# Solutions of a Class of Deviated-Advanced Nonlocal Problems for the Differential Inclusionx1(t)∈F(t,x(t))

Authors: A. M. A. El-Sayed; E. M. Hamdallah; Kh. W. Elkadeky
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We study the existence of solutions for deviated-advanced nonlocal and integral condition problems for the differential inclusionx1(t)∈F(t,x(t)).


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## Body

## 1. Introduction

Problems with nonlocal conditions have been extensively studied by several authors in the last two decades. The reader is referred to [1–12] and references therein. Consider the deviated-advanced nonlocal problem (1.1)dx(t)dt∈F(t,x(t)),a.e.t∈(0,1),(1.2)∑k=1makx(ϕ(τk))=α∑j=1nbjx(ψ(ηj)),ak,bj>0,
where τk,ηj∈(0,1), α>0 is a parameter, and ψ and ϕ are, respectively, deviated and advanced given functions.Our aim here is to study the existence of at least one absolutely continuous solutionx∈AC[0,1] for the problem (1.1)-(1.2) when the set-valued function F:R→P(R) is L1-Carathéodory.As an application, we deduce the existence of a solution for the nonlocal problem of the differential inclusion (1.1) with the deviated-advanced integral condition (1.3)∫01x(ϕ(s))ds=α∫01x(ψ(s))ds.It must be noticed that the following nonlocal and integral conditions are special cases of our nonlocal and integral conditions(1.4)x(ϕ(τ))=αx(ψ(η)),τ,η∈(0,1),∑k=1makx(ϕ(τk))=αx(ψ(η)),τk,η∈(0,1),∑k=1makx(ϕ(τk))=0,τk∈(0,1),∫01x(ϕ(s))ds=αx(ψ(η)),η∈(0,1),α∫01x(ψ(s))ds=x(ϕ(τ)),τ∈(0,1),∫01x(ϕ(s))ds=0,∫01x(ψ(s))ds=0.
As an example of the deviated function ϕ:(0,1)→(0,1), we have ϕ(t)=βt,β∈(0,1). As an example of the advanced function ψ:(0,1)→(0,1), we have ψ(t)=tβ,β∈(0,1).


## 2. Preliminaries

The following preliminaries are needed.Definition 2.1.
A set-valued functionF:[0,1]×R→P(R) is called L1-Carathéodory if(a)
t→F(t,x) is measurable for each x∈R,(b)
x→F(t,x) is upper semicontinuous for almost all t∈[0,1],(c)
there existsm∈L1([0,1],D),D⊂R such that(2.1)|F(t,x)|=sup⁡{|v|:v∈F(t,x)}≤m(t),for almost allt∈[0,1].Definition 2.2.
A single-valued functionf:[0,1]×R→R  is called L1-Carathéodory if(i)
t→f(t,x) is measurable for each x∈R,(ii)
x→f(t,x) is continuous for almost all t∈[0,1],(iii)
there existsm∈L1([0,1],D),D⊂R such that |f|≤m.Definition 2.3.
The set(2.2)SF(⋅,x(t))1={f∈([0,1],R):f(t,x)∈F(t,x(t))for a.e.t∈[0,1]}
is called the set of selections of the set-valued function F.Theorem 2.4.
For anyL1-Carathéodory set-valued function F, the set SF(·,x(t))1 is nonempty [1, 13].Theorem 2.5 (Carathéodory, [14]).
Letf:[0,1]×R→R be L1-Carathéodory. Then the problem
(2.3)dx(t)dt=f(t,x(t)),for a.e.t>0,x(0)=x0,
has at least one solution x∈AC[0,T].


