_{n}}{U_{n}} < 0 . According to (2.5), we have

    \begin{align*} U_{n+1}-\frac{V_{n+1}}{V_n}U_n& = (a_nU_n-b_nU_{n-1})-\frac{V_{n+1}}{V_n}U_n\\ & = \left(a_n-\frac{V_{n+1}}{V_n}\right)U_n+\left(a_n-\frac{V_{n+1}}{V_n}\right)\frac{V_{n}}{V_{n-1}}U_{n-1}-\left(a_n-\frac{V_{n+1}}{V_n}\right)\frac{V_{n}}{V_{n-1}}U_{n-1}-b_nU_{n-1}\\ & = \left(a_n-\frac{V_{n+1}}{V_n}\right)\left(U_n-\frac{V_{n}}{V_{n-1}}U_{n-1}\right)+\left[\left(a_n-\frac{V_{n+1}}{V_n}\right)\frac{V_{n}}{V_{n-1}}-b_n\right]U_{n-1}. \end{align*}

    Since a\in \lbrack \frac{1}{2}, 1) , it is easy to know that

    \begin{equation} 6+4a-2a^2 = -2(1-a)^2+8\geq \frac{15}{2},\; \; \; \; -5a^2+8a+2 = -5(a-4/5)^2+26/5\geq \frac{19}{4}, \notag \end{equation}

    and

    \begin{equation} a_n-\frac{V_{n+1}}{V_n} = \frac{2(n-1)^2+(6+4a-2a^2)(n-1)+(-5a^2+8a+2)}{2(n+1)(n+3)} \notag \end{equation}

    is positive for a\in \lbrack \frac{1}{2}, 1) when n\geq 1 . For a\in \lbrack \frac{1}{2}, 1) and n\geq 1 , we have that

    \begin{align*} \left(a_n-\frac{V_{n+1}}{V_n}\right)\frac{V_{n}}{V_{n-1}}-b_n = \frac{\delta(n)}{4n(n+1)(n+2)(n+3)} < 0, \end{align*}

    where

    \begin{align*} \delta(n) = -a^2(a-2)^2n^2+2(a^4-4a^3+6a^2-2)n+2(1-a)^2(3a^2-4a+2). \end{align*}

    In fact, \delta(n) is a quadratic function of n and is decreasing on (1, \infty) , it follows that

    \begin{equation} \begin{aligned} &-\frac{2(a^4-4a^3+6a^2-2)}{2(-a^2(a-2)^2)} = 1+\frac{2a^2-2}{a^2(a-2)^2} < 1,\\ &\delta(n) \leq \delta(2) = 2(a-1)(3a^3-7a^2+10a+2) < 0\; \; \; \hbox{for} \; n\geq2, \end{aligned} \end{equation} (2.6)

    which implies that

    \begin{equation} \left(a_n-\frac{V_{n+1}}{V_n}U_n\right)\frac{V_{n}}{V_{n-1}}-b_n < 0. \notag \end{equation}

    By induction, we conclude that U_{n+1}-\frac{V_{n+1}}{V_{n}}{U_{n}} < 0 for all n\geq1 . Therefore, the sequence \left\{\frac{{U_n}}{{V_n}}\right\}_{n\geq0} is decreasing. Consequently, the function f_2(r) is decreasing on (0, 1) . Moreover,

    \begin{equation} \lim\limits_{r \to 0^{+}}f_2(r) = \frac{a}{2-a},\; \; \; \; \; \; \lim\limits_{r \to 1^{-}}f_2(r) = 0. \notag \end{equation}

    This completes the proof.

    Lemma 2.7. For each a\in \lbrack\frac{1}{2}, 1) , we define the function h(r) on (0, 1) by

    \begin{equation*} h(r) = \frac{2\mathcal{E}_a(\mathcal{K}_a-\mathcal{E}_a)-{2(1-a)r^2}\mathcal{E}_a^2-{2(1-a)r'^2}(\mathcal{K}_a-\mathcal{E}_a)^2}{(\mathcal{K}_a-\mathcal{E}_a)(\mathcal{E}_a-r'^2\mathcal{K}_a)}. \end{equation*}

    Then, 2-a < h(r) < 2 .

    Proof. First of all, we prove the right-hand side inequality. To establish the desired result, we need to show the following inequality:

    \begin{equation*} 2\mathcal{E}_a(\mathcal{K}_a-\mathcal{E}_a)-{2(1-a)r^2}\mathcal{E}_a^2-{2(1-a)r'^2}(\mathcal{K}_a-\mathcal{E}_a)^2 < 2(\mathcal{K}_a-\mathcal{E}_a)(\mathcal{E}_a-r'^2\mathcal{K}_a), \end{equation*}

    which is equivalent to

    \begin{equation} -2(1-a)\mathcal{E}_a(\mathcal{E}_a-r'^2\mathcal{K}_a)+2ar'^2\mathcal{K}_a(\mathcal{K}_a-\mathcal{E}_a) < 0. \notag \end{equation}

    Denote that

    \begin{equation} h_1(r) = -2(1-a)\mathcal{E}_a(\mathcal{E}_a-r'^2\mathcal{K}_a)+2ar'^2\mathcal{K}_a(\mathcal{K}_a-\mathcal{E}_a). \notag \end{equation}

