On global attractivity of solutions of a functional-integral equation

We prove an existence theorem for a quadratic functional-integral equation of mixed type. The functional-integral equation studied below contains as special cases numerous integral equations encountered in nonlinear analysis. With help of a suitable measure of noncompactness, we show that the functional integral equation of mixed type has solutions being continuous and bounded on the interval [0,1) and those solutions are globally attractive.


Introduction
Quadratic integral equations are often applicable in the theory of radiative transfer, kinetic theory of gases, in the theory of neutron transport and in the traffic theory.Especially, the so-called quadratic integral equation of Chandrasekhar type can be very often encountered in many applications (cf.[1,2,3,6,7,8,9,10,13,14]). In this paper we study the functional integral equation of mixed type, namely x(t) = f t, x(t), t 0 u(t, s, x(s))ds, ∞ 0 v(t, s, x(s)) ds . (1) Eq.( 1) contains as special cases numerous integral and functional-integral equations encountered in nonlinear analysis.Also, the famous Chandrasekhar's integral equation is considered as a special case.
Using the technique associated with a suitable measure of noncompactness, we show that Eq.(1) has solutions being continuous and bounded on the interval [0, ∞) and those solutions are globally attractive.In fact, our result in this paper is motivated by the extension of the work of Hu and Yan [12].EJQTDE, 2007 No. 21, p. 1 This section collects some definitions and results which will be needed further on.Assume that (E, .) is a Banach space with zero element θ.Let B(x, r) denote the closed ball centered at x and with radius r.The symbol B r stands for the ball B(θ, r).
If X is a subset of E, then X and ConvX denote the closure and convex closure of X, respectively.Moreover, we denote by M E the family of all nonempty and bounded subsets of E and N E its subfamily consisting of all relatively compact subsets.
Next we give the concept of a measure of noncompactness [5]: ) 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 . 2) We recall the fixed point theorem due to Darbo [11].Before quoting this theorem, we need the following definition: Definition 2.2 Let M be a nonempty subset of a Banach space E, and T : M → E be a continuous operator that transforms bounded sets onto bounded ones.We say that T satisfies the Darbo condition (with constant c ≥ 0) with respect to a measure of noncompactness µ if for any bounded subset X of M we have µ(TX) ≤ c µ(X).
If T satisfies the Darbo condition with c < 1 then it is called a contraction operator with respect to µ. Theorem 2.1 Let Q be a nonempty, bounded, closed and convex subset of the space E and let H : Q → Q be a contraction with respect to the measure of noncompactness µ.
Then H has a fixed point in the set Q.
Remark 2.1 [5] Under the assumptions of the above theorem the set Fix H of fixed points of H belonging to Q is a member of ker µ.In fact, as µ(H(Fix H)) = µ(Fix H) ≤ c µ(Fix H) and 0 ≤ c < 1, we deduce that µ(Fix H) = 0.In what follows we will work in the Banach space BC(IR + ) consisting of all real functions defined, bounded and continuous on IR + .The space BC(IR + ) is equipped with the standard norm Now, we recollect the construction of the measure of noncompactness which will be used in the next section (see [4]) Let us fix a nonempty and bounded subset X of BC(IR + ) and let T be a positive number.For x ∈ X and ε ≥ 0 denote by ω T (x, ε) the modulus of continuity of the function x on the interval [0, T ], i.e., For a fixed number t ≥ 0 we denote Now, let us define the function µ on the family M BC(IR + ) by the following formula It can be shown [4] that the function µ is a measure of noncompactness on the space BC(IR + ).

Definition 2.3
The solution x(t) of Eq.( 1) is said to be globally attractive, if there are lim t→+∞ (x(t) − y(t)) = 0 for any solution y(t) of Eq.(1).

Main Results
In this section, we will study Eq.( 1) assuming that the following assumptions are satisfied: EJQTDE, 2007 No. 21, p. 3 a 1 ) f : IR + × IR + × IR × IR → IR is continuous and the function t → f (t, 0, 0, 0) is an element of the space BC(IR + ).
a 2 ) There exist continuous functions m 1 , m 2 : uniformly with respect to x ∈ BC(IR + ).Now, we are in a position to state and prove our main result in the paper Theorem 3.1 Let the assumptions a 1 ) − a 3 ) be satisfied.Then Eq.( 1) has at least one solution x ∈ BC(IR + ) which is globally attractive.
Proof: Denote by F the operator associated with the right-hand side of Eq.(1), i.e., equation (1) takes the form where Solving Eq.( 1) is equivalent to finding a fixed point of the operator F defined on the space BC(IR + ).
Clearly, in virtue of our assumptions the function F x is continuous on the interval IR + for any function x ∈ BC(IR + ).Moreover, from our assumptions we have where We claim that the operator F is continuous on B r .To establish this claim, let us fix ε > 0 and take arbitrary x, y ∈ B r such that x − y ≤ ε.Then, for t ∈ IR + , we have By assumptions a1)−a3) we choose T > 0 such that for t ≥ T the following inequalities hold m 1 (t) and Thus for t ≥ T , in view of ( 5)-( 7) we obtain On the other hand, taking into account the uniform continuity of the functions u = u(t, s, x) Thus, Hence, we established our claim.Now, let X be a nonempty set of B r .Then, for x, y ∈ X and t ∈ IR + , we get Therefore, thanks to assumption a 3 ).For any T > 0 and ε > 0. Choose a function x ∈ X and t, s ∈ [0, T] such that |t − s| ≤ ε, without loss of generality we may assume that t > s.Then, in view of our assumptions we have Hence From ( 8) and (9), keeping in mind the definition of the measure of noncompactness in the space BC(IR + ), we obtain The above obtained inequality together with the fact that k < 1 enable us to apply Theorem 2.1.Hence, we infer that Eq.( 1) has at least one solution x(t).On the other hand, for f (t, x, y, z) = h(t)+z, Eq.( 1) reduces to the well-known Urysohn integral equation Example 4.2 If f (t, x, y, z) = g(t, x, y), then Eq.( 1) becomes a functional-integral equation In [12], the authors proved the existence of solutions to Eq.( 11).These solutions continuous and bounded on the interval [0, ∞) and are globally attractive.
Remark 4.1 In order to apply our technique to Eq.( 12) we have to impose an additional condition that the characteristic function φ is continuous and satisfies φ(0) = 0.This condition that the kernel v(t, s, x) defined by v(t, s, x) = 0, s = 0, t ≥ 0, x ∈ IR t t+s φ(s) x, s = 0, t ≥ 0, x ∈ IR is continuous on I × I × IR in accordance with assumption a 3 ), see [6].