EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 5968 ISSN 1307-5543 – ejpam.com Published by New York Business Global Second Hankel Determinant for a Bi-univalent Function Subclass Based Gegenbauer (Ultraspherical) Polynomials Abdelbaset Zeyani1, Abdulmtalb Hussen2,∗ 1 Department of Mathematics and Statistics, Wichita State University, Wichita, KS, USA 2 School of Engineering, Math, and Technology, Navajo Technical University, Crownpoint NM, USA 3 Mathematics Department, College of Education, Al Zintan University, Dirj, Libya Abstract. In this paper, we aim to establish a new upper bound approximation for the second Hankel determinant utilizing a certain subclass of the class of normalized analytic and bi-univalent functions in the open unit disk U . These functions have inverses with a bi-univalent analytic con- tinuation to U and are associated with orthogonal polynomials; namely, Gegenbauer polynomials that satisfy subordination conditions on U. Finally, we introduce new essential results derived by specializing the parameter τ employed in our foundational finding. 2020 Mathematics Subject Classifications: 30C45 Key Words and Phrases: Gegenbauer (Ultraspherical) Polynomials, Bi-univalent Analytic Functions, Hankel Determinant 1. Introduction The nth - degree Gegenbauer (or ultraspherical) polynomials (GPS), denoted here by U (β) k (t), with parameter β at the point t are recursively defined by U (β) 0 (t) = 1, U (β) 1 (t) = 2βt, U (β) k (t) = 1 k [ 2t (k + β − 1) U (β) k−1(t)− (k + 2β − 2) U (β) k−2(t) ] , k ≥ 2. (1) These polynomials are orthogonal on the interval I = [−1, 1] with respect to the weight function (1 − t2) β− 1 2 , where β > −1 2 . That is, for any two GPS, U (β) k (t) and U (β) l (t), with k ̸= l, we have ∫ 1 −1 U (β) k (t) U (β) l (t) ( 1− t2 )β− 1 2 dt = 0, ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5968 Email addresses: abdelbaset.zeyani@wichita.edu (A. Zeyani), ahussen@navajotech.edu (A. Hussen) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 2 of 17 and with the condition that l = k, we have∫ 1 −1 ( U (β) k (t) )2 ( 1− t2 )β− 1 2 dt = √ π Γ(k + 2β − 1) 21−2β k! Γ(β) Γ(k + β − 1 2) . For β > 0, a generating function of GPS, G β (t, ζ), is defined by the form G β (t, ζ) = 1 (1− 2tζ + ζ2)β = ∞∑ k=0 U (β) k (t) ζk, (2) where t ∈ I, ζ is in the open unit disk U = {ζ : ζ ∈ C and |ζ| < 1}, and C is, as usual, the set of complex numbers. For a fixed t ∈ I, G β (t, ζ) is analytic in U that has a Taylor series expansion given by (2). Evidently, we see that G β (t, ζ) produces no values when β = 0. Therefore, the generating function of GPS is set to be of the form G0(t, ζ) = 1− log ( 1− 2 t ζ + ζ2 ) = ∞∑ k=0 U (0) k (t) ζk. (3) Note that U (β) k (t) are particular solutions of the Gegenbauer differential equation given by (1− t2) d 