## 3. Existence of Solution

Consider the following assumptions.(i)
F:[0,1]×R→P(R+) is  L1-Carathéodory.(ii)
(3.1)α∑j=1nbj≠∑k=1mak.(iii)
ϕ:(0,1)→(0,1),ϕ(t)≤t is a deviated continuous function.(iv)
ψ:(0,1)→(0,1),ψ(t)≥t is an advanced continuous function.Now we have the following lemma.Lemma 3.1.
Let assumptions (i)-(ii) be satisfied. The solution of the nonlocal problem (1.1)-(1.2) can be expressed by the integral equation
(3.2)x(t)=A(∑k=1mak∫0ϕ(τk)f(s,x(s))ds-α∑j=1nbj∫0ψ(ηj)f(s,x(s))ds)+∫0tf(s,x(s))ds,
where f(t,x)∈F(t,x),for allx∈R, and A=(α∑j=1nbj-∑k=1mak)-1.Proof.
From the assumption that the set-valued functionF:[0,1]×R→P(R+) is L1-Carathéodory, then (Theorem 2.4) there exists a single-valued selection f:[0,1]×R→R+ such that
(3.3)ddtx(t)=f(t,x)∈F(t,x),∀x∈R.
This selection f(t,x) is L1-Carathéodory.
Integrating (3.3), we get
(3.4)x(t)=x(0)+∫0tf(s,x(s))ds.
Let t=ϕ(τk). Then
(3.5)∑k=1makx(ϕ(τk))=∑k=1makx(0)+∑k=1mak∫0ϕ(τk)f(s,x(s))ds.
Let t=ψ(ηj). Then
(3.6)α∑j=1nbjx(ψ(ηj))=α∑j=1nbjx(0)+α∑j=1nbj∫0ψ(ηj)f(s,x(s))ds.
From (3.5) and (3.6), we obtain
(3.7)x(0)=A(∑k=1mak∫0ϕ(τk)f(s,x(s))ds-α∑j=1nbj∫0ψ(ηj)f(s,x(s))ds),
where A=(α∑j=1nbj-∑k=1mak)-1, α∑j=1nbj≠∑k=1mak.
Substituting (3.7) into (3.4), we obtain
(3.8)x(t)=A(∑k=1mak∫0ϕ(τk)f(s,x(s))ds-α∑j=1nbj∫0ψ(ηj)f(s,x(s))ds)+∫0tf(s,x(s))ds.
This proves that the solution of the nonlocal problem (1.1)-(1.2) can be expressed by the integral equation (3.2).For the existence of the solution, we have the following theorem.Theorem 3.2.
Assume that (i)–(iv) are satisfied. Then the integral equation (3.2) has at least one continuous solution x∈C[0,1].Proof.
Define a subsetQr⊂C[0,1] by
(3.9)Qr={x∈C[0,1]:|x(t)|≤r,r=AM(1+∑k=1mak+α∑j=1nbj)}.
Clearly, the set Qr is nonempty, closed, and convex.
LetH be an operator defined by
(3.10)(Hx)(t)=A(∑k=1mak∫0ϕ(τk)f(s,x(s))ds-α∑j=1nbj∫0ψ(ηj)f(s,x(s))ds)+∫0tf(s,x(s))ds.
Let x∈Qr. Let {xn(t)} be a sequence in Qr converging to x(t),xn(t)→x(t),for allt∈I. Then
(3.11)lim⁡n→∞(Hxn)(t)=A(∑k=1maklim⁡n→∞∫0ϕ(τk)f(s,xn(s))ds-α∑j=1nbjlim⁡n→∞∫0ψ(ηj)f(s,xn(s))ds)+lim⁡n→∞∫0tf(s,xn(s))ds,
By assumptions (i)-(ii) and the Lebesgue dominated convergence theorem, we deduce that
(3.12)lim⁡n→∞(Hxn)(t)=(Hx)(t).
Then H is continuous.
Now, lettingx∈Qr, (thenϕ(t)≤t and ψ(t)≥t), we obtain
(3.13)(Hx)(t)≤A(∑k=1mak∫0τkf(s,x(s))ds-α∑j=1nbj∫0ηjf(s,x(s))ds)+∫0tf(s,x(s))ds,|(Hx)(t)|≤A(∑k=1mak∫0τk|f(s,x(s))|ds+α∑j=1nbj∫0ηj|f(s,x(s))|ds)+∫0t|f(s,x(s))|ds≤A(∑k=1mak∫0τkm(s)ds+α∑j=1nbj∫0ηjm(s)ds)+∫0tm(s)ds≤A(∑k=1makM+α∑j=1nbjM)+M≤AM(1+∑k=1mak+α∑j=1nbj)≤r.
Then {Hx(t)} is uniformly bounded in Qr.
Also fort1,t2∈(0,1),t1<t2 such that |t2-t1|<δ, we have
(3.14)(Hx)(t2)-(Hx)(t1)=∫0t2f(s,x(s))ds-∫0t1f(s,x(s))ds,|(Hx)(t2)-(Hx)(t1)|≤∫t1t2|f(s,x(s))|ds≤∫t1t2m(s)ds,|(Hx)(t2)-(Hx)(t1)|≤ε.
Hence the class of functions {Hx(t)} is equicontinuous. By Arzela-Ascoli's theorem,   {Hx(t)}   is relatively compact.  Since all conditions of Schauder's theorem hold, then H has a fixed point in Qr.
Therefore the integral equation (3.2) has at least one continuous solution x∈C(0,1). Now,
(3.15)lim⁡t→0x(t)=Alim⁡t→0(∑k=1mak∫0ϕ(τk)f(s,x(s))ds-α∑j=1nbj∫0ψ(ηj)f(s,x(s))ds)+lim⁡t→0∫0tf(s,x(s))ds=A(∑k=1mak∫0ϕ(τk)f(s,x(s))ds-α∑j=1nbj∫0ψ(ηj)f(s,x(s))ds)=x(0).
Also
(3.16)x(1)=lim⁡t→1x(t)=A(∑k=1mak∫0ϕ(τk)f(s,x(s))ds-α∑j=1nbj∫0ϕ(ηj)f(s,x(s))ds)+∫01f(s,x(s))ds.
Then the integral equation (3.2) has at least one continuous solution x∈C[0,1].The following theorem proves the existence of at least one solution for the nonlocal problem(1.1)-(1.2).Theorem 3.3.
Let (i)–(iv) be satisfied. Then the nonlocal problem (1.1)-(1.2) has at least one solution x∈AC[0,1].Proof.
From Theorem3.2 and the integral equation (3.2), we deduce that there exists at least one solution, x∈AC[0,1], of the integral equation (3.2).
To complete the proof, we prove that the integral equation (3.2) satisfies nonlocal problem (1.1)-(1.2).
Differentiating (3.2), we get
(3.17)dxdt=f(t,x(t))∈F(t,x(t)),a.e.t∈(0,1).
Letting t=ϕ(τk) in (3.2), we obtain
(3.18)∑k=1makx(ϕ(τk))=∑k=1mak(A∑k=1mak+1)∫0ϕ(τk)f(s,x(s))ds-αA∑k=1mak∑j=1nbj∫0ψ(ηj)f(s,x(s))ds.
Also, letting t=ψ(ηj) in (3.2), we obtain
(3.19)α∑j=1nbjx(ψ(ηj))=αA∑j=1nbj∑k=1mak∫0ϕ(τk)f(s,x(s))ds+α∑j=1nbj(1-αA∑j=1nbj)∫0ψ(ηj)f(s,x(s))ds.
And from (3.19) from (3.18), we obtain
(3.20)∑k=1makx(ϕ(τk))=α∑j=1nbjx(ψ(ηj)).
This complete the proof of the equivalence between the nonlocal problem (1.1)-(1.2) and the integral equation (3.2).
This implies that there exists at least one absolutely continuous solutionx∈AC[0,1] of the nonlocal problem (1.1)-(1.2).


