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# Treatment of a Coupled System for Quadratic Functional Integral Equation on the Real Half-Line via Measure of Noncompactness

1
Faculty of Science, Alexandria University, Alexandria 21544, Egypt
2
Faculty of Science, Omar Al-Mukhtar University, Tripoli 22131, Libya
3
Faculty of Arts and Sciences, Department of Mathematics, The International University of Beirut, Beirut 1107, Lebanon
4
Faculty of Arts and Sciences, Department of Mathematics, Lebanese International University, Saida 1600, Lebanon
*
Author to whom correspondence should be addressed.
Foundations 2023, 3(1), 37-48; https://doi.org/10.3390/foundations3010004
Received: 10 October 2022 / Revised: 18 November 2022 / Accepted: 3 January 2023 / Published: 6 January 2023

## Abstract

:
This article is devoted to the solvability and the asymptotic stability of a coupled system of a functional integral equation on the real half-axis. Our consideration is located in the space of bounded continuous functions on $R +$$( B C ( R + ) ) .$ The main tool applied in this work is the technique associated with measures of noncompactness in $B C ( R + )$ by a given modulus of continuity. Next, we formulate and prove a sufficient condition for the solvability of that coupled system. We, additionally, provide an example and some particular cases to demonstrate the effectiveness and value of our results.

## 1. Preliminaries and Introduction

Measures of noncompactness are frequently employed in fixed-point theory, and they are especially useful in work on the concepts of differential equations, optimization theory, functional integral equations, and integral equations (see [1,2,3,4,5]).
Nonlinear integral equations are useful for describing many real-world phenomena and nonlinear analysis [4,5].
It is worthwhile mentioning that Darbo fixed-point theorem and the measures of noncompactness create a powerful and convenient technique which is very applicable in establishing theorems of existence for various types of operator equations (functional integral, integral, differential). For solvability on bounded domain, see [6,7,8].
Investigation on the real half-axis of the integral equations on different spaces of functions has received a great attention (see [9,10,11,12,13,14]).
In , measures of noncompactness in the space of functions which are defined, continuous and bounded on the real half-axis, and taking values in an arbitrary Banach space E, are constructed. One of the constructed measures of noncompactness is applied to prove the existence of solutions of an infinite system of quadratic integral equations in the space of functions defined, continuous and bounded on the real half-axis.
In addition, the solvability of an infinite system of integral equations of the Volterra–Hammerstein type in the space of functions defined, continuous and bounded on the real half-axis with values in the sequence space $l 1$ is discussed . Moreover, this result is extended to a wider class of considered infinite systems .
Motivated by these results, in this article, we discuss a coupled system of a functional integral equation, abbreviated by CSFIE
$x ( t ) = f 1 ( t , y ( t ) ) · g 1 t , ∫ 0 t ν 1 ( t , s , y ( s ) ) d s , t ≥ 0 y ( t ) = f 2 ( t , x ( t ) ) · g 2 t , ∫ 0 t ν 2 ( t , s , x ( s ) ) d s , t ≥ 0 ,$
and establish the existence of the solution of that coupled system on $R +$ utilizing Darbo’s fixed-point and the measure of noncompactness theorem. Furthermore, for the solution of (1), we study the asymptotic stability.
The present paper creates an essential extension of the investigations of the integral equation via the technique associated with measures of noncompactness on the real half line. However, we start by applying the technique associated with measures of noncompactness on a coupled system of integral equations in $B C ( R + ) .$
The following notations will be needed in our work. Assume that $B C ( R + )$ is the class of all continuous and bounded functions in $R +$. The norm of $f ∈ B C ( R + )$ is defined by
$∥ f ∥ = sup t ∈ R + | f ( t ) | .$
$x ∈ X ⊆ B C ( R + )$ and $ϵ ≥ 0$ are indicated by $w T ( x , ϵ )$; $T > 0$ is the modulus of continuity of the function x on the interval $[ 0 , T ]$, i.e.,
$w T ( x , ϵ ) = sup [ | x ( t ) − x ( s ) | : t , s ∈ [ 0 , T ] , | t − s | ≤ ϵ ] ,$
$w T ( X , ϵ ) = sup [ w T ( x , ϵ ) : x ∈ X ]$
and
$w 0 T ( X ) = lim ϵ → 0 w T ( X , ϵ ) , w 0 ( X ) = lim T → ∞ w 0 T ( X ) .$
$d i a m X ( t ) = sup { | x ( t ) − y ( t ) | , x , y ∈ X } , α ( X ) = lim t → ∞ sup d i a m X ( t )$
and the measure of noncompactness on $B C ( R + )$ is given by 
$μ ( X ) = w 0 ( X ) + α ( X ) = w 0 ( X ) + lim t → ∞ sup d i a m X ( t ) .$
Next, we state the Darbo fixed-point theorem .
Theorem 1.
Assume that $F : Q → Q$ is a continuous operator, and Q is a nonempty closed bounded convex subset of the space E with $μ ( F X ) ≤ k μ ( X )$ for any nonempty subset X of Q, where the constant $k ∈ [ 0 , 1 )$. Then, F has a fixed point in the set Q.
Now, let $E = B C ( R + ) × B C ( R + ) , X , Y ⊂ B C ( R + )$ and
$U = { u ∈ U : u = ( x , y ) , x ∈ X , y ∈ Y } = X × Y .$
Define the following modulus of continuity:
$ω ( x , ϵ ) = sup { | x ( t ) − x ( s ) | : t , s ∈ J = [ 0 , T ] , | t − s | ≤ ϵ } ,$
$ω ( y , ϵ ) = sup { | y ( t ) − y ( s ) | : t , s ∈ J = [ 0 , T ] , | t − s | ≤ ϵ }$
and
$ω ( u , ϵ ) = max { ω ( x , ϵ ) , ω ( y , ϵ ) } .$
Then,
$ω ( U , ϵ ) = sup { ω ( u , ϵ ) : u ∈ U } , ω 0 ( U ) = lim ϵ → 0 ω ( U , ϵ ) .$
$ω ( U ) = ω ( X × Y ) ≤ max { ω ( X , ϵ ) , ω ( Y , ϵ ) } ,$
$d i a m ( U ) = d i a m ( X × Y ) ≤ max { d i a m ( X ) , d i a m ( Y ) } ,$
$lim t → ∞ sup d i a m ( U ) ≤ max { lim t → ∞ sup d i a m ( X ) , lim t → ∞ sup d i a m ( Y ) } .$
and
$μ ( U ) = ω 0 ( U ) + α ( U ) = ω 0 ( U ) + lim t → ∞ sup d i a m U ( t ) .$

