• Tidak ada hasil yang ditemukan

Strong Differential Sandwich Results for Frasin Operator

N/A
N/A
Protected

Academic year: 2025

Membagikan "Strong Differential Sandwich Results for Frasin Operator"

Copied!
10
0
0

Teks penuh

(1)

https://doi.org/10.34198/ejms.3120.95104

Received: October 16, 2019; Accepted: December 13, 2019 2010 Mathematics Subject Classification: 30C45.

Keywords and phrases: strong subordinations, strong superordinations, convex function, best dominant, best subordinant.

Copyright © 2020 Abbas Kareem Wanas and B. A. Frasin. This is an open access article distributed under

Strong Differential Sandwich Results for Frasin Operator

Abbas Kareem Wanas1 and B. A. Frasin2

1 Department of Mathematics, College of Science, University of Al-Qadisiyah, Iraq e-mail: [email protected]

2 Department of Mathematics, Faculty of Science, Al al-Bayt University, Mafraq, Jordan e-mail: [email protected]

Abstract

In this paper, we define two new classes of analytic functions involving strong differential subordinations and superordination associated with Frasin operator. Further, we study some important properties of these classes.

1. Introduction

Let U denote the open unit disk of the complex plane U =

{

z∈C: z <1

}

,

{

:1

}

= z z

U C be the closed unit disk of the complex plane and let H

(

U×U

)

be the class of analytic functions in U×U. For a positive integer n and a∈C, let

[

a,n, ζ

] {

= f ∈H

(

U×U

)

: f

(

z,ζ

)

=a+an

( )

ζ zn +an+1

( )

ζ zn+1+..., H

}

,

, U

U z∈ ζ∈ where aj

( )

ζ are holomorphic functions in U for jn.

Let Aζ indicate the class of functions of the form:

(2)

(

,

) ( )

,

(

,

)

,

2

U U z z a z

z f

k

k ζ k ∈ ζ∈

+

=

ζ

=

(1.1) which are analytic in U×U and ak

( )

ζ are holomorphic functions in U for k ≥ 2.

Definition 1.1 [4]. Denote by Qζ the set of functions that are analytic and injective on U ×U\E

(

f, ζ

)

, where

(

,ζ

) {

= ∈∂ : lim

(

,ζ

)

=∞

}

,

f z U

r f

E

r z

and are such that fz

(

r,ζ

)

0 for r∈∂U×U\E

(

f, ζ

)

. The subclass of Qζ for which

( )

a

f 0, ζ = is denoted by Qζ

( )

a .

Definition 1.2 [5]. Let f

(

z

) (

, F z

)

analytic in U ×U. The function f

(

z

)

is said to be strongly subordinate to F

(

z, ζ

)

if there exists a function w analytic in U with

( )

0 =0

w and w

( )

z <1

(

zU

)

such that f

(

z, ζ

)

= F

(

w

( )

z , ζ

)

for all ζ∈U. In such a case we write f

(

z, ζ

)

≺≺ F

(

z,ζ

)

, zU,ζ∈U.

Remark 1.1 [5].

(1) Since f

(

z

)

is analytic in U ×U, for all ζ∈U and univalent in U, for all ,

U

ζ Definition 1.2 is equivalent to f

(

0,ζ

)

= F

(

0,ζ

)

for all ζ∈U and

(

U U

)

F

(

U U

)

.

f × ⊂ ×

(2) If f

(

z, ζ

)

= f

( )

z and F

(

z, ζ

)

= F

( )

z , the strong subordination becomes the usual notion of subordination.

If f

(

z

)

strongly subordinate to F

(

z,ζ

)

, then F

(

z

)

strongly superordinate to

(

z, ζ

)

.

f

Lemma 1.1 [3]. Let h

(

z

)

be a univalent with h

(

0,ζ

)

=a for every ζ∈U and let µ∈C\

{ }

0 with Re

( )

µ ≥ 0. If p∈H

[

a,1,ζ

]

and

(

z,

)

1 zp

(

z,

)

h

(

z,

) (

, z U, U

)

,

p z′ ζ ζ ∈ ζ∈

+ µ

ζ ≺≺ (1.2)

(3)

then

(

z,

)

q

(

z,

)

h

(

z,

) (

, z U, U

)

,

p ζ ≺≺ ζ ≺≺ ζ ∈ ζ∈

where q

(

z,ζ

)

=µzµ

0zh

(

t,ζ

)

tµ1dt is convex and it is the best dominant of (1.2).