    By differentiation, we obtain

    \begin{equation} \begin{aligned} h'_1(r) = &-2(1-a)\left[\frac{2(1-a)(\mathcal{E}_a-\mathcal{K}_a)}{r}(\mathcal{E}_a-r'^2\mathcal{K}_a)+2ar\mathcal{E}_a\mathcal{K}_a\right]\\ &+2a\left[-2r\mathcal{K}_a(\mathcal{K}_a-\mathcal{E}_a)+\frac{2(1-a)(\mathcal{E}_a-r'^2\mathcal{K}_a)}{r}(\mathcal{K}_a-\mathcal{E}_a)+2(1-a)r\mathcal{E}_a\mathcal{K}_a\right]\\ = &\frac{\mathcal{K}_a-\mathcal{E}_a}{r}\left[4(1-a)(\mathcal{E}_a-\mathcal{K}_a)+(4-8a)r^2\mathcal{K}_a\right] < 0. \end{aligned} \notag \end{equation}

    Therefore, h_1(r) is decreasing on (0, 1) and

    \begin{equation} h_1(r) < \lim\limits_{r \to 0^+}h(r) = 0, \notag \end{equation}

    which implies h(r) < 2 .

    Next, we prove h(r) > 2-a . This is equivalent to the following inequality.

    \begin{equation} \mathcal{E}_a[a(\mathcal{K}_a-\mathcal{E}_a)-(1-a)(\mathcal{E}_a-r'^2\mathcal{K}_a)]-[(1-a)(\mathcal{E}_a-r'^2\mathcal{K}_a)-ar'^2\mathcal{K}_a(\mathcal{K}_a-\mathcal{E}_a)] > 0. \notag \end{equation}

    Denote

    \begin{align*} F(r) = \mathcal{E}_a[a(\mathcal{K}_a-\mathcal{E}_a)-(1-a)(\mathcal{E}_a-r'^2\mathcal{K}_a)]-[(1-a)(\mathcal{E}_a-r'^2\mathcal{K}_a)-ar'^2\mathcal{K}_a(\mathcal{K}_a-\mathcal{E}_a)]. \end{align*}

    The derivative of F(r) yields

    \begin{aligned} F'(r)& = -2(1-a)\frac{\mathcal{K}_a-\mathcal{E}_a}{r} \left[ a(\mathcal{K}_a-\mathcal{E}_a)-(1-a)(\mathcal{E}_a-r'^2\mathcal{K}_a) \right]+\mathcal{E}_a\left[2a(1-a)\frac{r(\mathcal{E}_a-r'^2\mathcal{K}_a)}{r'^2}\right]\\ &-2r(\mathcal{K}_a-\mathcal{E}_a)\left[\frac{a(\mathcal{K}_a-\mathcal{E}_a)-(1-a)(\mathcal{E}_a-r'^2\mathcal{K}_a)}{r^2}+a\frac{\mathcal{E}_a-r'^2\mathcal{K}_a}{r^2}\right]\\ & = \frac{r(\mathcal{E}_a-r'^2\mathcal{K}_a-ar^2\mathcal{E}_a)}{r'^2}\left[2a\frac{\mathcal{E}_a-r'^2\mathcal{K}_a}{r}-2(2-a)\frac{r'^2(\mathcal{K}_a-\mathcal{E}_a)}{r^2}\cdot\frac{a(\mathcal{K}_a-\mathcal{E}_a)-(1-a)(\mathcal{E}_a-r'^2\mathcal{K}_a)}{\mathcal{E}_a-r'^2\mathcal{K}_a-ar^2\mathcal{E}_a}\right]. \notag \end{aligned}

    Note that (\mathcal{E}_a-r'^2\mathcal{K}_a-ar^2\mathcal{E}_a)/r'^2 is increasing from (0, 1) to (0, \infty) . In fact, by differentiation, we know

    \begin{equation} \left(\frac{\mathcal{E}_a-r'^2\mathcal{K}_a-ar^2\mathcal{E}_a}{r'^2}\right)' = \frac{2a(2-a)r(\mathcal{K}_a-\mathcal{E}_a)}{r'^4} > 0. \notag \end{equation}

    According to Lemma 2.2(1)(5) and Lemma 2.6, we have that F'(r) is increasing on (0, 1) and F'(r) > \lim\limits_{r\to0^+}F'(r) = 0 , which implies that F(r) is increasing on (0, 1) . Moreover,

    \begin{equation} F(r) > \lim\limits_{r\to0^+}F(r) = 0. \notag \end{equation}

    Thus, h(r) > 2-a . The proof is completed.

    For a\in\left[\frac{1}{2}, 1\right) , it is also found that h(r) is strictly increasing on (0, 1) .

    Lemma 2.8. For each a\in \lbrack\frac{1}{2}, 1) , r\in(0, 1) , we define the function f_4(r) by

    \begin{equation} f_4(r) = \frac{r'^{2a}(\mathcal{K}_a-\mathcal{E}_a)^2}{2\mathcal{E}_a-2ar^2\mathcal{E}_a-2r'^2\mathcal{K}_a}. \notag \end{equation}

    Then f_4(r) is strictly decreasing from (0, 1) to \left(0, \frac{(1-a)\pi}{2a(2-a)}\right) .

