Differential Subordination and Superordination for Multivalent Functions Involving a Generalized Differential Operator

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1 Differential Subordination Superordination for Multivalent Functions Involving a Generalized Differential Operator Waggas Galib Atshan, Najah Ali Jiben Al-Ziadi Departent of Matheatics, College of Coputer Science Inforation Technology, University of Al-Qadisiyah, Diwaniya, Iraq Departent of Matheatics, College of Science, University of Baghdad, Baghdad, Iraq ABSTRACT: In this paper, we deduce soe subordination superordination outcoes involving the generalized differential operator (a 1, b 1 ) for certain ultivalent analytic functions in the open unit disk. These outcoes are applied to obtain differential swich theores. KEYWORDS: Analytic function; ultivalent function; differential subordination; differential superordination; swich theore; generalized differential operator. I. INTRODUCTION LetH = H(U) sybolize the class of analytic functions in the open unit disk U = z C z < 1 let H[a, p] sybolize the subclass of the function f H of the shape: f z = a + a p + a p a C; p N = 1,2,. (1) Also, let A(p) be the subclass of H consisting of functions of the shape: f z = + a p+n +n p N = 1,2,. (2) n=1 Let f, g H, if there exists a Schwarz function w analytic in U with w 0 = 0 w z < 1 (z U) such that f z = g w z, the function f is invited subordinate to g, or g is invited superordinate to f, In such a case we write f g or f z g z (z U). If g is univalent in U, f g if only if f 0 = g(0) f(u) g(u). Let p, H φ r, s, t; z : C 3 U C. If p φ(p z, zp z, z 2 p z ; z) are univalent functions in U if p satisfies the second-order superordination z φ p z, zp z, z 2 p z ; z, (3) p is invited a solution of the differential superordination (3). (If f is subordinate to g, g is superordinate to f). An analytic function q is invited a subordinant of (3), if q p for all the function p satisfying (3). An univalent subordinant q that satisfies q q for all the subordinants q of (3) is invited the best subordinant. Recently, Miller Mocanu [1] gained conditions on the functions, q φ for which the following odulation holds: z φ p z, zp z, z 2 p z ; z q z p z. Now, x n denotes the Pochhaer sybol defined by Γ(x + n) 1, n = 0, x n = = Γ(x) x x 1 x + n 1, n = 1,2,3,. El-Yagubi Darus [2] defined a generalized differential operator, as follows: a 1, b 1 A(p) A(p) f z = + p + λ 1 + λ 2 n + b p + λ 2 n + b a 1 n a r n a p+n +n, (4) b 1 n b s n n! n=1 where, b, r, s N 0 = N 0, λ 2 λ 1 0 a i C, b q C \ 0, 1, 2,, i = 1,, r, q = 1,, s, r s + 1. Copyright to IJARSET

2 It follows fro (4) that λ 1 z = p + λ 2 n + b D +1,b p + λ 2 n pλ 1 + b. (5) It should be noted that the linear operator a 1, b 1 is a generalization of any other linear operators considered earlier. In particular: (1) For λ 2 = b = 0, the operator reduces to the operator was given by Selvaraj Karthikeyan [3]. (2) For = 0, the operator reduces to the operator was given by El-Ashwah [4]. (3) For = 0, p = 1, the operator reduces to the well-known operator introduced by Dziok Srivastava [5]. (4) For = 0, r = 2, s = 1 p = 1, we gain the operator which was given by Hohlov [6]. (5) For r = 1, s = 0, a 1 = 1, λ 1 = 1, λ 2 = b = 0 p = 1, we get the Salagean derivative operator [7]. The ain object of the present paper is to find sufficient conditions for certain noralized analytic functions f to satisfy q 1 z td +1,b q 1 z + 1 t q 2 z where q 1 q 2 are given univalent functions in U with q 1 0 = q 2 0 = 1. II. PRELIMINARIES In order to anifest our leading results, we require the following definition leas. q 2 z, Definition (1) [8]: Denote by Q the set of all functions f that are analytic injective on U E(f), where E f = ζ U: li z ζ f z = (6) are such that f ζ 0 for ζ U\E(f). Lea (1)[1]: Let q be a convex univalent function in U let α C, β C\ {0} with Re 1 + zq z q > ax 0, Re α z β. If p is analytic in U αp z + βzp z αq z + βzq z, (7) p q q is the best doinant of (7). Lea(2) [9]: Let q be univalent in the unit disk U let θ φ be analytic in a doain D containing q(u) with φ(w) 0 when w q u. Set Q z = zq z φ q z z = θ q z + Q(z). Suppose that (1) Q z is starlike univalent in U, (2) Re z z Q z > 0 for z U. If p is analytic in U, with p 0 = q 0, p(u) D θ p z + zp z φ p z θ q z + zq z φ q z, (8) p q q is the best doinant of (8). Lea (3) [1]: Let q be convex univalent in U let β C. Further assue that Re β > 0. If p H[q 0, 1] Q p z + βzp (z) is univalent in U, q z + βzq z p z + βzp z, (9) which iplies that q p q is the best subordinant of (9). Copyright to IJARSET

