One some differential subordination involving the Bessel-Struve kernel function

0.17 MB·

by  Saiful R Mondal

ONE SOME DIFFERENTIAL SUBORDINATION INVOLVING THE
BESSEL-STRUVE KERNEL FUNCTION
SAIFUL R. MONDAL AND MOHAMMED AL-DHUAIN
Abstract. In this article we study the inclusion properties of the Bessel-Struve kernel
functionsinthe Janowskiclass. Inparticular,wefindthe conditionsforwhichtheBessel-
Struve kernel functions maps the unit disk to right half plane. Some open problem in
6
1 this aspectarealsogiven. The thirdorderdifferentialsubordinationinvolvingthe Bessel-
0 Struve kernel is also considered. The results are derived by defining suitable classes of
2 admissible functions. One of the recurrence relation of the Bessel-Struve kernel play an

n important role to derive the main results.
a
J
8 1. Introduction and Preliminaries
2
1.1. Bessel-Struve Kernel functions. Consider the Bessel-Struve kernel function S
α,λ
]
defined on the unit disk ∆ = z : z < 1 as
V
{ | | }
C 1
S (z) := j (iλz) ih (iλz), α > , λ C, (1.1)
. α,λ α α
h − −2 ∈
t
a where, j (z) := 2αz−αΓ(α+1)J (z) and h (z) := 2αz−αΓ(α+1)H (z) are respectively
α α α α
m
known as the normalized Bessel functions and the normalized Struve functions of first
[ kind of index α. The Bessel-Struve transformation and Bessel-Struve kernel functions are

1 appeared in many article. In [13], Hamem et. al. studies an analogue of the Cowling-Price
v
theorem for the Bessel-Struve transform defined on real domain and also provide Hardy’s
0
2 type theorem associated with this transform. The Bessel-Struve intertwining operator on
9 C is considered in [10]. The fock space of the Bessel-Struve kernel functions is discussed
7
in [11].
0
. The kernel z S (z), λ C is the unique solution of the initial value problem
1 α,λ
7→ ∈
0
λΓ(α+1)
6 u(z) = λ2u(z), u(0) = 1,u′(0) = . (1.2)
1 Lα √πΓ(α+ 3)
: 2
v
Here , α > 1/2 is the Bessel-Struve operator given by
i α
X L −
d2u 2α+1 du du
r
a Lα(u(z)) := dz2(z)+ z (cid:18)dz(z)− dz(0)(cid:19). (1.3)
Now, the Bessel functions and the Struve functions of order α respectively have the
power series
∞ ( 1)n z 2n+α ∞ ( 1)n z 2n+α+1
J (z) = − 2 and H (z) := − 2 .
α n!Γ(α+(cid:0) n(cid:1)+1) α Γ n+α+(cid:0) 3(cid:1)Γ n+ 3
Xn=0 Xn=0 2 2
(cid:0) (cid:1) (cid:0) (cid:1)
2010 Mathematics Subject Classification. 30C45,33C10, 30C80, 40G05.
Key words and phrases. Bessel functions, Struve functions, Bessel-Struve kernel, Starlike, Close-to-
convex.
First author thanks the Deanship of Scientific Research at King Faisal University for funding this
∗
work under project number 150244.
1
2 SAIFUL R.MONDAL ANDMOHAMMED AL-DHUAIN
This implies that S possesses the power series
α,λ
∞ Γ(α+1)Γ n+1
S (z) := 2 λnzn. (1.4)
α,λ √πn!Γ n +(cid:0)α+(cid:1)1
Xn=0 2
(cid:0) (cid:1)
The kernel S also have the integral representation
α,λ
2Γ(α+1) 1 1
S (z) := (1 t2)α−2eλztdt. (1.5)
α,λ √πΓ α+ 1 Z −
2 0
(cid:0) (cid:1)
Now from (1.2) and (1.3) it is evident that S satisfy the differential equation
α
z2U′′(z)+(2α+1)zU′(z) zλ2U(z) = zM, (1.6)
−
where M = 2λΓ(α+1) √π Γ(α+ 1) −1.
2
It can be shown that(cid:0)S = S satis(cid:1)fy the recurrence relation
α,1 α
zS′ (z) = 2αS (z) 2αS (z). (1.7)
α α−1 − α
1.2. Differentialsubordinations. Inthissectionsadetailsintroductionabouttheclasses
of univalent functions theory, admissible functions and fundamental results about the dif-
ferential subordination of different orders are given. The differential subordinations and
its application are mainly encompassed the first order and second order differential sub-
ordination till the introduction of the third order differential technique by Antonino and
Miller [9].