    Proof. Let

    f_{41}(r) = r'^{2a}(\mathcal{K}_a-\mathcal{E}_a)^2,\; \; f_{42}(r) = 2\mathcal{E}_a-2ar^2\mathcal{E}_a-2r'^2\mathcal{K}_a.

    With Lemma 2.4 and f_{41}(0^+) = f_{42}(0^+) = 0 , we only prove the monotonicity of f'_{41}(r)/f'_{42}(r) . Then we have

    \begin{equation} f'_{41}(r) = \frac{r}{r'^{2-2a}}(\mathcal{K}_a-\mathcal{E}_a)[(4-2a)\mathcal{E}_a-2a\mathcal{K}_a], \notag \end{equation}
    \begin{equation} f'_{42}(r) = 4a(2-a)r(\mathcal{K}_a-\mathcal{E}_a), \notag \end{equation}
    \begin{equation} 4a(2-a)\frac{f'_{41}(r)}{f'_{42}(r)} = \frac{(4-2a)\mathcal{E}_a-\mathcal{K}_a}{r'^{2-2a}}\equiv f_{43}(r). \notag \end{equation}

    By differentiation, we see

    \begin{equation} f'_{43}(r) = 2(1-a)\frac{r\mathcal{K}_a}{r'^{4-2a}}\left[(4-4a)\frac{\mathcal{E}_a-r'^2\mathcal{K}_a}{r^2\mathcal{K}_a}-2a\right]. \notag \end{equation}

    With Lemma 2.2(2), we obtain

    (4-4a)\frac{\mathcal{E}_a-r'^2\mathcal{K}_a}{r^2\mathcal{K}_a}-2a < a(4-4a)-2a = 2a(1-2a)\leq 0.

    Thus, f_{43}(r) is strictly decreasing on (0, 1) , which shows f_4(r) is strictly decreasing. And

    \begin{equation} \lim\limits_{r \to 0^{+}}f_4(r) = \lim\limits_{r \to 0^{+}}\frac{f'_{41}(r)}{f'_{42}(r)} = \frac{(1-a)\pi}{2a(2-a)},\; \; \lim\limits_{r \to 1^{-}}f_4(r) = 0. \notag \end{equation}

    The proof is completed.

    Lemma 2.9. For each a\in \lbrack\frac{1}{2}, 1) , r\in(0, 1) , we define the function f_5(r) by

    \begin{equation} f_5(r) = \frac{\mathcal{E}_a(\mathcal{E}_a-r'^2\mathcal{K}_a)+r'^2\mathcal{K}_a(\mathcal{K}_a-\mathcal{E}_a)}{r^2r'^{2-2a}\mathcal{K}_a}. \notag \end{equation}

    Then f_5(r) is strictly increasing from (0, 1) to \left(\frac{\pi}{2}, +\infty\right) .

    Proof. Let

    f_{51}(r) = \mathcal{E}_a(\mathcal{E}_a-r'^2\mathcal{K}_a)+r'^2\mathcal{K}_a(\mathcal{K}_a-\mathcal{E}_a),\; \; f_{52}(r) = r^2r'^{2-2a}\mathcal{K}_a.

    Taking the derivative, we have

    \begin{equation} f'_{51}(r) = 2r\mathcal{K}_a(2\mathcal{E}_a-\mathcal{K}_a),\; \; f'_{52}(r) = \frac{r}{r'^{2a}}[2r'^2\mathcal{K}_a-2(1-a)(\mathcal{K}_a-\mathcal{E}_a)], \notag \end{equation}
    \begin{equation} f'_5(r) = \frac{f'_{51}(r)f_{52}(r)-f_{51}(r)f'_{52}(r)}{f^2_{52}(r)} = \frac{f_{53}(r)}{r^3r'^{4-2a}\mathcal{K}^2_a}, \notag \end{equation}

    where

    \begin{equation} f_{53}(r) = (\mathcal{K}_a-\mathcal{E}_a)\left[2a(\mathcal{E}^2_a-r'^2\mathcal{K}^2_a)-(4a-2)\mathcal{E}_a(\mathcal{E}_a-r'^2\mathcal{K}_a)\right]. \notag \end{equation}

    In fact, we see

    (2a(\mathcal{E}^2_a-r'^2\mathcal{K}^2_a)-(4a-2)\mathcal{E}_a(\mathcal{E}_a-r'^2\mathcal{K}_a))' = \frac{\mathcal{K}_a-\mathcal{E}_a}{r}[4a(\mathcal{K}_a-\mathcal{E}_a)+2(4a-2)(\mathcal{E}_a-r'^2\mathcal{K}_a)] > 0,

    which demonstrates f'_5(r) > 0 for r\in(0, 1) and f_5(r) is increasing on (0, 1) . Moreover,

    \begin{equation} \lim\limits_{r \to 0^{+}}f_5(r) = \frac{\mathcal{E}_a(\mathcal{E}_a-r'^2\mathcal{K}_a)/r^2+r'^2\mathcal{K}_a(\mathcal{K}_a-\mathcal{E}_a)/r^2}{r'^{2-2a}\mathcal{K}_a} = \frac{\pi}{2},\; \; \lim\limits_{r \to 1^{-}}f_5(r) = +\infty. \notag \end{equation}

    The proof is completed.

    Lemma 2.10. For each, a\in \lbrack\frac{1}{2}, 1) , r\in(0, 1) , h(r) is given as in Lemma 2.7. Then, h(r) is strictly increasing from (0, 1) to (2-a, 2) .

