# A Class of Symmetric Fractional Differential Operator Formed by Special Functions

Authors: Ibtisam Aldawish; Rabha W. Ibrahim; Suzan J. Obaiys
Journal: Journal of Mathematics
 (2022)

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## Abstract

In light of a certain sort of fractional calculus, a generalized symmetric fractional differential operator based on Raina’s function is built. The generalized operator is then used to create a formula for analytic functions of type normalized. We use the ideas of subordination and superordination to show a collection of inequalities using the suggested differential operator. The new Raina’s operator is also used to the generalized kinematic solutions (GKS). Using the concepts of subordination and superordination, we provide analytic solutions for GKS. As a consequence, a certain hypergeometric function provides the answer. A fractional coefficient differential operator is also created. The geometric and analytic properties of the object are being addressed. The symmetric differential operator in a complex domain is shown to be a generalized fractional differential operator. Finally, we explore the characteristics of the Raina’s symmetric differential operator.


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## Body

## 1. Introduction

Symmetry is both an abstract basis of attractiveness and an applied tool for resolving convoluted problems. As a consequence, symmetry is a well-known foundation in numerous fields of physics. Despite a well-developed abstract theory of analytic symmetry, symmetry in real-world complex networks has established little attention [1]. Many scientists in many domains of mathematical sciences have been interested in learning more about the theory of symmetric operators. A special class of symmetric operators is defined by using some special functions, which are satisfying the symmetric behavior. The Mittag–Leffler function and its extensions, including Raina’s functions, are solutions for all categories of fractional differential equations (see [2–8]).We examine how Raina’s function may utilize to expand a symmetric fractional differential operator in a complex domain in this research. A range of new normalized analytic functions are explained using the fractional symmetric operator. The idea of differential subordination and superordination is applied to study a collection of differential inequalities. The geometric behavior of the generalized kinematic solution (GKS), a family of analytic solutions, is also studied. A variety of applications employ the new convolution linear operator.


## 2. Methods and Techniques

We will go through the strategies we used in this part.


### 2.1. Geometric Concepts

We start by the following definition [9]:Concept 2.1. The analytic functionsψ1,ψ2 in U:=ξ∈ℂ:ξ<1 are subordinated ψ1≺ψ2 or(1)ψ1ξ≺ψ2ξ,ξ∈U.If for an analytic functionυ,υ≤ξ<1 owning(2)ψ1ξ=ψ2υξ,ξ∈U.Concept 2.2. Consider the subclass of analytic functionsΛ by(3)ψξ=ξ+∑n=2∞φnξn,η∈U,satisfyingψ0=0,ψ′0=1.Furthermore, the functionsψ1,ψ2∈Λ are called convoluted ψ1∗ψ2 if they admin the operation [10](4)ψ1∗ψ2ξ=ξ+∑n=2∞ϕnξn∗ξ+∑n=2∞φnξn=ξ+∑n=2∞ϕnφnξn.Concept 2.3. TheS∗ class of star-like functions and the C class of convex univalent functions are both related to the class of normalized analytic functions Λ. In addition, we require the class of analytic functions(5)P≔ϱ:ϱξ=1+ϱ1ξ+ϱ2ξ2+…,ξ∈U.


### 2.2. Modified Special Function

Special functions include integrals and the outputs of many different types of differential equations. Therefore, most integral sets include special duty descriptions, and these duties include the elementary integrals. Since symmetries are important in real life, the philosophy of special functions is tightly linked to various mathematical physics topics [11]. We will start with a well-known special function, the Mittag–Leffler function.Concept 2.4. The extended Mittag–Leffler function is formulated by the series [12](6)Tα,βυξ=∑n=0∞υnΓαn+βξnn!,such thatυn=Γυ+n/Γυ. Clearly, for υ=1, implies that(7)Tα,β1ξ=∑n=0∞ξnΓβ+αn.After that, we will go through Raina’s function.Concept 2.5. Raina’s function is determined by the power series as follows [12]:(8)ℐα,βυξ=∑n=0∞υnΓαn+βξn,ξ∈U,such thatα∈0,∞,β∈1,∞ and υ:=υ0,υ1,…,υn is a collection of real or complex numbers.Notice 2.6. We have the following well-known special cases:(i)
υn=1⟹Tα,β1ξ(ii)
υn=υn/n!⟹Tα,βυξ(iii)
α=1,β=1,υn=xnyn/sn⟹Θ12x,y;s;ξ=∑n=0∞xnyn/snξn/Γn+1Employing the functionalℐα,βυξ, we get the convolution operator, for ψ∈Λ(9)Iα,βυψξ=Γα+βυ1ℐα,βυ∗ψξ=Γα+βυ1υ0+ξ+∑n=2∞Γα+βυ1υnΓαn+βξn∗ξ+∑n=2∞anξn=ξ+∑n=2∞Γα+βΓβ+αnυnυ1anξn≔ξ+∑n=2∞ςnanξn,such that(10)ςn≔Γα+βΓβ+αnυnυ1ξ∈U,ψ∈Λ,α∈0,∞,β∈1,∞,υ=υ0,…,υn.Clearly,Iα,βυψξ∈Λ. From the above structure, the fractional differential operator can be viewed geometrically.Note that the operatorIα,βυψξ is a new type of the convoluted Carlson–Shaffer operator [13] satisfying α=β=1, and υn=1nyn/sn,∀n, with(11)Iα,βυψξ=∑n=0∞1nynΓn+1snanξn.Moreover, whenυn=Γβ+αn for all n≥1, we have the Sàlàgean operator [14]:(12)Iα,βυψξ=ξ+∑n=2∞nanξn.