Let denote the class of analytic functions f defined in the open unit disk ∆ normalized
A
by the conditions f(0) = 0 = f′(0) 1 and have the form
−
∞
f(z) = z + a zn+1. (1.8)
n+1
Xn=1
If f and g are analytic in ∆, then f is subordinate to g, written f(z) g(z), if there is
≺
an analytic self-map w of ∆ satisfying w(0) = 0 and f = g w. For 1 B < A 1, let
◦ − ≤ ≤
[A,B] be the class consisting of normalized analytic functions p(z) = 1+c z + in ∆
1
P ···
satisfying
1+Az
p(z) .
≺ 1+Bz
For instance, if 0 β < 1, then [1 2β, 1] is the class of functions p(z) = 1+c z+
1
≤ P − − ···
satisfying Rep(z) > β in ∆.
The class ∗[A,B] of Janowski starlike functions [14] consists of f satisfying
S ∈ A
zf′(z)
[A,B].
f(z) ∈ P
For 0 β < 1, ∗[1 2β, 1] := ∗(β) is the usual class of starlike functions of order
≤ S − − S
β; ∗[1 β,0] := ∗ = f : zf′(z)/f(z) 1 < 1 β , and ∗[β, β] := ∗[β] =
S − Sβ { ∈ A | − | − } S − S
f : zf′(z)/f(z) 1 < β zf′(z)/f(z) + 1 . These classes have been studied, for
{ ∈ A | − | | |}
example, in [1, 8]. A function f is said to be close-to-convex of order β [12, 17] if
∈ A
Re(zf′(z)/g(z)) > β for some g ∗ := ∗(0).
Let (∆) be the class of func∈tioSns whSich are analytic in ∆. For n N := 1,2,3,...
and a HC, consider the class [a,n] defined as ∈ { }
∈ H
[a,n] := f (∆) : f(z) = a+a zn +a zn+1 +... , (1.9)
n n+1
H { ∈ H }
and suppose that = [0,1].
0
H H
BESSEL-STRUVE KERNEL FUNCTION 3
The theory of the differential subordination in ∆ was introduced by Miller and Mocanu
[17] and subsequently many researcher either apply this concept to study the geometric
properties of analytic functions defined on ∆ and developed or reproduced several other
theory for subclasses of the univalent functions theory (See [2–7] and references their in).
Following result is important to show that the Bessel-Struve kernel have a relation with
Janowski class.
Lemma 1.1. [17, 19] Let Ω C, and Ψ : C3 ∆ C satisfy
⊂ × →
Ψ(iρ,σ,µ+iν;z) Ω
6∈
whenever z ∆, ρ real, σ (1+ρ2)/2 and σ+µ 0. If p is analytic in ∆ with p(0) = 1,
∈ ≤ − ≤
and Ψ(p(z),zp′(z),z2p′′(z);z) Ω for z ∆, then Rep(z) > 0 in ∆.
∈ ∈
Theory of the second-order differential subordinations in ∆ have extended to third-
order differential subordinations by Antonino and Miller [9]. They determined properties
of analytic functions p in ∆ that satisfy the following third order differential subordination:
ψ(p(z),zp′(z),z2p′′(z),z3p′′′(z);z) Ω,
{ } ⊂
where, z ∆, Ω is any set in C, and ψ : C4 ∆ C.
∈ × →
Next we write down few important definition and results from [9] which are required in
sequel.
Definition 1.1. [9, p. 440] Let ψ : C4 ∆ C and h be univalent in ∆. If p is analytic
× →
in ∆ and satisfies the third-order differential subordination
ψ(p(z),zp′(z),z2p′′(z),z3p′′′(z);z) h(z) (1.10)
≺
then p is called a solution of the differential subordination. A 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 (1.10). A dominant q˜ that satisfies q˜ q for all dominant of
≺ ≺
(1.10) is called the best dominant of (1.10). Note that the best dominant is unique up to a
rotation of ∆.
