Global Integrable Solution for a Nonlinear Functional Integral Inclusion

We study the global existence of positive integrable solution for the 
functional integral inclusion of fractional order 
 x ( t ) ∈ p ( t ) + F 1 ( t , I α f 2 ( t , x ( φ ( t ) ) ) , t ∈ ( 0 , 1 ) , α ≥ 0 , where F 1 ( t , x ( t ) ) is a set-valued function defined on ( 0 , 1 ) × R + .


Introduction
Consider the functional integral equation 1 0 k(t, s)x ϕ(t) ds, t ∈ (0, 1). (1) The authors (see [1]) proved the existence of monotonic integrable solution of (1), where the the function f (t, x) is assumed to be monotonic in both its arguments and satisfies the Caratheodory and growth conditions.
In [2], the author discusses the existence of solution of the functional integral equation k(t, s) f s, x ϕ(s) ds, t ∈ (0, 1). (2) In [3], the author omitted the condition of monotonicity of f (t, x) and proved the existence of integrable solution of (1).
Also he proved (see [4]) the existence of integrable solution of the functional integral equation where f (t, x) is assumed be satisfy the Caratheodory and growth conditions. Consider the functional integral equation The existence of the nonincreasing integrable solution of (4) was studied in [5], where f (t, x) is nondecreasing in both its arguments and satisfies the Caratheodory and growth conditions. Here we are concerned with the functional integral equation of fractional order we prove the global existence of positive integrable solution of (5), where f 1 (t, x) and f 2 (t, x) satisfy the Caratheodory and growth conditions. As a generalization of our results we study the global existence of positive integrable solution of the nonlinear functional integral inclusion of fractional order where the set-valued map F 1 : (0, 1) × R + → 2 R + has nonempty closed values.

Preliminaries
Let L 1 = L 1 (I) be the class of Lebesgue integrable functions on the interval I = [a, b], 0 ≤ a < b < ∞, and let Γ(·) be the gamma function.

Main Results
In this section we present our main result by proving the global existence of positive solution x ∈ L 1 for the functional integral equatoin (1). To facilitate our discussion, let us first state the following assumptions.
There exist four functions t → a i (t), t → b i (t) such that where a i (·) ∈ L 1 and b i (·) are measurable and bounded.
Define the operator T, Tx(t) = p(t) + f 1 t, I α f 2 t, x ϕ(t) , t ∈ (0, 1). (11) Now, we are in position to formulate and prove our main result. x r, r 0}. Then from assumptions (i) and (ii), we have, Hence the previous inequality means that the operator T maps B r into L 1 . Now, we will show that T is compact. To achieve this goal we will apply Theorem 2. So, let Ω be a bounded subset of B r . Then T(Ω) is bounded in L 1 ; that is, condition (i) of Theorem 2 is satisfied.
It remains to show that (Tx) h → Tx in L 1 as h → 0 uniformly with respect to Tx ∈ Ω. We have the following: Now f 1 , f 2 ∈ L 1 and I β f 2 ∈ L 1 , then (cf. [12]) Moreover, p(·) ∈ L 1 . So, we have for a.e t ∈ L 1 . Therefore, by Theorem 2 we deduce that T(Ω) is relatively compact; that is, T is compact operator. Set U = B r and D = X = L 1 (0, 1). Then in the view of assumption (iv) condition (A 2 ) of Theorem 1 does not hold. Theorem 1 implies that T has a fixed point. This completes the proof.
As an important consequence of the main result we can present the following Theorem 4. Let the assumptions of Theorem 3 be satisfied. The multifunction F 1 satisfies the following assumptions: (1) F 1 (t, x) are nonempty, closed and convex for all (t, x) ∈ (0, 1) × R + , (2) F 1 (t, ·) is lower semicontinuous from R + into R + , there exist a function a ∈ L 1 and a measurable and bounded function b such that |F 1 (t, x)| ≤ a(t) + b(t)|x| ∀t ∈ (0, 1), (16) Then there exists at least one positive solution x ∈ L 1 of the integral inclusion (2).