SOME DIFFERENTIAL SUBORDINATION AND

25

SOME DIFFERENTIAL SUBORDINATION AND SUPERORDINATION RESULTS ASSOCIATED WITH THE GENERALIZED BESSEL FUNCTIONS H. E. Darwish, A. Y. Lashin, and S. M. Sowileh* Department of Mathematics Faculty of Science, Mansoura University Mansoura, H. E. DARWISH, A.35516, Y. LASHIN, EGYPT. AND S. M. SOWILEH E-mail address: [email protected] Abstract. In this paper, we obtain many subordination and superordination 2 = Z0 , which involves the results, using a new operator Bkc k = p + b+1 2 normalized form of the generalized Bessel functions of …rst kind. Sandwichtype result is also obtained.

2010 Mathematics Subject Classi…cation: 30C45. Key words: Bessel functions; di¤erential subordination and superordination, sandwich-type result. 1. Introduction Let H=H(4) be the class of functions which are analytic in the open unit disk 4 = fz 2 C : jzj < 1g:

For a 2 C and n 2 N = f1; 2; :::g, let

H[a; n] = ff : f 2 H and f (z) = a + an z n + an+1 z n+1 + :::g;

with H0 = H[0; 1] and H1 =H[1; 1]: We denote by A the class of all normalized analytic functions in 4 of the form: X (1.1) f (z) = z + an+1 z n+1 (z 2 4): n 1

Let the functions f and g be in the class H. The function f is said to be subordinate to g; or g is superordinate to f , if there exists a Schwarz function w analytic in 4, with w(0) = 0 and jw(z)j < 1(z 2 4); such that f (z) = g(w(z)) (z 2 4): In such a case, we write f

g

or f (z)

g(z)

(z 2 4):

Furthermore, if the function g is univalent in 4, then we have the following equivalence (see, for details, [22]; see also [[5], [6], [7], [8], [9], [10], [11], [12], [13], [14] [18], [32], [34]]): f (z) 2

g(z) (z 2 4) () f (0) = g(0) and f (4)

g(4):

Let : C ! C and h(z) be univalent in 4. If p(z) is analytic in 4 and satis…es the …rst order di¤erential subordination: (1.2)

(p(z); zp0 (z); z) 1

00

h(z) (z 2 4);

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H. E. Darwish et al. Journal of Modern Science & Engineering Vol.2, No.2 (2018) 25-35 2

H . E. DARW ISH , A. Y. LASH IN , AND S. M . SOW ILEH

then p(z) is called a solution of the di¤erential subordination 1.2. A univalent function q(z) is called a dominant of the solutions of the di¤erential subordination 1.2, or more simply a dominant, if p(z) q(z) for all p(z) satisfying 1.2. A dominant qe that satis…es qe q for all dominants q of 1.2 is said to be the best dominant. If p(z) and (p(z); zp0 (z); z) are univalent in 4 and If p(z) satis…es the …rst order di¤erential superordination: (1.3)

h(z)

(p(z); zp0 (z); z) (z 2 4);

then p(z) is called a solution of the di¤erential superordination 1.3. An analytic function q(z) is called a subordinant of the solutions of the di¤erential superordination 1.3, or more simply a subordinant if q(z) p(z) for all p(z) satisfying 1.3. A univalent subordinant qe that satis…es q qe for all subordinants q of 1.3 is said to be the best subordinant 1.3. We recall that the generalized Bessel function of the …rst kind w = wp;b;c is de…ned as the particular solution of the second-order linear homogenous di¤erential equation (see for details [3] and [16]): (1.4)

z 2 w00 (z) + bzw0 (z) + cz 2

p2 + (1

b)p w(z) = 0;

(p; b; c; z) 2 C:

which is a natural generalization of Bessel’s equation. This function has the familiar representation as follows: n X z 2n+p ( c) (z 2 C); (1.5) w = wp;b;c = b+1 n! (p + n + 2 ) 2 n 0

