2014 Volume 22 No.2 GENERAL MATHEMATICS Daniel Florin

Transcription

2014
Volume 22
No.2
GENERAL MATHEMATICS
EDITOR-IN-CHIEF
Daniel Florin SOFONEA
ASSOCIATE EDITOR
Ana Maria ACU
HONORARY EDITOR
Dumitru ACU
EDITORIAL BOARD
Heinrich Begehr
Shigeyoshi Owa
Piergiulio Corsini
Detlef H. Mache
Aldo Peretti
Andrei Duma
Vijay Gupta
Dorin Andrica
Claudiu Kifor
Heiner Gonska
Dumitru Gaşpar
Malvina Baica
Vasile Berinde
Adrian Petruşel
SCIENTIFIC SECRETARY
Emil C. POPA
Nicuşor MINCULETE
Ioan ŢINCU
EDITORIAL OFFICE
DEPARTMENT OF MATHEMATICS AND INFORMATICS
GENERAL MATHEMATICS
Str.Dr. Ion Ratiu, no. 5-7 550012 - Sibiu, ROMANIA
Electronical version: http://depmath.ulbsibiu.ro/genmath/
Contents
R. Diaconu, On integral operators of analytic functions . . . . . . . . . . . . . 3
M. Birou, Some blending surfaces generated by univariate
interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
C. Selvaraj, C. Santhosh Moni, On the univalent condition for a
family of integral operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
V. Najjari, Centralizers of maximal tori in the classical group
SU (n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
A. Alb Lupaş, A note on differential superordinations using a
generalized Sălăgean and Ruscheweyh operators . . . . . . . . . . . . . . . . . . . . 43
M. K. Aouf, A. Shamandy, A. O. Mostafa and E. A. Adwan,
Subordination results for certain classes of analytic functions defined by
Dziok-Srivastava operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
M. Thongmoon, A Numerical solutions of differential transform method
for linear and nonlinear ordinary differential equations . . . . . . . . . . . . . . 69
B. Srivastava, On a generalization of Ramanujan’s seventh order mock
theta functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
R. M. El-Ashwah, M. K. Aouf, On certain class of meromorphic
multivalent functions with positive coefficients . . . . . . . . . . . . . . . . . . . . . . 91
R. Bucur, L. Andrei, D. Breaz, S. L. Georgescu, Fekete–Szegö
inequality for certain classes of analytic functions . . . . . . . . . . . . . . . . . . 107
General Mathematics Vol. 22, No. 2 (2014), 3–18
On integral operators of analytic functions 1
Radu Diaconu
Abstract
m,n
In the present paper we define two integral operators Fλ,k,γ
and
1 ,...,λl
m,n
defined using the differential operator IRλ,k . We introduce some
classes defined by these operators and we investigate properties of the
integral operators on these classes. Also, are obtained subordination results
m,n
for functions f ∈ T associated with the differential operator IRλ,k
.
Gm,n
λ,k,γ1 ,...,λl ,
2010 Mathematics Subject Classification: 30C45.
Key words and phrases: Analytic functions, Hadamard product, differential
operator, integral operator, starlike and convex functions, differential
subordination.
1
Introduction
Denote by U the unit disc of the complex plane, U = {z ∈ C : |z| < 1} and H(U )
the space of holomorphic functions in U .
Let An = {f ∈ H(U ) : f (z) = z + an+1 z n+1 + . . . , z ∈ U } with A1 = A. Denote
∞
∑
by T the subclass of A consisting the functions f of the form f (z) = z −
|aj | z j ,
j=2
z ∈ U.
For the functions f, g ∈ T , f (z) = z −
∞
∑
|aj | z and g (z) = z −
j
j=2
z ∈ U , the Hadamard product or convolution of f and g is defined by
(f ∗ g) (z) = z −
∞
∑
|aj | · |bj | z j , z ∈ U.
j=2
1
Received 15 February, 2014
Accepted for publication (in revised form) 12 June, 2014
3
∞
∑
j=2
|bj | z j ,
4
R. Diaconu
Definition 1 For f ∈ A, m ∈ N∪ {0}, λ, k ≥ 0, the operator I (m, λ, k) f (z) is
defined by the following infinite series
)
∞ (
∑
λ (j − 1) + k + 1 m
I (m, λ, k) f (z) = z +
aj z j .
k+1
j=2
Remark 1 It follows from the above definition that
I (0, λ, l) f (z) = f (z),
(k + 1) I (m + 1, λ, k) f (z) = (k + 1 − λ) I (m, λ, k) f (z)+λz (I (m, λ, k) f (z))′ , z ∈ U.
∑
j
Remark 2 If f ∈ T and f (z) = z − ∞
j=2 aj z , then
)
∞ (
∑
λ (j − 1) + k + 1 m
I (m, λ, k) f (z) = z −
aj z j .
k+1
j=2
Definition 2 (Ruscheweyh [9]) For f ∈ A and n ∈ N, the operator Rn is defined
by Rn : A → A,
R0 f (z) = f (z)
R1 f (z) = zf ′ (z)
...
f (z) = z (Rn f (z))′ + nRn f (z) , z ∈ U.
∑∞ (n+j−1)!
∑
j
n
j
Remark 3 If f ∈ A, f (z) = z + ∞
j=2 n!(j−1)! aj z
j=2 aj z , then R f (z) = z +
z ∈ U.
∑∞ (n+j−1)!
∑
j
n
j
If f ∈ T , f (z) = z − ∞
j=2 n!(j−1)! aj z z ∈ U .
j=2 aj z , then R f (z) = z −
(n + 1) R
n+1
Following [1] we can define a new differential operator
m,n
Definition 3 Let n, m ∈ N. Denote by IRλ,k
: A → A the operator given by the
Hadamard product of the generalized Sălăgean operator Dλm and the Ruscheweyh
operator Rn ,
(1)
m,n
IRλ,k
f (z) = (I (m, λ, k) ∗ Rn ) f (z) ,
for any z ∈ U and each nonnegative integers m, n.
∑
j
Remark 4 If f ∈ A and f (z) = z + ∞
j=2 aj z , then
)
∞ (
∑
λ (j − 1) + k + 1 m (n + j − 1)! 2 j
a z , z ∈ U.
SRm,n f (z) = z +
k+1
n! (j − 1)! j
j=2
If f ∈ T and f (z) = z −
SRm,n f (z) = z −
∑∞
j=2 aj z
j,
then
)
∞ (
∑
λ (j − 1) + k + 1 m (n + j − 1)! 2 j
a z , z ∈ U.
k+1
n! (j − 1)! j
j=2
On integral operators of analytic functions
5
Using simple computation one obtains the next result.
Proposition 1 For m, n ∈ N we have
(
)′
k + 1 − λ m,n
λ
m,n
IRλ,k f (z) +
z IRλ,k
f (z) .
k+1
k+1
m+1,n
IRλ,k
f (z) =
(2)
Proof. We have
m+1,n
IRλ,k
f
)
∞ (
∑
λ (j − 1) + k + 1 m+1 (n + j − 1)! 2 j
(z) = z +
a z
k+1
n! (j − 1)! j
j=2
)(
)
∞ (
∑
λ (j − 1) + k + 1 m (n + j − 1)! 2 j
λ (j − 1) + k + 1
=z+
a z
k+1
k+1
n! (j − 1)! j
j=2


)m
∞ (
∑
λ (j − 1) + k + 1
(n + j − 1)! 2 j 
k+1−λ
z+
a z
=
k+1
k+1
n! (j − 1)! j
j=2


)m
∞ (
∑
λ (j − 1) + k + 1
(n + j − 1)! 2 j 
λ 
z+
ja z
+
k+1
k+1
n! (j − 1)! j
j=2
(
)′
k + 1 − λ m,n
λ
m,n
=
IRλ,k f (z) +
z IRλ,k
f (z) .
k+1
k+1
Breaz and Breaz [3] and Breaz, Owa and Breaz [4] introduced and investigated
the following integral operators
)
(
)
∫ z(
f1 (t) γ1
fl (t) γl
Fγ1 ,...,γl (z) =
...
dt,
t
t
0
∫ z
( ′ )γ1 ( ′ )γl
f1 (t) ... fl (t) dt,
Gγ1 ,...,γl (z) =
0
where fi ∈ A, γi ∈ R, γi > 0, i ∈ {1, 2, ..., l}.
For functions fi ∈ T , γi ∈ T, i ∈ {1, 2, ..., l}, we define the integral operators
m,n
Fλ,k,γ
and Gm,n
λ,kγ1 ,...,γl as follows
1 ,...,γl
)γ1 ( m,n
)γl
∫ z ( m,n
IR
f
(t)
IR
f
(t)
1
l
λ,k
λ,k
m,n
(3)
Fλ,γ
(z) =
...
dt,
1 ,...,γl
t
t
0
∫
(4)
Gm,n
λ,kγ1 ,...,γl
(z) =
0
z
((
)′ )γ1
m,n
IRλ,k
f1 (t)
((
)′ )γl
m,n
... IRλ,k fl (t)
dt,
for z ∈ U , m, n ∈ N, λ ≥ 0.
m,n
m,n
By using the differential operator IRλ,k
f and the integral operators Fλ,k,γ
1 ,...,γl
and Gm,n
,
following
[2]
and
[5],
we
introduce
some
subclasses
of
analytic
λ,k,γ1 ,...,γl
functions f ∈ T .
6
R. Diaconu
Definition 4 A function f ∈ T is said to be in the class R (δ) if it satisfies the
inequality
 (
)′ 
m,n
z
IR
f
(z)
λ,k


Re 
 < δ, z ∈ U , δ > 1.
m,n
IRλ,k f (z)
Definition 5 A function f ∈ T is said to be in the class D (δ) if it satisfies the
inequality

(
)′′ 
m,n
z
IR
f
(z)
λ,k


Re 1 + (
)′  < δ, z ∈ U , δ > 1.
m,n
IRλ,k f (z)
Definition 6 A function f ∈ T is said to be in the class RA (β, µ) , for 0 ≤ β < 1
and 0 < µ ≤ 1, if it satisfies the inequality
(
)′
(
)′
m,n
m,n
z IRλ,k
z IRλ,k
f
(z)
f
(z)
−
1
−
1
<
µ
β
·
, z ∈ U.
m,n
m,n
IRλ,k
f (z)
IRλ,k
f (z)
Definition 7 A function f ∈ T is said to be in the class DA (β, µ) , for 0 ≤ β < 1
and 0 < µ ≤ 1, if it satisfies the inequality
(

)′′ (
)′′ 
m,n
m,n
z IRλ,k
f
(z)
z
IR
f
(z)
λ,k


)′ < µ β 1 + (
)′  + 1 , z ∈ U .
(
m,n
m,n
IRλ,k f (z)
IRλ,k f (z) Definition 8 A family of functions fi , i ∈ {1, 2, ..., l} is said to be in the class
LAF (λ, k, β, µ, γ1 , ..., γl ) if it satisfies the inequality
(

)′′ 
(
)′′ m,n
m,n
z
F
(z)
(z)
z
F
λ,k,γ1 ,...,γl
λ,k,γ1 ,...,γl


Re 1 + (
)′  ≥ β (
)′ + µ,
m,n
m,n
Fλ,k,γ1 ,...,γl (z)
Fλ,k,γ1 ,...,γl (z) m,n
for some β ≥ 0 and −1 ≤ µ ≤ 1, where Fλ,k,γ
is defined in (3).
1 ,...,γl
Definition 9 A family of functions fi , i ∈ {1, 2, ..., l} is said to be in the class
LAG (λ, k, β, µ, γ1 , ..., γl ) if it satisfies the inequality
(

(
)′′ 
)′′ m,n
m,n
z Gλ,k,γ ,...,γ (z) z Gλ,k,γ1 ,...,γl (z) 
1
l

Re 1 + (
)′  ≥ β (
)′ + µ,
m,n
Gm,n
Gλ,k,γ1 ,...,γl (z) λ,k,γ1 ,...,γl (z)
for some β ≥ 0 and −1 ≤ µ ≤ 1, where Gm,n
λ,k,γ1 ,...,γl is defined in (4).
7
∑
i j
Lemma 1 For fi (z) = z − ∞
j=2 aj z ∈ T , i ∈ {1, 2, ..., l}, we have


)′′
(
∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1
m,n
l
(z)
z Fλ,k,γ
−
a
z
∑
j
j=2
k+1
n!(j−2)!
1 ,...,γl


γi 
(
)′ =
,
(
)m
( )2
∑
m,n
λ(j−1)+k+1
(n+j−1)!
∞
i
j−1
Fλ,k,γ
(z)
i=1
z
1
−
a
j=2
j
k+1
n!(j−1)!
1 ,...,γl
m,n
is the integral operator given by (3).
where Fλ,k,γ
1 ,...,γl
∑
i j
Proof. Let fi (z) = z − ∞
j=2 aj z , for i ∈ {1, 2, ..., l}, then
)
∞ (
(
)′
∑
λ (j − 1) + k + 1 m (n + j − 1)! ( i )2 j−1
m,n
IRλ,k fi (z) = 1 −
j aj z ,
k+1
n! (j − 1)!
z ∈ U.
j=2
We obtain
(
so
m,n
Fλ,k,γ
1 ,...,γl
(
)′
(z) =
(
m,n
IRλ,k
f1 (z)
)γ1
z
(
...
m,n
IRλ,k
fl (z)
)γl
z
,
)′
)′′
(
m,n
m,n
(z)
(z)
=
E
F
Fλ,k,γ
1
λ,k,γ
,...,γ
,...,γ
1
1
l
l
z
+ ...
m,n
IRλ,k
f1 (z)
(
)′
z
m,n
+ El Fλ,k,γ
(z)
,
m,n
1 ,...,γl
IRλ,k fl (z)
where
Ei = γi
(
)′
m,n
m,n
z IRλ,k
fi (z) − IRλ,k
fi (z)
Next we calculate the expression
z2
, i ∈ {1, 2, ..., l}.
)′′
m,n
(z)
z Fλ,k,γ
1 ,...,γl
)′ .
(
m,n
Fλ,k,γ
,...,γ (z)
(
1
l
 (

)′′
)′
m,n
m,n
l
(z)
z Fλ,k,γ
z
IR
f
(z)
∑
λ,k i
1 ,...,γl


γi 
− 1 .
(
)′ =
m,n
m,n
IR
f
(z)
i
λ,k
Fλ,k,γ1 ,...,γl (z)
i=1
(
We find


(
)′′
∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j
m,n
l
z Fλ,k,γ
(z)
j
a
z
z
−
∑ 
j
j=2
k+1
n!(j−1)!
1 ,...,γl

γi 
− 1
)′ =
(
)m
( )2
(
∑
m,n
(n+j−1)!
λ(j−1)+k+1
∞
i
Fλ,k,γ1 ,...,γl (z)
i=1
z − j=2
zj
k+1
n!(j−1)! aj


∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j
l
z 
∑  − j=2
k+1
n!(j−2)! aj
=
γi 

(
)
(
)
m
2
∑
λ(j−1)+k+1
(n+j−1)!
i
j
i=1
z− ∞
a
z
j=2
j
k+1
n!(j−1)!


∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1
l
−
a
z
∑ 
j
j=2
k+1
n!(j−2)!

=
γi 
.
(
(
)2
)
m
∑∞ λ(j−1)+k+1
(n+j−1)!
i
j−1
i=1
1 − j=2
z
k+1
n!(j−1)! aj
8
R. Diaconu
∑
i j
Lemma 2 For fi (z) = z − ∞
j=2 aj z ∈ T , i ∈ {1, 2, ..., l}, we have


)′′
(
∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1
l
(z)
z Gm,n
j
a
z
∑
j
j=2
λ,k,γ1 ,...,γl
k+1
n!(j−2)!


γi 
(
)′ = −
,
(
)m
( )2
∑
λ(j−1)+k+1
(n+j−1)!
∞
i
j−1
Gm,n
(z)
i=1
1
−
j
a
z
λ,k,γ1 ,...,γl
j=2
j
k+1
n!(j−1)!
where Gm,n
λ,k,γ1 ,...,γl is the integral operator given by (4).
∑
i j
Proof. Let fi (z) = z − ∞
j=2 aj z , for i ∈ {1, 2, ..., l}. It follows that
(
)′ ((
)′ )γ1 ((
)′ )γl
m,n
m,n
m,n
... IRλ,k fl (z)
,
Gλ,k,γ1 ,...,γl (z) =
IRλ,k f1 (z)
so
(
)′′
(
m,n
l
)′′ ∑
(
)′ IRλ,k
fi (z)
Gm,n
=
γi Gm,n
(
)′ .
λ,k,γ1 ,...,γl (z)
λ,k,γ1 ,...,γl (z)
m,n
IRλ,k fi (z)
i=1
Next, we calculate the expression
(
)′′
z Gm,n
λ,k,γ1 ,...,γl (z)
)′ .
(
Gm,n
λ,k,γ ,...,γ (z)
1
l
 (
)′′
)′′ 
(
m,n
l
(z)
z
IR
f
(z)
z Gm,n
∑
λ,k i
λ,k,γ1 ,...,γl


γi  (
)′ =
(
)′ 
m,n
Gm,n
IRλ,k
fi (z)
i=1
λ,k,γ1 ,...,γl (z)


∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1
l
z
∑  − j=2
k+1
n!(j−2)! j aj

=
γi 
.
)
(
)
(
m
2
∑∞ λ(j−1)+k+1
(n+j−1)!
i
j−1
i=1
z
1 − j=2
k+1
n!(j−1)! j aj
Hence


)′′
(
∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1
m,n
l
z Gλ,k,γ1 ,...,γl (z)
z
∑ 
j=2
k+1
n!(j−2)! j aj

γi 
(
)′ = −
.
)
(
)
(
m
2
∑
(n+j−1)!
λ(j−1)+k+1
∞
i
j−1
Gm,n
(z)
i=1
1
−
j
a
z
λ,k,γ1 ,...,γl
j=2
j
k+1
n!(j−1)!
If f and g are analytic functions in U , we say that f is subordinate to g, written
f ≺ g, if there is a function w analytic in U , with w(0) = 0, |w(z)| < 1, for all z ∈ U
such that f (z) = g(w(z)) for all z ∈ U . If g is univalent, then f ≺ g if and only if
f (0) = g(0) and f (U ) ⊆ g(U ).
Let ψ : C3 × U → C and h an univalent function in U . If p is analytic in U and
satisfies the (second-order) differential subordination
(5)
ψ(p(z), zp′ (z), z 2 p′′ (z); z) ≺ h(z),
for z ∈ U,
then p is called a solution of the differential subordination. The univalent function q
is called a dominant of the solutions of the differential subordination, or more simply
a dominant, if p ≺ q for all p satisfying (5).
A dominant qe that satisfies qe ≺ q for all dominants q of (5) is said to be the best
dominant of (5). The best dominant is unique up to a rotation of U .
9
Lemma 3 [6] Let q (z) be univalent in U and let ϕ (z) be analytic in a domain
zq ′ (z)
containing q (U ). If ϕ(q(z))
is starlike, then
zψ ′ (z) ϕ (ψ (z)) ≺ zq ′ (z) ϕ (q (z)) , z ∈ U,
then ψ (z) ≺ q (z) and q (z) is the best dominant.
Lemma 4 [8] if p (z) and q (z) are analytic in U , q (z) is convex univalent, α, β
and γ are complex and γ ̸= 0. Further assume that
[
(
)]
α 2β
zq ′′ (z)
Re
+
q (z) + 1 +
> 0.
γ
γ
q (z)
If p (z) = 1 + cz + ... is analytic in U and satisfies
αp (z) + βp2 (z) + γzp′ (z) ≺ αq (z) + βq 2 (z) + γzq ′ (z) ,
then p (z) ≺ q (z) and q (z) is the best dominant.
Theorem 1 [7] Let q (z) be convex univalent and 0 < β ≤ 1,
(
)]
[
zq ′′ (z)
1−β
+ 2q (z) + 1 +
> 0.
Re
β
q (z)
If f ∈ A satisfies
zf ′ (z)
f ′′ (z)
+ βz 2 ′
≺ (1 − β) q (z) + βq 2 (z) + βzq ′ (z) ,
f (z)
f (z)
then
zf ′ (z)
f (z)
≺ q (z) and q (z) is the best dominant.
Theorem 2 [7] Let q (z) be analytic in U , q (0) = 1 and h (z) =
univalent in U . If f ∈ A satisfies
zq ′ (z)
q(z)
be starlike
(zf (z))′′
zf ′ (z)
−2
≺ h (z) ,
′
f (z)
f (z)
then
2
z 2 f ′ (z)
f 2 (z)
≺ q (z), and q (z) is the best dominant.
The classes LAF (λ, k, β, µ, γ1 , ..., γl ) and
LAG (λ, k, β, µ, γ1 , ..., γl )
In this section, we obtain sufficient conditions for a family of functions fi to belongs
to the classes LAF (λ, k, β, µ, γ1 , ..., γl ) and LAG (λ, k, β, µ, γ1 , ..., γl ) .
10
R. Diaconu
Theorem 3 Let the functions fi belongs to the class T , for i ∈ {1, 2, ..., l}. Then
fi ∈ LAF (λ, k, β, µ, γ1 , ..., γl ) for i ∈ {1, 2, ..., l} if and only if


∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2
l
a
∑
j
j=2
k+1
n!(j−2)!


(6)
γi (β + 1) 
∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2  ≤ 1 − µ,
i=1
1 − j=2
k+1
n!(j−1)! aj
where −1 ≤ µ < 1, β ≥ 0.
Proof. Suppose that (6) holds. It suffices to show that
(
 (
)′′ )′′ 
m,n
m,n
z Fλ,k,γ
(z)
z
F
(z)
λ,k,γ1 ,...,γl
1 ,...,γl


β (
)′ − Re  (
)′  ≤ 1 − µ,
m,n
m,n
Fλ,k,γ1 ,...,γl (z)
Fλ,k,γ1 ,...,γl (z) where −1 ≤ µ < 1.
We have
(
(
 (
)′′ )′′ )′′ 
m,n
m,n
m,n
z Fλ,k,γ
(z) 1 ,...,γl
z Fλ,k,γ1 ,...,γl (z)  z Fλ,k,γ1 ,...,γl (z) 
β (
≤
(β
+
1)
−Re
)′ )′ 
)′ .
(
 (
m,n
m,n
m,n
Fλ,k,γ1 ,...,γl (z)
Fλ,k,γ1 ,...,γl (z) Fλ,k,γ1 ,...,γl (z) Applying Lemma 1, we obtain
(
)′′ m,n
z Fλ,k,γ ,...,γ (z) 1
l
(β + 1) (
)′ m,n
Fλ,k,γ1 ,...,γl (z) 

∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1 l
∑
a
z
−
j
j=2
k+1
n!(j−2)!


= (β + 1) γi 
∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1 
1 − j=2
z
i=1
k+1
n!(j−1)! aj


)
(
)
(
∑∞ λ(j−1)+k+1 m (n+j−1)! i 2 j−1
l
|z|
∑
j=2
k+1
n!(j−2)! aj


≤ (β + 1)
γi 
)
(
)
(
∑∞ λ(j−1)+k+1 m (n+j−1)! i 2 j−1 
i=1
|z|
1 − j=2
k+1
n!(j−1)! aj


∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2
l
∑
j=2
k+1
n!(j−2)! aj


≤ (β + 1)
γi 
(
( )2  .
)
m
∑∞ λ(j−1)+k+1
(n+j−1)!
i
i=1
1 − j=2
k+1
n!(j−1)! aj
The last expression is bounded above 1 − µ and hence we have
(
 (
)′′ )′′ 
m,n
m,n
z Fλ,k,γ ,...,γ (z) 1
l
 z Fλ,k,γ1 ,...,γl (z) 
β (
)′ − Re  (
)′  ≤ 1 − µ,
m,n
m,n
Fλ,k,γ
(z)
Fλ,k,γ1 ,...,γl (z) 1 ,...,γl
or equivalently

11
(
)′′ m,n
z Fλ,k,γ ,...,γ (z) z
(z) 
1
l

Re 1 + (
)′  ≥ β (
)′ + µ.
m,n
m,n
Fλ,k,γ
(z)
Fλ,k,γ1 ,...,γl (z) 1 ,...,γl
(
m,n
Fλ,k,γ
1 ,...,γl
)′′ 
Hence fi ∈ LAF (λ, k, β, µ, γ1 , ..., γl ) .
Conversely, suppose fi ∈ LAF (λ, k, β, µ, γ1 , ..., γl ), for i ∈ {1, 2, ..., l}. From (6)
and Lemma 1, we obtain that


∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1
l
a
|z|
∑
j
j=2
k+1
n!(j−2)!


1−
γi 
∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1 
i=1
|z|
1 − j=2
k+1
n!(j−1)! aj
l
∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1 ∑ z
j=2
k+1
n!(j−2)! aj
≥β
γi ∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1 + µ
z
i=1 1 − j=2
k+1
n!(j−1)! aj


)
(
(
)
∑∞ λ(j−1)+k+1 m (n+j−1)! i 2 j−1
l
a
z
∑
j
j=2
k+1
n!(j−2)!


≥β
γi 
∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1  + µ,
i=1
1 − j=2
z
k+1
n!(j−1)! aj
which is equivalent to


∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1
l
a
z
∑
j
j=2
k+1
n!(j−2)!


γi β 
∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1 
i=1
1 − j=2
z
k+1
n!(j−1)! aj


)
(
)
(
∑∞ λ(j−1)+k+1 m (n+j−1)! i 2 j−1
l
a
z
∑ 
j
j=2
k+1
n!(j−2)!

+
γi 
 ≤ 1 − µ,
)
(
)2
(
m
∑∞ λ(j−1)+k+1
(n+j−1)!
i
j−1
i=1
1 − j=2
z
k+1
n!(j−1)! aj
which reduces to
l
∑
i=1


γi (β + 1) 
∑∞ ( λ(j−1)+k+1 )m
1−

( )2
(n+j−1)!
i
z j−1 
j=2
k+1
n!(j−2)! aj
∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2 j−1 
z
j=2
k+1
n!(j−1)! aj
≤ 1 − µ,
when z → 1− along the real axis, we obtain the inequalitiy (6).
Using the same technique as in the proof of Theorem 3 and applying Lemma 2,
we obtain
Theorem 4 Let fi belongs to the class T , for i ∈ {1, 2, ..., l}.
fi ∈ LAG (λ, k, β, µ, γ1 , ..., γl ) for i ∈ {1, 2, ..., l} if and only if


∑∞ ( λ(j−1)+k+1 )m (n+j−1)! ( i )2
l
j
a
∑
j
j=2
k+1
n!(j−2)!


