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On sharp third Hankel determinant for certain starlike functions

Neha Verma Department of Applied Mathematics, Delhi Technological University, Delhi–110042, India nehaverma1480@gmail.com  and  S. Sivaprasad Kumar Department of Applied Mathematics, Delhi Technological University, Delhi–110042, India spkumar@dce.ac.in
Abstract.

In this paper, we provide an estimation for the sharp bound of the third Hankel determinant of starlike functions of order α\alpha, where α\alpha ranges in the interval [0,1/6]∪{1/2}[0,1/6]\cup\{1/2\} and thereby extending the result of Rath et al. (Complex Anal Oper Theory: No. 65, 16(5), 8 pp 2022).

Key words and phrases:
Starlike, Sharp, Hankel determinant, Order alpha
2010 Mathematics Subject Classification:
30C45, 30C50

1. Introduction

Consider the set 𝒜\mathcal{A}, which comprises normalized analytic functions defined on the open unit disk 𝔻:={z∈ℂ:|z|<1}\mathbb{D}:=\{z\in\mathbb{C}:|z|<1\}. These functions are expressed in the form:

f​(z)=z+∑n=2∞an​zn.f(z)=z+\sum_{n=2}^{\infty}a_{n}z^{n}. (1.1)

Let 𝒮⊂𝒜\mathcal{S}\subset\mathcal{A}, where 𝒮\mathcal{S} represents the class of univalent functions, and 𝒫\mathcal{P} denotes the collection of analytic functions defined on 𝔻\mathbb{D} with a positive real part, expressed as p​(z)=1+∑n=1∞pn​znp(z)=1+\sum_{n=1}^{\infty}p_{n}z^{n}. Let hh and gg are two analytic functions, then we say hh is subordinated to gg, denoted as h≺gh\prec g, provided there exist a Schwarz function ww, adhering to two crucial conditions: w​(0)=0w(0)=0 and |w​(z)|≤|z||w(z)|\leq|z|, such that h​(z)=g​(w​(z))h(z)=g(w(z)).

In the year 1936, Robertson [15] introduced the class of starlike functions of order α\alpha, characterized as follows:

Definition 1.1.

[15] For 0≤α<10\leq\alpha<1, we say that a function f∈𝒜f\in\mathcal{A} is starlike of order α\alpha if and only if

Re⁡(z​f′​(z)f​(z))>α,z∈𝔻.\operatorname{Re}\bigg{(}\dfrac{zf^{\prime}(z)}{f(z)}\bigg{)}>\alpha,\quad z\in\mathbb{D}.

The class of all such functions is represented by 𝒮∗​(α)\mathcal{S}^{*}(\alpha).

In 1992, Ma and Minda [13] introduced a more general class of starlike functions through subordination, defined as follows:

𝒮∗​(φ)={f∈𝒜:z​f′​(z)f​(z)≺φ​(z)},\mathcal{S}^{*}(\varphi)=\bigg{\{}f\in\mathcal{A}:\dfrac{zf^{\prime}(z)}{f(z)}\prec\varphi(z)\bigg{\}},

where φ\varphi is an analytic univalent function such that Re⁡φ​(z)>0\operatorname{Re}\varphi(z)>0, φ​(𝔻)\varphi(\mathbb{D}) is symmetric about the real axis and starlike with respect to φ​(0)=1\varphi(0)=1 with φ′​(0)>0\varphi^{\prime}(0)>0. Through this concept, we can re-define the class 𝒮∗​(α)\mathcal{S}^{*}(\alpha) as:

𝒮∗​(α)={f∈𝒜:z​f′​(z)f​(z)≺1+(1−2​α)​z1−z,α∈[0,1)}.\mathcal{S}^{*}(\alpha)=\bigg{\{}f\in\mathcal{A}:\dfrac{zf^{\prime}(z)}{f(z)}\prec\frac{1+(1-2\alpha)z}{1-z},\quad\alpha\in[0,1)\bigg{\}}.

Note that 𝒮∗​(0)=𝒮∗\mathcal{S}^{*}(0)=\mathcal{S}^{*} and 𝒮∗​(φ)⊂𝒮∗​(α)\mathcal{S}^{*}(\varphi)\subset\mathcal{S}^{*}(\alpha) for some α\alpha depending upon the choice of φ\varphi.

The Bieberbach conjecture, as documented on [4, Page no. 17], has been a significant source of inspiration in the development of univalent function theory and in the formulation of coefficient problems. Building on this foundation, in 1966, Pommerenke [14] introduced the concept of q​t​hqth Hankel determinants, denoted as Hq​(n)H_{q}(n), where nn and qq are both natural numbers, associated with analytic functions as in (1.1), defined as follows:

Hq​(n)=|anan+1…an+q−1an+1an+2…an+q⋮⋮⋱⋮an+q−1an+q…an+2​q−2|.H_{q}(n)=\begin{vmatrix}a_{n}&a_{n+1}&\ldots&a_{n+q-1}\\ a_{n+1}&a_{n+2}&\ldots&a_{n+q}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n+q-1}&a_{n+q}&\ldots&a_{n+2q-2}\end{vmatrix}. (1.2)

By choosing specific values for both nn and qq, we can examine particular cases of this concept. For instance, when we set q=2q=2, we obtain the expression for the second-order Hankel determinant. Numerous studies have investigated and established sharp bounds for second-order Hankel determinants and other determinants within various subclasses of 𝒮\mathcal{S}, see [8, 7, 5] for more details. Now, if we choose q=3q=3 and n=1n=1 in (1.2), assuming a1:=1a_{1}:=1, we arrive at the expression for the Hankel determinant of order three, given by

H3​(1):=|​1a2a3a2a3a4a3a4a5​|=a3​(a2​a4−a32)−a4​(a4−a2​a3)+a5​(a3−a22).H_{3}(1):=\begin{vmatrix}$$1&a_{2}&a_{3}\\ a_{2}&a_{3}&a_{4}\\ a_{3}&a_{4}&a_{5}$$\end{vmatrix}=a_{3}(a_{2}a_{4}-a_{3}^{2})-a_{4}(a_{4}-a_{2}a_{3})+a_{5}(a_{3}-a_{2}^{2}). (1.3)

Determining the third-order Hankel determinant poses a greater challenge compared to the second-order, as evidenced in [22, 9]. We also list some of the sharp estimates for the third-order Hankel determinant concerning functions within the class 𝒮∗​(φ)\mathcal{S}^{*}(\varphi), considering various selections of φ​(z)\varphi(z) in Table 1. However, the sharp estimate of H3​(1)H_{3}(1) for 𝒮N​e∗\mathcal{S}_{Ne}^{*} is yet to be estimated.

Table 1. List of sharp third order Hankel determinants
Class Sharp bound Reference
𝒮∗:=𝒮∗​(0)\mathcal{S}^{*}:=\mathcal{S}^{*}(0) 4/9 [2, 6]
𝒮∗​(1/2)\mathcal{S}^{*}(1/2) 1/9 [11, 18]
𝒮ϱ∗:=𝒮∗​(1+z​ez)\mathcal{S}^{*}_{\varrho}:=\mathcal{S}^{*}(1+ze^{z}) 1/9 [19]
𝒮​ℒ∗:=𝒮∗​(1+z)\mathcal{SL}^{*}:=\mathcal{S}^{*}(\sqrt{1+z}) 1/36 [1]
𝒮e∗:=𝒮∗​(ez)\mathcal{S}^{*}_{e}:=\mathcal{S}^{*}(e^{z}) 1/9 [16]
𝒮ρ∗:=𝒮∗​(1+s​i​n​h−1​(z))\mathcal{S}^{*}_{\rho}:=\mathcal{S}^{*}(1+sinh^{-1}(z)) 1/9 [17]
𝒮N​e∗:=𝒮∗​(1+z−z3/3)\mathcal{S}_{Ne}^{*}:=\mathcal{S}^{*}(1+z-z^{3}/3) — —

For the class 𝒮∗​(α)\mathcal{S}^{*}(\alpha), Krishna and Ramreddy [7] computed the bound of the second order Hankel determinant, |a2​a4−a32|≤(1−α)2|a_{2}a_{4}-a_{3}^{2}|\leq(1-\alpha)^{2}, α∈[0,1/2]\alpha\in[0,1/2] while Xu and Fang [21] calculated the sharp bounds of the Fekete and Szegö functional |a3−λ​a22|≤(1−α)​max⁡{1,|3−2​α−4​λ​(1−α)|}|a_{3}-\lambda a_{2}^{2}|\leq(1-\alpha)\max\{1,|3-2\alpha-4\lambda(1-\alpha)|\}, λ∈ℂ\lambda\in\mathbb{C} and α∈[0,1)\alpha\in[0,1). We refer [3] for further information on Hankel determinants associated with the class 𝒮∗​(α)\mathcal{S}^{*}(\alpha).