### 2.3. Arguments

The following precursors are utilized to develop the results of this inquiry into differential subordination theory:Argument 2.7. (see [9]). Suppose that f1ξ and f2ξ are convex univalent in U with f10=f20. Then, for a fixed value η≠0,ℜη≥0, the subordination(13)f1ξ+1ηξf1′ξ≺f2ξ,gives(14)f1ξ≺f2ξ.Argument 2.8. (see [9]). Consider the class of holomorphic functions as follows:(15)Πb,n=φ:φξ=b+bnξn+bn+1ξn+1+⋯,whereb∈ℂ and n∈ℤ+. The condition ı∈ℝ implies(16)ℜφξ+ıξφ′ξ>0⇒ℜφξ>0.In addition, ifı>0 and φ∈Π1,n, then for η1,η2∈0,∞ satisfying(17)φξ+ıξφ′ξ≺1+ξ1−ξη1⇒φξ≺1+ξ1−ξη2.Argument 2.9 (see [15]). Let ℏ,φ∈Πb,n, where φ∈C and for w1,w2∈ℂ,w2≠0. Then, the subordination(18)w1ℏξ+w2ξℏ′ξ≺w1φξ+w2ξφ′ξ,yields(19)ℏξ≺φξ.Argument 2.10.  (see [16]). Let φ,ϕ∈Πb,n, where ϕ∈C and the functional φξ+vξφ′ξ is univalent for some positive fixed number v. Then the differential inequality(20)ϕξ+vξϕ′ξ≺φξ+vξφ′ξ,implies(21)ϕξ≺φξ.


## 2.1. Geometric Concepts

We start by the following definition [9]:Concept 2.1. The analytic functionsψ1,ψ2 in U:=ξ∈ℂ:ξ<1 are subordinated ψ1≺ψ2 or(1)ψ1ξ≺ψ2ξ,ξ∈U.If for an analytic functionυ,υ≤ξ<1 owning(2)ψ1ξ=ψ2υξ,ξ∈U.Concept 2.2. Consider the subclass of analytic functionsΛ by(3)ψξ=ξ+∑n=2∞φnξn,η∈U,satisfyingψ0=0,ψ′0=1.Furthermore, the functionsψ1,ψ2∈Λ are called convoluted ψ1∗ψ2 if they admin the operation [10](4)ψ1∗ψ2ξ=ξ+∑n=2∞ϕnξn∗ξ+∑n=2∞φnξn=ξ+∑n=2∞ϕnφnξn.Concept 2.3. TheS∗ class of star-like functions and the C class of convex univalent functions are both related to the class of normalized analytic functions Λ. In addition, we require the class of analytic functions(5)P≔ϱ:ϱξ=1+ϱ1ξ+ϱ2ξ2+…,ξ∈U.


## 2.2. Modified Special Function

Special functions include integrals and the outputs of many different types of differential equations. Therefore, most integral sets include special duty descriptions, and these duties include the elementary integrals. Since symmetries are important in real life, the philosophy of special functions is tightly linked to various mathematical physics topics [11]. We will start with a well-known special function, the Mittag–Leffler function.Concept 2.4. The extended Mittag–Leffler function is formulated by the series [12](6)Tα,βυξ=∑n=0∞υnΓαn+βξnn!,such thatυn=Γυ+n/Γυ. Clearly, for υ=1, implies that(7)Tα,β1ξ=∑n=0∞ξnΓβ+αn.After that, we will go through Raina’s function.Concept 2.5. Raina’s function is determined by the power series as follows [12]:(8)ℐα,βυξ=∑n=0∞υnΓαn+βξn,ξ∈U,such thatα∈0,∞,β∈1,∞ and υ:=υ0,υ1,…,υn is a collection of real or complex numbers.Notice 2.6. We have the following well-known special cases:(i)
υn=1⟹Tα,β1ξ(ii)
υn=υn/n!⟹Tα,βυξ(iii)
α=1,β=1,υn=xnyn/sn⟹Θ12x,y;s;ξ=∑n=0∞xnyn/snξn/Γn+1Employing the functionalℐα,βυξ, we get the convolution operator, for ψ∈Λ(9)Iα,βυψξ=Γα+βυ1ℐα,βυ∗ψξ=Γα+βυ1υ0+ξ+∑n=2∞Γα+βυ1υnΓαn+βξn∗ξ+∑n=2∞anξn=ξ+∑n=2∞Γα+βΓβ+αnυnυ1anξn≔ξ+∑n=2∞ςnanξn,such that(10)ςn≔Γα+βΓβ+αnυnυ1ξ∈U,ψ∈Λ,α∈0,∞,β∈1,∞,υ=υ0,…,υn.Clearly,Iα,βυψξ∈Λ. From the above structure, the fractional differential operator can be viewed geometrically.Note that the operatorIα,βυψξ is a new type of the convoluted Carlson–Shaffer operator [13] satisfying α=β=1, and υn=1nyn/sn,∀n, with(11)Iα,βυψξ=∑n=0∞1nynΓn+1snanξn.Moreover, whenυn=Γβ+αn for all n≥1, we have the Sàlàgean operator [14]:(12)Iα,βυψξ=ξ+∑n=2∞nanξn.