2 y dt2 − (2β + 1) t dy dt + k (k + 2β) y = 0, (4) and when setting (i) β = 1/2, equation (4) reduces to the Legendre differential equations, and the GPS reduce to the Legendre polynomials. (ii) β = 1, equation (4) reduces to the Chebyshev differential equations, and the GPS reduce to the Chebyshev polynomials of the second kind. Let A denote the class of all functions of the form f(ζ) = ζ + ∞∑ n=2 an ζ n, (ζ ∈ U), (5) which are analytic in U and normalized by these two conditions f(0) = 0 and f ′ (0) = 1. Moreover, let S be the subclass of A consisting of all normalized univalent functions of the form (5) which are also univalent in U. Two functions, f and g, are said to be subordinate ( f ≺ g ) if there is an analytic function h(ζ) (namely; a Schwarz function) in U, such that f(ζ) = g(h(ζ)) with h(0) = 0 and | h(ζ) | ≤ 1. Especially, if the function g is univalent in U, then the following equivalence is valid [1] f(ζ) ≺ g(ζ) ⇐⇒ f(0) = g(0) and f(U) ⊂ g(U). A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 3 of 17 The Koebe One-Quarter Theorem [2] states that the image of U under every function f ∈ S contains a disk of radius 1 4 and center at the origin; i.e., U 1 4 (0) ∈ f(U). Therefore, every univalent function f ∈ S has an inverse f −1 : f(U) → U which satisfies the following conditions: ( f −1 ◦ f )(ζ) = ζ ( ζ ∈ U ) and ( f ◦ f −1 )(η) = η ( |η| < r0(f); r0(f) ≥ 1 4 ) , where f −1 has the series expansion of the form f −1 (η) = η − a2 η 2 + (2a2 2 − a3) η 3 − (5a3 2 − 5a2a3 + a4) η 4 + · · · . (6) A function f ∈ A is said to be bi-univalent in U if both f and f −1 are univalent in U . Let Ξ be denoting the class of bi-univalent functions in U given by (5). Herein, we recall the following examples of functions in the bi-univalent function class Ξ that have apparently revived the study of bi-univalent functions in recent years: f1(ζ) = ζ 1− ζ , f2(ζ) = −log(1− ζ), and f3(ζ) = 1 2 log (1 + ζ 1− ζ ) , where their inverses are respectively given by f −1 1 (η) = η 1 + η , f −1 2 (η) = eη − 1 eη , and f −1 3 (η) = e2η − 1 e2η + 1 . However, the familiar Koebe function, K(ζ) = ζ (1−ζ)2 , is not a member of the bi-univalent function class Ξ since it maps the open unit disk U ⊂ C onto the set K(U) = C\(−∞,−1 4 ], which does not contain U (i.e., {η : η ∈ C and |η| ≤ 1 4} ⊆ K(U)) (see [3–11]). Other common univalent functions in S that are not members of Ξ are ϑ1(ζ) = ζ 1− ζ2 and ϑ2(ζ) = ζ − ζ2 2 . Historically speaking, certain subclasses of Ξ were introduced by Brannan and Taha (see [12]) similar to the familiar subclasses S∗(ε) and K(ε) of star-like and convex func- tions of order ε ∈ [ 0, 1) in the open unit disk U, which are respectively defined by S∗(ε) = { f : f ∈ S and ℜ { ζ f ′ (ζ) f(ζ) } > ε, ζ ∈ U } , and K(ε) = { f : f ∈ S and ℜ { 1 + ζ f ′′ (ζ) f ′ (ζ) } > ε, ζ ∈ U } . A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 4 of 17 Analogously, the bi-star-like and bi-convex function classes S∗ Ξ (ε) and KΞ(ε) of order ε ∈ [ 0, 1) in the open unit disk U, corresponding to S∗(ε) and K(ε), were introduced and studied, and non-sharp upper bound estimations of the initial Taylor-Maclaurin coefficients were obtained as well. In 1976, Noonan and Thomas defined the qth Hankel determinant of the function f ∈ A of the form (5) for integers n, q ∈ N = {1, 2, 3, . . . } by [13] H f (n, q) = ∣∣∣∣∣∣∣∣∣ an an+1 · · · an+q−1 an+1 an+2 · · · an+q ... ... ... ... an+q−1 an+q · · · an+2q−2 ∣∣∣∣∣∣∣∣∣ , a1 := 1. In particular, it is observed that, for n = 1, 2 and q = 2, the Hankel determinants are given by H f (1, 2) = ∣∣∣∣ a1 a2 a2 a3 ∣∣∣∣ = a3 − a2 2 and H f (2, 2) = ∣∣∣∣ a2 a3 a3 a4 ∣∣∣∣ = a2 a4 − a2 3 , which are respectively referred to as the well-known Fekete-Szegö functional and second Hankel determinant functional. Several other authors have considered the determinant H f (n, q) in their studies. In [14], for instance, Noor found that the rate of growth of H f (n, q) as n → ∞ occurs when functions f ∈ A of the form (5) with bounded boundary. The authors in [14] and [15], in particular, achieved sharp upper bounds on H f (2, 2) for several types of function classes. For f ∈ S in the open unit disk U, the authors in [16] obtained the sharp upper inequality for the functional H f (1, 2) that is given by |H f (1, 2) | = | a3−a2 2 | ≤ 1. Several authors have recently investigated the upper bounds of H f (1, 2) and Taylor-Maclaurin coefficients for various subclasses of bi-univalent functions (see, for examples, [17–30]). Furthermore, the subclass of S consisting of all functions whose derivatives have positive real part, introduced in [31], was considered by the authors of [32] in order to derive the sharp bounds for the functional H f (2, 2) that is given by |H f (2, 2) | = | a2a4 − a2 3 | ≤ 4 9 for each function belongs to that subclass. In addition, they discovered the sharp second Hankel determinant, H f (2, 2), in (see [32]) for star-like and convex function subclasses, S∗ and K, of S with bounds of |H f (2, 2) | = | a2a4−a2 3 | ≤ 1 8 and |H f (2, 2) | = | a2a4 − a2 3 | ≤ 1, respectively. In recent times, several researchers have explored upper bounds for the coefficients and Hankel determinant of functions within different subclasses of univalent functions (see, for examples, [33–36]). Definition 1. Let τ ∈ [0, 1] and t ∈ (12 , 1]. A function f ∈ Ξ of the form (5) is said to be in the class Ωβ Ξ(t, τ) with a nonzero real constant β if the following subordinations are satisfied τ ( 1 + ζ f ′′ (ζ) f ′(ζ) ) + (1− τ) ( ζ f ′ (ζ) f(ζ) ) ≺ G β (t, ζ) (7) and τ ( 1 + η g ′′ (η) g′(η) ) + (1− τ) ( η g ′ (η) g(η) ) ≺ G β (t, η), (8) A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 5 of 17 where the function g = f −1 is defined by (6) and G β is the GPS - generating function given by (2). Remark 1. [37] By setting τ = 0 in (1), we obtain the class Ωβ Ξ(t, 0) = Σβ Ξ(t) that consists of functions f ∈ Ξ satisfying the conditions ζ f ′ (ζ) f(ζ) ≺ G β (t, ζ) (9) and η g ′ (η) g(η) ≺ G β (t, η), (10) where the function g = f −1 is defined by (6). Remark 2. [37] By setting τ = 1 in (1), we obtain the class Ωβ Ξ(t, 1) = Dβ Ξ(t) that consists of functions f ∈ Ξ satisfying the conditions 1 + ζ f ′′ (ζ) f ′(ζ) ≺ G β (t, ζ) (11) and 1 + η g ′′ (η) g′(η) ≺ G β (t, η), (12) where the function g = f −1 is defined by (6). Let Q : U → C be the class of functions s(ζ) with positive real part consisting of all analytic functions satisfying the conditions that s(0) = 1 and ℜ (s(ζ)) > 0. To derive our desirable upper bounds estimation for the second Hankel determinant H f (2, 2) = a2a4−a2 3 associated with the class Ωβ Ξ(t, τ) in (1), we state the necessary lemmas: Lemma 1. [38] If the function s ∈ Q is defined by s(ζ) = 1 + ∞∑ k=1 s k ζk, (13) then | s k | ≤ 2, k = 1, 2, ... Lemma 2. [39] If the function s ∈ Q is of the form (13), then 2s2 = s2 1 + (4− s2 1 ) ξ (14) and 4s3 = s3 1 + 2(4− s2 1 ) s1 ξ − s1(4− s2 1 ) ξ2 + 2(4− s2 1 )(1− |ξ|2) ζ (15) for some ξ and ζ with | ξ | ≤ 1 and | ζ | ≤ 1. A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 6 of 17 Also by considering the class ∆ that consists of all analytic functions ω ∈ U satisfying the conditions that ω(0) = 0 and |ω(ζ) | < 1 for all ζ ∈ U, we state the following lemma: Lemma 3. [2] Let ω ∈ ∆ with ω(ζ) = ∑∞ k=1 ωk ζ k, ζ ∈ U. Then |ω1 | ≤ 1 and |ωk | ≤ 1− |ω1 | 2 for k ≥ 2. 2. Second Hankel Determinant Theorem 1. Let the function f ∈ Ξ of the form (5) be in the class Ωβ Ξ(t, τ) in (1). Then ∣∣ a2a4 − a2 3 ∣∣ ≤  T (2−, t) E1 ≥ 0 and E2 ≥ 0; max { 4β2 t2 (1+2 τ)2 , T (2−, t) } E1 > 0 and E2 < 0; 4β2 t2 (1+2τ)2 E1 ≤ 0 and E2 ≤ 0; max { T (c0 , t), T (2−, t) } E1 < 0 and E2 > 0, (16) where T (2−, t) = ( U (β) 1 (t) )2 (1 + 2 τ)2 + E1 + E2 3 (1 + 3 τ) (1 + τ)3 (1 + 2 τ)2 , (17) T (c0 , t) = ( U (β) 1 (t) )2 (1 + 2 τ)2 − E2 2 12 E1 (1 + 3 τ) (1 + τ)3 (1 + 2 τ)2 ; c0 = √ −2 E2 E1 , (18) E1 = 16 (1 + 2 τ)2 U (β) 1 (t) ∣∣∣∣∣ ( U (β) 3 (t) + U (β) 2 (t) + 1 4 U (β) 1 (t) ) (1 + τ)2 − ( U (β) 1 (t) )3∣∣∣∣∣ + ( U (β) 1 (t) )2[ 3 (1 + 3 τ)(1 + τ)3 − 12 (1 + τ)2(1 + 2 τ)2 ] − 2 (1 + τ)(1 + 2 τ) U (β) 1 (t) [ 3 (1 + 3 τ) ( U (β) 1 (t) )2 + 8 (1 + τ)(1 + 2 τ) U (β) 2 (t) ] , (19) E2 = 12 (1 + 2 τ)2(1 + τ)2 ( U (β) 1 (t) )2 + 6 (1 + τ)(1 + 2 τ)(1 + 3 τ) ( U (β) 1 (t) )3 + 16 (1 + τ)2(1 + 2 τ)2 U (β) 1 (t) U (β) 2 (t)− 6 (1 + 3 τ)(1 + τ)3 ( U (β) 1 (t) )2 , (20) and U (β) 1 (t), U (β) 2 (t), and U (β) 3 (t) are defined by (1). A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 