3 Lea(4) [9]: Let q be convex univalent in the unit disk U let θ φ be analytic in a doain D containing q(u). Suppose that (1) Re θ q z > 0 for z U, φ q z (2) Q z = zq (z)φ q z is starlike univalent in U. If p H[q 0, 1] Q, with p U D, θ p z + zp (z)φ p z is univalent in U θ q z + zq z φ q z φ p z + zp z φ p z, (10) q p q is the best subordinant of (10). III. SUBORDINATION RESULTS Theore (1): Let q(z) be convex univalent in U with q 0 = 1, η C/ 0, > 0 suppose that Re 1 + zq z q > ax 0, Re p + λ 2n + b. (11) z If 1 z = 1 + η q(z) is the best doinant of (13). a 1, b 1 f(z) Proof: Define the analytic function p(z) by η D +1,b (12) 1 z q z + p + λ 2 n + b zq z, (13) p z = Differentiating (15) logarithically with respect to z, we have zp (z) p(z) = p z D q z (14) (15). (16) Now, using the identity (5), we obtain the following zp (z) p(z) = p + λ 2n + b 1 D +1,b λ 1. Therefore, λ 1 p + λ 2 n + b zp z = 1 D +1,b. Thus, the subordination (13) is equivalent to p z + p + λ 2 n + b zp z q z + p + λ 2 n + b zq z. Applying Lea (1) with β = Putting q z = 1+Az 1+Bz p+λ 2 n+b α = 1, we obtain (14). ( 1 B < A 1) in Theore (1), we get the following result. Corollary (1): Let η C/{0} 1 B < A 1. Also, suppose that Re 1 Bz 1 + Bz > ax 0, Re p + λ 2n + b. Copyright to IJARSET

4 If f A(p) satisfies the following subordination condition: 1 z 1 + Az 1 + Bz + p + λ 2 n + b where 1 z given by (12), the function 1+Az is the best doinant. 1+Bz Taking A = 1 B = 1 in Corollary (1), we get the following result. 1 + Az 1 + Bz A B z 1 + Bz 2, Corollary (2): Let η C/{0} Suppose that Re 1 + z 1 z > ax 0, Re p + λ 2n + b. If f A(p) satisfies the following subordination: 1 z 1 + z 1 z + p + λ 2 n + b where 1 z given by (12), the function 1+z 1 z is the best doinant. 2z 1 z 2, 1 + z 1 z Theore (2): Let q z be univalent in U with q 0 = 1, q z 0 zq (z) is starlike in U, let, η C/ 0 u, v, ξ C. Let f A(p) suppose that f g satisfy the next two conditions: + (1 t) 0 z U, 0 t 1 (17) Re 1 + v 2ξ q z + η η q z 2 zq z q z + zq z q > 0 18 z If 2 z = u + v td +1,b + (1 t) +ξ td +1,b +η tz D +1,b + (1 t) + 1 t z + (1 t) q is the best doinant of (20). Proof: Define the analytic function p by 2 q(z) p. (19) 2 z u + vq z + ξ q z 2 + η zq z q z, (20) + (1 t) +1,b p z = td + (1 t) q z, (21). (22) Copyright to IJARSET

