General Mathematics Vol. 11, No. 1–2 (2003), 47–52
A new differential inequality I
Georgia Irina Oros
and
Adriana Cătaş
Dedicated to Professor dr. Gheorghe Micula on his 60th birthday
Abstract
We find conditions on the complex-valued functions A and B, in
the unit disc U such that the differential inequality
3
Re [A(z)p2 (z)+B(z)p(z)+α(zp0 (z)) −β(zp0 (z))2 +γ(zp0 (z))+δ] > 0
implies Re p(z) > 0, where α ≥ 0, β ≥ 0, γ ∈ R,
3
βn2 γn
δ < αn
+
8
4 + 2 and p ∈ H[1, n].
2000 Mathematical Subject Classification: 30C80
Keywords: differential subordination, dominant.
47
48
Georgia Irina Oros and Adriana Cătaş
1
Introduction and preliminaries
We let H[U ] denote the class of holomorphic functions in the unit
disc
U = {z ∈ C : | z |< 1}.
For a ∈ C and n ∈ N∗ we let
H[a, n] = {f ∈ H[U ],
f (z) = a + an z n + an+1 z n+1 + ..., z ∈ U }
and
An = {f ∈ H[U ],
f (z) = z + an+1 z n+1 + an+2 z n+2 + ..., z ∈ U }
with A1 = A.
In order to prove the new results we shall use the following lemma, which
is a particular form of Theorem 2.3.i[1,p.35].
Lemma A. [1,p.35]. Let ψ : C2 × U → C a function which satisfies
Re ψ(ρi, σ; z) ≤ 0,
where ρ, σ ∈ R,
2
σ ≤ −n
2 (1 + ρ ) z ∈ U
and n ≥ 1.
If p ∈ H[1, n] and
Re ψ (p(z), zp0 (z); z) > 0
then
Re p(z) > 0.
A new differential inequality I
2
49
Main results
3
βn2 γn
Theorem 1.Let α ≥ 0, β ≥ 0, γ ∈ R, δ < αn
+
8
4 + 2 and let
n be a positive integer. Suppose that the functions A, B : U → C satisfy
3
βn2 γn
(i ) Re A(z) > − 3αn
8 − 2 − 2
(1)
µ
2
(ii ) (ImB(z)) ≤ 4
µ
·
¶
3αn3 + βn2 + γn + Re A(z) ·
8
2
2
αn3 βn4 γn
+
+
−δ
8
4
2
¶
If p ∈ H[1, n] and
3
Re [A(z)p2 (z) + B(z)p(z) + α(zp0 (z)) −
−β(zp0 (z))2 + γ(zp0 (z)) + δ] > 0
(2)
then
Re p(z) > 0.
Proof. We let ψ : C2 × U → C be defined by
3
ψ(p(z), zp0 (z); z) = A(z)p2 (z) + B(z)p(z) + α(zp0 (z)) −
−β(zp0 (z))2 + γ(zp0 (z)) + δ] > 0
(3)
From (2) we have
50
(4)
Georgia Irina Oros and Adriana Cătaş
Re ψ(p(z), zp0 (z); z) > 0 for z ∈ U.
n2 (1 + ρ2 )2 ,
2
2
For ρ, σ ∈ R satisfying σ ≤ − n
(1
+
ρ
),
hence
−σ
≤
−
2
4
3
n
3
2 3
σ ≤ − 8 (1 + ρ ) and z ∈ U , by using (1) we obtain
Re ψ(ρi, σ; z) = Re [A(z)(ρi)2 + B(z)ρi+
+ασ 3 − βσ 2 + γσ + δ] = −ρ2 Re A(z) − ρ Im B(z)+
+ασ 3 − βσ 2 + γσ + δ ≤
αn3
(1 + ρ2 )3 −
8
γn
αn3 6
βn2
(1 + ρ2 )2 −
(1 + ρ2 ) + δ = −
ρ−
−
4
2
8
µ
¶
·µ
¶
3αn3 βn2
3αn3 βn2 γn
4
−
+
ρ −
+
+
+ Re A(z) ρ2 +
8
4
8
2
2
¸
αn3 βn2 γn
+ρ Im B(z) +
+
+
− δ ≤ 0.
8
4
2
≤ −ρ2 Re A(z) − ρ Im B(z) −
By using Lemma A we have that Re p(z) > 0.
If α = 0, then Theorem 1 can be rewritten as follows:
βn2 γn
Corollary 1. Let β ≥ 0, γ ∈ R, δ < 4 + 2 and let n be a positive
integer. Suppose that the functions A, B : U → C satisfy
βn2 γn
(i) Re A(z) > − 2 − 2
(5)
µ
2
(ii ) (ImB(z)) ≤ 4
¶µ
¶
βn2 γn
βn4 γn
2 + 2 + Re A(z)
4 + 2 −δ
A new differential inequality I
51
If p ∈ H[1, n] and
Re [A(z)p2 (z) + B(z)p(z) − β(zp0 (z))2 + γ(zp0 (z)) + δ] > 0
(6)
then
Re p(z) > 0.
If β = 0, then Theorem 1 can be rewritten as follows:
Corollary 2.
3
γn
Let α ≥ 0, γ ∈ R, δ < αn
8 + 2 and let n be a positive integer. Suppose
that the functions A, B : U → C satisfy
3
γn
(i ) Re A(z) > − 3αn
8 − 2
(7)
³
´³ 3
´
3
γn
αn + γn − δ
(ii ) (ImB(z))2 ≤ 4 3αn
+
+
Re
A(z)
8
2
8
2
If p ∈ H[1, n] and
3
Re [A(z)p2 (z) + B(z)p(z) + α(zp0 (z)) + γ(zp0 (z)) + δ] > 0
(8)
then
Re p(z) > 0.
52
Georgia Irina Oros and Adriana Cătaş
References
[1] S.S. Miller and P.T. Mocanu, Differential Subordinations. Theory and
Applications, Marcel Dekker Inc., New York, Basel, 2000.
University of Oradea
Department of Mathematics
Str. Armatei Române, nr. 5
3700 - Oradea, România.