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Research article

Differential subordination, superordination results associated with Pascal distribution

  • Received: 21 September 2022 Revised: 01 January 2023 Accepted: 05 January 2023 Published: 31 January 2023
  • MSC : 52A41, 32W50

  • This paper aims to study differential subordination and superordination preserving properties for certain analytic univalent functions with in the open unit disk. In the present investigation, we obtain some subordination and superordination results involving Pascal distribution series for certain normalized analytic functions in the open unit disk. Also we estimate the sandwich results for the same class.

    Citation: K. Saritha, K. Thilagavathi. Differential subordination, superordination results associated with Pascal distribution[J]. AIMS Mathematics, 2023, 8(4): 7856-7864. doi: 10.3934/math.2023395

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  • This paper aims to study differential subordination and superordination preserving properties for certain analytic univalent functions with in the open unit disk. In the present investigation, we obtain some subordination and superordination results involving Pascal distribution series for certain normalized analytic functions in the open unit disk. Also we estimate the sandwich results for the same class.



    Let be the family of holomorphic functions in Δ={z:|z|<1} and [t,n] be the subclass of involving the functions which can be defined by

    g(z)=t+tnzn+tn+1zn+1+..., (1.1)

    let Q be the subclass of involving the function defined by

    g(z)=z+n=2tnzn.

    Let β,h and consider θ(u,v,w,z):c3×ΔC. If β and θ(β(z),zβ(z),z2β(z),z) are univalent and if β satisfies the second order superordination,

    h(z)θ(β(z),zβ(z),z2β(z),z), (1.2)

    then β is a solution of the differential subordination [2]. (If g is subordinate to G, then G is superordinate to g). An holomorphic function α is called a subordinate if αβ for every β satisfying [2]. A univalent subordinant ¯α that satisfies α¯α for all subordinates [2] α(z) is called the best subordinant.

    Miller and Mocanu [5] found the conditions on h,α and θ it can be given by

    h(z)θ(β(z),zβ(z),z2β(z);z)α(z)β(z). (1.3)

    For two holomorphic functions

    λ(z)=z+n=2tnznandμ(z)=z+n=2rnzn,tn,rn0.

    The Hadamard product (or) convolution of λ and μ given below

    (λμ)(z)=z+n=2tnrnzn=(μλ)(z). (1.4)

    A variable X is said to have the Pascal distribution series if it takes the values 0, 1, 2, 3... with the probabilities

    (1α)r,αr(1α)r1!α2r(r+1)(1α)r2!,α3r(r+1)(r+2)(1α)r3!...,

    respectively where α, r are called the parameters and thus

    P(X=K)=(k+r1r1)αk(1α)r,k0,1,2,3,...

    Many essentially interesting proof techniques involving a power series, whose co-efficients are probabilities of the Pascal distribution series introduced by sheeza et al. [12] that is

    Qrα=z+k=2(k+r2r1)αk1(1α)rzk,zΔ,(r1,0α1).

    The first order differential subordination and superordination which was introduced and studied by Miller, Mocanu and Bulboaca [1,2,5]. Also recently studied by various authors for example Magesh and Murugusundaramoorthy [3,7,8,9], Magesh et al. [4], and Shanmugam et al. [11] and also obtained sandwich results for various classes of holomorphic functions.

    In the present article we determine some sufficient condition for the holomorphic function in Δ to satisfy

    α1(z)Qrα(z)α2(z), (1.5)

    where α1,α2 are given univalent functions in Δ with α1(0)=1,α2(0)=1.

    To prove our results we need the following lemmas and definitions.

    Lemma 2.1. [10] The function

    M(z,n)=t1(n)z+t2(n)z2+...witht1(n)0forn0andlimn|t1(n)|=

    is a subordination chain, if

    R{zM(z,n)zM(z,n)n}>0,zΔ,n0.

    Definition 2.2. [5] Denote by T, the set of all functions f that are holomorphic and one to one on ¯ΔE(f) where,

    E(f)={ζΔ:limzζf(z)=},

    and are such that f(ζ)0, for ζΔE(f).

    Lemma 2.3. [6] Let α be univalent in the unit disc Δ and ψ and θ be holomorphic in a domain D containing α(Δ) with θ(ω)0 when ωα(Δ). Set

    T(z)=zα(z)θ(α(z))andh(z)=ψ(α(z))+T(z),

    suppose that

    (1) T(z) is starlike univalent in Δ.

    (2) R{zh(z)T(z)}>0, for zΔ.

    If β is holomorphic with α(0)=β(0), β(Δ)D, and

    ψ(β(z))+zβ(z)θ(β(z))ψ(α(z))+zα(z)θ(α(z)), (2.1)

    then

    β(z)α(z),

    and α is th best dominant.

