Abstract
We discuss univalent solutions of boundary fractional differential equations in a complex domain. The fractional operators are taken in the sense of the SrivastavaOwa calculus in the unit disk. The existence of subsolutions and supersolutions (maximal and minimal) is established. The existence of a unique univalent solution is imposed. Applications are constructed by making use of a transformation formula for fractional derivatives as well as generalized fractional derivatives.
1 Introduction
Fractional calculus is the most significant branch of mathematical analysis that transacts
with the potential of covering real number powers or complex number powers of the
differentiation operator
Definition 1.1 For the function
where the multiplicity of
Definition 1.2 For the function
where the multiplicity of
Remark 1.1 From the above two definitions, we observe that
and
Later, the first author [9] modified these SrivastavaOwa operators into two fractional parameters. For a wealth of references on applications of SrivastavaOwa operators, see [1019].
In this paper, we study univalent solutions of boundary fractional differential equations in a complex domain. The fractional operators are considered in the sense of the SrivastavaOwa [8] differential operator
where
2 Preliminaries
Let
Definition 2.1 Let
Then every function
Definition 2.2 Let
Then every function
In this work, we utilize the generalized sets
and
When
Lemma 2.1Let Ψ be a positive, continuous and bounded function on ℂ.
1. Any subsolution for Ψ has a (Lipschitz) continuous extension to the closed unit disk
2. A function
where
3. If a sequence of subsolutions from
4. Let
Lemma 2.2 [[20], Lemma 2.7, Lemma 3.7]
Let Ψ be a positive, continuous and bounded function on ℂ. Assume that
Lemma 2.3If
Lemma 2.3 is a generalization of the result found in [21]. Thence, we cancel the proof.
Next result shows some properties of the set
Lemma 2.4Let Ψ be a positive, continuous and bounded function on ℂ.
1. Any univalent supersolutiongfor Ψ satisfies
2. A bounded univalent function
where
3. If a uniformly bounded sequence of univalent supersolutions for Ψ converges locally uniformly inU, then the limit functionfis again a univalent supersolution for Ψ.
4. Let
3 Main results
Our aim is to establish the largest univalent solution
Theorem 3.1Let Ψ be a positive continuous function on ℂ. Then there exists a unique univalent function
Proof By Lemma 2.1,
Assuming that
which involves
and for sufficiently small values of α, we have
Letting the function
we conclude that
and
Hence
Therefore, by virtue of the principle of subordination, this yields
and for all
This implies that
Now we proceed to show that the maximal subsolution
where φ is the harmonic function on
on U and
In addition, we conclude that
Consequently,
is identically equal to zero. Hence
Remark 3.1 Note that Theorem 3.1 can be introduced for the boundary problem
where f is analytic in U satisfying the Riemann mapping conditions.
As a consequence of our theorem, we have the following.
Corollary 3.1Let Ψ be a positive continuous function on ℂ. Then there exists a univalent function
Corollary 3.2LetDbe a simplyconnected region in ℂ and
Corollary 3.3Let Ψ be a positive continuous sublinear function on ℂ; i.e.,
Then there exists a univalent function
Next, we discuss the boundary problem for some functions
Now, in a manner similar to Theorem 3.1, we have the following theorem.
Theorem 3.2Consider the problem
where
and
Proof Set
where
which includes
As a consequence of Theorem 3.2 above, we have the following.
Corollary 3.4Let Ψ be a positive continuous function on ℂ. Then there exists a univalent function
Tremblay [24] studied a fractional calculus operator defined in terms of the RiemannLiouville
fractional differential operator. We extend this operator in the complex plane to
involve
where
Consequently, for the class consisting of analytic functions
Theorem 3.3Let
whereFis the hypergeometric function. The equality holds true for the Koebe function
Proof By De Branges’ theorem [25] (also known as Bieberbach conjecture, e.g., see Duren [26]), for
where
Finally, by letting the Koebe function
Moreover, we can prove the following theorem.
Theorem 3.4Let
whereFis a hypergeometric function.
Proof Let
Now, by applying the last assertion on
□
We obtain the following two corollaries by making use the above operator and, respectively,
letting
Corollary 3.5Consider the problem
where
and
Corollary 3.6Consider the problem
where
and
In the following example we demonstrate that, in view of Theorem 3.1, the above boundary
problems have univalent solutions in the unit disk with
Example 3.1 A computation implies
and for
we get
Thus, we have the boundary problems
and
It is clear that
By using the same technique as in the first part of Theorem 3.1, together with Lemma 2.2 and Lemma 2.4, we conclude the following.
Theorem 3.5Let Ψ be a positive, continuous and bounded function. Then there exists a unique function
Then the function
We call the function
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Both authors jointly worked on deriving the results and approved the final manuscript.
Acknowledgements
This work is supported by University of Malaya High Impact Research Grant no vote UM.C/625/HIR/MOHE/SC/13/2 from the Ministry of Higher Education Malaysia. The authors also would like to thank the referees for giving some suggestions for improving the work.
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