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
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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.
References