## 2.3. Arguments

The following precursors are utilized to develop the results of this inquiry into differential subordination theory:Argument 2.7. (see [9]). Suppose that f1ξ and f2ξ are convex univalent in U with f10=f20. Then, for a fixed value η≠0,ℜη≥0, the subordination(13)f1ξ+1ηξf1′ξ≺f2ξ,gives(14)f1ξ≺f2ξ.Argument 2.8. (see [9]). Consider the class of holomorphic functions as follows:(15)Πb,n=φ:φξ=b+bnξn+bn+1ξn+1+⋯,whereb∈ℂ and n∈ℤ+. The condition ı∈ℝ implies(16)ℜφξ+ıξφ′ξ>0⇒ℜφξ>0.In addition, ifı>0 and φ∈Π1,n, then for η1,η2∈0,∞ satisfying(17)φξ+ıξφ′ξ≺1+ξ1−ξη1⇒φξ≺1+ξ1−ξη2.Argument 2.9 (see [15]). Let ℏ,φ∈Πb,n, where φ∈C and for w1,w2∈ℂ,w2≠0. Then, the subordination(18)w1ℏξ+w2ξℏ′ξ≺w1φξ+w2ξφ′ξ,yields(19)ℏξ≺φξ.Argument 2.10.  (see [16]). Let φ,ϕ∈Πb,n, where ϕ∈C and the functional φξ+vξφ′ξ is univalent for some positive fixed number v. Then the differential inequality(20)ϕξ+vξϕ′ξ≺φξ+vξφ′ξ,implies(21)ϕξ≺φξ.


## 3. Consequences

The next class of normalized analytic functions is defined in this paper, and its features are investigated employing differential subordination and superordination theory.Concept 3.1. A functionψ∈Λ aims to be in the class Ωα,βμλ,ρ if it fulfills the inequality(22)1−λξIα,βυψξ+λIα,βυψξ′≺ρξ,ξ∈U,λ∈0,1,ρ0=1,α∈0,∞,β∈1,∞,wheneverρ∈C.Eventually, the convexity of the univalent function(23)ρξ=Aξ+1Bξ+1,implies that(24)ρ∈P≔ρ∈U:ρξ=1+∑n=1∞ρiξn.Consider the functionalΣψλ:U⟶U, as in the following structure:(25)Σψλξ≔1−λξIα,βυψξ+λIα,βμψξ′.Consequently, in view of Concept 3.1, we get the next inequality(26)Σψλξ≺Aξ+1Bξ+1,ξ∈U.We proceed to investigate the geometric possessions of the suggested operators.


### 3.1. Results of Subordination Formula

We begin with the following outcome.Proposition 1.
Assume thatψ∈Ωα,βμλ,ρ. If(27)ℜΣψλξ=ℜ1−λξIα,βυψξ+λIα,βυψξ′≔ℜ1+∑n=1∞ςn>0,then the upper bound of the coefficients ςn is determined by the probability measure dω:(28)ςn2≤∫02πe−inτdωτ.In addition, if(29)ℜeiτΣψλξ>0,ξ∈U,τ∈ℝ,then ψ∈Ωα,βμAξ+1/Bξ+1 and(30)Σψλξ=Aξ+1Bξ+1,ξ∈U.Proof.
Suppose that(31)ℜΣψλξ=ℜ1+∑n=1∞ςnξn>0.
Continuously, the Carathéodory positivist lemma entails(32)ςn≤2∫02πe−inτdωτ,where dω is a probability measure. Besides, if(33)ℜeiτΣψλξ>0,ξ∈U,τ∈ℝ,then according to Theorem 1.6 in [10] and for a fixed number τ∈ℝ, we have(34)Σψλξ=Aξ+1Bξ+1,ξ∈U.
Hence,ψ∈Ωα,βμλ,Aξ+1/Bξ+1.
The following findings reveal the functional sandwich theory’s required and adequate methodology.Proposition 2.
Suppose that(35)λξIα,βυψξ″+Iα,βυψξ′≺F2ξ+ξF2′ξ,where F20=1 and convex in U. Moreover, let Σψλξ be univalent in U with Σψ∈ΠF10,1∩D, where D indicates the class of all univalent analytic functions F having the limit limξ∈∂DF≠∞ and(36)F1ξ+ξF1′ξ≺λξIα,βυψξ″+Iα,βυψξ′.Then(37)F1ξ≺Σψλξ≺F2ξ,and F1ξ is the best subdominant and F2ξ is the best dominant.Proof.
Putting(38)Σψλξ=1−λξIα,βυψξ+λIα,βυψξ′,a calculation yields(39)Σψλξ+ξΣψλξ′=λIα,βυψξ′+ξλξIα,βυψξ″−λ−1Iα,βυψξ′+λ−1Iα,βυψξξ+1−λIα,βυψξξ=λξIα,βυψξ″+Iα,βυψξ′.
As a consequence, the double inequality produced is as follows:(40)F1ξ+ξF1′ξ≺Σψλξ+ξΣψλξ′≺F2ξ+ξF2′ξ.
Finally, Arguments 2.9 and 2.10 provide the required outcome. □Proposition 3.
Assume that(41)Σψλξ=1−λξIα,βυψξ+λIα,βυψξ′.
Then this leads to(42)Iα,βυψξ′ξε1+Iα,βυψξε1+3ε2+ε2ξIα,βυψξ″≺1+ξ1−ξγ1⇒Σψλξ≺1+ξ1−ξγ2,γ1>0,γ2>0,ε1=1−λ,ε2=λ>0.Proof.
A calculation gives that(43)Σψλξ+ξΣψλξ′=1−λξIα,βυψξ+λIα,βυψξ′+ξ1−λξIα,βυψξ+λIα,βυψξ′′=Iα,βυψξ′ξε1+ε1+3ε2Iα,βυψξ+ε2ξIα,βυψξ″≺1+ξ1−ξγ1.
In view of Argument 2.8, we obtain(44)Σψλξ≺1+ξ1−ξγ2.