    Proof. Let

    h_2(r) = \frac{2\mathcal{E}_a(\mathcal{K}_a-\mathcal{E}_a)-{2(1-a)r^2}\mathcal{E}_a^2-{2(1-a)r'^2}(\mathcal{K}_a-\mathcal{E}_a)^2}{\mathcal{K}_a-\mathcal{E}_a},\; \; h_3(r) = \mathcal{E}_a-r'^2\mathcal{K}_a.

    Clearly, h(r) = \frac{h_2(r)}{h_3(r)} and h_2(0^+) = h_3(0^+) = 0 . By differentiations,

    \begin{equation} \begin{aligned} h'_2(r)& = 2(1-a)\frac{2r'^2(\mathcal{K}_a-\mathcal{E}_a)^2(\mathcal{E}_a-r'^2\mathcal{K}_a)+r^2\mathcal{E}_a[2(1-a){\mathcal{E}_a}^2+(4a-2)r'^2\mathcal{E}_a\mathcal{K}_a-2ar'^2\mathcal{K}_a^2]}{rr'^2(\mathcal{K}_a-\mathcal{E}_a)^2},\\ h'_3(r)& = 2ar\mathcal{K}_a. \end{aligned} \notag \end{equation}

    Then,

    \begin{equation} \begin{aligned} \frac{h'_2(r)}{h'_3(r)} = &\frac{2(1-a)}{2a}\frac{2r'^2(\mathcal{K}_a-\mathcal{E}_a)^2(\mathcal{E}_a-r'^2\mathcal{K}_a)+r^2\mathcal{E}_a[2(1-a){\mathcal{E}_a}^2+(4a-2)r'^2\mathcal{E}_a\mathcal{K}_a-2ar'^2\mathcal{K}_a^2]}{r^2r'^2\mathcal{K}_a(\mathcal{K}_a-\mathcal{E}_a)^2}\\ = &\frac{1-a}{a}\left[\frac{2\mathcal{E}_a-2ar^2\mathcal{E}_a-2r'^2\mathcal{K}_a}{r'^{2a}(\mathcal{K}_a-\mathcal{E}_a)^2}\right]\left[\frac{\mathcal{E}_a(\mathcal{E}_a-r'^2\mathcal{K}_a)+r'^2\mathcal{K}_a(\mathcal{K}_a-\mathcal{E}_a)}{r^2r'^{2-2a}\mathcal{K}_a}\right]\\ = &\frac{1-a}{a}\frac{f_5(r)}{f_4(r)}. \end{aligned} \notag \end{equation}

    With Lemmas 2.8 and 2.9, we obtain that h(r) is strictly increasing on (0, 1) . Furthermore,

    \begin{equation} \lim\limits_{r \to 0^{+}}h(r) = 2-a,\; \; \; \; \lim\limits_{r \to 1^{-}}h(r) = 2. \notag \end{equation}

    This completes the proof.

    In this section, we present some of the main results of \mathcal{E}_a(r) .

    Theorem 3.1. Let a \in \left[\frac{1}{2}, 1\right) , p \in \mathbb{R} \setminus \left\{0\right\} , and for r \in (0, 1) , define

    \begin{equation} \mathcal{F}_{a,p}(r) = \frac{1 - \left[2\mathcal{E}_a(r)/\pi\right]^{\frac{p}{2(1-a)}}}{1 - r'^p}. \notag \end{equation}

    The monotonicity of \mathcal{F}_{a, p}(r) is as follows:

    (1) \mathcal{F}_{a, p}(r) is strictly increasing from (0, 1) to (1-a, 1-b) if and only if p \geq 2 , where

    b = \left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}.

    (2) \mathcal{F}_{a, p}(r) is strictly decreasing on (0, 1) if and only if p \leq 2-a . Moreover, if p\in(0, 2-a\rbrack , the range of \mathcal{F}_{a, p}(r) is (1-b, 1-a) , and the range is (0, 1-a) if p\in(-\infty, 0) .

    (3) If p \in (2-a, 2) , \mathcal{F}_{a, p}(r) is piecewise monotonic. To be precise, there exsists r_0 = r_0(a, p) \in (0, 1) such that \mathcal{F}_{a, p}(r) is strictly increasing on (0, r_0) and strictly decreasing on (r_0, 1) . Furthermore, for r \in (0, 1) , if p \in (2-a, p_0) , the range of \mathcal{F}_{a, p}(r) satisifies

    \begin{equation} 1-b < \mathcal{F}_{a,p}(r) \leq \sigma_0, \end{equation} (3.1)

    while

    \begin{equation} 1-a < \mathcal{F}_{a,p}(r) \leq \sigma_0, \end{equation} (3.2)

    if p \in \lbrack p_0, 2) , where

    p_0 = \frac{2(1-a)\log a}{\log(sin(\pi a)/(1-a)\pi)} \in (2-a,2),\; \; \sigma_0 = \mathcal{F}_{a,p}(r_0) > 1-a.

    Proof. For r\in (0, 1) ,

    \mathcal{F}_{a,p}(r) = \frac{1 - \left[2\mathcal{E}_a(r)/\pi\right]^{\frac{p}{2(1-a)}}}{1 - r'^p} = :\frac{\varphi_1(r)}{\varphi_2(r)}.