Definition 1.2. [9, p. 441] Let denote the set of functions Q that are analytic and
Q
univalent on the set ∆ E(q), where
\
E(q) = ζ ∂∆ : limq(z) = ,
{ ∈ z→ζ ∞}
and are such that
min q′(ζ) = ρ > 0
| |
for ζ ∆ E(q). Further, let the subclass of for which q(0) = a be denoted by (0) = .
0
∈ \ Q Q Q
Definition 1.3. [9, p. 449] Let Ω be any set in C, q and n N 1 . The class of
admissible function Ψ [Ω,q] consists of those function∈s ψQ: C4 ∆∈ \C{ t}hat satisfy the
n
× →
admissibility condition: ψ(r,s,t,u;z) / Ω, whenever
∈
t ζq′′(ζ)
r = q(ζ), s = mζq′(ζ), Re +1 mRe +1
(cid:18)s (cid:19) ≥ (cid:18) q′(ζ) (cid:19)
and
u ζ2q′′′(ζ)
Re m2Re +1
s ≥ (cid:18) q′(ζ) (cid:19)
(cid:16) (cid:17)
where z ∆, ζ ∆ E(q), and m n.
∈ ∈ \ ≥
4 SAIFUL R.MONDAL ANDMOHAMMED AL-DHUAIN
Theorem 1.1. [9, p. 449] Let p [a,n] with n 2. Also let q (a) and satisfy the
∈ H ≥ ∈ Q
following conditions:
ζ2q′′′(ζ) zp′(z)
Re 0 and m, (1.11)
(cid:18) q′(ζ) (cid:19) ≥ (cid:12) q′(ζ) (cid:12) ≤
(cid:12) (cid:12)
(cid:12) (cid:12)
where z ∆, ζ ∆ E(q), and m n. If Ω is any(cid:12)set in (cid:12)C, ψ Ψ [Ω,q] and
n
∈ ∈ \ ≥ ∈
ψ(p(z),zp′(z),z2p′′(z),z3p′′′(z);z) Ω,
∈
then p(z) q(z).
≺
Recently, Tang and Deniz [20] applied the above third-order differential subordination
concept to an operator involving the generalized Bessel functions by considering suitable
class of admissible functions.
In section 2, we obtain the sufficient condition for which the Bessel-Struve kernel func-
tions have relation with the functions of the Janowski class. The classes of admissible
functions applicable for the Besse-Struve kernel function in ∆ are introduced in Section 3
andthenresultsrelatedtothethirdorderdifferential subordinationsinvolving thisfunction
are derived.
2. Inclusion of Bessel-Struve kernel in the Janowski Class
Theorem 2.1. Let 1 B 3 2√2 0.171573. Suppose B < A 1, and λ, α R
− ≤ ≤ − ≈ ≤ ∈
satisfy
α max 0, |λ| λ(1+A)(1+B)+M(1+B)2 . (2.1)
≥ n 2 (cid:12) A−B (cid:12)o
(cid:12) (cid:12)
Further let A, B, α and λ satisfy either t(cid:12)he inequality (cid:12)
λ (1+B)2 1+B
4α2 4α λ(A+B)+2MB + λ(1+A)+M(1+B) +2α
−A B (cid:12) 1 B (cid:12) 1 B
− (cid:12) (cid:0) (cid:1) − (cid:0) (cid:1)(cid:12) −
λ2(1(cid:12) B2)(λ(1 A)+M(1 B))(λ(1+A)+M(1+B)) (cid:12)
(cid:12) (cid:12)
− − − , (2.2)
≥ (A B)2
−
whenever
4α λ(A+B)+2MB (1 B)+(1+B)2 λ(1+A)+M(1+B) 2λ3(1 B)(A B)
− ≥ − −
(cid:12) (cid:0) (cid:1) (cid:0) (cid:1)(cid:12) (2.3)
(cid:12) (cid:12)
or the inequality
1+B
4αλ λ(A+B)+2MB + λ2(1+A)(1+B)+λM(1+B)2 2
1 B
(cid:0) (cid:0) (cid:1) − (cid:0) (cid:1)(cid:1)
4((λ2(1 AB)+λM(1 B2))2 (λ2(1 A)(1 B)+λM(1 B)2)
≤ − − − − − −
(λ2(1+A)(1+B)+λM(1+B)2)) 4α2 +2α1+B λ2(1−AB)+λM(1−B2) 2 , (2.4)
1 B − A−B
(cid:16) (cid:17)
(cid:0) − (cid:1)
whenever
4α λ(A+B)+2MB (1 B)+(1+B)2 λ(1+A)+M(1+B) < 2λ3(1 B)(A B).
