New Developments on the Theory of Third-Order Differential Superordination Involving Gaussian Hypergeometric Function

Abstract

The present research aims to present new results regarding the fundamental problem of providing sufficient conditions for finding the best subordinant of a third-order differential superordination. A theorem revealing such conditions is first proved in a general context. As another aspect of novelty, the best subordinant is determined using the results of the first theorem for a third-order differential superordination involving the Gaussian hypergeometric function. Next, by applying the results obtained in the first proved theorem, the focus is shifted to proving the conditions for knowing the best subordinant of a particular third-order differential superordination. Such conditions are determined involving the properties of the subordination chains. This study is completed by providing means for determining the best subordinant for a particular third-order differential superordination involving convex functions. In a corollary, the conditions obtained are adapted to the special case when the convex functions involved have a more simple form.

1. Introduction

In 2011, José A. Antonino and Sanford S. Miller [1] broadened the framework of second-order differential subordinations originally developed by Sanford S. Miller and Petru T. Mocanu [2] to the context of third-order differential subordinations. The methods proposed by Antonino and Miller presented the opportunity to obtain interesting new results, and researchers started to follow this line of research [3,4]. The idea of extending the dual theory of differential superordination [5] to third-order differential superordination was implemented in 2014 [6], and new interesting results soon followed [7,8].

The following notations and terminology provide the overall background for the research.

The class of analytical functions is identified by $\mathcal{H}\left(U\right)$, where $U=\{z\in \mathbb{C}:|z|<1\}$, with $\overline{U}=\{z\in \mathbb{C}:|z|\le 1\}$ and $\partial U=\{z\in \mathbb{C}:|z|=1\}$.

With n being a positive integer and a being a complex number, the following important subclasses of $\mathcal{H}\left(U\right)$ are known:

$$H[a,n]=\{f\in \mathcal{H}\left(U\right):f\left(z\right)=a+a_n{z}^n+a_{n+1}{z}^{n+1}+\dots ,z\in U\},$$

denoting $H_0=H[0,1]$ and $H_1=H[1,1]$,

$$A_n=\{f\in \mathcal{H}\left(U\right):f\left(z\right)=z+a_{n+1}{z}^{n+1}+\dots ,z\in U\},$$

denoting $A_1=A$.

Notable subclasses of A are described as follows:

$$S=\{f\in A:f \mathrm{univalent},f\left(0\right)=0,f^{\prime }\left(0\right)=1\},$$

the class of univalent functions with $f\left(z\right)=z+a_2{z}^2+\dots$ and $z\in U$, and

$$K=\left\{f\in A:\mathrm{Re}\left({\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}+1}\right)>0,f\left(0\right)=0,f^{\prime }\left(0\right)\ne 0,z\in U\right\},$$

the class of convex functions.

The concepts of subordination and superordination are considered as given in [2,9].

Definition 1

([2,9]). Let f and F be members of $\mathcal{H}\left(U\right)$. The function f is said to be subordinate to F, or F is said to be superordinate to f, if there exists a function w analytic in U, with $w\left(0\right)=0$ and $\left|w\right(z\left)\right|<1$, $z\in U$ and such that $f\left(z\right)=F\left(w\right(z\left)\right)$; in such a case, we write $f\prec F$, $f\left(z\right)\prec F\left(z\right)$. If F is univalent, then $f\prec F$, if and only if $f\left(0\right)=F\left(0\right)$ and $f\left(U\right)\subset F\left(U\right)$.

A fundamental class of functions related to the theories of differential subordination and superordination is defined as follows:

Definition 2

([1]). Let Q denote the set of functions q that are analytic and univalent on $\overline{U}\setminus E\left(q\right)$, where

$$E\left(q\right)=\left\{\zeta \in \partial U;\underset{z\to \zeta }{lim}q\left(z\right)=\infty \right\},$$

and are such that for $|q^{\prime }\left(\zeta \right)|=\rho >0$ and for $q\in \partial U\setminus E\left(q\right)$. The subclass of Q for which $q\left(0\right)=a$ is denoted by $Q\left(a\right)$.

In their research [1], the authors have examined functions $\psi :{\mathbb{C}}^4\times U\to \mathbb{C}$ identifying the distinctive features of function p, leading to the following implication:

$$\left\{\psi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right):z\in U)\right\}\subset \Omega \Rightarrow p\left(U\right)\subset \Delta ,$$

(1)

with $\varphi \Delta$ sets in $\mathbb{C}$.

In [6], the authors have considered the dual problem of identifying the characteristics of function p, such that

$$\Omega \subset \left\{\phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right):z\in U)\right\}\Rightarrow \Delta \subset p\left(U\right),$$

(2)

with $\phi :{\mathbb{C}}^4\times U\to \mathbb{C}$ and $\Omega ,\Delta$ sets in $\mathbb{C}$.

Implication (2) can be expressed using the concept of superordination if $p,\phi \in S$, $\phi :{\mathbb{C}}^4\times U\to \mathbb{C}$ and $\Delta$ and $\Omega$ are simply connected domains, $\Delta ,\Omega \ne \mathbb{C}$. Considering this context, there exist a function q and conformal mapping of U into $\Omega$ such that $q\left(0\right)=p\left(0\right)$ and a function h and conformal mapping of U into $\Delta$ such that $h\left(0\right)=\phi \left(p\right(0),0,0,0;0)$. Then, (2) is equivalent to

$$h\left(z\right)\prec \left\{\phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right):z\in U)\right\}\Rightarrow q\left(z\right)\prec p\left(z\right).$$

(3)

For the third-order differential superordination

$$h\left(z\right)\prec \left\{\phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right):z\in U)\right\},$$

(4)

the function p is identified as a solution of the third-order differential superordination. The function $q\in \mathcal{H}\left(U\right)$ is designated as a subordinant of the solutions of the third-order differential superordination, or simply a subordinant, if $q\left(z\right)\prec p\left(z\right)$ for all functions p complying with (4). A subordinant $\tilde{q}\in S$ is regarded as the best subordinant of (4) if it satisfies $\tilde{q}\prec q\left(z\right)$ for all subordinants of (4).

A fundamental approach in the study of third-order differential superordination is to use the basic concept of admissible function as given in [10]. Using this approach, notable results were obtained by different authors examining appropriate classes of admissible functions involving generalized Bessel functions [7], fractional operators [8,11], the Srivastava–Attiya operator [12], linear operators [13,14], meromorphic functions [15] or Mittag-Leffler functions [16].