The purpose of this study is to establish the sharp bound of third order Hankel determinant for functions belonging to the class, 𝒮∗​(α)\mathcal{S}^{*}(\alpha). At the end of this paper, we demonstrate the validation of our main result by considering the class 𝒮∗​(α)\mathcal{S}^{*}(\alpha) specifically for the case when α=0\alpha=0, and we also present some relevant applications.

1.1. Preliminary

In this part of the section, we mention the initial coefficient bounds aia_{i} (i=2,3,4,5)(i=2,3,4,5) in terms of the Carathéodory coefficients and a lemma which will be used in our forthcoming results. Let f∈𝒮∗​(α)f\in\mathcal{S}^{*}(\alpha), then a Schwarz function w​(z)w(z) exists such that

z​f′​(z)f​(z)=1+(1−2​α)​w​(z)1−w​(z).\dfrac{zf^{\prime}(z)}{f(z)}=\frac{1+(1-2\alpha)w(z)}{1-w(z)}. (1.4)

Let p​(z)=1+∑n=2∞pn​zn∈𝒫p(z)=1+\sum_{n=2}^{\infty}p_{n}z^{n}\in\mathcal{P} and w​(z)=(p​(z)−1)/(p​(z)+1)w(z)=(p(z)-1)/(p(z)+1). The expressions of ai​(i=2,3,4,5)a_{i}(i=2,3,4,5) are obtained in terms of pj​(j=1,2,3,4)p_{j}(j=1,2,3,4) by substituting w​(z)w(z), p​(z)p(z), and f​(z)f(z) in equation (1.4) with suitable comparison of coefficients so that

a2=p1​(1−α),a_{2}=p_{1}(1-\alpha), (1.5)
a3=(1−α)2​(p2+p12​(1−α)),a_{3}=\dfrac{(1-\alpha)}{2}\bigg{(}p_{2}+p_{1}^{2}(1-\alpha)\bigg{)}, (1.6)
a4=(1−α)6​(2​p3+3​p1​p2​(1−α)+p13​(1−α)2)a_{4}=\dfrac{(1-\alpha)}{6}\bigg{(}2p_{3}+3p_{1}p_{2}(1-\alpha)+p_{1}^{3}(1-\alpha)^{2}\bigg{)} (1.7)

and

a5=(1−α)24​(6​p4+(1−α)​(3​p22+8​p1​p3)+(1−α)2​(6​p12​p2+p14​(1−α))).a_{5}=\dfrac{(1-\alpha)}{24}\bigg{(}6p_{4}+(1-\alpha)\bigg{(}3p_{2}^{2}+8p_{1}p_{3}\bigg{)}+(1-\alpha)^{2}\bigg{(}6p_{1}^{2}p_{2}+p_{1}^{4}(1-\alpha)\bigg{)}\bigg{)}. (1.8)

The formula for pjp_{j} (j=2,3,4)(j=2,3,4), which plays a significant role in finding the sharp bound of the Hankel determinant and has been prominently exploited in the main theorem, is contained in the Lemma 1.2 below.

Lemma 1.2.

[12, 10] Let p∈𝒫p\in\mathcal{P} has the form 1+∑n=1∞pn​zn.1+\sum_{n=1}^{\infty}p_{n}z^{n}. Then

2​p2=p12+γ​(4−p12),2p_{2}=p_{1}^{2}+\gamma(4-p_{1}^{2}),
4​p3=p13+2​p1​(4−p12)​γ−p1​(4−p12)​γ2+2​(4−p12)​(1−|γ|2)​η,4p_{3}=p_{1}^{3}+2p_{1}(4-p_{1}^{2})\gamma-p_{1}(4-p_{1}^{2}){\gamma}^{2}+2(4-p_{1}^{2})(1-|\gamma|^{2})\eta,

and

8​p4\displaystyle 8p_{4} =p14+(4−p12)γ(p12(γ2−3γ+3)+4γ)−4(4−p12)(1−|γ|2)(p1(γ−1)η\displaystyle=p_{1}^{4}+(4-p_{1}^{2})\gamma(p_{1}^{2}({\gamma}^{2}-3\gamma+3)+4\gamma)-4(4-p_{1}^{2})(1-|\gamma|^{2})(p_{1}(\gamma-1)\eta
+γ¯η2−(1−|η|2)ρ),\displaystyle\quad+\bar{\gamma}{\eta}^{2}-(1-|\eta|^{2})\rho),

for some γ\gamma, η\eta and ρ\rho such that |γ|≤1|\gamma|\leq 1, |η|≤1|\eta|\leq 1 and |ρ|≤1.|\rho|\leq 1.

2. Sharp H3​(1)H_{3}(1) for 𝒮∗​(α)\mathcal{S}^{*}(\alpha)

Recently, Kowalczyk et al. [6] and Banga and Kumar [2] obtained the sharp bound of the third-order Hankel determinant for functions in the class 𝒮∗:=𝒮∗​(0)\mathcal{S}^{*}:=\mathcal{S}^{*}(0), independently whereas Rath et al. [18] determined the sharp bound of H3​(1)H_{3}(1) for functions in the class 𝒮∗​(1/2)\mathcal{S}^{*}(1/2) and corrected the proof provided in [11]. In this section, we extend our analysis to calculate the sharp bound of H3​(1)H_{3}(1) for functions in the class 𝒮∗​(α)\mathcal{S}^{*}(\alpha) for some additional range of α\alpha. Below, is our main result.

Theorem 2.1.

Let f∈𝒮∗​(α)f\in\mathcal{S}^{*}(\alpha). Then

|H3​(1)|≤4​(1−α)29,α∈[0,1/6]∪{1/2}.|H_{3}(1)|\leq\frac{4(1-\alpha)^{2}}{9},\quad\alpha\in[0,1/6]\cup\{1/2\}. (2.1)

This result is sharp.

Proof.

Since, the class 𝒫\mathcal{P} is invariant under rotation, we have p1∈[0,2]p_{1}\in[0,2] and assume p1=:pp_{1}=:p. The expressions of aia_{i} (i=2,3,4,5)(i=2,3,4,5) from equations (1.5)-(1.8) are substituted in equation (1.3). We get

H3​(1)\displaystyle H_{3}(1) =(1−α)2144(−(1−α)4p6+3(1−α)3p4p2+8(1−α)2p3p3+24(1−α)pp2p3\displaystyle=\dfrac{(1-\alpha)^{2}}{144}\bigg{(}-(1-\alpha)^{4}p^{6}+3(1-\alpha)^{3}p^{4}p_{2}+8(1-\alpha)^{2}p^{3}p_{3}+24(1-\alpha)pp_{2}p_{3}
−18(1−α)p2p4−9(1−α)p23−9(1−α)2p2p22−16p32+18p2p4).\displaystyle\quad\quad\quad\quad\quad-18(1-\alpha)p^{2}p_{4}-9(1-\alpha)p_{2}^{3}-9(1-\alpha)^{2}p^{2}p_{2}^{2}-16p_{3}^{2}+18p_{2}p_{4}\bigg{)}.

After simplifying the calculations through Lemma 1.2, we obtain

H3​(1)=11152​(Δ1​(p,γ)+Δ2​(p,γ)​η+Δ3​(p,γ)​η2+ϕ​(p,γ,η)​ρ),forγ,η,ρ∈𝔻.H_{3}(1)=\dfrac{1}{1152}\bigg{(}\Delta_{1}(p,\gamma)+\Delta_{2}(p,\gamma)\eta+\Delta_{3}(p,\gamma){\eta}^{2}+\phi(p,\gamma,\eta)\rho\bigg{)},\quad\text{for}\quad\gamma,\eta,\rho\in\mathbb{D}.