− − −
(cid:12) (cid:0) (cid:1) (cid:0) (cid:1)(cid:12) (2.5)
(cid:12) (cid:12)
If (1+B)S (z) = (1+A), then S (z) [A,B].
α,λ α,λ
6 ∈ P
BESSEL-STRUVE KERNEL FUNCTION 5
Proof. Define the analytic function p : ∆ C by
→
(1 A) (1 B)S (z)
α,λ
p(z) := − − − ,
−(1+A) (1+B)S (z)
α,λ
−
Then
(1 A)+(1+A)p(z)
S (z) = − , (2.6)
α,λ
(1 B)+(1+B)p(z)
−
2(A B)p′(z)
S′ (z) = − , (2.7)
α,λ ((1 B)+(1+B)p(z))2
−
2(A B)((1 B)+(1+B)p(z))p′′(z) 4(1+B)(A B)p′2(z)
S′′ (z) = − − − − . (2.8)
α,λ ((1 B)+(1+B)p(z))3
−
Using (2.6)–(2.8), the Bessel-Struve differential equation (1.6) yields
z2p′′(z) 2(1+B) (zp′(z))2 +(2α+1)zp′(z)
− (1−B)+(1+B)p(z)
(λ2(1−A)+(1+A)p(z))((1−B)+(1+B)p(z))+λM((1−B)+(1+B)p(z))2 z = 0. (2.9)
−(cid:18) 2(A−B) (cid:19)
With Ω = 0 , define Ψ(r,s,t;z) by
{ }
Ψ(r,s,t;z) := t 2(1+B) s2 +(2α+1)s
− (1−B)+(1+B)r
(λ2(1−A)+(1+A)r)((1−B)+(1+B)r)+λM((1−B)+(1+B)r)2 z. (2.10)
−(cid:18) 2(A−B) (cid:19)
It follows from (2.9) that Ψ(p(z),zp′(z),z2p′′(z);z) Ω. To show Rep(z) > 0 for z ∆,
∈ ∈
from Lemma 1.1, it is sufficient to establish ReΨ(iρ,σ,µ+iν;z) < 0 in ∆ for any real ρ,
σ (1+ρ2)/2, and σ +µ 0.
≤ − ≤
With z = x+iy ∆, it readily follows from (2.10) that
∈
ReΨ(iρ,σ,µ+iν;z) = µ 2(1−B2) σ2 +(2α+1)σ + λ2(1−AB)+λM(1−B2) ρy
− (1−B)2+(1+B)2ρ2 (A−B)
(cid:16) (cid:17)
λ2((1−A)(1−B)−(1+A)(1+B)ρ2)+λM((1−B)2−(1+B)2ρ2) x. (2.11)
− 2(A−B)
(cid:16) (cid:17)
Since σ (1+ρ2)/2, and B [ 1,3 2√2],
≤ − ∈ − −
2(1 B2) 2(1 B2) (1+ρ2)2 1+B
− σ2 − .
(1 B)2 +(1+B)2ρ2 ≥ (1 B)2 +(1+B)2ρ2 4 ≥ 2(1 B)
− − −
Thus
ReΨ(iρ,σ,µ+iν;z)
2ασ λ2((1−A)(1−B)−(1+A)(1+B)ρ2)+λM((1−B)2−(1+B)2ρ2)x+ λ2(1−AB)+λM(1−B2)ρy 1+B
≤ − 2(A−B) A−B − 2(1−B)
α(1+ρ2) λ2((1−A)(1−B)−(1+A)(1+B)ρ2)+λM((1−B)2−(1+B)2ρ2)x+ λ2(1−AB)+λM(1−B2)ρy 1+B
≤ − − 2(A−B) (A−B) − 2(1−B)
= p ρ2 +q ρ+r := Q(ρ),
1 1 1
6 SAIFUL R.MONDAL ANDMOHAMMED AL-DHUAIN
where
λ2(1+A)(1+B)+λM(1+B)2
p = α+ x,
1
− 2(A B)
−
λ2(1 AB)+λM(1 B2)
q = − − y,
1
(A B)
−
1+B λ2(1 A)(1 B)+λM(1 B)2
r = α − − − x.
