Bulletin of Mathematical Analysis and Applications ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 2, Issue 3(2010), Pages 15-26. SUBORDINATION AND SUPERORDINATION FOR FUNCTIONS BASED ON DZIOK-SRIVASTAVA LINEAR OPERATOR RABHA W. IBRAHIM AND MASLINA DARUS Abstract. In this article, we obtain some subordination and superordination results involving Dziok-Srivastava linear operator and fractional integral operator for certain normalized analytic functions in the open unit disk. 1. Introduction and Preliminaries. Let β(U) denote the class of analytic functions in the unit disk π := {π§ ∈ β, β£π§β£ < 1}. For π positive integer and π ∈ β, let β[π, π] := {π ∈ π»(π ) : π (π§) = π + ππ π§ π + ππ+1 π§ π+1 + ..., π§ ∈ π }, and ππ = {π ∈ π»(π ) : π (π§) = π§ + ππ π§ π + ππ+1 π§ π+1 + ..., π§ ∈ π } with π1 = π. A function π ∈ β[π, π] is convex in π if it is univalent and π (π ) is convex. It is well-known that π is convex if and only if π (0) β= 0 and ℜ{1 + π§π ′′ (π§) } > 0, π§ ∈ π. π ′ (π§) Deο¬nition 1.1. [1] Denote by Q the set of all functions π (π§) that are analytic and injective on π − πΈ(π ) where πΈ(π ) := {π ∈ ∂π : ππππ§→π π (π§) = ∞} ′ and are such that π (π) β= 0 for π ∈ ∂π − πΈ(π ). Given two functions πΉ and πΊ in the unit disk π, the function πΉ is π π’ππππππππ‘π to πΊ, written πΉ βΊ πΊ, if πΊ is univalent, πΉ (0) = πΊ(0) and πΉ (π ) ⊂ πΊ(π ). Alternatively, given two functions πΉ and πΊ, which are analytic in π, the function πΉ is said to be subordinate to πΊ in π if there exists a function β, analytic in π with β(0) = 0 πππ β£β(π§)β£ < 1 π ππ πππ π§ ∈ π such that πΉ (π§) = πΊ(β(π§)) π ππ πππ π§ ∈ π. 2000 Mathematics Subject Classiο¬cation. 34G10, 26A33,30C45. Key words and phrases. Fractional calculus; Univalent solution; Subordination; Superordination. c β2010 Universiteti i PrishtineΜs, PrishtineΜ, KosoveΜ. Submitted May 6, 2010. Published June 9, 2010. The authors were partially supported by UKM-ST-06-FRGS0107-2009, MOHE Malaysia. 15 16 R. W. IBRAHIM, M. DARUS Let π : β2 → β and let β be univalent in π. If π is analytic in π and satisο¬es the diο¬erential subordination π(π(π§)), π§π′ (π§)) βΊ β(π§), then π is called a solution of the diο¬erential subordination. The univalent function π is called a dominant of the solutions of the diο¬erential subordination, if π βΊ π. If π and π(π(π§)), π§π′ (π§)) are univalent in π and satisfy the diο¬erential superordination β(π§) βΊ π(π(π§)), π§π′ (π§)), then π is called a solution of the diο¬erential superordination. An analytic function π is called subordinant of the solution of the diο¬erential superordination if π βΊ π. We shall need the following results: Lemma 1.1. [2] Let π be univalent in the unit disk π , and let π and π be analytic in a domain π· containing π(π ) with π(π€) β= 0 when π€ ∈ π(π ). Set π(π§) := π§π ′ (π§)π(π(π§)), β(π§) := π(π(π§)) + π(π§). Suppose that 1. π(π§) is starlike univalent in π , and ′ (π§) 2. ℜ π§β π(π§) > 0 for π§ ∈ π. If π(π(π§)) + π§π′ (π§)π(π(π§)) βΊ π(π(π§)) + π§π ′ (π§)π(π(π§)), then π(π§) βΊ π(π§) and π is the best dominant. Lemma 1.2. [3] Let π be convex univalent in the unit disk π and π and πΎ ∈ β with ′′ (π§) + ππΎ } > 0. If π(π§) is analytic in π and ππ(π§)+πΎπ§π′ (π§) βΊ ππ(π§)+πΎπ§π ′ (π§), ℜ{1+ π§ππ′ (π§) then π(π§) βΊ π(π§) and π is the best dominant. Lemma 1.3. [4] Let π be convex univalent in the unit disk π and π and π be analytic in a domain π· containing π(π ). Suppose that 1. π§π ′ (π§)π(π(π§)) is starlike univalent in π, and ′ (π(π§)) } > 0 for π§ ∈ π. 2. ℜ{ ππ(π(π§)) If π(π§) ∈ β[π(0), 1] ∩ Q, with π(π ) ⊆ π· and π(π(π§)) + π§π′ (π§)π(π§) is univalent in π and π(π(π§)) + π§π ′ (π§)π(π(π§)) βΊ π(π(π§)) + π§π′ (π§)π(π(π§)) then π(π§) βΊ π(π§) and π is the best subordinant. Lemma 1.4. [1] Let π be convex univalent in the unit disk π and πΎ ∈ β. Further, assume that ℜ{πΎ} > 0. If π(π§) ∈ β[π(0), 1] ∩ Q, with π(π§) + πΎπ§π′ (π§) is univalent in π then π(π§) + πΎπ§π ′ (π§) βΊ π(π§) + πΎπ§π′ (π§) implies π(π§) βΊ π(π§) and π is the best subordinant. ∑∞ ∑∞ For two functions π (π§) = π§ + π=2 ππ π§ π and π(π§) = π§ + π=2 ππ π§ π , the Hadamard product (or convolution) of π and π deο¬ned by ∞ ∑ (π ∗ π)(π§) := π§ + ππ ππ π§ π =: (π ∗ π )(π§). π=2 For πΌπ ∈ β (π = 1, 2, ..., π) and π½π ∈ ββ{0, −1, −2, ...} (π = 1, 2, ..., π), the generalized hypergeometric function π πΉπ (πΌ1 , ..., πΌπ ; π½1 , ...π½π ; π§) is deο¬ned by the inο¬nite series ∞ ∑ (πΌ1 )π ...(πΌπ )π π§ π π πΉπ (πΌ1 , ..., πΌπ ; π½1 , ..., π½π ; π§) := (π½1 )π ...(π½π )π π! π=0 (π ≤ π + 1 : π, π ∈ β0 := {0, 1, 2, ...}) where (π)π is the Pochhammer symbol deο¬ned by { Γ(π + π) 1, (π = 0); (π)π := = π(π + 1)(π + 2)...(π + π − 1), (π ∈ β). Γ(π) SUBORDINATION AND SUPERORDINATION 17 Corresponding to the function β(πΌ1 , ..., πΌπ ; π½1 , ...π½π ; π§) := π§π πΉπ (πΌ1 , ..., πΌπ ; π½1 , ...π½π ; π§), π the Dziok-Srivastava operator (see [5-7]) π»π (πΌ1 , ..., πΌπ ; π½1 , ...π½π ) is deο¬ned by the Hadamard product π π»π (πΌ1 , ..., πΌπ ; π½1 , ...π½π )π (π§) := β(πΌ1 , ..., πΌπ ; π½1 , ...π½π ; π§) ∗ π (π§) =π§+ ∞ ∑ (πΌ1 )π−1 ...(πΌπ )π−1 ππ π§ π (π½1 )π−1 ...(π½π )π−1 (π − 1)! π=2 π := π»π [πΌ1 ]π (π§). We can verify that π π π π§(π»π [πΌ1 ]π (π§))′ = πΌ1 π»π [πΌ1 + 1]π (π§) − (πΌ1 − 1)π»π [πΌ1 ]π (π§). Special cases of the Dziok-Srivastava linear operator include the Hohlov linear operator [8], the Carlson-Shaο¬er linear operator πΏ(π, π) [9], the Ruscheweyh derivative operator ππ [10], the generalized Bernardi-Libera-Livingston linear integral operator [11] and the Srivastava-Owa fractional derivative operator [12]: Deο¬nition 1.2. The fractional derivative of order πΌ is deο¬ned, for a function π by ∫ π§ π π (π) 1 π·π§πΌ π (π§) := ππ; 0 ≤ πΌ < 1, Γ(1 − πΌ) ππ§ 0 (π§ − π)πΌ where the function π is analytic in simply-connected region of the complex z-plane β containing the origin and the multiplicity of (π§ − π)−πΌ is removed by requiring πππ(π§ − π) to be real when(π§ − π) > 0. Deο¬nition 1.3. The