Abstract. For the Briot-Bouquet differential equations of the form given in [1] zu (z) u(z) = h(z),

Transcription

1 J. Korean Math. Soc. 43 (2006), No. 2, pp 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, Revised Novemer 15, Mathematics Suject Classification: 34A30, 30C35. Key words and phrases: differential equation, suordination, convex function, starlike function.

2 312 José A. Antonino 1975, Miller and Mocanu 6 in 1985 and Antonino and Romaguera 2 in 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,

3 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).

4 314 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 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.

5 Strong differential suordination and applications 315 Let g(z, ξ) e the function 1 + ξ g(z, ξ) = R z + z R 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 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

6 316 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.

7 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 (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

8 318 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.

9 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

10 320 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)

11 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),

12 322 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, , Strong differential suordination to Briot-Bouquet differential equations, J. Differential Equations 114 (1994), no 1, G. M. Goluzin, On the majorization principle in function theory, Doklady Akad. Nauk SSSR 42 (1935), D. J. Halleneck and S. T. Ruscheyh, Suordination y convex functions, Proc. Amer. Math. Soc. 52 (1975), S. S. Miller and P. T. Mocanu, Differential suordinations and univalent functions, Michigan Math. J. 28 (1981), no. 2, , Univalent Solutions of Briot-Bouquet Differential Equations, J. Differential Equations 56 (1985), no. 3, , A special differential suordinations its applications to univalency conditions, Currente Topics in Analytic Function Theory, World Scientific Pul. Co., Singapore, 1992, Ch. Pommeremke, Univalent Functions, Vanderhoeck and Ruprecht, Göttingen, R. M. Roinson, Univalent majorants, Trans. Amer. Math. Soc. 61 (1947), T. J. Suffridge, Some remarks on convex maps of the unit disc, Duke Math. J. 37 (1970), Departamento de Matemática Aplicada ETSICCP, Universidad Politécnica Valencia, Spain jantonin@mat.upv.es