Here

Δ1​(p,γ):\displaystyle\Delta_{1}(p,\gamma): =(1−α)2(α(1−2α)2(3−2α)p6−(2−15α+18α2)p2γ2(4−p2)2\displaystyle=(1-\alpha)^{2}\bigg{(}\alpha(1-2\alpha)^{2}(3-2\alpha)p^{6}-(2-15\alpha+18\alpha^{2})p^{2}{\gamma}^{2}(4-p^{2})^{2}
+p2​γ4​(4−p2)2−(10−15​α)​p2​γ3​(4−p2)2+36​α​γ3​(4−p2)2\displaystyle\quad+p^{2}{\gamma}^{4}(4-p^{2})^{2}-(10-15\alpha)p^{2}{\gamma}^{3}(4-p^{2})^{2}+36\alpha{\gamma}^{3}(4-p^{2})^{2}
+(3−12​α3+32​α2−19​α)​p4​γ​(4−p2)−9​(1−2​α)​p4​γ3​(4−p2)\displaystyle\quad+(3-12\alpha^{3}+32\alpha^{2}-19\alpha)p^{4}{\gamma}(4-p^{2})-9(1-2\alpha)p^{4}{\gamma}^{3}(4-p^{2})
+(3−16α2+2α)p4γ2(4−p2)−36(1−2α)p2γ2(4−p2)),\displaystyle\quad+(3-16\alpha^{2}+2\alpha)p^{4}{\gamma}^{2}(4-p^{2})-36(1-2\alpha)p^{2}{\gamma}^{2}(4-p^{2})\bigg{)},
Δ2​(p,γ):\displaystyle\Delta_{2}(p,\gamma): =4(1−|γ|2)(4−p2)(1−α)2((8α2−10α+3)p3+9(1−2α)p3γ\displaystyle=4(1-|\gamma|^{2})(4-p^{2})(1-\alpha)^{2}\bigg{(}(8\alpha^{2}-10\alpha+3)p^{3}+9(1-2\alpha)p^{3}{\gamma}
+(5−12α)pγ(4−p2)−pγ2(4−p2)),\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+(5-12\alpha)p\gamma(4-p^{2})-p{\gamma}^{2}(4-p^{2})\bigg{)},
Δ3​(p,γ):\displaystyle\Delta_{3}(p,\gamma): =4​(1−|γ|2)​(4−p2)​(1−α)2​(−8​(4−p2)−|γ|2​(4−p2)+9​(1−2​α)​p2​γ¯),\displaystyle=4(1-|\gamma|^{2})(4-p^{2})(1-\alpha)^{2}\bigg{(}-8(4-p^{2})-|\gamma|^{2}(4-p^{2})+9(1-2\alpha)p^{2}\bar{\gamma}\bigg{)},
ϕ​(p,γ,η):\displaystyle\phi(p,\gamma,\eta): =36​(1−|γ|2)​(4−p2)​(1−|η|2)​(1−α)2​((4−p2)​γ−(1−2​α)​p2).\displaystyle=36(1-|\gamma|^{2})(4-p^{2})(1-|\eta|^{2})(1-\alpha)^{2}\bigg{(}(4-p^{2})\gamma-(1-2\alpha)p^{2}\bigg{)}.

Assume x:=|γ|x:=|\gamma|, y:=|η|y:=|\eta| and since |ρ|≤1,|\rho|\leq 1, the above expression reduces to

|H3​(1)|≤11152​(|Δ1​(p,γ)|+|Δ2​(p,γ)|​y+|Δ3​(p,γ)|​y2+|ϕ​(p,γ,η)|)≤Z​(p,x,y),\displaystyle|H_{3}(1)|\leq\dfrac{1}{1152}\bigg{(}|\Delta_{1}(p,\gamma)|+|\Delta_{2}(p,\gamma)|y+|\Delta_{3}(p,\gamma)|y^{2}+|\phi(p,\gamma,\eta)|\bigg{)}\leq Z(p,x,y),

where

Z​(p,x,y)=11152​(z1​(p,x)+z2​(p,x)​y+z3​(p,x)​y2+z4​(p,x)​(1−y2))Z(p,x,y)=\dfrac{1}{1152}\bigg{(}z_{1}(p,x)+z_{2}(p,x)y+z_{3}(p,x)y^{2}+z_{4}(p,x)(1-y^{2})\bigg{)} (2.2)

with

z1​(p,x):\displaystyle z_{1}(p,x): =(1−α)2(α(1−2α)2(3−2α)p6+(2−15α+18α2)p2x2(4−p2)2\displaystyle=(1-\alpha)^{2}\bigg{(}\alpha(1-2\alpha)^{2}(3-2\alpha)p^{6}+(2-15\alpha+18\alpha^{2})p^{2}x^{2}(4-p^{2})^{2}
+p2​x4​(4−p2)2+(10−15​α)​p2​x3​(4−p2)2+36​α​x3​(4−p2)2\displaystyle\quad\quad\quad\quad\quad+p^{2}x^{4}(4-p^{2})^{2}+(10-15\alpha)p^{2}x^{3}(4-p^{2})^{2}+36\alpha x^{3}(4-p^{2})^{2}
+(3−12​α3+32​α2−19​α)​p4​x​(4−p2)+9​(1−2​α)​p4​x3​(4−p2)\displaystyle\quad\quad\quad\quad\quad+(3-12\alpha^{3}+32\alpha^{2}-19\alpha)p^{4}x(4-p^{2})+9(1-2\alpha)p^{4}x^{3}(4-p^{2})
+(3−16α2+2α)p4x2(4−p2)+36(1−2α)p2x2(4−p2)),\displaystyle\quad\quad\quad\quad\quad+(3-16\alpha^{2}+2\alpha)p^{4}x^{2}(4-p^{2})+36(1-2\alpha)p^{2}x^{2}(4-p^{2})\bigg{)},
z2​(p,x):\displaystyle z_{2}(p,x): =4(1−x2)(4−p2)(1−α)2((8α2−10α+3)p3+9(1−2α)p3x\displaystyle=4(1-x^{2})(4-p^{2})(1-\alpha)^{2}\bigg{(}(8\alpha^{2}-10\alpha+3)p^{3}+9(1-2\alpha)p^{3}x
+(5−12α)px(4−p2)+px2(4−p2)),\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+(5-12\alpha)px(4-p^{2})+px^{2}(4-p^{2})\bigg{)},
z3​(p,x):\displaystyle z_{3}(p,x): =4​(1−x2)​(4−p2)​(1−α)2​(8​(4−p2)+x2​(4−p2)+9​(1−2​α)​p2​x),\displaystyle=4(1-x^{2})(4-p^{2})(1-\alpha)^{2}\bigg{(}8(4-p^{2})+x^{2}(4-p^{2})+9(1-2\alpha)p^{2}x\bigg{)},
z4​(p,x):\displaystyle z_{4}(p,x): =36​(1−x2)​(4−p2)​(1−α)2​((4−p2)​x+(1−2​α)​p2).\displaystyle=36(1-x^{2})(4-p^{2})(1-\alpha)^{2}\bigg{(}(4-p^{2})x+(1-2\alpha)p^{2}\bigg{)}.

Note that for α∈[0,1/6]\alpha\in[0,1/6], all the factors involving α\alpha in |Δ1​(p,γ)||\Delta_{1}(p,\gamma)|, |Δ2​(p,γ)||\Delta_{2}(p,\gamma)|, |Δ3​(p,γ)||\Delta_{3}(p,\gamma)| and |ϕ​(p,γ,η)||\phi(p,\gamma,\eta)|, are positive as 1/61/6 is the smallest positive root of the equation 2−15​α+18​α2=02-15\alpha+18\alpha^{2}=0. We maximise Z​(p,x,y)Z(p,x,y) within the closed cuboid Y:[0,2]×[0,1]×[0,1]Y:[0,2]\times[0,1]\times[0,1], by finding the maximum values in the interior of YY, in the interior of the six faces and on the twelve edges.