5 Then p is analytic in U p 0 = 1, differentiating (22) logarithically with respect to z, we get zp z p z = tz D +1,b + 1 t z p. (23) + (1 t) By setting θ w = u + vw + ξw 2 φ w = η, (w C /{0}), w we see that θ w is analytic in C, φ w is analytic in C /{0} that φ w 0, w C /{0}. Also, we get Q z = zq z φ q z = η zq z, z U, q z z = θ q z + Q z = u + vq z + ξ q z 2 + η zq z q z. It is clear that Q z is starlike in U, that Re z z Q z = Re 1 + v 2ξ q z + η η q z 2 zq z q z + zq z q z > 0 z U. By aking use of (23), the hypothesis (20) can be equivalently written as θ p z + zp z φ p z φ q z + zq z φ q z, thus, by applying Lea (2), the proof is copleted. Theore (3): Let q z be univalent in U with q 0 = 1, let, η C/ 0 v, ξ C. Let f(z) A(p) suppose that f g satisfy the next two conditions: td + (1 t) 0 z U, 0 t 1 (24) And Re 1 + zq z q z > ax 0, Re v η z U. 25 If 3 z = td +1,b + 1 t v + η tz D +1,b + 1 t + 1 t q is the best doinant of (27). p + ξ (26) 3 z vq z + ηzq z + ξ, (27) + (1 t) Proof: Let the function p be defined on U by (16). Then a coputation shows that zp z = td +1,b + 1 t tz D +1,b + 1 t z + (1 t) By setting θ w = vw + ξ, φ w = η, (w C), we see that θ w, φ w are analytic in C that φ w 0. Also, we get Q z = zq z φ q z = ηzq z, z U, q z, (28) p. (29) Copyright to IJARSET

6 z = θ q z + Q z = vq z + ηzq z + ξ (z U). Fro the assuption (25) we see that Q z is starlike in U, that Re z z Q z = Re v η + zq z q + 1 > 0 z U, z, by using Lea (2) we deduce that the subordination (27) iplies p z q z, the function q is the best doinant of (27). IV. SUPERORDINATION RESULTS Theore (4): Let q be convex in U with q 0 = 1, > 0 Re η > 0. Let f A(p) satisfies H q 0, 1 Q. If the function 1 z given by (12) is univalent in U, q z + p + λ 2 n + b zq z 1 z, (30) q is the best subordinant of (30). q z Proof: Define the analytic function p(z) by D λ 1,λ2,p Copyright to IJARSET a 1,b 1 f z p z =. (32) Differentiating (32) logarithically with respect to z, we have zp z p z = p z D. (33) After soe coputations using the identity (5), fro (33), we have 1 z = p z + p + λ 2 n + b zp z, now, by using Lea (3), we get the desired result. Putting q z = 1+Az ( 1 B < A 1) in Theore (4), we get the following corollary. 1+Bz Corollary (3): Let 1 B < A 1, > 0 Re η > 0. Also let H q 0, 1 Q. If the function 1 z given by (12) is univalent in U, f A(p) satisfies the following superordination condition: 1 + Az 1 + Bz + A B z p + λ 2 n + b (1 + Bz) 2 1 z, 1 + Az 1 + Bz the function 1+Az is the best subordinant. 1+Bz Theore (5): Let q be convex univalent in U with q 0 = 1, q(z) 0 zq z u, v, ξ C. Further assue that q satisfies Re v + 2ξq z q z q z η q z 31 is starlike in U, let, η C/{0} > 0 z U. (34)

7 Let f(z) A(p) suppose that f(z) satisfies the next conditions: td + (1 t) 0 z U, 0 t 1 (35) td + (1 t) H q 0, 1 Q. (36) If the function 2 (z) given by (19) is univalent in U, u + vq z + ξ q z 2 + η zq z q z 2 z, (37) q z td +1,b q is the best subordinant of (37). + 1 t Copyright to IJARSET , (38) Proof: Let the function p(z) be defined on U by (22). Then a coputation shows that zp z p z = tz D +1,b + 1 t z p. (39) + (1 t) By setting θ w = u + vw + ξw 2 φ w = η, (w C/{0}), w we see that θ w is analytic in C, φ w is analytic in C/ 0 that φ w 0, w C/{0}. Also, we get Q z = zq z φ q z = η zq z, q z z U. It is observe that Q(z) is starlike in U, that Re θ q z = Re v + 2ξq z q(z)q z φ q z η > 0 z U. By aking use of (39) the hypothesis (37) can be equivalently written as θ q z + zq z φ q z θ p z + zp z φ p z, thus, by applying Lea (4), the proof is copleted. Using arguents siilar to those of the proof of Theore (3), by applying Lea (4), we obtain the following result. Theore (6): Let q be convex in U with q 0 = 1, let, η C/ 0 v, ξ C Re v η q z > 0. Let f A(p) suppose that f(z) satisfies the next conditions: td + (1 t) 0 z U, 0 t 1 (40) td + (1 t) H q 0, 1 Q. (41) If the function 3 z given by (26) is univalent in U, vq z + ηzq z + ξ 3 z, (42) +1,b q z td λ1,λ2,p q is the best subordinant of (42). a1,b 1 f z +(1 t)d λ a 1,λ2,p 1,b 1 f z V. SANDWICH RESULTS By cobining Theore (1) with Theore (4), we obtain the following swich theore: 43