fractional integral of order πΌ is deο¬ned, for a function π, by ∫ π§ 1 πΌπ§πΌ π (π§) := π (π)(π§ − π)πΌ−1 ππ; πΌ > 0, Γ(πΌ) 0 where the function π is analytic in simply-connected region of the complex π§-plane (β) containing the origin and the multiplicity of (π§ − π)πΌ−1 is removed by requiring πππ(π§ − π) to be real when(π§ − π) > 0. Remark 1.1. [12] π·π§πΌ {π§ π } = Γ(π + 1) {π§ π−πΌ }, π > −1; 0 ≤ πΌ < 1 Γ(π − πΌ + 1) and Γ(π + 1) {π§ π+πΌ }, π > −1; πΌ > 0. Γ(π + πΌ + 1) The main object of the present paper is to ο¬nd the suο¬cient conditions for certain normalized analytic functions π, π to satisfy πΌπ§πΌ {π§ π } = [ π πΌπ§πΌ π»π [πΌ1 ]π1 (π§) π πΌ πΌ π» π [πΌ1 ]π (π§) π πΌ πΌ π» π [πΌ1 ]π2 (π§) π ] βΊ[ π§ π ] βΊ[ π§ π ] ππΌ (π§) ππΌ (π§) ππΌ (π§) and π1 (π§) βΊ [ π πΌπ§πΌ π»π [πΌ1 ]π (π§) π ] βΊ π2 (π§), ππΌ (π§) β= 0, π§ ∈ π ππΌ (π§) 18 R. W. IBRAHIM, M. DARUS where π ≥ 1, π1 and π2 are given univalent functions in π. Also, we obtain the results as special cases. Further, in this paper, we study the existence of univalent solution for the fractional diο¬erential equation π π·π§πΌ ππΌ (π§)π’(π§) = π»π [πΌ1 ]π (π§), (1.1) subject to the initial condition π’(0) = 0, where π’ : π → β is an analytic function for all π§ ∈ π, π : π → ββ{0} is an analytic functions in π§ ∈ π and π : π → β is a univalent function in π. The existence is obtained by applying Schauder ο¬xed point theorem. Moreover, we discuss some properties of this solution involving fractional diο¬erential subordination. The following results are used in the sequel. Theorem 1.1. (Arzela-Ascoli) (see [13]) Let πΈ be a compact metric space and π(πΈ) be the Banach space of real or complex valued continuous functions normed by β₯π β₯ := π π’ππ‘∈πΈ β£π (π‘)β£. If π΄ = {ππ } is a sequence in π(πΈ) such that ππ is uniformly bounded and equicontinuous, then π΄ is compact. Let π be a subset of Banach space π and π΄ : π → π an operator. The operator π΄ is called πππππππ‘ on the set π if it carries every bounded subset of π into a compact set. If π΄ is continuous on π (that is, it maps bounded sets into bounded sets ) then it is said to be πππππππ‘πππ¦ ππππ‘πππ’ππ’π on π. Theorem 1.2. (Schauder) (see [14]) Let π be a Banach space, π ⊂ π a nonempty closed bounded convex subset and π : π → π is compact. Then π has a ο¬xed point. Recently, the subordination and superordination containing the Dziok-Srivastava linear operator are studied by many authors [15]. 2. Subordination and superordination. In this section, we study some important properties of the fractional diο¬erential and integral operators π·π§πΌ , πΌπ§πΌ , given by the authors [16] which are useful in the next results of the subordination and superordination. Theorem 2.1[16] For πΌ, ∈ (0, 1] and π is a continuous function, then (π§)πΌ−1 π π (0) + πΌπ§πΌ π·π (π§); π· = . Γ(πΌ) ππ§ 2 − πΌπ§πΌ π·π§πΌ π (π§) = π·π§πΌ πΌπ§πΌ π (π§) = π (π§). 