    Lemma 2.4. [2] Let α be convex univalent in the unit disk Δ and υ and ϱ be holomorphic in a domain D containing α(Δ). Suppose that

    (1) R{υ(α(z))ϱ(α(z))}>0, for zΔ.

    (2) ϕ(z)=zα(z)ϱ(α(z)) is starlike univalent in Δ.

    If β(z)[α(0),1]T with β(Δ)D and υ(α(z))+zβ(z)ϱ(β(z)) is univalent in Δ and

    υ(α(z))+zα(z)ϱ(α(z))υ(β(z))+zβ(z)ϱ(β(z)), (2.2)

    then

    α(z)β(z),

    and α is the best subordinant.

    To prove our following theorem need to using above Lemma 2.3.

    Theorem 3.1. Let QrαQ, ηiC(i=1,2,3),(η30),C, such that 0, α be convex univalent with α(0)=1, and assume that

    R{1η3+1(η1+η2α)η3+1+zαα(1+η2α(η1+η2α)2η3)+αα(η1(η1+η2α)2η3)zα(η2(η1+η2α)2η3)}, (3.1)

    which is greater than zero, zΔ.

    If gQ satisfies

    (ηi)31(g;Qrα)=(g,Qrα,η1,η2,η3)α(z)+zα(z)η1+η2α(z)+η3zα(z), (3.2)

    where

    (ηi)31(g;Qrα)=(Qrα(z)z)+zα(z)(Q(z))α(z)Q(z)(η1+η2(Qrα(z)z))Q(z)+η3[zα(z)(Q(z)Q(z))α(z)], (3.3)

    then

    (z+k=2(k+r2r1)αk1(1α)rzkz)α(z),

    and α is the best dominant.

    Proof. Define the function β by

    β(z)=(z+k=2(k+r2r1)αk1(1α)rzkz),(zΔ), (3.4)

    then the function β is holomorphic in Δ and β(0)=1. Therfore, by making use of (3.4), we obtain

    (Qrα(z)z)+z(Qrαz)(Q(z))α(z)Q(z)(η1+η2)(Qrα(z)z)Q(z)+η3[zα(z)(Q(z)Q(z))ηα(z)]=β(z)+zβ(z)η1+η2β(z)+η3zβ(z), (3.5)

    by using (3.5) in (3.2), we have

    β(z)+zβ(z)η1+η2β(z)+η3zβ(z)α(z)+zα(z)η1+η2α(z)+η3zα(z). (3.6)

    By setting

    ψ(α(z))=α(z)+zα(z)η1+η2α(z),

    and

    θ(z)=η3zα(z).

    This is easily observed that ψ(α(z)), θ(z) are holomorphic in c{0} and θ(z)0. Also we see that

    T(z)=zα(z)θ(α(z))=η3zα(z),

    and

    h(z)=ψ(α(z))+T(z)=α(z)+zα(z)η1+η2α(z)+η3zα(z).

    Here T(z) is starlike univalent in Δ and we get the result

    R{zh(z)T(z)}=R{1η3+1(η1+η2α)η3+1+...}>0.

    Hence the theorem.

    By taking

    α(z)=1+Az1+Bz(1BA1)

    in Theorem 4.1, we obtain the following corollary.

    Corollary 3.2. Let ηiC(i=1,2,3),(η30),C,s.t0 be convex univalent with α(0)=1 and (3.1) hold true. For g,QrαQ, let (Qrαz)H[1,1]T and (ηi)31(g;Qrα) defined in (3.3) be univalent in Δ satisfying

    (ηi)31(g;Qrα)1+Az1+Bz+z(AB)η1(1+Bz)2+η2(1+Az)(1+Bz)+η3z(AB(1+Bz)2),

    then

    (z+k=2(k+r2r1)αk1(1α)rzkz)1+Az1+Bz,

    and 1+Az1+Bz is the best dominant.

    Using Lemma 2.4 to prove the following theorem.

    Theorem 4.1. Let Qrα(z)Q, ηiC(i=1,2,3),(η30),C,s.t 0, α be convex univalent with α(0)=1, and assume that

    R{1η3+1η1+η2α)η3}0. (4.1)

    If gQ, Qrα(z)H,[α(0),1]T, Let (ηi)31(g;Qrα) be univalent in Δ and

    α(z)+zα(z)η1+η2α(z)+η3zα(z)(ηi)31(g;Qrα), (4.2)

    where (ηi)31(g;Qrα) is given in (3.3), then

    α(z)(Qrαz),

    and α is the best subordinant.