### 3.2. Fractional Differential Equation with Kinematic Solutions

We will use the generalized differential operator to continue our research in this section. A generalized formula for the kinematic solutions (GKS) is presented using the suggested operator. Kinematic behaviors describe the motion of an item with constant acceleration in a dynamic system.We aim to utilize the classΩα,βμλ,1+ξ/1−ξ to extend GKSs. We deal with the upper bound solution, for the fractional differential equation(45)1−λξIα,βυψξ+λIα,βυψξ′=Aξ+1Bξ+1,Iα,βυψ0=0,ς∈0,1,ξ∈U.The outcome of (45) is formulated as follows.Proposition 4.
Letξ∈Ωα,βυς,1+ξ/1−ξ. Then (45) has a solution expressed by(46)Iα,βυψξ=cξλ−1/λ+2ξ2Θ121,1+1/λ;2+1/λ;ξλ+1+λξλ+1+ξλ+1,where c is a constant and Θ12x,y,s;ξ is the hypergeometric function.Proof.
Letξ∈Ωα,βυς,1+ξ/1−ξ. Thus, we obtain(47)1−λξIα,βυψξ+λIα,βυψξ′=φξ+11−φξ,where χ≤ξ<1 and χ0=0. As a result, we get the integral formula(48)Iα,βυψξ=ξλ−1/λ∫0ξ−η1/λ−1χη+1λχη−1dη.
In view of Schwarz lemma, we getχξ=wξ,w=1 (see Theorem 5.34 in [17]). Therefore, by assuming χξ=ξ, we obtain the differential equation(49)1−λξIα,βυψξ+λIα,βυψξ′=1+ξ1−ξ.
If we reorganize the previous equation, we conclude that(50)Iα,βυψξ′+1−λλξIα,βυψξ=1λ1+ξ1−ξ.
Then multiplying by the functional(51)Tξ=exp∫1−λλξdξ=ξ1/λ−1,we obtain(52)ξ1/λ−1Iα,βυψξ′−Iα,βυψξ1−λξ1/λ−2λ=ξ1/λ−1λ1+ξ1−ξ.
As a result, we receive the solution(53)Iα,βυψξ=cξλ−1/λ+2ξ2Θ121,1+1/λ;2+1/λ;ξλ+1+λξλ+1+ξλ+1.Example 1.
Letψ∈Ωα,βυλ,1+ξ/1−ξ, where λ=1/2 and c=0. According to Proposition 5, we have(54)Iα,βυψξ=ξ2ξΘ121,1+1/λ,2+1/λ,ξλ+1+1,c=0=ξ4ξΘ121,3,4,ξ3+1,λ=12=ξ+ξ2+ξ3+ξ4+⋯+Oξ7,ξ<1.
Letψξ=ξ/1−ξ. Then(55)Iα,βυψξ=ξ+∑n=2∞Γα+βΓβ+αnυnυ1ξn.
Comparing the right sides of the above equations, we obtain thatυn=Γβ+αn,∀n. But ψξ=ξ/1−ξ is the optimal convex function in the open unit disk; thus, the operator Iα,βυψξ is convex whenever ψ is convex (see Figure 1).Figure 1
Plots of GKS of equation (45). (a)ψξ=ξ/1−ξ. (b)1+ξ/1−ξ. (c) Iα,βυψξ, when λ=0.5,c=0. (d) Iα,βυψξ, when λ=0.25,c=0. (e) Iα,βυψξ, when λ=0.5,c=1. (f) Iα,βυψξ, when λ=0.25,c=1.
(a)(b)(c)(d)(e)(f)