Case I:
We begin with every interior point of YY assuming (p,x,y)∈(0,2)×(0,1)×(0,1)(p,x,y)\in(0,2)\times(0,1)\times(0,1). We determine ∂Z/∂y\partial{Z}/\partial y to examine the points of maxima in the interior of YY. Thus

∂Z∂y\displaystyle\dfrac{\partial Z}{\partial y} =(4−p2)​(1−x2)​(1−α2)288(8(8−9x+x2)y−2p2(1−x)y(17−x−18α)\displaystyle=\dfrac{(4-p^{2})(1-x^{2})(1-\alpha^{2})}{288}\bigg{(}8(8-9x+x^{2})y-2p^{2}(1-x)y(17-x-18\alpha)
+p3​(3−x2−10​α+8​α2+x​(4−6​α))\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+p^{3}(3-x^{2}-10\alpha+8\alpha^{2}+x(4-6\alpha))
+4xp(5+x−12α)).\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+4xp(5+x-12\alpha)\bigg{)}.

Now, ∂Z/∂y=0\partial Z/\partial y=0 gives

y=y0:=4​x​p​(5+x−12​α)+p3​(3−x2−10​α+8​α2+x​(4−6​α))2​(1−x)​(−4​(8−x)+p2​(17−x−18​α)).y=y_{0}:=\dfrac{4xp(5+x-12\alpha)+p^{3}(3-x^{2}-10\alpha+8\alpha^{2}+x(4-6\alpha))}{2(1-x)(-4(8-x)+p^{2}(17-x-18\alpha))}.

The existence of critical points require that y0∈(0,1)y_{0}\in(0,1) and can only exist when

2​p2​(1−x)​(17−x−18​α)\displaystyle 2p^{2}(1-x)(17-x-18\alpha) >−p3​(−3+x2+10​α−8​α2−x​(4−6​α))\displaystyle>-p^{3}(-3+x^{2}+10\alpha-8\alpha^{2}-x(4-6\alpha))
+4​p​x​(5+x−12​α)+8​(1−x)​(8−x).\displaystyle\quad+4px(5+x-12\alpha)+8(1-x)(8-x). (2.3)

We try finding the solution satisfying the inequality (2.3) for the critical points. The possible range for which y0∈(0,1)y_{0}\in(0,1), is (0,(3+2​α)/9)×(p~,2)(0,(3+2\alpha)/9)\times(\tilde{p},2). Here

p~:\displaystyle\tilde{p}: =2​(17−18​α)p1~−24/3​(1−i​3)​(17−18​α)2p1~​(p2~+−256​(17−18​α)6+(p2~)2)1/3\displaystyle=\frac{2(17-18\alpha)}{\tilde{p_{1}}}-\frac{2^{4/3}(1-i\sqrt{3})(17-18\alpha)^{2}}{\tilde{p_{1}}(\tilde{p_{2}}+\sqrt{-256(17-18\alpha)^{6}+(\tilde{p_{2}})^{2}})^{1/3}}
−(1+i​3)​(p2~+−256​(17−18​α)6+(p2~)2)1/324/3​p1~\displaystyle\quad-\frac{(1+i\sqrt{3})(\tilde{p_{2}}+\sqrt{-256(17-18\alpha)^{6}+(\tilde{p_{2}})^{2}})^{1/3}}{2^{4/3}\tilde{p_{1}}}

where

p1~:\displaystyle\tilde{p_{1}}: =3​(3−10​α+8​α2);\displaystyle=3(3-10\alpha+8\alpha^{2});
p2~:\displaystyle\tilde{p_{2}}: =63056−146016​α+8640​α2+183168​α3−110592​α4.\displaystyle=63056-146016\alpha+8640\alpha^{2}+183168\alpha^{3}-110592\alpha^{4}.

Therefore, a calculation reveals that the maxima attained in the interior of YY at each such y0∈(0,1)y_{0}\in(0,1) is always less than 4​(1−α)2/94(1-\alpha)^{2}/9 for α∈[0,1/6]\alpha\in[0,1/6].

Case II:
The interior of six faces of the cuboid YY, is now under consideration, for the further calculations.
On p=0p=0, Z​(p,x,y)Z(p,x,y) turns into

s1​(x,y):=(1−α)2​((1−x2)​((8+x2)​y2+9​x​(1−y2))+9​x3​α)18,x,y∈(0,1).s_{1}(x,y):=\dfrac{(1-\alpha)^{2}((1-x^{2})((8+x^{2})y^{2}+9x(1-y^{2}))+9x^{3}\alpha)}{18},\quad x,y\in(0,1). (2.4)

Since

∂s1∂y=(1−x2)​(x+1)​(8−x)​(1−α)2​y9≠0,x,y∈(0,1).\dfrac{\partial s_{1}}{\partial y}=\dfrac{(1-x^{2})(x+1)(8-x)(1-\alpha)^{2}y}{9}\neq 0,\quad x,y\in(0,1).

Thus, s1s_{1} has no critical point in (0,1)×(0,1)(0,1)\times(0,1).

On p=2p=2, Z​(p,x,y)Z(p,x,y) reduces to

Z​(2,x,y):=α​(1−α)2​(1−2​α)2​(3−2​α)18,x,y∈(0,1).Z(2,x,y):=\dfrac{\alpha(1-\alpha)^{2}(1-2\alpha)^{2}(3-2\alpha)}{18},\quad x,y\in(0,1). (2.5)

On x=0x=0, Z​(p,x,y)Z(p,x,y) becomes

s2​(p,y):\displaystyle s_{2}(p,y): =(1−α)21152(α(3−2α)(1−2α)2p6+36(1−2α)p2(4−p2)(1−y2)\displaystyle=\dfrac{(1-\alpha)^{2}}{1152}\bigg{(}\alpha(3-2\alpha)(1-2\alpha)^{2}p^{6}+36(1-2\alpha)p^{2}(4-p^{2})(1-y^{2})
+32(4−p2)2y2+4(3−10α+8α2)p3y(4−p2))\displaystyle\quad\quad\quad\quad\quad\quad+32(4-p^{2})^{2}y^{2}+4(3-10\alpha+8\alpha^{2})p^{3}y(4-p^{2})\bigg{)} (2.6)

with p∈(0,2)p\in(0,2) and y∈(0,1)y\in(0,1). On solving ∂s2/∂p\partial s_{2}/\partial p and ∂s2/∂y\partial s_{2}/\partial y, to find the points of maxima. After resolving ∂s2/∂y=0,\partial s_{2}/\partial y=0, we get

y=p3​(3−10​α+8​α2)2​(17​p2−32−18​p2​α)(=:y0).y=\dfrac{p^{3}(3-10\alpha+8\alpha^{2})}{2(17p^{2}-32-18p^{2}\alpha)}(=:y_{0}). (2.7)

Upon calculations, we observe that to have y0∈(0,1)y_{0}\in(0,1) for the given range of yy, p=:p0>≈A(α)p=:p_{0}>\approx A(\alpha) (see Fig. 1) is needed with α∈[0,β0)\alpha\in[0,\beta_{0}). This β0∈[0,1)\beta_{0}\in[0,1) is the smallest positive root of −3+10​α−8​α2=0-3+10\alpha-8\alpha^{2}=0 and no such p∈(0,2)p\in(0,2) exists when α∈(β0,1)\alpha\in(\beta_{0},1). It is to be noted that the expression of A​(α)A(\alpha) is complex but the coefficient of its imaginary part for α∈[0,1/6]\alpha\in[0,1/6] is highly negative (of order 10−15)10^{-15}), which can be neglected and A​(α)A(\alpha) can be treated as a real number. Here,

A​(α)\displaystyle A(\alpha) :=13​(−3+10​α−8​α2)​(2​(18​α−17)−24/3​(1−i​3)​(18​α−17)2B−(1+i​3)​B24/3),\displaystyle:=\frac{1}{3(-3+10\alpha-8\alpha^{2})}\bigg{(}2(18\alpha-17)-\frac{2^{4/3}(1-i\sqrt{3})(18\alpha-17)^{2}}{B}-\frac{(1+i\sqrt{3})B}{2^{4/3}}\bigg{)},

with

B:=(C+−256​(18​α−17)6+C2)1/3B:=\bigg{(}C+\sqrt{-256(18\alpha-17)^{6}+C^{2}}\bigg{)}^{1/3}

and

C:=−63056+146016​α−8640​α2−183168​α3+110592​α4.C:=-63056+146016\alpha-8640\alpha^{2}-183168\alpha^{3}+110592\alpha^{4}.

