# A Third-Order Differential Equation and Starlikeness of a Double Integral Operator

Authors: Rosihan M. Ali; See Keong Lee; K. G. Subramanian; A. Swaminathan
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

Functionsf(z)=z+∑2∞anzn that are analytic in the unit disk and satisfy the differential equation f'(z)+αzf”(z)+γz2f'”(z)=g(z) are considered, where g is subordinated to a normalized convex univalent function h. These functions f are given by a double integral operator of the form f(z)=∫01∫01G(ztμsν)t-μs-νds dt with G' subordinated to h. The best dominant to all solutions of the differential equation is obtained. Starlikeness properties and various sharp estimates of these solutions are investigated for particular cases of the convex function h.


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## Body

## 1. Introduction

Let𝒜 denote the class of all analytic functions f defined in the open unit disk U∶={z∈ℂ:|z|<1} and normalized by the conditions f(0)=0, f′(0)=1. Further, let 𝒮 be the subclass of 𝒜 consisting of univalent functions, and let 𝒮* be its subclass of starlike functions. A starlike function f is characterized analytically by the condition Re(zf′(z)/f(z))>0 in U, that is, the domain f(U) is starlike with respect to origin. For two functions f(z)=z+a2z2+⋯ and g(z)=z+b2z2+⋯ in 𝒜, the Hadamard product (or convolution) of f and g is the function f*g defined by (1.1)(f*g)(z)=z+∑n=2∞anbnzn.Forf and g in 𝒜, a function f is subordinate to g, written as f(z)≺g(z), if there is an analytic function w satisfying w(0)=0 and |w(z)|<1, such that f(z)=g(w(z)), z∈U. When g is univalent in U, then f is subordinated to g which is equivalent to f(U)⊂g(U) and f(0)=g(0).In a recent paper, Miller and Mocanu [1] investigated starlikeness properties of functions f defined by double integral operators of the form(1.2)f(z)=∫01∫01W(s,t,z)dsdt.In this paper, conditions on a different kernelW are investigated from the perspective of starlikeness. Specifically, we consider functions f∈𝒜 given by the double integral operator of the form(1.3)f(z)=∫01∫01G(ztμsν)t-μs-νdsdt.
In this case, it follows that(1.4)f′(z)=∫01∫01g(ztμsν)dsdt,whereG′(z)=g(z). Further, the function f satisfies a third-order differential equation of the form(1.5)f′(z)+αzf′′(z)+γz2f′′′(z)=g(z)for appropriate parametersα and γ. The investigation of such functions f can be seen as an extension to the study of the class(1.6)R(α,h)={f∈A:f′(z)+αzf′′(z)≺h(z),z∈U}.The classR(α,h) or its variations for an appropriate function h have been investigated in several works; see, for example, [2–10] and more recently  [11, 12].


## 2. Results on Differential Subordination

We first recall the definition of best dominant solution of a differential subordination.Definition 2.1 ((dominant and best dominant) [13]).
LetΨ:ℂ3×U→ℂ, and let h be univalent in U.If p is analytic in U and satisfies the differential subordination
(2.1)Ψ(p(z),zp′(z),z2p′′(z))≺h(z),
then p is called a solution of the differential subordination. A univalent function q is called a dominant if p≺q for all p satisfying (2.1). A dominant q̃ that satisfies q̃≺q for all dominants q of (2.1) is said to be the best dominant of (2.1).In the following sequel, we will assume thath is an analytic convex function in U with h(0)=1. For α≥γ≥0, consider the third-order differential equation(2.2)f′(z)+αzf′′(z)+γz2f′′′(z)=g(z),g(z)≺h(z).
We will denote the class consisting of all solutions f∈𝒜 as R(α,γ,h), that is,(2.3)R(α,γ,h)={f∈A:f′(z)+αzf′′(z)+γz2f′′′(z)≺h(z),z∈U}.Let(2.4)μ=(α-γ)-(α-γ)2-4γ2,ν+μ=α-γ,μν=γ.
The discriminant is denoted by Δ∶=(α-γ)2-4γ. Note that Reμ≥0 and Reν≥0.We will rewrite the solution of(2.5)f′(z)+αzf′′(z)+γz2f′′′(z)=g(z)in its equivalent integral form(2.6)f′(z)=∫01∫01g(ztμsν)dsdt.It follows from  relations (2.4) that (2.7)g(z)=f′(z)+(μ(1+ν)+ν)zf′′(z)+μνz2f′′′(z)=νz1-1/ν(μz1+1/νf′′(z)+z1/νf′(z))′=νz1-1/ν(μz1+1/ν-1/μ(z1/μf′(z))′)′.Thus,(2.8)μz1+1/ν-1/μ(z1/μf′(z))′=1ν∫0zw1/ν-1g(w)dw.Making the substitutionw=zsν in the above integral and integrating again, a change of variables yields(2.9)f′(z)=∫01∫01g(ztμsν)dsdt.We will use the notationϕλ for(2.10)ϕλ(z)=∫01dt1-ztλ=∑n=0∞zn1+λn.
From [14] it is known that ϕλ is convex in U provided Reλ≥0.Theorem 2.2.
Letμ and ν be given by (2.4), and
(2.11)q(z)=∫01∫01h(ztμsν)dtds.
Then the function q(z)=(ϕν*ϕμ)*h(z) is convex. If f∈R(α,γ,h), then
(2.12)f′(z)≺q(z)≺h(z),
and q is the best dominant.Proof.
It follows from (2.10) that
(2.13)h(z)*ϕμ(z)=∫0111-ztμdt*h(z)=∫01h(ztμ)dt∶=k(z).
Thus,
(2.14)h(z)*(ϕμ(z)*ϕν(z))=k(z)*ϕν(z)=∫01k(zsν)ds=∫01∫01h(ztμsν)dtds=q(z).
Since the convolution of two convex functions is convex [15], the function q is convex. Let
(2.15)p(z)=f′(z)+νzf′′(z).
Then,
(2.16)p(z)+μzp′(z)=f′(z)+αzf′′(z)+γz2f′′′(z)≺h(z).
It is known from [16] that
(2.17)p(z)≺1μz1/μ∫0zζ1/μ-1h(ζ)dζ=(ϕμ*h)(z)≺h(z).
Similarly,
(2.18)p(z)=f′(z)+νzf′′(z)≺(ϕμ*h)(z)
implies
(2.19)f′(z)≺(ϕν*ϕμ*h)(z)=∑n=0∞zn(1+νn)(1+μn)*h(z)=(∫01∫01dtds1-ztμsν)*h(z)=∫01∫01h(ztμsν)dtds=q(z).
The differential chain(2.20)f′≺q≺ϕμ*h≺h
shows that q≺h. Since q(z)+αzq′(z)+γz2q′′(z)=h(z), the function
(2.21)Q(z)=∫0zq(w)dw
is a solution of the differential subordination f′(z)+αzf′′(z)+γz2f′′′(z)≺h(z), and thus q≺q̃ for all dominants q̃. Hence, q is the best dominant.Remark 2.3.
(1) Whenγ=0, then μ=0 and ν=α, and the above subordination reduces to the result of [16], that is,
(2.22)f′(z)+αzf′′(z)≺h(z)⟹f′(z)≺∫01h(ztα)dt.
(2) The above proof also reveals that(2.23)f∈R(α,γ,h)⟹f∈R(0,0,h),
that is, f′(z)≺h(z).Theorem 2.4.
Letμ, ν, and q be as given in Theorem 2.2. If f∈R(α,γ,h), then
(2.24)f(z)z≺∫01q(tz)dt=∫01∫01∫01h(zrsμtν)drdsdt.Proof.
Letp(z)=f(z)/z. Then
(2.25)p(z)+zp′(z)=f′(z)≺q(z).
With ϕ1 given by (2.10), this subordination implies
(2.26)p(z)=(ϕ1*(p+zp′))(z)≺(ϕ1*q)(z)=∫01q(tz)dt.In this paper, starlikeness properties will be investigated for functionsf given by a double integral operator of the form (1.3).