## 2. Main Result

Consider the coupled system of functional integral Equation (1) with the following assumptions:
(i)
$f i : R + × R → R$, $i = 1 , 2$ are continuous and $sup t ∈ R + | f i ( t , 0 ) | = f i *$.
(ii)
There exists a continuous function $m : R + → R +$ such that
$| f i ( t , x ) − f i ( t , y ) | ≤ m i ( t ) | x − y | , ∀ x , y ∈ R , t ∈ R + , i = 1 , 2 ,$
and $m i = sup t ∈ R + m i ( t ) .$
(iii)
$g i : R + × R → R , i = 1 , 2$ are continuous and satisfy the Lipschitz condition
$| g i ( t , x ) − g i ( s , y ) | ≤ l i ( | t − s | + | x − y | ) , l i > 0 , ∀ ( t , x ) , ( s , y ) ∈ R + × R$
$g i * = sup t ∈ R + | g i ( t , 0 ) | .$
(iv)
$ν i : R + × R + × R → R , i = 1 , 2$ are continuous and there exists a continuous function $k i ( t , s ) : R + × R + → R +$, such that $| ν i ( t , s , x ) | ≤ k i ( t , s ) , ∀ t , s ∈ R +$ and
$sup t ∈ R + ∫ 0 t k i ( t , s ) d s = k i , and lim t → ∞ ∫ 0 t k i ( t , s ) d s = 0 .$
(v)
Without loss of generality, we can write $m = max { m 1 , m 2 } , g * = max { g 1 * , g 2 * } ,$ $k = max { k 1 , k 2 }$, and $l = max { l 1 , l 2 } .$ Now there exists a positive constant C, such that $C = m g * + m l k < 1 .$
Remark 1.
From condition (ii) set $y = 0$, then
$| f i ( t , x ( s ) ) | − | f i ( t , 0 ) | ≤ | f i ( t , x ( s ) ) − f i ( t , 0 ) | ≤ m i ( t ) | x | , | f i ( t , x ( s ) ) | ≤ m i ( t ) | x | + | f i ( t , 0 ) | ≤ m i ( t ) | x | + f i * .$
Similarly, we have
$| g i ( t , x ( s ) ) | ≤ l i | x | + g i * .$
Theorem 2.
Assume that conditions (i)–(v) hold; then, the coupled system (1) has at least one solution $( x , y ) ∈ U ⊂ B C ( R + ) × B C ( R + )$.
Proof.
Suppose
$B r = { u = ( x , y ) ∈ U : ∥ u ∥ = max { | | x | | , | | y | | } ≤ r } .$
Define the operator A by
$A ( x , y ) ( t ) = ( A 1 y ( t ) , A 2 x ( t ) )$
where
$A 1 y ( t ) = f 1 ( t , y ( t ) ) · g 1 t , ∫ 0 t ν 1 ( t , s , y ( s ) ) d s$
$A 2 x ( t ) = f 2 ( t , x ( t ) ) · g 2 t , ∫ 0 t ν 2 ( t , s , x ( s ) ) d s .$
Let $u = ( x , y ) ∈ U$; from our assumptions, we can deduce that the function $A u$ is continues on U, and then we have
$| A 1 y ( t ) | = | f 1 ( t , y ( t ) ) . g 1 t , ∫ 0 t ν 1 ( t , s , y ( s ) ) d s | = | f 1 ( t , y ( t ) ) | . | g 1 t , ∫ 0 t ν 1 ( t , s , y ( s ) ) d s | ≤ ( m ( t ) | y ( t ) | + f 1 * ) ( g 1 * + l 1 | ∫ 0 t ν 1 ( t , s , y ( s ) ) d s | ) ≤ | y ( t ) | m 1 ( t ) ( g 1 * + l 1 ∫ 0 t k 1 ( t , s ) d s ) + ( g * + l 1 ∫ 0 t k 1 ( t , s ) d s ) f 1 * ≤ ∥ y ∥ ( m 1 g 1 * + m 1 l 1 k 1 ) + ( g 1 * + l 1 k 1 ) f 1 * ∥ A 1 y ∥ ≤ r 1 ( m 1 g 1 * + m 1 l 1 k 1 ) + ( g 1 * + l 1 k ) f 1 * = r 1 .$
Similar to the above calculation, we can conclude that
$∥ A 2 x ∥ ≤ r 2 ( m 2 g 2 * + m 2 l 2 k 2 ) + ( g 2 * + l 2 k 2 ) f 2 * = r 2 .$
Therefore,
