# Global Integrable Solution for a Nonlinear  Functional Integral Inclusion

Authors: A. M. A. El-Sayed; Sh. M. Al-Issa
Journal: SRX Mathematics
 (2010)

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## Abstract

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


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## Body

## 1. Introduction

Consider the functional integral equation(1)x(t)=g(t)+f(t,∫01k(t,s)x(φ(t)))ds,t∈(0,1).
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(2)x(t)=u(t)+∫0tk(t,s)f(s,x(φ(s)))ds,t∈(0,1).
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(3)x(t)=f1(t,r∫01k(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(4)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 [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(5)x(t)=p(t)+f1(t,Iαf2(t,x(φ(t)))),t∈(0,1),α≥0,
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
(6)x(t)∈p(t)+F1(t,Iαf2(t,x(φ(t)))),t∈(0,1),α≥0,
where the set-valued map F1:(0,1)×R+→2R+ has nonempty closed values.


## 2. Preliminaries

LetL1=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 functionf(·)∈L1(I) of order α∈R+ is defined by (cf. [6–9])
(7)Iaαf(t)=∫at(t-s)α-1Γ(α)f(s)ds.Theorem 1 (nonlinear alternative of Leray-Shauder type [10]).
LetU be an open subset of a convex set D in a Banach space X. Assume 0∈U and T∈C(U̅,D). Then either(A1)
T has a fixed point in U̅ or(A2)
there existsγ∈(0,1) and x∈∂U such that x=γTx.Theorem 2 (Kolmogorov compactness criterion [11]).
LetΩ⊆Lp(0,1),1≤P≤∞. If  (i)
Ω is bounded in Lp(0,1)(ii)
xh→x as h→0 uniformly with respect to x∈Ω, then Ω is relatively compact in Lp(0,1), where(8)xh(t)=1h∫0t+hx(s)ds.


## 3. Main Results

In this section we present our main result by proving the global existence of positive solutionx∈L1 for the functional integral equatoin (1).To facilitate our discussion, let us first state the following assumptions.(i)
p∈L1.(ii)
fi:(0,1)×R+→R+,i=1,2, satisfy Caratheodory condition that is, fi are measurabel in t for any x∈R+ and continuous in x for almost all t∈(0,1).There exist four functionst→ai(t),t→bi(t) such that
(9)|fi(t,x)|≤ai(t)+bi(t)|x|,i=1,2,∀t∈(0,1),x∈R,
where ai(·)∈L1 and bi(·) 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 solutionx(·)∈L1 to the equation
(10)x(t)=γ(p(t)+f1(t,Iαf2(t,x(φ(t))))),t∈(0,1),0<β<1,γ∈(0,1), satisfies ∥x∥≠r (r>0 is arbitrary but fixed).Define the operatorT,
(11)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 x∈L1.Proof.
Letx be an arbitrary element in the open set Br={x:∥x∥〈r,r〉0}.
Then from assumptions (i) and (ii), we have,
(12)∥Tx∥=∫01|(Tx)(t)|dt≤∫01|p(t)|dt+∫01|f1(t,Iαf2(t,x(φ(t))))|dt≤∫01|p(t)|dt+∫01a1(t)dt+∫01b(t)|Iαf2(t,x(φ(t)))|dt≤∫01|p(t)|dt+∫01|a1(t)|dt+∫01|b1(t)|Iα(a2(t)+b2(t)|x(φ(t))|)dt≤∥p∥+∥a1∥+b1∫01∫0t(t-s)α-1Γ(α)a2(s)dsdt+b1∫01∫0t(t-s)α-1Γ(α)|b2(s)||x(φ(s))|dsdt≤∥p∥+∥a1∥+b1∫01|a2(s)|∫s1(t-s)α-1Γ(α)dtds+b1b2∫01|x(φ(s))|∫s1(t-s)α-1Γ(α)dtds≤∥p∥+∥a1∥+b1∫01|a2(s)|(t-1)αΓ(α+1)ds+b1b2∫01|x(φ(s))|(t-1)αΓ(α+1)ds≤∥p∥+∥a1∥+b1∥a2∥Γ(α+1)+b1b2Γ(α+1)M∫01|x(φ(s))|φ′(s)ds≤∥p∥+∥a1∥+b1∥a2∥Γ(α+1)+b1b2Γ(α+1)M∫φ(0)φ(1)|x(u)|du≤∥p∥+∥a1∥+b1∥a2∥Γ(α+1)+b1b2Γ(α+1)M∫01|x(u)|du≤∥p∥+∥a1∥+b1∥a2∥Γ(α+1)+b1b2∥x∥Γ(α+1)M.
Hence the previous inequality means that the operatorT maps Br into L1.
Now, we will show thatT 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)h→Tx in L1 as h→0 uniformly with respect to Tx∈Ω.We have the following:
(13)∥(Tx)h-(Tx)∥=∫01|(Tx)h(t)-(Tx)(t)|dt=∫01|1h∫tt+h(Tx)h(τ)dτ-(Tx)(t)|dt=∫01|1h∫tt+h((Tx)h(τ)-(Tx)(t))dτ|dt≤∫011h∫tt+h|p(τ)-p(t)|dτdt+∫011h∫tt+h|f1(τ,Iβf2(τ,x(φ(τ))))-f1(t,Iβf2(t,x(φ(t))))|dτdt.
Nowf1,f2∈L1 and Iβf2∈L1, then (cf. [12])
(14)1h∫tt+h|f1(τ,Iβf2(τ,x(φ(τ))))-f1(t,Iβf2(t,x(φ(t))))|dτ→0.
Moreover, p(·)∈L1. So, we have
(15)1h∫tt+h|p(τ)-p(t)|dτ→0
for a.e t∈L1. Therefore, by Theorem 2 we deduce that T(Ω) is relatively compact; that is, T is compact operator.
SetU=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 followingTheorem 4.
Let the assumptions of Theorem3 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 functiona∈L1 and a measurable and bounded function b such that(16)|F1(t,x)|≤a(t)+b(t)|x|∀t∈(0,1),
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 Theorem3 are hold, then there exists a positive solution x∈L1 such that
(17)x(t)-p(t)=f1(t,Iβf2(t,x(φ(t))))∈F1(t,Iβf2(t,x(φ(t)))).



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