ROMAI J.

, 7, 1(2011), 77–86

ON THE EXISTENCE OF UNIVALENT

SOLUTIONS FOR FRACTIONAL INTEGRAL
EQUATION OF VOLTERRA TYPE IN
COMPLEX PLANE
Maslina Darus∗ , Rabha W. Ibrahim
School of Mathematical Sciences, Faculty of Science and Technology,
University Kebangsaan Malaysia, Malaysia
∗ corresponding author

maslina@ukm.my, rabhaibrahim@yahoo.com
Abstract In this article, we study the existence of univalent solutions for Volterra integral equa-
tion of fractional order in complex plane by using generalized Lipschitzian condition
and we discuss the extremal solutions by using the subordination concept. Moreover,
we establish the existence of solution for integral inclusion.
Acknowledgement: The work here was supported by MOHE grant: UKM-ST-06-
FRGS0107-2009.

Keywords: fractional calculus, set-valued function, integral inclusion, subordination; superordination,

univalent function, analytic function, Srivastava-Owa operators, unit disk.
2000 MSC: 34G10, 26A33, 30C45.
Received on January 10, 2010.

1. INTRODUCTION
Nowadays, there is increasing attention paid to fractional differential equations
and their applications in different research areas. So far there have been several fun-
damental works on the fractional derivative and fractional differential and integral
equations, see [8,11,14,16,17]. Recently, some interesting applications of the frac-
tional calculus in the theory of analytic functions have been developed. The classi-
cal definition of Riemann-Liouville in fractional calculus operators and their various
generalizations have fruitfully been applied in obtaining, for example, the characteri-
zation properties, coefficient estimates, distortion inequalities and convolution struc-
tures for various subclasses of analytic functions [19] and the works in the research
monographs [13]. In [19], Srivastava and Owa, gave definitions for fractional opera-
tors (derivative and integral) in the complex z-plane C as follows:
Definition 1.1. The fractional derivative of order α is defined, for a function f (z) by
Z z
α 1 d f (ζ)
Dz f (z) := dζ; 0 ≤ α < 1, (1)
Γ(1 − α) dz 0 (z − ζ)α

77
78 Maslina Darus, Rabha W. Ibrahim

where the function f (z) is analytic in simply-connected region of the complex z-plane
C containing the origin and the multiplicity of (z − ζ)−α is removed by requiring
log(z − ζ) to be real when (z − ζ) > 0.
Definition 1.2. The fractional integral of order α is defined, for a function f (z), by
Z z
α 1
Iz f (z) := f (ζ)(z − ζ)α−1 dζ; α > 0, (2)
Γ(α) 0
where the function f (z) is analytic in simply-connected region of the complex z-plane
(C) containing the origin and the multiplicity of (z − ζ)α−1 is removed by requiring
log(z − ζ) to be real when (z − ζ) > 0.
Remark 1.1. From Definition 1.1, we have D0z f (z) = f (0), limα→0 Izα f (z) = f (z) and
limα→0 D1−α
z f (z) = f 0 (z). Moreover,
Γ(µ + 1)
Dαz {zµ } = {zµ−α }, µ > −1, 0 ≤ α < 1
Γ(µ − α + 1)
and
Γ(µ + 1)
Izα {zµ } = {zµ+α }, µ > −1, α > 0.
Γ(µ + α + 1)
Lemma 1.1. [6] If α ∈ [0, 1) and f is a continuous function, then
(z)α−1 d
DIzα f (z) = f (0) + Izα D f (z); D = .
Γ(α) dz
In this work we investigate the existence and uniqueness of univalent solution for
fractional Volterra equation
Z z
λ
u(z) = f (ζ, u(ζ))(z − ζ)α−1 dζ + g(z); z ∈ U, (3)
Γ(α) 0
where λ is an arbitrary parameter, U := {z ∈ C : |z| < 1}, f : U × C → C is a
continuous function and g : U → C is a given increasing continuous function on U,
by using fixed point theorem for generalized contractions due to Pathak and Shahzad
[15]. Then, by using Covitz-Nadler Theorem [3], we establish the existence solution
for the integral inclusion
u(z) ∈ λIzα F(z, u(z)); z ∈ U, (4)
where F : U ×C → P(C) is compact closed set-valued function with nonempty values
in C, P(C) is the family of all nonempty subsets of C, Izα F(z, u(z)) is the definite
fraction integral for the set-valued function F of order α which is defined as
( Z z )
α 1 α−1
Iz F(z, u(z)) = f (ζ)(z − ζ) dζ : f (z, u) ∈ S F (u) ,
Γ(α) 0
where n o
S F (u) = f ∈ L1 (U, C) : f (z) ∈ F(z, u(z)), a.e. z ∈ U
denotes the set of selection of F.
 On the existence of univalent solutions for fractional integral equation... 79