## 4. Nonlocal Integral Condition

Letx∈[0,1] be a solution of the nonlocal problem (1.1)-(1.2). Let ak=tk-tk-1,τk∈(tk-1,tk)⊂(0,1). Also, let bj=tj-tj-1,ηj∈(tj-1,tj)⊂(0,1). Then the nonlocal condition (1.2) will be  (4.1)∑k=1m(tk-tk-1)x(ϕ(τk))=α∑j=1n(tj-tj-1)x(ψ(ηj)).
From the continuity of the solution x of the nonlocal condition (1.2) we obtain (4.2)lim⁡m→∞∑k=1m(tk-tk-1)x(ϕ(τk))=lim⁡n→∞α∑j=1n(tj-tj-1)x(ψ(ηj)).
That is, the nonlocal condition (1.2) is transformed to the integral condition (4.3)∫01x(ϕ(s))ds=α∫01x(ψ(s))ds,
and the solution of the integral equation (3.2) will be (4.4)x(t)=A*(∫01∫0ϕ(s)f(θ,x(θ))dθds-α∫01∫0ψ(s)f(θ,x(θ))dθds)+∫0tf(s,x(s))ds,A*=(α-1)-1.Now, we have the following theorem.Theorem 4.1.
Let assumptions (i)–(iv) of Theorem3.2 be satisfied. Then the nonlocal problem with the integral condition
(4.5)dx(t)dt=f(t,x(t))∈F(t,x(t)),for a.e.t∈(0,1),∫01x(ϕ(s))ds=α∫01x(ψ(s))ds
has at least one solution x∈AC[0,1] represented by (4.4).



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*Source: 476392-2011-06-18.xml*