(7)
γi (β + 1) 
(
)
(
)2  ≤ 1 − µ,
m
∑∞ λ(j−1)+k+1
(n+j−1)!
i
i=1
1 − j=2
k+1
n!(j−1)! j aj
where −1 ≤ µ < 1, β ≥ 0.
Then
12
3
R. Diaconu
m,n
Integral operators Fλ,k,γ
and Gm,n
λ,k,γ1 ,...,γl on the classes
1 ,...,γl
D (δ), R (δ), RA (β, µ) and DA (β, µ)
m,n
and
In this section some of the properties of the integral operators Fλ,k,γ
1 ,...,γl
m,n
Gλ,k,γ1 ,...,γl on the classes D (δ), R (δ), RA (β, µ) and DA (β, µ) are discussed.
m,n
(IR fi (z))′ < Mi .
Theorem 5 Let γi ∈ R with γi > 0, i ∈ {1, 2, ..., l}, fi ∈ T and IRλ,k
m,n
λ,k fi (z)
∑
m,n
If fi ∈ RA (βi , µi ), then Fλ,k,γ
(z) ∈ D (δ ′ ), where δ ′ = 1+ li=1 γi µi (βi Mi + 1) .
1 ,...,γl
m,n
Proof. It is clear from (3) that Fλ,k,γ
∈T.
1 ,...,γl
m,n
On differentiating Fλ,k,γ1 ,...,γl given by (3), we obtain
l
(
)′ ∏
m,n
Fλ,k,γ
(z)
=
1 ,...,γl
(8)
i=1
(
m,n
IRλ,k
fi (z)
)γi
z
.
Differentiating (8) logarithmically and multiplying by z, we get
 (

)′
(
)′′
m,n
m,n
l
z
IR
f
(z)
z Fλ,k,γ
(z)
∑
λ,k i
1 ,...,γl


γi 
− 1 ,
)′ =
(
m,n
m,n
IR
f
(z)
i
λ,k
Fλ,k,γ1 ,...,γl (z)
i=1
equivalently
(9)
 (

)′′
(
)′
m,n
m,n
l
(z)
z Fλ,k,γ
z
IR
f
(z)
∑ 
λ,k i
1 ,...,γl

1+ (
γi 
− 1 .
)′ = 1 +
m,n
m,n
IR
f
(z)
i
λ,k
Fλ,k,γ1 ,...,γl (z)
i=1
Taking real part of both sides of (9), we get

  (

(
)′′ 
)′
m,n
m,n
l
z Fλ,k,γ1 ,...,γl (z) 
∑   z IRλ,k fi (z)


Re 1 + (
γi Re 
− 1
)′  = 1 +
m,n
m,n
IRλ,k fi (z)
Fλ,k,γ1 ,...,γl (z)
i=1
(
)′
m,n
z IRλ,k
fi (z)
≤1+
γi −
1
.
m,n
IRλ,k
fi (z)
i=1
l
∑
Since fi ∈ RA (βi , µi ), for i ∈ {1, 2, ..., l}, we have

(
(
)′′ 
)′
m,n
m,n
l
z IRλ,k fi (z)
z Fλ,k,γ1 ,...,γl (z) 
∑

Re 1 + (
+ 1
γi µi βi
)′  < 1 +
m,n
m,n
IRλ,k fi (z)
Fλ,k,γ
(z)
i=1
1 ,...,γl
13
(
)′ m,n
l
l
IRλ,k
∑
f
(z)
∑
i
+
γ
µ
<
1
+
γi µi (βi Mi + 1) .
<1+
γi µi βi i
i
m,n
IRλ,k
fi (z) i=1
i=1
i=1
l
∑
As
∑l
i=1 γi µi (βi Mi
m,n
(z) ∈ D (δ ′ ), where
+ 1) > 0, Fλ,k,γ
1 ,...,γl
δ′ = 1 +
l
∑
γi µi (βi Mi + 1) .
i=1
By substituting l = 1, γ1 = γ, M1 = M , f1 = f , in Theorem 5, we obtain
′ (z) Corollary 1 Suppose γ ∈ R, with γ > 0, f ∈ T and ff (z)
< M , where M is fixed.
∫ z ( f (t) )γ
If f ∈ RA (β, µ), then 0
dt ∈ D (δ ′ ), where δ ′ = 1 + γµ (βM + 1) .
t
Theorem 6 Let γi > 0 and fi ∈ T∑for i ∈ {1, 2, ..., l}, with δi > 1.
m,n
Fλ,k,γ
(z) ∈ D (δ ′ ), where δ ′ = 1 + li=1 γi (δi − 1) .
1 ,...,γl
Then
Proof. From (9), we have

 (
(
)′′ 
)′ 
m,n
m,n
l
l
z
F
(z)
z
IR
f
(z)
λ,k,γ1 ,...,γl
λ,k i

 ∑

 ∑
Re 1 + (
γi Re 
−
γi + 1
)′  =

m,n
m,n
IRλ,k
fi (z)
Fλ,k,γ
(z)
i=1
i=1
1 ,...,γl
<
l
∑
i=1
γi δi −
l
∑
γi + 1 = 1 +
i=1
l
∑
γi (δi − 1) .
i=1
∑
m,n
(z) ∈ D (δ ′ ),
Since δi > 1, it is clear that li=1 γi (δi − 1) > 0 and hence Fλ,k,γ
1 ,...,γl
∑
where δ ′ = 1 + li=1 γi (δi − 1) .
Letting l = 1, γ1 = γ, δ1 = δ and f1 = f , in Theorem 6, we obtain
)γ
∫z(
Corollary 2 Suppose γ > 0, f ∈ R (δ) with δ > 1. Then 0 f (t)
dt ∈ D (δ ′ ),
t
where δ ′ = 1 + γ (δ − 1) .
Theorem 7 Let γi > 0 and fi ∈ D∑
(δi ), for i ∈ {1, 2, ..., l}, with δi > 1. Then
l
′ ), with δ ′ = 1 +
Gm,n
(z)
∈
D
(δ
i=1 γi (δi − 1) .
λ,k,γ1 ,...,γl
Proof. From the definition of Gm,n
λ,k,γ1 ,...,γl given by (4), we have
   (

(
)′′ 
)′′ 
m,n
m,n


l
l

 ∑
z Gλ,k,γ1 ,...,γl (z)  ∑

  z IRλ,k fi (z) 
Re 1 + (
γi Re 1+  (
γi+1
)′  =
)′  −


m,n

 i=1
Gm,n
IR
(z)
f
(z)
i=1
i
λ,k,γ1 ,...,γl
λ,k
<1+
l
∑
i=1
γi (δi − 1) .
14
R. Diaconu
∑
As δi > 1, it is clear that li=1 γi (δi − 1) > 0 and hence we get that Gm,n
λ,k,γ1 ,...,γl (z) ∈
∑l
′
′
D (δ ), where δ = 1 + i=1 γi (δi − 1) .
By substituting l = 1, γ1 = γ, δ1 = δ and f1 = f , in Theorem 7, we obtain the
following
∫z
Corollary 3 Let γ > 0 and f ∈ D (δ) with δ > 1. Then 0 (f ′ (t))γ dt ∈ D (δ ′ ),
where δ ′ = 1 + γ (δ − 1) .
m,n
(IRλ,k fi (z))′′ < Mi , i ∈
Theorem 8 Let fi ∈ DA (βi , µi ), γi ∈ R with γi > 0 and ′
m,n
fi (z)) (IRλ,k
∑l
′
′
{1, 2, ..., l}. Then Gm,n
i=1 γi µi [βi (1 + Mi ) + 1] .
λ,k,γ1 ,...,γl (z) ∈ D (δ ), where δ = 1+
Proof. From the definition of Gm,n
λ,k,γ1 ,...,γl given by (4), we have
(

)′′ (
)′′ 
m,n
m,n
l
z Gλ,k,γ1 ,...,γl (z)  ∑ z IRλ,k fi (z) 
γi (
Re 1 + (
)′  ≤
)′ <
m,n
m,n
Gλ,k,γ1 ,...,γl (z)
i=1
IRλ,k fi (z) 
(
)′′ 
m,n
z IRλ,k fi (z) 

1+
γi µi βi 1 + (
)′  + 1 <
m,n
IRλ,k fi (z)
i=1
(


)′′ m,n
l
l
l
z IRλ,k
f
(z)
∑
∑
i

 ∑
1+
γi µi βi 1 + (
γi µ i < 1 +
γi µi [βi (1 + Mi ) + 1] .
)′  +
m,n
i=1
i=1
i=1
IRλ,k fi (z) ∑
′
As li=1 γi µi [βi (1 + Mi ) + 1] > 0, we conclude that Gm,n
λ,k,γ1 ,...,γl (z) ∈ D (δ ), where
∑
δ ′ = 1 + li=1 γi µi [βi (1 + Mi ) + 1] .
By taking l = 1, γ1 = γ, M1 = 1, f1 = f , in Theorem 8, we obtain the following
corollary
′′ (z) Corollary 4 Let γ ∈ R with γ > 0, f ∈ DA (β, µ) and ff ′ (z)
< M , M is fixed.
∫z ′
γ
′
′
Then 0 (f (t)) dt ∈ D (δ ), where δ = 1 + γµ [β (1 + M ) + 1] .
l
∑
4
Subordination results for f ∈ T associated with the
m,n
operator IRλ,k
In this section we extend Theorem 1 and Theorem 2 for functions f ∈ T defined
through the operator given by (1).
Theorem 9 Let q (z) be convex and univalent, γ ̸= 0 and
{
(
)}
(k + 1) (1 − γ) 2 (k + 1)
zq ′′ (z)
Re
+
q (z) + 1 + ′
> 0.
λγ
λ
q (z)
If f ∈ T satisfies
(10)
(
m+1,n
IRλ,k
f (z)
m,n
IRλ,k
f (z)
1−γ+γ
m+2,n
IRλ,k
f (z)
m+1,n
IRλ,k
f
15
)
(z)
≺ (1 − γ) q (z) + γq 2 (z) +
λγ
zq ′ (z) ,
k+1
then
m+1,n
IRλ,k
f (z)
(11)
m,n
IRλ,k
f (z)
≺ q (z)
and q (z) is the best dominant.
Proof. Define the function
(12)
p (z) =
m+1,n
IRλ,k
f (z)
m,n
IRλ,k
f (z)
.
Logarithmic differential of (12) yields
(
)′
(
)′ 
m+1,n
m,n
m+1,n
m,n
m,n
IRλ,k f (z)  IRλ,k f (z) IRλ,k f (z) − IRλ,k f (z) IRλ,k f (z) 
p′ (z)
=