2. PRELIMINARIES
In this section, we introduce notations, definitions, and preliminary facts from set-
valued analysis which will be used throughout this paper. For further background and
details pertaining to this section we refer the reader to [1, 2, 4, 5].
Let (X, | . |) be a normed space, Pcl (X) = {Y ∈ P(X) : Y is closed}, Pb (X) = {Y ∈
P(X) : Y is bounded}, Pcp (X) = {Y ∈ P(X) : Y is compact}, Pc (X) = {Y ∈ P(X) :
Y is convex}, Pcl,c (X) = {Y ∈ P(X) : Y is closed and convex}, Pcp,c (X) = {Y ∈
P(X) : Y is compact and convex}. A set-valued function F : X → P(X) is called
convex (closed) valued if F(x) is convex
S (closed) for all x ∈ X. F is called bounded
valued on bounded set B if F(B) = x∈B F(x) is bounded in X for all B ∈ Pb (X) i.e.
sup x∈B {sup{|u| : u ∈ F(x)}} < ∞. F is called compact if for every M bounded subset
of X, F(M) is relatively compact.

For the single valued problem (3) we need the following definitions and prelimi-
naries.
Definition 2.1. [15] Let (X, d) be a metric space. A mapping T : Y ⊂ X → X is said
to be a metric a−contraction (or simply a−contraction) mapping if Y is T −invariant
and it satisfies the inequality d(T (x), T 2 (x)) ≤ a(x)d(x, T (x)) for all x in Y, where
T
a : Y → [0, 1) and sup x∈Y a(T x) = λ < 1. Further, if ∞ n
n=0 T (Y) is a singleton set,
where T n (Y) := T (T n−1 (Y)) for each n ∈ N and T 0 (Y) := idY , then T is said to be
strong a−contraction.
Definition 2.2. [10] A metric space (X, d) is called T −orbitally complete if T is a
self mapping of X and if any Cauchy subsequence {T ni x} in orbit O(x, T ), x ∈ X,
converges in X.
It is obvious that a complete metric space is orbitally complete with respect to any
self-mapping of X.
Lemma 2.1. (Pathak-Shahzad) [15] Let (X, d) be a metric space, and let T : X → X
be a strongly a−contraction mapping. Suppose T is orbitally complete. Then T has
a unique fixed point.
For the set valued problem (4) we need the following definitions and preliminaries.
Definition 2.3. A set-valued function F : U → P(C) is said to be measurable if for
any x ∈ X, the function z 7→ d(x, F(z)) = in f {|x − u| : u ∈ F(z)} is measurable.

Let A, B ∈ Pcl (X) and let a ∈ A. Then, by denoting

D(a, B) = in f {ka − bk : b ∈ B} and ρ(A, B) = sup {D(a, B) : a ∈ A} ,
the function H : Pcl (X) × Pcl,b (X) → R+ defined by
H(A, B) = max{ρ(A, B), ρ(B, A)}
80 Maslina Darus, Rabha W. Ibrahim

is a metric and is called the Hausdorff metric on X. Moreover (Pcl,b (X), H) is a metric
space and (Pcl (X), H) is a complete metric space (see [9]). It is clear that

H(0, C) = sup {kck : c ∈ C; C ∈ Pb (X)} .

Definition 2.4. A set-valued function F : C → Pcl (C) is called

(i) γ − Lipschitz if there exist γ > 0 such that

H(F(z), F(w)) ≤ γkz − wk, f or each z, w ∈ X;

the constant γ is called a Lipschitz constant.

(ii) contraction if and only if it is γ − Lipschitz with γ < 1.
Definition 2.5. A set-valued function F : U × C → Pcl (C) is called
(i) γ(z) − Lipschitz if and only if there exists γ ∈ L1 (U, R+ ) such that

H(F(z, x), F(z, y)) ≤ γ(z)kx − yk, f or each x, y ∈ X.

(ii) a contraction if and only if it is γ(z) − Lipschitz with kγk < 1.

Definition 2.6. A function f of L1 (U, X) is called ”integrably bounded“ if for every
ε > 0 there exists δ(ε) > 0 such that
Z
k f (t)kdt < ε
Ω

for every measurable Ω ⊂ U with m(Ω) < δ(ε).