    Clearly, we have \varphi_1(0) = \varphi_2(0) = 0 . By differentiation,

    \begin{equation} \begin{aligned} &\varphi'_1(r) = \frac{p}{2(1-a)}\left(\frac{2}{\pi}\right)^{\frac{p}{2(1-a)}}\mathcal{E}_a^{\frac{p}{2(1-a)}-1}\frac{2(1-a)(\mathcal{K}_a-\mathcal{E}_a)}{r},\\ &\varphi'_2(r) = prr'^{p-2}, \end{aligned} \notag \end{equation}

    and

    \begin{equation} \begin{aligned} \frac{\varphi'_1(r)}{\varphi'_2(r)} = \left(\frac{2}{\pi}\right)^{\frac{p}{2(1-a)}}\frac{\mathcal{E}_a^{{\frac{p}{2(1-a)}}-1}(\mathcal{K}_a-\mathcal{E}_a)}{r^2r'^{p-2}} = :\varphi_3(r){.} \end{aligned} \notag \end{equation}

    By differentiating \log \varphi_3(r), we obtain

    \begin{align} \frac{\varphi'_3(r)}{\varphi_3(r)}& = \frac{p}{2(1-a)}\frac{2(a-1)(\mathcal{K}_a-\mathcal{E}_a)}{r\mathcal{E}_a}+p\frac{r}{r'^2}-\frac{2}{r}+\frac{2(1-a)r\mathcal{E}_a}{r'^2(\mathcal{K}_a-\mathcal{E}_a)}+\frac{2(1-a)(\mathcal{K}_a-\mathcal{E}_a)}{r\mathcal{E}_a}-\frac{2r}{r'^2}\\ & = p\frac{\mathcal{E}_a-r'^2\mathcal{K}_a}{rr'^2\mathcal{E}_a}+\frac{2(1-a)r^2\mathcal{E}_a^2-2\mathcal{E}_a(\mathcal{K}_a-\mathcal{E}_a)+2(1-a)r'^2(\mathcal{K}_a-\mathcal{E}_a)^2}{rr'^2\mathcal{E}_a(\mathcal{K}_a-\mathcal{E}_a)}\\ & = \frac{\mathcal{E}_a-r'^2\mathcal{K}_a}{rr'^2\mathcal{E}_a}\left[p-\frac{2\mathcal{E}_a(\mathcal{K}_a-\mathcal{E}_a)-{2(1-a)r^2}\mathcal{E}_a^2-{2(1-a)r'^2}(\mathcal{K}_a-\mathcal{E}_a)^2}{(\mathcal{K}_a-\mathcal{E}_a)(\mathcal{E}_a-r'^2\mathcal{K}_a)}\right]\\ & = \frac{\mathcal{E}_a-r'^2\mathcal{K}_a}{rr'^2\mathcal{E}_a}(p-h(r)), \end{align} (3.3)

    where h(r) is defined as in Lemma 2.7. By Lemmas 2.2(2), 2.7, and 2.10, there are three cases to consider.

    (i) If p\geq 2. It follows from (3.3) that \varphi_3(r) is strictly increasing on (0, 1) , and so is \mathcal{F}_{a, p}(r) . Furthermore, in this case,

    \begin{equation} \mathcal{F}_{a,p}(0^+) = \lim\limits_{r \to 0^+}\frac{\varphi'_1(r)}{\varphi'_2(r)} = 1-a,\; \; \; \; \mathcal{F}_{a,p}(1^-) = 1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}. \notag \end{equation}

    (ii) If p\leq 2-a , as in the proof of case (i) , we know that \varphi_3(r) is strictly decreasing on (0, 1) , and so is \mathcal{F}_{a, p}(r) . Also, \mathcal{F}_{a, p}(0^+) = 1-a , and

    \begin{equation} \mathcal{F}_{a,p}(1^-) = \begin{cases} 0, &\text{for}\; p < 0, \\ 1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}, &\text{for}\; 0 < p\leq2-a. \end{cases} \notag \end{equation}

    (iii) If 2-a < p < 2 . According to Ramanujan's approximation (1.1), it shows that r'^c\mathcal{K}_a \to 0 ( r \to 1^{-} ) if c\geq 0 . With Lemma 2.2(4) and the equation

    H_{\varphi_1,\varphi_2}(r) = \frac{\varphi'_1}{\varphi'_2}\varphi_2-\varphi_1 = \varphi_2\varphi_3-\varphi_1,

    we obtain

    \begin{align} \lim\limits_{r \to 0^+}H_{\varphi_1,\varphi_2}(r) = 0,\; \; \lim\limits_{r \to 1^{-}}H_{\varphi_1,\varphi_2}(r) = \left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}-1 < 0. \end{align} (3.4)

