J. Korean Math. Soc. 43 (2006), No. 2, pp. 311 322 STRONG DIFFERENTIAL SUBORDINATION AND APPLICATIONS TO UNIVALENCY CONDITIONS José A. Antonino Astract. For the Briot-Bouquet differential equations of the form given in 1 u(z) + we can reduce them to zu (z) z f (z) αu(z) + β f(z) v(z) + F (z) v (z) v(z) = h(z), = g(z), where v(z) = αu(z) + β, h(z) = αg(z) + β and F (z) = f(z)/f (z). In this paper we are going to give conditions in order that if u and v satisfy, respectively, the equations (1) u(z) + F (z) u (z) u(z) = h(z), v(z) + G(z) v (z) v(z) = g(z) with certain conditions on the functions F and G applying the concept of strong suordination g h given in 2 y the author, implies that v u, where indicates suordination. 1. Introduction In 1935, Goluzin 3 considered the simple first order suordination zp (z) h(z). He showed that if h is convex then p(z) q(z), with zq (z) = h(z) (q(0) = 0) and this is the est dominant. Successive generalizations of this result have een done y Roinson 9 in 1947, Suffridge 10 in 1970, Hallemeck and Ruscheweyh 4 in Received Novemer 30, 2004. Revised Novemer 15, 2005. 2000 Mathematics Suject Classification: 34A30, 30C35. Key words and phrases: differential equation, suordination, convex function, starlike function.
312 José A. Antonino 1975, Miller and Mocanu 6 in 1985 and Antonino and Romaguera 2 in 1994. In all these results were compared functions p an q that were verifying the same differential relation. This is, in the Goluzin s result, for example, zp (z) = (z) and zq (z) = h(z) and (z) h(z), ut in oth cases it is verified the relation zf (z). In this paper we are going to see the case in which the functions p and q satisfy different differential equations. 2. Preliminaries Let F and G e analytic in the unit disk U. The function F is suordinate to G, written F G, if G is univalent, F (0) = G(0) and F (U) G(U). Let A n denote the set of functions f(z) = z + a n+1 z n+1 +, n 1 that are analytic in the unit disk U, and let A = A 1. For a C, a complex numer, and n a positive integer, let Ha, n denote the set of functions f(z) = a + a n z n +, n 1 that are analytic in U. Let F e analytic and univalent in U, with F (0) = 0. The class of F -convex functions, denoted y F K are those of f A for which Re F (z) f (z) f (z) + 1 > 0. The class of F -starlike functions, denoted y F S, are those of f A for which Re F (z) f (z) f(z) > 0. The class of closeto-f -convex functions, denoted y F C, are those of f A for which there is a g F S such that Re F (z) f (z) g(z) > 0. If F (z) = z, then we have the convex, starlike and close-to-convex functions, respectively. In this case, these sets are denoted y K, S*, C and S respectively. It is well known that K S C S. We will consider that in (1) F and G are analytic and univalent functions in U (G analytic in Ū) with F (0) = G(0) = 0, h(z) is an analytic convex function in U with h(0) 0, and we will designate for (2) g(z, ξ) v(z) + G(ξ) z v (z) ξ v(z) the analytic function in U Ū and for g(z) = g(z, z), with g (0) 0. We are going to enunciate two lemmas that will e used in this paper. Lemma 1. (5, Lemma 1) Let p H1, n and let q e analytic and univalent in U with p(0) = q(0). If p is not suordinate to q,
Strong differential suordination and applications 313 then there exist points z 0 U and ξ 0 U, and an m n for which p ( z < z 0 ) q(u), a) p(z 0 ) = q(ξ 0 ), and ) z 0 p (z 0 ) = mξ 0 q (ξ 0 ). The next lemma deals with the notion of a suordination chain. A function L(z, t), z U, t 0, is a suordination chain if L(z, t) is an analytic and univalent function of z for all t > 0 and is a continuously differentiale function of t on 0, for all z U, and L(z, s) L(z, t) when 0 s t. Lemma 2. (8, p.159) The function L(z, t) = a 1 (t)z +, with a 1 (t) 0 for all t 0 and lim t a 1 (t) =, is a suordination chain if and only if Re z z > 0 for z U and t 0. t 3. Differential suordination Definition 1. A function L(z, t, ξ, s), z U, ξ Ū, t 0, s > 0, is a set of suordination chains if for each ξ Ū and some s, L(z, t, ξ, s) is a suordination chain. Definition 2. Let H(z, ξ) e analytic in U Ū and let f(z) analytic and univalent in U. The function H(z, ξ) is strongly suordinate to f(z), written H(z, ξ) f(z), for ξ Ū, if the function of z, H(z, ξ), is suordinate to f(z). In the following theorems we will suppose that the functions u and v that satisfy the equations (1) are different from zero in U. Conditions for this to happen can e seen in 1. Theorem 1. Suppose that u(z) is an analytic and univalent solution of the differential equation (3) u(z) + F (z) u (z) u(z) = h(z) and that v(z) is an analytic function that satisfies the equation v(z) + G(z) v (z) v(z) = g(z).
314 José A. Antonino If a) g(z, ξ) h(z) and ) Re s G(z) z and s 1, then v u. F (z) z u (z) h (z)u(z) 0 z U Proof. Without loss of generality we can assume that the conditions of the theorem are satisfied on the closed disk Ū (or Ū Ū). In opposite case, we can replace u(z) y u r (z) = u(rz), v(z) y v r (z) = v(rz), F (z) y F r (z) = F (rz), g(z, ξ) y g r (z, ξ) = g(rz, ξ) and h(z) y h r (z) = h(rz), where 0 < r < 1. These new functions satisfy the conditions of the theorem on Ū (or Ū Ū). We would then prove that p r v r for all 0 < r < 1. By letting r 1, we would otain p(z) v(z). Suppose that a) and ) are satisfied, ut v is not suordinate to u. According to Lemma 1, there are points z 0 U and ξ 0 U, and m 1 such that v(z 0 ) = u(ξ 0 ) and z 0 v (z 0 ) = mξ 0 u (ξ 0 ). Using these results in (2) we otain (4) g(z 0, ξ 0 ) = v(z 0 ) + G(ξ 0) z 0 v (z 0 ) G(ξ = u(ξ ξ 0 v(z 0 ) 0 ) + mξ 0 ) u (ξ 0 ) 0 ξ 0 u(ξ 0 ). From (3), we have u(ξ 0 ) = h(ξ 0 ) F (ξ 0 ) u (ξ 0 ) u(ξ 0 ) and if we use this equation in (4) we have (5) g(z 0, ξ 0 ) = h(ξ 0 ) + m G(ξ 0) F (ξ 0) u (ξ 0 ) ξ0 ξ 0 ξ 0 h h (ξ (ξ 0 )u(ξ 0 ) 0 and ecause Re m G(ξ 0) F (ξ 0) u (ξ 0 ) ξ 0 ξ 0 h (ξ 0 )u(ξ 0 ) 0 and ξ 0 h (ξ 0 ) is an outward normal to the oundary of the convex domain h(u), we deduce that g(z 0, ξ 0 ) of (5) represents a complex outside of h(u). This contradicts g h, and we conclude that v u. Example 1. Let F = (M 2 R R 1)z with R > 2, R 2 < M < Mz 1 Mz 1 R + 1, and let h(z) = R + 1 + M. Since ω = M M z M z mapping U in ω < M, h(z) is a convex function. The differential equation u(z) + F (z) u (z) = h(z) is satisfied y the univalent function u(z) u(z) = MR M z.