Remark 2.1. [7] It is known that if the measurable set valued function F : U →
Pcp (X) is an integrably bounded set-valued operator, then the set S F is closed and
nonempty.
Lemma 2.2. (Covitz-Nadler) [3] Let (X, d) be a complete metric space. If G : X →
Pcl (X) is a contraction, then G has a fixed point.

3. SINGLE-VALUED PROBLEM
This section deals with the existence of a solution u(z) on U for the integral oper-
ator (3). Let B = C[U, C] be the Banach space of continuous functions from U to C,
endowed with max. norm. Define an operator T : B → B by
Z z
λ
(T u)(z) = f (ζ, u(ζ))(z − ζ)α−1 dζ + g(z); z ∈ U (5)
Γ(α) 0
with the following conditions:
T
(H1) ∞ n n
n=0 T (B) is a singleton set, where T (B) := T (T
n−1 (B)) for each n ∈ N and

T 0 (B) := idB .
 On the existence of univalent solutions for fractional integral equation... 81

(H2) | f (z, u) − f (z, v)| ≤ b(|u − v|)|u − v|, where b : R+ → R+ is an increasing function
with kbk < ∞.
(H3) 0 < | f (z, u)| < ∞, ∀ z ∈ U.
Theorem 3.1. Let the assumptions (H1-H3) hold. If |λ| < supu∈XΓ(α+1)
b(d(u,T (u))) , then the
integral equation (3) has a unique univalent solution u(z) on U.
Proof. For any u ∈ B and by using (H2) we obtain

d(T (u), T 2 (u)) = maxz∈U |(T u)(z) − T [(T u)(z)]|

|λ|
≤ b(maxz∈U |u(z) − (T u)(z)|) maxz∈U |u(z) − (T u)(z)||z|α
Γ(α + 1)
|λ|
< b(maxz∈U |u(z) − (T u)(z)|) maxz∈U |u(z) − (T u)(z)|
Γ(α + 1)
|λ|
= b(d(u, T (u))) d(u, T (u)).
Γ(α + 1)
Hence the mapping T is a a−contraction and in virtue of Lemma 2.1, the equation
(T u)(z) = u(z); i.e. eq.(3) has a solution in B. Further, by using (H1), the uniqueness
of u follows. Now for z1 , z2 ∈ U such that z1 , z2 ; and by assumption (H3) we can
pose that T is univalent function.

Consequently, we have the following results.

Corolar 3.1. Let the assumptions of Theorem 3.1 hold. Then the fractional initial
value problem
Dαz u(z) = f (z, u(z)), z ∈ U (6)
subject to the condition u(0) = 0 has a unique solution u(z) in U.
Proof. It is easy to show that problem (6) is equivalent to the integral equation
Z z
1
u(z) = f (ζ, u(ζ))(z − ζ)α−1 dζ
Γ(α) 0
hence in view of Theorem 3.1, with λ = 1 and g(z) = 0, problem (6) has a unique
solution.
Corolar 3.2. Let the assumptions of Theorem 3.1 hold. Then the fractional differen-
tial equation
Dα (u − T n−1 [u])(z) = f (z, u(z)); n − 1 < α ≤ n, z ∈ U, (7)
u(k) (0) = u0(k) ∈ C, k = 0, 1, ..., n − 1
where T n−1 [u] is the Taylor polynomial of order (n − 1) for u centered at 0, has a
unique solution u(z) in U.
82 Maslina Darus, Rabha W. Ibrahim

Proof. Problem (7) is equivalent to the integral equation

X
n−1 k Z z
z (k) 1
u(z) = u (0) + f (ζ, u(ζ))(z − ζ)α−1 dζ
k=0
k! Γ(α) 0

Pn−1 zk (k)
hence in view of Theorem 3.1, with λ = 1 and g(z) = k=0 k! u (0), problem (7) has
an unique solution.

4. SET-VALUED PROBLEM
This section concerns with the existence result for the integral inclusion (4). The
study will be made in view of fixed point theorem of set-valued function (Lemma
2.2).
Definition 4.1. A function u(.) ∈ C(U, C) is said to be a solution of (4) if there exists
a function f (z) ∈ F(z, u(z)), on U and u(z) = λIzα f (z).
Let us consider the following assumptions

