# Pascu-Type Analytic Functions by Using Mittag-Leffler Functions in Janowski Domain

Authors: Wali Khan Mashwan; Bakhtiar Ahmad; Muhammad Ghaffar Khan; Saima Mustafa; Sama Arjika; Bilal Khan
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

Keeping in view the various important applications of Mittag-Leffer functions in the fields of applied sciences, we introduce Pascu-type analytic functions utilizing the concept of Mittag-Leffler functions in the region of Janowski domain. Moreover, we investigate some useful properties of these functions such as sufficiency criteria, distortion and growth bounds, convex combination, radius of starlikeness, and some partial sum results.


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## Body

## 1. Introduction, Definitions, and Motivation

LetA denote the class of functions f that are analytic in the open unit disc D=z∈ℂ:z<1, and its Taylor series representation is as follows:(1)fz=z+∑k=2∞akzk,z∈D.Also, the most well-known subclass ofA is the class of univalent functions denoted by S.Further,S∗ represents the class of starlike functions in D, that is, f∈S and maps D on an starlike domain. Mathematically, it will satisfy the following condition:(2)ℜzf′zfz>0,z∈D.Now let us recall the familiar concept of subordinations; leth1 and h2 be two functions of class A, and we call that a function h1 is subordinated to h2 and symbolically represented as(3)h1≺h2orh1z≺h2z,if there exists Schwarz function u, with conditions u0=0 and uz<1 such that(4)h1z=h2uz.Further ifh2∈S, then the above definition is parallel to(5)h1z≺h2zz∈D⇔h10=h20,h1D⊂h2D.In 1973, Janowski [1] introduced the concepts of circular domain by introducing Janowski-type functions. A function gz analytic in open unit disc with properties that g0=1 is said to be in the class of Janowski denoted by PA,B if for −1≤B<A≤1,(6)gz≺1+Az1+Bz.Janowski showed thatgz maps D onto domain ΩA,B, where circular domain ΩA,B is defined by(7)ΩA,B=ω:ω−1−AB1−B2<A−B1−B2,forB≠−1,ω:ℜω>1−A2,forB=−1.The domainΩA,B, B≠−1, represents an open circular disc centered on real axis with D1=1−A/1−B and D2=1+A/1+B as diameter end points with 0<D1<1<D2. After introducing circular domain’s concept, several researchers focused on this domain and associated several subclasses to this domain and investigated some of its interesting geometric properties. For some recent works on Janowski-type domain, we refer the readers to [2–6].The use of special functions in the field of mathematics and applied sciences is very important. One special-type function with single parameterα is defined as(8)Eαz=∑k=0∞zkΓ1+αk,α,z∈ℂ,andℜα>0,and with two parameters α and β, its generalized form is(9)Eα,βz=∑k=0∞zkΓβ+αk,α,β,z∈ℂandℜα>0,ℜβ>0,which are known as Mittag-Leffler functions (see [7]). This was introduced by Mittag-Leffler [8] in connection with his method of summation of some divergent series. Mittag-Leffler in his paper [8] examined and investigated certain very interesting properties of functions defined in form (9).Mittag-Leffler functions naturally arise in solutions of fractional-order integral equations or fractional-order differential equations. Especially these type of functions can be seen in the investigations of the fractional generalization of the kinetic equation, Levy flights, random walks, superdiffusive transport, and in the study of complex systems. Mittag-Leffler functions, both ordinary and generalized functions, interpolate between a purely exponential law and power-law-like behavior of phenomena governed by ordinary kinetic equations and fractional counterparts (see [9] for more details). During the last 15 years, the interest of Mittag-Leffler functions is considerably increased among engineers and scientists due to their vast applications in several applied problems such as fluid flow, rheology, diffusive transport akin to diffusion, electric networks, probability, and probability distribution (for more applications, see [10, 11]). By giving specific values to α and β, we can get the following special cases easily.(1)
(10)E0z=11−z,z<1.(2)
(11)E1z=ez.(3)
(12)E2z=coshz,z∈ℂ.(4)
(13)E2−z2=cosz,z∈ℂ.(5)(14)E3z=12ez1/3+2e−1/2z1/3cos32z1/3,z∈ℂ.Now the function is defined by(15)Qα,βz=zΓβEα,βz=z+∑k=2∞ΓβΓβ+αk−1zk.Related to this, Elhaddad et al. [12] defined the operator Dμmα,βfz:A⟶A as follows:(16)Dμmα,βfz=z+∑k=2∞1+k−1μmΓβΓβ+αk−1zk,where m∈ℕ0=ℕ∪0, μ≥0.Now let us recall some recent Pascu-type works. In 2014, Vijaya et al. [13] defined Pascu-type class for harmonic functions with positive coefficient involving Salagean operator. After that, in 2015, Murugusundaramoorty et al. [14] found coefficient estimates for Pascu-type subclass of bi-univalent functions based on subordination. Recently, Vijaya [15] obtained coefficient bounds for subclass of Pascu-type bi-starlike functions in parabolic domain.Motivated from above and the work studied in [16–21], we now define a subfamily Sμm,tα,β,A,B of A by using the operator Dμm as follows.Definition 1.
With conditions−1≤B<A≤1, f∈A will be in the class Sμm,tα,β,A,B, if it satisfies(17)1−tDμm+1α,βfz+tDμm+2α,βfz1−tDμmα,βfz+tDμm+1α,βfz≺1+Az1+Bz,where the notation “≺” stands for the familiar concept of subordinations.It can easily be easily verified that a functionf∈A will be in the class Sμm,tα,β,A,B, if and only if(18)1−tDμm+1α,βfz+tDμm+2α,βfz/1−tDμmα,βfz+tDμm+1α,βfz−1B1−tDμm+1α,βfz+tDμm+2α,βfz/1−tDμmα,βfz+tDμm+1α,βfz−A<1.