Based on computations, ∂s2/∂p=0\partial s_{2}/\partial p=0 gives

0\displaystyle 0 =16​p​(9−18​α−y2​(25−18​α))−2​p2​y​(3−10​α+8​α2)​(5​p2−12)\displaystyle=16p(9-18\alpha-y^{2}(25-18\alpha))-2p^{2}y(3-10\alpha+8\alpha^{2})(5p^{2}-12)
+3​α​(1−2​α)2​(3−2​α)​p5−8​p3​(9−18​α−y2​(17−18​α)).\displaystyle\quad+3\alpha(1-2\alpha)^{2}(3-2\alpha)p^{5}-8p^{3}(9-18\alpha-y^{2}(17-18\alpha)). (2.8)

After substituting equation (2.7) into equation (2.8), we have

0\displaystyle 0 =p(49152(1−2α)−3072p2(25−68α+36α2)−p8(1−2α)2(153−1437α\displaystyle=p\bigg{(}49152(1-2\alpha)-3072p^{2}(25-68\alpha+36\alpha^{2})-p^{8}(1-2\alpha)^{2}(153-1437\alpha
+3118α2−2484α3+648α4)+16p4(2427−7890α+6020α2+616α3−1024α4)\displaystyle\quad+3118\alpha^{2}-2484\alpha^{3}+648\alpha^{4})+16p^{4}(2427-7890\alpha+6020\alpha^{2}+616\alpha^{3}-1024\alpha^{4})
−128p6(48−153α+20α2+340α3−352α4+96α5)).\displaystyle\quad\quad-128p^{6}(48-153\alpha+20\alpha^{2}+340\alpha^{3}-352\alpha^{4}+96\alpha^{5})\bigg{)}. (2.9)

A numerical calculation suggests that the solution of (2.9) in the interval (0,2)(0,2) is p≈B​(α)p\approx B(\alpha) whenever α∈[0,α2)\alpha\in[0,\alpha_{2}), where α2∈[0,1)\alpha_{2}\in[0,1) is the smallest positive root of 153−1437​α+3118​α2−2484​α3+648​α4=0153-1437\alpha+3118\alpha^{2}-2484\alpha^{3}+648\alpha^{4}=0, otherwise no such p∈(0,2)p\in(0,2) exists, see Fig. 1. Thus, s2s_{2} does not have any critical point in (0,2)×(0,1)(0,2)\times(0,1).
Here

B​(α):\displaystyle B(\alpha): =12[{1F(−3072+3648α+6016α2−9728α3+3072α4\displaystyle=\frac{1}{\sqrt{2}}\bigg{[}\bigg{\{}\frac{1}{F}\bigg{(}-3072+3648\alpha+6016\alpha^{2}-9728\alpha^{3}+3072\alpha^{4}
−4​F3​{1E2​(768​J2(1−2​α)2+2​G​E(1−2​α)+H​EI+I​E(1−2​α)2)}1/2\displaystyle\quad\quad\quad-\frac{4F}{\sqrt{3}}\bigg{\{}\frac{1}{E^{2}}\bigg{(}\frac{768J^{2}}{(1-2\alpha)^{2}}+\frac{2GE}{(1-2\alpha)}+\frac{HE}{I}+\frac{IE}{(1-2\alpha)^{2}}\bigg{)}\bigg{\}}^{1/2}
+4​F3{−1E3(−1536​J2​E(1−2​α)2−4​G​E2(1−2​α)+H​E2I+I​E2(1−2​α)2\displaystyle\quad\quad\quad+\frac{4F}{\sqrt{3}}\bigg{\{}\frac{-1}{E^{3}}\bigg{(}\frac{-1536J^{2}E}{(1-2\alpha)^{2}}-\frac{4GE^{2}}{(1-2\alpha)}+\frac{HE^{2}}{I}+\frac{IE^{2}}{(1-2\alpha)^{2}}
−K(1−2​α)3​{1E2​(768​J2(1−2​α)2+2​G​E(1−2​α)+H​EI+I​E(1−2​α)2)}1/2)}1/2)}1/2],\displaystyle\quad\quad\quad-\frac{K}{(1-2\alpha)^{3}\bigg{\{}\frac{1}{E^{2}}\bigg{(}\frac{768J^{2}}{(1-2\alpha)^{2}}+\frac{2GE}{(1-2\alpha)}+\frac{HE}{I}+\frac{IE}{(1-2\alpha)^{2}}\bigg{)}\bigg{\}}^{1/2}}\bigg{)}\bigg{\}}^{1/2}\bigg{)}\bigg{\}}^{1/2}\bigg{]},

with

E\displaystyle E :=153−1437​α+3118​α2−2484​α3+648​α4;\displaystyle:=153-1437\alpha+3118\alpha^{2}-2484\alpha^{3}+648\alpha^{4};
F\displaystyle F :=(1−2​α)​E;\displaystyle:=(1-2\alpha)E;
G\displaystyle G :=2427−3036​α−52​α2+512​α3;\displaystyle:=2427-3036\alpha-52\alpha^{2}+512\alpha^{3};
H\displaystyle H :=8217−173160​α+1260312​α2−2415264​α3+2091664​α4−1048576​α5+262144​α6;\displaystyle:=8217-173160\alpha+1260312\alpha^{2}-2415264\alpha^{3}+2091664\alpha^{4}-1048576\alpha^{5}+262144\alpha^{6};
I\displaystyle I :=(127065213−1886889978α+12579196752α2−49871499552α3+132494582880α4\displaystyle:=\bigg{(}127065213-1886889978\alpha+12579196752\alpha^{2}-49871499552\alpha^{3}+132494582880\alpha^{4}
−253944918720​α5+368411062528​α6−410152327680​α7+340236674304​α8\displaystyle\quad\quad-253944918720\alpha^{5}+368411062528\alpha^{6}-410152327680\alpha^{7}+340236674304\alpha^{8}
−196757891584​α9+71861010432​α10−14168358912​α11+1073741824​α12\displaystyle\quad\quad-196757891584\alpha^{9}+71861010432\alpha^{10}-14168358912\alpha^{11}+1073741824\alpha^{12}
+2886((1−2α)8(3−4α)2(3604621581−39763739565α+202125486510α2\displaystyle\quad\quad+288\sqrt{6}\bigg{(}(1-2\alpha)^{8}(3-4\alpha)^{2}(3604621581-39763739565\alpha+202125486510\alpha^{2}
−633657349224​α3+1436021769744​α4−2516421142080​α5+34829931648​α6\displaystyle\quad\quad-633657349224\alpha^{3}+1436021769744\alpha^{4}-2516421142080\alpha^{5}+34829931648\alpha^{6}
−3872882513280​α7+3466438619648​α8−2402201403136​α9+1198713174528​α10\displaystyle\quad\quad-3872882513280\alpha^{7}+3466438619648\alpha^{8}-2402201403136\alpha^{9}+1198713174528\alpha^{10}
−394099818496α11+75581358080α12−6442450944α13))1/2)1/3;\displaystyle\quad\quad-394099818496\alpha^{11}+75581358080\alpha^{12}-6442450944\alpha^{13})\bigg{)}^{1/2}\bigg{)}^{1/3};
J\displaystyle J :=−48+57​α+94​α2−152​α3+48​α4;\displaystyle:=-48+57\alpha+94\alpha^{2}-152\alpha^{3}+48\alpha^{4};

and

K\displaystyle K :=2883(15963705−129546873α+510658314α2−1308834456α3+2415583204α4\displaystyle:=288\sqrt{3}\bigg{(}15963705-129546873\alpha+510658314\alpha^{2}-1308834456\alpha^{3}+2415583204\alpha^{4}
−3321041560​α5+3420107120​α6−2619528992​α7+1464766656​α8\displaystyle\quad\quad\quad\quad\quad-3321041560\alpha^{5}+3420107120\alpha^{6}-2619528992\alpha^{7}+1464766656\alpha^{8}
−575732096α9+147709696α10−21284352α11+1179648α12).\displaystyle\quad\quad\quad\quad\quad-575732096\alpha^{9}+147709696\alpha^{10}-21284352\alpha^{11}+1179648\alpha^{12}\bigg{)}.
Refer to caption
Figure 1. Graphical representation of pp versus α\alpha. Here, B​(α)B(\alpha) (Red) and A​(α)A(\alpha) (blue) do not intersect for any choice of α\alpha. Dashed black line represents p=2p=2.