7 of 17 Proof. Suppose f ∈ Ωβ Ξ(t, τ) for some τ ∈ [0, 1]. Then from (7) and (8) we have τ ( 1 + ζ f ′′ (ζ) f ′(ζ) ) + (1− τ) ( ζ f ′ (ζ) f(ζ) ) ≺ G β (t, u(ζ)) (21) and τ ( 1 + η g ′′ (η) g′(η) ) + (1− τ) ( η g ′ (η) g(η) ) ≺ G β (t, v(η)), (22) where g = f −1 and u, v ∈ ∆ are given by u(ζ) = ∞∑ n=1 cn ζ n and v(η) = ∞∑ n=1 dn η n. Then by using G β (t, ζ) given in (2), we can write the right hand sides of (21) and (22) as follows: G β (t, u(ζ)) = 1 + U (β) 1 (t) c1 ζ + [ U (β) 1 (t) c2 + U (β) 2 (t) c21 ] ζ2 + [ U (β) 1 (t) c3 + 2U (β) 2 (t) c1 c2 + U (β) 3 (t) c31 ] ζ3 + · · · (23) and G β (t, u(η)) = 1 + U (β) 1 (t) d1 η + [ U (β) 1 (t) d2 + U (β) 2 (t) d21 ] η2 + [ U (β) 1 (t) d3 + 2U (β) 2 (t) d1 d2 + U (β) 3 (t) d31 ] η3 + · · · . (24) Therefore, (21) and (22) become τ [ 1 + 2 a2 ζ + (6 a3 − 4 a22) ζ 2 + 2 (4 a32 − 9 a2 a3 + 6a4)ζ 3 + · · · ] + (1− τ) [ 1 + a2 ζ + (2 a3 − a22) ζ 2 + (a32 − 3 a2 a3 + 3a4)ζ 3 + · · · ] = 1 + U (β) 1 (t) c1 ζ + [ U (β) 1 (t) c2 + U (β) 2 (t) c21 ] ζ2 + [ U (β) 1 (t) c3 + 2U (β) 2 (t) c1 c2 + U (β) 3 (t) c31 ] ζ3 + · · · , (25) A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 8 of 17 and τ [ 1− 2 a2 η + (8 a22 − 6 a3)η 2 + (−32 a32 + 42 a2 a3 − 12 a4) η 3 + · · · ] + (1− τ) [ 1− a2 η + (3 a22 − 2 a3) η 2 + (−10 a32 + 12 a2 a3 − 3 a4) η 3 + · · · ] = 1 + U (β) 1 (t) d1 η + [ U (β) 1 (t) d2 + U (β) 2 (t) d21 ] η2 + [ U (β) 1 (t) d3 + 2U (β) 2 (t) d1 d2 + U (β) 3 (t) d31 ] η3 + · · · (26) At this point, the corresponding coefficients in (25) and (26) can be equated to obtain (1 + τ) a2 = U (β) 1 (t) c1, (27) 2 (1 + 2 τ) a3 − (1 + 3τ) a22 = U (β) 1 (t) c2 + U (β) 2 (t) c21, (28) (1+7 τ) a32− 3 (1+5 τ) a2 a3+3 (1+3 τ) a4 = U (β) 1 (t) c3+2U (β) 2 (t) c1 c2+U (β) 3 (t) c31, (29) −(1 + τ) a2 = U (β) 1 (t) d1, (30) (3 + 5 τ) a22 − 2 (1 + 2 τ) a3 = U (β) 1 (t) d2 + U (β) 2 (t) d21, (31) and −2 (5 + 11 τ) a32 + 6 (2 + 5 τ) a2 a3 − 3 (1 + 3 τ) a4 = U (β) 1 (t) d3 + 2U (β) 2 (t) d1 d2 + U (β) 3 (t) d31. (32) From (27) and (30), we obtain that c1 = − d1 (33) and a2 = U (β) 1 (t) 1 + τ c1. (34) Upon subtracting (31) from (28), we have that a3 = a22 + U (β) 1 (t) (c2 − d2) 4 (1 + 2 τ) = ( U (β) 1 (t) )2 (1 + τ)2 c21 + U (β) 1 (t) (c2 − d2) 4 (1 + 2 τ) . (35) Furthermore, if we subtract (32) from (29), together with (27), (33) and (35), we have a4 = 5 ( U (β) 1 (t) )2 (c2 − d2) c1 8 (1 + τ) (1 + 2 τ) + U (β) 1 (t) (c3 − d3) 6(1 + 3τ) + U (β) 2 (t) (c2 + d2) c1 3 (1 + 3 τ) + [ U (β) 3 (t) 3 (1 + 3 τ) + 2 (1 + 4 τ) ( U (β) 1 (t) )3 3 (1 + τ)3 (1 + 3 τ) ] c31. (36) A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 9 of 17 Thus, when applying (27), (35), and (36), we can simply establish that a2 a4 − a23 = ( U (β) 1 (t) )3 (c2 − d2) c 2 1 8 (1 + τ)2 (1 + 2 τ) + ( U (β) 1 (t) )2 (c3 − d3) c1 6 (1 + τ) (1 + 3 τ) + U (β) 1 (t) U (β) 