Lemma 1.2 [4]. Let h

(

z

)

be a convex with h

(

0,ζ

)

=a for every ζ∈U and let

{ }

0

\

∈C

µ with Re

( )

µ ≥ 0. If p∈H

[

a,1, ζ

]

Qζ,

( )

(

ζ

)

ζ 1 ,

, zp z

z

p z is univalent

in U×U and

( ) ( )

1

(

,

)

,

(

,

)

,

,

, p z zp z z U U

z

h z′ ζ ∈ ζ∈

+ µ ζ

ζ ≺≺ (1.3)

then

(

z,

)

p

(

z,

)

,

(

z U, U

)

,

q ζ ≺≺ ζ ∈ ζ∈

where q

(

z, ζ

)

=µzµ

0zh

(

t, ζ

)

tµ1dt is convex and it is the best subordinant of (1.3).

For f ∈Aζ,m∈N,δ, j∈N0 =N∪

{ }

0 ,0≤ τ≤1. Frasin operator [2] Dmδ,τ :

ζ ζ →A

A (see [7]) is defined by

( ) ∑∞ ( ) ∑ ( ) ( ) ( )

=

δ

=

+

δ τ  ζ ∈ ζ∈



  − τ

 

−  + +

= ζ

2 1

, , 1 1 1 1 , , .

k

k k m

j

j

m j a z z U U

j k m

z z

f

D (1.4)

It is readily verified from (1.4) that

( ) (

τ z Dδ,τf

(

z

))

′ = Dδ+,1τf

(

z, ζ

) (

− 1−C

( ))

τ Dδ,τf

(

z, ζ

)

,

Cmj m z m mj m (1.5)

where

( ) ∑

=

( )

+ τ

 

= 

τ mj j j

mj

j C m

1

1 .

1

Special cases of this operator include the generalized Sălăgean operator [1] and the Sălăgean differential operator [6].

(4)

2. Main Results

Definition 2.1. Let ψ

(

z, ζ

)

be an analytic function in U×U with ψ

(

0,ζ

)

=1 for every ζ∈U and λ >0,m∈N, δ, j∈N0,0≤ τ≤1. A function f ∈Aζ is said to be in the class M

(

λ,δ, m, j,τ; ψ

)

if it satisfies the strong differential subordination

( ) ( )

( ) (

,

) (

,

) (

, ,

)

.

,

1 1 1

,

, D f z z z U U

C z

f D

z Cmj m mj m ψ ζ ∈ ζ∈





 ζ

τ + λ

 ζ



τ

− λ δ τ δ+τ ≺≺

A function f ∈Aζ is said to be in the class N

(

λ,δ,m, j,τ;ψ

)

if it satisfies the strong differential superordination

( )

( ) ( )

( ) (

,

) (

, ,

)

.

, 1 1

, , D ,1f z z U U

C z

f D z C

z m

mj m m

j

∈ ζ

 ∈



 ζ

τ + λ

 ζ



τ

− λ ζ

ψ ≺≺ δ τ δ+τ

Theorem 2.1. Let ψ

(

z, ζ

)

be a convex function in U×U with ψ

(

0,ζ

)

=1 for every ζ∈U and λ > 0. If M

(

λ,δ, m, j, τ; ψ

)

, then there exists a convex function

(

z

)

q such that q

(

z

)

≺≺ ψ

(

z

)

and fM

(

0, δ, m, j, τ; q

)

. Proof. Suppose that

( ) ( )

z z f z D

p m ζ

= ζ

δ τ ,

, ,

( ) ( ) ( ) ( )

=

δ

=

+  ζ ∈ ζ∈



  − τ

 

−  + +

=

2

1 1

1 , , .

1 1

1 1

k

k k m

j

j

j a z z U U

j

k m (2.1)

Then, pH

[

1,1,ζ

]

.

Since fM

(

λ,δ,m, j,τ;ψ

)

, then we have

( ) ( )

( ) (

,

) (

,

)

.

,

1 1 1

,

, ψ ζ





 ζ

τ + λ

 ζ



τ

− λ δ τ Dδ+τf z z

C z

f D

z C m m

j m m

j

≺≺ (2.2)

(5)

From (2.1) and (2.2), we get

( ) ( )

( ) (

,

) (

,

) (

,

) (

,

)

.