1
− − 2(1 B) −(cid:18) 2(A B) (cid:19)
− −
Condition (2.1) shows that
λ2(1+A)(1+B)+λM(1+B)2
p = α+ x
1
− 2(A B)
−
λ λ(1+A)(1+B)+M(1+B)2
< α | | < 0.
−(cid:18) − 2 (cid:12) A B (cid:12)(cid:19)
(cid:12) − (cid:12)
(cid:12) (cid:12)
Since max p ρ2 +q ρ+r = ((cid:12)4p r q2)/(4p ) for p < 0, it(cid:12) is clear that Q(ρ) < 0
ρ∈R { 1 1 1} 1 1 − 1 1 1
when
λ2(1 AB)+λM(1 B2) 2 λ2(1+A)(1+B)+λM(1+B)2
− − y2 < 2α+ x
(cid:18) A B (cid:19) (cid:18)− (cid:18) A B (cid:19) (cid:19)
− −
1+B λ2(1 A)(1 B)+λM(1 B)2
2α − − − x ,
×(cid:18)− − 1 B −(cid:18) A B (cid:19) (cid:19)
− −
x , y < 1. As y2 < 1 x2, the above condition holds whenever
| | | | −
λ2(1 AB)+λM(1 B2) 2
− − (1 x2)
(cid:18) A B (cid:19) −
−
λ2(1+A)(1+B)+λM(1+B)2
< 2α+ x
(cid:18)− (cid:18) A B (cid:19) (cid:19)
−
1+B λ2(1 A)(1 B)+λM(1 B)2
2α − − − x ,
×(cid:18)− − 1 B −(cid:18) A B (cid:19) (cid:19)
− −
that is, when
(λ2(1−AB)+λM(1−B2))2−(λ2(1−A)(1−B)+λM(1−B)2)(λ2(1+A)(1+B)+λM(1+B)2)x2
(A−B)2
+ 1 4αλ(λ(A+B)+2MB)) λ(1+B)2(λ(1+A)+M(1+B)) x
A−B − − 1−B
+4α(cid:0)2 +2α1+B λ2(1−AB)+λM(1−B2) 2 0. (cid:1) (2.12)
1−B − A−B ≥
(cid:0) (cid:1)
To establish inequality (2.12), consider the polynomial R given by
R(x) := mx2 +nx+r, x < 1,
| |
where
m := (λ2(1−AB)+λM(1−B2))2−(λ2(1−A)(1−B)+λM(1−B)2)(λ2(1+A)(1+B)+λM(1+B)2),
(A−B)2
= (λ2(1−AB)+λM(1−B2))2−λ2(1−B2)(λ(1−A)+M(1−B))(λ(1+A)+M(1+B)),
(A−B)2
n := λ 4α(λ(A+B)+2MB)) (1+B)2(λ(1+A)+M(1+B)) ,
A−B − − 1−B
r := 4α2 +(cid:0) 2α1+B λ2(1−AB)+λM(1−B2) 2. (cid:1)
1−B − A−B
(cid:0) (cid:1)
BESSEL-STRUVE KERNEL FUNCTION 7
The constraint (2.3) yields n 2 m , and thus R(x) m+r n . Now the inequality
| | ≥ | | ≥ −| |
(2.2) implies that
R(x)
m+r n
≥ −| |
= (λ2(1−AB)+λM(1−B2))2−λ2(1−B2)(λ(1−A)+M(1−B))(λ(1+A)+M(1+B)) +4α2 +2α1+B
(A−B)2 1−B
λ2(1−AB)+λM(1−B2) 2 |λ| 4α(λ(A+B)+2MB)) (1+B)2(λ(1+A)+M(1+B))
− A−B − A−B − − 1−B
=(cid:0)4α2 |λ| 4α((cid:1)λ(A+B)(cid:12)(cid:12)(cid:0)+2MB)) (1+B)2(λ(1+A)+M(1+B)) (cid:1)(cid:12)(cid:12)
− A−B − − 1−B
+2α1+B (cid:12)(cid:12)(cid:0)(λ2(1−A)(1−B)+λM(1−B)2)(λ2(1+A)(1+B)+λM(1+B)2 0. (cid:1)(cid:12)(cid:12)
1 B − (A−B)2 ≥
−
Now considers the case of the constraint (2.5), which is equivalent to n < 2m. Then
| |
the minimum of R occurs at x = n/(2m), and (2.4) yields
−
4mr n2
R(x) − 0.