Definition 3

([12]). Let Ω be a set in $\mathbb{C}$, $q\in H[a,n]$ and $q\left(0\right)\ne 0$. The class of admissible functions ${\varphi }_n[\Omega ,q]$ consists of those $\phi :{\mathbb{C}}^4\times U\to \mathbb{C}$ that satisfy the admissibility conditions.

$${\displaystyle \phi (r,s,t,u;\xi )\in \Omega ,r=q\left(z\right),s=\frac{z{q}^{\prime }\left(z\right)}{m},}$$

$${\displaystyle \mathrm{Re}\left({\displaystyle \frac{t}{s}+1}\right)\le \frac{1}{m}\left[{\displaystyle \mathrm{Re}\frac{z{q}^{″}\left(z\right)}{q^{\prime }\left(z\right)}+1}\right],}$$

$${\displaystyle \mathrm{Re}\frac{u}{s}\le \frac{1}{m^2}\mathrm{Re}\frac{z^2{q}^{‴}\left(z\right)}{q^{\prime }\left(z\right)},}$$

$$z\in U,\xi \in \partial U,m\ge n\ge 2.$$

The two dual theories of third-order differential subordination and superordination continue to develop nicely. Very recent results obtained using this approach can be found in papers like [17,18,19,20].

A new approach to third-order differential subordination has been taken in recent studies considering another fundamental concept of the theory of differential subordination, which is the best dominant of the differential subordination. In [21,22], methods for determining the dominant of a third-order differential subordination are described, and new techniques for determining a third-order differential subordination’s best dominant are also indicated.

The focus of this paper is on the same line of study adapted to the dual theory of third-order differential superordination. The results obtained here present the alternative to the approach that discusses the concept of the class of admissible functions. The present study aims to reveal new results concerning the problem of determining a subordinant for certain third-order differential superordinations and moreover provide the best subordinant for the third-order differential superordinations in the case when the third-order differential superordinations support such functions. The primary goal of the analysis performed in this work is to further extend the results established for the second-order differential superordinations by employing this second type of approach that has not yet been considered by other authors. This approach is expected to generate results that are relevant to the geometric theory of analytical functions by selecting certain noteworthy functions as the best subordinants considering their remarkable geometrical properties.

In light of the well-known univalence, convexity and starlikeness properties of the Gaussian hypergeometric function [23,24,25] as well as considering the recent findings for the Gaussian hypergeometric function in the context of geometric function theory [26,27] and its significance with regard to third-order differential subordination concepts employed in source–sink dynamics theory [28], this prominent function is implemented in this study to provide applications of the theoretical findings developed in this study.

2. Materials and Methods

Definition 4

([23]). Let $a,b,c\in \mathbb{C}$ and $c\ne 0,-1,-2,\dots$. The function

$$\begin{array}{cc}\hfill F(a,b,c;z)& {\displaystyle =1+\frac{ab}{c}·\frac{z}{1!}+\frac{a(a+1)b(b+1)}{c(c+1)}·\frac{z^2}{2!}+\dots }\hfill \\ \hfill & {\displaystyle =\sum _{k=0}^{\infty }\frac{{\left(a\right)}_k·{\left(b\right)}_k}{{\left(c\right)}_k}·\frac{z^k}{k!}}\hfill \\ \hfill & {\displaystyle =\frac{\Gamma \left(c\right)}{\Gamma \left(a\right)\Gamma \left(b\right)}\sum _{k=0}^{\infty }\frac{\Gamma (a+k)\Gamma (b+k)}{\Gamma (c+k)}·\frac{z^k}{k!},z\in U,}\hfill \end{array}$$

(5)

is called the Gaussian hypergeometric function, where ${\left(d\right)}_k$ is a Pochhammer symbol defined by

$${\displaystyle {\left(d\right)}_k=\frac{\Gamma (d+k)}{\Gamma \left(d\right)}=d(d+1)(d+2)\dots (d+k-1) with {\left(d\right)}_0=1,}$$

and

$$\Gamma \left(z\right)={\int }_0^{\infty }{e}^{-t}{t}^{z-2}dt,$$

with

$$\Gamma (z+1)=z·\Gamma \left(z\right),\Gamma \left(1\right)=1,\Gamma (n+1)=n!.$$

Regarding third-order differential subordinations and superordinations, the following lemma is a crucial tool for demonstrating the theorems in the following section. This result was stated by Miller and Mocanu, who extended a famous result known as Jack’s lemma [29] when they introduced the notion of differential subordination [30,31]. The result obtained by Miller and Mocanu gave a new method for the study within geometric function theory by showing what conditions are met when functions are not in a relation of subordination. It is considered a fundamental theorem on which the differential subordination theory was built, as shown by the famous book containing a comprehensive compilation of references for the topic [2]. This result was further adapted when the dual notion of differential superordination was introduced by Miller and Mocanu [5]. When the second-order differential subordination and superordination theories were extended to third-order differential subordination and superordination theories [1,6], the outcome was expanded upon and made compatible with those new theories. The form used for proving the new outcome of the present investigation is the following:

Lemma 1

([1,5,6]). Let $p\in Q\left(a\right)$, and let $q\left(z\right)=a+a_n{z}^n+a_{n+1}{z}^{n+1}+\dots$ by analytic in U, with $q\left(z\right)\not\equiv a$ and $n\ge 2$. If q is not subordinate to p, then there exist points $z_0=r_0{e}^{i{\theta }_0}\in U$ and ${\zeta }_0\in \partial U\setminus E\left(p\right)$ for which $q\left(U_{r_0}\right)\subset p\left(U\right)$ such that the following conditions are satisfied:

(i) 

$q\left(z_0\right)=p\left({\zeta }_0\right)$,

(ii) 

${\displaystyle \mathrm{Re}\frac{{\zeta }_0{q}^{″}\left({\zeta }_0\right)}{q^{\prime }\left({\zeta }_0\right)}\ge 0}$ and $\left|{\displaystyle \frac{z{p}^{\prime }\left(z\right)}{q^{\prime }\left(z\right)}}\right|\le n$.

When $z\in \overline{U_{r_0}}$ and $\zeta \in \partial U\setminus E\left(p\right)$ then there exists a real constant $m\ge n\ge 2$ such that:

(iii) 

$z_0{q}^{\prime }\left(z_0\right)=m{\zeta }_0{p}^{\prime }\left({\zeta }_0\right)$,

(iv) 

$\mathrm{Re}\left({\displaystyle \frac{z_0{q}^{″}\left(z_0\right)}{q^{\prime }\left(z_0\right)}+1}\right)\ge m\mathrm{Re}\left({\displaystyle \frac{{\zeta }_0{p}^{″}\left({\zeta }_0\right)}{p^{\prime }\left({\zeta }_0\right)}+1}\right)$,

(v) 

${\displaystyle \mathrm{Re}\frac{z_0^2{q}^{‴}\left(z_0\right)}{q^{\prime }\left(z_0\right)}\ge m^2\mathrm{Re}\frac{{\zeta }_0^2{p}^{‴}\left({\zeta }_0\right)}{p^{\prime }\left({\zeta }_0\right)}}$.