On x=1x=1, Z​(p,x,y)Z(p,x,y) reduces into

s3​(p,y):\displaystyle s_{3}(p,y): =(1−α)2576(288α+16p2(11−33α+9α2)−8p4(5−13α+5α2+3α3)\displaystyle=\dfrac{(1-\alpha)^{2}}{576}\bigg{(}288\alpha+16p^{2}(11-33\alpha+9\alpha^{2})-8p^{4}(5-13\alpha+5\alpha^{2}+3\alpha^{3})
−p6(1−4α+6α2−16α3+4α4)),p∈(0,2).\displaystyle\quad\quad\quad\quad\quad\quad-p^{6}(1-4\alpha+6\alpha^{2}-16\alpha^{3}+4\alpha^{4})\bigg{)},\quad p\in(0,2). (2.10)

While computing ∂s3/∂p=0\partial s_{3}/\partial p=0, p=:p0≈2L(α)p=:p_{0}\approx 2L(\alpha) for α∈[0,α0)∪(α0,α1)\alpha\in[0,\alpha_{0})\cup(\alpha_{0},\alpha_{1}), comes out to be the critical point, where α0∈[0,1)\alpha_{0}\in[0,1) is the smallest positive root of 1−4​α+6​α2−16​α3+4​α4=01-4\alpha+6\alpha^{2}-16\alpha^{3}+4\alpha^{4}=0 and α1(≈0.370803927)∈[0,1)\alpha_{1}(\approx 0.370803927)\in[0,1) (see Fig. 2) is the largest value so that p∈(0,2)p\in(0,2) otherwise no such real p∈(0,2)p\in(0,2) exists beyond this α1\alpha_{1}. Here

L​(α):=−10+26​α−10​α2−6​α3+M3​N;M:=133−751​α+1497​α2−1630​α3+1666​α4−708​α5+144​α6;N:=1−4​α+6​α2−16​α3+4​α4.}\displaystyle\left.\begin{array}[]{cc}&L(\alpha):=\sqrt{\dfrac{-10+26\alpha-10\alpha^{2}-6\alpha^{3}+M}{3N}};\\ &M:=\sqrt{133-751\alpha+1497\alpha^{2}-1630\alpha^{3}+1666\alpha^{4}-708\alpha^{5}+144\alpha^{6}};\\ &N:=1-4\alpha+6\alpha^{2}-16\alpha^{3}+4\alpha^{4}.\end{array}\right\} (2.11)

Undergoing simple calculations, s3s_{3} achieves its maximum value, approximately equals P​(α)P(\alpha), [0,α0)∪(α0,α1)[0,\alpha_{0})\cup(\alpha_{0},\alpha_{1}) at p0p_{0}. Here

P(α):=(1−α)2486(243α−18​(11−33​α+9​α2)​(10−26​α+10​α2+6​α3−M)N−12​(5−13​α+5​α2+3​α3)​(−10+26​α−10​α2−6​α3+M)2N2−2​(−10+26​α−10​α2−6​α3+M)3N2).}\displaystyle\left.\begin{array}[]{cc}&P(\alpha):=\dfrac{(1-\alpha)^{2}}{486}\bigg{(}243\alpha-\frac{18(11-33\alpha+9\alpha^{2})(10-26\alpha+10\alpha^{2}+6\alpha^{3}-M)}{N}\\ &\quad\quad\quad\quad\quad\quad\quad\quad-\frac{12(5-13\alpha+5\alpha^{2}+3\alpha^{3})(-10+26\alpha-10\alpha^{2}-6\alpha^{3}+M)^{2}}{N^{2}}\\ &\quad\quad\quad\quad-\frac{2(-10+26\alpha-10\alpha^{2}-6\alpha^{3}+M)^{3}}{N^{2}}\bigg{)}.\end{array}\right\} (2.12)
Refer to caption
Figure 2. Graphical representation of p0p_{0} versus α\alpha. Here, Re⁡(p0)\operatorname{Re}(p_{0}) (green) and Im⁡(p0)\operatorname{Im}(p_{0}) (red) represent the value of p0p_{0} at different α\alpha, where α1\alpha_{1} (blue circle) is the point at which p0p_{0} transforms from completely real to imaginary. Dashed black line represents p0=2p_{0}=2.

On y=0y=0, Z​(p,x,y)Z(p,x,y) can be seen as

s4​(p,x):\displaystyle s_{4}(p,x): =(1−α)21152(576(x−x3(1−α))+16p2(9−18x+x4+x3(28−33α)\displaystyle=\dfrac{(1-\alpha)^{2}}{1152}\bigg{(}576\bigg{(}x-x^{3}(1-\alpha)\bigg{)}+16p^{2}\bigg{(}9-18x+x^{4}+x^{3}(28-33\alpha)
−18α+x2(2−15α+18α2))−4p4(9+2x4+x3(20−21α)\displaystyle\quad\quad\quad\quad\quad-18\alpha+x^{2}(2-15\alpha+18\alpha^{2})\bigg{)}-4p^{4}\bigg{(}9+2x^{4}+x^{3}(20-21\alpha)
−18α+x2(1−32α+52α2)+x(−12+19α−32α2+12α3))\displaystyle\quad\quad\quad\quad\quad-18\alpha+x^{2}(1-32\alpha+52\alpha^{2})+x(-12+19\alpha-32\alpha^{2}+12\alpha^{3})\bigg{)}
+p6(x4+α(1−2α)2(3−2α)+x3(1+3α)\displaystyle\quad\quad\quad\quad\quad+p^{6}\bigg{(}x^{4}+\alpha(1-2\alpha)^{2}(3-2\alpha)+x^{3}(1+3\alpha)
−x2(1+17α−34α2)+x(−3+19α−32α2+12α3))).\displaystyle\quad\quad\quad\quad\quad-x^{2}(1+17\alpha-34\alpha^{2})+x(-3+19\alpha-32\alpha^{2}+12\alpha^{3})\bigg{)}\bigg{)}.

Furthermore, through some calculations, such as

∂s4∂x\displaystyle\dfrac{\partial s_{4}}{\partial x} =(1−α)21152(576(1−3x2(1−α))−16p2(18−4x3−3x2(28−33α)\displaystyle=\dfrac{(1-\alpha)^{2}}{1152}\bigg{(}576\bigg{(}1-3x^{2}(1-\alpha)\bigg{)}-16p^{2}\bigg{(}18-4x^{3}-3x^{2}(28-33\alpha)
−2x(2−15α+18α2))+p6(4x3+3x2(1+3α)−2x(1+17α\displaystyle\quad\quad\quad\quad\quad\quad-2x(2-15\alpha+18\alpha^{2})\bigg{)}+p^{6}\bigg{(}4x^{3}+3x^{2}(1+3\alpha)-2x(1+17\alpha
−34α2)−3+19α−32α2+12α3)−4p4(8x3+3x2(20−21α)\displaystyle\quad\quad\quad\quad\quad\quad-34\alpha^{2})-3+19\alpha-32\alpha^{2}+12\alpha^{3}\bigg{)}-4p^{4}\bigg{(}8x^{3}+3x^{2}(20-21\alpha)
+2x(1−32α+52α2)−12+19α−32α2+12α3))\displaystyle\quad\quad\quad\quad\quad\quad+2x(1-32\alpha+52\alpha^{2})-12+19\alpha-32\alpha^{2}+12\alpha^{3}\bigg{)}\bigg{)}

and

∂s4∂p\displaystyle\dfrac{\partial s_{4}}{\partial p} =(1−α)21152(32p(9−18x+x4+x3(28−33α)−18α+x2(2−15α+18α2))\displaystyle=\dfrac{(1-\alpha)^{2}}{1152}\bigg{(}32p\bigg{(}9-18x+x^{4}+x^{3}(28-33\alpha)-18\alpha+x^{2}(2-15\alpha+18\alpha^{2})\bigg{)}
−16p3(9+2x4+x3(20−21α)−18α+x2(1−32α+52α2)\displaystyle\quad\quad\quad\quad\quad\quad-16p^{3}\bigg{(}9+2x^{4}+x^{3}(20-21\alpha)-18\alpha+x^{2}(1-32\alpha+52\alpha^{2})
+x(−12+19α−32α2+12α3))+6p5(x4+α(1−2α)2(3−2α)\displaystyle\quad\quad\quad\quad\quad\quad+x(-12+19\alpha-32\alpha^{2}+12\alpha^{3})\bigg{)}+6p^{5}\bigg{(}x^{4}+\alpha(1-2\alpha)^{2}(3-2\alpha)
+x3(1+3α)−x2(1+17α−34α2)+x(−3+19α−32α2+12α3))),\displaystyle\quad\quad\quad\quad\quad\quad+x^{3}(1+3\alpha)-x^{2}(1+17\alpha-34\alpha^{2})+x(-3+19\alpha-32\alpha^{2}+12\alpha^{3})\bigg{)}\bigg{)},

indicates that there does not exist any common solution for the system of equations ∂s4/∂x=0\partial s_{4}/\partial x=0 and ∂s4/∂p=0\partial s_{4}/\partial p=0, thus, s4s_{4} has no critical points in (0,2)×(0,1)(0,2)\times(0,1).