$∥ A u ∥ U = ∥ A ( x , y ) ∥ U = ∥ ( A 1 y , A 2 x ) ∥ U = max { ∥ A 1 y ∥ , ∥ A 2 x ∥ } = r .$
Then, the operator A is bounded on U and $A u ∈ B r$ and
$∥ A ( x ) ∥ ≤ 2 ( C r + A * ) = r , r = 2 A * 1 − 2 C$
where $A * = ( g * + l k ) f * < ∞$. This proves that the operator $A : B r → B r .$
Now, we show that A is continuous on the ball $B r$.
Let $ϵ > 0$ be given, take $y 1 , y 2 ∈ Y$ such that $∥ y 1 − y 2 ∥ ≤ ϵ$, then
$| A 1 y 1 ( t ) − A 1 y 2 ( t ) | = | f 1 ( t , y 1 ( t ) ) g 1 t , ∫ 0 t ν 1 ( t , s , y 2 ( s ) ) d s − f 1 ( t , y 2 ( t ) ) g 1 t , ∫ 0 t ν 1 ( t , s , y 2 ( s ) ) d s | ≤ | f 1 ( t , y 1 ( t ) ) g 1 t , ∫ 0 t ν 1 ( t , s , y 1 ( s ) ) d s − f 1 ( t , y 2 ( t ) ) g 1 t , ∫ 0 t ν 1 ( t , s , y 1 ( s ) ) d s | + | f 1 ( t , y 2 ( t ) ) g 1 t , ∫ 0 t ν 1 ( t , s , y 1 ( s ) ) d s − f 1 ( t , y 2 ( t ) ) g 1 t , ∫ 0 t ν 1 ( t , s , y 2 ( s ) ) d s | ≤ | f 1 ( t , y 1 ( t ) ) − f 1 ( t , y 2 ( t ) ) | | g 1 t , ∫ 0 t ν 1 ( t , s , y 1 ( s ) ) d s | + | f 1 ( t , y 2 ( t ) ) | | g 1 t , ∫ 0 t ν 1 ( t , s , y 1 ( s ) ) d s − g 1 t , ∫ 0 t ν 1 ( t , s , y 2 ( s ) ) d s | ≤ m 1 ( t ) | y 1 ( t ) − y 2 ( t ) | ( g * + l 1 ∫ 0 t | k 1 ( t , s ) | d s )$
$+ ( m 1 ( t ) | y 2 ( t ) | + f * ) l 1 ∫ 0 t | ν 1 ( t , s , y 1 ( s ) ) − ν 1 ( t , s , y 2 ( s ) ) | d s$
$≤ m 1 g * | y 1 ( t ) − y 2 ( t ) | + m 1 l 1 k 1 | y 1 ( t ) − y 2 ( t ) | + ( m 1 | y 2 ( t ) | + f 1 * ) k 1 ∫ 0 t | ν 1 ( t , s , y 1 ( s ) ) − ν 1 ( t , s , y 2 ( s ) ) | d s ≤ ( m 1 g 1 * + m 1 l 1 k 1 ) ϵ + ( m 1 r 1 + f 1 * ) l 1 ∫ 0 t | ν 1 ( t , s , y 1 ( s ) ) − ν 1 ( t , s , y 2 ( s ) ) | d s .$
$≤ ( m 1 g 1 * + m 1 l 1 k 1 ) ϵ + 2 ( m 1 r 1 + f 1 * ) l 1 ∫ 0 t k 1 ( t , s ) d s .$
Select $T > 0$ such that the following inequality holds for $t > T$.
$2 r 1 l 1 m 1 ∫ 0 t k 1 ( t , s ) d s ≤ ( 1 − C ) ϵ 2$
$2 f * k 1 ∫ 0 t k ( t , s ) d s ≤ ( 1 − C ) ϵ 2 .$
Take into account the following two situations.
($i *$)
$t ≥ T$. In light of (4) and (5), we obtain
$| A 1 y 1 ( t ) − A 1 y 2 ( t ) | ≤ ( m 1 g 1 * + m 1 l 1 k 1 ) ϵ + ( 1 − C ) ϵ 2 + ( 1 − C ) ϵ 2 = ϵ .$
($i i *$)
$t ≤ T$. In this instance, let us take a look at the function w = $w = w ( ϵ )$ given by
$w ( ϵ ) = sup { | ν 1 ( t , s , y 1 ) − ν 1 ( t , s , y 2 ) | : t , s ∈ [ 0 , T ] , x , y ∈ [ − r , r ] , | y 1 − y 2 | < ϵ } .$
Then, from the uniform continuity of the function $ν 1 = ν 1 ( t , s , x )$ on the set $[ 0 , T ] × [ 0 , T ] × [ − r , r ]$, we deduce that $w ( ϵ ) → 0$ as $ϵ → 0$.
Thus, from (3), we obtain
$| A 1 y 1 ( t ) − A 1 y 2 ( t ) | ≤ ( m 1 g 1 * + l 1 k 1 ) ϵ + ( 1 r m 1 + f 1 * ) T w ( ϵ ) .$
Finally, from the two cases ($i *$), ($i i *$) and the above established facts, we can deduce that the operator $A 1$ is continuous on Y.