The method of the subordination chains is often associated with studies on differential subordination and superordination theories in conjunction to Lemma 1. The following result due to Pommerenke [9] gives a condition that characterizes the notion of the subordination chain and has been intensively used in the research associated with differential subordination and superordination theories, as shown in the landmark books on the topics [2,10].

Definition 5

([2,9,10]). A function $L(z,t)=a_1\left(t\right)z+a_2\left(t\right)z^2+\dots$ with $a_1\left(t\right)\ne 0$ for $t\ge 0$ and $z\in U$ is a subordination chain if $L(·,t)$ is analytic and univalent in U for all $t\ge 0$, $L(·,t)$ and is continuously differentiable on ${\mathbb{R}}^{+}$ for all $z\in U$ and $L(z,s)\prec L(z,t)$ when $0\le s\le t$.

Lemma 2

([2,9,10]). The function $L(z,t)=a_1\left(t\right)z+a_2\left(t\right)z^2+\dots$ with $a_1\left(t\right)\ne 0$ for $t\ge 0$ and

$$\underset{t\to \infty }{lim}\left|a_1\left(t\right)\right|=\infty ,$$

is a subordination chain if

$${\displaystyle \mathrm{Re}\frac{{\displaystyle z\frac{\partial L(z,t)}{\partial z}}}{{\displaystyle \frac{\partial L(z,t)}{\partial t}}}>0,z\in U,t\ge 0.}$$

The theorems established throughout this investigation and discussed in the following section of this paper offer previously unconsidered third-order differential superordination extensions of several classical findings known for the second-order differential superordination theory. The first proved theorem establishes conditions for finding subordinants of the third-order differential superordination in the general case. It is also proved that if such a subordinant is the solution of a certain differential equation, which corresponds to the investigated third-order differential superordination, it is regarded as the best subordinant. A noteworthy corollary is obtained when the Gaussian hypergeometric function is employed as an application for the results of Theorem 1. The second theorem establishes the best subordinant for a certain third-order differential superordination using the method of the subordination chains. The proof of this theorem uses the results established in Theorem 1. The third theorem uses the results proved in Theorem 2 for obtaining conditions for a particular third-order differential superordination involving convex functions to admit the best subordinant. The conditions obtained in Theorem 3 are interpreted in a corollary considering a more simple form for the functions involved in the third-order differential superordination considered.

3. Results

The first theorem presented in this study provides the best subordinant of a third-order differential superordination. This result is expected to generate new outcomes in the field of third-order differential superordinations since it gives sufficient conditions for determining the best subordinant of a certain third-order differential superordination. This affirmation is supported by the interesting application of this result seen in the corollary that follows it and by the fact that the results of this theorem are further used for obtaining the findings shown in the next theorem.

Theorem 1.

Let $h\in \mathcal{H}\left(U\right)$, $h,q\in \mathcal{H}[a,n]$ and $n\ge 2$ and consider $\phi :{\mathbb{C}}^4\times \overline{U}\to \mathbb{C}$ and $p\in Q\left(a\right)$. Suppose that

$${\displaystyle \mathrm{Re}\frac{\zeta q^{″}\left(\zeta \right)}{q^{\prime }\left(\zeta \right)}\ge 0and\left|{\displaystyle \frac{z{p}^{\prime }\left(z\right)}{q^{\prime }\left(\zeta \right)}}\right|\le n,}$$

(6)

$$\left\{\phi (q\left(z\right),\nu z{q}^{\prime }\left(z\right),z^2{q}^{″}\left(z\right),z^3{q}^{‴}\left(z\right);z\in U)\right\}\subset h\left(U\right),$$

(7)

with $z\in U$, $\zeta \in \partial U\setminus E\left(q\right)$, $n\ge 2$ and ${\displaystyle 0<\nu \le \frac{1}{n}}$.

If $p\in Q\left(a\right)$ and $\phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right);z\in U)$ are univalent functions in U, then

$$h\left(z\right)\prec \phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right);z\in U),$$

(8)

which implies

$$q\left(z\right)\prec p\left(z\right),z\in U.$$

Moreover, if the differential equation

$$\phi (q\left(z\right),z{q}^{\prime }\left(z\right),z^2{q}^{″}\left(z\right),z^3{q}^{‴}\left(z\right);z\in U)=h\left(z\right),$$

(9)

has a univalent solution $q\in Q\left(a\right)$, then q is the best subordinant of (8).

Proof. 

For the $z=z_0\in U$ relation, (7) becomes

$$\phi (q\left(z_0\right),\nu z_0{q}^{\prime }\left(z_0\right),z_0^2{q}^{″}\left(z_0\right),z_0^3{q}^{‴}\left(z_0\right);z_0)\in h\left(U\right).$$

(10)

Taking ${\displaystyle \nu =\frac{1}{m}\le \frac{1}{n}}$, relation (10) turns into

$$\phi \left({\displaystyle q\left(z_0\right),\frac{z_0{q}^{\prime }\left(z_0\right)}{m},z_0^2{q}^{″}\left(z_0\right),z_0^3{q}^{‴}\left(z_0\right);z_0}\right)\in h\left(U\right).$$

Since $h\left(0\right)=\phi (a,0,0,0;0)$ and $\phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right);z\in U)$ is univalent in U, considering subordination (8) and Definition 1 we obtain

$$h\left(U\right)\subset \left\{\phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right);z\in U)\right\},$$

and

$$\left\{\phi (p\left(\zeta \right),\zeta p^{\prime }\left(\zeta \right),{\zeta }^2{p}^{″}\left(\zeta \right),{\zeta }^3{p}^{‴}\left(\zeta \right);\zeta \in \partial U\setminus E\left(q\right))\right\}\not\subset h\left(U\right).$$

(11)

The set on the left side of the inclusion represents the border of U through function $\phi$.

For $\zeta ={\zeta }_0\in \partial U\setminus E\left(q\right)$ with $|{\zeta }_0|=1$, relation (11) becomes

$$\phi (p\left({\zeta }_0\right),{\zeta }_0{p}^{\prime }\left({\zeta }_0\right),{\zeta }_0^2{p}^{″}\left({\zeta }_0\right),{\zeta }_0^3{p}^{‴}\left({\zeta }_0\right);{\zeta }_0)\notin h\left(U\right).$$

(12)

In order to prove the assertion of the theorem, Lemma 1 will be invoked.