Here stands for the Euler’s gamma function: The series in Equation 1.5 permits the study of Bessel, modi…ed Bessel, spherical Bessel, modi…ed spherical Bessel and ultraspherical Bessel functions all together. Solutions of 1.4 are referred to as the generalized Bessel function of order p: The particular solution given by 1.5 is called the generalized Bessel function of the …rst kind of order p: Although the series de…ned above is convergent everywhere, the functionwp;b;c is generally not univalent in 4. It is worth mentioning that, in particular: (1) If b = c = 1 in 1.5, then we have wp;1;1 (z) = Jp (z); where Jp (z) denotes the familiar Bessel function of the …rst kind of order p de…ned by (see [33]; see also [3]) n X ( 1) z 2n+p (1.6) Jp (z) = (z 2 C): n! (p + n + 1) 2 n 0

(2) If b = 1 and c =

1 in 1.5; then we have wp;1;

1 (z)

= Ip (z);

where the modi…ed Bessel function Ip (z) of the …rst kind of order p de…ned by (see [33]; see also [3]) X z 2n+p 1 (z 2 C): (1.7) Ip (z) = n! (p + n + 1) 2 n 0

(3) If b = 2 and c = 1 in 1.5, then we have p p wp;2;1= 2jp = ;

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H. E. Darwish et al. Journal of Modern Science & Engineering Vol.2, No.2 (2018) 25-35 3

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where jp is the spherical Bessel function of the …rst kind of order p, de…ned by (see [3]) p X n ( 1) z 2n+p (1.8) jp (z) = p (z 2 C): 3 2 n 0 n! (p + n + 2 ) 2 Now, consider the function 'p;b;c (z) : 4 ! C, de…ned by the transformation 'p;b;c (z) = 2p

(1.9)

p+

b+1 2

z1

p 2

p wp;b;c ( z):

Using the general Pochhammer symbol (or Appell) ( )# de…ned in terms of the Euler gamma function for ; # 2 C; by ( )# =

( + #) = ( )

1;

( + 1)

( +n

# = 0; 2 C= f0g ; # = n 2 N, 2 C

1);

and (0)0 = 1; we obtain for the function 'p;b;c the following representation (1.10)

'p;b;c = z +

X ( c)n z n+1 4n (k)n n!

k =p+

n 1

b+1 2 = Z0 2

;

where Z0 = f0; 1; 2; 3; g = Z [ f0g : For simplicity, we write 'k;c (z) = 'p;b;c (z): For f; g 2 A, where f is given by 1.1 and g is de…ned by g(z) = z + X bn+1 z n+1 ; then the convolution (or Hadamard product) f g of the functions n 1

f and g is de…ned by (f

g)(z) = z +

X

an+1 bn+1 z n+1 = (g f (z);

n 1

z 2 4.

Note that f g 2 A. Now, we introduce a new operator Bkc : A ! A, which is de…ned by the Hadamard product (1.11) Bkc = 'k;c (z) f (z) = z +

X ( c)n an+1 z n+1 4n (k)n n!

k =p+

n 1

b+1 2 = Z0 2

:

It is not di¢ cult to verify from the de…nition 1.11 that (1.12)

0

c z Bk+1 f (z) = kBkc f (z)

(k

c 1)Bk+1 f (z):

In fact, the function Bkc f (z) given by 1.11 is an elementary transform of the generalized hypergeometric function, that is, we have c z) f (z): Bkc f (z) = z0 F1 (k; 4 By giving speci…c values to the parameters b and c; we obtain some new operators: (i) Putting b = c = 1 in 1.11, or 1.12 we have the operator Jp : A !A related with Bessel function, de…ned by (1.13) h X ( 1)n an+1 z n+1 p i p Jp f (z) = 'p;1;1 (z) f (z) = 2p (p + 1) z 1 2 Jp ( z) f (z) = z+ ; 4n (p + 1)n n! n 1

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H. E. Darwish et al. Journal of Modern Science & Engineering Vol.2, No.2 (2018) 25-35 4

H . E. DARW ISH , A. Y. LASH IN , AND S. M . SOW ILEH

which satis…es the recurrence relation 0

z [Jp+1 f (z)] = (p + 1)Jp f (z)

pJp+1 f (z):

(ii) Setting b = 1 and c = 1 in 1.11 or 1.12 we have the operator Ip : A !A related with modi…ed Bessel function, de…ned by (1.14) h X an+1 z n+1 p i p Ip f (z) = 'p;1; 1 (z) f (z) = 2p (p + 1) z 1 2 Ip ( z) f (z) = z+ ; 4n (p + 1)n n! n 1

which satis…es the recursive relation 0

z [Ip+1 f (z)] = (p + 1)Ip f (z)

pIp+1 f (z):