Similarly, we can conclude that the operator $A 1$ is continuous and for any $x 1 , x 2 ∈ X$
$| A 2 x 1 ( t ) − A 2 x 2 ( t ) | ≤ ( m 1 g 1 * + l 1 k 1 ) ϵ + ( r 1 m 1 + f 1 * ) T w ( ϵ ) .$
Therefore,
$∥ A ( x 1 , y 1 ) − A ( x 2 , y 2 ) ∥ U = ∥ ( A 1 y 1 , A 2 x 1 ) − ( A 1 y 2 , A 2 x 2 ) ∥ = ∥ ( A 1 y 1 − A 1 y 2 , A 2 x 1 − A 2 x 2 ) ∥ = max { ∥ A 1 y 1 ( t ) − A 1 y 2 ( t ) ∥ , ∥ A 2 x 1 ( t ) − A 2 x 2 ( t ) ∥ } ≤ ( m 1 g 1 * + l 1 k 1 ) ϵ + ( r 1 m 1 + f 1 * ) T w ( ϵ ) .$
Then, the operator A is continuous on the ball $B r$.
Now, for any $u 1 , u 2 ∈ U$ and fixed $t ≥ 0$, we obtain
$| A 1 y 1 ( t ) − A 1 y 2 ( t ) | ≤ m 1 g 1 * | y 1 ( t ) − y 2 ( t ) | + m 1 l 1 k 1 | y 1 ( t ) − y 2 ( t ) | + ( m 1 | y 2 ( t ) | + f 1 * ) l 1 ∫ 0 t | ν 1 ( t , s , y 1 ( s ) ) − ν 1 ( t , s , y 2 ( s ) ) | d s ≤ m 1 g 1 * | y 1 ( t ) − y 2 ( t ) | + m 1 l 1 k 1 | y 1 ( t ) − y 2 ( t ) | + 2 ( m 1 r 1 + f 1 * ) l 1 ∫ 0 t k 1 ( t , s ) d s . ≤ ( m 1 g 1 * + m 1 k 1 l 1 ) d i a m Y ( t ) + 2 ( m 1 r 1 + f 1 * ) l 1 ∫ 0 t k 1 ( t , s ) d s .$
Hence, we obtain
$d i a m ( A 1 Y ) ( t ) ≤ ( m 1 g 1 * + m 1 k 1 l 1 ) d i a m Y ( t ) + 2 ( m 1 r 1 + f 1 * ) l 1 ∫ 0 t k 1 ( t , s ) d s .$
As performed above, we can conclude that for any $u 1 , u 2 ∈ U$, and fixed $t ≥ 0$, we obtain
$d i a m ( A 2 X ) ( t ) ≤ ( m 1 g 2 * + m 2 k 2 l 2 ) d i a m X ( t ) + 2 ( m 2 r 2 + f 2 * ) l 2 ∫ 0 t k 2 ( t , s ) d s .$
Therefore,
$d i a m A ( X , Y ) ( t ) = max { d i a m ( A 1 Y ) ( t ) , d i a m ( A 2 X ) ( t ) } .$
Hence,
$d i a m A U ( t ) ≤ ( m g * + m k k 1 ) max { ( d i a m Y ( t ) , d i a m X ( t ) ) } + 2 m r + f * ) l [ ∫ 0 t k 1 ( t , s ) d s + ∫ 0 t k 2 ( t , s ) d s ] ≤ ( m g * + m k k 1 ) d i a m U ( t ) + 2 ( m r + f * ) l [ ∫ 0 t k 1 ( t , s ) d s + ∫ 0 t k 2 ( t , s ) d s ]$
and
$lim t → ∞ sup d i a m A U ( t ) ≤ ( m g * + m k k 1 ) lim t → ∞ sup d i a m U ( t ) .$
Then,
$lim t → ∞ sup d i a m A U ( t ) ≤ C lim t → ∞ sup d i a m U ( t ) .$
Let $T > 0$ and $ϵ > 0$ be given. Let $( x , y ) ∈ U$ and $t , s ∈ [ 0 , T ]$ such that $s ≤ t$ and $| t − s | ≤ ϵ$, then
$| A 1 y ( t ) − A 1 y ( s ) | = | f 1 ( t , y ( t ) ) g 1 t , ∫ 0 t ν 1 ( t , τ , y ( τ ) ) d τ − f 1 ( s , y ( s ) ) g 1 s , ∫ 0 s ν 1 ( s , τ , y ( τ ) ) d τ | ≤ | f 1 ( t , y ( t ) ) g 1 t , ∫ 0 t ν 1 ( t , τ , y ( τ ) ) d τ − f 1 ( s , y ( s ) ) g 1 t , ∫ 0 t ν 1 ( t , τ , y ( τ ) ) d τ + f 1 ( s , y ( s ) ) g 1 t , ∫ 0 t ν 1 ( t , τ , y ( τ ) ) d τ − f 1 ( s , y ( s ) ) g 1 s , ∫ 0 s ν 1 ( s , τ , y ( τ ) ) d τ | ≤ | f 1 ( t , y ( t ) ) − f 1 ( s , y ( s ) ) | g 1 t , ∫ 0 t ν 1 ( t , τ , y ( τ ) ) d τ + | f 1 ( s , y ( s ) ) | | g 1 t , ∫ 0 t ν 1 ( t , τ , y ( τ ) ) d τ − g 1 s , ∫ 0 s ν 1 ( s , τ , y ( τ ) ) d τ | ≤ ( | f 1 ( t , y ( t ) ) − f 1 ( t , y ( s ) ) | + | f 1 ( t , y ( s ) ) − f 