1 − π·πΌπ§πΌ π (π§) = But, ο¬rst we consider the subordination results involving Dziok-Srivastava linear operator and fractional integral operator as the following: Theorem 2.2. Let π, π be analytic in π. [ π πΌπ§πΌ π»π [πΌ1 ]π(π§) ππΌ (π§) that subordination [ β= 0 and π πΌπ§πΌ π»π [πΌ1 ]π(π§) π ] ππΌ (π§) πΌ πΌ π» π [πΌ1 ]π(π§) π ′ π§([ π§ πππΌ (π§) ] ) be univalent in π such be starlike univalent in π. If the π πΌπ§πΌ π»π π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ π§π′ (π§) [πΌ1 ]π (π§) π ] {1 + π( π§πΌ ππ − )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) SUBORDINATION AND SUPERORDINATION βΊ[ π πΌπ§πΌ π»π [πΌ1 ]π(π§) π π§πΌ πΌ [π» π [πΌ1 ]π(π§)]′ π§π′ (π§) ] {1 + π( π§πΌ ππ − } ππΌ (π§) πΌπ§ π»π [πΌ1 ]π(π§) π(π§) holds and ℜ{ π§πΊ′ (π§) π§πΊ(π§)ππΌ (π§) + (π − 1) πΌ π } > 0, π§ ∈ π, πΊ(π§) πΌπ§ π»π [πΌ1 ]π(π§) where πΊ(π§) := [ Then [ and [ 19 π πΌπ§πΌ π»π [πΌ1 ]π π ] ππΌ π πΌπ§πΌ π»π [πΌ1 ]π(π§) ′ ]. ππΌ (π§) π πΌπ§πΌ π»π [πΌ1 ]π (π§) π πΌ πΌ π» π [πΌ1 ]π(π§) π ] βΊ[ π§ π ] ππΌ (π§) ππΌ (π§) is the best dominant. Proof. Setting π(π§) := [ π [πΌ1 ]π (π§) π πΌ πΌ π» π [πΌ1 ]π(π§) π πΌπ§πΌ π»π ] , π(π§) := [ π§ π ] . ππΌ (π§) ππΌ (π§) Our aim is to apply Lemma 1.1. First we show that ℜ{1 + ℜ{1 + π§π ′′ (π§) π ′ (π§) } > 0. π§πΊ′ (π§) πΊ(π§)ππΌ (π§) π§π ′′ (π§) } = ℜ{1 + + (π − 1) πΌ π } > 0. ′ π (π§) πΊ(π§) πΌπ§ π»π [πΌ1 ]π(π§) Assume that π(π) := π πππ π(π) := 1, it can easily be observed that π, π are analytic in β. Also, we let π(π§) := π§π ′ (π§)π(π§) = π§π ′ (π§), β(π§) := π(π(π§)) + π(π§) = π(π§) + π§π ′ (π§). By the assumptions of the theorem we ο¬nd that π is starlike univalent in π and that π§β′ (π§) π§π ′′ (π§) ℜ{ } = ℜ{2 + ′ } > 0. π(π§) π (π§) By using Theorem 2.1, a computation shows π π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ π§π′ (π§) πΌπ§πΌ π»π [πΌ1 ]π (π§) π ] {1 + π( π§πΌ ππ − )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) πΌ πΌ π» π [πΌ1 ]π(π§) π π§πΌ πΌ [π» π [πΌ1 ]π(π§)]′ π§π′ (π§) βΊ[ π§ π ] {1 + π( π§πΌ ππ )} )− ππΌ (π§) πΌπ§ π»π [πΌ1 ]π(π§) π(π§) = π(π§) + π§π ′ (π§). π(π§) + π§π′ (π§) = [ Thus in view of Lemma 1.1, π(π§) βΊ π(π§) and π is the best dominant. Corollary 2.1. Let π, π be analytic in π. [ πΌ πΌ πΏ(π,π)π π§([ π§ ππΌ ]π )′ πΌπ§πΌ πΏ(π,π)π π ] ππΌ be univalent in π and be starlike univalent in π. If the subordination [ π§πΌ πΌ [πΏ(π, π)π (π§)]′ π§π′ (π§) πΌπ§πΌ πΏ(π, π)π (π§) π ] {1 + π( π§πΌ − )} ππΌ (π§) πΌπ§ πΏ(π, π)π (π§) π(π§) βΊ[ πΌπ§πΌ πΏ(π, π)π(π§) π π§πΌ πΌ [πΏ(π, π)π(π§)]′ π§π′ (π§) ] {1 + π( π§πΌ − )} ππΌ (π§) πΌπ§ πΏ(π, π)π(π§) π(π§) 20 R. W. IBRAHIM, M. DARUS holds and ℜ{ π§πΊ′ (π§) π§πΊ(π§)ππΌ (π§) + (π − 1) πΌ } > 0, π§ ∈ π, πΊ(π§) πΌπ§ πΏ(π, π)π(π§) where πΊ(π§) := [ Then [ πΌπ§πΌ πΏ(π,π)π π ] ππΌ and [ πΌπ§πΌ πΏ(π, π)π(π§) ′ ]. ππΌ (π§) πΌπ§πΌ πΏ(π, π)π (π§) π πΌ πΌ πΏ(π, π)π(π§) π ] βΊ[ π§ ] ππΌ (π§) ππΌ (π§) is the best dominant. Proof. By putting π = 2, π = 1, πΌ1 = π, πΌ2 = 1 and π½1 = π in Theorem 2.2. Corollary 2.2. Let π, π be analytic in π, [ π πΌπ§πΌ π»π [πΌ1 ]π(π§) ππΌ (π§) [ β= 0 and πΌπ§πΌ π (π§) π