On y=1y=1, Z​(p,x,y)Z(p,x,y) reduces to

s5​(p,x):\displaystyle s_{5}(p,x): =(1−α)21152(64px(1−x2)(5+x−12α)+64(8−7x2−x4+9α)\displaystyle=\dfrac{(1-\alpha)^{2}}{1152}\bigg{(}64px(1-x^{2})(5+x-12\alpha)+64(8-7x^{2}-x^{4}+9\alpha)
+16p3(1−x2)(3−x−2x2−10α+6xα+8α2)−2p6(1−4α\displaystyle\quad\quad\quad\quad\quad+16p^{3}(1-x^{2})(3-x-2x^{2}-10\alpha+6x\alpha+8\alpha^{2})-2p^{6}(1-4\alpha
−16α3+4α4)+16p2(6+14x2+2x4+x(9−18α)−66α\displaystyle\quad\quad\quad\quad\quad-16\alpha^{3}+4\alpha^{4})+16p^{2}\bigg{(}6+14x^{2}+2x^{4}+x(9-18\alpha)-66\alpha
+18α2−9x3(1−2α))+4p5(1−x2)(x2−3+10α−8α2\displaystyle\quad\quad\quad\quad\quad+18\alpha^{2}-9x^{3}(1-2\alpha)\bigg{)}+4p^{5}(1-x^{2})\bigg{(}x^{2}-3+10\alpha-8\alpha^{2}
−x(4−6α))−4p4(7x2+x4+x(9−18α)−9x3(1−2α)\displaystyle\quad\quad\quad\quad\quad-x(4-6\alpha)\bigg{)}-4p^{4}\bigg{(}7x^{2}+x^{4}+x(9-18\alpha)-9x^{3}(1-2\alpha)
+4(3−13α+5α2+3α3))).\displaystyle\quad\quad\quad\quad\quad+4(3-13\alpha+5\alpha^{2}+3\alpha^{3})\bigg{)}\bigg{)}.

We note that the equations ∂s5/∂x=0\partial s_{5}/\partial x=0 and ∂s5/∂p=0\partial s_{5}/\partial p=0 possess no common solution in (0,2)×(0,1).(0,2)\times(0,1).

Case III: Now, we determine the maximum values that Z​(p,x,y)Z(p,x,y) may obtain on the edges of the cuboid YY.
From equation (2.6), we have

Z​(p,0,0)=r1​(p):=p2(1−α)2(1−2α)(144−36p2+p4α(3−8α+4α2)1152.Z(p,0,0)=r_{1}(p):=\frac{p^{2}(1-\alpha)^{2}(1-2\alpha)(144-36p^{2}+p^{4}\alpha(3-8\alpha+4\alpha^{2})}{1152}.

Here, we consider the following three subcases for different choices of α\alpha.

  1. (1)

    For α=0\alpha=0, r1​(p)r_{1}(p) reduces to p2​(4−p2)/32p^{2}(4-p^{2})/32 and r1′​(p)=0r_{1}^{\prime}(p)=0 for p=0p=0, the point of minima and p=2p=\sqrt{2}, the point of maxima. Therefore

    Z​(p,0,0)≤18,p∈[0,2].Z(p,0,0)\leq\frac{1}{8},\quad p\in[0,2].
  2. (2)

    For α=1/2\alpha=1/2, r1​(p)=0r_{1}(p)=0.

  3. (3)

    For α=(0,1/2)∪(1/2,1),\alpha=(0,1/2)\cup(1/2,1), r1′​(p)=p​(1−α)2​(1−2​α)​(48−24​p2+p4​α​(3−8​α+4​α2))=0r^{\prime}_{1}(p)=p(1-\alpha)^{2}(1-2\alpha)(48-24p^{2}+p^{4}\alpha(3-8\alpha+4\alpha^{2}))=0 for p=0p=0 and p=2(((3−R(α))/(3α−8α2+4α3))1/2p=2\bigg{(}((3-R(\alpha))/(3\alpha-8\alpha^{2}+4\alpha^{3})\bigg{)}^{1/2} as the points of minima and maxima respectively. So,

    Z​(p,0,0)≤R0​(α):=(1−α)2​(3−R​(α))​(−3+6​α−16​α2+8​α3+R​(α))6​(3−2​α)2​α2​(1−2​α),Z(p,0,0)\leq R_{0}(\alpha):=\frac{(1-\alpha)^{2}(3-R(\alpha))(-3+6\alpha-16\alpha^{2}+8\alpha^{3}+R(\alpha))}{6(3-2\alpha)^{2}\alpha^{2}(1-2\alpha)},

    with R​(α):=3​(3−3​α+8​α2−4​α3)R(\alpha):=\sqrt{3(3-3\alpha+8\alpha^{2}-4\alpha^{3})}.

Now, equation (2.6) at y=1,y=1, implies that Z​(p,0,1)=r2​(p):=(1−α)2​(32​(4−p2)2+α​(1−2​α)2​(3−2​α)​p6+4​p3​(4−p2)​(3−10​α+8​α2))/1152.Z(p,0,1)=r_{2}(p):=(1-\alpha)^{2}(32(4-p^{2})^{2}+\alpha(1-2\alpha)^{2}(3-2\alpha)p^{6}+4p^{3}(4-p^{2})(3-10\alpha+8\alpha^{2}))/1152. Note that r2′​(p)r_{2}^{\prime}(p) is a decreasing function in [0,2][0,2] and hence p=0p=0 becomes the point of maxima. Thus

Z​(p,0,1)≤4​(1−α)29,p∈[0,2].Z(p,0,1)\leq\dfrac{4(1-\alpha)^{2}}{9},\quad p\in[0,2].

Through calculations, equation (2.6) shows that Z​(0,0,y)Z(0,0,y) attains its maximum value at y=1,y=1, which implies that

Z​(0,0,y)≤4​(1−α)29,y∈[0,1].Z(0,0,y)\leq\dfrac{4(1-\alpha)^{2}}{9},\quad y\in[0,1].

Since, the equation (2.10) is free from yy, we have

Z​(p,1,1)=Z​(p,1,0)=r3​(p):\displaystyle Z(p,1,1)=Z(p,1,0)=r_{3}(p): =(1−α)2576(288α+16p2(11−33α+9α2)\displaystyle=\frac{(1-\alpha)^{2}}{576}\bigg{(}288\alpha+16p^{2}(11-33\alpha+9\alpha^{2})
−8​p4​(5−13​α+5​α2+3​α3)\displaystyle\quad\quad\quad\quad\quad\quad-8p^{4}(5-13\alpha+5\alpha^{2}+3\alpha^{3})
−p6(1−4α+6α2−16α3+4α4)).\displaystyle\quad\quad\quad\quad\quad\quad-p^{6}(1-4\alpha+6\alpha^{2}-16\alpha^{3}+4\alpha^{4})\bigg{)}.