Assume that $q\nprec p$. In this case, Lemma 1 assures that there exist points $z_0\in U$ and ${\zeta }_0\in \partial U\setminus E\left(q\right)$, satisfying (i)–(v):

$$q\left(z_0\right)=p\left({\zeta }_0\right),z_0{q}^{\prime }\left(z_0\right)=m{\zeta }_0{p}^{\prime }\left({\zeta }_0\right),t={\zeta }_0^2{p}^{″}\left({\zeta }_0\right),u={\zeta }_0^3{p}^{‴}\left({\zeta }_0\right).$$

(13)

The admissibility conditions given in Definition 3 can be written using $r=q\left(z_0\right)$, ${\displaystyle s=\frac{z_0{q}^{\prime }\left(z_0\right)}{m}}$, t and u, as seen above, as follows:

$$\phi \left({\displaystyle q\left(z_0\right),\frac{z_0{q}^{\prime }\left(z_0\right)}{m},{\zeta }_0^2{p}^{″}\left({\zeta }_0\right),{\zeta }_0^3{p}^{‴}\left({\zeta }_0\right);{\zeta }_0}\right)\in h\left(U\right).$$

(14)

Using the equalities given by (13) and relation (14), we obtain the following:

$$\phi (p\left({\zeta }_0\right),{\zeta }_0{p}^{\prime }\left({\zeta }_0\right),{\zeta }_0^2{p}^{″}\left({\zeta }_0\right),{\zeta }_0^3{p}^{‴}\left({\zeta }_0\right);{\zeta }_0)\in h\left(U\right).$$

(15)

Since (15) contradicts (12), the conclusion is that the assumption that $q\nprec p$ is false; hence,

$$q\left(z\right)\prec p\left(z\right),z\in U.$$

Function q is a univalent solution of the differential Equation (9); hence, q is regarded as the best subordinant of the third-order differential superordination (8). □

Remark 1.

Consider $h\in \mathcal{H}\left(U\right)$, $q\left(z\right)=1+z^2+z^3$ and $q\in K$ and take $p\left(z\right)=F(a,b,c;z)$, the Gaussian hypergeometric function given by (5) in Definition 4. It has been proved in [26], Corollary 2, that $p\in K$. The following corollary can be given for Theorem 1.

Corollary 1.

Let $h\in \mathcal{H}\left(U\right)$, $q\left(z\right)=1+z^2+z^3$ and $q\in K$. Consider $\phi :{\mathbb{C}}^4\times \overline{U}\to \mathbb{C}$, $\phi \in {\varphi }_n[h\left(U\right);q]$ and $p\left(z\right)=F(a,b,c;z)$ given by (5). Suppose that

$${\displaystyle \mathrm{Re}\frac{\zeta q^{″}\left(\zeta \right)}{q^{\prime }\left(\zeta \right)}\ge 0and\left|{\displaystyle \frac{z{\left(F(a,b,c;z)\right)}^{\prime }}{q^{\prime }\left(\zeta \right)}}\right|\le n,}$$

$$\phi (1+z^2+z^3,\nu (2z^2+3z^3),2z^2+6z^3,6z^3)\in h\left(U\right),$$

with $z\in U$, $\zeta \in \partial U$, ${\displaystyle 0<\nu \le \frac{1}{n}<1}$.

If $p\left(z\right)=F(a,b,c;z)$ and $\phi (F(a,b,c;z),z{\left(F(a,b,c;z)\right)}^{\prime },z^2{\left(F(a,b,c;z)\right)}^{″},z^3{\left(F(a,b,c;z)\right)}^{‴})$ are univalent functions in U, then

$$h\left(z\right)\prec \phi (F(a,b,c;z),z{\left(F(a,b,c;z)\right)}^{\prime },z^2{\left(F(a,b,c;z)\right)}^{″},z^3{\left(F(a,b,c;z)\right)}^{‴}),$$

(16)

which implies

$$1+z^2+z^3\prec F(a,b,c;z),z\in U.$$

Moreover, if

$$\phi (1+z^2+z^3,2z^2+3z^3,2z^2+6z^3,6z^3)=h\left(z\right),$$

(17)

then $q\left(z\right)=1+z^2+z^3$ is the best subordinant for the third-order differential superordination (16).

Proof. 

We begin by showing that $q\in K$. We need to evaluate $\mathrm{Re}\left({\displaystyle \frac{z{q}^{″}\left(z\right)}{q^{\prime }\left(z\right)}+1}\right)$. We have

$$q^{\prime }\left(z\right)=2z+3z^2,q^{″}\left(z\right)=2+6z,$$

hence,

$${\displaystyle \mathrm{Re}\left({\displaystyle \frac{z{q}^{″}\left(z\right)}{q^{\prime }\left(z\right)}+1}\right)=\mathrm{Re}\left({\displaystyle 1+\frac{2+6z}{2+3z}}\right)=2+3\mathrm{Re}\frac{z}{2+3z}=2+3·\frac{2\rho \cos \alpha +3{\rho }^2}{4+12\rho \cos \alpha +9{\rho }^2},}$$

where $z=\rho (\cos \alpha +i\sin \alpha )$.

When $\rho \to 1$, we obtain the following:

$${\displaystyle \mathrm{Re}\left({\displaystyle \frac{z{q}^{″}\left(z\right)}{q^{\prime }\left(z\right)}+1}\right)=\underset{\rho \to 1}{lim}\left[{\displaystyle 2+3·\frac{2\rho \cos \alpha +3{\rho }^2}{4+12\rho \cos \alpha +9{\rho }^2}}\right]=2+3·\frac{3+2\cos \alpha }{13+12\cos \alpha }>0.}$$

$\mathrm{Re}\left({\displaystyle \frac{z{q}^{″}\left(z\right)}{q^{\prime }\left(z\right)}+1}\right)>0$ implies that $q\in K$; hence, it is univalent.

Next, we evaluate ${\displaystyle \mathrm{Re}\frac{\zeta q^{″}\left(\zeta \right)}{q^{\prime }\left(\zeta \right)}}$.

$${\displaystyle \mathrm{Re}\frac{\zeta q^{″}\left(\zeta \right)}{q^{\prime }\left(\zeta \right)}=\mathrm{Re}\frac{2+6\zeta }{2+3\zeta }=1+\frac{3(3+2\cos \theta )}{13+12\cos \theta }\ge 0,}$$

where $\zeta =R(\cos \theta +i\sin \theta )$.

It has been proved in [26], Corollary 2, that the Gaussian hypergeometric function is convex and, hence, univalent.

With the conditions required by Theorem 1 satisfied, considering

$$q\left(z\right)=1+z^2+z^3\mathrm{and}p\left(z\right)=F(a,b,c;z),$$

given by (5), we conclude that

$$1+z^2+z^3\prec F(a,b,c;z),z\in U.$$

Since $q\left(z\right)=1+z^2+z^3$ is a univalent solution of Equation (17), q is regarded as the best subordinant of the third-order differential superordination (16). □

Example 1.