### 3.3. Symmetric Differential Operator

The Raina’s convoluted operator is assumed to present an extended symmetric differential operator.(56)ℳℓ0Iα,βυψξ=Iα,βυψξℳℓ1Iα,βυψξ=ℓξIα,βυψξ′−1−ℓξIα,βυψ−ξ′=ℓξ+∑n=2∞nanςnξn−1−ℓ−ξ+∑n=2∞n−1nanςnξn=ξ+∑n=2∞nℓ−1−ℓ−1nanςnξnℳℓ2Iα,βυψξ=ℳℓ1ℳℓ1Iα,βυψξ=ξ+∑n=2∞nℓ−1−ℓ−1n2anςnξn⋮ℳℓkIα,βυψξ=ℳℓ1ℳℓk−1Iα,βυψξ=ξ+∑n=2∞nℓ−1−ℓ−1nkςnanξn.Whenςn=1,∀n, we have the symmetric operator in [18]. Moreover, when ςn=1 and ℓ=1, we receive the Sàlàgean integral operator [14].The following classes will be studied:Concept. Letψ∈Λ. Then, we define the subclass of star-like functions:(i)
ψ∈Sα,β,ℓ∗υ,kℏ if and only if there occurs a convex function ℏ∈C satisfying the subordination(57)ξℳℓkIα,βυψξ′ℳℓkIα,βυψξ≺ℏξ.(ii)
ψ∈Jℓ♮A,B,k if and only if(58)1+1♮2ℳℓk+1Iα,βυψξℳℓkIα,βυψξ−ℳℓkIα,βυψ−ξ≺1+Aξ1+Bξ,ξ∈U,−1≤B<A≤1,k=1,2,…,♮∈ℂ∖0,ℓ∈0,1.Proposition 5.
Considerψ∈Sα,β,ℓ∗υ,kℏ. Then(59)ℳℓk+1Iα,βυψξ≺ξe∫0ξℏyz−1/zdz,where yξ is analytic in U with y0=0 and yξ<1. Additionally, for ξ=ℵ, ℳℓk+1Iα,βυψξ satisfies the inequality(60)exp∫01ℏy−ℵ−1ℵdℵ≤ℳℓk+1Iα,βυψξξ≤exp∫01ℏyℵ−1ℵdℵ.Proof.
Becauseψ∈Sα,β,ℓ∗μ,kℏ, then we conclude that(61)ξℳℓk+1Iα,βυψξ′ℳℓk+1Iα,βυψξ≺ℏξ,ξ∈U.
This leads to the existence of a Schwarz function withy0=0 and yξ<1 such that(62)ξℳℓk+1Iα,βυψξ′ℳℓk+1Iα,βυψξ=ℏyξ,ξ∈U,which implies that(63)ℳℓk+1Iα,βυψξ′ℳℓk+1Iα,βυψξ−1ξ=ℏyξ−1ξ.
Integration implies that(64)logℳℓk+1Iα,βυψξ−logξ=∫0ξℏyz−1zdz.
A computation brings(65)logℳℓk+1Iα,βυψξξ=∫0ξℏyz−1zdz.
The subordination yields(66)ℳℓk+1Iα,βυψξ≺ξexp∫0ξℏyz−1zdz.
Moreover, the disk is mapped byℏξ. When we apply 0<ξ<ℵ<1 to an area that is convex and symmetric with respect to the real axis, we get(67)ℏ−ℵξ≤ℜℏyℵξ≤ℏℵξ,ℵ∈0,1,which brings(68)ℏ−ℵ≤ℏ−ℵξ,ℏℵξ≤ℏℵ,∫01ℏy−ℵξ−1ℵdℵ≤ℜ∫01ℏyℵ−1ℵdℵ≤∫01ℏyℵξ−1ℵdℵ.
Employing equation (65), we obtain(69)∫01ℏy−ℵξ−1ℵdℵ≤logℳℓk+1Iα,βυψξξ≤∫01ℏyℵξ−1ℵdℵ.
As a result, we get the inequality(70)exp∫01ℏy−ℵξ−1ℵdℵ≤ℳℓk+1Iα,βυψξξ≤exp∫01ℏyℵξ−1ℵdℵ.
Hence, we receive(71)exp∫01ℏy−ℵ−1ℵdℵ≤ℳℓk+1Iα,βυψξξ≤exp∫01ℏyℵ−1ℵdℵ.Proposition 6.
Suppose thatψ∈Jℓ♮A,B,k then the odd function(72)Lξ=12ψξ−ψ−ξ,ξ∈U,fulfills the inequality(73)1+1♮ℳℓk+1Iα,βυLξℳℓkIα,βυLξ−1≺1+Aξ1+Bξ,ℜξLξ′Lξ≥1−η21+η2,ξ=η<1,ξ∈U,−1≤B<A≤1,k=1,2,…,♮∈ℂ∖0,ℓ∈0,1.Proof.
By the conditionψ∈Jℓ♮A,B,k, we obtain the existence of a function G∈JA,B such that(74)♮Gξ−1=2ℳℓk+1Iα,βυψξℳℓkIα,βυψξ−ℳℓkIα,βμψ−ξ,♮G−ξ−1=−2ℳℓk+1Iα,βυψ−ξℳℓkIα,βυψξ−ℳℓkIα,βυψ−ξ.
This leads to(75)1+1♮ℳℓk+1Iα,βυLξℳℓkIα,βυLξ−1=Gξ+G−ξ2.
Also, because(76)Gξ≺1+Aξ1+Bξ,where 1+Aξ/1+Bξ is univalent, then the above subordination yields(77)1+1♮ℳℓk+1Iα,βυLξℳℓkIα,βυLξ−1≺1+Aξ1+Bξ.
Additionally, the functionLξ is star-like in U, which gives the inequality(78)ξLξ′Lξ≺1−ξ21+ξ2.
As a consequence, we confirm the existence of Schwarz function℘∈U,℘ξ≤ξ<1,℘0=0 such that(79)ϒξ≔ξLξ′Lξ≺1−℘ξ21+℘ξ2,which yields that there is ξ,ξ=η<1 such that(80)℘2ξ=1−ϒξ1+ϒξ,ξ∈U.
By rearranging the above inequality, we receive(81)1−ϒξ1+ϒξ=℘ξ2≤ξ2.
Hence, we have the following conclusion:(82)ϒξ−1+ξ41−ξ42≤4ξ41−ξ42,or(83)ϒξ−1+ξ41−ξ4≤2ξ21−ξ4.
This yields(84)ℜϒξ≥1−η21+η2,ξ=η<1.
As a result, we obtain the next outcomes.Corollary 1 (see [18])..
Letμn=Γαn+β in Proposition 6. Then(85)1+1♮ℳℓk+1Iα,βυLξℳℓkIα,βυLξ−1=1+1♮ℳℓk+1LξℳℓkLξ−1≺1+Aξ1+Bξ.Corollary 2 (see [19])..
Letℓ=1 and υn=Γαn+β in Theorem 3.9. Then(86)1+1♮ℳ1k+1Iα,βυLξℳ1kIα,βυLξ−1=1+1♮ℳ1k+1Lξℳ1kLξ−1≺1+Aξ1+Bξ.Corollary 3 (see [20])..
Letℓ=1,k=1 and υn=Γαn+β in Theorem 3.9. Then(87)1+1♮ℳ12Iα,βμLξℳ11Iα,βμLξ−1=1+1♮ℳ12Lξℳ11Lξ−1≺1+Aξ1+Bξ.Corollary 4.
Letk=0,ℓ=1 and υn=Γαn+β in Theorem 3.9. Then(88)1+1♮ℳ11Iα,βυLξIα,βυLξ−1≺1+Aξ1+Bξ.