Now, r3′​(p)=32​p​(11−33​α+9​α2)−32​p3​(5−13​α+5​α2+3​α3)−6​p5​(1−4​α+6​α2−16​α3+4​α4)=0r_{3}^{\prime}(p)=32p(11-33\alpha+9\alpha^{2})-32p^{3}(5-13\alpha+5\alpha^{2}+3\alpha^{3})-6p^{5}(1-4\alpha+6\alpha^{2}-16\alpha^{3}+4\alpha^{4})=0 when p=δ1:=0p=\delta_{1}:=0 and p=δ2:=2​L​(α)p=\delta_{2}:=2L(\alpha) for α∈[0,α0)∪(α0,α1)\alpha\in[0,\alpha_{0})\cup(\alpha_{0},\alpha_{1}), as the points of minima and maxima respectively, in the interval [0,2][0,2]. The justification of P​(α)P(\alpha), α0\alpha_{0} and α1\alpha_{1} are provided above through equation (2.11) and (2.12). Thus, from equation (2.10),

Z​(p,1,1)=Z​(p,1,0)≤P​(α),p∈[0,2]andα∈[0,α0)∪(α0,α1).Z(p,1,1)=Z(p,1,0)\leq P(\alpha),\quad p\in[0,2]\quad\text{and}\quad\alpha\in[0,\alpha_{0})\cup(\alpha_{0},\alpha_{1}).

Consider equation (2.10) at p=0p=0, we get

Z​(0,1,y)=α​(1−α)22.Z(0,1,y)=\frac{\alpha(1-\alpha)^{2}}{2}.

Equation (2.5) indicates that

Z​(2,1,y)=Z​(2,0,y)=Z​(2,x,0)=Z​(2,x,1)=α​(1−2​α)2​(1−α)2​(3−2​α)18.Z(2,1,y)=Z(2,0,y)=Z(2,x,0)=Z(2,x,1)=\dfrac{\alpha(1-2\alpha)^{2}(1-\alpha)^{2}(3-2\alpha)}{18}.

Using equation (2.4), Z​(0,x,1)=r4​(x):=(1−α)2​(8−7​x2−x4+9​x3​α)/18.Z(0,x,1)=r_{4}(x):=(1-\alpha)^{2}(8-7x^{2}-x^{4}+9x^{3}\alpha)/18. Upon calculations, we see that r4r_{4} is a decreasing function of xx in [0,1][0,1] and therefore x=0x=0 is the point of maxima. Hence

Z​(0,x,1)≤4​(1−α)29,x∈[0,1].Z(0,x,1)\leq\dfrac{4(1-\alpha)^{2}}{9},\quad x\in[0,1].

On again using equation (2.4), Z​(0,x,0)=r5​(x):=x​(1−(1−α)​x2)​(1−α)2/2.Z(0,x,0)=r_{5}(x):=x(1-(1-\alpha)x^{2})(1-\alpha)^{2}/2. Moreover, r5′​(x)=0r_{5}^{\prime}(x)=0 when x=δ3:=1/3​(1−α).x=\delta_{3}:=1/\sqrt{3(1-\alpha)}. Observe that r5​(x)r_{5}(x) increases in [0,δ3)[0,\delta_{3}) and decreases in (δ3,1].(\delta_{3},1]. Hence,

Z​(0,x,0)≤(1−α)23​3​(1−α),x∈[0,1].Z(0,x,0)\leq\frac{(1-\alpha)^{2}}{3\sqrt{3(1-\alpha)}},\quad x\in[0,1].

We also provide a graphical representation of six upper-bounds (u.b) of H3​(1)H_{3}(1) in Fig. 3. Given all the cases, the sharp inequality |H3​(1)|≤4​(1−α)2/9|H_{3}(1)|\leq 4(1-\alpha)^{2}/9, holds for every α∈[0,1/6]∪{1/2}\alpha\in[0,1/6]\cup\{1/2\}.

Refer to caption
Figure 3. Graph of six upper-bounds (u.b) versus α\alpha. The upper-bounds (u.b) of H3​(1)H_{3}(1) are 4​(1−α)2/94(1-\alpha)^{2}/9 (red), (1−α)2/(3​3​(1−α))(1-\alpha)^{2}/(3\sqrt{3(1-\alpha)}) (green), α​(1−2​α)2​(1−α)2​(3−2​α)/18\alpha(1-2\alpha)^{2}(1-\alpha)^{2}(3-2\alpha)/18 (blue), α​(1−2​α)2/2\alpha(1-2\alpha)^{2}/2 (black), P​(α)P(\alpha) (pink) and R0​(α)R_{0}(\alpha) (cyan) for α∈[0,1/6]\alpha\in[0,1/6].

Let the function f0∈𝒮∗​(α):𝔻→ℂf_{0}\in\mathcal{S}^{*}(\alpha):\mathbb{D}\rightarrow\mathbb{C}, be defined as

f0​(z)=z​exp⁡(∫0z1+(1−2​α)​t31−t3−1t​𝑑t)=z+2​(1−α)3​z4+(1−α)​(5−2​α)9​z7+⋯,f_{0}(z)=z\exp\bigg{(}\int_{0}^{z}\dfrac{\frac{1+(1-2\alpha)t^{3}}{1-t^{3}}-1}{t}dt\bigg{)}=z+\dfrac{2(1-\alpha)}{3}z^{4}+\dfrac{(1-\alpha)(5-2\alpha)}{9}z^{7}+\cdots,

with f0​(0)=0f_{0}(0)=0 and f0′​(0)=1f_{0}^{\prime}(0)=1, plays the role of an extremal function for the inequality presented in equation (2.1) with a2=a3=a5=0a_{2}=a_{3}=a_{5}=0 and a4=2​(1−α)/3a_{4}=2(1-\alpha)/3. ∎

Now, we provide remarks which incorporate the bound of |H3​(1)||H_{3}(1)| for the class 𝒮∗\mathcal{S}^{*} and 𝒮∗​(1/2)\mathcal{S}^{*}(1/2), which are subclasses of 𝒮∗​(α)\mathcal{S}^{*}(\alpha), given as follows:

Remark 2.2.

On substituting α=0\alpha=0 in Theorem 2.1, 𝒮∗​(0)=𝒮∗\mathcal{S}^{*}(0)=\mathcal{S}^{*} and from equation (2.1), we get |H3​(1)|≤4/9|H_{3}(1)|\leq 4/9. This bound is sharp and coincides with that of Kowalczyk et al.[6] and Banga and Kumar [2].

Remark 2.3.

On substituting α=1/2\alpha=1/2 in Theorem 2.1, 𝒮∗​(1/2)\mathcal{S}^{*}(1/2) and from equation (2.1), we get |H3​(1)|≤1/9|H_{3}(1)|\leq 1/9. This bound is sharp and coincides with that of Rath et al.[18].

For some already known sharp bounds of H3​(1)H_{3}(1), regarding various choices of φ​(z)\varphi(z), See Table 1. We note that the same bound is not available for φ​(z):=1+z−z3/3\varphi(z):=1+z-z^{3}/3. Hence, as an application of Theorem 2.1, we provide a better bound of |H3​(1)||H_{3}(1)| for functions belonging to the class, 𝒮N​e∗:=𝒮∗​(1+z−z3/3)\mathcal{S}_{Ne}^{*}:=\mathcal{S}^{*}(1+z-z^{3}/3).

Corollary 2.4.

If f∈𝒮N​e∗f\in\mathcal{S}_{Ne}^{*}. Then |H3​(1)|≤32/81≈0.395062|H_{3}(1)|\leq 32/81\approx 0.395062.

Proof.

From [20], we have

min|z|=r⁡Re⁡(φ​(z))={1−r+13​r3,r≤1/31−13​(1+r2)3/2,r≥1/3.\min_{|z|=r}\operatorname{Re}(\varphi(z))=\begin{cases}1-r+\frac{1}{3}r^{3},&r\leq 1/\sqrt{3}\\ 1-\frac{1}{3}(1+r^{2})^{3/2},&r\geq 1/\sqrt{3}.\end{cases}

We note that α=min|z|=r⁡Re⁡(φ​(z))=1−2​2/3\alpha=\min_{|z|=r}\operatorname{Re}(\varphi(z))=1-2\sqrt{2}/3 as rr tends to 11. Now, substitution of α=1−2​2/3≈0.057191∈[0,1/6]\alpha=1-2\sqrt{2}/3\approx 0.057191\in[0,1/6] in equation (2.1) implies that |H3​(1)|≤32/81≈0.395062|H_{3}(1)|\leq 32/81\approx 0.395062. ∎

Open Problem:
We have attempted to provide the sharp bound of H3​(1)H_{3}(1) for functions, f∈𝒮∗​(α)f\in\mathcal{S}^{*}(\alpha) for α∈[0,1/6]∪{1/2}\alpha\in[0,1/6]\cup\{1/2\} in Theorem 2.1. Further, this result is still open for the remaining range of α\alpha in [0,1)[0,1).

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