ππΌ (π§) ] {1 + πΌπΌπ π§([ ππ§πΌ ]π )′ π§πΌ πΌ [π (π§)]′ π( π§πΌ πΌπ§ π (π§) − πΌπ§πΌ π π ππΌ ] be univalent in π such that be starlike univalent in π. If the subordination π§π′ (π§) π§π′ (π§) πΌ πΌ π(π§) π π§πΌ πΌ [π(π§)]′ )} βΊ [ π§ ] {1 + π( π§πΌ − )} π(π§) ππΌ (π§) πΌπ§ π(π§) π(π§) holds and π§πΊ′ (π§) π§πΊ(π§)ππΌ (π§) πΌ πΌ π(π§) ′ ℜ{ + (π − 1) } > 0, π§ ∈ π, π€βπππ πΊ(π§) := [ π§ ]. πΌ πΊ(π§) πΌπ§ π(π§) ππΌ (π§) Then [ and [ πΌπ§πΌ π π ππΌ ] πΌ πΌ π(π§) π πΌπ§πΌ π (π§) π ] βΊ[ π§ ] ππΌ (π§) ππΌ (π§) is the best dominant. Proof. By putting π = 1, π = 0, πΌ1 = 1, in Theorem 2.2. Theorem 2.3. Let π, π be analytic in π, π be convex univalent in π with ℜ{1 + π§π ′′ (π§) π ′ (π§) [ + πΎ1 }, πΎ ∈ β and [ π πΌπ§πΌ π»π [πΌ1 ]π π ] ππΌ be analytic in π. If the subordination π πΌπ§πΌ π»π [πΌ1 ]π (π§) π π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ π§π′ (π§) ] {1 + ππΎ( π§πΌ ππ − )} βΊ π(π§) + πΎπ§π ′ (π§) ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) holds. Then [ π πΌπ§πΌ π»π [πΌ1 ]π (π§) π ] βΊ π(π§) ππΌ (π§) and π is the best dominant. Proof. Setting π πΌπ§πΌ π»π [πΌ1 ]π (π§) π ] . ππΌ (π§) Our aim is to applied Lemma 1.2. Let π := 1, since π(π§) := [ π πΌπ§πΌ π»π [πΌ1 ]π (π§) π πΌ πΌ π» π [πΌ1 ]π (π§) π ′ ] + πΎπ§([ π§ π ] ) ππΌ (π§) ππΌ (π§) πΌ πΌ π» π [πΌ1 ]π (π§) π π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ π§π′ (π§) =[ π§ π ] {1 + ππΎ( π§πΌ ππ − )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) βΊ π(π§) + πΎπ§π ′ (π§) π(π§) + πΎπ§π′ (π§) = [ SUBORDINATION AND SUPERORDINATION 21 then, in view of Lemma 1.2, π(π§) βΊ π(π§) and π is the best dominant. 1+π΄π§ π Corollary 2.3. Let π, π be analytic in π, −1 ≤ π΅ ≤ π΄ ≤ 1, π(π§) := [ 1+π΅π§ ] with ℜ{1 + [ π§π ′′ (π§) π ′ (π§) + πΎ1 }, πΎ ∈ β and [ π πΌπ§πΌ π»π [πΌ1 ]π π ] ππΌ be analytic in π. If the subordination π πΌπ§πΌ π»π [πΌ1 ]π (π§) π π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ π§π′ (π§) 1 + π΄π§ π ππΎπ§(π΄ − π΅) ] {1+π( π§πΌ ππ − )} βΊ [ ] {1+ } ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) 1 + π΅π§ (1 + π΄π§)(1 + π΅π§) holds. Then [ π πΌπ§πΌ π»π [πΌ1 ]π (π§) π 1 + π΄π§ π ] βΊ[ ] , −1 ≤ π΅ < π΄ ≤ 1 ππΌ (π§) 1 + π΅π§ 1+π΄π§ π ] is the best dominant. and [ 1+π΅π§ Next, applying Lemma 1.3 and Lemma 1.4 respectively, to obtain the following theorems. Theorem 2.4. π such that π πΌπ§πΌ π»π [πΌ1 ]π(π§) ππΌ (π§) πΌ πΌ π» π [πΌ1 ]π π ′ (π§[ π§ π ] ) ππΌ [ β= 0, convex univalent in univalent in π and be univalent in π. If the subordination π [πΌ1 ]π(π§) π π§πΌ πΌ [π» π [πΌ1 ]π(π§)]′ π§π′ (π§) πΌπ§πΌ π»π ] {1 + π( π§πΌ ππ − )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π(π§) π(π§) βΊ[ π [πΌ1 ]π (π§) π πΌπ§πΌ π»π π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ π§π′ (π§) ] {1 + π( π§πΌ ππ − )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) πΌ−1 holds and [ π§Γ(πΌ) ]π [ π π»π [πΌ1 ]π (π§) π ] ππΌ (π§) [ and [ π πΌπ§πΌ π»π [πΌ1 ]π π ] be ππΌ π πΌπ§πΌ π»π [πΌ1 ]π π ′ ] ) be starlike π§([ ππΌ Let π, π be analytic in