(iii) Choosing b = 2 and c = 1 in 1.11 or 1.12 we obtain the operator Sp : A !A related with spherical Bessel function, de…ned by (1.15) X ( 1)an+1 z n+1 p p 1 1 3 2 2p+ 2 ; Sp f (z) = z 1 2 jp ( z) f (z) = z + p+ 2 4n (p + 23 )n n! n 1

which satis…es the recurrence relation 3 1 0 z [Sp+1 f (z)] = (p + )Sp f (z) (p + )Sp+1 f (z): 2 2 The generalized Bessel function is a recent topic of study in Geometric Function Theory (e.g. see the work of [1], [2], [3], [4], [29], [30], [31]). Motivated by results on connections between various subclasses of analytic univalent functions by using hypergeometric functions (see [21], [24], [26], [27], [28]). We also mention that very recently, Deniz et al. [17] and Deniz [15] were interested on the univalence and convexity of some integral operators de…ned by normalized form of the generalized Bessel functions of the …rst kind given by 1.10. In the present paper, we obtain some subordination and superordination results, c by using Bessel function of …rst kind BK ; which is de…ned by 1.11, together with some sandwich-type theorems. To prove our main results, we need the following de…nitions and Lemmas. De…nition 1. [23] We denote by F the class of functions q(z) that are analytic and injective on 4nE(q); where E(q) =

2 @4 : lim q(z) = 1 ; z!

and are such that q 0 ( ) 6= 0 for

2 @4nE(q):

Further let the subclass of F for which q(0) = a be denoted by F(a), F(0) F(1) F1 :

F0 and

De…nition 2. [23] A function L(z; t) : 4 [0; 1) ! C is said to be a subordination(or a loewner) chain if L(:; t) is analytic and univalent in 4 for all t 0; L(z; :) is continously di¤ erentiable on [0; 1) for all z 2 4 and L(z; t1 ) L(z; t2 ) for all 0 t1 t2 :

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H. E. Darwish et al. Journal of Modern Science & Engineering Vol.2, No.2 (2018) 25-35 5

SO M E SU BO RD IN ATIO N AN D SU PERO RD IN ATIO N

2. A Set of Lemmas The following lemmas will be required in our present investigation. Lemma 1. [25] The function L(z; t) : 4 [0; 1) ! C of the form L(z; t) = a1 (t)z + a2 (t)z 2 + ::: with (a1 (t) 6= 0; t subordination chain if and only if ( z@L(z;t) ) @z @L(z;t) @t

Re

Lemma 2. condition:

>0

0 ) and lim ja1 (t)j = 1 is a

(z 2 4; 0

t!1

t < 1):

[19] Suppose that the function H : C2 ! C satis…es the following RefH(is; t)g

0

for all real s and for all t

n(1 + s2 )=2;

(n 2 N):

If the function p(z) = 1 + pn z n + pn+1 z n+1 ::: is analytic in 4 and RefH(p(z); zp0 (z))g > 0 (z 2 4);

then,

Refp(z)g > 0 (z 2 4): Lemma 3. [20] Let

;

2 C with

6= 0 and h 2 H(4) with h(0) = c: If

Ref h(z) + g > 0 (z 2 4); then, the solution of the following di¤ erential equation q(z) +

zq 0 (z) q(z) +

= h(z)

(z 2 4; q(0) = c)

is analytic in U and satis…es the following inequality: Ref q(z) + g > 0 (z 2 4): Lemma 4. [22] Let p 2 F(a) and let q(z) = a + an z n + an+1 z n+1 + ::: be analytic in 4 with

q(z) 6= a and n

1:

If the function q is not subordinate to p, then there exist two points

such that q(4r0 )

z0 = r0 ei 2 4 and

0

2 @4nE(q);

p(4); q(z0 ) = p( 0 ) and z0 q 0 (z0 ) = m 0 p0 ( 0 )

(m

n):

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H. E. Darwish et al. Journal of Modern Science & Engineering Vol.2, No.2 (2018) 25-35 6

H . E. DARW ISH , A. Y. LASH IN , AND S. M . SOW ILEH

Lemma 5. [23] Let q 2 H[a; 1] and ' : C2 ! C: Also set ' (q(z); zq 0 (z)) := h(z)

(z 2 4):