≥ 4m ≥
Evidently Ψ satisfies the hypothesis of Lemma 1.1, and thus Re p(z) > 0, that is,
(1 A) (1 B)S (z) 1+z
α
− − − .
−(1+A) (1+B)S (z) ≺ 1 z
α
− −
Hence there exists an analytic self-map w of ∆ with w(0) = 0 such that
(1 A) (1 B)S (z) 1+w(z)
α
− − − = ,
−(1+A) (1+B)S (z) 1 w(z)
α
− −
which implies that S (z) (1+Az)/(1+Bz). (cid:3)
α
≺
Considering λ = A = B = 1, following result can be obtain from Theorem 2.1.
−
Corollary 2.1. For α [0,α ] [3/2, ), Re(S (z)) > 0. Here α = 0.5 is the positive
0 α 0
∈ ∪ ∞
root of the identity 4αΓ(α+1) = √πΓ(α+1/2).
This result along with the recurrence relation (1.7) gives that
zS′ (z)+2αS (z)
Re α α > 0.
(cid:18) 2α (cid:19)
Inparticular, the functionzS (z) isclose-to-convex functions with respect to z, andhence
1/2
it is univalent.
Theorem 2.2. Let 3 2√2 B < A 1 and λ, α R satisfy
− ≤ ≤ ∈
λ λ(1+A)(1+B)+M(1+B)2
α max 0, | | . (2.13)
≥ (cid:26) 2 (cid:12) A B (cid:12)(cid:27)
(cid:12) − (cid:12)
(cid:12) (cid:12)
suppose A, B, α and λ satisfy eithe(cid:12)r the inequality (cid:12)
(α2(A B)2 λ(A B) α(λ(A+B)+2MB)+ 4B(1−B)(λ(1+A)+M(1+B))
− − − (1+B)2
+8αB(1−B)(A−B)2 1 λ2(cid:12)(cid:12)(1 A)(1 B)+λM(1 B)2 λ2(1+A)(1+B)+λM(cid:12)(cid:12) (1+B)2
(1+B)3 ≥ 4 − − −
(cid:0) (cid:1)(cid:0) (cid:1)
(2.14)
8 SAIFUL R.MONDAL ANDMOHAMMED AL-DHUAIN
whenever
4Bλ(1 B)
(A B) αλ(λ(A+B)+2MB)+ − λ(1+A)+M(1+B)
− (cid:12) (1+B)2 (cid:12)
(cid:12) (cid:0) (cid:1)(cid:12)
λ2 (cid:12) (cid:12)
λ(cid:12)(1 AB)+M(1 B2) 2 (1 B2) λ(1 A)+M(1 B) (cid:12)λ(1+A)+M(1+B) ,
≥ 2 − − − − − −
(cid:12) (cid:12)
(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1)
(cid:12) (cid:12)
(2.15)
(cid:12) (cid:12)
or the inequality
2
4B(1 B)
α λ(A+B)+2MB + − λ(1+A)+M(1+B)
(cid:18) (cid:18) (cid:19) (1+B)2 (cid:18) (cid:19)(cid:19)
λ(1 AB)+M(1 B2) 2 (1 B2) λ(1 A)+M(1 B)
≤ − − − − − −
(cid:0) (cid:1) 1 B(cid:0) λ2(1 AB)+λM(cid:1) (1 B2) 2
λ(1+A)+M(1+B) α2 +8αB − − − (2.16)
(cid:18) (1+B)3 −(cid:18) 2(A B) (cid:19) (cid:19)
(cid:0) (cid:1) −
whenever
4Bλ(1 B)
(A B) αλ(λ(A+B)+2MB)+ − λ(1+A)+M(1+B)
− (cid:12) (1+B)2 (cid:12)
(cid:12) (cid:0) (cid:1)(cid:12)
λ2 (cid:12) (cid:12)
λ(cid:12)(1 AB)+M(1 B2) 2 (1 B2) λ(1 A)+M(1 B) (cid:12)λ(1+A)+M(1+B) ,
≤ 2 (cid:12) − − − − − − (cid:12)
(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1)
(cid:12) (2.17) (cid:12)
(cid:12) (cid:12)
If (1+B)S (z) = (1+A), then S (z) [A,B].