1 ( s , y ( s ) ) | ) ( g * + l 1 ∫ 0 t k 1 ( t , τ ) d τ ) + ( m 1 ( s ) | y | + f 1 * ) | ( l 1 ( ∫ 0 t | ν 1 ( t , τ , y ( τ ) ) − ν 1 ( s , τ , y ( τ ) ) | d τ ) + ∫ s t | ν 1 ( t , τ , y ( τ ) ) | ) ≤ ( m 1 ( t ) | y ( t ) − y ( s ) | + | f 1 ( t , y ( s ) ) − f 1 ( s , y ( s ) ) | ) ( g 1 * + l 1 ∫ 0 t k 1 ( t , τ ) d τ ) + ( m 1 ( s ) | y ( s ) | + f 1 * ) ( l 1 ( ∫ 0 t | ν 1 ( t , τ , y ( τ ) ) − ν 1 ( s , τ , y ( τ ) ) | d τ ) + ∫ s t | ν 1 ( t , τ , y ( τ ) ) d τ )$
$≤ ( m 1 | y ( t ) − y ( s ) | + w r T ( f 1 , ϵ ) ) ( g 1 * + l 1 ∫ 0 t k 1 ( t , τ ) d τ ) + m r 1 l 1 ∫ s t k 1 ( s , τ ) d τ + f 1 * l 1 ∫ s t k 1 ( t , τ ) d τ + l 1 T [ m 1 r 1 + f 1 * ] ) w r − T ( ν 1 , ϵ )$
where
$w r T ( f i , ϵ ) = sup { | f i t , y ) − f i ( s , y ) : t , s ∈ [ 0 , T ] , | t − s | ≤ ϵ , | y | ≤ r , i = 1 , 2 . }$
$w r − T ( ν i , ϵ ) = sup { | ν i ( t , τ , y ) − ν i ( s , τ , y ) : | t − s | ≤ ϵ , τ ∈ [ 0 , T ] , | y | ≤ r , i = 1 , 2 . } .$
Hence, we deduce that
$w ( A 1 Y , ϵ ) ≤ ( m w ( Y , ϵ ) + w r T ( f 1 , ϵ ) ) ( g 1 * + l 1 ∫ 0 τ | k 1 ( s , τ ) | d τ ) + ( ϵ m 1 r l 1 sup { | k 1 ( t , τ ) | : τ ∈ [ 0 , T ] } + ϵ l 1 f 1 * sup { | k 1 ( t , τ ) | : τ ∈ [ 0 , T ] } + k 1 T ( m 1 r 1 + f 1 * ) w r − T ( ν 1 , ϵ )$
Through a similar method, we obtain
$w ( A 2 X , ϵ ) ≤ ( m 2 w T ( X , ϵ ) + w r T ( f 2 , ϵ ) ) ( g 2 * + l 2 ∫ 0 τ | k 2 ( s , τ ) | d τ ) + ( ϵ m 2 r 2 l 2 sup { | k 2 ( t , τ ) | : τ ∈ [ 0 , T ] } + ϵ k 2 f 2 * sup { | k 2 ( t , τ ) | : τ ∈ [ 0 , T ] } + l 2 T ( m 2 r 2 + f 2 * ) w r − T ( ν 2 , ϵ ) .$
Therefore,
$w ( A U , ϵ ) = max { w ( A 1 Y , ϵ ) , w ( A 2 X , ϵ ) } = ( m w ( U , ϵ ) + w r T ( f , ϵ ) ) ( g * + l ∫ 0 τ | k ( s , τ ) | d τ ) + ( ϵ m r l sup { | k ( t , τ ) | : τ ∈ [ 0 , T ] } + ϵ l f * sup { | k ( t , τ ) | : τ ∈ [ 0 , T ] } + l T ( m r + f * ) w r − T ( ν 1 , ϵ ) + ( ϵ m r l sup { | k ( t , τ ) | : τ ∈ [ 0 , T ] } + ϵ l f * sup { | k ( t , τ ) | : τ ∈ [ 0 , T ] } + l T ( m r + f * ) w r − T ( ν 2 , ϵ ) .$
From the uniform continuity of the functions $f i = f i ( t , x )$ and the functions $ν i = ν i ( t , s , x ) , i = 1 , 2$ on the set $[ 0 , T ] × [ − r , r ]$, we deduce that $w r ( f i , ϵ ) → 0$ and $w r − T ( ν i , ϵ ) → 0$ as $ϵ → 0$. Consequently, we obtain
$w 0 ( A U ) ≤ m w 0 ( U ) ( g * + l ∫ s t k ( t , τ ) d τ ) ,$
$w 0 ( A U ) ≤ ( m g * + m k l ) w 0 T ( U )$
and as $T → ∞$, we have
$w 0 ( A U ) ≤ C w 0 ( U ) .$
Now, from the estimations (9) and (10) and the definition of $μ$ on U, we obtain
$μ ( A U ) ≤ C μ ( U ) .$
Since all the requirements of Theorem 2 are met, then A has a fixed point $( x , y ) ∈ U$. Consequently, the coupled system of quadratic functional integral Equation (1) has at least one solution $( x , y ) ∈ E , x , y ∈ B C ( R + )$. □