(
)2
m+1,n
p (z)
IRλ,k
f (z)
IRm,n f (z)
λ,k
=
(
)′
m+1,n
IRλ,k
f (z)
m+1,n
IRλ,k
f (z)
−
(
)′
m,n
IRλ,k
f (z)
m,n
IRλ,k
f (z)
.
So,
(13)
zp′ (z)
p (z)
=
(
)′
m+1,n
z IRλ,k
f (z)
m+1,n
IRλ,k
f (z)
−
(
)′
m,n
z IRλ,k
f (z)
m,n
IRλ,k
f (z)
.
By using (2) in (13), we get
m+2,n
m+1,n
zp′ (z)
k + 1 IRλ,k f (z) k + 1 − λ k + 1 IRλ,k f (z) k + 1 − λ
=
−
−
+
.
m,n
m+1,n
p (z)
λ IRλ,k
λ
λ
λ
IRλ,k
f (z)
f (z)
We obtain
zp′ (z)
k+1
=
p (z)
λ
(
m+2,n
IRλ,k
f (z)
m+1,n
IRλ,k
f (z)
)
− p (z)
and
(14)
[ ′
]
λ
zp (z) k + 1
=
+
p (z) .
m+1,n
k + 1 p (z)
λ
IRλ,k
f (z)
m+2,n
IRλ,k
f (z)
16
R. Diaconu
Hence from (14)
(15)
m+1,n
IRλ,k
f (z)
[
m+2,n
IRλ,k
f (z)
]
1 − γ + γ m+1,n
m,n
IRλ,k
f (z)
IRλ,k f (z)
]
[
λ zp′ (z)
+ γp (z)
= p (z) 1 − γ + γ
k + 1 p (z)
γλ
= (1 − γ) p (z) + γp2 (z) +
zp′ (z) .
k+1
In view of (15), the subordination (10) becomes
(1 − γ) p (z) + γp2 (z) +
γλ
γλ
zp′ (z) ≺ (1 − γ) q (z) + γq 2 (z) +
zq ′ (z) .
k+1
k+1
Applying Lemma 4, we obtain that
m+1,n
IRλ,k
f (z)
m,n
IRλ,k
f (z)
≺ q (z)
Theorem 10 Let q (z) be univalent in U , q (0) ̸= 0, γ ̸= 0 and
univalent in U . If f (z) ∈ T satisfies
(16)
m+2,n
IRλ,k
f (z)
m+1,n
IRλ,k
f (z)
−γ
m+1,n
IRλ,k
f (z)
m,n
IRλ,k
f (z)
≺
zq ′ (z)
q(z)
be starlike and
λ zq ′ (z) k (γ − 1) − 1
−
,
k + 1 q (z)
k+1
then
(17)
m+1,n
z γ−1 IRλ,k
f (z)
(
)γ ≺ q (z)
m,n
IRλ,k f (z)
Proof. Define the analytic function p in U by
(18)
m+1,n
z γ−1 IRλ,k
f (z)
)γ .
p (z) = (
m,n
IRλ,k
f (z)
Logarithmic differentiation of (18) yields
(
)′
m+1,n
m+1,n
z γ−2 (γ − 1) IRλ,k
f (z) + z γ−1 IRλ,k
f (z)
(
)γ
p′ (z) =
m,n
IRλ,k
f (z)
(
)′
m+1,n
m,n
γz γ−1 IRλ,k
f (z) IRλ,k
f (z)
(
)γ
−
.
m,n
m,n
IRλ,k
f (z) IRλ,k
f (z)
17
So,
(19)
zp′ (z)
p (z)
= (γ − 1) +
(
)′
m+1,n
z IRλ,k
f (z)
m+1,n
IRλ,k
f (z)
−γ
(
)′
m,n
z IRλ,k
f (z)
m,n
IRλ,k
f (z)
.
Using (2) in (19) we get
m+2,n
m+1,n
zp′ (z)
k (γ − 1) − 1 k + 1 IRλ,k f (z) γ (k + 1) IRλ,k f (z)
=
+
−
,
m,n
m+1,n
p (z)
λ
λ IRλ,k
λ
IRλ,k
f (z)
f (z)
which is equivalent to
m+2,n
IRλ,k
f (z)
m+1,n
IRλ,k
f (z)
−γ
m+1,n
IRλ,k
f (z)
m,n
IRλ,k
f (z)
By hypothesis (16), we obtain
zp′ (z)
p(z)
≺
=
λ zp′ (z) k (γ − 1) − 1
−
.
k + 1 p (z)
k+1
zq ′ (z)
q(z) .
From Lemma 3 we obtain
m+1,n
z γ−1 IRλ,k
f (z)
(
)γ ≺ q (z)
m,n
IRλ,k
f (z)
Acknowledgement. This work was suported by the strategic project POSDRU/159/1.5/S/138963 - ”PERFORM”.
References
[1] L. Andrei, On some differential sandwich theorems using a generalized Sălăgean
operator and Ruscheweyh operator, Journal of Computational Analysis and
Applications, Vol. 18, 2015 (to appear).
[2] L. Andrei, A. Alb Lupaş, Certain integral operators of analytic functions,
submitted 2014.
[3] D. Breaz, N. Breaz, Two integral operators, Studia Universitatis Babes-Bolyai,
Cluj-Napoca, 3 (2002), 13-21.
[4] D. Breaz, S. Owa, N. Breaz, A new integral univalent operator, Acta Univ.
Apul., 16 (2008), 11-16.
[5] R. Diaconu, Some integral operators of analytic functions, submitted 2014.
[6] S.S. Miller, P.T. Mocanu, Differential Subordinations: Theory and Applications,
Series on Monographs nad Textbooks in Pure and Applied Mathematics, No.
225, Marcel Dekker, New York and Basel, 2000.
18
R. Diaconu
[7] V. Ravichandran, Certain applications of first order differential subordination,
Far East J. Math. Sci., 12 (2004), 41-51.
[8] V. Ravichandran, M. Darus, M. Hussain Khan, K.G. Subramanian, Differential
subordination associated with linear operators defined for multivalent functions,
Acta Mathematica Vietnamica, 30 (2) (2005), 113-121.
[9] St. Ruscheweyh, New criteria for univalent functions, Proc. Amet. Math. Soc.,
49(1975), 109-115.
[10] G. St. Sălăgean, Subclasses of univalent functions, Lecture Notes in Math.,
Springer Verlag, Berlin, 1013(1983), 362-372.
Radu Diaconu
University of Pitesti
Department of Mathematics
Str. Targul din Vale, no.1
110040 - Pitesti, Romania
e-mail: radudyaconu@yahoo.com
Some blending surfaces generated by univariate
interpolation 1
Marius Birou
Abstract
In this article we construct the blending surfaces using univariate
interpolation functions. The surfaces match a given family of curves.
Remarks about the parabolic points of these surfaces and an example
are given.
2010 Mathematics Subject Classification: 41A05, 65D17
Key words and phrases: blending interpolation, surfaces, parabolic
points, Lagrange interpolation
1
Introduction
The blending surfaces are the surfaces which contain some given contours
(curves, segments, points). They have been created by Coons S.A. [4]. In
some previous papers (see [2, 3]), there were constructed the blending surfaces
which stay on a rectangle or a triangle (the border of the surfaces domain)
and having a fixed height in the point (0, 0). In this paper we use univariate
interpolation function to get surfaces which match some given curves. The
surfaces stay on the one of these curve (the surfaces domain are bounded by
this curve). We assume that the point (0, 0) belongs to the surfaces domains
1
Received 15 March, 2010
Accepted for publication (in revised form) 29 June, 2010
19
20
M. Birou
and we fix the height of the surfaces in this point. These surfaces can be used
in civil engineering (roof surfaces) or in Computer Aided Geometric Design
(CAGD). The position of the parabolic points are studied because the maximal
stress holds in these points (see [2, 3, 6]). We give an example of surfaces which
match given circles using Lagrange interpolation. Other examples of surfaces
which match circles can be found in [1].
2
Construction of the surfaces
Let 0 = y0 < y1 < ... < yn−1 < yn = a and f : [0, a] → R a function with the
properties
(1)
f (0) = h > 0, f (a) = 0,
f (yj ) = hj > 0, j = 1, ..., n − 1.
We consider a univariate interpolation function of the form
(Φn f )(y) =
n
∑
φj (y)f (yj )
j=0
where φj are C 2 functions which satisfy the conditions
{
1, if i = j
φj (yi ) = δij =
0, if i ̸= j
for i, j = 0, ..., n.
Taking into account (1) we obtain
(2)
(Φn f )(y) = φ0 (y)h +
n−1
∑
φj (y)hj .
j=1
The function (2) has the properties
(Φn f )(0) = h, (Φn f )(a) = 0,
(Φn f )(yj ) = hj , j = 1, ..., n − 1.
Let u(X, Y ) a bivariate positive C 2 function such that the curves
Cj : u(X, Y ) = yj
Some blending surfaces generated by univariate interpolation
21
for j = 1, . . . , n are simple and closed. Let Dj the domain bounded by the
curve Cj , for j = 1, . . . , n. We assume
(0, 0) ∈ D1 ⊂ D2 ⊂ . . . ⊂ Dn = D
and that the curve u(X, Y ) = 0 is reduced to the point (0, 0). We also assume
that D is a star convex set respect to the point (0, 0) and the function u is
homogeneous of order one, i.e.
u(tX, tY ) = t u(X, Y ), (X, Y ) ∈ D, t > 0.
If we make the substitution
y = u(X, Y )
in (2), we obtain the bivariate function
(3)
Fe (X, Y ) = (Φn f )(u(X, Y )) =
n−1
∑
= φ0 (u(X, Y ))h +
φj (u(X, Y ))hj , (X, Y ) ∈ D.
j=1
The function Fe from (3) has the properties
Fe|∂D = 0,
Fe|Cj = hj , j = 1, ..., n − 1,
Fe (0, 0) = h.
It follows that the surfaces Z = Fe (X, Y ) match the curve u(X, Y ) = a, Z = 0
(the surfaces stay on the border of the domain D), the curves u(X, Y ) = yj ,
Z = hj for j = 1, ..., n − 1 and the height of the surfaces in the point (0, 0) is
h.
3
The parabolic points
It is important to study the position of the parabolic points. For the roof
surfaces, the maximal stress holds in these points (see [2], [6]). It is preferable
to avoid the parabolic points or all the points of the surface to be of parabolic
22
M. Birou
type. We can give conditions on parameters hj to control the position of the
parabolic points.
If we note h = h0 the function Fe is given by
Fe (X, Y ) =
n−1
∑
φj (u(X, Y ))hj , (X, Y ) ∈ D.
j=0
The second partial derivatives of the function Fe can be expressed by
FeXX (X, Y ) =
(4)
= (uX (X, Y
))2
n−1
∑
φ′′j (u(X, Y
))hj + uXX (X, Y )
j=0
n−1
∑
φ′j (u(X, Y ))hj ,
j=0
(5)
FeXY (X, Y ) =
= uX (X, Y )uY (X, Y )
n−1
∑
φ′′j (u(X, Y
))hj + uXY (X, Y )
j=0
n−1
∑
φ′j (u(X, Y ))hj ,
j=0
FeY Y (X, Y ) =
(6)
= (uY (X, Y ))2
n−1
∑
φ′′j (u(X, Y ))hj + uY Y (X, Y )
j=0
n−1
∑
φ′j (u(X, Y ))hj .
j=0
The parabolic points (X, Y ) ∈ D of the surfaces Z = Fe (X, Y ) satisfy the
condition
(7)
FeXX (X, Y )FeY Y (X, Y ) − (FeXY (X, Y ))2 = 0.
From (7) and (4)-(6) we have
A(X, Y )B(X, Y )(u2X (X, Y )uY Y (X, Y ) + u2Y (X, Y )uXX (X, Y )−
−2uX (X, Y )uY (X, Y )uXY (X, Y )) + (A(X, Y ))2 ×
×(uXX (X, Y )uY Y (X, Y ) − u2XY (X, Y )) = 0
with
A(X, Y ) =
B(X, Y ) =
n−1
∑
j=0
n−1
∑
j=0
φ′j (u(X, Y ))hj
φ′′j (u(X, Y ))hj .
23
Taking into account that the function u is homogeneous of order one, we obtain
uXY (X, Y )(u(X, Y ))2 A(X, Y )B(X, Y )
= 0.
XY
We consider the following two equations with unknown y
n−1
∑
(8)
φ′j (y)hj = 0
j=0
and
n−1
∑
(9)
φ′′j (y)hj = 0.
j=0
If
uXY (X, Y )(u(X, Y ))2
̸= 0
XY
for every (X, Y ) ∈ D and the equations (8) and (9) have no solutions in the
interval [0, a] then the surfaces generated by function Fe have no parabolic
points in D. If y = y ∈ (0, a] is a solution for one of the equations (8) or (9),
the surfaces have parabolic points of whose set of projections on XOY plane
is the curve u(X, Y ) = y.
4
Example
We choose the Lagrange univariate interpolation function, i.e.
n
∏
y − yi
φj (y) =
yj − yi
i=0
i̸=j
for j = 0, ..., n and
u(X, Y ) =
√
X 2 + Y 2.
It follows that the domain D is the disk {(X, Y ) ∈ R2 |X 2 + Y 2 ≤ a2 } and the
curves Cj are the circles {(X, Y ) ∈ R2 |X 2 + Y 2 = yj2 }, for j = 1, . . . , n.
For n = 1 we obtain the function
√
a − X2 + Y 2
e
(10)
F (X, Y ) =
h, (X, Y ) ∈ D.
a
24
M. Birou
It has the properties
We have
Fe |∂D = 0, Fe(0, 0) = h.
FeXX (X, Y )FeY Y (X, Y ) − (FeXY (X, Y ))2 = 0
for every (X, Y ) ∈ D. All the points of the surface Z = Fe (X, Y ) are parabolic
points.
2
1
0
-1
-2
4
3
2
1
0
-2
-1
0
1
2
Figure 1: The surface Z = Fe (X, Y ), with Fe given by (10) for a = 2 and h = 4.
For n = 2 we take y1 = b ∈ (0, a) and we get the function
(11)
=
(b−a)h+ah1
2
ab(b−a) (X
It has the properties
+ Y 2) −
Fe (X, Y ) =
√
X 2 + Y 2 + h, (X, Y ) ∈ D.
ab(b−a)
(b2 −a2 )h+a2 h1
Fe |∂D = 0, Fe(0, 0) = h,
Fe|X 2 +Y 2 =b2 = h1 .
Next we study the position of the parabolic points of the surface Z =
e
F (X, Y ), with Fe given by (11). We have
uXY (X, Y )(u(X, Y ))2
−1
=√
̸= 0
XY
X2 + Y 2
for (X, Y ) ̸= (0, 0). The equations (8) and (9) become
(12)
2((b − a)h + ah1 )y − ((b2 − a2 )h + a2 h1 ) = 0,
(13)
(b − a)h + ah1 = 0.
−b )h−a h1
Let y = (a
2((a−b)h−ah1 ) , a1 =
the following cases:
2
2
2
(a−b)2 h
,
a2
a2 =
(a−b)h
,
a
a3 =
(a2 −b2 )h
.
a2
We have
25
2
1
0
-1
-2
2
4
2
1
2
1
0
1
0
-1
3
0
-1
-2
4
-1
-2
4
-2
4
2
3
3
3
1
2
0
2
1
-2
-1
2
1
1
0
0
-2
-2
0
0
-2
-1
-1
-1
0
0
1
0
1
2
1
2
(a) h1 = 0.5
(b) h1 = 1
1
2
(c) h1 = 1.5
2
(d) h1 = 2
2
1
0
-1
2
-2
4
2
1
1
0
0
-1
-1
3
-2
-2
4
4
2
3
1
3
2
2
1
0
-2
1
0
-1
0
-2
0
-2
-1
-1
0
1
0
1
2
(e) h1 = 2.5
1
2
(f) h1 = 3
2
(g) h1 = 3.7
Figure 2: The surface Z = Fe(X, Y ) with Fe given by (11) for a = 2, b = 1,
h = 4.
i) If h1 = a2 then the relation (13) holds. It follows that all the points of
the surface are parabolic points.
ii) If a1 < h1 < a2 then the equation (12) has the solution y = y > a. The
surface has no parabolic points.
iii) If a2 < h1 < a3 then the equation (12) has the solution y = y < 0. The
surface has no parabolic points.
iv) If h1 = a1 then the equation (12) has the solution y = a. The parabolic
points are situated on the border of domain D.
v) If h1 = a3 then the equation (12) has the solution y = 0. It follows that
the surface has no parabolic points.
vi) If 0 < h1 < a1 or a3 < h1 < h then the equation (12) has the solution
y = y in interval (0, a). It follows that the surface has parabolic points
of which projections are situated on a circle, inside of domain D.
26
M. Birou
References
[1] M. Birou, Blending surfaces on circular domains generated by Birkhoff
interpolation, Autom. Comput. Appl. Math, 18, 2009, No 1, 97-106.
[2] Gh. Coman, I. Gânscă, L. Tâmbulea, Some new roof-surfaces generated
by blending interpolation technique, Studia Univ. Babeş-Bolyai Math., 36,
1991, No 1, 119-130.
[3] Gh. Coman, I. Gânscă, L. Tâmbulea, Surfaces generated by blending interpolation, Studia Univ. Babeş-Bolyai Math., 38, 1993, No. 3, 39-48.
[4] S.A. Coons, Surface for computer aided design of space forms, Project
MAC, Design Div., Dep of Mech. Engineering, MIT, 1964.
[5] G. Farin, Curves and surfaces for Computer Aided Geometric Design,
Academic Press, Inc Harcourt Brace Jovanovich, Publishers, Boston, San
Diego, New York, London, Sidney, Tokyo, Toronto, 1990.
[6] M. Mihailescu, I. Horvath. Velaroidal shells for covering universal industrial halls, Acta Tech. Acad. Hungaricae, 85, 1977, 135-145.
Marius Birou
Technical University of Cluj Napoca
Str. Memorandumului 28, 400114 Cluj-Napoca, Romania
e-mail: marius.birou@math.utcluj.ro
On the univalent condition for a family of integral
operators 1
C.Selvaraj, C.Santhosh Moni
Abstract
The purpose of the paper is to introduce a general integral operator. For
this general integral operator we study some interesting univalence properties,
to which a number of univalent conditions would follow upon specializing the
parameters involved.
Key words and phrases: Analytic functions, univalent functions, integral
operators, Schwarz Lemma.
1
Introduction, definitions and preliminaries
Let A1 = A denote the class of functions of the form
(1)
f (z) = z +
∞
∑
an z n
an ≥ 0,
n=2
which are analytic in the open disc U = {z : z ∈ C | z |< 1} and S be the class of
function f (z) ∈ A which are univalent in U.
For αj ∈ C (j = 1, 2, . . . , q) and βj ∈ C \ {0, −1, −2, . . .} (j = 1, 2, . . . , s), the
generalized hypergeometric function q Fs (α1 , . . . , αq ; β1 , . . . , βs ; z) is defined by the
infinite series
q Fs (α1 , α2 , . . . , αq ; β1 , β2 , . . . , βs ; z) =
∞
∑
(α1 )n . . . (αq )n z n
(β1 )n . . . (βs )n n!
n=0
1
Received 30 April, 2010
Accepted for publication (in revised form) 17 February, 2012
27
28
(q ≤ s + 1; q, s ∈ N0 = {0, 1, 2, . . .}; z ∈ U ),
where (x)n is the Pochhammer symbol defined by
Γ(x + n)
(x)n =
=
Γ(x)
{
1
if n = 0
x(x + 1)(x + 2) . . . (x + n − 1)
if n ∈ N = {1, 2, , . . .}.
Corresponding to a function h(α1 , . . . , αq ; β1 , . . . , βs ; z) defined by
h(α1 , . . . , αq ; β1 , . . . , βs ; z) := z q Fs (α1 , α2 , . . . , αq ; β1 , β2 , . . . , βs ; z)
the Dziok and Srivastava operator[6] H(α1 , . . . , αq ; β1 , . . . , βs ; z)f (z) is defined by
the Hadamard product
H(α1 , . . . , αq ; β1 , . . . , βs ; z)f (z) := h(α1 , . . . , αq ; β1 , . . . , βs ; z) ∗ f (z)
∞
∑
(α1 )n−1 . . . (αq )n−1 an z n
=z+
(β1 )n−1 . . . (βs )n−1 (n − 1)!
(2)
n=2
For convenience, we write
Hsq (α1 , β1 )f := H(α1 , . . . , αq ; β1 , . . . , βs ; z)f (z).
The linear operator Hsq (α1 , β1 )f includes (as its special cases) various other
linear operators which were introduced and studied by Hohlov, Carlson and Shaffer,
Ruscheweyh. For details see [1].
Using Dziok-Srivastava operator, we now introduce the following:
For n ∈ N∪{0} and γi , γ2 , . . . , γn , δ ∈ C\{0, −1, −2, . . .}, we define the integral
operator Fγi ,δ (α1 , β1 ; z) : An −→ A by
(3)
{ ∫
Fγi ,δ (α1 , β1 ; z) = δ
t
0
)
n ( q
∏
Hs (α1 , β1 )fi (t) γi
1
z
δ−1
i=1
t
}1
δ
dt
.
where fi ∈ A and Hsq (α1 , β1 )f is the Dziok-Srivastava operator.
Remark 1 It is interesting to note that the integral operator Fγi ,δ (α1 , β1 ; z) generalizes many operators which were introduced and studied recently. Here we list a few
of them
On the univalent condition for a family of integral operators
29
1. Let q = 2, s = 1, α1 = β1 , α2 = 1, γi = 1/(α − 1) and δ = n(α − 1) + 1, then
the operator Fγi ,δ (α1 , β1 ; z) reduces to an integral operator
[
(4)
∫
z
Fn,α (z) = (n(α − 1) + 1)
(
f1 (t)
)α−1
(
. . . fn (t)
)α−1
]
1
(n(α−1)+1)
dt.
,
0
studied recently by D.Breaz et.al. [4].
2. Let q = 2, s = 1, α1 = β1 , α2 = 1, γi = 1/αi and δ = 1, then the operator
Fγi ,δ (α1 , β1 ; z) reduces to an operator
∫
(5)
z
(
Fα (z) =
0
f1 (t)
t
)α1
(
...
fn (t)
t
)αn
dt
recently introduced by D.Breaz and N.Breaz [2].
3. Let q = 2, s = 1, α1 = 2, β1 = 1, α2 = 1, γi = 1/αi and δ = 1, then the
operator Fγi ,δ (α1 , β1 ; z) reduces to an operator
∫
(6)
Gα (z) =
z
(
)α
( ′ )α
′
f1 (t) 1 . . . fn (t) n dt,
0
recently introduced and studied by D.Breaz and H.Ozlem Guney [3].
Apart from the above several well-known and new integral operators will follow as a
special case on specializing the parameters.
We now state the following lemma which we need to establish our results in the
sequel.
Theorem 1 [8] If f ∈ A satisfies the condition
(7)
2 ′
z f (z)
f 2 (z) − 1 ≤ 1,
for all z ∈ U,
then the function f (z) is univalent in U.
Theorem 2 (Schwartz Lemma) Let f ∈ A satisfy the condition | f (z) |≤ 1, for all
z ∈ U, then
| f (z) |≤| z |, for all z ∈ U,
and equality holds only if f (z) = ϵz, where | ϵ |= 1.
Pescar [9] has proved the following result
30
Theorem 3 [9]Let δ ∈ C, Reδ > 0 and c ∈ C (| c |≤ 1; c ̸= −1). If f ∈ A satisfies
(
) zf ′′ (z) 2δ
2δ
c | z | + 1− | z |
≤ 1 (z ∈ U ),
δf ′ (z) then the function
[ ∫
Fδ (z) = δ
z
t
δ−1
]1
δ
f (t) dt = z + . . .
′
0
is analytic and univalent in U.
Theorem 4 [10]Let f ∈ A satisfies the condition (7). Also let
(
[
])
3
α∈R
α ∈ 1,
and c ∈ C.
2
If
| c |≤
3 − 2α
α
then the function
(c ̸= −1)
and
( ∫
Hα (z) = α
z
[
| f (z) |≤ 1 (z ∈ U)
]α−1
g(t)
dt
)1
α
0
belongs to S.
2
Main Results
Theorem 5 Let each of the functions Hsq (α1 , β1 )fi ∈ A (i ∈ {1, 2, . . . , n}) satisfy
the inequality (7). Also for M ≥ 1, let γi , δ be complex numbers such that
Re δ ≥
n
∑
2M + 1
|γi |
and c ∈ C.
i=1
If
(8)
| c |≤ 1 −
n
1 ∑ 2M + 1
Reδ
|γi |
i=1
and
| Hsq (α1 , β1 )fi (z) |≤ M (z ∈ U; i ∈ {1, 2, . . . , n}),
then the function Fγi ,δ (α1 , β1 ; z) defined by (3) is univalent.
Proof. From the definition of the operator Hsq (α1 , β1 )f (z), it can easily seen that
Hsq (α1 , β1 )f (z)
̸= 0
z
(z ∈ U)
31
and moreover for z = 0, we have
(
Hsq (α1 , β1 )f1 (z)
z
)
1
γ1
(
...
Hsq (α1 , β1 )fn (z)
z
)
1
γn
= 1.
We define the function
∫
h(z) =
n (
z∏
0 i=1
so that, obviously
Hsq (α1 , β1 )fi (t)
t
)1
γi
)
n ( q
∏
Hs (α1 , β1 )fi (z) γi
dt,
1
′
h (z) =
z
i=1
−1
and
zh (z) ∑ 1
=
γi
h′ (z)
′′
(9)
n
i=1
(
)
′
z(Hsq (α1 , β1 )fi (z))
−
1
.
Hsq (α1 , β1 )fi (z)
So, from (9) we have
(
)
′′
′
n
∑
2δ
2δ
zh
(z)
1 z(Hsq (α1 , β1 )fi (z))
2δ
c|z| + (1 − |z| ) ′
= c|z| + (1 − |z|2δ ) 1
−
1
q
δ
γi
Hs (α1 , β1 )fi (z)
δh (z)
i=1
(
)
′
n
1 ∑ z 2 (Hsq (α1 , β1 )fi (z)) |Hsq (α1 , β1 )fi (z)|
+1 .
≤ |c| +
]2 [ q
|δ|
|z|
Hs (α1 , β1 )fi (z)
i=1
From the hypothesis, we have
| Hsq (α1 , β1 )fi (z) |≤ M (z ∈ U; i ∈ {1, 2, . . . , n}),
then by using the inequality (7), we obtain
′′
n
∑
2δ
2M + 1
c|z| + (1 − |z|2δ ) zh ′ (z) ≤ |c| + 1
Re δ
| γi |
δh (z)
(z ∈ U),
i=1
which, in the light of the hypothesis (8), yields
′′
2δ
c|z| + (1 − |z|2δ ) zh ′ (z) ≤ 1 (z ∈ U).
δh (z) Finally, by applying Theorem 3, we conclude that the function Fγi ,δ (α1 , β1 ; z) defined
by (3) is in the univalent function class S. This completes the proof of Theorem 5.
32
3
Some Extensions of Univalence Conditions
First of all, upon setting q = 2, s = 1, α1 = β1 , α2 = 1, γi = 1/(α − 1) and
δ = n(α − 1) + 1 (α a real number), we obtain
Corollary 1 [4]Let M ≥ 1 and suppose that each of the functions fi ∈ A (i=1, . . . , n)
satisfies the inequality (7). Also let
(
[
])
(2M + 1)n
α∈R
α ∈ 1,
and c ∈ C
(2M + 1)n − 1
If
(
| c |≤ 1 +
)
1−α
(2M + 1)n
α
and
| fi (z) |≤ M (z ∈ U; i ∈ {1, . . . , n}),
then the function Fn,α (z) defined by (4) is in the univalent function class S.
Remark 2 Corollary 1 provides an extension of Theorem 4 due to Pescar [10], if
we let n = 1.
Next we set q = 2, s = 1, α1 = β1 , α2 = 1, δ = 1 and γi =
we obtain the following
1
αi
(αi a real number),
Corollary 2 Let M ≥ 1 and suppose that each of the functions fi ∈ A (i=1, . . . , n)
satisfies the inequality (7). Also let
(
[
])
(2M + 1)n
α∈R
α ∈ 1,
and c ∈ C
(2M + 1)n − 1
If
| c |≤ 1 + α(2M + 1)n
and
| fi (z) |≤ M (z ∈ U; i ∈ {1, . . . , n}),
then the function Fα (z) defined by (5) is in the univalent function class S.
Remark 3 Many other interesting corollaries and results can be obtained by
specializing the parameters in Theorem 5, for example see [4, 5, 7].
References
[1] R. Aghalary et al., Subordinations for analytic functions defined by the DziokSrivastava linear operator, Appl. Math. Comput. 187 (2007), no. 1, 13–19.
[2] D. Breaz and N. Breaz, Two integral operators, Studia Univ. Babeş-Bolyai
Math. 47 (2002), no. 3, 13–19.
33
[3] Daniel Breaz and H.Ozlem Guney, The integral operators of the calsses Sα∗ (b)
and Cα (b), J. Math. Inequal. 2 (2008), no. 1, 97–100.
[4] D. Breaz, et al., An extension of the univalent condition for a family of operators, Appl. Math. Lett. (2008), doi:10.1016/j.aml.2007.11.008.
[5] Daniel Breaz and H. Özlem Güney, On the univalence criterion of a general
integral operator, Journal of Inequalities and Applications, vol. 2008, Article ID
702715, 8 pages, 2008. doi:10.1155/2008/702715
[6] J. Dziok and H. M. Srivastava, Classes of analytic functions associated with the
generalized hypergeometric function, Appl. Math. Comput., 103(1999), 1-13.
[7] Georgia Irina Oros, Gheorghe Oros, and Daniel Breaz, Sufficient conditions for
univalence of an integral operator, Journal of Inequalities and Applications, vol.
2008, Article ID 127645, 7 pages, 2008. doi:10.1155/2008/127645
[8] S. Ozaki and M. Nunokawa, The Schwarzian derivative and univalent functions,
Proc. Amer. Math. Soc. 33 (1972), 392–394.
[9] V. Pescar, A new generalization of Ahlfors’s and Becker’s criterion of univalence , Bull. Malaysian Math. Soc. (2) 19 (1996), no. 2, 53–54.
[10] V. Pescar, On the univalence of some integral operators, J. Indian Acad. Math.
27 (2005), no. 2, 239–243.
C.Selvaraj
Presidency College
Chennai-600 005, Tamilnadu, India
e-mail: pamc9439@yahoo.co.in
C.Santhosh Moni
R.M.K.Engineering College
R.S.M.Nagar, Kavaraipettai-601206,Tamilnadu,India
e-mail: csmrmk@yahoo.com
Centralizers of maximal tori in the classical
group SU (n) 1
Vadoud Najjari
Abstract
In this paper after reviewing maximal tori in a Lie group, we define some
new families of matrices, also some notations and a few conventions for them.
Then after defining homomorphism ΦM (from T k to T n ), we find the centralizers
of ΦM such that M consist of only one PM-block. In general, matrices in the
centralizer of ΦM won’t be in SU (n) however, we will find conjugates that are.
Also we characterize the maximal tori of the classical group SU (n), and finally
we calculate centralizers of this classical group.
2010 Mathematics Subject Classification: 53D22, 11F23, 11F99.
Key words and phrases: Centralizer, maximal torus, classical groups.
1
Introduction
Brocker, Tom Dieck, and Tolman ([2], [4]) characterized the finite subgroups of
centralizers of tori in the real symplectic group SP (2n, R) and then Schmah [8],
characterized centralizers of all tori in SP (2n, R). Najjari [7], calculated centralizers
of maximal tori in the classical group SO(2n, R). This paper tries to calculate the
centralizers of maximal tori of the classical group SU (n), which may be of interest
in its own right.
2
spacial case of the centralizers
In this section we study maximal tori in a Lie group. let G be a Lie group.
Definition 2.1. A torus T is maximal if T ⊂ U ⊂ G and U is a torus, then T = U .
1
Received 18 May, 2010
Accepted for publication (in revised form) 6 December, 2010
35
36
V. Najjari
Theorem 2.2. Any torus is contained in a maximal torus.
Proof. See [3].
Lemma 2.3. If G is connected, then any element of G is conjugate to an element
in T .
Proof. See [3].
Lemma 2.4. In a connected group, any two maximal tori are conjugate.
Proof. See [3].
In the following the maximal tori of the classical group SU (n) and U (n) will be
shown.
Proposition 2.5. The maximal torus of U (n) is,
 iθ

e 1
0


..
T = {
}.
.
iθ
0
e 1
Proof. Firstly, we observe that T is a torus. Then, we set Tj = subgroup of T
consisting of matrices with 1 in jth diagonal entry. Let A be an element of SU (n)
which commutes with T. Then, if ti ∈ Tj we get,
tj Aej = Atj ej = Aej .
Thus, Aej is left fixed by Tj and so Aej = λj ej . Since A ∈ U (n) all its eigenvalues
are complex numbers of absolute value 1 so λj = eiϕj for some ϕj . Thus,
 iϕ

e 1
0


..
A=
.
.
0
eiϕ1
And therefore, A ∈ T, that means, T is a maximal abelian subgroup of U (n).
Proposition 2.6. The maximal torus of SU (n) is,
 iθ

e 1
0


..
T = {
 : θ1 + ... + θn = 0}.
.
iθ
1
0
e
Proof. To see that this is a torus, we use the isomorphism given
 iθ

 i(θ −θ )
e 1
0
e 1 n
0



..
..

→
.
.
iθ
1
0
e
0
ei(θ1 −θn )
by



Centralizers of maximal tori in the classical group SU (n)
37
which maps T onto the maximal torus in U (n − 1). The argument given for U (n)
now shows that T is a maximal abelian subgroup in the case n ≥ 3. The group
SU (2) has already been treated as a spacial case.
We now define some notations. Firstly, a few conventions, we will consider the
standard Tn to be subgroups of Cn in the usual way,
T n = {(z1 , z2 , ..., zn ) ∈ Cn : |zi | = 1, ∀i}.
We denote by exp the map from Rn to T n given by t → exp(2πit) (componentwise exponentiation). Define the map diag : Cn −→ M at(n, C) by,

 

z1
0
z1

 

..
diag  ...  = 
.
.
zn
0
zn
We will identify M at(n, C) with its representation in M at(2n, R) induced by,
(
)
a −b
a + ib 7−→
.
b a
All homomorphisms from T k to T n are of the form exp(t) 7−→ exp(M t) for some
n × k matrix M with integer entries. For all such matrices M , get
φM : T k −→ Tn
exp(t) 7−→ diag(exp(M t))
and let ΦM = Im(φM ). All tori in Tn are of this form. Note that ΦM is a K-torus
if and only if M has rank k (though φM need not be faithful).
Lemma 2.7. If M1 and M2 be n × k matrices with integer entries and M2 = QM1 ,
where Q is a permutation matrix, then Z(ΦM2 ) = QZ(ΦM1 )Q−1 .
Proof. See [8].
Definition 2.8. A matrix M is in PM-block form ( the ”PM” stands for ”Plus or
minus”) if it can be divided horizontally into blocks,


Block1
 Block2 



,
..


.
Blockr
satisfying the following conditions,
(
)
A
• Each block can be subdivided horizontally into sub-blocks ,
called the
B
top half and the bottom half, such that A is nonempty, all rows in A are equal,
and if B is nonempty, then all rows in B are equal and each row in B equals
-1 times each row in A.
38
V. Najjari
• No row is equal to ±1 times a row from a different block.
• if there is a zero block, it is the bottom one (Block r).
Each block satisfying these conditions will be called a PM-block.
Definition 2.9. For any sets of matrices S1 ⊂ M at(n1 , C) and S2 ⊂ M at(n2 , C)
define,
(
)
A1 0
S1 ×l S2 = {
: A1 ∈ S1 andA2 ∈ S2 }.
0 A2

