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On the Existence of Locally Attractive Solutions of a Nonlinear Quadratic Volterra Integral Equation of Fractional Order

Abstract

The authors employs a hybrid fixed point theorem involving the multiplication of two operators for proving an existence result of locally attractive solutions of a nonlinear quadratic Volterra integral equation of fractional (arbitrary) order. Investigations will be carried out in the Banach space of real functions which are defined, continuous, and bounded on the real half axis .

1. Introduction

The theory of differential and integral equations of fractional order has recently received a lot of attention and now constitutes a significant branch of nonlinear analysis. Numerous research papers and monographs have appeared devoted to differential and integral equations of fractional order (cf., e.g., [1â€“6]). These papers contain various types of existence results for equations of fractional order.

In this paper, we study the existence of locally attractive solutions of the following nonlinear quadratic Volterra integral equation of fractional order:

(11)

for all and , in the space of real functions defined, continuous, and bounded on an unbounded interval.

It is worthwhile mentioning that up to now integral equations of fractional order have only been studied in the space of real functions defined on a bounded interval. The result obtained in this paper generalizes several ones obtained earlier by many authors.

In fact, our result in this paper is motivated by the extension of the work of Hu and Yan [7]. Also, We proceed and generalize the results obtained in the papers [8, 9].

2. Notations, Definitions, and Auxiliary Facts

Denote by the space of Lebesgue integrable functions on the interval , which is equipped with the standard norm. Let and let be a fixed number. The Riemann-Liouville fractional integral of order of the function is defined by the formula:

(2.1)

where denotes the gamma function.

It may be shown that the fractional integral operator, transforms the space into itself and has some other properties (see [10â€“12]).

Let be the space of continuous and bounded real-valued functions on and let be a subset of . Let be an operator and consider the following operator equation in , namely,

(2.2)

for all . Below we give different characterizations of the solutions for the operator equation (2.2) on . We need the following definitions in the sequel.

Definition 2.1.

We say that solutions of (2.2) are locally attractive if there exists an and an such that for all solutions and of (2.2) belonging to we have that:

(2.3)

Definition 2.2.

An operator is called Lipschitz if there exists a constant such that for all . The constant is called the Lipschitz constant of on .

Definition 2.3 (Dugundji and Granas [13]).

An operator on a Banach space into itself is called compact if for any bounded subset of , is a relatively compact subset of . If is continuous and compact, then it is called completely continuous on .

We seek the solutions of (1.1) in the space of continuous and bounded real-valued functions defined on . Define a standard supremum norm and a multiplication "" in by

(2.4)

Clearly, becomes a Banach space with respect to the above norm and the multiplication in it. By we denote the space of Lebesgue integrable functions on with the norm defined by

(2.5)

We employ a hybrid fixed point theorem of Dhage [14] for proving the existence result.

Theorem 2.4 (Dhage [14]).

Let be a closed-convex and bounded subset of the Banach space and let be two operators satisfying:

1. (a)

is Lipschitz with the Lipschitz constant ,

2. (b)

is completely continuous,

3. (c)

for all , and

4. (d)

where .

Then the operator equation

(2.6)

has a solution and the set of all solutions is compact in .

3. Existence Result

We consider the following set of hypotheses in the sequel.

(H1) The function is continuous, and there exists a bounded function with bound satisfying

(3.1)

for all and .

(H2) The function defined by is bounded with

(3.2)

(H3) The function is continuous and .

(H4) The function is continuous. Moreover, there exist a function being continuous on and a function being continuous on with and such that

(3.3)

for all such that and for all .

For further purposes let us define the function by putting

(3.4)

Obviously the function is continuous on .

In what follows we will assume additionally that the following conditions are satisfied.

(H5) The functions defined by the formulas

(3.5)

are bounded on and vanish at infinity, that is, .

Remark 3.1.

Note that if the hypotheses and hold, then there exist constants and such that:

(3.6)

Theorem 3.2.

Assume that the hypotheses hold. Furthermore, if , where and are defined in Remark 3.1, then (1.1) has at least one solution in the space . Moreover, solutions of (1.1) are locally attractive on .

Proof.