## 3. Asymptotic Stability

We can now deduce from the proof of Theorem 2 the following corollary.
Corollary 1.
The solution $u ∈ U$ of the coupled system of quadratic functional integral Equation (1) is asymptotically stable; that is to say,$ϵ > 0$, there exists $T ( ϵ ) > 0$ and $r > 0$, such that, if any two solutions to the coupled system of the quadratic functional integral Equation (1) are $( x , y ) , ( x 1 , y 1 ) ∈ U$, then
$| | ( x , y ) − ( x 1 , y 1 ) | | ≤ ϵ , t ≥ T ( ϵ ) .$
This implies that
$| x ( t ) − x 1 ( t ) | ≤ ϵ a n d | y ( t ) − y 1 ( t ) | ≤ ϵ , t ≥ T ( ϵ ) .$
Proof.
Let $( x , y ) , ( x 1 , y 1 ) ∈ U$ be any two solutions of the coupled system of quadratic functional integral Equation (1). Using assumptions of Theorem 2 and by a similar way to how relations (6) and (7) are estimated, we have
$| y ( t ) − y 1 ( t ) | = | A 1 y ( t ) − A 1 y 1 ( t ) | ≤ ( m 1 g 1 * + l 1 k 1 ) ϵ + ( r 1 m 1 + f 1 * ) T w ( ϵ ) , and | x ( t ) − x 1 ( t ) | = | A 2 x ( t ) − A 2 x 1 ( t ) | ≤ ( m 2 g 2 * + l 2 k 2 ) ϵ + ( r 2 m 2 + f 2 * ) T w ( ϵ ) . Hence ∥ ( x , y ) − ( x 1 , y 1 ) ∥ U = ∥ ( A 1 y , A 2 x ) − ( A 1 y 1 , A 2 x 1 ) ∥ = ∥ ( A 1 y − A 1 y 1 , A 2 x − A 2 x 1 ) ∥ = max { ∥ A 1 y ( t ) − A 1 y 1 ( t ) ∥ , ∥ A 2 x ( t ) − A 2 x 1 ( t ) ∥ } ≤ ( m g * + k 1 k ) ϵ + ( r m + f * ) T w ( ϵ ) ,$
for $t > T ( ϵ )$. □

## 4. Particular Cases and Example

In this section, we demonstrate some particular systems, which are deduced from Theorem 2.
• Let $f i ( t , x ) = 1 , i = 1 , 2$; then, the coupled system (1) takes the form
$x ( t ) = g 1 t , ∫ 0 t ν 1 ( t , s , y ( s ) ) d s , t ≥ 0 y ( t ) = g 2 t , ∫ 0 t ν 2 ( t , s , x ( s ) ) d s , t ≥ 0 .$
Based on conditions (iii)–(v) of Theorem 2, then(1) has at least one asymptotically stable solution $x ∈ B C ( R + )$.
Moreover, when $g i ( t , x ) = x , i = 1 , 2 .$ Then, we have a coupled system of Urysohn integral equations
$x ( t ) = ∫ 0 t ν 1 ( t , s , y ( s ) ) d s , t ≥ 0 y ( t ) = ∫ 0 t ν 2 ( t , s , x ( s ) ) d s , t ≥ 0 ,$
• Let $g i ( t , x ) = x , i = 1 , 2$, then, the coupled system (1) takes the form
$x ( t ) = f 1 ( t , y ( t ) ) · ∫ 0 t ν 1 ( t , s , y ( s ) ) d s , t ≥ 0 y ( t ) = f 2 ( t , x ( t ) ) · ∫ 0 t ν 2 ( t , s , x ( s ) ) d s , t ≥ 0 ,$
under the conditions of Theorem 2, then the coupled system of quadratic integral Equation (1) has at least one asymptotically stable solution $( x , y ) ∈ B C ( R + ) × B C ( R + )$.
• Let $g ( t , x ) = 1 + x$, in (1), we have
$x ( t ) = f 1 ( t , y ( t ) ) 1 + ∫ 0 t ν 1 ( t , s , y ( s ) ) d s , t ≥ 0 y ( t ) = f 2 ( t , x ( t ) ) 1 + ∫ 0 t ν 2 ( t , s , x ( s ) ) d s , t ≥ 0 ,$
which under the assumptions of Theorem 2, has at least one asymptotically stable solution $( x , y ) ∈ B C ( R + ) × B C ( R + )$.
Example: Consider the following coupled system of functional equations
$x ( t ) = arctan ( t + x ( t ) ) ( 1 + t ) 2 sin ( t + ∫ 0 t t ( 2 t − s ) sin ( | y ( s ) | ) 1 + t 4 d s ) , t ≥ 0 y ( t ) = a r c c o t ( t + x ( t ) ) ( 1 + t ) 2 cos ( t + ∫ 0 t t sin ( | x ( s ) | ) 4 π ( t 2 + 1 ) ( s + 1 ) d s ) , t ≥ 0 .$
Now, we study the solvability of a coupled system of functional Equation (11) on the space $B C ( R + ) × B C ( R + )$. Take into account that this coupled system of functional equations is a specific instance of system (1) with
$g 1 ( t , x ( t ) ) = 1 1 + t sin ( t + x ( t ) ) , g 2 ( t , x ( t ) ) = 1 1 + t cos ( t + x ( t ) ) , f 1 ( t , x ( t ) ) = arctan ( t + x ( t ) ) 1 + t , f 2 ( t , x ( t ) ) = arccot ( t + x ( t ) ) 1 + t , ν 1 ( t , s , x ( s ) ) = t ( 2 t − s ) sin ( | x ( t ) | ) 1 + t 4 , ν 2 ( t , s , x ( s ) ) = sin ( | x ( t ) | ) 4 π ( t 2 3 + 1 ) ( t − s + 1 ) .$
Obviously, functions $f i , ( i = 1 , 2 )$ are mutually continuous. Currently, for any $x , y ∈ R + and t ∈ R +$
$| f i ( t , x ( t ) ) − f i ( t , y ( t ) ) | ≤ 1 2 | x ( t ) − y ( t ) | .$
This indicates that condition $( v )$ is satisfied with $f * = π 2$ and $m = 1 2 ,$ where $f 1 ( t , 0 ) = t 1 + t 2 arctan ( t ) , f 2 ( t , 0 ) = 1 1 + t arccot ( t ) .$ However, we also have
$| g i ( t , x ( t ) ) − g i ( t , y ( t ) ) | ≤ | x ( t ) − y ( t ) | 2 ,$
with $l = 1 2$, $g 1 ( t , 0 ) = 1 1 + t sin ( t ) ,$ $g 2 ( t , 0 ) = 1 1 + t cos ( t )$ and $g * = 1 2$. Further, observe that $ν i ( t , s , x ) ( i = 1 , 2 )$ fulfills condition (iv), with
$| ν 1 ( t , s , x ( s ) ) | ≤ t 2 π ( t 2 + 1 ) ( s + 1 ) ,$
and
$| ν 2 ( t , s , x ( s ) ) | ≤ 1 4 π ( t 2 3 + 1 ) ( t − s + 1 )$
This indicates that we can insert $k 1 ( t , s ) = t 2 π ( t 2 + 1 ) ( s + 1 ) ,$ and $k 2 ( t , s ) = 1 4 π ( t 3 4 + 1 ) ( t − s + 1 ) .$ To verify the assumption (iv), notice that
$lim t → ∞ ∫ 0 t k 1 ( t , s ) = lim t → ∞ ∫ 0 t t 2 π ( t 2 + 1 ) ( s + 1 ) d s = lim t → ∞ t ln t + 1 2 π · t 2 + 1 = 0 ,$
and
$lim t → ∞ ∫ 0 t k 2 ( t , s ) = lim t → ∞ ∫ 0 t 1 4 π ( t 2 3 + 1 ) ( t − s + 1 ) d s = lim t → ∞ t − ln t + 1 2 π · ( t 2 3 + 1 ) = 0 .$
Moreover, we have $k = 0.0906987 .$
Finally, let us pay attention to the fact that the inequality of Theorem 2 has the form
$C = m g * + m k l ≃ 0.272674675 < 1 ,$
consequently, all the requirements of Theorem 2 have been met. As a result, the coupled system (11) has at least one asymptotically stable solution in the space $B C ( R + ) × B C ( R + )$.