Strong differential suordination and applications 315 Let g(z, ξ) e the function 1 + ξ g(z, ξ) = R + 1 + z + z R + 1 + z, where G(ξ) = ξ +ξ 2, is strongly suordinated to h(z) and the differential equation v(z) + (1 + z)z v (z) = g(z) is satisfied y v(z) = R + 1 + z. v(z) Moreover, Re s G(z) F (z) u (z) z z h (z)u(z) = Re s(1 + z) (M 2 R 1) M z M(M 2 > 0, s 1. 1) According to Theorem 1, v u since it is verified directly. If the function h(z) is univalent ut is not convex, then the conditions of the previous theorem can e modified to have an analogous result. Theorem 2. Suppose that u(z) is an analytic and univalent solution of the differential equation (6) u(z) + F (z) u (z) u(z) = h(z) and that v(z) is an analytic function that satisfies the equation If v(z) + G(z) v (z) v(z) = g(z). a) g(z, ξ) h(z), ) P (z) = z u is starlike, and u c) Re s G(ξ) F (ξ) u (z) ξ ξ h > 0, (z, ξ) U Ū and s 1, (z)u(z) then v u. Proof. Let (7) L(z, t, ξ, s) = h(z) + t s G(ξ) F (ξ) P (z) ξ ξ e an analytic function in U for all t 0, every ξ Ū and s 1, and it is continuously differentiale on 0, respect of t, for all z U, ξ Ū
316 José A. Antonino and s 1. Routine calculations allow to otain t 0, for each ξ and s fixed z = h (z) + t s G(ξ) F (ξ) P (z), ξ ξ and we can calculate z z t from ) and c), Re z z t = s G(ξ) F (ξ) P (z), ξ ξ 1 = s G(ξ) F (ξ) ξ ξ z t > 0. Moreover, =a 1 (t) = h (0) + t z=0 = h (0) 1 + t u (z) h (z)u(z) s G(ξ) F (ξ) ξ ξ s G(ξ) F (ξ) ξ ξ + tz P (z) P (z), u (0) u(0)h (0) h (0) 0, u (0) u(0)h (0) since t 0 and from c), Re s G(ξ) F (ξ) u (0) ξ ξ u(0)h (0) > 0, also lim a 1 (t) = for every ξ and s fixed. From Lemma 2, we t conclude that L(z, t, ξ, s) is a suordination chain for ξ and s fixed, that we have designated as set of suordination chains; we oserve that h(z) = L(z, 0, ξ, s). Suppose that v is not suordinate to u. Then we can apply a similar argument to that of Theorem 1 to estalish that g(z 0, ξ 0 ) = h(ξ 0 ) + t m G(ξ 0) F (ξ 0) P (ξ ξ 0 ξ 0 ). 0 From (6), with z = ξ = ξ 0 and s = m, we have that g(z 0, ξ 0 ) = L(ξ 0, t, ξ 0, m) and since L(ξ 0, 0, ξ 0, m) L(z, t, ξ 0, m) and L(ξ 0, 0, ξ 0, m) / h(u), g(z, ξ) is not suordinated to h(z) in opposition to what we have supposed. Hence, v u.
Strong differential suordination and applications 317 Example 2. Let F (z), h(z) and u(z) e as in Example 1. Oviously h(z) is univalent. Let G(z) = a z(r + z) and (R + ξ)z g(z, ξ) = R + z + (a ) R + z, where 0 < < a, 0 < < R and 2 < M < R + 1. Let the differential equation v(z) + G(z) v (z) v(z) = g(z), e satisfied y v(z) = R + z. We are going to see that the conditions of Theorem 2 are satisfied. 1) g(z, ξ) h(z). It is sufficient to see that R + ξ 1 + z + (a )z R + z < M. But R + ξ 1 + z + (a )z + R + z 1 + + (a )R and fixing = R R/10 and taking (a ) sufficiently small for that (a ) R + R < 1 10, then 1+R/10+1/10 < M < R + 1, this is, 11+R < 10M < 10 R + 1 which is compatile for R < 9. 2) P (z) is starlike. As z P (z) P (z) = 1 + z is sufficient to see that z M z M z 1 M 1 < 1. But since M > 2 the inequality is satisfied. 3) Re s G(ξ) F (ξ) u (z) ξ ξ h (z)u(z) > 0. For that Re s a (R + ξ) + R + 1 M 2 M z M(M 2 > 0 is sufficient to see that Re s a 1) (R + ξ) + R + 1 M 2 (M z) > 0. In this expression let s put ξ = re iϕ and z = ρe iθ, 0 < r < 1, 0 < ρ < 1 and