## 3.1. Results of Subordination Formula

We begin with the following outcome.Proposition 1.
Assume thatψ∈Ωα,βμλ,ρ. If(27)ℜΣψλξ=ℜ1−λξIα,βυψξ+λIα,βυψξ′≔ℜ1+∑n=1∞ςn>0,then the upper bound of the coefficients ςn is determined by the probability measure dω:(28)ςn2≤∫02πe−inτdωτ.In addition, if(29)ℜeiτΣψλξ>0,ξ∈U,τ∈ℝ,then ψ∈Ωα,βμAξ+1/Bξ+1 and(30)Σψλξ=Aξ+1Bξ+1,ξ∈U.Proof.
Suppose that(31)ℜΣψλξ=ℜ1+∑n=1∞ςnξn>0.
Continuously, the Carathéodory positivist lemma entails(32)ςn≤2∫02πe−inτdωτ,where dω is a probability measure. Besides, if(33)ℜeiτΣψλξ>0,ξ∈U,τ∈ℝ,then according to Theorem 1.6 in [10] and for a fixed number τ∈ℝ, we have(34)Σψλξ=Aξ+1Bξ+1,ξ∈U.
Hence,ψ∈Ωα,βμλ,Aξ+1/Bξ+1.
The following findings reveal the functional sandwich theory’s required and adequate methodology.Proposition 2.
Suppose that(35)λξIα,βυψξ″+Iα,βυψξ′≺F2ξ+ξF2′ξ,where F20=1 and convex in U. Moreover, let Σψλξ be univalent in U with Σψ∈ΠF10,1∩D, where D indicates the class of all univalent analytic functions F having the limit limξ∈∂DF≠∞ and(36)F1ξ+ξF1′ξ≺λξIα,βυψξ″+Iα,βυψξ′.Then(37)F1ξ≺Σψλξ≺F2ξ,and F1ξ is the best subdominant and F2ξ is the best dominant.Proof.
Putting(38)Σψλξ=1−λξIα,βυψξ+λIα,βυψξ′,a calculation yields(39)Σψλξ+ξΣψλξ′=λIα,βυψξ′+ξλξIα,βυψξ″−λ−1Iα,βυψξ′+λ−1Iα,βυψξξ+1−λIα,βυψξξ=λξIα,βυψξ″+Iα,βυψξ′.
As a consequence, the double inequality produced is as follows:(40)F1ξ+ξF1′ξ≺Σψλξ+ξΣψλξ′≺F2ξ+ξF2′ξ.
Finally, Arguments 2.9 and 2.10 provide the required outcome. □Proposition 3.
Assume that(41)Σψλξ=1−λξIα,βυψξ+λIα,βυψξ′.
Then this leads to(42)Iα,βυψξ′ξε1+Iα,βυψξε1+3ε2+ε2ξIα,βυψξ″≺1+ξ1−ξγ1⇒Σψλξ≺1+ξ1−ξγ2,γ1>0,γ2>0,ε1=1−λ,ε2=λ>0.Proof.
A calculation gives that(43)Σψλξ+ξΣψλξ′=1−λξIα,βυψξ+λIα,βυψξ′+ξ1−λξIα,βυψξ+λIα,βυψξ′′=Iα,βυψξ′ξε1+ε1+3ε2Iα,βυψξ+ε2ξIα,βυψξ″≺1+ξ1−ξγ1.
In view of Argument 2.8, we obtain(44)Σψλξ≺1+ξ1−ξγ2.