## 5. Conclusions

Coupled systems of differential and integral equations have been addressed by many authors and in different classes of functions; for example, see [16,17,18,19,20,21,22,23].
The investigations in this work continue those contained in papers [11,12,13]. In particular, in this paper, we use a technique associated with measures of noncompactness in $B C ( R + )$ by a given modulus of continuity, to establish the solvability of a coupled system of integral equations.
We discussed the solvability and asymptotic stability of that coupled system of functional integral equation on the real half-axis. Our investigation is lying in the space of bounded continuous functions on $R +$$( B C ( R + ) ) .$ We started by applying the technique associated with measures of noncompactness on a coupled system of functional integral equation in $B C ( R + ) .$ Finally, some particular coupled systems of the well-known Uryshon integral equations, a coupled system of functional equations and an example are illustrated.

## Author Contributions

These authors contributed equally to this work. All authors have read and agreed to the published version of the manuscript.

## Funding

This research received no external funding.

Not applicable.

Not applicable.

Not applicable.

## Acknowledgments

The authors are thankful to the referees for remarks and suggestions for the improvement of this paper.

## Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

El-Sayed, A.M.A.; Omar, Y.M.Y.; Hashem, H.H.G.; Al-Issa, S.M. Treatment of a Coupled System for Quadratic Functional Integral Equation on the Real Half-Line via Measure of Noncompactness. Foundations 2023, 3, 37-48. https://doi.org/10.3390/foundations3010004

AMA Style

El-Sayed AMA, Omar YMY, Hashem HHG, Al-Issa SM. Treatment of a Coupled System for Quadratic Functional Integral Equation on the Real Half-Line via Measure of Noncompactness. Foundations. 2023; 3(1):37-48. https://doi.org/10.3390/foundations3010004

Chicago/Turabian Style

El-Sayed, Ahmed M. A., Yasmin M. Y. Omar, Hind H. G. Hashem, and Shorouk M. Al-Issa. 2023. "Treatment of a Coupled System for Quadratic Functional Integral Equation on the Real Half-Line via Measure of Noncompactness" Foundations 3, no. 1: 37-48. https://doi.org/10.3390/foundations3010004