Take $p\left(z\right)=F(-2,2+i,1+i;z)=1-(3-i)z+(2-i)z^2$. Since function $F(a,b,c;z)$ given by (5) is a convex function and hence univalent, which is proved in [26], using Corollary 1 we can write the following:

Let $h\in \mathcal{H}\left(U\right)$, $q\left(z\right)=1+z^2+z^3$ and $q\in K$ and consider $\phi :{\mathbb{C}}^4\times U\to \mathbb{C}$,

$$\phi (1+z^2+z^3,\nu (2z^2+3z^3),2z^2+6z^3,6z^3)\in h\left(U\right),p\left(z\right)=1-(3-i)z+(2-i)z^2,$$

with $p\left(0\right)=q\left(0\right)=1$ and $p\in S$.

Since $\phi (1-(3-i)z+(2-i)z^2,z[-3+i+2(1-i)z],2(1-i)z^2)$ is univalent in U,

$$h\left(z\right)\prec \phi (1-(3-i)z+(2-i)z^2,z[-3+i+2(1-i)z],z(1-i)z^2),$$

(18)

which implies

$$1+z^2+z^3\prec 1-(3-i)z+(2-i)z^2.$$

Since $q\left(z\right)=1+z^2+z^3$ is a univalent solution of the equation

$$\phi (1+z^2+z^3,2z^2+3z^3,2z^2+6z^2,6z^3)=h\left(z\right),$$

we conclude that function $q\left(z\right)=1+z^2+z^3$ is the best subordinant of (18).

For the proof of the next result, the notion of the subordination chain is used as it was previously for similar research in [27].

The next theorem states certain conditions that need to be satisfied in order to obtain the best subordinant of a given third-order differential superordination. The knowledge found in Theorem 1 will also be applied for completing the proof of this new result.

Theorem 2.

Consider the functions $p\in Q\left(a\right)$, $\phi :{\mathbb{C}}^4\times \overline{U}\to \mathbb{C}$, $h:\mathbb{C}\to \mathbb{C}$ and $q\in \mathcal{H}[a,n]\cap Q$, $n\ge 2$, a univalent solution of the differential equation

$$h\left(z\right)=\phi (q\left(z\right),z{q}^{\prime }\left(z\right),z^2{q}^{″}\left(z\right),z^3{q}^{‴}\left(z\right);z\in U),$$

(19)

satisfying the stipulations

$${\displaystyle \mathrm{Re}\frac{\zeta q^{″}\left(\zeta \right)}{q^{\prime }\left(\zeta \right)}\ge 0and\left|{\displaystyle \frac{z{q}^{\prime }\left(z\right)}{p^{\prime }\left(\zeta \right)}}\right|\le n,}$$

where $z\in U$, $\zeta \in \partial U\setminus E\left(q\right)$, $n\ge 2$ and

$$L(z,t)=\phi (q\left(z\right),tz{q}^{\prime }\left(z\right),z^2{q}^{″}\left(z\right),z^3{q}^{‴}\left(z\right);z\in U),t\ge 0,$$

(20)

is a subordination chain.

If $p\left(z\right)$ and $\phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right);z\in U)$ are univalent, then

$$h\left(z\right)\prec \phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right);z\in U),$$

(21)

which implies

$$q\left(z\right)\prec p\left(z\right),z\in U,$$

with function q being regarded as the best subordinant of the third-order differential superordination (21).

Proof. 

Replacing $z=0$ in (19), we write $h\left(0\right)=\phi (a,0,0,0;0)$. Since $L(z,t)$ is a subordination chain, from Definition 5 we conclude that $L(z,t)\in \mathcal{H}\left(U\right)$ and $L(z,t)\in S$ for any $t\in [0,\infty ]$. Hence, we have

$$L(z,t)\prec L(z,s),0\le t\le s.$$

(22)

For $t=1$, using (20), we write

$$L(z,1)=\phi (q\left(z\right),z{q}^{\prime }\left(z\right),z^2{q}^{″}\left(z\right),z^3{q}^{‴}\left(z\right);z\in U)=h\left(z\right).$$

If $s=1$, relation (22) becomes

$$L(z,t)\prec \phi (q\left(z\right),z{q}^{\prime }\left(z\right),z^2{q}^{″}\left(z\right),z^3{q}^{‴}\left(z\right);z\in U)=h\left(z\right).$$

(23)

Taking $z=0$, relation (20) becomes

$$L(0,t)=\phi \left(q\right(0),0,0,0;0)=\phi (a,0,0,0;0).$$

Since $L(0,t)=h\left(0\right)$ and $h\left(z\right)\in S$, $L(z,t)\in S$, subordination (23) is equivalent to

$$\left\{\phi (q\left(z\right),tz{q}^{\prime }\left(z\right),z^2{q}^{″}\left(z\right),z^3{q}^{‴}\left(z\right);z\in U)\right\}\subset h\left(U\right),$$

(24)

according to Definition 1.

Considering relation (24), the third-order differential superordination (21) and Theorem 1, we conclude that

$$q\left(z\right)\prec p\left(z\right),$$

with function $q\left(z\right)$ being regarded as the best subordinant. □

Remark 2.

Following the proofs of the results seen so far, it can be remarked that if a univalent solution is given for the differential equation that corresponds to the investigated third-order differential superordination, the best subordinant is actually known.

Using the results proved in Theorem 2, the next theorem provides the means for finding the best subordinant of a particular third-order differential superordination.

Theorem 3.

Let $q\in K$, consider $\alpha ,\beta \in \mathcal{H}\left(D\right)$, where $D\supset q\left(U\right)$, and take $\gamma \in \mathcal{H}\left(\mathbb{C}\right)$ and $p\in \mathcal{H}[a,n]\cap Q$.