(H4) F : U × C → Pcl,cp (C) such that

(i) (z, .) 7→ F(z, u) is measurable for each u ∈ C,
(ii) F is integrably bounded for all u ∈ C, such that 0 < |F(z, u)| < h(z), h ∈
L1 (U, R+ ), where khk < ∞.
(iii) F is `(z) − Lipschitz, i.e. H(F(z, u), F(z, v)) ≤ `(z)ku − vkB where ` : U → R+ .
|λ|k`k
Theorem 4.1. Let the assumption (H4) hold. If Γ(α+1) < 1, then the integral inclusion
(4) has at least one locally univalent solution u(z) on U.
Proof. In virtue of Remark 2.1, for each u(z) in B, the set S F (u) is nonempty (H4-i).
Define a set-valued function operator N : C(U, C) → P(C(U, C)) by
Z z
λ
Nu = {u ∈ C(U, C) : u(z) = f (ζ)(z − ζ)α−1 dζ, f ∈ S F (u)}. (8)
Γ(α) 0
Our aim is to apply Lemma 2.2. The proof is performed in three steps

Step 1: N(u) ∈ Pcl (B) for each u ∈ B = C(U, C). Let {um }m≥0 ∈ N(u) such that
um → eu in B. Then e
u ∈ B and there exists fm ∈ S F (u) such that for all z ∈ U
Z z
λ
um (z) = fm (ζ)(z − ζ)α−1 dζ.
Γ(α) 0
Using the fact that F has closed values, we get that fm converges to f ∈ S F (u). Then
for each z ∈ U Z z
λ
um (z) → eu(z) = f (ζ)(z − ζ)α−1 dζ.
Γ(α) 0
 On the existence of univalent solutions for fractional integral equation... 83

So e
u ∈ N(u).

Step 2: There exists γ < 1 such that

H(N(u), N(v)) ≤ γku − vkB , f or each u, v ∈ B.

Let u, v ∈ C(U, C). Then by (H4-iii) there exists f ∈ F satisfies

| f (t, u) − f (t, v)| ≤ `(z)ku − vkB

then for h1 (z) ∈ N(u) where

Z z
λ
h1 (z) = f (ζ, u)(z − ζ)α−1 dζ
Γ(α) 0

And for h2 (z) ∈ N(v) where

Z z
λ
h2 (z) = f (ζ, v)(z − ζ)α−1 dζ
Γ(α) 0

we have
Z z
|λ|
|h1 (t) − h2 (t)| ≤ | f (ζ, u) − f (ζ, v)||(z − ζ)|α−1 dζ
Γ(α) 0
|λ|`(z)
≤ ku − vkB |z|α
Γ(α + 1)
|λ|k`k
< ku − vkB
Γ(α + 1)

Denotes that
|λ|k`k
γ := .
Γ(α + 1)
It follows that

H(N(u), N(v)) ≤ γku − vkB , f or each u, v ∈ B,

where γ < 1. Implies that N is a contraction set-valued mapping. Then in view of

Lemma 2.2, N has a fixed point which is corresponding to a solution of inclusion (4).

Step 3: For z1 , 0, z2 , 0 ∈ U such that z1 , z2 and by assumption (H4-ii) we can

verify that N is univalent function. Consequently, we have every solution u(z), z ∈ U
for the integral inclusion (4) is univalent. Hence the proof.
84 Maslina Darus, Rabha W. Ibrahim

5. EXISTENCE OF EXTREMAL SOLUTIONS

In this section, we define the lower and upper solutions of the problem (3) by using
the concept of the subordination and superordination.
Given two functions F(z) and G(z), that are analytic in U, the function F(z) is said
to be subordination to G(z) in U if there exists a function h(z), analytic in U with
h(0) = 0 and |h(z)| < 1 f or all z ∈ U such that F(z) = G(h(z)) f or all z ∈ U.
Let φ : C2 → C and let h be univalent in U. If p is analytic in U and satisfies the
differential subordination φ(p(z)), zp0 (z)) ≺ h(z) then p is called a solution of the dif-
ferential subordination. The univalent function q is called a dominant of the solutions
of the differential subordination if p ≺ q. If p and φ(p(z)), zp0 (z)) are univalent in U
and satisfy the differential superordination h(z) ≺ φ(p(z)), zp0 (z)) then p is called a
solution of the differential superordination. An analytic function q is called subordi-
nant of the solution of the differential superordination if q ≺ p. Let H be the class of
functions analytic in U and H[a, n] be the subclass of H consisting of functions of
the form f (z) = a + an zn + an+1 zn+1 + ... . Let A be the subclass of H consisting of
functions of the form f (z) = z + a2 z2 + ... .
Definition 5.1 (12). Denote by Q the set of all functions f (z) that are analytic and
injective on U − E( f ) where E( f ) := {ζ ∈ ∂U : limz→ζ f (z) = ∞} and are such that
f 0 (ζ) , 0 for ζ ∈ ∂U − E( f ).
Lemma 5.1 (18). Let q(z) be convex univalent in the unit disk U and ψ and γ ∈ C
00 (z)
with <{1+ zqq0 (z) + ψγ } > 0. If p(z) is analytic in U and ψp(z)+γzp0 (z) ≺ ψq(z)+γzq0 (z),
then p(z) ≺ q(z) and q is the best dominant.
Lemma 5.2 (12). Let q(z) be convex univalent in the unit disk U and γ ∈ C. Further,
assume that <{γ} > 0. If p(z) ∈ H[q(0), 1] ∩ Q, with p(z) + γzp0 (z) is univalent
in U then q(z) + γzq0 (z) ≺ p(z) + γzp0 (z) implies q(z) ≺ p(z) and q(z) is the best
subordinant.
Definition 5.2. A function u(.) ∈ C[U, C] is said to be a lower solution of (3) if there
exist f, g such that f ∈ F(z, u) and g ∈ G(z, u) a.e. on U and
u(z) ≺ λIzα f (z, u(z)) + g(z), ∀z ∈ U.
A function u(.) ∈ C[U, C] is said to be a upper solution of (3) if there exist f, g such
that f ∈ F(z, u) and g ∈ G(z, u) a.e. on U and
λIzα f (z, u(z)) + g(z) ≺ u(z), ∀z ∈ U.
Thus we have the following results.
Theorem 5.1. Let u(z) be convex univalent in U. Assume that
( )
zu00 (z) 1
< 1+ 0 + > 0, γ , 0 (9)
u (z) γ
 On the existence of univalent solutions for fractional integral equation... 85

