American Journal of Theoretical and Applied Statistics
Volume 6, Issue 5-1, September 2017, Pages: 13-22

Solvability of Some Nonlinear Integral Functional Equations

Mahmoud M. El-Borai, Wagdy G. El-Sayed, Noura N. Khalefa

Department of Mathematics, Faculty of Science, Alexandria University, Alexandria, Egypt

Email address:

(M. M. El-Borai)
(W. G. El-Sayed)
(N. N. Khalefa)

To cite this article:

Mahmoud M. El-Borai, Wagdy G. El-Sayed, Noura N. Khalefa. Solvability of Some Nonlinear Integral Functional Equations. American Journal of Theoretical and Applied Statistics. Special Issue: Statistical Distributions and Modeling in Applied Mathematics. Vol. 6, No. 5-1, 2017, pp. 13-22. doi: 10.11648/j.ajtas.s.2017060501.13

Received: February 11, 2017; Accepted: February 15, 2017; Published: February 28, 2017


Abstract: This paper discussed some existence theorems for nonlinear functional integral equations in the space L^1 of Lebesgue integrable functions,by using the Darbo fixed point theorem associated with the Hausdorff measure of noncompactness. Also, as an application, we discuss the existence of solutions for some nonlinear integral equations with fractional order

Keywords: Superposition Operator, Carathe'odory Conditions, Measure of Noncompactness, Fixed Point Theorem


1. Introduction

The subject of nonlinear integral equations considered as an important branch of mathematics because it is used for solving many problems such as physics, chemistry [4, 20].

In this paper we will use the technique of measures of noncompactness and Darbo fixed point theorem to prove the existence theorem for a nonlinear integral equation in the spaces. Also, as an application, we discuss the existence of solutions for some nonlinear integral equations with fractional order, which extends to some previous results in literature [2].

2. Notation and Auxiliary Facts

Let  be the field of real numbers,  be the interval [0,) and be the space of Lebesgue integerable functions on a measureable subset [0,) of with the standard norm.

One of the most important operators studied in nonlinear functional analysis is the so-called superposition operator [1].

Assume that a function  satisfies Carathodory conditions, i.e. it is measurable in t for anyand continuous in  for almost all .

Then to every function  being measurable on  we may assign the function

The operator  in such a way is called the superposition operator generated by the function.

We have the following theorem due to Appell and Zabrejko [1].

Theorem 2.1

The superposition operator  generated by the function maps continuously the space  if and only if  and , where and .

Next, we will mention a desired theorems concerning the compactness in measure of a subset

Of  [2].

Theorem 2.2

Let X be a bounded subset of  consisting of functions which are almost everywhere nondecreasing (or nonincreasing) on the interval [0,. Then X is compact in measure.

Furthermore, we recall few facts about the convolution operator [19].

Let be a given function. Then for any function , the integral

,

exists for almost every Moreover, the function  belongs to the space.

Thus is a linear operator which maps the space  intoand is also bounded since

, for every; so, it will be continuous.

Hence the norm of the convolution operator is majored by

In the sequel, we have the following theorem due to Krzyz [18].

Theorem 2.3

Assume that  is measurable on  such that the integral operator,

,

maps  into itself, then  transforms the set of nonincreasing functions from  into itself if and only if for any , the following implication is true.

.

In the case space we will use the following corollary

Corollary 2.1

Let  be ameasurable function generated the Fredholm operator K acting from  into, if for every  and for all  the implication holds,

.

Finally, we give a short note on measures of noncompactness and fixed point theorem.

Let  be an arbitrary Banach space with and the zero element

Let also  be a nonempty and bounded subset of  and  be a closed ball in E centered atand radius.

The Hausdorff measure of noncompactness  [3] is defined as

The De Blasi measure of weak noncompactness [2] is defined as

De Blasi measure can be expressed in the space  in a very useful formula, given by Appell and De Pascale [2]

Another measure was defined in the space  [2]. For any , let

,

and

Where meas  denotes the lebesgue measure of sub set

put

.

Then we have the following theorem [2], which connects between the two measures

and .

Theorem 2.4

Let X be a nonempty, bounded and compact in measure subset of  then

.

In the case space we have the following theorems [2].

Theorem 2.5

Let  be abounded subset of  and suppose that there is a family of measurable subset, of the interval  such that

And for every

Then the set  is compact in measure.

Theorem 2.6

Let 𝑋be an arbitrary nonempty and bounded subset of . If  is compact in measure then

As an application of measures of noncompactness, we recall the fixed point theorem due to Darbo [5].

Theorem 2.7

Let  be a non-empty, bounded, closed and convex subset of  and let  be a continuous transformation which is a contraction with respect to the measure of noncompactness, i.e there exist  such that  for any nonempty subset  of, then A has at least one fixed point.

3. Existence Solutions in

Now we will discuss the solvability for the following nonlinear integral equation

(1)

in the space.

We shall treat equation (3.1) under the following assumptions which are listed below:

(i) , almost everywhere positive and nonincreasing in

(ii)  are nonincreasing functions on with respect to  and  satisfy

Caratheodory conditions and there are two functions and two Constants,

such that

, for all  and

(iii)  are measurable with respect to and ,, and for all

and , the following condition is satisfied

Note that, from the assumption (iii) we deduce that the operator  is bounded with norm.

(iv) .

Then we can prove the following theorem

Theorem 3.1

Let the assumptions (i) (iv) be satisfied, then the equation (1) has at least one solution, being almost everywhere non increasing on

Proof

Consider the operator H

Then, the equation (3.1) takes the form

First, let

Then using our assumption (i)(iii), we have

From the last estimate, the space into itself using theorem (2.1)

Moreover, using the estimate (1), we see that the operator  transforms the ball  into itself, where:

Let be subset of consisting of all functions being are almost everywhere positive and non-increasing on

Note that  is non-empty, bounded, closed, convex subset of

Moreover, in view of Theorem (2.2) the set is compact in measure.