Suppose that the following stipulations are met:

$${\displaystyle \mathrm{Re}\frac{\zeta q^{″}\left(\zeta \right)}{q^{\prime }\left(\zeta \right)}\ge 0and\left|{\displaystyle \frac{z{p}^{\prime }\left(z\right)}{q^{\prime }\left(\zeta \right)}}\right|\le n,}$$

when $z\in U$, $\zeta \in \partial U$, $n\ge 2$,

$$\mathrm{Re}\left[{\displaystyle \frac{{\alpha }^{\prime }\left(q\left(z\right)\right)+{\beta }^{\prime }\left(q\left(z\right)\right)·\gamma \left(\nu z{q}^{\prime }\left(z\right)\right)}{\beta \left(q\left(z\right)\right)·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)}+\frac{2z{q}^{″}\left(z\right)+4z^2{q}^{‴}\left(z\right)+z^3{q}^{⁗}\left(z\right)}{q^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right)·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)}}\right]\ge 0,$$

(25)

when $z\in U$, $\nu >0$ and

$$h\left(z\right)=\alpha \left(q\left(z\right)\right)+\beta \left(q\left(z\right)\right)·\gamma \left(z{q}^{\prime }\left(z\right)\right)+z^2{q}^{″}\left(z\right)+z^3{q}^{‴}\left(z\right)\in \mathcal{H}\left(U\right).$$

(26)

If functions $p\in \mathcal{H}[a,n]\cap Q$ with $p\left(0\right)=q\left(0\right)$ and $\alpha \left(p\left(z\right)\right)+\beta \left(p\left(z\right)\right)·\gamma \left(z{p}^{\prime }\left(z\right)\right)+z^2{p}^{″}\left(z\right)+z^3{p}^{‴}\left(z\right)$ are univalent in U, then

$$h\left(z\right)\prec \alpha \left(p\left(z\right)\right)+\beta \left(p\left(z\right)\right)·\gamma \left(z{p}^{\prime }\left(z\right)\right)+z^2{p}^{″}\left(z\right)+z^3{p}^{‴}\left(z\right),$$

(27)

which implies

$$q\left(z\right)\prec p\left(z\right),z\in U,$$

with function $q\left(z\right)$ regarded as the best subordinant.

Proof. 

Let $\alpha ,\beta \in \mathcal{H}\left(U\right)$ be defined as follows:

$$\begin{array}{cc}\hfill & \alpha :D\to \mathbb{C},\alpha \left(r\right)=r,\hfill \\ \hfill & \beta :D\to \mathbb{C},\beta \left(r\right)=r,\hfill \\ \hfill & \gamma :\mathbb{C}\to \mathbb{C},\gamma \left(s\right)=s,\hfill \end{array}$$

when $r,s\in \mathbb{C}$.

Take $\phi :{\mathbb{C}}^4\times \overline{U}\to \mathbb{C}$ defined by

$$\phi (r,s,t,u)=\alpha \left(r\right)+\beta \left(r\right)·\gamma \left(s\right)+t+u,$$

(28)

where $r,s,t,u\in \mathbb{C}$.

Taking $r=p\left(z\right)$, $s=z{p}^{\prime }\left(z\right)$, $t=z^2{p}^{″}\left(z\right)$ and $u=z^3{p}^{‴}\left(z\right)$, (28) becomes

$$\phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right))=\alpha \left(p\left(z\right)\right)+\beta \left(p\left(z\right)\right)·\gamma \left(z{p}^{\prime }\left(z\right)\right)+z^2{p}^{″}\left(z\right)+z^3{p}^{‴}\left(z\right),$$

(29)

$\phi \in S$.

Using (29), the third-order differential superordination (27) can be written as follows:

$$h\left(z\right)\prec \phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right)).$$

(30)

For the proof to be finalized, we must define the function $L:U\times [0,\infty ]\to \mathbb{C}$:

$$\begin{array}{cc}\hfill L(z,\nu )& =a_1\left(\nu \right)z+a_2\left(\nu \right)z^2+\dots \hfill \\ \hfill & =\alpha \left(q\left(z\right)\right)+\beta \left(q\left(z\right)\right)·\gamma \left(\nu z{q}^{\prime }\left(z\right)\right)+z^2{q}^{″}\left(z\right)+z^3{q}^{‴}\left(z\right).\hfill \end{array}$$

(31)

We show that the function given by (31) is a subordination chain. For that, we evaluate the following:

$$\begin{array}{cc}\hfill {\displaystyle \frac{\partial L(z,\nu )}{\partial z}}& =a_1\left(\nu \right)+2a_2\left(\nu \right)z+\dots \hfill \\ \hfill & =q^{\prime }\left(z\right)·{\alpha }^{\prime }\left(q\left(z\right)\right)+q^{\prime }\left(z\right)·{\beta }^{\prime }\left(q\left(z\right)\right)·\gamma \left(\nu z{q}^{\prime }\left(z\right)\right)\hfill \\ \hfill & +\beta \left(q\left(z\right)\right)·[\nu q^{\prime }\left(z\right)+\nu z{q}^{″}\left(z\right)]·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)\hfill \\ \hfill & +2z{q}^{″}\left(z\right)+z^2{q}^{‴}\left(z\right)+3z^2{q}^{‴}\left(z\right)+z^3{q}^{IV}\left(z\right).\hfill \end{array}$$

(32)

Taking $z=0$, relation (32) becomes

$$\begin{array}{cc}\hfill {\displaystyle \frac{\partial L(0,\nu )}{\partial z}}& =a_1\left(\nu \right)=q^{\prime }\left(0\right)·{\alpha }^{\prime }\left(q\left(0\right)\right)+q^{\prime }\left(0\right)·{\beta }^{\prime }\left(q\left(0\right)\right)·\gamma \left(0\right)+\beta \left(q\left(0\right)\right)·\nu q^{\prime }\left(0\right)·{\gamma }^{\prime }\left(0\right)\hfill \\ \hfill & =q^{\prime }\left(0\right)·{\gamma }^{\prime }\left(0\right)·\beta \left(q\left(0\right)\right)\left({\displaystyle \nu +\frac{{\alpha }^{\prime }\left(q\left(0\right)\right)+{\beta }^{\prime }\left(q\left(0\right)\right)·\gamma \left(0\right)}{{\gamma }^{\prime }\left(0\right)·\beta \left(q\left(0\right)\right)}}\right).\hfill \end{array}$$

Since $q\in K$, we know that $q^{\prime }\left(0\right)\ne 0$, and from (25), $\beta \left(q\right(0\left)\right)·\gamma \left(0\right)\ne 0$, we deduce the following:

$${\displaystyle a_1\left(\nu \right)=\frac{\partial L(0,z)}{\partial z}=q^{\prime }\left(0\right)·{\gamma }^{\prime }\left(0\right)·\beta \left(q\left(0\right)\right)\left({\displaystyle \nu +\frac{{\alpha }^{\prime }\left(q\left(0\right)\right)+{\beta }^{\prime }\left(q\left(0\right)\right)·\gamma \left(0\right)}{{\gamma }^{\prime }\left(0\right)·\beta \left(q\left(0\right)\right)}}\right).}$$

Then, it follows that

$$\underset{\nu \to \infty }{lim}\left|a_1\left(\nu \right)\right|=\underset{\nu \to \infty }{lim}\left|q^{\prime }\left(0\right)·{\gamma }^{\prime }\left(0\right)·\beta \left(q\left(0\right)\right)\left({\displaystyle \nu +\frac{{\alpha }^{\prime }\left(q\left(0\right)\right)+{\beta }^{\prime }\left(q\left(0\right)\right)·\gamma \left(0\right)}{{\gamma }^{\prime }\left(0\right)·\beta \left(q\left(0\right)\right)}}\right)\right|=\infty ,$$