and for f, g ∈ A the subordination

λIzα f (z, u(z)) + λγzIzα f 0 (z, u(z)) + [g(z) + γzg0 (z)] ≺ u(z) + γzu0 (z)

holds. Then
λIzα f (z, u(z)) + g(z) ≺ u(z) (10)
and u(z) is the best dominant.
Proof. Our aim is to apply Lemma 5.1. We set p(z) := λIzα f (z, u(z)) + g(z) and ψ = 1.
Then by Lemma 1.1, we obtain

p(z) + γzp0 (z) = λIzα f (z, u(z)) + g(z) + γz[λIzα f (z, u(z)) + g(z)]0
= λIzα f (z, u(z)) + g(z) + γλzIzα f 0 (z, u(z)) + γzg0 (z)
≺ u(z) + γzu0 (z) := q(z) + γzq0 (z).

Hence the relation (10) follows by an application of Lemma 5.1.

Theorem 5.2. Let u(z) be convex univalent in the unit disk U. Assume that

<{γ} > 0, γ , 0 ∈ C. (11)

If λIzα f (z, u(z)) + g(z) ∈ H[0, 1] ∩ Q where f, g ∈ A,

λIzα f (z, u(z)) + λγzIzα f 0 (z, u(z)) + [g(z) + γzg0 (z)]

is univalent is U and the subordination

u(z) + γzu0 (z) ≺ λ[Izα f (z, u(z)) + γzIzα f 0 (z, u(z))] + [g(z) + γzg0 (z)]

holds, then
u(z) ≺ λIzα f (z, u(z)) + g(z) (12)
and u(z) is the best subordinant.
Proof. Our aim is to apply Lemma 5.2. We set p(z) := λIzα f (z, u(z)) + g(z). By
computation shows the following subordination

q(z) + γzq0 (z) := u(z) + γzu0 (z)

≺ λ[Izα f (z, u(z)) + γzIzα f 0 (z, u(z))] + [g(z) + γzg0 (z)]
= λIzα f (z, u(z)) + g(z) + γz[λIzα f (z, u(z)) + g(z)]0
= p(z) + γzp0 (z).

Then (12) follows by an application of Lemma 5.2.

We combine Theorems 5.1 and 5.2 in order to get the following sandwich theorem
86 Maslina Darus, Rabha W. Ibrahim

Theorem 5.3. Let u(z), u(z) be convex univalent in the unit disk U satisfy (11) and
(9) respectively. If λIzα f (z, u(z)) + g(z) ∈ H[0, 1] ∩ Q where f, g ∈ A,
λ[Izα f (z, u(z)) + γzIzα f 0 (z, u(z))] + [g(z) + γzg0 (z)]
is univalent in U and the subordination
u(z) + γzu0 (z) ≺ λ[Izα f (z, u(z)) + γzIzα f 0 (z, u(z))] + [g(z) + γzg0 (z)] ≺ u(z) + γzu0 (z)
holds, then u(z) ≺ λIzα f (z, u(z)) + g(z) ≺ u(z), and u(z) is the best subordinant, while
u(z) is the best dominant.

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