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 . This concept was harnessed in geometric function theory (GFT). It was applied to derive different types of differential and integral operators mapping the class of univalent functions and its subclasses into themselves. Hohlov [1,2] imposed sufficient conditions that guaranteed such mappings for the operators defined by means of the Hadamard product (or convolution) with Gauss hypergeometric functions. This was further extended by Kiryakova and Saigo [3] and Kiryakova [4,5] to the operators of the generalized fractional calculus (GFC) consisting of product functions of the Gaussian function, generalized hypergeometric functions, Gfunctions, Wright functions and FoxWright generalized functions as well as rendering integral representations by means of Fox Hfunctions and the Meijer Gfunction. These techniques can be used to display sufficient conditions that guarantee mapping of univalent functions (or, respectively, of convex functions) into univalent functions. For example, for the case of DziokSrivastava operator see [6], and for an extension to the Wright functions see [7] which is concerned with the SrivastavaWright operator. With the help of operators introduced in [6] and [7] one can establish univalence criteria for a large number of operators in GFT and GFC and for many of their special cases such as operators of the classical fractional calculus. Srivastava and Owa [8] generalized the definitions of fractional operators as follows.
Definition 1.1 For the function analytic in a simplyconnected region of the complex zplane ℂ containing the origin and for , the fractional derivative of order α is defined by
where the multiplicity of is removed by requiring to be real when . Moreover, when , we have .
Definition 1.2 For the function analytic in a simplyconnected region of the complex zplane ℂ containing the origin and for , the fractional integral of order α is defined by
where the multiplicity of is removed by requiring to be real when .
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 is the open unit disk and f is analytic in U satisfying the Riemann mapping conditions. The existence of subsolutions and supersolutions (minimal and maximal) is established. The existence of a unique univalent solution is introduced. Applications are also constructed by making use of some transformation formula for fractional derivatives. Equation (1) is a generalization of Beurling problem.
2 Preliminaries
Let be the set of analytic functions f on the unit disk U normalized by and . And let be the subset of normalized by and . We denote by the set of all univalent functions .
Definition 2.1 Let be a positive, continuous and bounded function and the set
Then every function is a subsolution for Ψ. If f is univalent, then it is called a univalent subsolution for Ψ.
Definition 2.2 Let be a positive, continuous and bounded function and the set
Then every function is a supersolution for Ψ. If f is univalent, then it is called a univalent supersolution for Ψ.
In this work, we utilize the generalized sets
and
When and , the above sets reduce to [20]. The next result shows some properties of . This is ultimately Lemma 2.3 from [20], therefore we omit the proof.
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. The setis uniformly bounded onand equicontinuous on.
2. A functionwith a continuous extension toUis a subsolution for Ψ if and only if
3. If a sequence of subsolutions fromconverges locally uniformly inUto a function (algebra of a holomorphic function that satisfies, ), then.
4. Letand letbe a positive, continuous and bounded function with. Then, for allsufficiently close to 1, the function, , is a subsolution for.
Lemma 2.2 [[20], Lemma 2.7, Lemma 3.7]
Let Ψ be a positive, continuous and bounded function on ℂ. Assume that, are two subsolutions for Ψ (univalent supersolutions for Ψ). Then the upper ofandis also a subsolution for Ψ (a univalent supersolution for Ψ).
Lemma 2.3Ifis a solution to problem (1), thenfhas 0 as its unique critical point.
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 , which basically is a generalization of [[20], Lemma 3.3]. So we skip the proof.
Lemma 2.4Let Ψ be a positive, continuous and bounded function on ℂ.
1. Any univalent supersolutiongfor Ψ satisfies.
2. A bounded univalent functionbelongs toif and only if
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. Letbe a positive, continuous and bounded function with, and letgbe a univalent supersolution for Ψ. Ifgis bounded, then, for allsufficiently close to 1, the function, , is a univalent supersolution for Φ.
3 Main results
Our aim is to establish the largest univalent solution and the smallest univalent solution . We are able to state and prove the following theorem.
Theorem 3.1Let Ψ be a positive continuous function on ℂ. Then there exists a unique univalent function (algebra unit disk), , such thatand. Furthermore, the maximal subsolutionis a solution.
Proof By Lemma 2.1, is nonempty and bounded in (). Assume that such that
Assuming that is the upper of f and and that . In view of Lemma 2.2, , we get . On the other hand, we have
which involves . Set . Then h is a welldefined holomorphic function on U such that . By letting
and for sufficiently small values of α, we have
Letting the function be defined by
we conclude that
and
Hence
Therefore, by virtue of the principle of subordination, this yields
Now we proceed to show that the maximal subsolution is a solution in . Define a function
where φ is the harmonic function on whose boundary values are logΨ (see [[22], p.402]). From the definition of Φ, we may conclude that is the maximal subsolution of Φ. Let be a solution for (1) having a unique critical point at 0 (Lemma 2.3). Obviously, and the two functions and are continuous on , harmonic on U and coincide on ∂U. Thus
on U and
In addition, we conclude that
Consequently,
is identically equal to zero. Hence is a solution. This completes the proof. □
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 functionsatisfyingandsuch thatis a solution to (1).
Corollary 3.2LetDbe a simplyconnected region in ℂ andbe an analytic function on the open unit disk. If Ψ is a positive convex function, then there exists a unique univalent functionsatisfyingandsuch thatis a solution to (1).
Corollary 3.3Let Ψ be a positive continuous sublinear function on ℂ; i.e.,
Then there exists a univalent functionsatisfyingandsuch thatis a solution to (1).
Next, we discuss the boundary problem for some functions , where . From [[23], Theorem 1.3], for , we define the fractional transform
Now, in a manner similar to Theorem 3.1, we have the following theorem.
Theorem 3.2Consider the problem
where , andis analytic in a simplyconnected region. Ifand Ψ is a continuous function on ℂ, then there exists a unique univalent function, , such that, where
Proof Set . It is clear that and . Also is nonempty and bounded in because . Let such that
where is as in Theorem 3.1. Next, assume that is the upper of P and and that , where . Thus . On the other hand, we may have
which includes . Now the proof is complete by proceeding with a similar manner to that of the first part of the proof of Theorem 3.1. □
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 functionsatisfyingand, wheresolves problem (2).
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 as follows:
where
Consequently, for the class consisting of analytic functions that are univalent in U, we have the following upper bound of the operator .
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 , we have . Therefore
where is the Pochhammer symbol defined by
Finally, by letting the Koebe function for in (3), we can show that the result is sharp. Hence the proof. □
Moreover, we can prove the following theorem.
Theorem 3.4Let. If, and, then we have the sharp bound
whereFis a hypergeometric function.
Proof Let and . Then we may use , . If , then we put . Otherwise, we let . From the above two cases, we conclude that , . Therefore
Now, by applying the last assertion on , we conclude
□
We obtain the following two corollaries by making use the above operator and, respectively, letting and .
Corollary 3.5Consider the problem
whereis analytic in a simplyconnected region. If, and Ψ is a continuous function on ℂ, then there exists a unique univalent function, , such that, , , where
Corollary 3.6Consider the problem
whereis analytic in a simplyconnected region. Ifand Ψ is a continuous function on ℂ, then there exists a unique univalent function, , such that, , , where
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 , as well as , which satisfies , .
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 functionsuch that
Then the functionmapsUconformally onto, where
We call the function of Theorem 3.5 the minimal univalent supersolution for Ψ.
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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