M1


Note that if M =  ... , then ΦM = ΦM1 ×l ΦM2 ×l ... ×l ΦMr .
Mr

Proposition 2.10. If M has integer entries and is in PM-block form, with PMblocks M1 , M2 , ..., Mr , then
Z(ΦM ) = Z(ΦM1 ) ×l Z(ΦM2 ) ×l ... ×l Z(ΦMr )
.
Proof. See [8].
Note that for every n×n matrix M , there exists some row-permutation matrix Q
such that QM is in PM-block form. Thus Lemma 2.7 and the preceding proposition
reduce our problem to one of finding the centralizers of ΦM for matrices M consisting
of only one PM-block.
3
Computation of the centralizers
We now find the centralizers of ΦM such that M consist of only one PM-block. In
general, matrices in the centralizer of ΦM won’t be in SU (n) however, we will find
conjugates that are. We need to define some new families of matrices.(See [8]).
Definition 3.1. For any p < n and q = n − p, let Fp,q ∈ SU (n) be the matrix in
which the upper left hand block is the 2p × 2p identity matrix, the remainder of the
diagonal consists of alternating 1′ s and −1′ s, and the rest of the matrix is zero.
T = F −1 .
Note that Fp,q = Fp,q
p,q
Lemma 3.2. Let M be the n × 1 matrix (m, m, · · · , m, −m, −m, · · · , −m)T for
some non-zero integer m, where there are p entries of m and q entries of −m. Then
−1 ⊂ GL(n, C).
Fp,q Z(ΦM )Fp,q
Proof. See [8].
39
Definition 3.3. For any p ≤ n and q = n − p, let Ip,q be the complex n × n matrix,
(
)
Ip
0
0 −Iq
where Ip is the p × p identity matrix.
−1 = −iI .
Note that, Fp,q Jn Fp,q
p,q
Definition 3.4. Let U (p, q) = {A ∈ GL(n, C)|A∗ Ip,q A = Ip,q }. The matrices in
there groups are called pseudounitary.
Note that U (0, n) = U (0, n) = U (n), are the unitary groups.
−1 .
Lemma 3.5. The pseudounitary group U (p, q) = GL(n, C) ∩ Fp,q SU (n)Fp,q
Proof. Note that this is a special case of the standard result that U (n) = GL(n, C)∩
SP (2n, R), (see [6]).
−1 U (p, q)F .
Definition 3.6. For every p and q, let W (p, q) = Fp,q
p,q
Note that W (n, 0) = W (0, n) = U (n).
Theorem 3.7. Let M be an n × k matrix, with integer entries, in PM-block form.
Let r be the number of PM-blocks in M , and let ni be the number of rows in the ith
PM-block, pi the number of rows in the top half and qi = ni − pi the number of rows
in the bottom half. If all rows of M are nonzero, then,
Z(ΦM ) = W (p1 , q1 ) ×l ... ×l W (pr , qr ).
And if M does contain some zero rows, then
Z(ΦM ) = W (p1 , q1 ) ×l ... ×l W (pr−1 , qr−1 ) ×l SU (n).
The proof will follow from the following result (also see [8]).
Proposition 3.8. Suppose M is a nonzero n × k matrix with integer entries which
consists of only one PM-block with p rows in the top half and q rows in the bottom
half. Then Z(ΦM ) = W (p, q).
Proof. The rank of M is 1, so ΦM is also the image of a faithful homomorphism
from T1 to SU (n). So without loss of generality, we will assume that M has only
one column, in other words, that M = (m, ..., m, −m, ..., −m)T for some nonzero
integer m. We can now easily prove that Z(ΦM ) ⊂ W (p, q). Lemma 3.2 shows
−1 ⊂ GL(n, C). Hence by Lemma 3.5 we have, F Z(Φ )F −1 ⊂
that, Fp,q Z(ΦM )Fp,q
p,q
M
p,q
−1 U (p, q)F , which equals W (p, q) by definition.
U (p, q), So Z(ΦM ) ⊂ Fp,q
p,q
In order to prove the inclusion, W (p, q) ⊂ Z(ΦM ), note that for every t ∈ R,
( 2πimt
)
e
Ip
0
−1
−1
Fp,q φM (exp(t))Fp,q = Fp,q
Fp,q
0
e−2πimt Ip
40
V. Najjari
= e2πimt In ,
−1 ∈
which commutes with everything in GL(n, C). Let B ∈ W (p, q). Then Fp,q BFp,q
−1 commutes with every element of F Φ F −1 ,
GL(n, C) by definition. So Fp,q BFp,q
p,q M p,q
and hence B commutes with every element of ΦM . Therefore W (p, q) ⊂ Z(ΦM ).
Proof. (proof of Theorem 3.7). Let the PM-blocks of M be M1 , ..., Mr . By proposition 2.10,
Z(ΦM ) = Z(ΦM1 ) ×l ... ×l Z(ΦMr ).
For each nonzero Mi , Proposition 3.8 shows that Z(ΦMi ) = W (pi , qi ). If all of
the PM-blocks are nonzero, then
Z(ΦM ) = W (p1 , q1 ) ×l ... ×l W (pr , qr ).
By the definition of PM-block form, the only block that can be zero is the last one,
Mr . Suppose that Mr is zero, Then ΦMr = {I2n }. So Z(ΦMr ) = SU (n), and hence,
Z(ΦM ) = W (p1 , q1 ) ×l ... ×l W (pr−1 , qr−1 ) ×l SU (n).
References
[1] T. Brocker and Dieck, Representation of Compact Lie Groups, New Mexico,
USA, 1991.
[2] T. Brocker and T. Tom Dieck, Representation of compact Groups, SpringerVerlag, 1985.
[3] H. D. Fegan, Introduction to compact lie groups, World Scientific, 1991.
[4] E. Lerman and S. Tolman, Hamiltonian torus actions on symplectic orbifolds
and toric varieties, Trans. Amer. Math. Soc. 349(1997), 4201-4230.
[5] J. E. Marsden and T. S. Ratiu, Introduction to Mechanics and Symmetry, 2nd
ed, Springer-Verlag, 1999.
[6] D. Mcduff and D. Salamon, Introduction to Symplectic Topology,
University Press, December 1995.
Oxford
[7] V. Najjari. Centralizers of maximal tori in the classical group SO(2n, R),
General Mathematics, (4) 2012, 57–62.
[8] T. Schmah, Torus action on symplectic orbi-spaces, Proc. Amer. Math. Soc.
129, 2000.
[9] S.F. Singer, J. Talvacchina and N. Watson, Nontoric hamiltonian circle actions
on four dimensional symplectic orbifolds, Proc. Amer. Math. Soc. 127, 1999.
[10] N. Watson, Symplectic vector orbi-spaces with torus actions,
Haverford College, 1991.
Senior Paper,
Vadoud Najjari
Islamic Azad University
Maragheh Branch, Maragheh, Iran
e-mail: vnajjari@iau-maragheh.ac.ir; vnajjari@gazi.edu.tr
41
A note on differential superordinations using a
generalized Sălăgean and Ruscheweyh operators 1
Alina Alb Lupaş
Abstract
In the present paper we define a new operator using the generalized Sălăgean
and Ruscheweyh operators. Denote by DRλm the Hadamard product of the
generalized Sălăgean operator Dλm and the Ruscheweyh operator Rm , given
by DRλm : An → An , DRλm f (z) = (Dλm ∗ Rm ) f (z) and An = {f ∈ H(U ),
f (z) = z + an+1 z n+1 + . . . , z ∈ U } is the class of normalized analytic functions.
We study some differential superordinations regarding the operator DRλm .
2010 Mathematics Subject Classification: 30C45, 30A20, 34A40.
Key words and phrases: Differential superordination, convex function, best
subordinant, differential operator.
1
Introduction
Denote by U the unit disc of the complex plane U = {z ∈ C : |z| < 1} and H(U )
the space of holomorphic functions in U .
Let
An = {f ∈ H(U ), f (z) = z + an+1 z n+1 + . . . , z ∈ U }
and
H[a, n] = {f ∈ H(U ), f (z) = a + an z n + an+1 z n+1 + . . . , z ∈ U }
for a ∈ C and n ∈ N.
If f and g are analytic functions in U , we say that f is superordinate to g, written
g ≺ f , if there is a function w analytic in U , with w(0) = 0, |w(z)| < 1, for all z ∈ U
such that g(z) = f (w(z)) for all z ∈ U . If f is univalent, then g ≺ f if and only if
f (0) = g(0) and g(U ) ⊆ f (U ).
1
Received 8 March, 2010
Accepted for publication (in revised form) 19 July, 2010
43
44
A. Alb Lupaş
Let ψ : C2 × U → C and h analytic in U . If p and ψ (p (z) , zp′ (z) ; z) are
univalent in U and satisfies the (first-order) differential superordination
h(z) ≺ ψ(p(z), zp′ (z); z),
(1)
for z ∈ U,
then p is called a solution of the differential superordination. The analytic function
q is called a subordinant of the solutions of the differential superordination, or more
simply a subordinant, if q ≺ p for all p satisfying (1).
An univalent subordinant qe that satisfies q ≺ qe for all subordinants q of (1) is
said to be the best subordinant of (1). The best subordinant is unique up to a
rotation of U .
Definition 1 (Al Oboudi [4]) For f ∈ An , λ ≥ 0 and m, n ∈ N, the operator Dλm is
defined by Dλm : An → An ,
Dλ0 f (z) = f (z)
Dλ1 f (z) = (1 − λ) f (z) + λzf ′ (z) = Dλ f (z)
...
Dλm f (z)
(
)
= (1 − λ) Dλm−1 f (z) + λz (Dλm f (z))′ = Dλ Dλm−1 f (z) ,
for z ∈ U.
Remark 1 If f ∈ An and f (z) = z +
Dλm f
(z) = z +
∞
∑
∑∞
j=n+1 aj z
j,
then
[1 + (j − 1) λ]m aj z j , for z ∈ U.
j=n+1
Remark 2 For λ = 1 in the above definition we obtain the Sălăgean differential
operator [7].
Definition 2 (Ruscheweyh [6]) For f ∈ An , m, n ∈ N, the operator Rm is defined
by Rm : An → An ,
R0 f (z) = f (z)
R1 f (z) = zf ′ (z)
...
(m + 1) R
m+1
f (z) = z (Rm f (z))′ + mRm f (z) ,
Remark 3 If f ∈ An , f (z) = z +
∑∞
Rm f (z) = z +
j=n+1 aj z
∞
∑
j=n+1
j,
then
m
Cm+j−1
aj z j , z ∈ U.
z ∈ U.
A note on differential superordinations ...
45
Definition 3 ([5]) We denote by Q the set of functions that are analytic and
injective on U \E (f ), where E (f ) = {ζ ∈ ∂U : lim f (z) = ∞}, and f ′ (ζ) ̸= 0
z→ζ
for ζ ∈ ∂U \E (f ). The subclass of Q for which f (0) = a is denoted by Q (a).
We will use the following lemmas.
Lemma 1 (Miller and Mocanu [5]) Let h be a convex function with h(0) = a, and
let γ ∈ C\{0} be a complex number with Re γ ≥ 0. If p ∈ H[a, n] ∩ Q, p(z) + γ1 zp′ (z)
is univalent in U and
h(z) ≺ p(z) +
1 ′
zp (z),
γ
for z ∈ U,
then
where q(z) = nzγγ/n
best subordinant.
∫z
0
q(z) ≺ p(z),
for z ∈ U,
h(t)tγ/n−1 dt, for z ∈ U. The function q is convex and is the
Lemma 2 (Miller and Mocanu [5]) Let q be a convex function in U and let
h(z) = q(z) + γ1 zq ′ (z), for z ∈ U, where Re γ ≥ 0.
If p ∈ H [a, n] ∩ Q, p(z) + γ1 zp′ (z) is univalent in U and
q(z) +
1
1 ′
zq (z) ≺ p(z) + zp′ (z) ,
γ
γ
for z ∈ U,
then
where q(z) =
2
γ
nz γ/n
∫z
0
q(z) ≺ p(z),
for z ∈ U,
h(t)tγ/n−1 dt, for z ∈ U. The function q is the best subordinant.
Main Results
Definition 4 ([2]) Let λ ≥ 0 and m, n ∈ N. Denote by DRλm the operator given by
the Hadamard product (the convolution product) of the generalized Sălăgean operator
Dλm and the Ruscheweyh operator Rm , DRλm : An → An ,
DRλm f (z) = (Dλm ∗ Rm ) f (z) .
Remark 4 If f ∈ An and f (z) = z +
DRλm f (z) = z +
∞
∑
∑∞
j=n+1 aj z
j,
then
m
Cm+j−1
[1 + (j − 1) λ]m a2j z j , for z ∈ U.
j=n+1
Remark 5 For λ = 1 we obtain the Hadamard product SRm [1] of the Sălăgean
operator S m and Ruscheweyh operator Rm .
46
A. Alb Lupaş
Theorem 1 Let h be a convex function, h(0) = 1. Let λ ≥ 0, m, n ∈ N, f ∈ An and
m(1−λ)
m+1
suppose that (mλ+1)z
DRλm+1 f (z)− (mλ+1)z
DRλm f (z) is univalent and (DRλm f (z))′ ∈
H [1, n] ∩ Q. If
h(z) ≺
(2)
m+1
m (1 − λ)
DRλm+1 f (z) −
DRλm f (z) ,
(mλ + 1) z
(mλ + 1) z
for z ∈ U,
then
q(z) ≺ (DRλm f (z))′ ,
where q(z) =
subordinant.
∫z
mλ+1
mλ+1
nλz nλ
0
h (t) t
(m−n)λ+1
nλ
for z ∈ U,
dt. The function q is convex and it is the best
Proof. With notation
∑
p (z) = (DRλm f (z))′ = 1 + ∞
Cm
[1 + (j − 1) λ]m ja2j z j−1 t and p (0) = 1,
∑∞j=n+1 jm+j−1
we obtain for f (z) = z + j=n+1 aj z ,
)
(
m+1
m
p (z)+zp′ (z) = m+1
f (z)− m − 1 + λ1 (DRλm f (z))′ − m(1−λ)
λz DRλ
λz DRλ f (z) and
m(1−λ)
λ
m+1
DRλm+1 f (z) − (mλ+1)z
DRλm f (z) . Evidently
p (z) + mλ+1
zp′ (z) = (mλ+1)z
p ∈ H[1, n].
Then (2) becomes
h(z) ≺ p (z) +
λ
zp′ (z) ,
mλ + 1
for z ∈ U.
By using Lemma 1 for γ = m + λ1 , we have
q(z) ≺ p(z),
where q(z) =
subordinant.
for z ∈ U,
∫z
mλ+1
mλ+1
nλz nλ
0
h (t) t
i.e.
(m−n)λ+1
nλ
q(z) ≺ (DRλm f (z))′ ,
for z ∈ U,
dt. The function q is convex and it is the best
Corollary 1 ([3]) Let h be a convex function, h(0) = 1. Let m, n ∈ N, f ∈ An
m
and suppose that z1 SRm+1 f (z) + m+1
z (SRm f (z))′′ is univalent and (SRm f (z))′ ∈
H [1, n] ∩ Q. If
(3)
h(z) ≺
1
m
SRm+1 f (z) +
z (SRm f (z))′′ ,
z
m+1
for z ∈ U,
then
q(z) ≺ (SRm f (z))′ ,
where q(z) =
nant.
1
1
nz n
∫z
0
for z ∈ U,
h(t)t n −1 dt. The function q is convex and it is the best subordi1
47
λ
Theorem 2 Let q be convex in U and let h be defined by h (z) = q (z)+ mλ+1
zq ′ (z),
m(1−λ)
m+1
λ ≥ 0, m, n ∈ N. If f ∈ An , suppose that (mλ+1)z
DRλm+1 f (z)− (mλ+1)z
DRλm f (z) is
′
univalent and (DRλm f (z)) ∈ H [1, n]∩Q and satisfies the differential superordination
(4) h(z) = q (z)+
λ
m+1
m (1 − λ)
zq ′ (z) ≺
DRλm+1 f (z)−
DRλm f (z) ,
mλ + 1
(mλ + 1) z
(mλ + 1) z
for z ∈ U, then
where q(z) =
q(z) ≺ (DRλm f (z))′ ,
∫z
mλ+1
mλ+1
nλz nλ
h (t) t
0
(m−n)λ+1
nλ
for z ∈ U,
dt. The function q is the best subordinant.
Proof.
∑
m
m
2 j−1 .
Let p (z) = (DRλm f (z))′ = 1 + ∞
j=n+1 Cm+j−1 [1 + (j − 1) λ] jaj z
Differentiating, we obtain
(
)
m+1
n
p (z) + zp′ (z) = m+1
f (z) − m − 1 + λ1 (DRλm f (z))′ − m(1−λ)
λz DRλ
λz DRλ f (z) and
m(1−λ)
λ
m+1
p (z) + mλ+1
zp′ (z) = (mλ+1)z
DRλm+1 f (z) − (mλ+1)z
DRλm f (z) , for z ∈ U and (4)
becomes
q(z) +
λ
λ
zq ′ (z) ≺ p(z) +
zp′ (z) ,
mλ + 1
mλ + 1
for z ∈ U.
Using Lemma 2 for γ = m + λ1 , we have q(z) ≺ p(z), for z ∈ U, i.e.
∫
mλ + 1
q(z) =
nλz
z
h (t) t
mλ+1
nλ
(m−n)λ+1
nλ
dt ≺ (DRλm f (z))′ , for z ∈ U,
0
and q is the best subordinant.
Corollary 2 ([3]) Let q be convex in U and let h be defined by h (z) = q (z)+zq ′ (z) .
m
If m, n ∈ N, f ∈ An , suppose that z1 SRm+1 f (z) + m+1
z·
′′
′
m
m
(SR f (z)) is univalent, (SR f (z)) ∈ H [1, n] ∩ Q and satisfies the differential
superordination
(5)
h(z) = q (z) + zq ′ (z) ≺
then
where q(z) =
1
1
nz n
∫z
0
1
m
SRm+1 f (z) +
z (SRm f (z))′′ ,
z
m+1
q(z) ≺ (SRm f (z))′ ,
for z ∈ U,
for z ∈ U,
h(t)t n −1 dt. The function q is the best subordinant.
1
Theorem 3 Let h be a convex function, h(0) = 1. Let λ ≥ 0, m, n ∈ N, f ∈ An
m f (z)
DRλ
and suppose that (DRλm f (z))′ is univalent and
∈ H [1, n] ∩ Q. If
z
(6)
h(z) ≺ (DRλm f (z))′ ,
for z ∈ U,
48
A. Alb Lupaş
then
q(z) ≺
where q(z) =
nant.
1
1
nz n
∫z
0
DRλm f (z)
,
z
for z ∈ U,
Proof. Consider
∑
m 2 j
m
z+ ∞
DRλm f (z)
j=n+1 Cm+j−1 [1 + (j − 1) λ] aj z
=
p (z) =
z
z
∞
∑
m
=1+
Cm+j−1
[1 + (j − 1) λ]m a2j z j−1 .
j=n+1
Evidently p ∈ H[1, n].
We have p (z) + zp′ (z) = (DRλm f (z))′ , for z ∈ U .
Then (6) becomes
h(z) ≺ p(z) + zp′ (z),
for z ∈ U.
By using Lemma 1 for γ = 1, we have
q(z) ≺ p(z),
where q(z) =
dinant.
1
1
nz n
∫z
0
for z ∈ U,
i.e.
q(z) ≺
DRλm f (z)
,
z
for z ∈ U,
h(t)t n −1 dt. The function q is convex and it is the best subor1
Corollary 3 ([3]) Let h be a convex function, h(0) = 1. Let m, n ∈ N, f ∈ An and
m
suppose that (SRm f (z))′ is univalent and SR zf (z) ∈ H [1, n] ∩ Q. If
h(z) ≺ (SRm f (z))′ ,
(7)
then
q(z) ≺
where q(z) =
nant.
1
1
nz n
∫z
0
SRm f (z)
,
z
for z ∈ U,
for z ∈ U,
Theorem 4 Let q be convex in U and let h be defined by h (z) = q (z)+zq ′ (z) . If λ ≥
0, m, n ∈ N, f
∈ An , suppose that (DRλm f (z))′ is univalent,
m f (z)
DRλ
z
∈ H [1, n] ∩ Q and satisfies the differential superordination
(8)
h(z) = q (z) + zq ′ (z) ≺ (DRλm f (z))′ ,
then
q(z) ≺
where q(z) =
1
1
nz n
∫z
0
DRλm f (z)
,
z
for z ∈ U,
for z ∈ U,
1
49
Proof. Let
∑
m 2 j
m
z+ ∞
DRλm f (z)
j=n+1 Cm+j−1 [1 + (j − 1) λ] aj z
p (z) =
=
z
z
∞
∑
m
=1+
Cm+j−1
[1 + (j − 1) λ]m a2j z j−1 .
j=n+1
Differentiating, we obtain p(z) + zp′ (z) = (DRλm f (z))′ , for z ∈ U and (8)
becomes
q(z) + zq ′ (z) ≺ p(z) + zp′ (z) , for z ∈ U.
Using Lemma 2 for γ = 1, we have
q(z) ≺ p(z), for z ∈ U,
i.e.
∫
1
q(z) =
1
nz n
z
h(t)t n −1 dt ≺
1
0
DRλm f (z)
,
z
for z ∈ U, and q is the best subordinant.
Corollary 4 ([3]) Let q be convex in U and let h be defined by h (z) = q (z)+zq ′ (z) .
m
If m, n ∈ N, f ∈ An , suppose that (SRm f (z))′ is univalent, SR zf (z) ∈ H [1, n] ∩ Q
and satisfies the differential superordination
h(z) = q (z) + zq ′ (z) ≺ (SRm f (z))′ ,
(9)
then
q(z) ≺
where q(z) =
1
1
nz n
∫z
0
SRm f (z)
,
z
for z ∈ U,
for z ∈ U,
1
Theorem 5 Let h(z) = 1+(2β−1)z
be a convex function in U , where 0 ≤ β < 1.
1+z
Let λ ≥ 0, m, n ∈ N, f ∈ An and suppose that (DRλm f (z))′ is univalent and
m f (z)
DRλ
z
∈ H [1, n] ∩ Q. If
h(z) ≺ (DRλm f (z))′ ,
(10)
for z ∈ U,
then
DRλm f (z)
,
for z ∈ U,
z
∫ z t n1 −1
where q is given by q(z) = 2β − 1 + 2(1−β)
1
0 1+t dt, for z ∈ U. The function q is
nz n
convex and it is the best subordinant.
q(z) ≺
Proof. Following the same steps as in the proof of Theorem 3 and considering
m f (z)
DRλ
p(z) =
, the differential superordination (10) becomes
z
h(z) =
1 + (2β − 1)z
≺ p(z) + zp′ (z),
1+z
for z ∈ U.
50
A. Alb Lupaş
By using Lemma 1 for γ = 1, we have q(z) ≺ p(z), i.e.,
∫ z
∫ z
1
1
1
1
1 + (2β − 1) t
−1
q(z) =
h (t) t n dt =
t n −1
dt
1
1
1+t
nz n 0
nz n 0
1
∫
DRλm f (z)
2(1 − β) z t n −1
= 2β − 1 +
dt ≺
, for z ∈ U.
1
z
0 1+t
nz n
The function q is convex and it is the best subordinant.
Theorem 6 Let (
h be a convex
)′ function, h(0) = 1. Let λ ≥ 0, m, n ∈ N, f ∈ An
m+1
DRm+1 f (z)
zDRλ f (z)
is univalent and DRλm f (z) ∈ H [1, n] ∩ Q. If
and suppose that
DRm f (z)
λ
λ
(
h(z) ≺
(11)
zDRλm+1 f (z)
DRλm f (z)
then
q(z) ≺
where q(z) =
dinant.
1
1
nz n
∫z
0
)′
,
DRλm+1 f (z)
,
DRλm f (z)
for z ∈ U,
for z ∈ U,
h (t) t n −1 dt. The function q is convex and it is the best subor1
Proof. Consider
∑
m+1 2 j
m+1
z+ ∞
aj z
DRλm+1 f (z)
j=n+1 Cm+j [1 + (j − 1) λ]
∑
=
p (z) =
∞
m 2 j
m
m
DRλ f (z)
z + j=n+1 Cm+j−1 [1 + (j − 1) λ] aj z
∑∞
m+1
1 + j=n+1 Cm+j
[1 + (j − 1) λ]m+1 a2j z j−1
∑
=
m 2 j−1 .
m
1+ ∞
j=n+1 Cm+j−1 [1 + (j − 1) λ] aj z
′
′
(DRλm+1 f (z))
(DRm f (z))
We have p′ (z) = DR
− p (z) · DRλm f (z) . Then
m f (z)
λ
(
p (z) + zp′ (z) =
λ
zDRλm+1 f (z)
DRλm f (z)
)′
.
Then (11) becomes
h(z) ≺ p(z) + zp′ (z),
for z ∈ U.
By using Lemma 1 for γ = 1, we have
q(z) ≺ p(z),
where q(z) =
dinant.
1
1
nz n
∫z
0
for z ∈ U,
i.e.
DRλm+1 f (z)
,
q(z) ≺
DRλm f (z)
for z ∈ U,
51
Corollary 5(([3]) Let h )be a convex function, h(0) = 1. Let m, n ∈ N, f ∈ An and
m+1 f (z) ′
m+1 f (z)
suppose that zSR
is univalent and SR
SRm f (z)
SRm f (z) ∈ H [1, n] ∩ Q. If
(
h(z) ≺
(12)
zSRm+1 f (z)
SRm f (z)
then
q(z) ≺
where q(z) =
dinant.
1
1
nz n
∫z
0
)′
for z ∈ U,
,
SRm+1 f (z)
,
SRm f (z)
for z ∈ U,
Theorem 7 Let q be convex in U and let(h be defined)by h (z) = q (z) + zq ′ (z) . If
′
m+1
m+1
zDRλ
f (z)
DRλ
f (z)
λ ≥ 0, m, n ∈ N, f ∈ An , suppose that
is
univalent,
m
DR f (z)
DRm f (z) ∈
λ
λ
H [1, n] ∩ Q and satisfies the differential superordination
)′
(
zDRλm+1 f (z)
′
(13)
h(z) = q (z) + zq (z) ≺
,
DRλm f (z)
then
q(z) ≺
where q(z) =
1
1
nz n
∫z
0
DRλm+1 f (z)
,
DRλm f (z)
for z ∈ U,
for z ∈ U,
h (t) t n −1 dt. The function q is the best subordinant.
Proof. Let p (z) =
1
m+1
DRλ
f (z)
m f (z)
DRλ
∑
1+ ∞
C m+1 [1+(j−1)λ]m+1 a2j z j−1
∑j=n+1 m+j
m 2 j−1 .
m
1+ ∞
j=n+1 Cm+j−1 [1+(j−1)λ] aj z
=
∑
C m+1 [1+(j−1)λ]m+1 a2j z j
z+ ∞
∑j=n+1 m+j
m 2 j
m
z+ ∞
j=n+1 Cm+j−1 [1+(j−1)λ] aj z
=
(
)′
m+1
zDRλ
f (z)
′
Differentiating, we obtain p(z) + zp (z) =
, for z ∈ U and (13)
DRm f (z)
λ
becomes
q(z) + zq ′ (z) ≺ p(z) + zp′ (z) ,
for z ∈ U.
Using Lemma 2 for γ = 1, we have
q(z) ≺ p(z), for z ∈ U,
i.e.
q(z) =
∫
1
z
h (t) t
1
nz n
0
1
−1
n
DRλm+1 f (z)
dt ≺
,
DRλm f (z)
for z ∈ U, and q is the best subordinant.
Corollary 6 ([3]) Let q be convex (in U and let)h be defined by h (z) = q (z)+zq ′ (z) .
m+1
′
m+1
f (z)
f (z)
If m, n ∈ N, f ∈ An , suppose that zSR
is univalent, SR
SRm f (z)
SRm f (z) ∈ H [1, n] ∩
Q and satisfies the differential superordination
(
)′
zSRm+1 f (z)
′
(14)
h(z) = q (z) + zq (z) ≺
, for z ∈ U,
SRm f (z)
52
A. Alb Lupaş
then
q(z) ≺
where q(z) =
1
1
nz n
∫z
0
SRm+1 f (z)
,
SRm f (z)
for z ∈ U,
h (t) t n −1 dt. The function q is the best subordinant.
1
1+(2β−1)z
1+z
be a convex function in U , where 0 ≤ β < 1. Let
(
)′
m+1
m+1
f (z)
f (z)
zDRλ
DRλ
λ ≥ 0, m, n ∈ N, f ∈ An and suppose that
is
univalent,
m
DR f (z)
DRm f (z) ∈
Theorem 8 Let h(z) =
λ
λ
H [1, n] ∩ Q. If
(
h(z) ≺
(15)
then
zDRλm+1 f (z)
DRλm f (z)
)′
DRλm+1 f (z)
,
q(z) ≺
DRλm f (z)
where q is given by q(z) = 2β − 1 + 2(1−β)
1
nz n
convex and it is the best subordinant.
for z ∈ U,
,
∫z
for z ∈ U,
1
t n −1
0 1+t dt,
for z ∈ U. The function q is
Proof. Following the same steps as in the proof of Theorem 6 and considering
p(z) =
m+1
DRλ
f (z)
m f (z) ,
DRλ
the differential superordination (15) becomes
h(z) =
1 + (2β − 1)z
≺ p(z) + zp′ (z),
1+z
for z ∈ U.
By using Lemma 1 for γ = 1, we have q(z) ≺ p(z), i.e.,
∫ z
∫ z
1
1
1
1
1 + (2β − 1) t
−1
n
q(z) =
h (t) t
t n −1
dt =
dt
1
1
1+t
nz n 0
nz n 0
1
∫
DRλm+1 f (z)
2(1 − β) z t n −1
= 2β − 1 +
dt
≺
, for z ∈ U.
1
DRλm f (z)
0 1+t
nz n
The function q is convex and it is the best subordinant.
References
[1] Alina Alb Lupaş, A note on differential subordinations using Sălăgean and
Ruscheweyh operators, Romai Journal, Proceedings of CAIM 2009, (to appear).
[2] Alina Alb Lupaş, A note on differential subordinations using a generalized
Sălăgean operator and Ruscheweyh operator, Acta Universitatis Apulensis, Proceedings of ICTAMI 2009, Alba Iulia, 29-35.
53
[3] Alina Alb Lupaş, A note on differential superordinations using Sălăgean and
Ruscheweyh operators, Acta Universitatis Apulensis, No. 24. 2010, 201-209.
[4] F.M. Al-Oboudi, On univalent functions defined by a generalized Sălăgean
operator, Ind. J. Math. Math. Sci., 2004, no.25-28, 1429-1436.
[5] S.S. Miller, P.T. Mocanu, Subordinants of Differential Superordinations,
Complex Variables, vol. 48, no. 10, 2003, 815-826.
[6] St. Ruscheweyh, New criteria for univalent functions, Proc. Amer. Math. Soc.,
49 (1975), 109-115.
[7] G.St. Sălăgean, Subclasses of univalent functions, Lecture Notes in Math.,
Springer Verlag, Berlin, 1013 (1983), 362-372.
Alina Alb Lupaş
University of Oradea
Department of Mathematics and Computer Science
1 Universitatii Street, 410087 Oradea, Romania
E-mail: dalb@uoradea.ro
Subordination results for certain classes of analytic
functions defined by Dziok-Srivastava operator 1
M. K. Aouf, A. Shamandy, A. O. Mostafa and E. A. Adwan
Abstract
In this paper, we drive several interesting subordination results of a certain
classes of analytic functions defined by Dziok-Srivastava operator.
Key words and phrases: Analytic functions, starlike functions, convex
functions, Hadamard products, Dziok-Srivastava operator, subordination, factor
sequence.
1
Introduction
Let A denote the class of functions of the form
(1.1)
f (z) = z +
∞
∑
an z n ,
n=2
which are analytic in the open unit disc U = {z : |z| < 1}. Let K(α) and S ∗ (α)
denote the subclasses of A which are, respectively, convex and starlike functions of
order α, 0 ≤ α < 1. For convenience, we write K(0) = K and S ∗ (0) = S ∗ (see [24]
).
Let M (β) be the subclasses of A consisting of functions f (z) which satisfy the
inequality:
{
(1.2)
Re
′
zf (z)
f (z)
1
}
<β
Received 10 June, 2010
Accepted for publication (in revised form) 21 October, 2010
55
56
M.K.Aouf, A.Shamandy, A.O.Mostafa, E.A.Adwan
′
for some β(β > 1). Replacing f (z) in (1.2) by zf (z) we have the condition:
{
}
zf ′′ (z)
Re 1 + ′
<β
f (z)
required for the function f to be in the subclass N (β) (β > 1) (see [17], [19], [21]
and [23] ).
The Hadamard product (or convolution) (f ∗g)(z) of the functions f (z) and g(z),
∞
∑
bn z n , is defined
that is, if f (z) is given by (1.1) and g(z) is given by g(z) = z +
n=2
by:
(f ∗ g)(z) = z +
∞
∑
an bn z n = (g ∗ f )(z).
n=2
If f and g are analytic functions in U , we say that f is subordinate to g, written
f ≺ g if there exists a Schwarz function w, which (by definition) is analytic in U
with w(0) = 0 and |w(z)| < 1 for all z ∈ U, such that f (z) = g(w(z)), z ∈
U. Furthermore, if the function g is univalent in U, then we have the following
equivalence (cf., e.g., [4] and [13]):
f (z) ≺ g(z) ⇔ f (0) = g(0) and f (U ) ⊂ g(U ).
Definition 1 ( Subordinating Factor Sequence ) [25]. A sequence {bn }∞
n=1 of complex
numbers is said to be a subordinating factor sequence if, whenever f of the form (1.1)
is analytic, univalent and convex in U, we have the subordination given by
(1.3)
∞
∑
bn an z n ≺ f (z)
(z ∈ U ; a1 = 1).
n=1
For positive real parameters α1 , ..., αq and β1 , ..., βs (βj ∈ C\Z − ,
= 0, −1, −2, ..., j = 1, 2, ..., s) , the generalized hypergeometric function
q Fs (α1 , ..., αq ; β1 , ..., βs ; z) is defined by
Z−
q Fs (α1 , ....., αq ; β1 , ....., βs ; z)
=
∞
∑
(α1 )n ...(αq )n n
z
(β1 )n ...(βs )n n!
n=0
(q ≤ s + 1; s, q ∈ N0 = N ∪ {0}, N = {1, 2, .........}; z ∈ U ),
where (θ)n , is the Pochhammer symbol defined in terms of the Gamma function
Γ, by
{
Γ(θ + n)
1
(n = 0)
(θ)n =
=
θ(θ + 1)....(θ + n − 1)
(n ∈ N ).
Γ(θ)
For the function h(α1 , ...., αq ; β1 , ...βs ; z) = zq Fs (α1 , ....., αq ; β1 , ....., βs ; z), the DziokSrivastava linear operator ( see [8] and [9] ) Hq,s (α1 , ....., αq ; β1 , ..., βs ) : A −→ A, is
defined by the Hadamard product as follows:
Subordination results for certain classes of analytic functions...
57
Hq,s (α1 , ....., αq ; β1 , ....., βs )f (z) = h(α1 , ...., αq ; β1 , ...βs ; z) ∗ f (z)
∞
∑
=z+
Ψn an z n (z ∈ U ),
(1.4)
n=2
where
(1.5)
Ψn =
(α1 )n−1 .......(αq )n−1
.
(β1 )n−1 ...(βs )n−1 (n − 1)!
For brevity, we write
(1.6)
Hq,s (α1 , ....., αq ; β1 , ....., βs ; z)f (z) = Hq,s (α1 )f (z).
For 0 ≤ λ ≤ 1, 0 ≤ α < 1, αi > 0, i = 1, .....q, βj > 0, j = 1, ..., s, q, s ∈ N0 and for
all z ∈ U, let SHq,s ([α1 ]; λ, α) denote the subclass of A consisting of functions f (z)
of the form (1.1) and satisfying the analytic criterion:
{
(1.7)
Re
′
}
′′
z (Hq,s (α1 )f (z)) + λz 2 (Hq,s (α1 )f (z))
(1 − λ) (Hq,s (α1 )f (z)) + λz (Hq,s (α1 )f (z))
> α,
′
and for 0 ≤ λ ≤ 1, β > 1, αi > 0, i = 1, .....q, βj > 0, j = 1, ..., s, q, s ∈ N0 and for all
z ∈ U, let M Hq,s ([α1 ]; λ, β) denote the subclass of A consisting of functions f (z) of
the form (1.1) and satisfying the analytic criterion:
{
(1.8)
Re
′
}
′′
′
< β.
We note that for suitable choices of q, s, λ, α and β, we obtain the following subclasses
studied by various authors.
(1) For q = 2, s = 1, α1 = a (a > 0) , α2 = 1 and β1 = c (c > 0) in (1.7) and (1.8),
the class SH2,1 ([a, 1; c]; λ, α) reduces to the class SL(a, c; λ, α)
{
=
{
f ∈ A : Re
′
}
′′
z (L(a.c)f (z)) + λz 2 (L(a.c)f (z))
(1 − λ) (L(a.c)f (z)) + λz (L(a.c)f (z))
> α, 0 ≤ λ ≤ 1,
′
0 ≤ α < 1, a > 0, c > 0, z ∈ U } , (see [14], with β = 0)
and the class M H2,1 ([a, 1; c]; λ, β) reduces to the class M L(a, c; λ, β)
{
=
{
f ∈ A : Re
′
z (L(a.c)f (z)) + λz 2 (L(a.c)f (z))
}
′′
′
(1 − λ) (L(a.c)f (z)) + λz (L(a.c)f (z))
β > 1, a > 0, c > 0, z ∈ U } ,
< β, 0 ≤ λ ≤ 1,
58
where L(a, c) is the Carlson - Shaffer operator ( see [5] );
(2) For q = 2, s = 1, α1 = γ + 1 (γ > −1) and α2 = β1 = 1 in (1.7) and (1.8), the
class SH2,1 ([γ + 1, 1; 1]; λ, α) reduces to the class SW (γ; λ, α)
{
{
f ∈ A : Re
=
′
z (Dγ f (z)) + λz 2 (Dγ f (z))
}
′′
(1 − λ) (Dγ f (z)) + λz (Dγ f (z))
> α, 0 ≤ λ ≤ 1,
′
0 ≤ α < 1, γ > −1, z ∈ U } ,
and the class M H2,1 ([γ + 1, 1; 1]; λ, β) reduces to the class M W (γ; λ, β)
{
{
f ∈ A : Re
=
′
}
′′
z (Dγ f (z)) + λz 2 (Dγ f (z))
′
(1 − λ) (Dγ f (z)) + λz (Dγ f (z))
< β, 0 ≤ λ ≤ 1,
β > 1, γ > −1, z ∈ U } ,
where Dγ (γ > −1) is the Ruscheweyh derivative operator ( see [22] );
(3) For q = 2, s = 1, α1 = v + 1 (v > −1) , α2 = 1 and β1 = v + 2 in (1.7) and (1.8),
the class SH2,1 ([v + 1, 1; v + 2]; λ, α) reduces to the class Sζ(v; λ, α)
{
{
f ∈ A : Re
=
′
′′
z (Jv f (z)) + λz 2 (Jv f (z))
(1 − λ) (Jv f (z)) + λz(Jv f (z))′
}
> α, 0 ≤ λ ≤ 1,
0 ≤ α < 1, v > −1, z ∈ U } ,
and the class M H2,1 ([v + 1, 1; v + 2]; λ, β) reduces to the class M ζ(v, λ, β)
{
{
f ∈ A : Re
=
′
′′
z (Jv f (z)) + λz 2 (Jv f (z))
(1 − λ) (Jv f (z)) + λz(Jv f (z))′
}
< β, 0 ≤ λ ≤ 1,
β > 1, v > −1, z ∈ U } ,
where Jv f (z) is the generalized Bernardi - Libera - Livingston operator ( see [3],
[11] and [12] );