,

1 1 1

,

, = ζ +λ ′ ζ ψ ζ





 ζ

τ + λ

 ζ



τ

− λ δ τ Dδ+τf z p z zp z z

C z

f D

z C m m z

j m m

j

≺≺

An application of Lemma 1.1 with 1,

= λ

µ yields

(

z, ζ

)

q

(

z, ζ

)

ψ

(

z, ζ

)

.

p ≺≺ ≺≺

By using (2.1), we obtain

( )

(

,

) (

,

)

,

, , ζ ζ ψ ζ

δ τ

z z

z q z f Dm

≺≺

≺≺

where

(

z ζ

)

= λzλ

zψ

( )

t ζ tλ dt q

0

1 1 1

1 , , is convex and it is the best dominant.

Theorem 2.2. Let ψ

(

z, ζ

)

be a convex function in U×U with ψ

(

0,ζ

)

=1 for every ζ∈U and λ > 0. If f ∈N

(

λ, δ, m, j, τ; ψ

)

, δ τ

(

ζ

)

[

ζ

]

ζ

z Q z f Dm

, ∩ 1 , , 1

, H

and

( ) ( )

( ) ( )





 ζ

τ + λ

 ζ



τ

− λ δ τ , δ+τ ,

1 1 1

,

, D f z

C z

f D

z C m m

j m m

j

is univalent in U×U, then there exists a convex function q

(

z

)

such that

(

0, , m, j, ; q

)

.

f ∈N δ τ

Proof. Suppose that the function p

(

z, ζ

)

be defined by (2.1). It is evident that

[

1,1, ζ

]

ζ.

Q

p H ∩ After a short calculation and considering f ∈N

(

λ,δ,m, j,τ;ψ

)

, we can conclude that

(

,ζ

) (

, ζ

)

+λ ′

(

,ζ

)

. ψ z ≺≺ p z zpz z
(6)

An application of Lemma 1.2 with 1,

= λ

µ yields

(

z,ζ

)

p

(

z,ζ

)

.

q ≺≺

In view of (2.1), we obtain

( ) ( )

, ,

, ,

z z f z D

q m ζ

ζ ≺≺ δ τ where

(

z ζ

)

= λzλ

zψ

( )

t ζ tλ dt q

0

1 1 1

1 , ,

is convex and it is the best subordinant.

If we combine the results of Theorem 2.1 and Theorem 2.2, we obtain the following strong differential “sandwich theorem”.

Theorem 2.3. Let ψ1

(

z, ζ

)

and ψ2

(

z, ζ

)

be convex functions in U×U with

(

0,

)

2

(

0,

)

1

1 ζ =ψ ζ =

ψ for every ζ∈U and λ >0. If f ∈M

(

λ,δ, m, j, τ; ψ1

)

(

λ,δ,m, j,τ;ψ2

)

,

N

( )

[ ]

ζ

δ τ ζ ∈ ζ

z Q z f Dm

, ∩ 1 , , 1

, H and

( ) ( )

( ) ( )





 ζ

τ + λ

 ζ



τ

− λ δ τ , δ+τ ,

1 1 1

,

, D f z

C z

f D

z C m m

j m m

j

is univalent in U ×U, then

(

0, , m, j, ; q1

) (

0, , m, j, ; q2

)

,

f ∈M δ τ ∩N δ τ

where

(

z ζ

)

= λzλ

zψ

( )

t ζ tλ dt q

0

1 1 1 1

1 1 ,

, and

(

z ζ

)

= λzλ

zψ

( )

t ζ tλ dt q

0

1 1 2

1

2 1 , .

, The functions q and 1 q are convex. 2

(7)

Theorem 2.4. Let ψ

(

z, ζ

)

be a convex function in U×U with ψ

(

0,ζ

)

=1 for every ζ∈U and

(

,

)

2

( )

, ,

(

, ,Re

( )

2

)

.