0

If L(z; t) = '(q(z); tzq (z)) is a subordination chain and p 2 H[a; 1] \ F(a); then. '(p(z); zp0 (z))

h(z)

(z 2 4):

implies that q(z) p(z) (z 2 4): Furthermore, if '(q(z); zq (z)) = h(z) has a univalent solution q 2 F(a); then q is the best subordinant. 0

In this article, we investigate the subordination and superordination preserving c properties of the Bessel function BK de…ned by 1.11, with the Sandwich-type Theorems. 3. Di¤erential subordination, Superordination and sandwich type B c (g)(z) B c (g)(z) + kz theorem for (1 ) k 1z

Theorem 1. Let f; g 2 A and suppose that (3.1)

Re 1 +

z

00 0

(z) (z)

>

;

where (z) := (1

)

Bkc

(k > 1; 0 and (3.2)

=

)2 + (k

(1

Bkc (g)(z) ; z z < 1; z 2 4);

1 (g)(z)

1)2 (1 )2 4(1 )(k 1)

+

(k

1)2

(0
0 (z 2 4): Taking the logarithmic di¤erentiation on both sides of the second equation in 3.5 and using the equation 1.12 we obtain (3.7)

(k

1) (z) := (k

)zG0 (z):

1)G(z) + (1

Now, by di¤erentiating both sides of 3.7, we obtain the relationship: 1+

z

(3.8)

00

(z) zG00 (z) zq 0 (z) = 1 + + 0 G0 (z) q(z) + (k 1)=(1 (z) = q(z) +

zq 0 (z) q(z) + (k 1)=(1

)

h(z):

)

We also note from 3.1 that Re h(z) +

(k 1

1)

>0

(z 2 4);

Moreover, by using Lemma 3, we conclude that the di¤erential equation 3.8 has a solution q 2 H(4) with q(0) = h(0) = 1: Let v (3.9) H(u; v) = u + + ; u + (k 1)=(1 ) where

is given by 3.2. From 3.1, 3.8 and 3.9, we have RefH(q(z); zq 0 (z))g > 0

(z 2 4):

Now, we proceed to show that (3.10)

RefH(iv; t)g

0

1 (1 + v 2 ) : 2

v 2 R; t

Indeed, from 3.9, we have RefH(iv; t)g = Re iv +

(3.11) where (3.12)

=

(k j(k

1)t=(1

2

1)=(1

M (v) :=

k 1

) ) + ivj

1

2

t 1)=(1

iv + (k

+

M (v)

+

v2

)

2 j(k (k 1

1)

2

1)=(1 (k 1

1)

2;

) + ivj

1 :

For given by 3.2, we can prove easily that the expression M (v) is given by 3.12 is greater than or equal to zero. Hence, from 3.9, we see that 3.10 holds true. Thus using Lemma 2, we conclude that Refq(z)g > 0 0

(z 2 4):

Moreover, we see that the condition G (0) 6= 0 is satis…ed. Hence, the function G de…ned by 3.5 is convex in U: Next, we prove that the subordination condition 3.3 implies that (3.13)

F (z)

G(z)

(z 2 4);

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H. E. Darwish et al. Journal of Modern Science & Engineering Vol.2, No.2 (2018) 25-35 8

H . E. DARW ISH , A. Y. LASH IN , AND S. M . SOW ILEH

for the functions F and G de…ned by 3.5. Without loss of generality, we can assume that G is analytic and univalent on U and G0 ( ) 6= 0 ( 2 @4): For this purpose, we consider the function L(z; t) given by L(z; t) := G(z) +

(1 (k

)(1 + t) 0 zG (z) (k > 1; 0 1)

We note that (k @L(z; t) = G0 (0) @z z=0

t < 1; 0

1) + (1 )(1 + t) 6= 0 (k > 1; 0 (k 1)

< 1; z 2 4) : t < 1; 0

< 1) :

This shows that the function L(z; t) =

1 (t)z

+ :::

satis…es the condition 1 (t)

Furthermore, we have z@L(z; t)=@z Re @L(z; t)=@t

= Re

6= 0 (0 (k 1

1)

t < 1): + (1 + t) 1 +

zG00 (z) G0 (z)

> 0:

Therefore, by virtue of Lemma 1, L(z; t) is a subordination chain. We observe from the de…nition of subordination chain that L( ; t) 2 = L(4; 0) = (4)