## 3. Applications

First, we consider a class of convex univalent functionsh so that h(U) is symmetric with respect to the real axis. Denote by R(α,γ,A,B) the class(3.1)R(α,γ,A,B)={f∈A:f′(z)+αzf′′(z)+γz2f′′′(z)≺1+Az1+Bz,z∈U},where-1≤B<A≤1, and let h(z;A,B)=(1+Az)/(1+Bz). When A=1-2β and B=-1, let hβ(z)∶=h(z;1-2β,-1). The class of R(α,γ,hβ) is of particular significance, and we will simply denote it by (3.2)R(α,γ,hβ)∶=R(α,γ,β)={f∈A:f′(z)+αzf′′(z)+γz2f′′′(z)≺1+(1-2β)z1-z,z∈U}.Equivalently,(3.3)R(α,γ,β)={f∈A:Re(f′(z)+αzf′′(z)+γz2f′′′(z))>β}.The following result is an immediate consequence of Theorems2.2 and 2.4.Theorem 3.1.
Under the assumptions of Theorem2.2, if
(3.4)f′(z)+αzf′′(z)+γz2f′′′(z)≺1+Az1+Bz,
then
(3.5)f′(z)≺{q(z;A,B)≺1+Az1+Bz,ifB≠0,q(z;A)≺1+Az,ifB=0,
where
(3.6)q(z;A,B)∶=1+(A-B)∑n=1∞(-B)n-1zn(1+μn)(1+νn),q(z;A)∶=1+Az(1+α)
is the best dominant. Further,
(3.7)f(z)z≺AB-A-BB∫01∫01∫01dsdtdu1+Bzutμsν=1+(A-B)∑n=1∞(-B)n-1zn(1+n)(1+μn)(1+νn)
if B≠0, and
(3.8)f(z)z≺1+Az2(1+α)
if B=0.


## 4. Starlikeness Property

Starlikeness properties of functions given by a double integral operator are investigated in this section. The following result will be required.Lemma 4.1 (see [5]).
Iff∈𝒜 satisfies
(4.1)Re(f′(z)+αzf′′(z))>(-1/α)∫01t1/α-1((1-t)/(1+t))dt1-1/α∫01t1/α-1((1-t)/(1+t))dt,z∈U,
for α≥1/3, then f∈S*. This result is sharp.Theorem 4.2.
Letμ and ν be given by (2.4)  with Δ≥0 and ν≥1/3.  If
(4.2)f(z)=∫01∫01G(ztμsν)t-μs-νdsdt,
where G′(z)≺hβ(z)=h(z;1-2β,-1), and β satisfies
(4.3)β=1-12(1-(1/ν)∫01t1/ν-1((1-t)/(1+t))dt)(1-∫01(dt/(1+tμ))),
then f∈S*.Proof.
The functionf satisfies
(4.4)f′(z)=∫01∫01g(ztμsν)dsdt,G′(z)=g(z)≺hβ(z),
and thus
(4.5)f′(z)+αzf′′(z)+γz2f′′′(z)=g(z)≺hβ(z).
Now, Rehβ(z)>β also implies that Reg(z)>β, and so
(4.6)Re(f′(z)+αzf′′(z)+γz2f′′′(z))>β,β<1.
It follows from the proof of Theorem 2.2 that
(4.7)f′(z)+νzf′′(z)≺(ϕμ*hβ)(z)∶=qμ(z),
where
(4.8)qμ(z)=2β-1+2(1-β)∫01dt1-ztμ.
Since
(4.9)Reqμ(z)>2β-1+2(1-β)∫01dt1+tμ,
an application of Lemma 4.1 yields the result.Corollary 4.3.
Letα≥3 and
(4.10)Re(f′(z)+αzf′′(z)+α-12z2f′′′(z))>β,β<1.
If β satisfies
(4.11)β=1-12(1-log⁡2)(1-(2/(α-1))∫01t2/(α-1)-1((1-t)/(1+t))dt),
then f∈S*.Proof.
In this case,μ=1, ν=(α-1)/2, and the result now follows from Theorem 4.2.Example 4.4.
If(4.12)Re(f′(z)+3zf′′(z)+z2f′′′(z))>β
and β satisfies
(4.13)β=4(1-log⁡2)2-14(1-log⁡2)2≈-1.65509,
then f∈S*.Theorem 4.5.
Letf,g∈R(α,γ,β) and let μ and ν be given by (2.4)  with Δ≥0.  If β satisfies
(4.14)β=1-14(1-∫01∫01∫01(dsdtdu/(1+utμsν))),
then f*g∈R(α,γ,β).Proof.
Clearly,(4.15)(f*g)′(z)+αz(f*g)′′(z)+γz2(f*g)′′′(z)=((f′+αzf′′+γz2f′′′)*gz)(z).
Since f∈R(α,γ,β), substituting A=1-2β and B=-1 in (3.7) gives
(4.16)Reg(z)z>2β-1+2(1-β)∫01∫01∫01dsdtdu1+utμsν=12.
Hence, it follows that
(4.17)Re((f*g)′(z)+αz(f*g)′′(z)+γz2(f*g)′′′(z))>β.



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