α α
6 ∈ P
Proof. Proceeding similarly as in the proof of Theorem 2.1, consider ReΨ(iρ,σ,µ+iν;z)
as given in (2.11). For σ (1+ρ2)/2, ρ R, and B 3 2√2,
≤ − ∈ ≥ −
2(1 B2) 2(1 B2) (1+ρ2)2 8B(1 B)
− σ2 − − .
(1 B)2 +(1+B)2ρ2 ≥ (1 B)2 +(1+B)2ρ2 4 ≥ (1+B)3
− −
With z = x+iy ∆, and µ+σ < 0, it follows that
∈
ReΨ(iρ,σ,µ+iν;z)
α(1+ρ2) 8B(1−B) + λ2(1−AB)+λM(1−B2) ρy
≤ − − (1+B)3 A−B
(cid:16) (cid:17)
1 λ2((1 A)(1 B) (1+A)(1+B)ρ2)+λM((1 B)2 (1+B)2ρ2) x
− 2(A−B) − − − − −
(cid:0) (cid:1)
= p ρ2 +q ρ+r := Q (ρ),
2 2 2 1
where
p = α+ λ2(1+A)(1+B)+λM(1+B)2x,
2 − 2(A−B)
q = λ2(1−AB)+λM(1−B2)y,
2 A−B
r = α 8B 1−B λ2(1−A)(1−B)+λM(1−B)2 x.
2 − − (1+B)3 −(cid:18) 2(A−B) (cid:19)
In the proof of Theorem 2.1, it can be observed that the constraint (2.15) implies p < 0.
2
Thus Q (ρ) < 0 for all ρ R provided q2 4p r , that is,
1 ∈ 2 ≤ 2 2
λ2(1−AB)+λM(1−B2) 2y2 2α+ λ2(1+A)(1+B)+λM(1+B)2x
A−B ≤ − A−B
(cid:16) (cid:17) (cid:16) (cid:17)
2α 16B 1−B λ2(1−A)(1−B)+λM(1−B)2x ,
× − − (1+B)3 − A−B
(cid:16) (cid:17)
BESSEL-STRUVE KERNEL FUNCTION 9
x , y < 1. With y2 < 1 x2, it is sufficient to show
| | | | −
λ2(1−AB)+λM(1−B2) 2(1 x2) 2α+ λ2(1+A)(1+B)+λM(1+B)2x
A−B − ≤ − A−B
(cid:16) (cid:17) (cid:16) (cid:17)
2α 16B(1−B) λ2(1−A)(1−B)+λM(1−B)2x ,
× − − (1+B)3 − A−B
(cid:16) (cid:17)
for x < 1. The above inequality is equivalent to showing
| |
R (x) := m x2 +n x+r 0, (2.18)
1 1 1 1
≥
where
1
m := − λ2(1 A)(1 B)+λM(1 B)2 λ2(1+A)(1+B)+λM(1+B)2
1 (A B)2 − − −
− (cid:0)(cid:0) (cid:1)(cid:0) (cid:1)(cid:1)
λ2(1 AB)+λM(1 B2) 2
+ − − ,
(cid:18) (A B) (cid:19)
−
1 16B(1 B)
n := − 4αλ(λ(A+B)+2MB + − λ2(1+A)+λM(1+B) ,
1 A B (1+B)2
− (cid:0) (cid:1) (cid:0) (cid:1)
B(1 B) λ2(1 AB)+λM(1 B2) 2
r := 4α2 +32α − − − .
1
(1+B)3 −(cid:18) A B (cid:19)
−
If (2.15) holds, then n 2 m . Since R is increasing, then R (x) m + r
1 1 1 1 1 1
| | ≥ | | ≥ −
n , which is nonnegative from (2.14). On the other hand, if (2.16) holds, then n <
1 1
| | | |
2 m , R (x) (4m r n2)/4m , and (2.17) implies R (x) 0. Either case establishes
(2|.118|). 1 ≥ 1 1 − 1 1 1 ≥ (cid:3)
Remark 2.1. A graphical experiment using mathematica shows that Re(S (z)) > 0 for all
α
α 0 and z ∆. But our computation restrict on [0,1/2] [3/2, ). Thus the result is
≥ ∈ ∪ ∞
open for α (0.5,1.5).