( 2 @4; 0

t < 1):

Now, suppose that F is not subordinate to G; then by Lemma 4, there exists points z0 2 4 and 0 2 @4; such that F (z0 ) = G( 0 ) and z0 F 0 (z0 ) = (1 + t) 0 G0 ( 0 ) (0

Hence, we have L( 0 ; t) = G( 0 ) +

(1 (k

)(1 + t) 1)

Bkc

0 0G ( 0)

= F (z0 ) +

(1 (k

t < 1):

) z0 F 0 (z0 ) 1)

Bkc f (z0 ) 2 (4); z0 z0 by virtue of the subordination condition 3.3. This contradicts the above observation L( 0 ; t) 2 = (4): Therefore, the subordination condition 3.3 must imply the subordination given by 3.13. Considering F (z) = G(z); we see that the function G is the best dominant. This evidently completes the proof of Theorem 1 = (1

)

1 f (z0 )

+

Theorem 2. Let f; g 2 A and suppose that Re 1 +

00

z

0

(z) (z)

>

;

where (z) := (1

)

Bkc

(k > 1; 0 and

is given by 3.2, and (1

)

Bkc

Bkc g(z) z z < 1; z 2 4); 1 g(z)

1 f (z)

z

+

+

Bkc f (z) z

;

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H. E. Darwish et al. Journal of Modern Science & Engineering Vol.2, No.2 (2018) 25-35 SO M E SU BO RD IN ATIO N AN D SU PERO RD IN ATIO N

is univalent in 4 and reltaion (3.14)

(z)

Bkc (f )(z) z

(1

)

9

2 H[1; 1] \ F: Then, the following superordination Bkc

1 (f )(z)

z

Bkc (f )(z) (z 2 4); z

+

implies that Bkc (g)(z) Bkc (f )(z) (z 2 4); z z c B (g)(z) and the function k z is the best subordinant. Proof. The …rst part of the proof is similar to that of Theorem 1 and so we will use the same notation as in the proof of Theorem 1. Now, let us de…ne the functions F and G by 3.5. We …rst note that, if the function q is de…ned by 3.6, using 3.7, then we obtain 1 zG0 (z) := '(G(z); zG0 (z)): (3.15) (z) = G(z) + (k 1) Then using the same method as in the proof of Theorem 1, we can prove that Refq(z)g > 0

(z 2 4);

that is, G de…ned by 3.5 is convex (univalent) in 4. Next, we prove that the subordination condition 3.14 implies that (3.16)

G(z)

F (z)

(z 2 4);

for the functions F and G de…ned by 3.5. Now consider the function L(z; t) de…ned by (1 )t 0 L(z; t) := G(z) + zG (z) (z 2 4; k > 1; 0 t < 1): (k 1) As G is convex and (k 1)=(1 ) > 0; we can prove easily that L(z; t) is a subordination chain as in the proof of Theorem 1. Therefore according to Lemma 5, we conclud that the superordination condition 3.14 must imply the superordination given by 3.16. Furthermore, as the di¤erential equation 3.15 has the univalent solution G, it is the best subordinant of the given di¤erential superordination. Therefore, we complete the proof of Theorem 2. If we combine Theorems 1 and 2, then we obtain the following sandwich-type Theorem. Theorem 3. Let f; g 2 A ( = 1; 2) and suppose that (3.17)

(

< 1+

00

z 0

(z) (z)

)

>

;

where (z) := (1

)

(k > 1; 0

Bkc

Bkc g (z) z z < 1; z 2 4);

1g

(z)

+

;

34

H. E. Darwish et al. Journal of Modern Science & Engineering Vol.2, No.2 (2018) 25-35 10

where

H . E. DARW ISH , A. Y. LASH IN , AND S. M . SOW ILEH

is given by 3.2, and (1

is univalent in 4 and 1 (z)

)

Bkc (f )(z) z

(1

)

Bkc

1 f (z)

z

+

Bkc f (z) z

2 H[1; 1] \ F: Then, the following reltaion

Bkc

1 f (z)

z

+

Bkc f (z) z

'2 (z) (z 2 4);

implies that Bkc g1 (z) z and the functions best dominant.

Bkc f (z) z

Bkc g1 (z) B c g (z) ; k z2 z

Bkc g2 (z) (z 2 4); z

are respectively, the best subordinant and the

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11

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