8 Theore (7): Let q 1 q 2 be two convex functions in U, q 1 0 = q 2 0 = 1 q 2 satisfies (11), > 0, η C with Re η > 0. If f A(p) such that D λ 1,λ2,p a 1,b 1 f z H 1,1 Q, 1 (z) is univalent in U satisfies q 1 z + p + λ 2 n + b zq 1 z 1 z q 2 z + p + λ 2 n + b zq 2 z, (44) where 1 (z) is given by (12), q 1 z q 2 z, where q 1 q 2 are, respectively, the best subordinant the best doinant of (44). By cobining Theore (2) with Theore (5), we obtain the following swich theore: Theore (8): Let q i be two convex functions in U, such that q i 0 = 1, q i z 0 zq i z q i z i = 1,2 is starlike in U, let, η C/{0} u, v, ξ C. Further assue that q 1 satisfies (34), q 2 satisfies (18). Let f A(p) suppose that f satisfies the next conditions: + (1 t) 0 z U, 0 t 1, + (1 t) H 1,1 Q. If the function 2 z given by (19) is univalent in U, u + vq 1 z + ξ q 1 z 2 + η zq 1 z q 1 z 2 z u + vq 2 z + ξ q 2 z 2 + η zq 2 z q 2 z, (45) q 1 z td +1,b + 1 t q 2 z, where q 1 q 2 are, respectively, the best subordinant the best doinant of (45). By cobining Theore (3) with Theore (6), we obtain the following swich theore: Theore (9): Let q 1 q 2 be two convex functions in U, with q 1 0 = q 2 0 = 1, let, η C/{0} v, ξ C with Re v η q 1 z > 0 q 2 satisfies (25). Let f A(p) suppose that f satisfies the next conditions: + (1 t) 0 z U, 0 t 1, + (1 t) H 1,1 Q. If the function 3 z given by (26) is univalent in U, vq 1 z + ηzq 1 z + ξ 3 z vq 2 z + ηzq 2 z + ξ, (46) q 1 z td +1,b + 1 t where q 1 q 2 are, respectively, the best subordinant the best doinant of (46). REFERENCES q 2 z, [1] S. S. Miller P. T. Mocanu, Subordinants of differential superordinations, Coplex Variables, 48(10), 2003, pp: Copyright to IJARSET

9 [2] E. El-Yagubi M. Darus, A study on a class of p-valent functions associated with generalized hypergeoetric functions, VladikavkazskiiMateaticheskiiZhurnal, 17(1), 2015, pp: [3] C. Selvaraj K. R. Karthikeyan, Differential subordination superordination for certain subclasses of analytic functions, Far East Journal of Matheatical Sciences, 29(2), 2008, pp: [4] R. M. El-Ashwah, Majorization properties for subclass of analytic p-valent functions defined by the generalized hypergeoetric function, Tasui Oxford Journal of Matheatical Sciences, 28(4), 2012, pp: [5] J. Dziok H. M. Srivastava, Classes of analytic functions associated with the generalized hypergeoetric function, Applied Matheatics Coputation, 103(1), 1999, pp: [6] J. E. Hohlov, Operators operations on the class of univalent functions, IzvestiyaVysshikhUchebnykhZavedeniiMateatika, 10(197), 1978, pp: [7] G. S. Salagean, Subclasses of univalent functions, Coplex Analysis-Fifth Roanian-Finnish Seinar, Part 1 (Bucharest, 1981), Lecture Notes in Math., vol.1013, Springer, Berlin,1983, 1981, pp: [8] S. S. Miller P. T. Mocanu, Differential subordinations: Theory Applications, Series on Monographs Textbooks in Pure Applied Matheatics Vol. 225, Marcel Dekker Inc., New York Basel, [9] T. Bulboacӑ, Classes of first order differential superordinations, deonstration atheatica, 35(2), 2002, pp: Copyright to IJARSET