    Together with (3.3), (3.4), Lemmas 2.7 and 2.10, and the formulas

    \begin{align*} \mathcal{F}'_{a,p}(r) = \left(\frac{\varphi_1}{\varphi_2}\right)' = \frac{\varphi'_2}{\varphi^2_2}H_{\varphi_1,\varphi_2}(r), \\\notag H'_{\varphi_1,\varphi_2}(r) = \left(\frac{\varphi'_1}{\varphi'_2}\right)'\varphi_2 = \varphi'_3(r)\varphi_2(r), \end{align*}

    which follows from (2.2) and (2.3), it shows that there exists r_0\in(0, 1) such that H_{\varphi_1, \varphi_2}(r) > 0 for r \in (0, r_0) and H_{\varphi_1, \varphi_2}(r) < 0 for r \in (r_0, 1) . Thus, \mathcal{F}_{a, p}(r) is strictly increasing on (0, r_0) and strictly decreasing on (r_0, 1) . Therefore, for all r\in(0, 1) , we get

    \begin{equation} \mathcal{F}_{a,p}(r)\leq \mathcal{F}_{a,p}(r_0) = \sigma_0. \notag \end{equation}

    In fact, \mathcal{F}_{a, p}(r_0)\geq \mathcal{F}_{a, p}(r) > \text{max}{\left\{\mathcal{F}_{a, p}(0^+), \mathcal{F}_{a, p}(1^-)\right\}} . It follows from Lemma 2.5 that

    p_0 = \frac{2(1-a)\log a}{\log(sin(\pi a)/(1-a)\pi)} \in (2-a,2),

    which makes p_0 the unique root of

    1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{{\frac{p}{2(1-a)}}} = 1-a

    on (2-a, 2) and p \mapsto 1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{{\frac{p}{2(1-a)}}} is strictly increasing on (-\infty, \infty) . Hence we have \mathcal{F}_{a, p}(0^+) \geq \mathcal{F}_{a, p}(1^-) if p\in(2-a, p_0\rbrack and \mathcal{F}_{a, p}(0^+) < \mathcal{F}_{a, p}(1^-) if p \in (p_0, 2) . Consequently, the range of \mathcal{F}_{a, p}(r) in case 3 is valid. The proof is completed.

    Figure 1 shows the monotonicity of \mathcal{F}_{a, p} with a = 0.7 as an example.

    Figure 1.  Monotonicity of \mathcal{F}_{a, p} with a = 0.7 as an example.

    Applying the property of \mathcal{F}_{a, p}(r) from Theorem 3.1, we obtain our main result.

    Theorem 3.2. For a \in \lbrack \frac{1}{2}, 1) , let \mu, \nu \in [0, 1] and p_0, \sigma_0 be given as in Theorem 3.1. Then for any fixed p\in \mathbb{R} , the double inequality

    \begin{equation} \frac{\pi}{2}H^{2(1-a)}_p(1,r';\mu)\leq\mathcal{E}_a\leq\frac{\pi}{2}H^{2(1-a)}_p(1,r';\nu) \end{equation} (3.5)

    holds for all r\in (0, 1) with the equality only for certain values of r if and only if \mu \leq \mu(a, p) and \nu \geq \nu(a, p) , where \mu(a, p) and \nu(a, p) satisfy

    \begin{equation} \mu(a,p)= \begin{cases} a,&\mathit{\text{$p\in (-\infty,0) \cup (0,2-a\rbrack$}},\\ 1-\sigma_0,&\mathit{\text{$p\in (2-a,2)$}},\\ b,&\mathit{\text{$p\in\lbrack2,+\infty)$}}, \end{cases} \nu(a,p)= \begin{cases} 1,&\mathit{\text{$p\in (-\infty,0)$}},\\ b,&\mathit{\text{$p\in (0,p_0)$}},\\ a,&\mathit{\text{$p\in\lbrack p_0,+\infty)$}}, \end{cases} \end{equation} (3.6)

    where

    b = \left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{{\frac{p}{2(1-a)}}}.

    Particularly, for p = 0 , (3.5) holds if and only if \mu \leq 1-2(1-a)^2 and \nu \geq 1 .

    Proof. First we consider the case of p \neq 0 , by (1.5), the inequality (3.5) is equivalent to

    \begin{equation} \mu < 1-\mathcal{F}_{a,p}(r) < \nu, \end{equation} (3.7)

    where F_{a, p}(r) is defined as in Theorem 3.1. It follows from Theorem 3.1 that we immediately conclude the best possible constants \mu = \mu(a, p) and \nu = \nu(a, p) in (3.6).

    For p = 0 , we define the function T(r) on (0, 1) by

    \begin{equation} T(r) = \frac{\log(2\mathcal{E}_a/\pi)}{\log r'} = :\frac{T_1(r)}{T_2(r)}. \notag \end{equation}

    Obviously, we see that T_1(0^+) = T_2(0^+) = 0 . By differentiation, we have

    \begin{equation} \frac{T'_1(r)}{T'_2(r)} = 2(1-a)\frac{r'^2(\mathcal{K}_a-\mathcal{E}_a)}{r^2\mathcal{E}_a}. \notag \end{equation}

    Together with Lemma 2.2(3), this implies \frac{T'_1(r)}{T'_2(r)} is strictly decreasing on (0, 1), and by Lemma 2.4, T(r) shares the same monotonicity. Clearly, T(1^-) = 0 and

    \begin{equation} T(0^+) = \lim\limits_{r\to 0^+}\frac{T'_1(r)}{T'_2(r)} = 2(1-a)^2, \notag \end{equation}

    which indicates 1-2(1-a)^2 < 1-T(r) < 1 for r\in(0, 1) . As a result, Eq (1.5) demonstrates that the inequality

    \begin{equation} \frac{\pi}{2}H^{2(1-a)}_p(1,r';\mu) < \mathcal{E}_a(r) < \frac{\pi}{2}H^{2(1-a)}_p(1,r';\nu) \notag \end{equation}

    holds for all r\in(0, 1) if and only if \mu\leq 1-2(1-a)^2 and \nu \geq 1 .