(4) For q = 2, s = 1, α1 = 2, α2 = 1 and β1 = 2 − µ (µ ̸= 2, 3, ....) in (1.7) and (1.8),
the class SH2,1 ([2, 1; 2 − µ]; λ, α) reduces to the class SF(µ; λ, α)
{
=
f ∈ A : Re
{
′
}
′′
z (Ωµz f (z)) + λz 2 (Ωµz f (z))
(1 − λ)(Ωµz f (z)) + λz (Ωµz f (z))
′
> α, , 0 ≤ λ ≤ 1,
0 ≤ α < 1, µ ̸= 2, 3, ...., z ∈ U } ,
and the class M H2,1 ([2, 1; 2 − µ]; λ, β) reduces to the class M F(µ; λ, β)
{
{
f ∈ A : Re
=
′
59
}
′′
z (Ωµz f (z)) + λz 2 (Ωµz f (z))
(1 − λ)(Ωµz f (z)) + λz (Ωµz f (z))
< β, 0 ≤ λ ≤ 1,
′
β > 1, µ ̸= 2, 3, ...., z ∈ U } ,
where Ωµz f (z) is the Srivastava - Owa fractional derivative operator ( see [18] and
[20] );
(5) For q = 2, s = 1, α1 = µ (µ > 0) , α2 = 1 and β1 = γ + 1 (γ > −1) in (1.7) and
(1.8), the class SH2,1 ([µ, 1; γ + 1]; λ, α) reduces to the class SΥ(µ, γ; λ, α)
{
{
f ∈ A : Re
=
′
z (Iγ,µ f (z)) + λz 2 (Iγ,µ f (z))
}
′′
> α, 0 ≤ λ ≤ 1,
′
(1 − λ)(Iγ,µ f (z)) + λz (Iγ,µ f (z))
0 ≤ α < 1, µ > 0, γ > −1, z ∈ U } ,
and the class M H2,1 ([µ, 1; γ + 1]; λ, β) reduces to the class M Υ(µ, γ; λ, β)
{
=
{
f ∈ A : Re
′
}
′′
z (Iγ,µ f (z)) + λz 2 (Iγ,µ f (z))
′
(1 − λ)(Iγ,µ f (z)) + λz (Iγ,µ f (z))
< β, 0 ≤ λ ≤ 1,
β > 1, µ > 0, γ > −1, z ∈ U } ,
where Iγ,µ f (z) is the Choi-Saigo-Srivastava operator ( see [7] );
(6) For q = 2, s = 1, α1 = 2, α2 = 1 and β1 = k + 1 (k > −1) in (1.7) and (1.8), the
class SH2,1 ([2, 1; k + 1]; λ, α) reduces to the class SA(µ; λ, α)
{
=
{
f ∈ A : Re
′
z (Ik f (z)) + λz 2 (Ik f (z))
}
′′
(1 − λ)(Ik f (z)) + λz (Ik f (z))
> α, 0 ≤ λ ≤ 1,
′
0 ≤ α < 1, k > −1, z ∈ U } ,
and the class M H2,1 ([2, 1; k + 1]; λ, β) reduces to the class M A(µ, γ; λ, β)
{
=
{
f ∈ A : Re
′
}
′′
z (Ik f (z)) + λz 2 (Ik f (z))
′
(1 − λ)(Ik f (z)) + λz (Ik f (z))
< β, 0 ≤ λ ≤ 1,
β > 1, k > −1, z ∈ U } ,
where Ik f (z) is the Noor integral operator ( see [16] );
(7) For q = 2, s = 1, α1 = c (c > 0) , α2 = γ + 1 (γ > −1) and β1 = a (a > 0) in (1.7)
and (1.8), the class SH2,1 ([c, γ + 1; a]; λ, α) reduces to the class Sz(c, a, γ; λ, α)
60
{
=
{
f ∈ A : Re
′
z (I γ (a, c)f (z)) + λz 2 (I γ (a, c)f (z))
}
′′
> α, 0 ≤ λ ≤ 1,
′
(1 − λ)(I γ (a, c)f (z)) + λz (I γ (a, c)f (z))
0 ≤ α < 1, a > 0, c > 0, γ > −1, z ∈ U } ,
and the class M H2,1 ([c, γ + 1; a]; λ, β) reduces to the class M z(µ, γ; λ, β)
{
=
{
f ∈ A : Re
′
}
′′
z (I γ (a, c)f (z)) + λz 2 (I γ (a, c)f (z))
′
(1 − λ)(I γ (a, c)f (z)) + λz (I γ (a, c)f (z))
< β, 0 ≤ λ ≤ 1,
β > 1, a > 0, c > 0, γ > −1, z ∈ U } ,
where I γ (a, c)f (z) is the Cho-Kwon-Srivastava operator ( see [6] );
(8) For q = 2 and s = α1 = α2 = β1 = 1 in (1.7) and (1.8), the class SH2,1 ([1, 1;
1]; λ, α) reduces to the class T (λ, α) (0 ≤ α < 1, 0 ≤ λ ≤ 1) (see [1] ) and the class
M H2,1 ([1, 1; 1]; λ, β) reduces to the class M T (λ, β)
{
=
{
f ∈ A : Re
′
′′
zf (z) + λz 2 f (z)
(1 − λ)f (z) + λzf ′ (z)
}
< β, 0 ≤ λ ≤ 1,
β > 1, z ∈ U } ,
the class M T (0, β) which for λ = 0 reduces to the class M (β) (β > 1) (see [17] )
and the class M T (1, β) which for λ = 1 reduces to the class N (β) (β > 1) (see [17] );
(9) For λ = 0 in (1.7) and (1.8), the class SH2,1 ([α1 ]; 0, α) reduces to the class
Sq,s ([α1 ]; α, 0) (see [2], with β = 0) and the class M H([α1 ]; 0, β) reduces to the
class M Sq,s ([α1 ]; β)
{
{
}
}
′
z (Hq,s (α1 )f (z))
= f ∈ A : Re
< β, β > 1, z ∈ U .
(Hq,s (α1 )f (z))
2
Main results
Unless otherwise mentioned, we shall assume in the reminder of this paper that, the
parameters α1 , ..., αq and β1 , ..., βs are positive real numbers, 0 ≤ λ ≤ 1, 0 ≤ α <
1, β > 1, n ≥ 2, z ∈ U and Ψn is defined by (1.5).
To prove our main results we need the following lemmas.
Lemma 1 [25]. The sequence {bn }∞
n=1 is a subordinating factor sequence if and only
if
{
(2.1)
Re 1 + 2
∞
∑
n=1
}
bn z
n
>0,
(z ∈ U ).
61
Lemma 2. Let the function f defined by (1.1) satisfy the following condition:
∞
∑
(2.2)
(n − α)[1 + λ(n − 1)]Ψn |an | ≤ 1 − α.
n=2
Then f ∈ SHq,s ([α1 ]; λ, α).
Proof. Assume that the inequality (2.2) holds true. Then we find that
z (H (α )f (z))′ + λz 2 (H (α )f (z))′′
q,s 1
q,s 1
−
1
(1 − λ) (Hq,s (α1 )f (z)) + λz (Hq,s (α1 )f (z))′
∞
∑
≤
(n − 1)[1 + λ(n − 1)]Ψn |an | |z|n−1
n=2
1−
∞
∑
[1 + λ(n − 1)]Ψn |an | |z|n−1
n=2
∞
∑
≤
(n − 1)[1 + λ(n − 1)]Ψn |an |
n=2
1−
∞
∑
[1 + λ(n − 1)]Ψn |an |
≤ 1 − α.
n=2
This shows that the values of the function
′
(2.3)
Φ(z) =
′′
′
lie in a circle which is centered at w = 1 and whose radius is 1 − α. Hence f satisfies
the condition (2.2). This completes the proof of Lemma 2.
Corollary 1. Let the function f (z) defined by (1.1) be in the class SHq,s ([α1 ];
λ, α), then
|an | ≤
(2.4)
1−α
(n ≥ 2).
(n − α)[1 + λ(n − 1)]Ψn
The result is sharp for the function
(2.5)
f (z) = z +
1−α
z n (n ≥ 2).
(n − α)[1 + λ(n − 1)]Ψn
Lemma 3. Let the function f defined by (1.1) satisfy the following condition:
(2.6)
∞
∑
[1 + λ(n − 1)] [(n − 1) + |n − 2β + 1|] Ψn |an | ≤ 2(β − 1).
n=2
Then f ∈ M Hq,s ([α1 ]; λ, β).
62
Proof. Assume that the inequality (2.6) holds true. Then we find that
z (H (α )f (z))′ + λz 2 (H (α )f (z))′′
q,s 1
q,s 1
− 1
(1 − λ) (Hq,s (α1 )f (z)) + λz (Hq,s (α1 )f (z))′
z (H (α )f (z))′ + λz 2 (H (α )f (z))′′
q,s 1
q,s 1
≤
′ − (2β − 1)
∞
∑
(n − 1)[1 + λ(n − 1)]Ψn an n=2
= ∞
∑
1+
[1 + λ(n − 1)]Ψn an n=2
∞
∑
2(β − 1) +
|n − 2α + 1| [1 + λ(n − 1)]Ψn an n=2
.
≤ ∞
∑
1+
[1 + λ(n − 1)]Ψn an
n=2
The last expression is bounded above by 1 if
∞
∑
(n − 1)[1 + λ(n − 1)]Ψn |an | ≤ 2(β − 1) +
n=2
∞
∑
|n − 2β + 1| [1 + λ(n − 1)]Ψn |an | ,
n=2
which is equivalent to condition (2.6). This completes the proof of Lemma 3.
Corollary 2.
Let the function f (z) defined by (1.1) be in the class
M Hq,s ([α1 ]; λ, β), then
(2.7)
|an | ≤
2(β − 1)
(n ≥ 2).
[1 + λ(n − 1)] [(n − 1) + |n − 2β + 1|] Ψn
The result is sharp for the function
(2.8)
f (z) = z +
2(β − 1)
z n (n ≥ 2).
[1 + λ(n − 1)] [(n − 1) + |n − 2β + 1|] Ψn
∗ ([α ]; λ, α) denote the class of functions f (z) ∈ A whose coefficients
Let SHq,s
1
∗ ([α ]; λ, α) ⊆ SH ([α ]; λ, α) and
satisfy the condition (2.2). We note that SHq,s
1
q,s
1
∗
let M Hq,s ([α1 ]; λ, β) denote the class of functions f (z) ∈ A whose coefficients satisfy
∗ ([α ]; λ, β) ⊆ M H ([α ]; λ, β).
the condition (2.6). We note that M Hq,s
1
q,s
1
∗ ([α ]; λ, α), Ψ ≥ Ψ > 0 (n ≥ 2) . Then for every
Theorem 1. Let f ∈ SHq,s
1
n
2
function g ∈ K, we have
(2.9)
(2 − α) (1 + λ) Ψ2
(f ∗ g)(z) ≺ g(z)
2[(2 − α) (1 + λ) Ψ2 + (1 − α)]
63
and
Re{f (z)} > −
(2.10)
(2 − α) (1 + λ) Ψ2 + (1 − α)
.
(2 − α) (1 + λ) Ψ2
(2 − α) (1 + λ) Ψ2
is the best estimate.
2[(2 − α) (1 + λ) Ψ2 + (1 − α)]
∞
∑
∗ ([α ]; λ, α) and let g(z) = z +
Proof. Let f ∈ SHq,s
cn z n ∈ K. Then we have
1
The constant
n=2
(2 − α) (1 + λ) Ψ2
(f ∗ g)(z)
2[(2 − α) (1 + λ) Ψ2 + (1 − α)]
(
)
∞
∑
(2 − α) (1 + λ) Ψ2
n
=
z+
an cn z .
2[(2 − α) (1 + λ) Ψ2 + (1 − α)]
n=2
(2.11)
Thus, by Definition 1, the subordination result (2.9) will hold true if the sequence
{
(2.12)
(2 − α) (1 + λ) Ψ2
an
2[(2 − α) (1 + λ) Ψ2 + (1 − α)]
}∞
,
n=1
is a subordinating factor sequence, with a1 = 1. In view of Lemma 1, this is equivalent to the following inequality:
{
(2.13)
Re 1 +
∞
∑
n=1
(2 − α) (1 + λ) Ψ2
an z n
(2 − α) (1 + λ) Ψ2 + (1 − α)
}
> 0.
Now, since
(n − α)[1 + λ(n − 1)]Ψn ,
is an increasing function of n (n ≥ 2), we have
{
}
∞
∑
(2 − α) (1 + λ) Ψ2
n
Re 1 +
an z
n=1 (2 − α) (1 + λ) Ψ2 + (1 − α)
{
= Re 1 +
≥1−
(2−α)(1+λ)Ψ2
(2−α)(1+λ)Ψ2 +(1−α) z
(2−α)(1+λ)Ψ2
(2−α)(1+λ)Ψ2 +(1−α) r
>1−
1
+ (2−α)(1+λ)Ψ
2 +(1−α)
1
− ( (2−α)(1+λ)Ψ
2 +(1−α)
(2−α)(1+λ)Ψ2
(2−α)(1+λ)Ψ2 +(1−α) r
= 1 − r > 0 (|z| = r < 1),
−
∞
∑
∞
∑
}
(2 − α) (1 + λ) Ψ2 an z
n
n=2
(n − α)[1 + λ(n − 1)]Ψn |an | rn )
n=2
(1−α)
(2−α)(1+λ)Ψ2 +(1−α) r
64
where we have used the assertion (2.2) of Lemma 2. Thus (2.13) holds true in U. This
proves the inequality (2.9). The inequality (2.10) follows from (2.9) by taking the
∞
∑
z
convex function g(z) = 1−z
=z+
z n ∈ K.
n=2
To prove the sharpness of the constant
∗ ([α ]; λ, α) given by
f0 (z) ∈ SHq,s
1
f0 (z) = z −
(2.14)
(2−α)(1+λ)Ψ2
2[(2−α)(1+λ)Ψ2 +(1−α)] ,
we consider the function
(1 − α)
z2.
(2 − α) (1 + λ) Ψ2
Thus from (2.9), we have
(2 − α) (1 + λ) Ψ2
z
f0 (z) ≺
.
2[(2 − α) (1 + λ) Ψ2 + (1 − α)]
1−z
(2.15)
Moreover, it can easily be verified for the function f0 (z) given by (2.14) that
(2.16)
{
min Re
|z|≤r
}
(2 − α) (1 + λ) Ψ2
1
f0 (z) = − .
2[(2 − α) (1 + λ) Ψ2 + (1 − α)]
2
(2−α)(1+λ)Ψ2
is the best possible. This comThis show that the constant 2[(2−α)(1+λ)Ψ
2 +(1−α)]
pletes the proof of Theorem 1.
Putting q = 2, s = 1, α1 = a (a > 0) , α2 = 1 and β1 = c (c > 0) in Lemma 2 and
Theorem 1, we obtain the following corollary:
Corollary 3. Let f defined by (1.1) be in the class SL∗ (a, c; λ, α) and satisfy the
condition
∞
∑
(n − α)[1 + λ(n − 1)]
n=2
(a)n−1
|an | ≤ 1 − α.
(c)n−1
Then for every function g ∈ K, we have
a(2 − α) (1 + λ)
(f ∗ g)(z) ≺ g(z),
2[a(2 − α) (1 + λ) + c(1 − α)]
and
Re{f (z)} > −
a(2 − α) (1 + λ) + c(1 − α)
.
a(2 − α) (1 + λ)
a(2 − α) (1 + λ)
2[a(2 − α) (1 + λ) + c(1 − α)]
Remark 1.
(1) Putting q = 2, s = α1 = α2 = β1 = 1 and λ = 0 in Lemma 2 and Theorem
1, we obtain the result obtained by Frasin [10, Corollary 2.2];
The constant
65
1, we obtain the result obtained by Frasin [10, Corollary 2.5];
(3) Putting λ = 0 in Lemma 2 and Theorem 1, we obtain the result obtained by
Murugusndaramoorthy et. al. [15, Theorem 2.1, with k = 0];
(4) Specializing the parameters q, s, α1 , ..., αq and β1 , ..., βs , in Lemma 2 and
Theorem 1, we obtain the corresponding results for the corresponding classes (2-9)
defined in the introduction.
Similarly, we can prove the following theorem.
∗ ([α ]; λ, β), Ψ ≥ Ψ > 0 (n ≥ 2) . Then for every
Theorem 2. Let f ∈ M Hq,s
1
n
2
function g ∈ K, we have
(2 − β) (1 + λ) Ψ2
(f ∗ g)(z) ≺ g(z)
2[(2 − β) (1 + λ) Ψ2 + (β − 1)]
(2.17)
and
Re{f (z)} > −
(2.18)
(2 − β) (1 + λ) Ψ2 + (β − 1)
.
(2 − β) (1 + λ) Ψ2
(2 − β) (1 + λ) Ψ2
2[(2 − β) (1 + λ) Ψ2 + (β − 1)]
The constant
Putting q = 2, s = 1, α1 = a (a > 0) , α2 = 1 and β1 = c (c > 0) in Lemma 3 and
Theorem 2, we obtain the following corollary:
Corollary 4. Let f defined by (1.1) be in the class M L∗ (a, c; λ, β) and satisfy the
condition
∞
∑
[1 + λ(n − 1)] [(n − 1) + |n − 2β + 1|]
n=2
(a)n−1
|an | ≤ 2(β − 1).
(c)n−1
Then for every function g ∈ K, we have
a(2 − β) (1 + λ)
(f ∗ g)(z) ≺ g(z),
2[a(2 − β) (1 + λ) + c(β − 1)]
and
Re{f (z)} > −
a(2 − β) (1 + λ) + c(β − 1)
.
a(2 − β) (1 + λ)
a(2 − β) (1 + λ)
2[a(2 − β) (1 + λ) + c(β − 1)]
Remark 2.
2, we obtain the result obtained by Srivastava and Attiya [23, Corollary 2];
The constant
66
2, we obtain the result obtained by Srivastava and Attiya [23, Corollary 4];
(3) Specializing the parameters q, s, α1 , ..., αq and β1 , ..., βs , in Lemma 3 and
Theorem 2, we obtain the corresponding results for the corresponding classes (2-9)
defined in the introduction.
References
[1] O. Altintaş, On a subclass of certain starlike functions with negative coefficients,
Math. Japon., 36 (1991), 489-495.
[2] M. K. Aouf and G. Murugusundaramoorthy, On a subclass of uniformly convex functions defined by the Dziok-Srivastava operator, Austral. J. Math. Anal.
Appl., 5 (2008), no 1, Art. 3 , 1-17.
[3] S.D. Bernardi, Convex and starlike univalent functions, Trans. Amer. Math.
Soc., 135 (1969), 429–449.
[4] T. Bulboaca, Differential Subordinations and Superordinations, Recent Results,
House of Scientific Book Publ., Cluj-Napoca, 2005.
[5] B. C. Carlson and D. B. Shaffer, Starlike and prestarlike hypergeometric functions, J. Math. Anal. Appl., 15 (1984), 737-745.
[6] N. E. Cho, O. S. Kwon and H. M. Srivastava, Inclusion relationships and argument properties for certain subclasses of multivalent functions associated with
a family of linear operators, J. Math. Anal. Appl., 292(2004), 470-483.
[7] J. H. Choi, M. Saigo and H. M. Srivastava, Some inclusion properties of a
certain family of integral operators, J. Math. Anal. Appl., 276(2002), 432-445.
[8] J. Dziok and H.M. Srivastava, Classes of analytic functions with the generalized
hypergeometric function, Appl. Math. Comput., 103 (1999), 1-13.
[9] J. Dziok and H.M. Srivastava, Certain subclasses of analytic functions associated with the generalized hypergeometric function, Integral Transform. Spec.
Funct., 14 (2003), 7-18.
[10] B. A. Frasin, Subordination results for a class of analytic functions defined by
a linear operator, J. Inequal. Pure Appl. Math., 7 (2006), no. 4, Art. 134, 1-7.
[11] R. J. Libera, Some classes of regular univalent functions, Proc. Amer. Math.
Soc., 16 (1969), 755-758.
[12] A. E. Livingston,On the radius of univalence of certain analytic functions, Proc.
Amer. Math. Soc., 17 (1966), 352-357.
67
[13] S.S. Miller and P.T. Mocanu,Differential Subordinations: Theory and Applications, in: Series on Monographs and Textbooks in Pure and Appl. Math., 255,
Marcel Dekker, Inc, New York, 2000.
[14] G. Murugusundaramoorthy, T. Rosy and K. Muthunagai, Carlson-Shaffer operator and their applications to certain subclass of uniformly convex functions,
General Math., 15 (2007), 131-143.
[15] G. Murugusundaramoorthy, T. Rosy and K. Muthunagai, Subordination results
and integral means for k-uniformly starlike functions, Acta Univ. Apul., 21
(2010), 171-183.
[16] K. I. Noor, On new classes of integral operators, J. Natur. Geom., 16(1999),
71-80.
[17] J. Nishiwaki and S. Owa, Coefficient inequalities for certain analytic functions,
Internat. J. Math. Math. Sci., 29 (2002), no. 5, 285-290.
[18] S. Owa, On the distortion theorems I, Kyungpook Math. J., 18 (1978), 53-59.
[19] S. Owa and J. Nishiwaki, Coefficient estimates for certain classes of analytic
functions, J. Inequal. Pure Appl. Math., 3 (2002), no. 5, Art. 72, 1-12.
[20] S. Owa and H. M. Srivastava, Univalent and starlike generalized hypergeometric
functions, Canad. J. Math., 39 (1987), 1057-1077.
[21] S. Owa and H. M. Srivastava, Some generalized convolution properties associated
with certain subclasses of analytic functions, J. Inequal. Pure Appl. Math., 3
(2002), no. 3, Art.42, 1-27.
[22] St. Ruscheweyh, New criteria for univalent functions, Proc. Amer. Math. Soc.,
49 (1975), 109-115.
[23] H. M. Srivastava and A .A . Attiya, Some subordination results associated with
certain subclasses of analytic functions, J. Inequal. Pure Appl. Math., 5 (2004),
no. 4, Art.82, 1-14.
[24] H.M. Srivastava and S. Owa (Editors), Current Topics in Analytic Function
Theory, World Scientific Publishing Company, Singapore, New Jersey, London,
Hong Kong, 1992.
[25] H.S. Wilf, Subordinating factor sequence for convex maps of the unit circle,
Proc. Amer. Math. Soc., 12 (1961), 689-693.
M. K. Aouf
MansouraUniversity
Faculty of Science
Mansoura 35516, Egypt
mkaouf127@yahoo.com
68
A. Shamandy
MansouraUniversity
Faculty of Science
shamandy16@hotmail.com
A. O. Mostafa
MansouraUniversity
Faculty of Science
adelaeg254@yahoo.com
E. A. Adwan
MansouraUniversity
Faculty of Science
eman.a2009@yahoo.com
Numerical solutions of differential transform method for
linear and nonlinear ordinary differential equations 1
Montri Thongmoon
Abstract
The differential transform method is one of the approximate methods which
can be easily applied to many linear and nonlinear problems and is capable of
reducing the size of computational work. Exact solutions can also be achieved by
the known forms of the series solutions. In this paper, we present the definition
and operation of the one-dimensional differential transform and investigate the
particular exact solutions of some ordinary differential equation that usually
arise in mathematical biology by one-dimensional differential transform method.
The numerical results of the present method are presented and compared with
the exact solutions.
2010 Mathematics Subject Classification: 35A24, 35Qxx.
Key words and phrases:Differential Transform Method, Ordinary Differential
Equations.
1
Introduction
The differential transformation method is a numerical method based on Taylor expansion. This method constructs an analytical solution in the form of a polynomial.
The concept of differential transform method was first proposed and applied to solve
linear and nonlinear initial value problems in electric circuit analysis by [1]. Chen
and Liu had applied this method to solve two-boundary-value problems [2]. Jang,
Chen and Liu apply the two-dimensional differential transform method to solve partial differential equations [3]. Yu and Chen apply the differential transformation
method to the optimization of the rectangular fins with variable thermal parameters [4, 5]. Not like the traditional high order Taylor series method which requires
a lot of symbolic computations, the differential transform method is an iterative
1
Received 25 September, 2009
69
70
M. Thongmoon
procedure for obtaining Taylor series solutions. This method will not consume too
much computer time when applying to nonlinear or parameter varying systems. This
method gives an analytical solution in the form of a polynomial. But, It is different from Taylor series method that requires computation the high order derivatives.
The differential transform method is an iterative procedure that is described by the
transformed equations of original functions for solution of differential equations.
In this paper, three ordinary differential equation problems are considered by
differential transformation technique, a closed form series solution or an approximate solution can be obtained and the numerical solutions are compared with the
exact solutions.
2
One-Dimensional Differential Transform
As in Refs. [6, 7, 8], the basic definition of the differential transformation are introduced as the follows:
Definition 1 If x(t) is analytic in the domain T , then it will be differentiated continuously with respect to time t,
(1)
∂ k x(t)
= ϕ(t, k)
∂tk
for all t ∈ T
for t = ti , then ϕ(t, k) = ϕ(ti , k), where k belongs to the set of non-negative integer,
denoted as the K-domain. Therefore, Eq.(1) can be written as:
[ k
]
∂ x(t)
(2)
X(k) = ϕ(ti , k) =
,
∂tk t=ti
where X(k) is called the spectrum of x(t) at t = ti .
Definition 2 If x(t) can be expressed by Taylor’s series, then x(t) can be represented
as:
]
∞ [
∑
(t − ti )k
(3)
x(t) =
X(k).
k!
k=0
Eq.(3) is called the inverse of x(t), with the symbol D denoting the differential
transformation process. Upon combining Eq.(2) and Eq.(3), we obtain
]
∞ [
∑
(t − ti )k
(4)
x(t) =
X(k) ≡ D−1 X(k).
k!
k=0
Using the differential transformation, a differential equation in the domain of interest
can be transformed to an algebraic equation in the K-domain and the x(t) can be
obtained by finite term Taylor’s series plus a remainder, as
]
n [
∑
(t − ti )k
(5)
x(t) =
X(k) + Rn+1 (t).
k!
k=0
Numerical solutions of differential transform method ...
71
where Rn+1 (t) is the remainder.
The fundamental mathematics operations performed by differential transformation
are presented as the following theorems.
Theorem 1 If c(t) = u(t) ± v(t), then
C(k) = U (k) ± V (k).
(6)
Theorem 2 If c(t) = αu(t), then
(7)
C(k) = αU (k)
where α is a constant.
Theorem 3 If c(t) =
(8)
then
C(k) = (k + 1)U (k + 1).
Theorem 4 If c(t) =
(9)
∂
∂t u(t),
∂r
∂tr u(t),
then
C(k) = (k + 1)(k + 2) · · · (k + r)U (k + r).
Theorem 5 If c(t) = u(t)v(t), then
(10)
C(k) =
k
∑
U (r)V (k − r).
r=0
Theorem 6 If c(t) = tn , then
C(k) = δ(k − n)
(11)
{
where
δ(k − n) =
1; k = n
0; k ̸= n
Corollary 1 If c(t) = αtn , then
(12)
C(k) = αδ(k − n)
{
where
δ(k − n) =
Theorem 7 If c(t) = eλt , then C(k) =
1; k = n
0; k ̸= n
λk
k! .
Theorem 8 If c(t) = (1 + t)m , then C(k) =
m(m−1)...(m−k+1)
.
k!
Theorem 9 If c(t) = sin(ωt + α), then C(k) =
ωk
k!
Theorem 10 If c(t) = cos(ωt + α), then C(k) =
sin( πk
2! + α).
ωk
k!
cos( πk
2! + α).
72
3
3.1
M. Thongmoon
Numerical Examples
Example 1.
Consider the simple linear initial value problem
dc
= −250c
dt
(13)
with the initial condition
(14)
c(0) = 1.
Solution. Taking one-dimensional differential transform of (13), we obtain
(15)
(k + 1)C(k + 1) = −250C(k).
Rearranging (15) we obtain
(16)
C(k + 1) =
−250
C(k).
k+1
Applying the differential transform to an initial condition we have
(17)
C(0) = 1.
From (16) and (17), we can calculate the values of C(k + 1) for all k = 0, 1, 2, . . . , 10.
k
0
k+1
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
C(k + 1)
-250
(−250)2
2
(−250)3
3!
(−250)4
4!
(−250)5
5!
(−250)6
6!
(−250)7
7!
(−250)8
8!
(−250)9
9!
(−250)10
10!
(−250)11
11!
Table 1. The values of C(k + 1) for all k = 0, 1, 2, . . . , 10.
Substituting all values of C(k) for k = 0, 1, 2, . . . , 11 into (5). We have
(18)
c(t) =
11
∑
k=0
C(k)tk .
18
1
73
Numerical solutions
x 10
0
value of c
−1
DTM
Exact solution
−2
−3
−4
−5
−6
1
2
3
4
5
6
7
value of t
8
9
10
11
Figure 1: Numerical solutions of Differential transform and Exact solutions when
0 ≤ t ≤ 1 and ∆t = 0.1
Therefore we have the numerical solution of this example as the following equation:
c(t) = 1 − 250t +
2502 2
2 t
−
2503 3
3! t
+
2504 4
4! t
−
2505 5
5! t
(19)
6
6
+ 250
6! t −
2507 7
7! t
+
2508 8
8! t
−
2509 9
9! t
+
25010 10
10! t
−
25011 11
11! t .
This equation is closed to the exact solution c(t) = e−250t . The numerical and exact
solutions are presented as following figure.
3.2
Example 2.
Consider the ordinary differential equation
(20)
d2 c
dc
− 2 − 3c = 2et − 10 sin t
dt2
dt
with the initial conditions
(21)
c(0) = 2; c′ (0) = 4.
[
]
2
1
πk
(22) (k + 1)(k + 2)C(k + 2) − 2(k + 1)C(k + 1) − 3C(k) =
− 10
[sin( )] .
k!
k!
2!
(23)
[
(
)
]
1
2
1
πk
C(k + 2) =
− 10
[sin( )] + 2(k + 1)C(k + 1) + 3C(k) .
(k + 1)(k + 2) k!
k!
2!
74
M. Thongmoon
(24)
C(0) = 2; C(1) = 4.
k
0
1
2
3
4
5
k+2
2
3
4
5
6
7
C(k + 2)
8
6
61
12
47
15
14
9
4
15
(25)
c(t) =
7
∑
C(k)tk .
k=0
(26)
c(t) = 2 + 4t + 8t2 + 6t3 +
4
61 4 47 5 14 6
t + t + t + t7 .
12
15
9
15
This equation is closed to the exact solution c(t) = e3t + 21 e−t −
The numerical and exact solutions are presented in Figure 2.
3.3
et
2
+ cos t + sin t.
Example 3.
Consider the nonlinear ordinary differential equation
d2 c
+ e−2t c3 = 2et
dt2
(27)
with the initial conditions
c(0) = c′ (0) = 1.
(28)
(29)
(k + 1)(k + 2)C(k + 2) +
(−2)k
2
δ(k − 3) = .
k!
k!
(30)
[
]
2
(−2)k
1
−
δ(k − 3) .
C(k + 2) =
(k + 1)(k + 2) k!
k!
75
Numerical solutions
35
DTM
Exact solution
30
value of c
25
20
15
10
5
0
1
2
3
4
5
6
7
value of t
8
9
10
11
0 ≤ t ≤ 1 and ∆t = 0.1
(31)
C(0) = C(1) = 1.
k
0
1
2
3
4
5
6
7
8
k+2
2
3
4
5
6
7
8
9
10
C(k + 2)
1
1
3(1!)
1
6(2!)
1
2(3!)
1
15(4!)
1
21(5!)
1
28(6!)
1
36(7!)
1
45(8!)
(32)
c(t) =
10
∑
k=0
C(k)tk .
76
M. Thongmoon
Numerical solutions
3.5
DTM
Exact solution
value of c
3
2.5
2
1.5
1
1
2
3
4
5
6
7
value of t
8
9
10
11
0 ≤ t ≤ 1 and ∆t = 0.1
c(t) = 1 + t + t2 +
1 3
3(1!) t
+
1 4
6(2!) t
1
7
21(5!) t
+
1
8
28(6!) t
+
1 5
2(3!) t
(33)
1
t6 +
+ 15(4!)
+
1
9
36(7!) t
+
1
10
45(8!) t .
This equation is closed to the exact solution c(t) = et . The numerical and exact
solutions are presented in Figure 3.
4
Conclusion
One-dimensional differential transform have been applied to linear and non-linear
ordinary differential equations. The numerical examples have been presented to
show that the approach is promising and the research is worth to continue in this
direction. Using the differential transform method, the solution of the ordinary
differential equations can be obtained in Taylors series form. All the calculations in
the method are very easy. The calculated results are quite reliable. Therefore, this
method can be applied to many complicated linear and non-linear ODEs
Acknowledgment
The author would like to thanks the referee for his valuable suggestions that improved the presentation of the paper.
77
References
[1] X. Zhou, Differential Transformation and its Applications for Electrical Circuits,
Huazhong University Press, Wuhan, China, 1986 (in Chinese).
[2] C.L. Chen, Y.C. Liu, Solution of two-boundary-value problems using the differential transformation method, Journal of Optimization Theory and Application
99, (1998) 23-35.
[3] M.J. Jang, C.L. Chen, Y.C. Liu, Two-dimensional differential transform for
partial differential equations, Applied Mathematics and Computation 121, (2001)
261-270.
[4] L.T. Yu, C.K. Chen, The Solution of the Blasius Equation by the Differential
Transformation Method, Math. Comput. Modeling Vol. 28, No. 1, (1998) 101-111.
[5] L.T. Yu, C.K. Chen, Application of Taylor transformation to Optimize Rectangular Fins with Variable Thermal Parameters, Applied Mathematical Modeling
22, (1998) 11-21.
[6] C.L. Chen, Y.C. Liu, Differential transformation technique for steady nonlinear
heat conduction problems, Applied Mathematics and Computation 95, (1998) 155164.
[7] M.J. Jang, C.L. Chen, Analysis of the response of a strongly nonlinear damped
system using a differential transformation technique, Applied Mathematics and
Computation 88, (1997) 137-151.
[8] I. H. Abdel-Halim Hassan, Differential transformation technique for solving
higher-order initial value problems, Applied Mathematics and Computation 154,
(2004) 299-311.
Montri Thongmoon
Mahasarakham University
Faculty of Science
Mahasarakham, Thailand, 45140
e-mail: montri.t@msu.ac.th
On a generalization of Ramanujan’s seventh order mock
theta functions 1
Bhaskar Srivastava
Abstract
We have given a generalization of Ramanujan’s seventh order mock theta
functions and shown that they are Fq -functions. Relations between the generalized functions are shown via difference operator. We also establish relations
between partial fifth order mock theta functions χ0,N (q) and χ1,N (q) and partial
seventh order mock theta functions F0,N (q) and F2,N (q). q-Integral representation and multibasic expansions are given for these generalized functions.
2010 Mathematics Subject Classification: 33D15.
Key words and phrases: Mock Theta Functions, q-Integral, Multibasic
Hypergeometric Series.
1
Introduction
Mock theta functions were first studied and named by S. Ramanujan. In his last
letter to G. H. Hardy [9, pp. 354-355] four month before he died, he wrote his results.
These results formed the basis on which the study of these functions proceeded.
Ramanujan included in the letter four separate classes of mock theta functions : one
class of third order, two of fifth order and one of seventh order. There have been a
number of deep results obtained about the third order functions. However, the fifth
and the seventh order functions have been more of a problem. The seventh order
specially were mysterious functions. Andrews’ [1] remarkable paper proved certain
Hecke type identities which paved the way for proving “Mock Theta Conjectures”,
Hickerson [6,7].
Generalized Ramanujan’s seventh order mock theta functions are defined as :
1
Received 9 July, 2011
79
80
(1)
(2)
(3)
B. Srivastava
F0 (t, z, α; q) =
F1 (t, z, α; q) =
1
(t)∞
1
(t)∞
F2 (t, z, α; q) =
∞
∑
2 −3n+nα 2n
z (z;q)
(t)n q n
n=0
∞
∑

