On the existence of continuous solutions of a nonlinear quadratic fractional integral equation

We prove an existence theorem for a nonlinear quadratic integral equation of fractional order, in the Banach space of real functions deﬁned and continuous on closed interval. This equation contains as a special cases numerous integral equation studied by other authors. Finally, we give an example for indicating the natural realizations of our abstract result presented in this paper.


Introduction
Several problems in many branches of science can be modeled as an integral equation such as in physics, engineering, biology,...ect see for example ([4], [6], [7], [14]). Quadratic integral equations have many useful applications in describing problems in the real world. For example, quadratic integral equations are applicable in the kinetic theory of gases, in the theory of neutron transport and in the theory of radiative transfer. Especially the so called quadratic integral equation of Chandrsekher type can be seen in many application(cf.( [9], [10] [26], [12], [25]). On the other hand the fractional calculus topic is enjoying growing interest of engineers and scientists, see ( [16], [17], [19], [20]) . The most used techniques to study the existence for solutions of an integral equation are based on fixed point argument for example, Banach fixed point theorem was used by many authors in order to establish existence results for different kind of integral equations (see, for example [ [8], [28]]). But in the absence of compactness and the Lipschitz condition, the above mentioned theorem cannot be applied and in this case, the measure of non-compactness can be useful. Many existence results for integral equations were established using fixed point theorems involving measure of non-compactness. For most details, we refer the reader to the references therein and to ([1], [2], [3], [11], [16] , [21], [27]). In this paper we deal with quadratic integral equation of fractional order with respect to another function. Using fixed point theorem via measure of non-compactness Due to Darbo are the main tool in carrying out our proof.

Preliminaries
The main purpose of this paper to study the following integral equation where α ∈ (0, 1), 0 ≤ a ≤ T 1 , f : I → R, g : I × R → R, u : I × I × R × R → R, and m : I → R, from the view of the theory of existence of solutions. Eq(1) can be written in the form where I α a,m is the fractional integral of orderα with respect to the function m defined by which in particular cases we have: (i) If m(s) = ln s and a > 0 then I α a,m is the Hadamard fractional integral,  [11] A mapping µ : M E → [0, ∞) is said to be a measure of noncompactness in E if it satisfies the following conditions: (1) the family kerµ = {X ∈ M E : µ(X) = 0} is nonempty and kerµ ⊂ N E , where kerµ is called the kernel of the measure µ.
is said to be a comparison function if it satisfies the following properties: Examples of comparison functions are We denote the set of comparison functions by Φ as in [15] [15]): Let C be a nonempty, bounded, closed and convex subset of Banach space E. Let U : C → C be a continuous transformation such that Where φ ∈ Φ and µ is a measure of non-compactness in E. Then U has at least one fixed point.
For given functions φ, ψ : [0, 1) → R and a real number λ, define the operator T max and T by

Lemma 2.3
The following properties hold: We will use through this paper the following notations.
We define the mapping Ω : It was proved in [11] that the mapping µ : is a measure of non-compactness in the Banach space E. In this section, we will investigate the existence theorem of our integral equation (1) under suitable assumptions:

Main result
Consider the following assumptions: (a 1 ) f : I → R; is continuous nondecreasing and nonnegative function on I, (ii) φ 1 is nondecreasing and continuous.
(iii) there exist δ ≥ 0 and θ ∈Φ such that for each x ∈ C(I) and t ∈ I , (a 6 ) u : There exists a nondecreasing function φ : There exists a positive constant r 0 such that Proof. Let us consider the operator F defined on E = C([a, T 1 ], R) by At first,we show that the operator F : C([a, From assumption (a 1 )(a 2 )(a 4 ) We have just to prove U maps E into itself, that is U x : [a, T 1 ] → R is continuous for To show the continuity of U x at the point a. Let {t n } be a sequence in[a, T 1 ] such that t n → a + as n → ∞ using the above result,for all n ∈ N , we have letting n → ∞ in the above inequality and using the continuity of m, we get which prove the continuity of U x at point a. Now let t ∈ (a, T 1 ] be fixed and {t n } be a sequence in(a, T 1 ] such that t n → t as n → ∞ . With out restrictions we suppose that t n ≥ t for n large enough. We have Then U x is continuous at t, then as consequence, U x ∈ E for all x ∈ E and F : E → E is well defined. For arbitrary fixed x ∈ E and t ∈ [a, T 1 ] we have Then From assumption(a 9 ), the fact that the functions φ 1 , φ are nondecreasing, and the above inequality, we infer that the operator F maps B(0, r 0 ) into itself, where Now we will prove that the operator F : B(0, r 0 ) → B(0, r 0 ) is continuous, in order to prove this, it is sufficient to show thatf, g, T, U are continuous on B(0, r 0 ). From assumption (a 1 ) the function f is continuous. Firstly we show that g is continuous on B(0, r 0 ). To do this, let{x n } ⊂ B(0, r 0 ) and x ∈ B(0, r 0 ) such that x n − x → 0 as n → ∞ , and we want to prove that gx n − gx → 0 as n → ∞ . In fact, for all t ∈ [a, T ], we have Then Letting n → ∞ in the above inequality, using the continuity of the function φ 1 , and the fact that φ 1 (0) = 0, we get Next,for the continuity of U on B(0, r 0 ). To do this, we fix a real number ρ > 0 and take arbitrary functions where β u (ρ) = sup{| u(t, s, x 2 , y 2 ) − u(t, s, x 1 , y 1 ) |: t, s ∈ I, x 1 , x 2 , y 1 , y 2 ∈ [0, r 0 ]}, Thus, By the uniform continuity of the function u on I × I × [0, r 0 ] × [0, r 0 ] we have that β u (ρ) → 0 as ρ + → 0 and, therefore, Thus U is continuous on B(0, r 0 ), and consequently, F is continuous. Now let us take a nonempty set W ⊂ B(0, r 0 ).

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