1 0 ζ ∈ ζ∈ ε >−

+

= ε

ζ ε+

tεf t dt z U U z

z

G z (2.3)

If f ∈M

(

1,δ,m, j,τ;ψ

)

, then there exists a convex function q

(

z

)

such that

(

z

)

ψ

(

z, ζ

)

q ≺≺ and G∈M

(

1,δ, m, j,τ; q

)

. Proof. Suppose that

(

z,

) (

D , G

(

z,

))

,

(

z U, U

)

.

p ζ = mδ τ ζ ′z ∈ ζ∈ (2.4)

Then, pH

[

1,1,ζ

]

. From (2.3), we have

(

ζ

) (

= ε+

) ∫ ε ( ζ)

+

ε z

dt t f t z

G z

0

1 , 2 , . (2.5)

Differentiating both sides of (2.5) with respect to z, we get

(

ε+2

) (

f z

) (

= ε+1

) (

G z

)

+zGz

(

z, ζ

)

and

(

ε+2

)

Dmδ,τf

(

z, ζ

) (

= ε+1

)

Dmδ,τG

(

z,ζ

)

+ z

(

Dmδ,τG

(

z, ζ

))

z. Differentiating the last relation with respect to z, we have

( ( )) ( ( )) ( (

,

))

2

2 , 1

, , ,

, z m z m z

m f z D G z D G z

D ζ ″

+ + ε ζ ′

′ =

ζ δ τ δ τ

δ τ (2.6)

Since f ∈M

(

1, δ,m, j, τ;ψ

)

, then we have

( ) [ (

,

) (

1

( )) (

,

)] (

,

)

.

1

, 1

, ζ − − τ ζ ψ ζ

τ

δ τ +

δ τf z C D f z z

D z

C m

m j m m

j

≺≺ (2.7)

Now, from (1.5), (2.7) is equivalent to

(

Dmδ,τf

(

z,ζ

))

z ≺≺ ψ

(

z,ζ

)

. (2.8)
(8)

From (2.6) and (2.8), we get

( ( )) ( (

,

)) (

,

)

.

2

, 1 , 2

, ζ ″ ψ ζ

+ + ε

ζ ′ δ τ

δ τG z D G z z

Dm z m z ≺≺ (2.9)

Replacing (2.4) in (2.9), we obtain

( ) (

,

) (

,

)

.

2

, 1 ′ ζ ψ ζ

+ + ε

ζ zp z z

z

p z ≺≺

An application of Lemma 1.1 with µ = ε+2, yields

(

z, ζ

)

q

(

z, ζ

)

ψ

(

z, ζ

)

.

p ≺≺ ≺≺

By using (2.4), we obtain

(

Dmδ,τG

(

z,ζ

))

z ≺≺ q

(

z,ζ

)

≺≺ ψ

(

z,ζ

)

, where

(

z ζ

) (

= ε+

)

z(ε+ )

zψ

( )

t ζ tε+ dt q

0

1

2 ,

2 ,

is convex and it is the best dominant.

Theorem 2.5. Let ψ

(

z, ζ

)

be a convex function in U×U with ψ

(

0,ζ

)

=1 for every ζ∈U and G

(

z

)

is given by (2.3). If f ∈ N

(

1,δ,m, j,τ;ψ

)

,

(

Dmδ,τG

(

z

))

z ∈H

[

1,1,ζ

]

Qζ and

( ) [ (

ζ

) (

− −

( ))

τ

(

ζ

)]

τ

δ τ +

δ τ , 1 ,

1

1 ,

, f z C D f z

D z

C m

mj m m

j

is univalent in U×U, then there exists a convex function q

(

z

)

such that

(

1, , m, j, ; q

)

.

G∈N δ τ

Proof. Suppose that the function p

(

z, ζ

)

be defined by (2.4). It is evident that

[

1,1, ζ

]

ζ.

Q

p H ∩ After a short calculation and considering f ∈ N

(

1,δ,m, j,τ;ψ

)

, we can conclude that

( ) ( ) (

,

)

.

2 , 1

, ′ ζ

+ + ε ζ ζ

ψ z ≺≺ p z zpz z

(9)

An application of Lemma 1.2 with µ = ε+2, yields

(

z,ζ

)

p

(

z,ζ

)

.

q ≺≺

By using (2.4), we obtain

(

z,

) (

Dm, G

(

z,

))

z ,

q ζ ≺≺ δ τ ζ ′

where

(

z ζ

) (

= ε+

)

z(ε+ )

zψ

( )

t ζ tε+ dt q

0

1

2 ,

2 ,

is convex and it is the best subordinant.