Next, by taking,

Then  is almost everywhere positive and non-decreasing on  and consequently  is also of the same type (in virtue of the assumption (iii) and theorem (2.3))

Further, the assumption (ii) permits us to deduce that,

is almost everywhere positive and non-decreasing on , this fact together with assertion  gives that self-mapping of the set since the Operator is continuous and  is continuous in view theorem (2.1), we conclude that  maps continuously  into .

Finally, assume that is non-empty subset of  and is fixed, then for an arbitrary  and for a set meas we obtain

Where, since

meas

Then, the last inequality gives

(2)

Further, more fixing  we arrive at the following estimate

Since  the above inequality gives

(3)

Hence, combining (2) and (3) we get

Where  denotes the measure of noncompactness, since is compact in measure, then by using Theorem (2.4) the last inequality together with the assumption (iv), enable us to apply Theorem (2.7), which proves the existence of a fixed point for the operator  in

4. Nonlinear Integral Equations with Fractional Order

In this section we will discuss solvability for the following nonlinear integral equation with fractional order in .

(4)

We shall treat equation (4.1) under the following assumptions which are listed below

(i)  and almost everywhere positive and nondecreasing in .

(ii)  are nondecreasing functions on with respect to and  satisfy Caratheodory conditions, there are two functions  and two constants such that for all  and  

(iii)is a measurable with respect to  and and  is bounded with norm

Note that, for all  and  we have

(iv)

Then we can prove the following theorem,

Theorem 4.1

Let the assumptions (i)-(iv) are satisfied, then the equation (4.1) has at least one solution, being almost everywhere non-decreasing on

Proof

Consider the operator

Then, the equation (4.1) takes the form

First, let

Then using our assumption (i)-(iii) we have,

 ]

(5)

From the last estimate we deduce that the operator maps continuously, the space  into itself using theorem (2.1).

Moreover, using the estimate (1), we see that the operator  transforms the ball  into itself where:

Let  be subset of consisting of all functions being are almost everywhere positive and non-increasing on

Note that  is nonempty, bounded, closed, convex subset of

Moreover, in view of Theorem (2.2) the set is compact in measure.

Next, by taking then  is almost everywhere positive and non-increasing on and consequently  is also of the same type (in virtue of the assumption (iii) and Theorem (2.3).

Further, the assumption (ii) permits us to deduce that:

Is also almost everywhere positive and non-decreasing on , this fact together with assertion, gives that self – mapping of the set

Since the operator  is continuous and  is continuous in view theorem (2.1), we conclude that mapps continuous  into .

Note, that

,

Let J=

=

Then

 then

Finally, assume that is nonempty subset of  and is fixed, then for an arbitrary  and for a set meas we obtain

Where, since

meas

Then, the above inequality gives

(6)

Further, more fixing  we arrive at the following estimate

Since  the above in quality gives

(7)

Hence, combining (2) and (3) we get

Where  denotes the measure of noncompactness, since  is compact in measure, then by using Theorem (2.4), The last inequality together with the assumption (iv), enable us to apply Theorem (2.7), Which proves the existence of a fixed point for the operator  in

In the same way, we will discuss the solvability of the nonlinear integral equation with fractional order

(8)

in the space .

We shall treat the equation (4.2) under the following assumptions which are listed below

(i) almost everywhere positive and nondecreasing in (0,1),

(ii)  are nondecreasing functions with respect to and , satisfy Caratheodory conditions and there are two functions  and two constants such that for all  and  

(iii)is a measurable with respect to and and  (From assumption (iii), we see that  is continuous and so it is bounded with norm ).

Also, for all and  we have

(iv)

Then we can prove the following theorem,

Theorem 4.2

Let the assumptions (i)-(iv) are satisfied, then the equation (4) has at least one solution, being almost everywhere non-decreasing on (0,1).

Proof

Consider the operator

Then, the equation (4.2) takes the form

First, let

Then using our assumption (i)-(iii) we have,

From the last estimate we deduce that the operator maps continuously, the space  into itself using theorem (2.1).

Moreover, using the estimate (1), we see that the operator  transforms the ball  into itself where:

Let be subset of consisting of all functions being are almost everywhere positive and non-increasing on

Note that  is nonempty, bounded, closed, convex subset of

Moreover, in view of Theorem (2.5) the set is compact in measure.

Next, by taking then  is almost everywhere positive and nonincreasing on (0, 1) and consequently  is also of the same type (in virtue of the assumption (iii) and Theorem (2.1).

Further, the assumption (ii) permits us to deduce that the operator

is also almost everywhere positive and nondecreasing on (0, 1), this fact together with assertion, gives that self – mapping of the set

Since the operator  is continuous and  is continuous in view Theorem (2.1), we conclude that mapps continuous  into.

Note, that ,  then

 then

Finally, assume that is nonempty subset of  and is fixed, then for an arbitrary  and for a set meas  we obtain

Where , since

meas

Then the above inequality gives

Where  is the De Blasi measure of noncompactness:

Since  is compact in measure, then by using Theorem (2.6(, we can write the last inequality in the form

This inequality together with the assumption (vi) enables us to apply Theorem (2.8), which proves the existence of a fixed point for the operator in

5. Conclusion

In this work, we determined the sufficient conditions under which the existence theorem of a nonlinear integral equation with convolution kernel is proved in the space . Also, the same situation is proved for a nonlinear integral equation with fractional order in the spaces  and .

Acknowledgment

We would like to thank the referees for their careful reading of the paper and their valuable comments.


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