318 José A. Antonino ϕ, θ 0, 2π. Re s a (R + re iϕ ) + R + 1 M 2 (M ρe iθ ) = s a (R + r cos ϕ) + R + 1 M 2 (M ρ cos θ) + s a ρr sin ϕ sin θ s a (R ) + R + 1 M 2 (M 1) s(a ) + (R + 1 M 2 )(M 1) = s a RM R M + (R + 1 M 2 )(M 1). As a > 0, R + 1 M 2 > 0 and M 1 > 0, in order that the latter expression is > 0, is sufficient that M(R ) R > 0 = M > R R. If = R/10, for example, then 10/9 < M < R + 1 and it would e enough to take 3 < R < 9. The conditions of Theorem 2 can e estalished and, consequently v u. Corollary 2.1. Let u(z) + F (z) u (z) = h(z) and suppose that the u(z) conditions a) and ) of Theorem 1 or the conditions a), ) and c) of Theorem 2 are verified, with (8) g(z, ξ) = G (z) + G(ξ) z f (z) ξ f (z), where f(z) is an analytic function in U such that G(z) f (z) 0 in U. f(z) Then (9) G(z) f (z) f(z) u(z). Proof. The equation v(z) + G(z) v (z) v(z) = g(z), with g(z) = G (z) + G(z) f (z) f (z) has the solution v(z) = G(z)f (z) and using Theorem 1 or f(z) Theorem 2, as correspond, we otain the conclusion.
Strong differential suordination and applications 319 If Re h(z) > 0 under certain conditions Re u(z) > 0 (see 1) and from (7) with ξ = z, f(z) is G-convex and from (8) G-starlike. This is, in this context, a G-convex function is G-starlike. This concept has een introduced and used y the author in other papers. If in the equations (1) we replace F y γf and G y γg (γ > 0) the conclusions of Theorem 1 and Theorem 2 are not changed. In the following theorems we will apply this new version. Theorem 3. Let u(z) + γf (z) u (z) = h(z) and suppose that the u(z) conditions a) and ) of Theorem 1 or the conditions a), ) and c) of Theorem 2 are verified, with g(z, ξ) = G(ξ) z H (z) ξ H(z), where G(z) A n, γ > 0 and H(z) A n. If f(z) is defined as γ z H 1/γ (t) (10) f(z) = γg(t) dt which is different from zero in U {0}, then f(z) A n and (11) G(z) f (z) f(z) u(z). 0 Proof. From condition a) of the Theorem 1 or Theorem 2, we are assuming that H(z) G(z) 0 in U. If in the equation (z) γg(z)v v(z) + v(z) = G(z) H (z) H(z) we set g(z) = (z) G(z)H H(z) The function γg(z) v (z) v(z) v(z) = z 0 then g(z) H1, n and we otain + v(z) = g(z). H 1/γ H 1/γ (t) γg(t) dt is a nonzero analytic solution of this equation and satisfies the conditions of Theorem 1 or Theorem 2. Consequently, v u. From (9) and from the latter expression, we have that f(z) = H(z)/v γ (z) and f A n. By
320 José A. Antonino logarithmically differentiating G(z) f (z) f(z) = (z) G(z)H H(z) (z) γg(z)v = v(z) u(z). v(z) This completes the proof. A simple computation using (9) leads to ( ) (1 γ)g(z) f (z) f(z) + γ zg(ξ) G(z) f (z) ξg(z) f (z) + G (z) = g(z, ξ) and designating y J(γG, f; G) the term on the left, we have that J(γ z G(ξ), f(z); G(z)) ξ = ( G(z) γ z ) f ξ G(ξ) (z) f(z) + γ zg(ξ) ξ ( ) f (z) f (z) + G (z) G(z) and, more generally, J (γ zξ ) H(ξ), f(z), G(z) = γ z ( f ) ( ξ H(ξ) (z) f (z) + G (z) + G(z) γ z ) f G(z) ξ H(ξ) (z) f(z). Theorem 4 can e written now in the following form: If the condition ) of Theorem 1 is satisfied (or the conditions ) and c) of Theorem 2), then J(γ z ξ G(ξ), f(z); G(z) = g(z, ξ) h = G(z)f (z) f(z) u(z). In 5 the authors have investigated a form of this result. Our following theorem contains this result. Theorem 4. Let s suppose that the conditions ) of Theorem 1 or ) and c) of Theorem 2 are verified for the functions u(z), h(z), G(z). Let g(z, ξ) e defined of the following form: for all f A n such that G(z)f (z)/f(z) is analytic and nonzero function in U g(z, ξ) = J(γ z G(ξ), f(z); G(z)) ξ and if (12) K(z) = G(z) exp z 0 u(t) G (t) dt, G(t)
