1. Introduction

The study of fractional calculus that grows out of traditional concepts of the calculus derivative and integral operators. Several authors were introduced many different forms of noninteger differential operators and discussed varous results on existence, uniqueness; and qualitative and quantative properties of solutions for fractional differential equations, the reader referred to [7, 11, 12, 16, 23] and the monographs: Samko et al. (1993); Miller et al. (1993); Podlubny (1999); Hilfer (2000); Kilbas et al. (2006); Cresson (2007); Diethem K. (2010); Katugampola (2011) and Abbas S. et al. (2012).

Podlubny I. [20], studied the existence and uniqueness of an initial value problem:

(1.1)

(1.2)

where denotes the real space and denotes Riemann-Liouville fractional derivative operator.

Recently, Chinchane V. L. and Pachpatte D. B. [2] have discussed the uniqueness of solution of fractional differential equation with the Riemann-Liouville derivative. Existence and uniqueness of an implicit fractional differential equations via the Liouville-Caputo derivative have studied by authors in [17] using the fixed point concepts. Kucche et al. [10] investigated existence, uniqueness, continuous dependence and estimates of solutions for an implicit fractional differential equations.

Motivated by the above mentioned works in this manuscript, we discuss the existence and uniqueness of the solution for the following implicit fractional differential equations with Riemann-Liouville derivative:

(1.3)

(1.4)

where denotes Riemann-Liouville fractional derivative operator and is real conitnuous valued function on into denotes the real space.

Furthermore, our intention is to extend the results presented by Chinchane V. L. and Pachpatte D. B. to nonlinear implicit fractional differential equation.

The paper is organized as follows. In Section 2, some definitions, lemmas and preliminary results are intoduced to be used in the sequel. Section 3 will involve the assumptions and main result of existence and uniqueness by fixed point theorerm. Finally Section 4 deal the results of uniqueness for the problem (1.3)-(1.4) via inequalities.

2. Preliminaries

Let us recall some definitions and concepts of the fractional calculus [9, 10, 15, 17, 20, 21] and state the few results which are used throuhout this paper.

Definition 2.1. The fractional derivative of order of a continuous function is given by

(2.1)

provided that the right side is pointwise defined on

Definition 2.2. The fractional primitive of order of a function is given as follows

(2.2)

provided that the right side is pointwise defined on where

In [19], Medved introduced a special class of nonlinear functions and developed a method to estimate solution for nonlinear integral inequalities with singular kernel. The functions of such class are defined as follows:

Definition 2.3. Let be a real number and . The function w: satisfies the following condition

(2.3)

for all , where is a continuous, nonnegative function.

Remark 2.1. If then

(2.4)

for any i.e the condition (2.3) is satisfies with . For , where the function satisfies the condition (2.3) with and see [14].

Lemma 2.1 [19] Let , be nondecreasing function, on and

(2.5)

for where is constant. Then following hold:

(i) Suppose and if satisfies the condition (2.3) with then

(2.6)

for where

(2.7)

where is gamma function, is the inverse of and is such that for all

(ii) Let and satisfies the condition (2.3) with where i. e. Let be as in part (i). Then

(2.8)

for where

(2.9)

(2.10)

and is such that for all

Lemma 2.2 [19] Let and

(2.11)

for where is constant. Then following hold:

(i) Suppose then

(2.12)

(ii) If for some then

(2.13)

for where is defined as in (2.10),

For detail proof of above two theorems see [16].

Lemma 2.3 ([3, 18], p. 152) (Bihari inequality) Let and be nonnegative defined on let be continuous nondecreasing function defined on and on If

(2.14)

for where is nonnegative constant, for

(2.15)

where

and is the inverse fucntion of and is chosen so that for all laying in the interval

Lemma 2.4 Let be a non-empty complete metric space with a contraction mapping Then admits a unique fixed point in

3. Existence and Uniqueness

In this section, we prove existence and uniqueness result for the problem (1.3)-(1.4). We first note that if is an absolutely continuous function satisfying (1.3)-(1.4), then

(3.1)

Theorem 3.1 Assume that there exist such that

(3.2)

for each then the problem (1.3)-(1.4) has unique solution

Proof. Consider a function and defined by

(3.3)

Let Then we have

(3.4)

Hence

(3.5)

for each From the Banach fixed point theorem, Lemma 2.4, there exists a unique such that Therefore

(3.6)

Set

(3.7)

This implies that and therefore

This shows that the function satisfies the problem (1.3)-(1.4) and uniqueness of the solution follows from the unique existence of This completes the proof of the theorem.

4. Uniqueness via Inequalites

In this section, we discuss the uniqueness of solution of the initial value problem (1.3)-(1.4).

Theorem 4.1 If is continuous and satisfies condition

(4.1)

whre is positive constant, and is a continuous nondecreasing function on with and

(4.2)

then the problem (1.3)-(1.4) has unique solution on

Proof. Let and be two solutions of the problem (1.3)-(1.4). Then we have

(4.3)

and

(4.4)

Hence we have

(4.5)

But by hypothesis (4.1) for any and any

This implies

(4.6)

Using above estimation in (4.5), we get

(4.7)

Now an application of Lemma 2.3 to (4.7) which yields

(4.8)

where is primitive for We shall prove that the right-hand side of (4.8) tends toward zero as As is independent of it follows that which we need. Let us remark that condition (4.2) implies no matter how we choose the primitive of Thus as Consequently, in the inequality (4.8), the right-hand side tends toward zero. This completes the proof of the theorem.

Theorem 4.2 If the function is continuous and satisfies the conditionz

(4.9)

for some positive constant and then the initial value problem (1.3)-(1.4) has unique solution in the interval

Proof. Let and be two solutions of the problem (1.3)-(1.4). Then we have

(4.10)

and

(4.11)

Therefore, using these (4.10), (4.11) and hypothesis (4.9), we have

(4.12)

But by again hypothesis (4.9) for any and any

This implies

(4.13)

Using (4.13) in (4.12), we obtain

(3.14)

Now, (a) suppose that then applying Leema 2.2 (i) to (4.14), we have

(4.15)

for Since was arbitrary, as the inequality (4.13) implies that on

(b) Let for some Then by Lemma 2.2 (ii) to (4.14), again we have,

(4.16)

for where is defined by (2.10). Since was arbitrary in (4.16), implies that as This completes the proof of the theorem.