    This completes the proof.

    Figure 2 shows the sharpness of the bound with a = 0.7 as an example.

    Figure 2.  Sharp bound for \mathcal{E}_a with a = 0.7 as an example.

    Remark 3.1. For a = \frac{1}{2} , we see that (3.5) holds if the parameters satisfy the conditions given in Theorem 3.2. This conclusion has been proved in [15].

    In this section, by applying Theorem 3.2, we present several sharp bounds of weighted Hölder mean for \mathcal{E}_a .

    Note that for the case of \mu(a, p) = \nu(a, p) = a , the best bounds of \mathcal{E}_a are attained at p = 2-a and p = p_0 , which will be proved in the following corollary.

    Corollary 4.1. Let a\in \lbrack\frac{1}{2}, 1) and p_1, p_2\in \mathbb{R} . Then the inequality

    \begin{equation} \frac{\pi}{2}H^{2(1-a)}_{p_1}(1,r';a) < \mathcal{E}_a(r) < \frac{\pi}{2}H^{2(1-a)}_{p_2}(1,r';a) \end{equation} (4.1)

    holds for all r\in (0, 1) with the best possible constants p_1 = 2-a and p_2 = p_0 , where p_0 is given as in Theorem 3.1.

    Proof. For a\in [\frac{1}{2}, 1) , we consider (\mu, p) = (a, 2-a) and (\nu, p) = (a, p_0) satisfying (3.6), which yields (4.1) upon substitution into (3.5).

    To establish that a and p_0 are the best possible constants, we observe that the Hölder mean is monotonically increasing with respect to p . Consequently, it suffices to analyze the case of 2-a < p < p_0 .

    According to Theorem 3.2, the inequality

    \begin{equation} \frac{\pi}{2}H^{2(1-a)}_p(1,r';1-\sigma_0)\leq \mathcal{E}_a\leq \frac{\pi}{2}H^{2(1-a)}_p(1,r';b) \end{equation} (4.2)

    holds for all r\in (0, 1) , where 1-\sigma_0 and b are sharp, with b given as in Theorem 3.2. From Theorem 3.1, together with the monotonicity of \omega \mapsto H_p(1, r';\omega) , we have 1-\sigma_0 < a < b for p\in(2-a, p_0) , implying

    \begin{equation} \frac{\pi}{2}H^{2(1-a)}_p(1,r';1-\sigma_0)\leq \frac{\pi}{2}H^{2(1-a)}_p(1,r';a) \leq \frac{\pi}{2}H^{2(1-a)}_p(1,r';b). \notag \end{equation}

    Therefore, considering the sharpness of 1-\sigma_0 and b in inequality (4.2), we conclude that there exist two numbers r_1, r_2 \in (0, 1) such that

    \begin{equation} \frac{\pi}{2}H^{2(1-a)}_p(1,{r'_1};a) > \mathcal{E}_a(r_1),\; \; \frac{\pi}{2}H^{2(1-a)}_p(1,{r'_2};a) < \mathcal{E}_a(r_2). \notag \end{equation}

    Thus, the proof is completed.

    Figure 3 demonstrates that the sharp bounds of \mathcal{E}_a are attained at p_1 = 2-a and p_2 = p_0 with a = 0.7 as an example.

    Figure 3.  Best constants for (4.1) with a = 0.7 as an example.

    Furthermore, it is observed that computing the lower bound in (3.6) for the case \mu(a, p) = 1-\sigma_0 is challenging, while the case \nu(a, p) = 1 is trivial. Thus, we propose using \mu(a, p) = b for p\in [2, \infty) and \nu(a, p) = b for p\in (0, p_0) to establish new bounds. The specific inequality is as follows.

    Corollary 4.2. Inequality

    \begin{align} & \frac{\pi}{2}\left\{\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{{\frac{1}{1-a}}}+\left[1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{{\frac{1}{1-a}}}\right]r'^2\right\}^{{1-a}}\\ < &\mathcal{E}_a < \frac{\pi}{2}\left\{\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{{\frac{p_0}{2(1-a)}}}+\left[1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{{\frac{p_0}{2(1-a)}}}\right]r'^{p_0}\right\}^{{\frac{2(1-a)}{p_0}}} \end{align} (4.3)

    holds for r\in (0, 1) .

    Lemma 4.3. Let a\in\lbrack\frac{1}{2}, 1) ,

    \begin{align} \Delta(p,r)& = H^{2(1-a)}_p(1,r';b) \\ & = \left\{\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{{\frac{p}{2(1-a)}}}+\left[1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{{\frac{p}{2(1-a)}}}\right]r'^p\right\}^{{\frac{2(1-a)}{p}}}. \end{align}

    Then, the function \Delta(p, r) with respect to p is strictly decreasing on (0, \infty) for r\in (0, 1) .