π,[ π πΌπ§πΌ π»π [πΌ1 ]π π ] ππΌ ∈ β[0, 1] ∩ Q . Then π [πΌ1 ]π(π§) π πΌπ§πΌ π»π πΌ πΌ π» π [πΌ1 ]π (π§) π ] βΊ[ π§ π ] ππΌ (π§) ππΌ (π§) is the best subordinant. Proof. Setting π(π§) := [ π πΌπ§πΌ π»π [πΌ1 ]π (π§) π πΌ πΌ π» π [πΌ1 ]π(π§) π ] , π(π§) := [ π§ π ] . ππΌ (π§) ππΌ (π§) Our aim is to apply Lemma 1.3. By taking π(π) := π πππ π(π) := 1, it can easily observed that π, π are analytic in β. Thus ℜ{ π′ (π(π§)) } = 1 > 0. π(π(π§)) Now we must show that π(π§) + π§π ′ (π§) βΊ π(π§) + π§π′ (π§). 22 R. W. IBRAHIM, M. DARUS a computation shows that π πΌπ§πΌ π»π [πΌ1 ]π(π§) π π§πΌ πΌ [π» π [πΌ1 ]π(π§)]′ π§π′ (π§) ] {1 + π( π§πΌ ππ − )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π(π§) π(π§) πΌ πΌ π» π [πΌ1 ]π (π§) π π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ π§π′ (π§) − βΊ[ π§ π ] {1 + π( π§πΌ ππ )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) = π(π§) + π§π′ (π§). π(π§) + π§π ′ (π§) = [ Thus in view of Lemma 1.3, π(π§) βΊ π(π§) and π is the best subordinant. Theorem 2.5. Let π, π be analytic in π, π be convex univalent in π, [ β[0, 1] ∩ Q and [ π πΌπ§πΌ π»π [πΌ1 ]π (π§) π ] ππΌ (π§) ∈ π π§π′ (π§) πΌπ§πΌ π»π [πΌ1 ]π (π§) π π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ ] {1 + ππΎ( π§πΌ ππ − )}, ℜ{πΎ} > 0, ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) be univalent in π. If the subordination πΌ πΌ π» π [πΌ1 ]π (π§) π π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ π§π′ (π§) π(π§) + πΎπ§π ′ (π§) βΊ [ π§ π ] {1 + ππΎ( π§πΌ ππ − )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) holds. Then π(π§) βΊ [ π πΌπ§πΌ π»π [πΌ1 ]π (π§) π ] ππΌ (π§) and π is the best subordinant. Proof. Setting π [πΌ1 ]π (π§) π πΌπ§πΌ π»π ] . ππΌ (π§) Our aim is to apply Lemma 1.4. Since π(π§) := [ π πΌπ§πΌ π»π [πΌ1 ]π(π§) π π§π′ (π§) π§πΌ πΌ [π» π [πΌ1 ]π(π§)]′ − ] {1 + ππΎ( π§πΌ ππ )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π(π§) π(π§) π§π′ (π§) πΌ πΌ π» π [πΌ1 ]π (π§) π π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ βΊ[ π§ π ] {1 + ππΎ( π§πΌ ππ − )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) = π(π§) + πΎπ§π′ (π§) π(π§) + πΎπ§π ′ (π§) = [ then, in view of Lemma 1.4, π(π§) βΊ π(π§) and π is the best subordinant. Combining the results of diο¬erential subordination and superordination, we state the following sandwich theorems. π πΌπ§πΌ π»π [πΌ1 ]π1 π ] be convex univalent in π ππΌ πΌ π πΌ π πΌ π πΌπ§ π»π [πΌ1 ]π(π§) πΌπ§ π»π [πΌ1 ]π1 π ′ πΌπ§ π» π [πΌ1 ]π2 π ′ that β= 0, π§([ ] ) , π§([ ] ) be starlike univalent ππΌ (π§) ππΌ ππΌ πΌ π πΌ π πΌ π» [πΌ1 ]π π ′ πΌπ§ π»π [πΌ1 ]π2 π and let (π§[ π§ π ] ),[ ] be univalent in π. If the subordination ππΌ ππΌ Theorem 2.6. Let π, π1 , π2 be analytic in π, [ such in π [ π πΌπ§πΌ π»π [πΌ1 ]π1 (π§) π π§πΌ πΌ [π» π [πΌ1 ]π1 (π§)]′ π§π′ (π§) ] {1 + π( π§πΌ ππ − )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π1 (π§) π(π§) βΊ[ π πΌπ§πΌ π»π [πΌ1 ]π (π§) π π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ π§π′ (π§) ] {1 + π( π§πΌ ππ − )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) SUBORDINATION AND SUPERORDINATION βΊ[ πΌ−1 holds, [ π§Γ(πΌ) ]π [ π πΌπ§πΌ π»π [πΌ1 ]π2 (π§) π π§πΌ πΌ [π» π [πΌ1 ]π2 (π§)]′ π§π′ (π§) ] {1 + π( π§πΌ ππ − )} ππΌ (π§) πΌπ§ π»π [πΌ1 ]π2 (π§) π(π§) π π»π [πΌ1 ]π (π§) π ] ππΌ (π§) ℜ{ ∈ β[0, 1] ∩ Q and π§πΊ′2 (π§) π§πΊ2 (π§)ππΌ (π§) + (π − 1) πΌ π } > 0, π§ ∈ π, πΊ2 (π§) πΌπ§ π»π [πΌ1 ]π2 (π§) where πΊ2 (π§) := [ Then [ 23 π πΌπ§πΌ π»π [πΌ1 ]π2 (π§) ′ ]. ππΌ (π§) π πΌπ§πΌ π»π [πΌ1 ]π1 (π§) π πΌ πΌ π» π [πΌ1 ]π (π§) π πΌ πΌ π» π [πΌ1 ]π2 (π§) π ] βΊ[ π§ π ] βΊ[ π§ π ] ππΌ (π§) ππΌ (π§) ππΌ (π§) π πΌ πΌ π»π [πΌ1 ]π1 π ] ππΌ and [ π§ inant. and [ π πΌπ§πΌ π»π [πΌ1 ]π2 π ] ππΌ are respectively the best subordinant and dom- Theorem 2.7. Let π, π1 , π2 ∈ π, π1 , π2 be convex univalent in π, with ℜ{1 + π§π2′′ (π§) π2′ (π§) + πΎ1 }, πΎ ∈ β, [ [ π πΌπ§πΌ π»π [πΌ1 ]π π ] ππΌ ∈ β[0, 1] ∩ Q, and analytic in π and π [πΌ1 ]π π πΌπ§πΌ π»π π§π′ π§πΌ πΌ [π» π [πΌ1 ]π ]′ − )}, ℜ{πΎ} > 0, ] {1 + π( π§πΌ ππ ππΌ πΌπ§ π»π [πΌ1 ]π π be univalent in π. If the subordination π§πΌ πΌ [π» π [πΌ1 ]π (π§)]′ π§π′ (π§) πΌ πΌ π» π [πΌ1 ]π (π§) π ] {1+ππΎ( π§πΌ ππ − )} βΊ π2 (π§)+πΎπ§π2′ (π§) π1 (π§)+πΎπ§π1′ (π§) βΊ [ π§ π ππΌ (π§) πΌπ§ π»π [πΌ1 ]π (π§) π(π§) holds. Then π [πΌ1 ]π (π§) π πΌπ§πΌ π»π ] βΊ π2 (π§) ππΌ (π§) and π1 , π2 are respectively the best subordinant and the best dominant. π1 (π§) βΊ [ 3. Existence of univalent solution. Let β¬ := π[π, β] be a Banach space of all continuous functions on π endowed with the sup. norm β₯π’β₯ := π π’ππ§∈π β£π’(π§)β£. By using the properties in Theorem 2.1, we can easily obtain the following result: Lemma 3.1. If the function π ∈ π, then the initial value problem (1.1) is equivalent to the nonlinear integral equation ∫ π§ 1 (π§ − π)πΌ−1 π π’(π§) = π»π [πΌ1 ]π (π)ππ. (3.1) ππΌ (π§) 0 Γ(πΌ) In other words, every solution of the equation (3.1) is also a solution of the initial value problem (1.1) and vice versa. Theorem 3.1.(Existence) Assume that β£ππΌ1(π§)β£ ≤ π ; π > 0. Then there exists a univalent function π’ : π → β solving the problem (1.1). Proof. Deο¬ne an operator π : β → β ∫ π§ 1 (π§ − π)πΌ−1 π π»π [πΌ1 ]π (π)ππ. (3.2) (π π’)(π§) := ππΌ (π§) 0 Γ(πΌ) 24 R. W. IBRAHIM, M. DARUS (πΌ1 )π−1 ...(πΌπ )π−1 1 Denotes π΅π := (π½ . Our aim is to apply Theorem 2.1. First we 1 )π−1 ...(π½π )π−1 (π−1)! show that π is bounded operator: ∫ π§ (π§ − π)πΌ−1 π 1 β£(π π’)(π§)β£ = β£ π»π [πΌ1 ]π (π)ππβ£ ππΌ (π§) 0 Γ(πΌ) ∫ π§ (π§ − π)πΌ−1 π 1 β£β£ π»π [πΌ1 ]π (π)ππβ£ ≤β£ ππΌ (π§) 0 Γ(πΌ) ∫ π§ ∞ ∑ (π§ − π)πΌ−1 < π (1 + π΅π β£ππ β£)β£ ππβ£ Γ(πΌ) 0 π=2 = π (1 + ∞ ∑ π΅π β£ππ β£) π=2 β£π§ πΌ β£ Γ(πΌ + 1) ∑∞ π (1 + π=2 π΅π β£ππ β£) < Γ(πΌ + 1) Thus we obtain that β₯π β₯ < ∑∞ π (1 + π=2 π΅π β£ππ β£) := π Γ(πΌ + 1) that is π : π΅π → π΅π . Then π maps π΅π into itself. Now we proceed to prove that π is equicontinuous. For π§1 , π§2 ∈ π such that π§1 β= π§2 , β£π§2 − π§1 β£ < πΏ, πΏ > 0 Then for all π’ ∈ π, where ∑∞ π (1 + π=2 π΅π β£ππ β£) := π, π > 0}, π := {π’ ∈ β, : β£π’β£ ≤ Γ(πΌ + 1) we obtain β£(π π’)(π§1 ) − (π π’)(π§2 )β£ ∫π§ ∑∞ ≤ π (1 + π=2 π΅π β£ππ β£)β£ 0 1 ∫π§ ∑∞ ≤ π (1 + π=2 π΅π β£ππ β£)β£ 0 1 = < < ∑∞ (π§1 −π)πΌ−1 ππ Γ(πΌ) ∫ π§2 0 (π§2 −π)πΌ−1 ππβ£ Γ(πΌ) [(π§1 −π)πΌ−1 −(π§2 −π)πΌ−1 ] ππ Γ(πΌ) π=2 π΅π β£ππ β£) β£[2(π§2 − π§1 )πΌ Γ(πΌ+1) ∑ 2π (1+ ∞ π=2 π΅π β£ππ β£) β£π§2 − π§1 β£πΌ Γ(πΌ+1) ∑∞ 2π (1+ π=2 π΅π β£ππ β£) πΌ πΏ , Γ(πΌ+1) π (1+ − + ∫ π§2 π§1 (π§2 −π)πΌ−1 ππβ£ Γ(πΌ) + π§2πΌ − π§1πΌ ]β£ which is independent on π’. Hence π is an equicontinuous mapping on π. By the assumption of the theorem we can show that π is a univalent function (see [17]). The Arzela-Ascoli theorem yields that every sequence of functions from π (π) has got a uniformly convergent subsequence, and therefore π (π) is relatively compact. Schauder, s ο¬xed point theorem asserts that π has a ο¬xed point. By construction, a ο¬xed point of π is a univalent solution of the initial value problem (1.1). The next theorems show the relation between univalent solutions and the subordination for a class of fractional diο¬erential problem. Theorem 3.2. Let the assumptions of Theorem 2.6 be satisο¬ed. Then univalent solutions π’1 , π’, π’2 , of the problem π·π§πΌ π’(π§) = πΉ (π§, π’(π§)), (3.3) SUBORDINATION AND SUPERORDINATION 25 subject to the initial condition π’(0) = 0, where π’ : π → β is an analytic function for all π§ ∈ π and πΉ : π × β → β, is an analytic functions in π§ ∈ π, are satisfying the subordination π’1 βΊ π’ βΊ π’2 . Proof. Setting π = 1 and let πΉ (π§, π’1 (π§)) := and πΉ (π§, π’2 (π§)) := π π»π [πΌ1 ]π2 (π§) ππΌ (π§) π π»π [πΌ1 ]π1 (π§) , ππΌ (π§) πΉ (π§, π’(π§)) := π π»π [πΌ1 ]π (π§) , ππΌ (π§) where ππΌ (π§) β= 0, ∀ π§ ∈ π. Theorem 3.3. Let the assumptions of Theorem 2.7 be satisο¬ed. Then every univalent solution π’(π§) of the problem (3.3) satisο¬es the subordination π1 (π§) βΊ π’(π§) βΊ π2 (π§), where π1 (π§) and π2 (π§) are univalent function in π. Proof. Setting π = 1, πΉ (π§, π’(π§)) := π π»π [πΌ1 ]π (π§) . ππΌ (π§) Acknowledgement: The authors would like to thank the anonymous referee for the informative and creative comments given to the article. References [1] S.S.Miller and P.T.Mocanu, Subordinants of diο¬erential superordinations, Complex Variables, 48(10)(2003), 815-826. [2] S.S.Miller and P.T.Mocanu, Diο¬erential Subordinantions: Theory and Applications. Pure and Applied Mathematics No.225 Dekker, New York, (2000). [3] T.N.Shanmugam and V.Ravichangran and S.Sivasubramanian, Diο¬erential sandwich theorems for some subclasses of analytic functions, Austral. J. Math, Anal.Appl.3(1)(2006),1-11. [4] T.Bulboaca, Classes of ο¬rst-order diο¬erential superordinations, Demonstr.Math. 35(2)(2002),287-292. 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Ibrahim and Maslina Darus School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia Bangi 43600 Selangor D.E., Malaysia E-mail address: saibahmath@yahoo.com E-mail address: maslina@ukm.my (corresponding author)