MATHEMATICSSRX Mathematics2090-1127Scholarly Research Exchange89198210.3814/2010/891982891982Research ArticleGlobal Integrable Solution for a Nonlinear Functional Integral InclusionEl-SayedA. M. A.1Al-IssaSh. M.21Faculty of ScienceAlexandria UniversityAlexandriaEgyptalex.edu.eg2Faculty of ScienceBeirut Arab UniversityBeirutLebanonbau.edu.lb20102610200920102082009141020092010Copyright © 2010This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We study the global existence of positive integrable solution for the functional integral inclusion of fractional order x(t)p(t)+F1(t,Iαf2(t,x(φ(t))), t(0,1), α0, where F1(t,x(t)) is a set-valued function defined on (0,1)×R+.

1. Introduction

Consider the functional integral equation

x(t)=g(t)+f(t,01k(t,s)x(φ(t)))ds,t(0,1). The authors (see ) 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 , the author discusses the existence of solution of the functional integral equation

x(t)=u(t)+0tk(t,s)f(s,x(φ(s)))ds,t(0,1). In , the author omitted the condition of monotonicity of f(t,x) and proved the existence of integrable solution of (1).

Also he proved (see ) the existence of integrable solution of the functional integral equation

x(t)=f1(t,r01k(t,s)f2(s,x(s))ds),t(0,1), where f(t,x) is assumed be satisfy the Caratheodory and growth conditions.

Consider the functional integral equation x(t)=0tk1(t,s)f(s,0sk2(t,θ)x(φ(θ))dθ),t(0,1). The existence of the nonincreasing integrable solution of (4) was studied in , 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

x(t)=p(t)+f1(t,Iαf2(t,x(φ(t)))),t(0,1),α0, 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 x(t)p(t)+F1(t,Iαf2(t,x(φ(t)))),t(0,1),α0, where the set-valued map F1:(0,1)×R+2R+ has nonempty closed values.

2. Preliminaries

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

Definition 1.

The fractional integral of the function f(·)L1(I) of order αR+ is defined by (cf. ) Iaαf(t)=at(t-s)α-1Γ(α)f(s)ds.

Theorem 1 (nonlinear alternative of Leray-Shauder type [<xref ref-type="bibr" rid="B7">10</xref>]).

Let U be an open subset of a convex set D in a Banach space X. Assume 0U and TC(U̅,D). Then either

T has a fixed point in U̅ or

there exists γ(0,1) and xU such that x=γTx.

Theorem 2 (Kolmogorov compactness criterion [<xref ref-type="bibr" rid="B6">11</xref>]).

Let ΩLp(0,1),1P. If

Ω is bounded in Lp(0,1)

xhx as h0 uniformly with respect to xΩ, then Ω is relatively compact in Lp(0,1), where

xh(t)=1h0t+hx(s)ds.

3. Main Results

In this section we present our main result by proving the global existence of positive solution xL1 for the functional integral equatoin (1).

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

pL1.

fi:(0,1)×R+R+,i=1,2, satisfy Caratheodory condition that is, fi are measurabel in t for any xR+ and continuous in x for almost all t(0,1).

There exist four functions tai(t),tbi(t) such that |fi(t,x)|ai(t)+bi(t)|x|,i=1,2,t(0,1),xR, where ai(·)L1 and bi(·) are measurable and bounded.

ϕ:(0,1)(0,1) is absolutely continuous, and there exists a constant M>0 such that φ(t)M,t(0,1).

Assume that every solution x(·)L1 to the equation x(t)=γ(p(t)+f1(t,Iαf2(t,x(φ(t))))),t(0,1),0<β<1,γ(0,1), satisfies xr (r>0 is arbitrary but fixed).

Define the operator T, Tx(t)=p(t)+f1(t,Iαf2(t,x(φ(t)))),t(0,1). 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 xL1.

Proof.

Let x be an arbitrary element in the open set Br={x:xr,r0}.

Then from assumptions (i) and (ii), we have,

Tx=01|(Tx)(t)|dt01|p(t)|dt+01|f1(t,Iαf2(t,x(φ(t))))|dt01|p(t)|dt+01a1(t)dt+01b(t)|Iαf2(t,x(φ(t)))|dt01|p(t)|dt+01|a1(t)|dt+01|b1(t)|Iα(a2(t)+b2(t)|x(φ(t))|)dtp+a1+b1010t(t-s)α-1Γ(α)a2(s)dsdt+b1010t(t-s)α-1Γ(α)|b2(s)||x(φ(s))|dsdtp+a1+b101|a2(s)|s1(t-s)α-1Γ(α)dtds+b1b201|x(φ(s))|s1(t-s)α-1Γ(α)dtdsp+a1+b101|a2(s)|(t-1)αΓ(α+1)ds+b1b201|x(φ(s))|(t-1)αΓ(α+1)dsp+a1+b1a2Γ(α+1)+b1b2Γ(α+1)M01|x(φ(s))|φ(s)dsp+a1+b1a2Γ(α+1)+b1b2Γ(α+1)Mφ(0)φ(1)|x(u)|dup+a1+b1a2Γ(α+1)+b1b2Γ(α+1)M01|x(u)|dup+a1+b1a2Γ(α+1)+b1b2xΓ(α+1)M.

Hence the previous inequality means that the operator T maps Br 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 Br. Then T(Ω) is bounded in L1; that is, condition (i) of Theorem 2 is satisfied.

It remains to show that (Tx)hTx in L1 as h0 uniformly with respect to TxΩ.  We have the following: (Tx)h-(Tx)=01|(Tx)h(t)-(Tx)(t)|dt=01|1htt+h(Tx)h(τ)dτ-(Tx)(t)|dt=01|1htt+h((Tx)h(τ)-(Tx)(t))dτ|dt011htt+h|p(τ)-p(t)|dτdt+011htt+h|f1(τ,Iβf2(τ,x(φ(τ))))-f1(t,Iβf2(t,x(φ(t))))|dτdt.

Now f1,f2L1 and Iβf2L1, then (cf. ) 1htt+h|f1(τ,Iβf2(τ,x(φ(τ))))-f1(t,Iβf2(t,x(φ(t))))|dτ0. Moreover, p(·)L1. So, we have 1htt+h|p(τ)-p(t)|dτ0 for a.e tL1. Therefore, by Theorem 2 we deduce that T(Ω) is relatively compact; that is, T is compact operator.

Set U=Br 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 F1:[0,1]×R+2R+ 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:

F1(t,x) are nonempty, closed and convex for all (t,x)(0,1)×R+,

F1(t,·) is lower semicontinuous from R+ into R+,

F1(·,·) is measurable,

there exist a function aL1 and a measurable and bounded function b such that

|F1(t,x)|a(t)+b(t)|x|t(0,1), Then there exists at least one positive solution xL1 of the integral inclusion  (2).

Proof.

By conditions (1)(4) (see ) 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 xL1 such that x(t)-p(t)=f1(t,Iβf2(t,x(φ(t))))F1(t,Iβf2(t,x(φ(t)))).

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