## 2. The Main Results and Their Consequences

In this section, we start with sufficiency criteria for this newly defined class.Theorem 1.
Letf∈A of form (1) be in the class Sμm,tα,β,A,B if and only if the inequality(19)∑k=2∞Λkσkak≤B−Ais satisfied, where σk=Γβ/Γαk−1+β and Λk=1+k−1ηk1+λk−1η1+Bk−1η+B−A.Proof.
To show thatf∈Sμm,tα,β,A,B, it will be enough if we show that inequality (18) holds true, i.e.,(20)Kz−HzBKz−AHz<1,where(21)Kz=1−tDμm+1α,βfz+tDμm+2α,βfz=z+∑k=2∞1+k−1ηk+11+λk−1ησkakzk,Hz=1−tDμmα,βfz+tDμm+1α,βfz=z+∑k=2∞1+k−1ηk1+λk−1ησkakzk.
Now consider(22)Kz−HzBKz−AHz=∑k=2∞1+k−1ηk1+λk−1ηk−1ησkakzkB−Az+∑k=2∞1+k−1ηk1+λk−1ηBk−1η+B−Aσkakzk=∑k=2∞1+k−1ηk1+λk−1ηk−1ησkakB−A−∑k=2∞1+k−1ηk1+λk−1ηBk−1η+B−Aσkak<1.
Here we have used inequality (19), and this completes the direct part of the proof.
For converse part, letf∈Sμm,tα,β,A,B be given by (1); then, from (18), we have for z∈D,(23)1−tzDμm+1α,βfz+tDμm+2α,βfz/1−tzDμmα,βfz+tDμm+1α,βfz−1B1−tzDμm+1α,βfz+tDμm+2α,βfz/1−tzDμmα,βfz+tDμm+1α,βfz−A=∑k=2∞1+k−1ηk1+λk−1ηk−1ησkakzkB−Az+∑k=2∞1+k−1ηk1+λk−1ηBk−1η+B−Aσkakzk.
Sinceℜez≤z, we have(24)ℜe∑k=2∞1+k−1ηk1+λk−1ηk−1ησkakzkB−A+∑k=2∞1+k−1ηk1+λk−1ηBk−1η+B−Aσkakzk<1.
Now we choose values ofz on the real axis so that 1−tzDμm+1α,βfz+tDμm+2α,βfz/1−tzDμmα,βfz+tDμm+1α,βfz is real. Upon clearing the denominator in (2) and letting z⟶1− through real values, we obtain (19).Example 1.
For the function(25)fz=z+∑k=2∞B−AΛkσkxkzk,where σk=Γβ/Γαk−1+β and Λk=1+k−1ηk1+λk−1η1+Bk−1η+B−A, we haveHence,f∈Sμm,tα,β,A,B, and the result is sharp for this function.(26)∑k=2∞Λkσkak=sumn=2∞ΛkσkB−AΛkσkxk=B−A∑n=2∞xn=B−A.Corollary 1.
Letf∈A of form (1) be in the class Sμm,tα,β,A,B. Then,(27)ak≤B−AΛkσk,where Λk and σk are given above for k≥2.In the following, we discuss the growth and distortion theorems for our new class of functions.Theorem 2.
Iff is in the class Sμm,tα,β,A,B and has form (1). Then, for z=r, we have(28)r−B−AΛ2σ2r2≤fz≤r+B−AΛ2σ2r2,(29)1−2B−AΛ2σ2r≤f′z≤1+2B−AΛ2σ2r.
These inequalities are sharp for the function defined as(30)fz=z−B−AΛ2σ2z2.Proof.
First, we want to prove inequality (28). For this, consider(31)fz=z+∑k=2∞akzk≤z+∑k=2∞akzk,since for z=r<1 we have rn≤r2 for n≥2 and(32)fz≤r+r2∑k=2∞ak.
Similarly,(33)fz≥r−r2∑k=2∞ak.
Now (19) implies that(34)∑k=2∞Λnσkak≤B−A.
But(35)σ2Λ2∑k=2∞ak≤∑k=2∞Λkσkak.
Hence,(36)∑k=2∞σ2Λ2ak≤B−A,which gives(37)∑k=2∞ak≤B−AΛ2σ2.
Now by putting this value in (32) and (33), we get the required result. The proof of (29) is similar to (28), so it is omitted.Theorem 3.
The classSμm,tα,β,A,B is closed under convex combination.Proof.
To show the classSμm,tα,β,A,B is closed under convex combination, it is enough to show that μf1z+1−μf2z∈Sμm,tα,β,A,B, where f1 and f2∈Sμm,tα,β,A,B and μ∈0,1. Since(38)μf1z+1−μf2z=z+∑k=2∞μak,1+1−μak,2zk,by Theorem 1, we have(39)∑k=2∞Λkσkμak,1+1−μak,2≤μ∑k=2∞Λkσkak,1+1−μ∑k=2∞Λkσkak,2≤μB−A+1−μB−A=B−A.
Hence, in light of Theorem1, it is the required result.Theorem 4.