z2

;q 
q
2n
n
,
2 −3n+nα 2n
z (z;q)n−1


(t)n q n
n=0
1
(t)∞


∞
∑
z2
;q 
q
2n−1
,
2
(t)n q n −2n+nα z 2n (z;q)n


z2
n=0

;q 
q
.
2n+1
Also by simple computation it can be (
verified that
) F0 (t, z, α; q) satisfies the func2
z
q 2−α 1 −
(1 + z)
q
tional equation F0 (tq, zq, α; q) =
(F0 (t, z, α; q) − 1). The
z2
other two functions satisfy similar equations. For t = 0 and α = 1 they reduce
to the generalized functions of Choi [4] in which he has shown a relationship between basic bilateral hypergeometric series and the generalized functions. For t = 0
and α = 1 and taking z = q l , l any positive integer, and using mathematical induction the functions F0 (0, q l , 1; q), F1 (0, q l , 1; q) and F2 (0, q l , 1; q) satisfy Ramanujan’s
description of mock theta functions. Taking t = 0, α = 1 and z = q the generalized
functions reduce to the seventh order mock theta functions of Ramanujan :
(4)
∞
∑
F0 (q) =
n=0
(5)
F1 (q) =
2
qn
(q n+1 ;q)
∞
∑
n=1
(6)
F2 (q) =
∞
∑
n=0
,
n
2
qn
(q n ;q)n
,
2 +n
qn
(q n+1 ;q)n+1
.
The idea of giving this generalization is two fold− one is to show that these
generalized functions of Ramanujan’s seventh order mock theta functions belong to
the Fq -class of functions (this is done in section 4 and the definition of q-analogue
of F-functions is given in section 3) and secondly to show that F0 (t, z, α; q) and
F2 (t, z, α; q) are related by q-difference operator. This is done in section 5. We may
point out that Ramanujan did not give any relationship between his seventh order
mock theta functions.
On a generalization of Ramanujan’s seventh order mock theta functions
81
We define partial mock theta function of fifth and seventh order by taking the
summation from 0 to N in place of 0 to ∞ and put N in the suffix to indicate the
summation. In the last section− section 9− we show that the partial fifth order
mock theta functions χ0,N (q) and χ1,N (q) are related to partial seventh order mock
theta functions F0,N (q) and F2,N (q) , respectively.
In section 6 we represent these generalized function as q-Integral and in section
7 develop multibasic expansion for these functions.
2
Basic Facts
We use the following usual basic hypergeometric notations :
For q k < 1,
(
a; q k
)
n
=
n−1
∏ (
j=0
(
and
(
)
1 − aq kj , n ≥ 1,
a; q k
a; q k
)
∞
=
)
0
=1 ,
∞ (
∏
1 − aq kj
)
.
j=0
For convenience we shall write
(
a1 , a 2 , . . . , am ; q k
)
n
(
) (
)
(
)
= a1 ; q k n a2 ; q k n . . . am ; q k n .
When k = 1, we usually write (a)n and (a)∞ instead of (a; q)n and (a; q)∞
respectively.
We use the following notation for multibasic hypergeometric series :
[
ϕ
a1 , · · · · · · , ar : c1,1 , · · · · · · , c1,r1 : · · · · · · : cm,1 , · · · · · · , cm,rm
; q, q1 , · · · , qm ; z
b1 , · · · · · · , bs : e1,1 , · · · · · · , e1,s1 : · · · · · · : em,1 , · · · · · · , em,sm
]1+s−r
[
∞ (a , · · · · · · , a ; q)
∑
n2 −n
1
r
n n
z (−1)n q 2
n=0
( (q, b1 , · · · · · · , bs ; q))n [
]sj −rj
m
n2 −n
∏ cj,1 , · · · · · · , cj,rj ; qj n
(
) (−1)n qj 2
.
j=1 ej,1 , · · · · · · , ej,sj ; qj n
=
A generalized basic hypergeometric series with base q1 is defined
]
a1 , · · · · · · , aA ; b1 , · · · · · · , bA−1 ; q1 , z
(a1 ; q1 )n · · · · · · (aA ; q1 )n z n
=
, |z| < 1.
n=0 (b1 ; q1 )n · · · · · · (bA−1 ; q1 )n (q1 ; q1 )n
A ϕA−1
∞
∑
[
]
82
3
B. Srivastava
Definition of Fq -functions
Truesdell [11] in his book unified the theory of special functions and named the
functions which satisfy the functional equation
∂
F (z, α) = F (z, α + 1) ,
∂z
F -functions.
The q-analogue of this is: Functions which satisfy the functional equation
Dq,z F (z, α) = F (z, α + 1) ,
where
zDq,z F (z, α) = F (z, α) − F (zq, α) .
are called Fq -Functions.
4
The generalized functions are Fq -Functions
In this section we show that these generalized functions are Fq -Functions.
Theorem 1 F0 (t, z, α; q), F1 (t, z, α; q) and F2 (t, z, α; q) are Fq -functions.
Proof.
We give the proof only for F0 (t, z, α; q) only. The proof for other functions being
similar is omitted.
Applying the difference operator Dq,t on F0 (t, z, α; q), we have
tDq,t F0 (t, z, α; q) = F0 (t, z, α; q) − F0 (tq, z, α; q)
=
=
1
(t)∞
1
(t)∞
∞
∑
2 −3n+nα 2n
z (z;q)n


(t)n q n
n=0
∞
∑
n=0

z2
;q 
q
2n
2 −3n+nα 2n
z (z;q)n


(t)n q n

z2
;q 
q
2n
=
t
(t)∞
∞
∑
n=0
−
1
(tq)∞
∞
∑
2
(tq)n q n −3n+nα z 2n (z;q)n


z2
n=0

;q 
q
−
(t)n q n
1
(t)∞
∞
∑
(t)n q n
n=0
2 −3n+nα 2n
z (z;q)n (1−tq n )



z2
;q 
q
2n
2 −3n+n(α+1) 2n
z (z;q)n



z2
;q 
q
2n
2n
,
83
so
Dq,t F0 (t, z, α; q) = F0 (t, z, α + 1; q) .
(7)
Hence F0 (t, z, α; q) is a Fq -function.
Similarly it can be shown that F1 (t, z, α; q) and F2 (t, z, α; q) are Fq -functions.
5
Relations between the generalized seventh order mock
theta functions
Now by (7)
Dq,t F0 (t, z, α; q) =
=
1
(t)∞
∞
∑
1
(t)∞
∞
∑
2 −2n+nα 2n
z (z;q)
(t)n q n
n=0
2
(t)n q n −2n+nα z 2n (z;q)n


z2
n=0

;q 


z2
q

n
(1−z 2 q2n−1 )
;q 
2n+1
2
∞
∑
z2
(t)n q n +nα z 2n (z;q)n


− (t)1
q ∞ n=0  z 2 
;q
q
2n+1
q
2n+1
z2 2
= F2 (t, z, α; q) − Dq,t F2 (t, z, α; q) .
q
6
q-Integral representation for the generalized functions
The q-integral was defined by Thomae and Jackson [5, p. 19] as
∫1
f (t) dq t = (1 − q)
0
∞
∑
f (q n ) q n .
n=0
(1 − q)−1 ∫1 t−1
w
(wq; q)∞ F0 (0, z, aw; q)dq w .
(q; q)∞ 0
(1 − q)−1 ∫1 t−1
(ii)F1 (q t , z, α; q) =
w
(wq; q)∞ F1 (0, z, aw; q)dq w .
(q; q)∞ 0
(1 − q)−1 ∫1 t−1
(iii)F2 (q t , z, α; q) =
w
(wq; q)∞ F2 (0, z, aw; q)dq w .
(q; q)∞ 0
Theorem 2 (i)F0 (q t , z, α; q) =
Proof.
We give the proof of Theorem 2 (i) only. The proofs of the others are on the
same line, so omitted.
Limiting case of q-beta integral [5, eq.(1.11.7), p.19 ] is
84
B. Srivastava
∫ x−1
1
(1 − q)−1 ∞
=
t
(tq; q)∞ dq t .
x
(q ; q)∞
(q; q)∞ 0
(8)
Now
F0 (t, z, α; q) =
1
(t)∞
∞
∑
2 −3n+nα 2n
z (z;q)
(t)n q n
n=0


z2

;q 
q
2n
n
.
Replacing t by q t and q α by a , we have
2
∞ ( )
1 ∑ q t n q n −3n+nα z 2n (z; q)n
( 2 )
F0 (q , z, α; q) = t
(q )∞
z
n=0
;q
q
2n
∞ n2 −3n+nα 2n
∑
q
z (z; q)n
( 2 )
=
z
n=0
;q
(q n+t )∞
q
2n
1
2
∞
∑ q n −3n+nα z 2n (z; q) (1 − q)−1 ∫
n
( 2 )
=
wn+t−1 (wq; q)∞ dq wby (8)
(q; q)∞
z
n=0
;q
0
q
2n
∫1
∞ n2 −3n 2n
∑
q
z (z; q) (aw)n
(1 − q)−1
t−1
( 2 )n
=
w
(wq; q)∞
dq w.
(q; q)∞
z
n=0
;q
0
q
2n
t
(9)
But
∞
∑
F0 (0, z, α; q) =
2 −3n+nα 2n
z (z;q)n


qn
n=0
and since q α = a,
∞
∑
F0 (0, z, a; q) =
n=0
Hence
(10)