## 3.2. Fractional Differential Equation with Kinematic Solutions

We will use the generalized differential operator to continue our research in this section. A generalized formula for the kinematic solutions (GKS) is presented using the suggested operator. Kinematic behaviors describe the motion of an item with constant acceleration in a dynamic system.We aim to utilize the classΩα,βμλ,1+ξ/1−ξ to extend GKSs. We deal with the upper bound solution, for the fractional differential equation(45)1−λξIα,βυψξ+λIα,βυψξ′=Aξ+1Bξ+1,Iα,βυψ0=0,ς∈0,1,ξ∈U.The outcome of (45) is formulated as follows.Proposition 4.
Letξ∈Ωα,βυς,1+ξ/1−ξ. Then (45) has a solution expressed by(46)Iα,βυψξ=cξλ−1/λ+2ξ2Θ121,1+1/λ;2+1/λ;ξλ+1+λξλ+1+ξλ+1,where c is a constant and Θ12x,y,s;ξ is the hypergeometric function.Proof.
Letξ∈Ωα,βυς,1+ξ/1−ξ. Thus, we obtain(47)1−λξIα,βυψξ+λIα,βυψξ′=φξ+11−φξ,where χ≤ξ<1 and χ0=0. As a result, we get the integral formula(48)Iα,βυψξ=ξλ−1/λ∫0ξ−η1/λ−1χη+1λχη−1dη.
In view of Schwarz lemma, we getχξ=wξ,w=1 (see Theorem 5.34 in [17]). Therefore, by assuming χξ=ξ, we obtain the differential equation(49)1−λξIα,βυψξ+λIα,βυψξ′=1+ξ1−ξ.
If we reorganize the previous equation, we conclude that(50)Iα,βυψξ′+1−λλξIα,βυψξ=1λ1+ξ1−ξ.
Then multiplying by the functional(51)Tξ=exp∫1−λλξdξ=ξ1/λ−1,we obtain(52)ξ1/λ−1Iα,βυψξ′−Iα,βυψξ1−λξ1/λ−2λ=ξ1/λ−1λ1+ξ1−ξ.
As a result, we receive the solution(53)Iα,βυψξ=cξλ−1/λ+2ξ2Θ121,1+1/λ;2+1/λ;ξλ+1+λξλ+1+ξλ+1.Example 1.
Letψ∈Ωα,βυλ,1+ξ/1−ξ, where λ=1/2 and c=0. According to Proposition 5, we have(54)Iα,βυψξ=ξ2ξΘ121,1+1/λ,2+1/λ,ξλ+1+1,c=0=ξ4ξΘ121,3,4,ξ3+1,λ=12=ξ+ξ2+ξ3+ξ4+⋯+Oξ7,ξ<1.
Letψξ=ξ/1−ξ. Then(55)Iα,βυψξ=ξ+∑n=2∞Γα+βΓβ+αnυnυ1ξn.
Comparing the right sides of the above equations, we obtain thatυn=Γβ+αn,∀n. But ψξ=ξ/1−ξ is the optimal convex function in the open unit disk; thus, the operator Iα,βυψξ is convex whenever ψ is convex (see Figure 1).Figure 1
Plots of GKS of equation (45). (a)ψξ=ξ/1−ξ. (b)1+ξ/1−ξ. (c) Iα,βυψξ, when λ=0.5,c=0. (d) Iα,βυψξ, when λ=0.25,c=0. (e) Iα,βυψξ, when λ=0.5,c=1. (f) Iα,βυψξ, when λ=0.25,c=1.
(a)(b)(c)(d)(e)(f)