∈
3. Third order differential subordination for S
α
In this section we introduce an admissible class Φ [Ω,q] as follows:
g
Definition 3.1. Let Ω be a set in C and q . The class of admissible function
0 0
Φ [Ω,q] consists of those functions φ : C4 ∆∈ Q C∩tHhat satisfy the following admissibility
g
× →
condition
φ(β ,β ,β ,β ;z) / Ω
1 2 3 4
∈
whenever
mζq′(ζ)+(α+1)q(ζ)
β = q(ζ) β = ,
1 2
α+1
Re 4α(α+1)β3+8α(α+1)β2−(4α2+8α+1)β1 +1 mRe ζq′′(ζ) +1 ,
2(α+1)(β2−β1) ≥ q′(ζ)
(cid:16) (cid:17) (cid:16) (cid:17)
and
Re 8α(α2−1)β4−4α(α+1)(6α−1)β3+2(α+1)(36α2−12α−1)β2+(40α3+16α2−18α−6)β1 m2Re ζ2q′′(ζ) ,
2(α+1)(β2−β1) ≥ q′(ζ)
(cid:16) (cid:17) (cid:16) (cid:17)
where z ∆, α > 1, ζ ∂∆ E(q) and m 2.
∈ ∈ \ ≥
Our first result give the sufficient conditions for the inclusion of S in the admissible
α
class Φ [Ω,q]. In this purpose, let define g (z) := zS (z). Then a calculation along with
g α α
(1.7), yields the recurrence relation
zg′ (z) = 2αg (z)+(1 2α)g (z), (3.1)
α α−1 − α
10 SAIFUL R.MONDAL ANDMOHAMMED AL-DHUAIN
which play the main role in this article. Now we will state and proof our main results on
differential subordination involving S .
α
Theorem 3.1. Let φ Φ [Ω,q]. Suppose that q satisfy the following inequalities:
g 0
∈ ∈ Q
ζq′′(ζ) g (z)
α
Re 0, m. (3.2)
(cid:18) q′(ζ) (cid:19) ≥ (cid:12)q′(ζ)(cid:12) ≤
(cid:12) (cid:12)
(cid:12) (cid:12)
For all z ∆ and α > 1, if
(cid:12) (cid:12)
∈
φ(g (z),g (z),g (z),g (z);z Ω, (3.3)
α+1 α α−1 α−2
{ } ⊂
then g (z) q(z).
α+1
≺
Proof. Define the analytic function p in ∆ as
p(z) := g (z), α > 1. (3.4)
α+1
Differentiate (3.4) with respect to z. Then an application of (3.1) for α+1 yields
zp′(z)+(2α+1)p(z)
g (z) = . (3.5)
α
2(α+1)
Differentiate both side of (3.5) with respect to z and then multiply by z gives
z2p′′(z)+(2α+1)zp′(z)
zg′ (z) = . (3.6)
α 2(α+1)
Again use of (3.1) implies
z2p′′(z)+4αzp′(z)+(4α2 1)p(z)
g (z) = − . (3.7)
α−1
4α(α+1)
Similarly, it can be shown that
g (z) = z3p′′′(z)+(6α−1)z2p′′(z)+(12α2−8α−1)zp′(z)+(4α2−1)(2α−3)p(z). (3.8)
α−2 8α(α2−1)
Now consider the four transformation β : C4 C, i = 1,2,3,4, as follows:
i
7→
(i) β (r,s,t,u) = r
1
s+(α+1)r
(ii) β (r,s,t,u) =
2
α+1
t+4αs+(4α2 1)r
(iii) β (r,s,t,u) = −
3
4α(α+1)
u+(6α 1)t+(12α2 8α 1)s+(2α 3)(4α2 1)r
(iv) β (r,s,t,u) = − − − − − .
4
8α(α2 1)
−
Define ψ : C4 C as
→
ψ(r,s,t,u;z) := φ(β ,β ,β ,β ;z). (3.9)
1 2 3 4
Then clearly
ψ(p(z),zp′(z),z2p′′(z),z3p′′′(z);z) = φ(g (z),g (z),g (z),g (z);z).
α+1 α α−1 α−2
From (i)-(iv), it follows that
s = 2(α+1)(β β )
2 1
−
t = 4α(α+1)β +8α(α+1)β (4α2 +8α+1)β
3 2 1
−
u = 8α(α2 1)β 4α(α+1)(6α 1)β +2(α+1)(36α2 12α 1)β
4 3 2
− − − − −
+(40α3+16α2 18α 6)β .
1
− −