(33)

$${\displaystyle \frac{\partial L(z,\nu )}{\partial \nu }=z·q^{\prime }\left(z\right)·\beta \left(q\left(0\right)\right)·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right).}$$

We now evaluate the following

$$\begin{array}{cc}\hfill {\displaystyle \mathrm{Re}\frac{{\displaystyle \frac{z\partial L(z,\nu )}{\partial z}}}{{\displaystyle \frac{\partial L(z,\nu )}{\partial \nu }}}}& =\mathrm{Re}\left[{\displaystyle \frac{z{q}^{\prime }\left(z\right)·{\alpha }^{\prime }\left(q\left(z\right)\right)}{z{q}^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right)·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)}+\frac{z{q}^{\prime }\left(z\right)·{\beta }^{\prime }\left(q\left(z\right)\right)·\gamma \left(\nu z{q}^{\prime }\left(z\right)\right)}{z{q}^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right)·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)}}\right.\hfill \\ \hfill & +\left.{\displaystyle \frac{\gamma z\beta \left(q\left(z\right)\right)[q^{\prime }\left(z\right)+z{q}^{″}\left(z\right)]·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)}{z{q}^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right)·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)}+\frac{2z{q}^{″}\left(z\right)+z^2{q}^{″}\left(z\right)+3z^2{q}^{‴}\left(z\right)+z^3{q}^{IV}\left(z\right)}{z{q}^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right)·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)}}\right]\hfill \\ \hfill & =\mathrm{Re}\left[{\displaystyle \frac{{\alpha }^{\prime }\left(q\left(z\right)\right)}{\beta \left(q\left(z\right)\right)·\gamma \left(\nu z{q}^{\prime }\left(z\right)\right)}+\frac{{\beta }^{\prime }\left(q\left(z\right)\right)·\gamma \left(\nu z{q}^{\prime }\left(z\right)\right)}{\beta \left(q\left(z\right)\right)·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)}+\nu +\nu ·\frac{z{q}^{″}\left(z\right)}{q^{\prime }\left(z\right)}}\right.\hfill \\ \hfill & \left.{\displaystyle +\frac{2z{q}^{″}\left(z\right)+4z^2{q}^{‴}\left(z\right)+z^3{q}^{IV}\left(z\right)}{q^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right)·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)}}\right].\hfill \end{array}$$

Since $q\in K$, using (25), the result is that

$${\displaystyle \mathrm{Re}\frac{{\displaystyle \frac{z\partial L(z,\nu )}{\partial z}}}{{\displaystyle \frac{\partial L(z,\nu )}{\partial \nu }}}\ge 0.}$$

(34)

Considering (33) and (34), according to Lemma 2, we have that $L(z,\nu )$ is a subordination chain.

If in (31) we take $\nu =1$, it follows that

$$L(z,1)=\alpha \left(q\left(z\right)\right)+\beta \left(q\left(z\right)\right)·\gamma \left(z{q}^{\prime }\left(z\right)\right)+z^2{q}^{″}\left(z\right)+z^3{q}^{‴}\left(z\right)=h\left(z\right).$$

(35)

Replacing (35) in (30), we obtain the following:

$$\begin{array}{cc}\hfill h\left(z\right)& =\alpha \left(q\left(z\right)\right)+\beta \left(q\left(z\right)\right)·\gamma \left(z{q}^{\prime }\left(z\right)\right)+z^2{q}^{″}\left(z\right)+z^3{q}^{‴}\left(z\right)\hfill \\ \hfill & \prec \phi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right),z^3{p}^{‴}\left(z\right)).\hfill \end{array}$$

(36)

Taking $r=q\left(z\right)$, $s=\nu z{q}^{\prime }\left(z\right)$, $t=z^2{q}^{″}\left(z\right)$, $u=z^3{q}^{‴}\left(z\right)$, we write

$$\begin{array}{cc}\hfill \phi (q\left(z\right),\nu z{q}^{\prime }\left(z\right),z^2{q}^{″}\left(z\right),z^3{q}^{‴}\left(z\right))& =\alpha \left(q\left(z\right)\right)+\beta \left(q\left(z\right)\right)·\gamma \left(\nu z{q}^{\prime }\left(z\right)\right)+z^2{q}^{″}\left(z\right)+z^3{q}^{‴}\left(z\right)\hfill \\ \hfill & =L(z,\nu ),\hfill \end{array}$$

(37)

and we deduce that function $\phi (q\left(z\right),\nu z{q}^{\prime }\left(z\right),z^2{q}^{″}\left(z\right),z^3{q}^{‴}\left(z\right))$ is a subordination chain.

Using (36) and (37) and applying Theorem 2, we obtain

$$q\left(z\right)\prec p\left(z\right),$$

with function $q\left(z\right)$ regarded as the best subordinant. □

Remark 3.

Considering $\beta \left(w\right)=1$ and $w\in D$, the following corollary can be given for Theorem 3.

Corollary 2.

Let $q\in K$, consider $\alpha \in \mathcal{H}\left(D\right)$, where $D\supset q\left(U\right)$ is a domain, and take $\gamma \in \mathcal{H}\left(\mathbb{C}\right)$ and $p\in \mathcal{H}[a,n]\cap Q$.

Suppose that the following stipulations are met:

$${\displaystyle \mathrm{Re}\frac{\zeta q^{″}\left(\zeta \right)}{q^{\prime }\left(\zeta \right)}\ge 0and\left|{\displaystyle \frac{z{p}^{\prime }\left(z\right)}{q^{\prime }\left(\zeta \right)}}\right|\le n,}$$

when $z\in U$, $\zeta \in \partial U$ and $n\ge 2$,

$$\mathrm{Re}\left[{\displaystyle \frac{{\alpha }^{\prime }\left(q\left(z\right)\right)}{{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)}+\frac{2z{q}^{″}\left(z\right)+4z^2{q}^{‴}\left(z\right)+z^3{q}^{⁗}\left(z\right)}{q^{\prime }\left(z\right)·{\gamma }^{\prime }\left(\nu z{q}^{\prime }\left(z\right)\right)}}\right]\ge 0,$$

when $z\in U$, $\nu >0$ and

$$h\left(z\right)=\alpha \left(q\left(z\right)\right)+\gamma \left(z{q}^{\prime }\left(z\right)\right)+z^2{q}^{″}\left(z\right)+z^3{q}^{‴}\left(z\right)\in \mathcal{H}\left(U\right).$$

If functions $p\in \mathcal{H}[a,n]\cap Q$ with $p\left(0\right)=q\left(0\right)$ and $\alpha \left(p\left(z\right)\right)+\gamma \left(z{p}^{\prime }\left(z\right)\right)+z^2{p}^{″}\left(z\right)+z^3{p}^{‴}\left(z\right)$ are univalent in U, then

$$h\left(z\right)\prec \alpha \left(p\left(z\right)\right)+\gamma \left(z{p}^{\prime }\left(z\right)\right)+z^2{p}^{″}\left(z\right)+z^3{p}^{‴}\left(z\right),$$

which implies

$$q\left(z\right)\prec p\left(z\right),$$

with function $q\left(z\right)$ regarded as the best subordinant.