2 (t) (c2 + d2) c 2 1 3 (1 + τ) (1 + 3 τ) − ( U (β) 1 (t) )2 (c2 − d2) 2 16 (1 + 2 τ)2 + U (β) 1 (t) [ U (β) 3 (t) (1 + τ)2 − ( U (β) 1 (t) )3] c41 3 (1 + 3 τ) (1 + τ)3 . (37) Next, according to Lemma (2), we now have that c2 − d2 = 4− c2 2 (x− y), (38) c2 + d2 = c21 + 4− c2 2 (x+ y), (39) and c3 − d3 = c31 2 + (4− c2) c1 2 (x+ y)− (4− c21) c1 4 (x2 + y2) + 4− c21 2 [( 1− |x|2 ) ζ − ( 1− |y|2 ) η ] , (40) for some x, y, ζ, and η with |x| ≤ 1, |y| ≤ 1, |ζ| ≤ 1, and |η| ≤ 1. Then, by substituting (38), (39), and (40) into (37), we obtain that ∣∣∣ a2 a4 − a23 ∣∣∣ ≤ U (β) 1 (t) ∣∣∣(U (β) 3 (t) + U (β) 2 (t) + 1 4 U (β) 1 (t) ) (1 + τ)2 − ( U (β) 1 (t) )3∣∣∣ 3 (1 + 3 τ) (1 + τ)3 c41 + ( U (β) 1 (t) )2 (4− c21) c1 6 (1 + τ) (1 + 3 τ) + [(U (β) 1 (t) )3 (4− c21) c 2 1 16 (1 + τ)2 (1 + 2 τ) + ( U (β) 1 (t) )2 (4− c21) c 2 1 12 (1 + τ) (1 + 3 τ) + U (β) 1 (t) U (β) 2 (t) (4− c21) c 2 1 6 (1 + τ) (1 + 3 τ) ](∣∣x∣∣+ ∣∣y∣∣) + [(U (β) 1 (t) )2 (4− c21) c 2 1 24 (1 + τ) (1 + 3 τ) − ( U (β) 1 (t) )2 (4− c21) c1 12 (1 + τ) (1 + 3 τ) ](∣∣x∣∣2 + ∣∣y∣∣2) + [(U (β) 1 (t) )2 ( 4− c21 )2 64 (1 + 2 τ)2 ](∣∣x∣∣+ ∣∣y∣∣)2. (41) Lemma (1) allows us to assume, without any loss of generality, that c ∈ [0, 2] where c = |c1 |. Thus, for δ1 = | x | ≤ 1 and δ2 = | y | ≤ 1, we can rewrite (41) to be in the A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 10 of 17 following form:∣∣∣ a2 a4 − a23 ∣∣∣ ≤ Υ1 +Υ2 (δ1 + δ2) + Υ3 ( δ21 + δ22 ) +Υ4 ( δ1 + δ2 )2 = φ(δ1, δ2), (42) where Υ1 = U (β) 1 (t) ∣∣∣(U (β) 3 (t) + U (β) 2 (t) + 1 4 U (β) 1 (t) ) (1 + τ)2 − ( U (β) 1 (t) )3∣∣∣ 3 (1 + 3 τ) (1 + τ)3 c4+ ( U (β) 1 (t) )2 (4− c2) c 6 (1 + τ) (1 + 3 τ) ≥ 0, (43) Υ2 = [(U (β) 1 (t) )3 (4− c2) c2 16 (1 + τ)2 (1 + 2 τ) + ( U (β) 1 (t) )2 (4− c2) c2 12 (1 + τ) (1 + 3 τ) + U (β) 1 (t) U (β) 2 (t) (4− c2) c2 6 (1 + τ) (1 + 3 τ) ] ≥ 0, (44) Υ3 = [(U (β) 1 (t) )2 (4− c2) (c− 2) c 24 (1 + τ) (1 + 3 τ) ] ≤ 0, (45) and Υ4 = [(U (β) 1 (t) )2 ( 4− c2 )2 64 (1 + 2 τ)2 ] ≥ 0. (46) Now, we have to maximize the function φ(δ1, δ2) in (42) on the closed square S = [0, 1]× [0, 1] by investigating the maximum values of φ(δ1, δ2) in accordance with 0 < c < 2, c = 0, and c = 2. For the case that 0 < c < 2, since Υ3 < 0 and Υ3 + 2Υ4 > 0 for all t ∈ (12 , 1), we deduce that φ δ1δ1 φ δ2δ2 − φ2 δ1δ2 < 0, for all δ1, δ2 ∈ S. Therefore, as a result of this, the function φ cannot have a local maximum in the interior of the square S. Now, we will explore the maximum value of φ on the boundary of S. (1) for δ1 = 0 and 0 ≤ δ2 ≤ 1 (similarly, for δ2 = 0 and 0 ≤ δ1 ≤ 1), φ(δ1, δ2) takes the form ψ1(δ2) := φ(0, δ2) = Υ1 +Υ2 δ2 + (Υ3 +Υ4) δ 2 2. Next, we will separately discuss the following two cases. Case (i) : When Υ3 +Υ4 ≥ 0, for 0 < δ2 < 1, for any fixed c ∈ (0, 2), and for all t ∈ (12 , 1), it is obvious that ψ ′ 1 (δ2) = Υ2 + 2 (Υ3 +Υ4) δ2 > 0. Case (ii) : When Υ3 + Υ4 < 0 and since Υ2 + 2 (Υ3 + Υ4) ≥ 0, for 0 < δ2 < 1, for any fixed c ∈ (0, 2), and for all t ∈ (12 , 1), it is obvi- ous that Υ2 + 2 (Υ3 + Υ4) < Υ2 + 2 (Υ3 + Υ4) δ2 < Υ2 and thus ψ ′ 1 (δ2) = Υ2 + 2 (Υ3 +Υ4) δ2 > 0 . A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 11 of 17 In both cases, ψ1(δ2) is an increasing function and; therefore, for any fixed c ∈ (0, 2) and t ∈ (12 , 1), the maximum value of ψ1(δ2) occurs at δ2 = 1 and max δ2 { ψ1(δ2) } = ψ1(1) = Υ1 +Υ2 +Υ3 +Υ4. (47) For c = 0 and c = 2, we respectively obtain that φ(δ1, δ2) = Υ4 ∣∣∣ c=0 = ( U (β) 1 (t) )2 4 (1 + 2 τ)2 ( δ1 + δ2 )2 and φ(δ1, δ2) = Υ1 ∣∣∣ c=2 = 16 U (β) 1 (t) ∣∣∣(U (β) 3 (t) + U (β) 2 (t) + 1 4 U (β) 1 (t) ) (1 + τ)2 − ( U (β) 1 (t) )3∣∣∣ 3 (1 + 3 τ) (1 + τ)3 . (48) By taking equation (48) and the two mentioned cases in account, for 0 ≤ δ2 < 1, for any fixed c ∈ [0, 2], and for all t ∈ (12 , 1), the maximum value of ψ1(δ2) is max δ2 { ψ1(δ2) } = ψ1(1) = Υ1 +Υ2 +Υ3 +Υ4. (2) for δ1 = 1 and 0 ≤ δ2 ≤ 1 (similarly, for δ2 = 1 and 0 ≤ δ1 ≤ 1), φ(δ1, δ2) takes the form ψ2(δ2) := φ(1, δ2) = (Υ3 +Υ4) δ 2 2 + (Υ2 + 2Υ4) δ2 +Υ1 +Υ2 +Υ3 +Υ4. Analogous to the previously mentioned cases of Υ3 +Υ4, we conclude that max δ2 { ψ2(δ2) } = ψ2(1) = Υ1 + 2Υ2 + 2Υ3 + 4Υ4. (49) Since ψ1(1) ≤ ψ2(1) for c ∈ (0, 2) and t ∈ (12 , 1), we see that max δ1 , δ2 { φ(δ1, δ2) } = φ(1, 1) (50) on the boundary of S. Therefore, the maximum value of φ(δ1, δ2) occurs at δ1 = 1 and δ2 = 1 in the closed square S. Now, for a fixed value of t, let T : [0, 2] → R be the function defined by T (c, t) = max δ1 , δ2 ( φ(δ1, δ2) ) = φ(1, 1) = Υ1 + 2Υ2 + 2Υ3 + 4Υ4 (51) Upon substituting the expressions of Υ1, Υ2, Υ3, and Υ4 into (51), we obtain that T (c, t) = ( U (β) 1 (t) )2 (1 + 2 τ)2 + E1 c4 + 4 E2 c2 48 (1 + 3 τ) (1 + τ)3 (1 + 2 τ)2 , (52) A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 12 of 17 where E1 = 16 (1 + 2 τ)2 U (β) 1 (t) ∣∣∣∣∣(U (β) 3 (t) + U (β) 2 (t) + 1 4 U (β) 1 (t) ) (1 + τ)2 − ( U (β) 1 (t) )3∣∣∣∣∣ + ( U (β) 1 (t) )2 [ 3 (1 + 3 τ) (1 + τ)3 − 12 (1 + τ)2 (1 + 2 τ)2 ] − 2 (1 + τ) (1 + 2 τ) U (β) 1 (t) [ 3 (1 + 3 τ) ( U (β) 1 (t) )2 + 8 (1 + τ) (1 + 2 τ)U (β) 2 (t) ] (53) and E2 = 12 (1 + 2 τ)2 (1 + τ)2 ( U (β) 1 (t) )2 + 6 (1 + τ) (1 + 2 τ) (1 + 3 τ) ( U (β) 1 (t) )3 + 16 (1 + τ)2 (1 + 2 τ)2 U (β) 1 (t) U (β) 2 (t)− 6 (1 + 3 τ) (1 + τ)3 ( U (β) 1 (t) )2 . (54) By assuming that the function T (c, t) has a maximum value at an interior point 0 < c < 2, we obtain d T d c = E1 c3 + 2 E2 c 12 (1 + 3 τ) (1 + τ)3 (1 + 2 τ)2 . (55) With some calculations, we can examine the sign of d T d c taking into account the following four cases. (i) Suppose that E1 ≥ 0 and E2 ≥ 0, then d T d c ≥ 0; indicating that T (c, t) is an increasing function. Therefore, we get that max 0< c<2 { T (c, t) } = T (2−, t) = ( U (β) 1 (t) )2 (1 + 2 τ)2 + E1 + E2 3 (1 + 3 τ) (1 + τ)3 (1 + 2 τ)2 , (56) which means: max 0< c<2 { max S { φ(δ1, δ2) }} = T (2−, t). (ii) Suppose that E1 > 0 and E2 < 0, then c0 = √ −2 E2 E1 is a critical value of the function T (c, t). By assuming c0 ∈ (0, 2), we get that d2 T d c2 ∣∣∣ c=c0 > 0, that is, c = c0 is a local minimum value of T (c, t). Thus, the function T (c, t) can not possess a local maximum. A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 13 of 17 (iii) Suppose that E1 ≤ 0 and E2 ≤ 0, then d T d c ≤ 0; indicating that T (c, t) is a decreasing function. Thus, max 0< c<2 { T (c, t) } = T (0+, t) = 4Υ4 = ( U (β) 1 (t) )2 (1 + 2 τ)2 . (57) (iv) Suppose that E1 < 0 and E2 > 0, then c0 is a critical value of the function T (c, t). By assuming c0 ∈ (0, 2), we obtain that d2 T d c2 ∣∣∣ c=c0 < 0, which means that the function T (c, t) has a local maximum occurring at c = c0 . Thus, max 0< c<2 { T (c, t) } = T (c0 , t), (58) where T (c0 , t) = ( U (β) 1 (t) )2 (1 + 2 τ)2 − E2 2 12 E1 (1 + 3 τ) (1 + τ)3 (1 + 2 τ)2 . Therefore, the proof of the above Theorem is evidently completed. Ultimately, we introduce two essential corollaries that obtained from the classes Σβ Ξ(t) and Λβ Ξ(t). Corollary 1. Let f ∈ Ξ of the form (5) be in the class Ωβ Ξ(t, 0) = Λβ Ξ(t). Then ∣∣ a2a4 − a2 3 ∣∣ ≤  T (2−, t) E∗ 1 ≥ 0 and E∗ 2 ≥ 0; max t { 4β2 t2, T (2−, t) } E∗ 1 > 0 and E∗ 2 < 0; 4β2 t2 E∗ 1 ≤ 0 and E∗ 2 ≤ 0; max t { T (c0 , t), T (2−, t) } E∗ 1 < 0 and E∗ 2 > 0, (59) where T (2−, t) = 4β2 t2 + E∗ 1 + E∗ 2 3 , (60) T (c0 , t) = 4β2 t2 − E∗2 2 12 E∗ 1 ; c0 = √ −2 E∗ 2 E∗ 1 , (61) E∗ 1 = 16 U (β) 1 (t) ∣∣∣∣∣U (β) 3 (t) + U (β) 2 (t) + 1 4 U (β) 1 (t)− ( U (β) 1 (t) )3∣∣∣∣∣ − 9 ( U (β) 1 (t) )2 − 2 U (β) 1 (t) [ 3 ( U (β) 1 (t) )2 + 8 U (β) 2 (t) ] , (62) A. Zeyani, A. Hussen / Eur. J. Pure Appl. Math, 18 (2) (2025), 5968 14 of 17 E∗ 2 = 2 U (β) 1 (t) [ 3 U (β) 1 (t) + 3 ( U (β) 1 (t) )2 + 8 U (β) 2 (t) ] , (63) and U (β) 1 (t), U (β) 2 (t), and U (β) 3 (t) are defined by (1). Corollary 2. Let f ∈ Ξ of the form (5) be in the class Ωβ Ξ(t, 1) = Σβ Ξ(t). Then ∣∣ a2a4 − a2 3 ∣∣ ≤  T (2−, t) D∗ 1 ≥ 0 and D∗ 2 ≥ 0; max t { 4β2 t2 9 , T (2−, t) } D∗ 1 > 0 and D∗ 2 < 0; 4β2 t2 9 D∗ 1 ≤ 0 and D∗ 2 ≤ 0; max t { T (c0 , t), T (2−, t) } D∗ 1 < 0 and D∗ 2 > 0, (64) where T (2−, t) = 4β2 t2 9 + D∗ 1 +D∗ 2 864 , (65) T (c0 , t) = 4β2 t2 9 − D∗2 2 3456D∗ 1 , c0 = √ −2D∗ 2 D∗ 1 , (66) D∗ 1 = 144 U (β) 1 (t) ∣∣∣∣∣4(U (β) 3 (t) + U (β) 2 (t) + 1 4 U (β) 1 (t) ) − ( U (β) 1 (t) )3∣∣∣∣∣ − 336 ( U (β) 1 (t) )2 − 144 U (β) 1 (t) [( U (β) 1 (t) )2 + 4 U (β) 2 (t) ] , (67) D∗ 2 = 48 U (β) 1 (t) [ 5 U (β) 1 (t) + 3 ( U (β) 1 (t) )2 + 12 U (β) 2 (t) ] , (68) and U (β) 1 (t), U (β) 2 (t), and U (β) 3 (t) are defined by (1). 3. 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