    Proof. The function β is defined by

    β(z)=(Qrαz), (4.3)

    simplify above equation, we get

    (ηi)31(g;Qrα)=β(z)+zβ(z)η1+η2β(z)+η3zβ(z),

    then

    α(z)+zα(z)η1+η2α(z)+η3zα(z)β(z)+zβ(z)η1+η2β(z)+η3zβ(z).

    By setting

    υ(z)=α(z)+zα(z)η1+η2α(z)andϱ(z)=η3zα(z).

    Here υ(α(z)) is holomorphic in C. Also ϱ(z) is holomorphic in C{0} and ϱ(z)0. Consider,

    M(z,n)=υ(α(z))+ϱ(α(z))nzα(z)=α(z)+zα(z)η1+η2α(z)+η3nzα(z)=t1(n)z+t2(n)z+...,

    differentiating the above equation with respect to z and n, we have

    M(z,n)z=α(z)+(η1+η2α)[zα+α]zα(η2α)(η1+η2α)2+η3nzα(z)=t1(n)z+t2(n)z+...,M(z,n)n=η3zα(z),

    and

    M(0,n)z=α(0)+η1α+η2αα(η1+η2α)2.

    From the univalence of α we have α(0)0 and α(0)=1, it follows that t1(n)0 for n0 and limn|t1(n)|=.

    A simple compution yields,

    R{zM(z,n)zM(z,n)n}=R{1η3+1(η1+η2α)η3}.

    Using the fact that α is convex univalent function in Δ and η30 we have

    R{zM(z,n)zM(z,n)n}>0,ifR{1η3+1(η1+η2α)η3}>0,zΔ,n0.

    Hence the theorem.

    By taking

    α(z)=1+Az1+Bz(1BA1)

    in Theorem 4.1 we obtain the following corollary.

    Corollary 4.2. Let ηiC(i=1,2,3),(η30),C,s.t0 be convex univalent with α(0)=1 and (4.1) hold true. For g,QrαQ, let (Qrαz)H[1,1]T and (ηi)31(g;Qrα) defined in (3.3) be univalent in Δ satisfying

    1+Az1+Bz+z(AB)η1(1+Bz)2+η2(1+Az)(1+Bz)+η3z(AB(1+Bz)2)(ηi)31(g;Qrα),

    then

    1+Az1+Bz(z+k=2(k+r2r1)αk1(1α)rzkz),

    and 1+Az1+Bz is the best subordinant.

    To obtain the sandwich results get from combining the subordination results and superordination results

    Theorem 5.1. Let α1 and α2 be convex univalent in Δ, ηiC(i=1,2,3),(η30),C,s.t0 and let α2 satisfying (3.1) and α1 satisfying (4.1). For g,QrαQ, let (Qrαz)H[1,1]T and (ηi)31(g;Qrα) defined in (3.3) be univalent in Δ satisfying

    α1(z)+zα1(z)η1+η2α1(z)+η3zα1(z)(ηi)31,(g;Qrα)α2(z)+zα2(z)η1+η2α2(z)+η3zα2(z),

    then

    α1(z)(Qrαz)α2(z),

    and α1, α2 are respectively best subordinant and best dominant.

    Hence the proof of the theorem. By taking

    α1(z)=1+A1z1+B1z,(1B1A11)

    and

    α2(z)=1+A2z1+B2z,(1B2A21)

    in Theorem 5.1, we obtain the following result.

    Corollary 5.2. For g,QrαQ, let (Qrαz)H[1,1]T and (ηi)31(g;Qrα) defined in (3.3) be univalent in Δ satisfying

    1+A1z1+B1z+z(A1B1)η1(1+B1z)2+η2(1+A1z)(1+B1z)+η3z(A1B1(1+B1z)2)(ηi)31(g;Qrα)1+A2z1+B2z+z(A2B2)η1(1+B2z)2+η2(1+A2z)(1+B2z)+η3z(A2B2(1+B2z)2),

    then

    1+A1z1+B1z(Qrαz)A2B2(1+B2z)2,

    and 1+A1z1+B1z, 1+A2z1+B2z are respectively the best subordinant and best dominant.

    This paper deals with the applications of the differential subordination and superordination results involving Pascal distribution series. In addition we found the sandwich results to be in the class of holomorphic functions. Many interesting particular cases of the main theorems are emphazied in the form of corollaries. Furthermore to illustrate the results of application in various classes of analytic function. We anticipate that differential subordination and superordination will be important in several fields related to mathematics, science and technology.

    The authors would like to thanks the reviewers for the deep comments to improve our work. Also, we express our thanks to editorial office for their advice.

    The authors declare no conflicts of interest.



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