We study the global existence of positive integrable solution for the functional integral inclusion of fractional order x(t) ∈ p(t) + F₁(t, I^αf₂(t, x(φ(t))), t ∈ (0, 1), α ≥ 0, where F₁(t, x(t)) is a set-valued function defined on (0, 1) × R⁺.

1. Introduction

Consider the functional integral equation

()

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

()

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 f₁(t, x) and f₂(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

has nonempty closed values.

2. Preliminaries

Let L¹ = L¹(I) be the class of Lebesgue integrable functions on the interval I = [a, b], 0 ≤ a < b < ∞, and let Γ(·) be the gamma function.

Definition 1. The fractional integral of the function f(·) ∈ L¹(I) of order α ∈ R⁺ is defined by (cf. [6–9])

()

Theorem 1 (nonlinear alternative of Leray-Shauder type [10]). Let U be an open subset of a convex set D in a Banach space X. Assume 0 ∈ U and . Then either

A₁ T has a fixed point in or
A₂ there exists γ ∈ (0,1) and x ∈ ∂U such that x = γTx.

Theorem 2 (Kolmogorov compactness criterion [11]). Let Ω⊆L^p(0,1), 1 ≤ P ≤ ∞. If

(i)
Ω is bounded in L^p(0, 1)
(ii)
x_h → x as h → 0 uniformly with respect to x ∈ Ω, then Ω is relatively compact in L^p(0, 1), where

()

3. Main Results

In this section we present our main result by proving the global existence of positive solution x ∈ L¹ for the functional integral equatoin (1).

To facilitate our discussion, let us first state the following assumptions.

(i)
p ∈ L¹.
(ii)
f_i : (0,1) × R⁺ → R⁺, i = 1,2, satisfy Caratheodory condition that is, f_i are measurabel in t for any x ∈ R⁺ and continuous in x for almost all t ∈ (0,1).
There exist four functions t → a_i(t), t → b_i(t) such that
()
where a_i(·) ∈ L¹ and b_i(·) are measurable and bounded.

(iii)
ϕ : (0,1) → (0,1) is absolutely continuous, and there exists a constant M > 0 such that φ^′(t) ≥ M, ∀t ∈ (0,1).
(iv)
Assume that every solution x(·) ∈ L¹ to the equation
()
t ∈ (0,1), 0 < β < 1, γ ∈ (0,1), satisfies ∥x∥ ≠ r (r > 0 is arbitrary but fixed).

Define the operator T,

()

Now, we are in position to formulate and prove our main result.

Theorem 3. Let the assumptions (i)–(iv) satisfied. Then (1) has at least one positive solution x ∈ L₁.

Proof. Let x be an arbitrary element in the open set B_r = {x : ∥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 L¹.

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 L¹; that is, condition (i) of Theorem 2 is satisfied.

It remains to show that (Tx)_h → Tx in L¹ as h → 0 uniformly with respect to Tx ∈ Ω. We have the following:

()

Now f₁, f₂ ∈ L¹ and I^βf₂ ∈ L¹, then (cf. [12])

()

Moreover, p(·) ∈ L¹. So, we have

()

for a.e t ∈ L¹. 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 = L¹(0,1). Then in the view of assumption (iv) condition (A₂) of Theorem 1 does not hold. Theorem 1 implies that T has a fixed point. This completes the proof.

4. Integral Inclusion

Consider now the integral inclusion (2), where has nonempty closed convex values.

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 F₁ satisfies the following assumptions:

(1)
F₁(t, x) are nonempty, closed and convex for all (t, x)∈(0,1) × R⁺,
(2)
F₁(t, ·) is lower semicontinuous from R⁺ into R⁺,
(3)
F₁(·, ·) is measurable,
(4)
there exist a function a ∈ L₁ and a measurable and bounded function b such that

()

Then there exists at least one positive solution x ∈ L¹ of the integral inclusion (2).

Proof. By conditions (1)–(4) (see [13–16]) we can find a selection function f₁ (Caratheodory function) f₁ : (0,1) × R⁺ → R⁺ such that f₁(t, x) ∈ F₁(t, x) for all (t, x)∈(0,1) × R⁺, this function satisfies condition (ii) of Theorem 3.

Clearly all assumptions of Theorem 3 are hold, then there exists a positive solution x ∈ L¹ such that

()

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Global Integrable Solution for a Nonlinear Functional Integral Inclusion

Abstract

1. Introduction

2. Preliminaries

3. Main Results

4. Integral Inclusion

References

References

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Global Integrable Solution for a Nonlinear Functional Integral Inclusion

Abstract

1. Introduction

2. Preliminaries

3. Main Results

4. Integral Inclusion

References

References

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