## 3.3. Symmetric Differential Operator

The Raina’s convoluted operator is assumed to present an extended symmetric differential operator.(56)ℳℓ0Iα,βυψξ=Iα,βυψξℳℓ1Iα,βυψξ=ℓξIα,βυψξ′−1−ℓξIα,βυψ−ξ′=ℓξ+∑n=2∞nanςnξn−1−ℓ−ξ+∑n=2∞n−1nanςnξn=ξ+∑n=2∞nℓ−1−ℓ−1nanςnξnℳℓ2Iα,βυψξ=ℳℓ1ℳℓ1Iα,βυψξ=ξ+∑n=2∞nℓ−1−ℓ−1n2anςnξn⋮ℳℓkIα,βυψξ=ℳℓ1ℳℓk−1Iα,βυψξ=ξ+∑n=2∞nℓ−1−ℓ−1nkςnanξn.Whenςn=1,∀n, we have the symmetric operator in [18]. Moreover, when ςn=1 and ℓ=1, we receive the Sàlàgean integral operator [14].The following classes will be studied:Concept. Letψ∈Λ. Then, we define the subclass of star-like functions:(i)
ψ∈Sα,β,ℓ∗υ,kℏ if and only if there occurs a convex function ℏ∈C satisfying the subordination(57)ξℳℓkIα,βυψξ′ℳℓkIα,βυψξ≺ℏξ.(ii)
ψ∈Jℓ♮A,B,k if and only if(58)1+1♮2ℳℓk+1Iα,βυψξℳℓkIα,βυψξ−ℳℓkIα,βυψ−ξ≺1+Aξ1+Bξ,ξ∈U,−1≤B<A≤1,k=1,2,…,♮∈ℂ∖0,ℓ∈0,1.Proposition 5.
Considerψ∈Sα,β,ℓ∗υ,kℏ. Then(59)ℳℓk+1Iα,βυψξ≺ξe∫0ξℏyz−1/zdz,where yξ is analytic in U with y0=0 and yξ<1. Additionally, for ξ=ℵ, ℳℓk+1Iα,βυψξ satisfies the inequality(60)exp∫01ℏy−ℵ−1ℵdℵ≤ℳℓk+1Iα,βυψξξ≤exp∫01ℏyℵ−1ℵdℵ.Proof.
Becauseψ∈Sα,β,ℓ∗μ,kℏ, then we conclude that(61)ξℳℓk+1Iα,βυψξ′ℳℓk+1Iα,βυψξ≺ℏξ,ξ∈U.
This leads to the existence of a Schwarz function withy0=0 and yξ<1 such that(62)ξℳℓk+1Iα,βυψξ′ℳℓk+1Iα,βυψξ=ℏyξ,ξ∈U,which implies that(63)ℳℓk+1Iα,βυψξ′ℳℓk+1Iα,βυψξ−1ξ=ℏyξ−1ξ.
Integration implies that(64)logℳℓk+1Iα,βυψξ−logξ=∫0ξℏyz−1zdz.
A computation brings(65)logℳℓk+1Iα,βυψξξ=∫0ξℏyz−1zdz.
The subordination yields(66)ℳℓk+1Iα,βυψξ≺ξexp∫0ξℏyz−1zdz.
Moreover, the disk is mapped byℏξ. When we apply 0<ξ<ℵ<1 to an area that is convex and symmetric with respect to the real axis, we get(67)ℏ−ℵξ≤ℜℏyℵξ≤ℏℵξ,ℵ∈0,1,which brings(68)ℏ−ℵ≤ℏ−ℵξ,ℏℵξ≤ℏℵ,∫01ℏy−ℵξ−1ℵdℵ≤ℜ∫01ℏyℵ−1ℵdℵ≤∫01ℏyℵξ−1ℵdℵ.
Employing equation (65), we obtain(69)∫01ℏy−ℵξ−1ℵdℵ≤logℳℓk+1Iα,βυψξξ≤∫01ℏyℵξ−1ℵdℵ.
As a result, we get the inequality(70)exp∫01ℏy−ℵξ−1ℵdℵ≤ℳℓk+1Iα,βυψξξ≤exp∫01ℏyℵξ−1ℵdℵ.
Hence, we receive(71)exp∫01ℏy−ℵ−1ℵdℵ≤ℳℓk+1Iα,βυψξξ≤exp∫01ℏyℵ−1ℵdℵ.Proposition 6.
Suppose thatψ∈Jℓ♮A,B,k then the odd function(72)Lξ=12ψξ−ψ−ξ,ξ∈U,fulfills the inequality(73)1+1♮ℳℓk+1Iα,βυLξℳℓkIα,βυLξ−1≺1+Aξ1+Bξ,ℜξLξ′Lξ≥1−η21+η2,ξ=η<1,ξ∈U,−1≤B<A≤1,k=1,2,…,♮∈ℂ∖0,ℓ∈0,1.Proof.
By the conditionψ∈Jℓ♮A,B,k, we obtain the existence of a function G∈JA,B such that(74)♮Gξ−1=2ℳℓk+1Iα,βυψξℳℓkIα,βυψξ−ℳℓkIα,βμψ−ξ,♮G−ξ−1=−2ℳℓk+1Iα,βυψ−ξℳℓkIα,βυψξ−ℳℓkIα,βυψ−ξ.
This leads to(75)1+1♮ℳℓk+1Iα,βυLξℳℓkIα,βυLξ−1=Gξ+G−ξ2.
Also, because(76)Gξ≺1+Aξ1+Bξ,where 1+Aξ/1+Bξ is univalent, then the above subordination yields(77)1+1♮ℳℓk+1Iα,βυLξℳℓkIα,βυLξ−1≺1+Aξ1+Bξ.
Additionally, the functionLξ is star-like in U, which gives the inequality(78)ξLξ′Lξ≺1−ξ21+ξ2.
As a consequence, we confirm the existence of Schwarz function℘∈U,℘ξ≤ξ<1,℘0=0 such that(79)ϒξ≔ξLξ′Lξ≺1−℘ξ21+℘ξ2,which yields that there is ξ,ξ=η<1 such that(80)℘2ξ=1−ϒξ1+ϒξ,ξ∈U.
By rearranging the above inequality, we receive(81)1−ϒξ1+ϒξ=℘ξ2≤ξ2.
Hence, we have the following conclusion:(82)ϒξ−1+ξ41−ξ42≤4ξ41−ξ42,or(83)ϒξ−1+ξ41−ξ4≤2ξ21−ξ4.
This yields(84)ℜϒξ≥1−η21+η2,ξ=η<1.
As a result, we obtain the next outcomes.Corollary 1 (see [18])..
Letμn=Γαn+β in Proposition 6. Then(85)1+1♮ℳℓk+1Iα,βυLξℳℓkIα,βυLξ−1=1+1♮ℳℓk+1LξℳℓkLξ−1≺1+Aξ1+Bξ.Corollary 2 (see [19])..
Letℓ=1 and υn=Γαn+β in Theorem 3.9. Then(86)1+1♮ℳ1k+1Iα,βυLξℳ1kIα,βυLξ−1=1+1♮ℳ1k+1Lξℳ1kLξ−1≺1+Aξ1+Bξ.Corollary 3 (see [20])..
Letℓ=1,k=1 and υn=Γαn+β in Theorem 3.9. Then(87)1+1♮ℳ12Iα,βμLξℳ11Iα,βμLξ−1=1+1♮ℳ12Lξℳ11Lξ−1≺1+Aξ1+Bξ.Corollary 4.
Letk=0,ℓ=1 and υn=Γαn+β in Theorem 3.9. Then(88)1+1♮ℳ11Iα,βυLξIα,βυLξ−1≺1+Aξ1+Bξ.


## 4. Conclusion

The preceding study used symmetric derivative and Jackson’s calculus to generalize Raina’s transformations inU. We used the suggested linear convolution operator on the normalized subclass. The operator is utilized to analyze the outcome of a specific form of GKS, which is utilized as an application. The hypergeometric function was used to determine the behavior of solutions. We further stressed that the answer belongs to the normalized analytic functions category.



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