z2
;q 
q
2n
2 −3n 2n
z (z;q)
qn


z2
q

n
n (a)
.
;q 
2n
F0 (0, z, aw; q) =
∞
∑
n=0
2 −3n 2n
z (z;q)n (aw)n


qn

z2
;q 
q
2n
.
85
By (10), we can write (9) as
F0 (q t , z, α; q) =
(1 − q)−1 ∫1 t−1
w
(wq; q)∞ F0 (0, z, aw; q)dq w ,
(q; q)∞ 0
which proves (i).
7
Multibasic Expansions
Using the summation formula [5, eq.(3.6.7), p. 71] and [8, Lemma 10, p. 57], we
have the multibasic expansion
(11)
( a )
)
∞
1 − bpk q −k (a, b; q)k c, ; q q k ∑
bc ) k
( aq ) ( ap
αm+k
(1 − a) (1 − b) q, ; q
, bcp; q
k=0
k=0
k
( aq b ) k c
∞ (ap, bp; p)
;q
∑
m cq,
bc
(
)
(
)m αm .
=
aq
ap
;q
, bcp; p
m=0 q,
b
c
m
m
∞
∑
(
1 − apk q k
)(
Corollary 1 Letting q → q 3 and c → ∞ in (11), we have
(12)
)(
)
3k2 +3k ∞
∞ (
∑
1 − apk q 3k 1 − bpk q −3k (a, b; p)k q 2 ∑
)
(
αm+k
3
k2 +k
aq
k
3
3
k=0
k=0
b p 2
(1 − a) (1 − b) q ,
;q
b
k
∞
∑
3m2 +3m
(ap, bp; p)m q 2
)
(
=
αm
m2 +m
aq 3 3
m
m=0 q 3 ,
;q
b q 2
b
m
Theorem 3 The generalized functions have the following expressions as a multibasic hypergeometric series.
(
)(
)
2
∞ 1 − tq 4k−1
1 − q −2k+2 (t; q)k−1 (z; q)k q k −3k+kα z 2k
1 ∑
( 2 )
(i)F0 (t, z, α; q) =
(t)∞ k=0
z
(1 − q k+2 )
;q
q
2k
[
]
q, zq k : 0, 0 : tq 3k , q 3k+3 :
2
3
2
α−2
×ϕ
; q, q , q ; z q
.
q k+3 : z 2 q 2k−1 , z 2 q 2k : 0, 0 :
(
)(
)
2
∞ 1−tq 4k−1
1−q −2k+2 (t; q)k−1 (z; q)k−1 q k −3k+kα z 2k
1 ∑
( 2 )
(ii)F1 (t, z, α; q) =
(t)∞ k=0
z
k+2
(1 − q )
;q
q
2k−1
86
B. Srivastava
[
]
q, zq k−1 : 0, 0 : tq 3k , q 3k+3 :
2
3
2
α−2
×ϕ
; q, q , q ; z q
.
q k+3 : z 2 q 2k−2 , z 2 q 2k−1 : 0, 0 :
(
)
(
)
2
∞ 1−tq 4k−1
1−q −2k+2 (t; q)k−1 (z; q)k q k −2k+kα z 2k
1 ∑
( 2 )
(iii)F2 (t, z, α; q) =
(t)∞ k=0
z
(1 − q k+2 )
;q
q
2k+1
[
]
q, zq k : 0, 0 : tq 3k , q 3k+3 :
2
3
2
α−1
×ϕ
; q, q , q ; z q
.
q k+3 : z 2 q 2k , z 2 q 2k+1 : 0, 0 :
Proof.
We shall give the proof of (i) only, for others we will state the value of the
parameters.
(
) ( 3)
mα−2m q 3 ; q 3
q
t; q m (z; q)m z 2m
t
m
2
( 2 )
Taking a = , b = q , p = q and αm =
in
q
z
(q 3 ; q)m
;q
q
2m
(12), we have
(13)
(
1−
)(
q −2k+2
)
(
t 2
,q ;q
q
)
3k2 +3k
1−
q 2
k
(
)
k2 +5k
t
k=0
1−
(1 − q 2 ) (t, q 3 ; q 3 )k q 2
q
( 3 3)
(
)
∞
∑ q ; q m+k t; q 3 m+k (z; q)m+k z 2(m+k) q (m+k)α−2(m+k)
( 2 )
×
z
3
k=0
;q
(q ; q)m+k
q
2m+2k
∞ m2 −3m+mα
2m
∑
q
(t; q) (z; q)m z
( 2 )m
=
z
n=0
;q
q
2m
∞
∑
tq 4k−1
The right hand side of (13) is equal to
(t; q)∞ F0 (t, z, α; q) .
The left hand side of (13) is equal to
(
)(
)
2
∞ 1 − tq 4k−1
∑
( 2 )
z
k=0
k+2
(1 − q )
;q
( 3k+3 3 ) ( 3k 3 ) ( kq ) 2k2m mα−2m
∞ q
; q m tq ; q m zq ; q m z q
∑
×
k+3 ; q) (z 2 q 2k−1 ; q 2 ) (z 2 q 2k ; q 2 )
k=0 ( (q
m
)m(
) m
2
∞ 1 − tq 4k−1
∑
( 2 )
=
z
k=0
k+2
(1 − q )
;q
q
2k
[
]
q, zq k : 0, 0 : tq 3k , q 3k+3 :
2
3
2
α−2
×ϕ
; q, q , q ; z q
,
q k+3 : z 2 q 2k−1 , z 2 q 2k : 0, 0 :
87
which proves (i).
Proof of (ii)
(
) (
)
q mα−2m q 3 ; q 3 m t; q 3 m (z; q)m−1 z 2m
t
2
(ii) Taking a = , b = q , p = q and αm =
q
(q 3 ; q)m (z 2 q; q)2m−1
in (12), we have the Theorem 3 (ii).
Proof of (iii)
(
) (
)
q mα−m q 3 ; q 3 m t; q 3 m (z; q)m z 2m
t
2
(iii) Taking a = , b = q , p = q and αm =
q
(q 3 ; q)m (z 2 q; q)2m+1
in (12), we have the Theorem 3 (iii).
8
Definition of Partial Mock Theta Functions of fifth
and seventh order
By taking the partial sums of the series defining the mock theta functions from 0 to
N , we have the partial mock theta functions. We put N in the suffix to denote the
partial sums from 0 to N . Thus the partial mock theta functions of fifth order are
(14)
χ0,N (q) =
N q n (q; q)
∑
n ,
n=0 (q; q)2n
(15)
χ1,N (q) =
N q n (q; q)
∑
n
,
(q;
q)
n=0
2n+1
and seventh order are
N q n (q; q)
∑
n ,
F0,N (q) =
n=0 (q; q)2n
2
(16)
N q n +n (q; q)
∑
n .
(q;
q)
n=0
2n+1
2
F2,N (q) =
(17)
9
Relations between Ramanujan’s seventh order mock
theta functions and fifth order mock theta functions
We shall use the following summation formula [10, p. 166]
(18)
p
∑
αr βr = βp+1
r=0
p
∑
αr +
r=0
p
∑
m=0
(βm − βm+1 )
m
∑
αr .
r=0
88
B. Srivastava
Theorem 4 (i) F0,p (q) = q p(p+1) χ0,p (q) +
(ii) F0 (q) =
∞ (
∑
q m(m−1)
−
q m(m+1)
)
p (
∑
)
q m(m−1) − q m(m+1) χ0,m (q) .
m=0
χ0,m (q) .
m=0
)
p (
∑
2
2
2
(iii) F2,p (q) = q (p+1) χ1,p (q) +
q m − q (m+1) χ1,m (q) .
m=0
)
∞ (
∑
2
2
(iv) F2 (q) =
q m − q (m+1) χ1,m (q) .
m=0
Proof of (i)
Taking αr =
(19)
q n (q; q)n
, βr = q n(n−1) in (18), we have
(q; q)2n
F0,p (q) = q p(p+1) χ0,p (q) +
p (
∑
)
q m(m−1) − q m(m+1) χ0,m (q) ,
m=0
which proves the Theorem 4(i).
Proof of (ii)
Letting p → ∞ in (19), we have
F0 (q) =
(20)
∞ (
∑
)
q m(m−1) − q m(m+1) χ0,m (q) ,
m=0
which proves the Theorem 4(ii).
Proof of (iii)
q n (q; q)n
2
Taking αr =
, βr = q n in (18), we have
(q; q)2n+1
(21)
)
p (
∑
2
2
q m − q (m+1) χ1,m (q) ,
2
F2,p (q) = q (p+1) χ1,p (q) +
m=0
which proves the Theorem 4(iii).
Proof of (iv)
Taking p → ∞ in (21), we have
(22)
F2 (q) =
∞ (
∑
2
2
q m − q (m+1)
)
χ1,m (q) ,
m=0
which proves the Theorem 4(iv).
References
[1] G. E. Andrews, The fifth and seventh order mock theta functions, Trans. Amer.
Math. Soc., vol. 293, 1986, 113-134.
89
[2] G. E. Andrews, Mock theta functions, Proc. Sympos. Pure Math., vol. 49, Part
2, 1989, 283-298.
[3] Y. -S. Choi, Tenth order mock theta functions in Ramanujan’s Lost Notebook,
Invent. Math. Soc., vol. 136, 1999, 497-569.
[4] Y. -S. Choi, The basic bilateral hypergeometric series and mock theta functions,
Ramanujan J., vol. 24, 2011, 345-386.
[5] G. Gasper and M. Rahman, Basic Hypergeometric Series, Cambridge University
Press, Cambridge, 1990.
[6] D. Hickerson, A proof of the mock theta conjectures, Invent. Math., vol. 94,
1988, 639-660.
[7] D. Hickerson,, On the seventh order mock theta functions, Invent. Math., vol.
94, 1988, 661-677.
[8] E. D. Rainville, Special Functions, Chelsea Publishing Company, Bronx, New
York, 1960.
[9] S. Ramanujan, Collected Paper, Cambridge University Press, 1927, reprinted
by Chelsea New York, 1962.
[10] B. Srivastava, Ramanujan’s mock theta functions, Math. J. Okayama Univ., vol.
47, 2005, 163-174.
[11] C. Truesdell, An Essay Toward a Unified Theory of Special Functions, Princeton
University Press, Princeton, 1948.
Bhaskar Srivastava
University of Lucknow
Department of Mathematics and Astronomy
Lucknow, India
e-mail: bhaskarsrivastav@yahoo.com
On certain class of meromorphic multivalent functions
with positive coefficients 1
R. M. El-Ashwah, M. K. Aouf
Abstract
∑∗ In this paper we introduce a new class of meromorphic multivalent functions
present paper is
p (k, α, β) with positive coefficients. The main object of the∑
∗
to prove some properties for functions belonging to the class p (k, α, β).
Key words and phrases: Analytic, meromorphic, positive coefficients,
Hadamard product.
1
Introduction
Let
∑
p
denote the class of functions of the form :
f (z) = z −p +
(1)
∞
∑
an+p−1 z n+p−1 (p ∈ N = {1, 2, ...}) ,
n=1
which are analytic and p-valent in the punctured disc U ∗ = {z : z ∈ C and 0 < |z| <
1} = U \{0}.
∑
For a function f (z) in the class p , we define
Ip0 f (z) = f (z),
′
Ip1 f (z) = zf (z) + (p + 1)z −p ,
′
Ip2 f (z) = z(Ip1 f (z)) + (p + 1)z −p ,
1
Received 07 July, 2010
Accepted for publication (in revised form) 17 September, 2012
91
92
and for k = 1, 2, 3, ...
′
Ipk f (z) = z(Ipk−1 f (z)) + (p + 1)z −p
= z −p +
(2)
∞
∑
(n + p − 1)k an+p−1 z n+p−1 .
n=1
The operator I1k = I k was considered by Frasin and Darus [2].
With the help of the differential operator Ipk , we define the class
follows:
Definition 1 A function f (z) ∈
satisfies the following inequaltiy :
(3)
∑
p
is said to be in the class
∑
p (k, α, β) as
∑
p (k, α, β)
if it
′
z(Ipk f (z))
+ p k
Ip f (z)
< β (p ∈ N; k ∈ N0 = N ∪ {0}),
z(Ipk f (z))′
I k f (z) + 2α − p p
for some α(0 ≤ α < p), β(0 < β ≤ 1) and for all z ∈ U ∗ .
∑
∑
Let ∗p denote the subclass of p consisting of functions of the form:
(4)
f (z) = z
−p
+
∞
∑
|an+p−1 | z n+p−1 .
n=1
We now write
(5)
∑∗
p
(k, α, β) =
∑
p
(k, α, β) ∩
∑∗
p
.
We note that:
∑
∑
(i) ∗1 (k, α, 1) = ∗d [k, α] ( Frasin and Darus [2]);
∑∗
∑∗
(ii)
p (0, α, β) =
p [α, β], is the class of meromorphically p-valent starlike
functions of order α and type β with positive coefficients, studied by Aouf [1] and
Mogra [3] ;
∑
∑
(iii) ∗p (0, α, 1) = ∗p [α], is the class of meromorphically p-valent starlike functions of order α with positive coefficients (see Aouf [1] and Mogra [3] );
∑
∑
(iv) ∗1 (0, α, β) = ∗d [α, β], is the class of meromorphically starlike functions of
order α and type β with positive coefficients, studied by Mogra et al. [4]
2
93
Coefficient estimates
We
∑ begin by proving a sufficient condition (involving coefficient bounds) for the class
p (k, α, β).
Theorem 1 Let the function f (z) defined by (1) be analytic in U ∗ . If
(6)
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1] |an+p−1 | ≤ 2β(p − α)
n=1
(0 ≤ α < p ; 0 < β ≤ 1 ; p ∈ N; k ∈ N0 ),
∑
then f (z) ∈ p (k, α, β) .
Proof. Suppose that (6) holds true for all admissible values of α, β, p and k. Consider
the expression
′
′
′
H(Ipk , Ipk )(z) = z(Ipk f (z)) + pIpk f (z) − β z(Ipk f (z)) + (2α − p)Ipk f (z)
∞
∑
= (n + p − 1)k (n + 2p − 1)an+p−1 z n+p−1 n=1
∞
2(α − p) ∑
k
n+p−1 −β +
(n
+
p
−
1)
(n
+
2α
−
1)a
z
,
n+p−1
zp
(7)
n=1
which readily yields
′
|z|p H(Ipk , Ipk )(z) ≤
∞
∑
{
−β
(n + p − 1)k (n + 2p − 1) |an+p−1 | |z|n+2p−1
n=1
2(p − α) −
∞
∑
}
n+2p−1
(n + p − 1) (n + 2α − 1) |an+p−1 | |z|
k
n=1
=
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1] |an+p−1 | |z|n+2p−1
n=1
−2β(p − α).
(8)
Letting |z| → 1− in (8), we have
′
H(Ipk , Ipk )(z) ≤
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1] |an+p−1 |
n=1
94
−2β(p − α) ≤ 0,
(9)
by the inequality (9). Hence it follows that
′
z(Ipk f (z))
+ p k
Ip f (z)
<β
k
z(Ip f (z))′
I k f (z) + 2α − p (10)
(z ∈ U ∗ ),
p
∑
which ( in view of Definition 1.1 ) proves that f (z) ∈ p (k, α, β).
∑
Next we ∑
give a necessary and sufficient condition for function f (z) ∈ ∗p to be
in the class ∗p (k, α, β).
Theorem 2 Let the function f (z) defined by (4) be analytic in U ∗ . Then
∑
f (z) ∈ ∗p (k, α, β) if and only if (6) is satisfied.
Proof. In view of Theorem 2.1, it suffices to show that the ∑
” only if ” part. Let us
assume that the function f (z) defined by (4) is in the class ∗p (k, α, β). Then
′
z(Ipk f (z))
+ p k
Ip f (z)
k
z(Ip f (z))′
I k f (z) + 2α − p p
∞
∑
≤
(n + p − 1)k (n + 2p − 1) |an+p−1 | |z|n+2p−1
n=1
2(p − α) −
∞
∑
n+2p−1
(n + p − 1)k (n + 2α − 1) |an+p−1 | |z|
< β (z ∈ U ∗ ).
n=1
Using the fact that Re(z) ≤ |z| for all z , we thus have
(11)

∞
∑


k (n + 2p − 1) |a
n+2p−1


(n
+
p
−
1)
|
z
n+p−1


n=1
∑∞
ℜ
< β (z ∈ U ∗ ).
k (n + 2α − 1) |a
n+2p−1 

2(p
−
α)
−
(n
+
p
−
1)
|
z
n+p−1


n=1


′
z(Ipk f (z))
Now choose the values of z on the real axis so that
is real. Upon clearing
Ipk f (z)
the denominator in (11) and letting z → 1− through real values, we obtain
∞
∑
(n + p − 1)k (n + 2p − 1) |an+p−1 | ≤
n=1
{
2(p − α) −
β
∞
∑
95
}
(n + p − 1) (n + 2α − 1) |an+p−1 | ,
k
n=1
or
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1] |an+p−1 |
n=1
≤ 2β (p − α),
(12)
which proves Theorem 2.2.
Corollary 1 Let the function f (z) defined by (4) be in the class
Then
(13)
|an+p−1 | ≤
(n + p −
1)k
∑∗
p (k, α, β).
2β(p − α)
(p, n ∈ N; k ∈ N0 ).
[(1 + β)n + (2α − 1)β + 2p − 1]
The result is sharp for the function f (z) given by
fn+p−1 (z) = z −p +
2β(p − α)
z n+p−1
[(1 + β)n + (2α − 1)β + 2p − 1]
(p, n ∈ N; k ∈ N0 ).
(14)
3
(n + p −
1)k
Distortion theorem
Theorem 3 If a function f (z) defined by (4) be in the class
{
∑∗
p (k, α, β) ,
}
(p + j − 1)!
β(p − α)
p!
2p
− k
.
r
r−(p+j)
(p − 1)!
p (p + αβ) (p − j)!
≤ f (j) (z) ≤
{
(15)
}
(p + j − 1)!
β(p − α)
p!
2p
+ k
.
r
r−(p+j)
(p − 1)!
p (p + αβ) (p − j)!
(0 < |z| = r < 1; 0 ≤ α < p; 0 < β ≤ 1; p ∈ N; j, k ∈ N0 ; p > j).
The result is sharp for the function f (z) given by
(16)
f (z) = z −p +
β(p − α) p
z
pk (p + αβ)
iπ
(p ∈ N; k ∈ N0 ; z = r; z = re 2p ).
then
96
Proof. In view of Theorem 2.2, we have
∞
2pk (p + αβ) ∑
(n + p − 1)! |an+p−1 |
p!
n=1
≤
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1] |an+p−1 | ≤ 2β(p − α),
n=1
which yields
∞
∑
(17)
(n + p − 1)! |an+p−1 | ≤
n=1
β(p − α)p!
(p ∈ N; k ∈ N0 ).
pk (p + αβ)
Now, by differentiating both sides of (4) j times with respect to z, we have
∞
f (j) (z) = (−1)j
(p + j − 1)! −(p+j) ∑ (n + p − 1)!
z
+
|an+p−1 | z n+p−1−j
(p − 1)!
(n + p − 1 − j)!
n=1
(p ∈ N; j ∈ N0 ; p > j),
(18)
and Theorem 3.1 follows easily from (17) and (18).
Finally, it is easy to see that the bounds in (15) are attained for the function
f (z) given by (16).
4
Radius of convexity
∑
Theorem 4 Let the function f (z) defined by (4) be in the class ∗p (k, α, β). Then
f (z) is meromorphically p-valent convex of order δ(0 ≤ δ < p) in 0 < |z| < r =
r(k, p, α, β, δ) , where
r(k, p, α, β, δ) =
{
(19)
inf
n
(n + p − 1)k−1 [(1 + β)n + (2α − 1)β + 2p − 1]p(p − δ)
2β(p − α)(n + 3p − 1 − δ)
The result is sharp.
Proof. It is sufficient to show that
(zf ′ (z))′ + pf ′ (z) ≤p−δ
f ′ (z)
1
} n+2p−1
(n ∈ N).
f or 0 < |z| < r(k, p, α, β, δ) .
Note that
(zf ′ (z))′ + pf ′ (z) ∑∞ (n + p − 1)(n + 2p − 1) |a
n+p−2 n=1
n+p−1 | z
∑
=
n+p−2 pz −p−1 + ∞
f ′ (z)
n=1 (n + p − 1) |an+p−1 | z
≤
∑∞
97
+ p − 1)(n + 2p − 1) |an+p−1 | |z|n+2p−1
.
∞
∑
p−
(n + p − 1) |an+p−1 | |z|n+2p−1
n=1 (n
n=1
Thus
′
′
′
(zf (z)) + pf (z)
≤ p − δ,
f ′ (z)
if
∞
∑
(n + p − 1)(n + 3p − 1 − δ)
(20)
p(p − δ)
n=1
|an+p−1 | |z|n+2p−1 ≤ 1.
But Theorem 2.2 ensures that
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]
(21)
2β(p − α)
n=1
|an+p−1 | ≤ 1.
In view of (21), it follows that (20) will be true if
(n + p − 1)(n + 3p − 1 − δ) n+2p−1
|z|
≤
p(p − δ)
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]
2β(p − α)
(22)
or, if
(23)
{
|z| ≤
(n ∈ N)
(n + p − 1)k−1 [(1 + β)n + (2α − 1)β + 2p − 1]p(p − δ)
2β(p − α)(n + 3p − 1 − δ)
1
} n+2p−1
(n ∈ N).
Setting |z| = r(k, p, α, β, δ) in (23), the result follows.
The result (19) is sharp with the extremal function fn+p−1 (z) given by (14).
5
Closure theorems
Theorem 5 Let
(24)
fp−1 (z) =
1
zp
and
fn+p−1 (z) = z −p +
(n + p −
1)k [(1
2β(p − α)
z n+p−1
+ β)n + (2α − 1)β + 2p − 1]
98
(p, n ∈ N; k ∈ N0 ).
(25)
Then f (z) is in the class
∑∗
p (k, α, β)
(26)
f (z) =
∞
∑
if and only if
λn+p−1 fn+p−1 (z),
n=0
where
λn+p−1 ≥ 0 and
Proof. Let f (z) =
1 . Then
∞
∑
λn+p−1 = 1.
n=0
∑∞
n=0 λn+p−1 fn+p−1 (z),
f (z) =
∞
∑
where λn+p−1 ≥ 0 and
∑∞
n=0 λn+p−1
=
λn+p−1 fn+p−1 (z)
n=0
= z −p +
∞
∑
n=1
Then
(n + p −
2β(p − α)λn+p−1
z n+p−1 .
+ β)n + (2α − 1)β + 2p − 1]
1)k [(1
∞
∑
n=1
.
(n + p −
2β(p − α)λn+p−1
.
+ β)n + (2α − 1)β + 2p − 1]
1)k [(1
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]
2β(p − α)
=
∞
∑
λn+p−1 = 1 − λp−1 ≤ 1,
n=1
∑
α, β).
which shows that f (z) ∈ ∗p (k,∑
Conversely, suppose f (z) ∈ ∗p (k, α, β). Then
|an+p−1 | ≤
(n + p −
1)k [(1
2β(p − α)
(p, n ∈ N; k ∈ N0 ),
+ β)n + (2α − 1)β + 2p − 1]
setting
λn+p−1 =
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]
|an+p−1 | (p, n ∈ N; k ∈ N0 )
2β(p − α)
and
λp−1 = 1 −
it follows that f (z) =
5.1.
∑∞
∞
∑
λn+p−1 ,
n=1
n=1 λn+p−1 fn+p−1 (z).
This completes the proof of Theorem
Theorem 6 The class
∑∗
p (k, α, β)
99
is closed under convex linear combinations.
Proof. Let the functions
(27)
fj (z) = z
−p
+
∞
∑
|an+p−1,j | z n+p−1
(j = 1, 2),
n=1
be in the class
∑∗
p (k, α, β). It
is sufficient to show that the function h(z) defined by
h(z) = (1 − t)f1 (z) + tf2 (z) (0 ≤ t ≤ 1),
∑
is also in the class ∗p (k, α, β). Since
(28)
(29)
h(z) = z
−p
+
∞
∑
[(1 − t) |an+p−1,1 | + t |an+p−1,2 |] z n+p−1 ,
n=1
then with the aid of Theorem 2.2, we have
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1] [(1 − t) |an+p−1,1 | + t |an+p−1,2 |]
n=1
≤ (1 − t)
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1] |an+p−1,1 |
n=1
+t
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1] |an+p−1,2 |
n=1
≤ 2β(p − α)(1 − t) + 2β(p − α)t = 2β(p − α),
∑
which shows that h(z) ∈ ∗p (k, α, β). Hence we have proved Theorem 5.2.
6
Integral transforms
∑
In this section we consider integral transforms of function in the class ∗p (k, α, β).
In order to derive our results here, we need the following Lemma given by Aouf [1].
Lemma 1 [1]. Let the function f (z) ∈
0 ≤ δ < p, if and only if
(30)
∞
∑
∑∗
p
be given by (4). Then f (z) ∈
(n + p − 1 + δ) |an+p−1 | ≤ (p − δ).
n=1
∑∗
p [δ],
100
∑
Theorem 7 If f (z) is in the class ∗p (k, α, β), then the integral transforms
∫ 1
(31)
Fc+p−1 (z) = c
uc+p−1 f (uz)du, 0 < c < ∞,
≤ θ < p, where
{ k
}
p (p + αβ)(c + 2p) − β(p − α)c
θ = θ(k, p, α, β, c) = p
.
pk (p + αβ)(c + 2p) + β(p − α)c
are in the class
(32)
0
∑∗
p [θ], 0
The result is the best possible for the function f (z) given by
β(p − α) p
z (p ∈ N; k ∈ N0 ).
pk (p + αβ)
∑
∑∗
n+p−1 ∈
Proof. Suppose f (z) = z −p + ∞
n=1 |an+p−1 | z
p (k, α, β). Then we have
∫ 1
Fc+p−1 (z) = c
uc+p−1 f (uz)du
f (z) = z −p +
(33)
0
=z
−p
+
∞
∑
n=1
c |an+p−1 |
z n+p−1 .
n + c + 2p − 1
In view of Lemma 6.1, it is sufficient to show that
∞
∑
n+p−1+θ
(34)
Since f (z) ∈
∑∗
n=1
p (k, α, β),
p−θ
.
c |an+r−1 |
≤ 1.
n + c + 2p − 1
we have
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]
2β(p − α)
n=1
|an+p−1 | ≤ 1.
Thus (34) will be satisfied if
(n + p − 1 + θ)c
≤
(p − θ)(n + c + 2p − 1)
(35)
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]
for each n,
2β(p − α)
or
(36)
θ≤
p(n+p−1)k [(1+β)n+(2α−1)β+2p−1](n+c+2p−1)−(n+p−1)2β(p−α)c
.
(n+p−1)k [(1+β)n+(2α−1)β+2p−1](n+c+2p−1)+2β(p−α)c
Since the right hand side of (36) is an increasing function of n, putting n = 1 in
(36), we get
}
{ k
p (p + αβ)(c + 2p) − β(p − α)c
.
θ≤p
pk (p + αβ)(c + 2p) + β(p − α)c
Hence we have proved Theorem 6.1.
Putting k = 0 in Theorem 6.1, we obtain :
101
∑
Corollary 2∑If f (z) is in the class ∗p [α, β], then the integral transforms (31) are
in the class ∗p [θ], 0 ≤ θ < p, where
{
}
(p + αβ)(c + 2p) − β(p − α)c
(37)
θ = θ(p, α, β, c) = p
.
(p + αβ)(c + 2p) + β(p − α)c
The result is the best possible for the function f (z) given by
β(p − α) p
z (p ∈ N).
p + αβ
f (z) = z −p +
(38)
Remark 1 We observe that Corollary 6.2 improves the result obtained by Mogra
[3].
7
Convolution properties
∑
For the functions fj (z)(j = 1, 2) defined by (27) belonging to the class ∗p , we denote
by (f1 ∗ f2 )(z) the Hadamard product (or convolution) of the functions f1 (z) and
f2 (z), that is,
(f1 ∗ f2 )(z) = z −p +
(39)
∞
∑
|an+p−1,1 | · |an+p−1,2 | z n+p−1 .
n=1
Theorem
8 Let the functions f∑
j (z)(j = 1, 2) defined by (27) be in the class
∑∗
∗
(k,
α,
β)
.
Then
(f
∗
f
)(z)
∈
1
2
p (k, γ, β), where
p
{
γ =p 1−
(40)
β(1 + β)(p − α)2
pk (p + αβ)2 + β 2 (p − α)2
}
.
The result is sharp for the functions fj (z)(j = 1, 2) given by
fj (z) = z −p +
(41)
β(p − α) p
z
pk (p + αβ)
(j = 1, 2; p ∈ N; k ∈ N0 ).
Proof. Employing the technique used earlier by Schild and Silverman [5], we need
to find the largest γ such that
(42)
∞
∑
(n + p − 1)k [(1 + β)n + (2γ − 1)β + 2p − 1]
2β(p − γ)
n=1
(fj (z) ∈
Since fj (z) ∈
(43)
∑∗
p (k, α, β)(j
∑∗
p
(k, α, β) (j = 1, 2)).
= 1, 2), we readily see that
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]
n=1
|an+p−1,1 | . |an+p−1,2 | ≤ 1,
2β(p − α)
|an+p−1,j | ≤ 1 (j = 1, 2).
102
Therefore, by Cauchy-Schwarz inequality, we obtain
(44)
∞
√
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]
2β(p − α)
n=1
|an+p−1,1 | . |an+p−1,2 | ≤ 1.
Thus we only need to show that
[(1 + β)n + (2γ − 1)β + 2p − 1]
|an+p−1,1 | . |an+p−1,2 |
(p − γ)
(45) ≤
[(1 + β)n + (2α − 1)β + 2p − 1]
(p − α)
√
|an+p−1,1 | . |an+p−1,2 | (p, n ∈ N; k ∈ N0 ),
or, equivalently, that
√
|an+p−1,1 | . |an+p−1,2 | ≤
(46)
(p − γ)[(1 + β)n + (2α − 1)β + 2p − 1]
(p, n ∈ N; k ∈ N0 ).
(p − α)[(1 + β)n + (2γ − 1)β + 2p − 1]
Hence, in light of the inequality (44), it is sufficient to prove that
(n + p −
(47)
1)k [(1
2β(p − α)
≤
+ β)n + (2α − 1)β + 2p − 1]
(p − γ)[(1 + β)n + (2α − 1)β + 2p − 1]
(p, n ∈ N; k ∈ N0 ).
(p − α)[(1 + β)n + (2γ − 1)β + 2p − 1]
It follows from (47) that
γ≤p
(48)
−
2β(1 + β)(p − α)2 (n + 2p − 1)
(p, n ∈ N; k ∈ N0 ).
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]2 + 4β 2 (p − α)2
Now, defining the function G(n) by
G(n) = p−
(49)
2β(1 + β)(p − α)2 (n + 2p − 1)
(p, n ∈ N; k ∈ N0 ),
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]2 + 4β 2 (p − α)2
we see that G(n) is an increasing function of n. Therefore, we conclude that
(50)
{
γ ≤ G(1) = p 1 −
β(1 + β)(p − α)2
pk (p + αβ)2 + β 2 (p − α)2
which evidently completes the proof of Theorem 7.1.
}
,
103
Remark 2 We observe that Theorem 7.1 with k = 0 improves the result obtained
by Mogra [3].
Using arguments similar to those in the proof of Theorem 7.1, we obtain the
following result:
∑
Theorem 9 Let the function f1 (z) defined by (27) be in the class ∗p (k, α, β).
∑
Suppose also that ∑
the function f2 (z) defined by (27) be in the class ∗p (k, φ, β).
Then (f1 ∗ f2 )(z) p∗ (k, δ, β), where
}
{
β(1 + β)(p − α)(p − φ)
.
(51)
δ =p 1− k
p (p + αβ)(p + φβ) + β 2 (p − α)(p − φ)
The result is sharp for the functions fj (z)(j = 1, 2) given by
(52)
f1 (z) = z −p +
β(p − α) p
z (p ∈ N; k ∈ N0 )
pk (p + αβ)
f2 (z) = z −p +
β(p − φ) p
z (p ∈ N; k ∈ N0 ).
pk (p + φβ)
and
Remark 3 We observe that Theorem 7.2 with k = 0 improves the result obtained
by Mogra [3].
Theorem
∑∗ 10 Let the functions fj (z)(j = 1, 2) defined by (27) be in the
class p (k, α, β). Then the function h(z) defined by
h(z) = z −p +
(53)
∞
∑
(|an+p−1,1 |2 + |an+p−1,2 |2 )z n+p−1 ;
n=1
belongs to the class
∑∗
p (k, ζ, β),
{
ζ =p 1−
(54)
where
2β(1 + β)(p − α)2
pk (p + αβ)2 + 2β 2 (p − α)2
}
.
The result is sharp for the functions fj (z)(j = 1, 2) given already by (41).
Proof. Noting that
]2
∞ [
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]
|an+p−1,j |2
2β(p − α)
n=1
(55)
≤
[
∞
∑
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]
n=1
2β(p − α)
]2
|an+p−1,j |
≤ 1 (j = 1, 2),
104
∑
for fj (z) ∈ ∗p (k, α, β)(j = 1, 2), we have
(56)
[
]2
∞
∑
1 (n+p−1)k [(1+β)n+(2α−1)β +2p−1] (
n=1
2
2β(p−α)
)
|an+p−1,1 |2 +|an+p−1,2 |2 ≤ 1.
Therefore, we have to find the largest ζ such that
[(1 + β)n + (2ζ − 1)β + 2p − 1]
≤
(p − ζ)
1 (n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]2
(n ∈ N),
2
2β(p − α)2
(57)
that is, that
(58) ζ ≤ p−
4β(1 + β)(n + 2p − 1)(p − α)2
(n ∈ N).
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]2 + 8β 2 (p − α)2
Now, defining a function D(n) by
D(n) = −p
(59)
4β(1 + β)(n + 2p − 1)(p − α)2
(n + p − 1)k [(1 + β)n + (2α − 1)β + 2p − 1]2 + 8β 2 (p − α)2
(n ∈ N).
We observe that D(n) is an increasing function of n(n ∈ N ). Thus, we conclude that
(60)
{
ζ ≤ D(1) = p 1 −
2β(1 + β)(p − α)2
pk (p + αβ)2 + 2β 2 (p − α)2
}
,
which completes the proof of Theorem 7.3.
Putting k = 0 in Theorem 7.3,we obtain the following result.
Corollary
3 Let
∑∗ the functions fj (z)(j = 1, 2) defined by (27) be in the class
∑∗
(0,
α,
β)
=
p (α, β). Then the function h(z) defined by (53) belongs to the class
∑p∗
p (ζ, β), where
(61)
{
ζ =p 1−
2β(1 + β)(p − α)2
(p + αβ)2 + 2β 2 (p − α)2
}
.
105
References
[1] M. K. Aouf, Certain classes of meromorphic multivalent functions with positive
coefficients, Math. Comput. Modelling, 47(2008), no. 3-4, 328-340.
[2] B. A. Farsin and M. Darus, On certain meromorphic functions with positive
coefficients, South. Asian Bull. Math. 28(2004), 615- 623.
[3] M. L. Mogra, Meromorphic multivalent functions with positive coefficients I,
Math. Japon. 35 (1990) , no.1, 1-11.
[4] M. L. Mogra, T. R. Reddy and O. P. Juneja, Meromorphic univalent functions
with positive coefficients, Bull. Austral. Math. Soc. 32 (1985), 161-176.
[5] A. Schild and H. Silverman, Convolution of univalent functions with negative
coefficients, Ann. Univ. Mariae Curie-Sklodowska Sect. A 29(1975), 99-107.
R. M. El-Ashwah
Damietta University
Faculty of Science
New Damietta 34517, Egypt
e-mail: relashwah@yahoo.com
M. K. Aouf
Mansoura University
Faculty of Science
e-mail: mkaouf127@yuhoo.com
Fekete–Szegö inequality for certain classes of analytic
functions 1
Roberta Bucur, Loriana Andrei, Daniel Breaz, Sabina Ligia Georgescu
Abstract
In this paper we introduce new classes of analytic functions and we obtain
sharp upper bounds of the functional |a3 −ζa22 |, for analytic functions belonging
to these classes.
Key words and phrases: analytic, starlike, convex, coefficient bounds,
Fekete–Szegö functional, differential operator.
1
Introduction
Let A be the class of analytic functions of the form
(1.1)
f (z) = z +
∞
∑
aj z j ,
z∈U
j=2
in the open unit disk U = {z : |z| < 1}.
By S we denote the subclass of A consisting of univalent functions.
Principle of Subordination (see [7]): If f and g are two analytic functions in U ,
we say that f is subordinate to g, written as f ≺ g, if there exists a Schwarz function
w analytic in U, with w(0) = 0 and |w(z)| < 1, such that f (z) = g(w(z)), for all
z ∈ U. In particular, if the function g is univalent in U , the above subordination is
equivalent to f (0) = g(0) and f (U ) ⊂ g(U ).
Let B denote the class of Schwarz functions w(z) of the form
(1.2)
w(z) =
∞
∑
cj z j ,
z ∈ U,
j=1
1
Received 10 September, 2014
Accepted for publication (in revised form) 20 October, 2014
107
108
R. Bucur, L. Andrei, D. Breaz, S. L. Georgescu
which are analytic in the unit disk U = {z : |z| < 1} and satisfying the condition
w(0) = 0 and |w(z)| < 1.
We state the following well-known results for the class B.
Lemma 1.1. (Schwarz lemma) If w(z) ∈ B, then |w(z)| ≤ |z| and |c1 | ≤ 1 are
obtained.
Lemma 1.2. [8] If w(z) ∈ B, then |c2 | ≤ 1 − |c1 |2 .
A function f ∈ A is said to be in the class R0 if it satisfies
(
)
f (z)
(1.3)
Re
> 0, z ∈ U.
z
The class R0 is a particular case of the class of close to star functions defined by
Reade [12]. The class R0 and its subclasses were studied by several authors including
MacGregor [5].
Let R be the class of functions f ∈ A and satisfying
( ′
)
Re f (z) > 0,
(1.4)
z ∈ U.
The class R was introduced by Noshiro [9] and Warschawski [14] and it was shown
by them that R is a class of univalent functions. The class R and its subclasses were
investigated by many authors (see [2], [3]).
A function f ∈ A is said to be in the class Q (α, β, γ) with α, β > 0 and 0 ≤ γ <
α + β ≤ 1, if it satisfies the condition
(
)
f (z)
′
Re α
+ βf (z) ≥γ,
z
(1.5)
z ∈ U.
This class was considered and studied by Zhi- Gang Wang, Chun-yi gao and ShaoMou yuan [10]. In [11], D. Vamshee Krishna and T. Ramreddy studied the second
Hankel determinant for the function f ∈ A belonging to the class Q (α, β, γ), by
using Toeplitz determinants.
In the present investigation we introduce the following subclass of A:


(1.6)
R (α, β) =

(
f ∈ A; Re
′
(α+β)zf (z)+βz 2 f ′′ (z)
′
αf (z)+βzf (z)
)
> 0,



α, β > 0, 0 < α + β ≤ 1, z ∈ U
or, equivalently
{
(1.7)
R (α, β) =
f ∈ A;
′
(α+β)zf (z)+βz 2 f ′′ (z)
′
αf (z)+βzf (z)
≺
1+z
1−z ,
α, β > 0, 0 < α + β ≤ 1, z ∈ U
}
.
,
Fekete–Szegö inequality for certain classes of analytic functions
109
For α = 1 and β = 0, R (1, 0) is the class of starlike functions, and for α = 0 and
β = 1, R (0, 1) is the class of convex functions. These classes were investigated and
studied by Ma and Minda [6]. They have obtained the Fekete-Szego inequality for
the functions belonging to the class of starlike and convex functions.
For α + β = 1, Gagandeep Singh and Gurcharanjit Singh [13] studied the third
Hankel determinant for the class R (1 − β, β) .
We also introduce the following subclass of the class R (α, β):
{
(1.8)
R (α, β; A, B) =
f ∈ A;
′
(α+β)zf (z)+βz 2 f ′′ (z)
′
αf (z)+βzf (z)
≺
1+Az
1+Bz ,
−1 ≤ B < A ≤ 1, α, β > 0, 0 < α + β ≤ 1, z ∈ U
}
,
In the present investigation, we will establish Fekete-Szegö inequality for the
functions belonging to the new classes R (α, β) and R (α, β; A, B). Many other
authors obtained Fekete–Szegö inequalities for different classes of functions (see [1]).
Some consequences of the main results are also mentioned.
Unless otherwise mentioned, we assume throughout this paper that 0 < δ ≤
1, −1 ≤ B < A ≤ 1, α, β > 0, 0 < α + β ≤ 1 and z ∈ U .
2
Main results
Theorem 2.1. Let f (z) ∈ R(α, β). Then:
i). If ζ is complex,
(2.1)
|a3 −
ζa22 |
≤



(α+β)
(α+3β)
4(α+β)2
|σ
(α+2β)2
if |σ − ζ|≤ γ;
− ζ|
if |σ − ζ| ≥ γ.
ii). If ζ is real,
(2.2)
|a3 − ζa22 | ≤







4(α+β)2
(σ − ζ)
(α+2β)2
(α+β)
(α+3β)
4(α+β)2
(ζ − σ)
(α+2β)2
if ζ≤ σ − γ;
if σ − γ ≤ ζ ≤σ + γ;
if ζ≥ σ + γ;
where
(2.3)
σ=
3 (α + 2β)2
,
4 (α + 3β) (α + β)
γ=
(α + 2β)2
.
4 (α + β) (α + 3β)
and
(2.4)
110
Proof. By definition of R(α, β), we obtain
′
(α + β) zf (z) + βz 2 f ′′ (z)
1 + w(z)
=
,
′
1 − w(z)
αf (z) + βzf (z)
(2.5)
where w(z) ∈ B. Expanding the series from (2.5), we have
(
)
1 + w(z)
= 1 + 2c1 z + 2 c2 + c21 z 2 + . . . ,
1 − w(z)
(2.6)
and
(2.7)
′
(α + β) z + 2a2 (α + 2β) z 2 + 3a3 (α + 3β) z 3 + ...
(α + β) zf (z) + βz 2 f ′′ (z)
=
.
′
(α + β) z + a2 (α + 2β) z 2 + a3 (α + 3β) z 3 + ...
αf (z) + βzf (z)
From (2.6) and (2.7), we get
(α + β) z + 2a2 (α + 2β) z 2 + 3a3 (α + 3β) z 3 + ... =
(2.8)
[
(
)
][
]
1 + 2c1 z + 2 c2 + c21 z 2 + . . . (α + β) z + a2 (α + 2β) z 2 + a3 (α + 3β) z 3 + ... .
Identifying terms, we obtain:
(2.9)
a2 =
2 (α + β)
c1 ,
α + 2β
and
(2.10)
a3 =
]
(α + β) [ 2
3c1 + c2 .
(α + 3β)
So,
]
4 (α + β)2 2
(α + β) [ 2
3c1 + c2 − ζ
c1 .
(α + 3β)
(α + 2β)2
|a3 − ζa22 | =
(2.11)
Perform calculations, we obtain
(2.12)
|a3 −
ζa22 |
(α + β)
4 (α + β)2
=
c2 +
(α + 3β)
(α + 2β)2
{
3 (α + 2β)2
−ζ
4 (α + 3β) (α + β)
By setting
σ=
3 (α + 2β)2
,
4 (α + 3β) (α + β)
we get
|a3 − ζa22 | =
(α + β)
4 (α + β)2
c2 +
(σ − ζ) c21 .
(α + 3β)
(α + 2β)2
}
c21 .
So,
(2.13)
|a3 − ζa22 | ≤
(α + β)
4 (α + β)2
|c2 | +
|σ − ζ||c1 |2 .
(α + 3β)
(α + 2β)2
Therefore, by using Lemma 1.2, (2.13) becomes
(2.14)
|a3 − ζa22 | ≤
4 (α + β)2
(α + β)
[|σ − ζ| − γ] |c1 |2 ,
+
(α + 3β)
(α + 2β)2
where γ is given by (2.4). If |σ − ζ| ≤ γ, then from (2.14) we get
|a3 − ζa22 | ≤
(2.15)
(α + β)
.
(α + 3β)
If |σ − ζ| ≥ γ, then again from (2.14) we obtain
|a3 − ζa22 | ≤
(2.16)
4 (α + β)2
|σ − ζ|.
(α + 2β)2
Now, we discuss the case when ζ is real.
Case1. For ζ ≤ σ, from (2.14) we have
(2.17)
|a3 −
ζa22 |
4 (α + β)2
(α + β)
2
+
≤
2 [(σ − γ) − ζ] |c1 | .
(α + 3β)
(α + 2β)
If ζ ≤ σ − γ, then (2.17) becomes
(2.18)
|a3 −
ζa22 |
4 (α + β)2
≤
(σ − ζ) .
(α + 2β)2
If σ − γ ≤ ζ, then (2.17) becomes
|a3 − ζa22 | ≤
(2.19)
(α + β)
.
(α + 3β)
Case 2. For ζ ≥ σ, from (2.14) we have
(2.20)
|a3 − ζa22 | ≤
(α + β)
4 (α + β)2
+
[ζ − (σ + γ)] |c1 |2 .
(α + 3β)
(α + 2β)2
If ζ ≤ σ + γ, then (2.20) becomes
(2.21)
|a3 − ζa22 | ≤
(α + β)
.
(α + 3β)
111
112
If ζ ≥ σ + γ, then (2.20) becomes
|a3 − ζa22 | ≤
(2.22)
4 (α + β)2
(ζ − σ) ,
(α + 2β)2
and the proof is completed.
Putting α = 1 and β = 0 in Theorem 2.1, we obtain the following result
Corollary 2.1. Let f (z) ∈ R(1, 0). Then:
{
1
2
(2.23)
|a3 − ζa2 | ≤
|3 − 4ζ|
if | 34 − ζ|≤ 41 ;
if | 34 − ζ| ≥ 41 .
ii). If ζ is real,

 3 − 4ζ
1
|a3 − ζa22 | ≤

4ζ − 3
(2.24)
if ζ≤ 12 ;
if 21 ≤ ζ ≤ 1;
if ζ≥ 1.
This result coincides with that of Keogh and Merkes [4].
Corollary 2.2. Let f (z) ∈ R(0, 1). Then:
{ 1
2
3
(2.25)
|a3 − ζa2 | ≤
|1 − ζ|
if |1 − ζ|≤ 13 ;
if |1 − ζ| ≥ 13 .
ii). If ζ is real,
|a3 − ζa22 | ≤
(2.26)

 1−ζ

1
3
ζ −1
if ζ≤ 23 ;
if 32 ≤ ζ ≤ 43 ;
if ζ≥ 43 .
This result coincides with that of Keogh and Merkes [4].
Theorem 2.2. Let f (z) ∈ R(α, β; A, B). Then:
(2.27)
|a3 −
ζa22 |
≤



(A−B)(α+β)
2(α+3β)
(A−B)2 (α+β)2
|σ
(α+2β)2
if |σ − ζ|≤ γ;
− ζ|
if |σ − ζ| ≥ γ.
113
ii). If ζ is real,
(2.28)
|a3 − ζa22 | ≤







(A−B)2 (α+β)2
(σ − ζ)
(α+2β)2
(A−B)(α+β)
2(α+3β)
(A−B)2 (α+β)2
(ζ − σ)
(α+2β)2
if ζ≤ σ − γ;
if σ − γ ≤ ζ ≤σ + γ;
if ζ≥ σ + γ.
where
(2.29)
σ=
(A − 2B) (α + 2β)2
,
2 (α + 3β) (A − B) (α + β)
γ=
(α + 2β)2
.
2 (A − B) (α + β) (α + 3β)
and
(2.30)
Proof. By definition of R(α, β; A, B), we obtain
′
(α + β) zf (z) + βz 2 f ′′ (z)
1 + Aw(z)
=
,
′
1 + Bw(z)
αf (z) + βzf (z)
(2.31)
where w(z) ∈ B. Expanding the series from (2.31), we have
(2.32)
(
)
1 + Aw(z)
= 1 + (A − B) c1 z + (A − B) c2 − Bc21 z 2 + . . .
1 + Bw(z)
and
(2.33)
′
(α + β) z + 2a2 (α + 2β) z 2 + 3a3 (α + 3β) z 3 + ...
(α + β) zf (z) + βz 2 f ′′ (z)
=
.
′
(α + β) z + a2 (α + 2β) z 2 + a3 (α + 3β) z 3 + ...
αf (z) + βzf (z)
Identifying terms, we obtain:
(2.34)
a2 =
(A − B) (α + β)
c1 ,
α + 2β
and
(2.35)
a3 =
]
(A − B) (α + β) [
(A − 2B) c21 + c2 .
2 (α + 3β)
So,
(2.36) |a3 − ζa22 | =
]
(A − B)2 (α + β)2 2
(A − B) (α + β) [
(A − 2B) c21 + c2 − ζ
c1 .
2 (α + 3β)
(α + 2β)2
114
Perform calculations, we obtain
(2.37)
{
}
(A − B) (α + β)
(A − B)2 (α + β)2
(A − 2B) (α + 2β)2
2
|a3 −ζa2 | =
c2 +
− ζ c21 .
2
2 (α + 3β)
2
(α
+
3β)
(A
−
B)
(α
+
β)
(α + 2β)
By setting
σ=
(A − 2B) (α + 2β)2
,
2 (α + 3β) (A − B) (α + β)
we get
(A − B) (α + β)
(A − B)2 (α + β)2
c2 +
(σ − ζ) c21 .
2 (α + 3β)
(α + 2β)2
|a3 − ζa22 | =
So,
(2.38)
|a3 − ζa22 | ≤
(A − B) (α + β)
(A − B)2 (α + β)2
|c2 | +
|σ − ζ||c1 |2 .
2 (α + 3β)
(α + 2β)2
Therefore, by using Lemma 1.2, (2.38) becomes
(2.39)
|a3 − ζa22 | ≤
(A − B) (α + β) (A − B)2 (α + β)2
+
[|σ − ζ| − γ] |c1 |2 ,
2 (α + 3β)
(α + 2β)2
where γ is given by (2.30). If |σ − ζ| ≤ γ, then from (2.39) we get
|a3 − ζa22 | ≤
(2.40)
(A − B) (α + β)
.
2 (α + 3β)
If |σ − ζ| ≥ γ, then again from (2.39) we obtain
|a3 − ζa22 | ≤
(2.41)
(A − B)2 (α + β)2
|σ − ζ|.
(α + 2β)2
Now, we discuss the case when ζ is real.
Case1. For ζ ≤ σ, from (2.39) we have
(2.42)
|a3 − ζa22 | ≤
(A − B) (α + β) (A − B)2 (α + β)2
+
[(σ − γ) − ζ] |c1 |2 .
2 (α + 3β)
(α + 2β)2
If ζ ≤ σ − γ, then (2.42) becomes
(2.43)
|a3 − ζa22 | ≤
(A − B)2 (α + β)2
(σ − ζ) .
(α + 2β)2
If σ − γ ≤ ζ, then (2.42) becomes
|a3 − ζa22 | ≤
(2.44)
(A − B) (α + β)
.
2 (α + 3β)
Case 2. For ζ ≥ σ, from (2.39) we have
(2.45)
|a3 − ζa22 | ≤
(A − B) (α + β) (A − B)2 (α + β)2
+
[ζ − (σ + γ)] |c1 |2 .
2
2 (α + 3β)
(α + 2β)
If ζ ≤ σ + γ, then (2.45) becomes
|a3 − ζa22 | ≤
(2.46)
(A − B) (α + β)
.
2 (α + 3β)
If ζ ≥ σ + γ, then (2.45) becomes
|a3 − ζa22 | ≤
(2.47)
(A − B)2 (α + β)2
(ζ − σ) ,
(α + 2β)2
and the proof is completed.
Corollary 2.3. Let f (z) ∈ R1 (1, 0; A, B). Then:
{
|a3 − ζa22 | ≤
(2.48)
(A−B)
2
if |σ − ζ|≤ γ;
if |σ − ζ| ≥ γ.
(A − B)2 |σ − ζ|
ii). If ζ is real,
(2.49)
|a3 − ζa22 | ≤

2

 (A − B) (σ − ζ)
(A−B)
2

 (A − B)2 (ζ − σ)
if ζ≤ σ − γ ;
if σ − γ ≤ ζ ≤σ + γ;
if ζ≥ σ + γ;
where
(2.50)
σ=
(A − 2B)
,
2 (A − B)
γ=
1
.
2 (A − B)
and
(2.51)
115
116
Corollary 2.4. Let f (z) ∈ R1 (0, 1; A, B). Then:
{
|a3 − ζa22 | ≤
(2.52)
(A−B)
6
(A−B)2
|σ
4
− ζ|
if |σ − ζ|≤ γ;
if |σ − ζ| ≥ γ.
ii). If ζ is real,
(2.53)
|a3 −
ζa22 |
≤





(A−B)2
(σ − ζ)
4
(A−B)
6
(A−B)2
(ζ − σ)
4
if ζ≤ σ − γ ;
if σ − γ ≤ ζ ≤σ + γ;
if ζ≥ σ + γ;
where
(2.54)
σ=
2 (A − 2B)
,
3 (A − B)
γ=
2
.
3 (A − B)
and
(2.55)
References
[1] R.Bucur, L.Andrei, D.Breaz, Coeffcient Bounds and Fekete-Szego Problem for
a Class of Analytic Functions Defined by Using a New Differential Operator,
Applied Mathematical Sciences, Vol. 9, 2015, no. 28, 1355 - 1368.
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Roberta Bucur
Department of Mathematics, University of Piteş
Târgul din Vale Street No.1, 110040 Piteşti, Romania
e-mail: roberta bucur@yahoo.com
Loriana Andrei
Department of Mathematics and Computer Science, University of Oradea
1 Universitatii Street, 410087 Oradea, Romania
e-mail: lori andrei@yahoo.com
Daniel Breaz
Department of Mathematics,”1 Decembrie 1918” University of Alba Iulia
N. Iorga Street,No. 11-13, 510009 Alba Iulia,Romania
e-mail: dbreaz@uab.ro
Sabina Ligia Georgescu
National College ”Unirea” of Targu Mures
Mihai Viteazul str., No. 17, Târgu Mureş, Romania
e-mail: sabinageorgescu1997@gmail.com

2014 Volume 22 No.2 GENERAL MATHEMATICS Daniel Florin

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