    Proof. By differentiating \log\Delta(p, r) :

    \begin{equation} \frac{1}{\Delta(p,r)}\frac{\partial\Delta(p,r)}{\partial p} = -\frac{\tilde{\Delta}(p,r'^p)}{p^2\psi(p,r'^p)}, \end{equation} (4.4)

    where

    \begin{align*} \psi(p,x) = &\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}+\left[1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}\right]x, \end{align*}

    and

    \begin{align*} \tilde{\Delta}(p,x)& = 2(1-a)\psi(p,x)\log(\psi(p,x))-p(1-x)\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}\log\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)\\ &-2(1-a)\left[1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}\right]x\log x. \end{align*}

    Differentiating \tilde{\Delta}(p, x) with respect to x yields

    \begin{equation} \begin{aligned} \frac{\partial\tilde{\Delta}(p,x)}{\partial x}& = 2(1-a)\left[1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}\right]\log\frac{\psi(p,x)}{x}\\ &+p\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}\log\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)\triangleq\Delta_0(p,x). \end{aligned} \notag \end{equation}

    Give the observation that \Delta_0(p, x) is strictly decreasing for x\in(0, 1) . In fact,

    \begin{equation} \begin{aligned} \frac{\partial\Delta_0(p,x)}{\partial x} = -2(1-a)\frac{\left[1-\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}\right]\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}}{x\psi(p,x)} < 0. \end{aligned} \notag \end{equation}

    And

    \begin{equation} \Delta_0(p,0^+) = \infty,\; \; \; \; \; \; \Delta_0(p,1^-) = p\left(\frac{\sin(\pi a)}{(1-a)\pi}\right)^{\frac{p}{2(1-a)}}\log\left(\frac{\sin(\pi a)}{(1-a)\pi}\right) < 0 \notag \end{equation}

    indicate that \tilde{\Delta}(p, x) first strictly increases on (0, x_0) and then strictly decreases on (x_0, 1) for some x_0\in (0, 1) . Note that for p > 0 , it is observed that

    \begin{equation} \tilde{\Delta}(p,0^+) = \tilde{\Delta}(p,1^-) = 0. \end{equation} (4.5)

    Hence, \tilde{\Delta}(p, x) > 0 for x \in (0, 1) .

    Consequently, monotonicity of \Delta(p, r) with respect to p follows from (4.4).

    Remark 4.1. Following Lemma 4.3 and inequality (3.5), we observe that

    \begin{equation} \begin{cases} \mathcal{E}_a > \frac{\pi}{2}H^{2(1-a)}_2(1,r';b^{{\frac{1}{1-a}}}_1)\geq \frac{\pi}{2}H^{2(1-a)}_p(1,r';b^{{\frac{p}{2(1-a)}}}_1), &if \text{ $p\in\lbrack2,\infty)$},\\ \mathcal{E}_a < \frac{\pi}{2}H^{2(1-a)}_{p_0}(1,r';b^{{\frac{p_0}{2(1-a)}}}_1)\leq \frac{\pi}{2}H^{2(1-a)}_p(1,r';b^{{\frac{p}{2(1-a)}}}_1), &if \text{ $p\in(0,p_0\rbrack$}{,} \end{cases} \end{equation} (4.6)

    where

    b_1 = \frac{sin(\pi a)}{(1-a)\pi}.

    According to the proof of (3.2), if p\in(p_0, 2) , it follows that

    \begin{equation} 1-\sigma_0 < b < a. \notag \end{equation}

    Therefore, it results in

    \begin{equation*} \frac{\pi}{2}H^{2(1-a)}_{p}(1,r';1-\sigma_0) < \frac{\pi}{2}H^{2(1-a)}_{p}(1,r';b) < \frac{\pi}{2}H^{2(1-a)}_{p}(1,r';a) \end{equation*}

    by the monotonicity of H^{2(1-a)}_p(1, r';\zeta) with respect to \zeta .Theorem 3.2 presents that, for p\in(p_0, 2) , 1-\sigma_0 is sharp weight of H^{2(1-a)}_p(1, r';\zeta) as the lower bound of \mathcal{E}_a , while a is sharp weight as the upper bound of \mathcal{E}_a .

    Hence, as a bound of \mathcal{E}_a , H^{2(1-a)}_p(1, r';b) can attain the best upper bound at p = p_0 and the best lower bound at p = 2 by (4.6).

    In this article, we have proved the monotonicity of \mathcal{F}_{a, p}(r) , where \mathcal{F}_{a, p}(r) is given as in Theorem 3.1. Moreover, we find the sharp weighted Hölder mean approximating \mathcal{E}_a :

    \begin{equation} \frac{\pi}{2}H^{2(1-a)}_p(1,r';\mu)\leq\mathcal{E}_a\leq\frac{\pi}{2}H^{2(1-a)}_p(1,r';\nu) \notag \end{equation}

    holds for all r\in (0, 1) if and only if \mu \leq \mu(a, p) and \nu \geq \nu(a, p) , where \mu(a, p) and \nu(a, p) are given as in (3.6). Besides, we derive several bounds of \mathcal{E}_a in terms of weights and power, which are given by Corollary 4.1, Corollary 4.2, and Remark 4.1. These conclusions provide an extension of the work of [15].

    Zixuan Wang: Investigation, Writing – original draft. Chuanlong Sun: Validation. Tiren Huang: Writing – review & editing. All authors have read and approved the final version of the manuscript for publication.

    The authors declare that they have not used artificial intelligence (AI) tools in the creation of this article.

    The authors declare no conflicts of interest regarding the publication for the paper.



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