Since function $F(a,b,c;z)$ given by (5) is a convex function and hence univalent, which is proved in [26], using Corollary 2 we can write the following:

Example 2.

Let $q\in K$, $q\left(z\right)=1+z+\frac{z^2}{2}$, consider $\alpha \in \mathcal{H}\left(D\right)$, where $D\supset q\left(U\right)$ is a domain and take $\gamma \in \mathcal{H}\left(\mathbb{C}\right)$ and $p=F(-2,3-i,2-i;z)=1-\frac{2(7+i)}{5}z+z^2$.

Suppose that the following stipulations are met:

$${\displaystyle \mathrm{Re}\frac{\zeta q^{″}\left(\zeta \right)}{q^{\prime }\left(\zeta \right)}\ge 0and\left|{\displaystyle \frac{z{F}^{\prime }(-2,3-i,2-i;z)}{1+\zeta }}\right|\le n,}$$

when $z\in U$, $\zeta \in \partial U$, $n\ge 2$,

$$\mathrm{Re}\left[{\displaystyle \frac{{\alpha }^{\prime }(1+z+\frac{z^2}{2})}{{\gamma }^{\prime }\left(\nu z(1+z)\right)}+\frac{2z}{(1+z){\gamma }^{\prime }\left(\nu z(1+z)\right)}}\right]\ge 0,$$

when $z\in U$, $\nu >0$ and

$$h\left(z\right)=\alpha (1+z+\frac{z^2}{2})+\gamma (z+z^2)+z^2\in \mathcal{H}\left(U\right),z\in U.$$

If functions $p=F(-2,3-i,2-i;z)=1-\frac{2(7+i)}{5}z+z^2$, with $p\left(0\right)=q\left(0\right)=1$ and $\alpha (1-\frac{2(7+i)}{5}z+z^2)+\gamma (z+z^2)+z^2$ are univalent in U, then

$$\alpha (1+z+\frac{z^2}{2})+\gamma (z+z^2)+z^2\prec \alpha (1-\frac{2(7+i)}{5}z+z^2)+\gamma (z+z^2)+z^2,$$

which implies

$$1+z+\frac{z^2}{2}\prec 1-\frac{2(7+i)}{5}z+z^2,$$

with function $q\left(z\right)=1+z+\frac{z^2}{2}$ regarded as the best subordinant.

Indeed,

$${\displaystyle \mathrm{Re}\frac{\zeta q^{″}\left(\zeta \right)}{q^{\prime }\left(\zeta \right)}=\mathrm{Re}\frac{\zeta }{1+\zeta }=\frac{1}{2}}>0and\left|{\displaystyle \frac{z(-\frac{2(7+i)}{5})+2z}{1+\zeta }}\right|\le n,n\ge 2.$$

$q\left(z\right)=1+z+\frac{z^2}{2}\in K$; hence, it is univalent and results from evaluating the following:

$${\displaystyle \mathrm{Re}(\frac{q^{″}\left(z\right)}{q^{\prime }\left(z\right)}+1)=\mathrm{Re}(\frac{z}{1+z}+1)=1+\mathrm{Re}\frac{z}{1+z}=1+\frac{1}{2}=\frac{3}{2}}>0.$$

Since $q\left(z\right)=1+z+\frac{z^2}{2}$ is the univalent solution of the equation $h\left(z\right)=\alpha (1+z+\frac{z^2}{2})+\gamma (z+z^2)+z^2$, we conclude that it is the best dominant.

4. Discussion

The research presented in this paper provides altogether novel developments for the extension of a number of outcomes from the well-known second-order differential superordination theory to the third-order differential superordination theory. The theorems presented in this paper exhibit the third-order differential superordination results that were obtained using a different method than the one imposed by the use of the class of admissible functions, i.e., by identifying subordinants and the best subordinants for the third-order differential superordinations under consideration. The results proved in Theorem 1 generate an interesting corollary when the Gaussian hypergeometric function is included among the functions employed and are further used in the proof of Theorem 2. Theorem 2 is proved utilizing the results given by Theorem 1 and the concept of the subordination chain, which is familiar to the studies regarding differential subordination and superordination theories. The results contained in Theorem 2 are further used for the proof of Theorem 3, with the two theorems leading to the conclusion that determining the best subordinant of a certain third-order differential superordination is guaranteed if the solution to the differential equation corresponding to the third-order differential superordination is found and that function is a subordinant for the third-order differential superordination.

As for the future directions of study, the results obtained in this paper could be extended to the newer theories of strong differential superordination [32] and fuzzy differential superordination [33] for obtaining third-order strong differential superordinations or third-order fuzzy differential superordinations. The Gaussian hypergeometric function could also be replaced for applications in order to obtain particular third-order differential superordinations, which could be interpreted for obtaining geometric properties for different classes of univalent functions. Moreover, considering the recent results concerning fourth-order differential superordinations obtained using a linear operator [34], the results obtained here for the Gaussian hypergeometric function could be further extended to higher-order differential superordinations.

5. Conclusions

The research contained in this paper brings new knowledge to the field of differential superordination. The well-known results established for second-order differential superordinations are extended in order to enrich the new line of research initiated for the study of higher-order differential superordinations. There are two major directions for studying third-order differential superordination. The first was largely explored by many authors and focuses on the fundamental concept of admissible function. Following this direction, the new results are obtained by defining suitable classes of admissible functions in order to develop interesting third-order differential superordinations. An alternative direction focuses on the other major direction for studying third-order differential superordinations, the fundamental concept of the best subordinant of the third-order differential superordination. This second direction has not yet been extensively investigated. Hence, the new outcome that resulted from present study is relevant to this direction of research, increases the visibility of the subject among the researchers in the field and could spark new ideas for further exploiting ideas concerning conditions that guarantee the existence of the best subordinants for the third-order differential superordinations. Since source–sink dynamics theory has applications of third-order differential subordination theory as seen in [28], the results of the study given here could be modified to fit this theory. Building on concepts from [35], applications for fluid mechanics could also emerge.