Let(40)fkz=z+1Λkσkzk,with f1z=z; then, f∈Sμm,tα,β,A,B if and only if(41)fz=∑k=1∞γkfkz,forγk≥0,with property that ∑k=1∞γk=1.Proof.
Let (41) hold. Then,(42)∑k=2∞Λkσkak=∑k=2∞Λkσk1Λkσkγk≤B−A1−γ1≤B−A.
Hence, by Theorem1, fz∈Sμm,tα,β,A,B. Conversely, suppose fz∈Sμm,tα,β,A,B; then, we have(43)γk=Λkσkak,fork≥2,and γ1=1−∑k=2∞γk. Then, function is of the form given by (40), and thus this completes the proof.Theorem 5.
Letfz∈Sμm,tα,β,A,B; then, f is starlike function of order δ in z<r∗, where(44)r∗=1−δΛkσkn−δB−A1/n−1.Proof.
We know thatf is starlike function of order δ if and only if(45)zf′z−fzzf′z−1−2δfz<1.
Using (8) and upon simplification yields(46)∑k=2∞n−δ1−δakzn−1≤1.
In light of Theorem1, we can easily obtain(47)∑k=2∞ΛkσkB−Aak≤1,which is possible if(48)∑k=2∞n−δ1−δakzn−1<∑k=2∞ΛkσkB−Aak,which implies that(49)z<1−δΛkσkn−δB−A1/n−1.
Hence, we obtain the desired result.Theorem 6.
Iff is in the class Sμm,tα,β,A,B and has form (1). Then,(50)ℜfzfmz≥1−1cm+1,(51)ℜfmzfz≥cm+11+cm+1,where(52)ck=ΛkσkB−A,ck≥.1fork=2,3,…,mck+1fork=m+1,….Proof.
First we prove (50); for this, let us write(53)ck+1fzfmz−1−1cm+1=1+ck+1∑k=m+1∞akzk−11+∑k=2makzk−1=1+wz1−wz,where(54)wz=ck+1∑k=m+1∞akzk−12+2∑k=2makzk−1+ck+1∑k=m+1∞akzk−1,wz≤ck+1∑k=m+1∞ak2−2∑k=2mak−ck+1∑k=m+1∞ak≤1,if and only if(55)∑k=2mak+ck+1∑k=m+1∞ak≤1.
In order to prove the inequality in (50), it is enough to show that the left-hand side of (55) is bounded above by the following sum:(56)∑k=2∞ckak,and is equivalent to(57)∑k=2m1−ckak+ck+1∑k=m+1∞cm+1−ckak≥0,which is true for inequality (50).
Next to derive inequality (51), let us write(58)1+ck+1fmzfz−1+1cm+1=1−1+ck+1∑k=m+1∞akzk−11+∑k=2makzk−1=1+wz1−wz,where(59)wz=−1+ck+1∑k=m+1∞akzk−12+2∑k=2makzk−1−1+ck+1∑k=m+1∞akzk−1,wz≤1+ck+1∑k=m+1∞ak2−2∑k=2mak−1+ck+1∑k=m+1∞ak≤1is equivalent to the following inequality:(60)∑k=2mak+ck+1∑k=m+1∞ak≤1.
Finally, we can see that the left-hand side of this inequality is bounded above by the following sum:(61)∑k=2∞ckak,which is equivalent to(62)∑k=2mck−1ak+ck+1∑k=m+1∞ck−cm+1−1ak≥0,which is true for inequality (51).Theorem 7.
Iff is in the class Sμm,tα,β,A,B and has form (1), then(63)ℜf′zfm′z≥1−m+1cm+1,ℜfm′zf′z≥cm+11+m+cm+1,where(64)ck=ΛkσkB−A,ck≥1fork=2,3,…,mck+1fork=m+1,….Proof.
The proof is quite the same as Theorem6, so it is omitted.


## 3. Conclusion

The present article is the motivation of some applications of Mittag-Leffer functions in various fields of applied sciences. A subclass of analytic functions is introduced using the concepts of Pascu-type and Mittag-Leffler functions in the region of Janowski domain. Some investigations are discussed with aid of some useful properties such as sufficiency criteria, distortion and growth bounds, convex combination, radius of starlikeness, and some partial sum results. This idea can be extended to various other classes like meromorphic functions, bi-univalent functions, and harmonic functions.



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