Strong differential suordination and applications 321 then (13) J(γ z G(ξ), f(z); G(z)) J(γF, K; G) = ξ (14) G(z) f (z) f(z) G(z)K (z) K(z) = u(z). Proof. From (12), we otain u(z) = G(z)K (z)/k(z) and sustituting in h(z) = u(z) + γf (z)u (z)/u(z) we have that ( h(z) = (G(z) γf (z)) K (z) K ) (z) + γf (z) K(z) K (z) + G (z) G(z) = J(γF, K; G). Since u(z) is univalent, also is univalent G(z)K (z)/k(z), and the suordination (14) are well defined. If we set g(z, z) = g(z) = J(γG, f; G), g(z) H1, n. Using this g(z) in Theorem 1 or Theorem 2, the equation γg(z) v + v = J(γG, f; G), has a nonzero analytic solution v(z) = v G(z)f (z)/f(z) that satisfies G(z)f (z)/f(z) G(z)k (z)/k(z). Example 3. Let u(z) + F (z) u (z) u(z) = h(z), v(z) + G(z) v (z) v(z) = g(z), with F (z) = az (a 1, 0), h(z) = (1 + az) 2 + a 2 z /(1+az), G(z) = z + az 2. The first one is satisfied y u = 1/(1 + az). If a < 1/26, then h(z) is a convex function and the conditions a) and ) of Theorem 1 are satisfied. If g(z) = 1 + n z n, with n a 2, then (z) h(z) and we can apply Theorem 1. We are going to give an application of the previous results to see how conditions of univalency can e otained. Example 4. Let h(z) = 1 + mz, u(z) = e z, F (z) = 1 + mz e z (m 1) and G(z) = z. If a = min z U Re ez 1 z > 0, then for 0 < m < 1 + a the conditions of Theorem 1 are verified with s 1. If f A and 1 + z f (z) f (z) h(z),
322 José A. Antonino then y Corollary 1, zf (z)/f(z) u(z). This means that z f (z) f (z) < m = Re z f (z) f(z) > 0. That is, f is starlike and therefore univalent. References 1 J. A. Antonino and S. Romaguera, Analytic and univalent solutions of Briot- Bouquet differential equations, Math. Japon. 36 (1991), no. 3, 447 449. 2, Strong differential suordination to Briot-Bouquet differential equations, J. Differential Equations 114 (1994), no 1, 101 105. 3 G. M. Goluzin, On the majorization principle in function theory, Doklady Akad. Nauk SSSR 42 (1935), 647 650. 4 D. J. Halleneck and S. T. Ruscheyh, Suordination y convex functions, Proc. Amer. Math. Soc. 52 (1975), 191 195. 5 S. S. Miller and P. T. Mocanu, Differential suordinations and univalent functions, Michigan Math. J. 28 (1981), no. 2, 157 172. 6, Univalent Solutions of Briot-Bouquet Differential Equations, J. Differential Equations 56 (1985), no. 3, 297 309. 7, A special differential suordinations its applications to univalency conditions, Currente Topics in Analytic Function Theory, World Scientific Pul. Co., Singapore, 1992, 171 185. 8 Ch. Pommeremke, Univalent Functions, Vanderhoeck and Ruprecht, Göttingen, 1975. 9 R. M. Roinson, Univalent majorants, Trans. Amer. Math. Soc. 61 (1947), 1 35. 10 T. J. Suffridge, Some remarks on convex maps of the unit disc, Duke Math. J. 37 (1970), 775 777. Departamento de Matemática Aplicada ETSICCP, Universidad Politécnica 46071 Valencia, Spain E-mail: jantonin@mat.upv.es