If we combine the results of Theorem 2.4 and Theorem 2.5, we obtain the following strong differential “sandwich theorem”.

Theorem 2.6. Let ψ1

(

z, ζ

)

and ψ2

(

z, ζ

)

be convex functions in U×U with

(

,

)

2

(

,

)

1

1 ζ =ψ ζ =

ψ z z for every ζ∈U and G

(

z

)

is given by (2.3). If f

(

1,δ, m, j,τ; ψ1

)

N

(

1,δ, m, j,τ; ψ2

)

,

M ∩

(

Dmδ,τG

(

z

))

z∈H

[

1,1,ζ

]

Qζ and

( ) [ (

ζ

) (

− −

( ))

τ

(

ζ

)]

τ

δ τ +

δ τ , 1 ,

1

1 ,

, f z C D f z

D z

C m

mj m m

j

is univalent in U ×U, then

(

1, , m, j, ; q1

) (

1, ,m, j, ; q2

)

,

f ∈M δ τ ∩N δ τ

where

(

z ζ

) (

= ε+

)

z(ε+ )

zψ

( )

t ζ tε+ dt q

0

1 1

2

1 , 2 ,

and

(

z ζ

) (

= ε+

)

z(ε+ )

zψ

( )

t ζ tε+ dt q

0 2 1

2 , 2 2 , .

The functions q and 1 q are convex. 2

(10)

References

[1] F. M. Al-Oboudi, On univalent functions defined by a generalized Sălăgean operator, Int.

J. Math. Math. Sci. 27 (2004), 1429-1436. https://doi.org/10.1155/S0161171204108090

[2] B. A. Frasin, A new differential operator of analytic functions involving binomial series, Bol. Soc. Paran. Mat. 38(5) (2020), 205-213. https://doi.org/10.5269/bspm.v38i5.40188

[3] S. S. Miller and P. T. Mocanu, Differential Subordinations: Theory and Applications, Series on Monographs and Textbooks in Pure and Applied Mathematics, Vol. 225, Marcel Dekker Inc., New York and Basel, 2000.

[4] G. I. Oros, Strong differential superordination, Acta Univ. Apulensis 19(2009), 101-106.

[5] G. I. Oros and Gh. Oros, Strong differential subordination, Turk. J. Math. 33(3) (2009), 249-257.

[6] G. S. Salagean, Subclasses of univalent functions, Lecture Notes in Math. 1013, Springer Verlag, Berlin, 1983, pp. 362-372. https://doi.org/10.1007/BFb0066543

[7] F. Yousef, T. Al-Hawary and G. Murugusundaramoorthy, Fekete-Szegö functional problems for some subclasses of bi-univalent functions defined by Frasin differential operator, Afrika Matematika 30(3-4) (2019), 495-503.

https://doi.org/10.1007/s13370-019-00662-7

Referensi

Dokumen terkait

SATORU MURAKAMI † and TOSHIKI NAITO, NGUYEN VAN MINH ‡ Department of Applied Mathematics, Okayama University of Science,.. Okayama 700-0005, Japan

Sri Gemawati : Department of Mathematics, Faculty of Mathematics and Natural Sciences, University of Riau, Bina Widya Campus, Pekanbaru 28293, IndonesiaM.

of Mosul, Mosul, IRAQ 4Physics Department, College of Science and Humanities in Al-Kharj, Prince Sattam bin AbdulAziz University, Al-Kharj 11942, Saudi Arabia A R T I C L E I N F O

Hassan Department of Microbiology, College of Medicine, AL-Nahrain University, Baghdad, Iraq E-mail: [email protected] Phone number: +91-964-07802559522 Abstract

v Thesis Title Three-Dimensional Differential Transform Method for Solving Partial Differential Equations Author Chalong Sawaddee Degree Master of Science Applied Mathematics Thesis

Touhid Bhuiyan Chairman Professor and Director Department of Computer Science and Engineering Faculty of Science and Information Technology Daffodil International University Dr..

Majeed3 1Applied Physics Branch, Applied Sciences Department, University of Technology, Baghdad, Iraq 2Department of Laser and Optoelectronic Engineering, College of Engineering,

Monaquel Mathematics Department, Faculty of Science, King Abdul Aziz University, Jeddah, Saudi Arabia Abstract-Our main purpose of this paper is to find the corresponding set of