Set . Consider the closed ball in centered at origin 0 and of radius , where .

Let us define two operators and on by

(3.7)

for all .

According to the hypothesis , the operator is well defined and the function is continuous and bounded on . Also, since the function is continuous on , the function is continuous and bounded in view of hypothesis . Therefore and define the operators . We will show that and satisfy the requirements of Theorem 2.4 on .

The operator is a Lipschitz operator on . In fact, let be arbitrary. Then by hypothesis , we get

(3.8)

for all . Taking the supremum over ,

(3.9)

for all . This shows that is a Lipschitz on with the Lipschitz constant .

Next, we show that is a continuous and compact operator on . First we show that is continuous on . To do this, let us fix arbitrary and take such that . Then we get

(3.10)

Since is continuous on , then it is bounded on , and there exists a nonnegative constant, say , such that . Hence, in view of hypothesis , we infer that there exists such that for . Thus, for we derive that

(3.11)

Furthermore, let us assume that . Then, evaluating similarly to the above we obtain the following estimate:

(3.12)

where ,â€‰â€‰,â€‰â€‰.

Therefore, from the uniform continuity of the function on the set we derive that as . Hence, from the above-established facts we conclude that the operator maps the ball continuously into itself.

Now, we show that is compact on . It is enough to show that every sequence in has a Cauchy subsequence. In view of hypotheses and , we infer that:

(3.13)

for all . Taking the supremum over , we obtain for all . This shows that is a uniformly bounded sequence in . We show that it is also equicontinuous. Let be given. Since , there is constant such that for all .

Let be arbitrary. If , then we have

(3.14)

From the uniform continuity of the function on and the function in , we get as .

If , then we have

(3.15)

as .

Similarly, if with , then we have

(3.16)

Note that if , then and . Therefore from the above obtained estimates, it follows that:

(3.17)

As a result, as . Hence is an equicontinuous sequence of functions in . Now an application of the ArzelÃ¡-Ascoli theorem yields that has a uniformly convergent subsequence on the compact subset of . Without loss of generality, call the subsequence of the sequence itself.

We show that is Cauchy sequence in . Now as for all . Then for given there exists an such that for , then we have

(3.18)

This shows that is Cauchy. Since is complete, then converges to a point in . As is closed, converges to a point in . Hence, is relatively compact and consequently is a continuous and compact operator on .

Next, we show that for all . Let be arbitrary, then

(3.19)

for all . Taking the supremum over , we obtain for all . Hence hypothesis of Theorem 2.4 holds.

Also we have

(3.20)

and therefore . Now we apply Theorem 2.4 to conclude that (1.1) has a solution on

Finally, we show the local attractivity of the solutions for (1.1). Let and be any two solutions of (1.1) in defined on , then we get

(3.21)

for all . Since , and , for , there are real numbers , and such that for , for all and for all . If we choose , then from the above inequality it follows that for , where . This completes the proof.

4. An Example

In this section we provide an example illustrating the main existence result contained in Theorem 3.2.

Example 4.1.

Consider the following quadratic Volterra integral equation of fractional order:

(4.1)

where .

Observe that the above equation is a special case of (1.1). Indeed, if we put and

(4.2)

Then we can easily check that the assumptions of Theorem 3.2 are satisfied. In fact, we have that the function is continuous and satisfies assumption , where and as in assumption . We have that the function is continuous and it is easily seen that as , thus assumption is satisfied. Next, let us notice that the function satisfies assumption , where , and . Thus . To check that assumption is satisfied let us observe that the functions appearing in that assumption take the form:

(4.3)

Thus it is easily seen that as . Finally, let us note that in Remark 3.1 there are two constants such that . It is also easy to check that , and . Then . Hence, taking into account that (cf. [4]), all the assumptions of Theorem 3.2 are satisfied and (4.1) has a solution in the space . Moreover, solutions of (4.1) are uniformly locally attractive in the sense of Definition 2.1.

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Abbas, M. On the Existence of Locally Attractive Solutions of a Nonlinear Quadratic Volterra Integral Equation of Fractional Order. Adv Differ Equ 2010, 127093 (2010). https://doi.org/10.1155/2010/127093