# On a Multipoint Boundary Value Problem for a Fractional Order Differential Inclusion on an Infinite Interval

Authors: Nemat Nyamoradi; Dumitru Baleanu; Ravi P. Agarwal
Journal: Advances in Mathematical Physics
 (2013)

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## Abstract

We investigate the existence of solutions for the following multipoint boundary value problem of a fractional order differential inclusionD0+αut+Ft,ut,u′t∋0,0<t<+∞,u0=u′0=0,Dα-1u+∞-∑i=1m-2‍βiuξi=0, where D0+α is the standard Riemann-Liouville fractional derivative, 2<α<3,0<ξ1<ξ2<⋯<ξm-2<+∞, satisfies 0<∑i=1m-2‍βiξiα-1<Γ(α),  and  F:[0,+∞)×ℝ×ℝ→𝒫(ℝ) is a set-valued map. Several results are obtained by using suitable fixed point theorems when the right hand side has convex or nonconvex values.


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## Body

## 1. Introduction

In this paper, we will consider the existence of solutions for the following multipoint boundary value problem of a fractional order differential inclusion(1)D0+αu(t)+F(t,u(t),u′(t))∋0,0<t<+∞,u(0)=u′(0)=0,Dα-1u(+∞)-∑i=1m-2‍βiu(ξi)=0,
where D0+α is the standard Riemann-Liouville fractional derivative, 2<α<3, 0<ξ1<ξ2<⋯<ξm-2<+∞, satisfies 0<∑i=1m-2‍βiξiα-1<Γ(α), and F:[0,+∞)×ℝ×ℝ→𝒫(ℝ) is a set-valued map.The present paper is motivated by a recent paper of Liang and Zhang [1], where it is considered problem (1) with F single valued, and several existence results are provided.Fractional differential equations have been of great interest recently. This is because of both the intensive development of the theory of fractional calculus itself and the applications of such constructions in various scientific fields such as physics, mechanics, chemistry, and engineering. For details, see [2–4] and the references therein.The existence of solutions of initial value problems for fractional order differential equations has been studied in the literature [5–17] and the references therein. The study of fractional differential inclusions was initiated by El-Sayed and Ibrahim [18]. Also, recently, several qualitative results for fractional differential inclusions were obtained in [19–23] and the references therein.The aim here is to establish existence results for problem (1) when the right hand side is convex as well as nonconvex valued. In the first result (Theorem 21), we consider the case when the right hand side has convex values and prove an existence result via nonlinear alternative for Kakutani maps. In the second result (Theorem 25), we will use the fixed point theorem for contraction multivalued maps according to Covitz and Nadler. The paper is organized as follows.In Section2 we recall some preliminary facts that we need in the sequel, and in Section 3 we prove our main results. Finally, in Section 4, an example is given to demonstrate the application of one of our main results.


## 2. Preliminaries

In this section, we present some notations and preliminary lemmas that will be used in the proof of the main result.Let(X,d) be a metric space with the corresponding norm ∥·∥ and let I=[0,+∞). We denote by ℒ(I) the σ-algebra of all Lebesgue measurable subsets of I, by ℬ(X) the family of all nonempty subsets of X, and by 𝒫(X) the family of all Borel subsets of X. If A⊂I then χA:I→{0,1} denotes the characteristic function of A. For any subset A⊂X, we denote by A- the closure of A.Recall that the Pompeiu-Hausdorff distance of the closed subsetsA,B⊂X is defined by the following:
(2)dH(A,B)=max{d*(A,B),d*(B,A)},d*(A,B)=sup{d(a,B),a∈A},
where d(x,B)=infy∈Bd(x,y). Define
(3)𝒫(X)={Y⊂X:Y≠∅},𝒫b(X)={Y∈𝒫(X):Ybounded},𝒫cl(X)={Y∈𝒫(X):Yclosed},𝒫cp(X)={Y∈𝒫(X):Ycompact},𝒫cv(X)={Y∈𝒫(X):Yconvex}.Also, we denote byC(I,X) the Banach space of all continuous functions x:[0,+∞)→X endowed with the norm |x|c=supt∈[0,+∞)|x(t)| and by L1([0,+∞),X) the Banach space of all (Bochner) integrable functions x:[0,+∞)→X endowed with the norm |x|1=∫[0,+∞)‍|x(t)|dt.Let(X,d1) and (Y,d2) be two metric spaces. If T:X→𝒫(X) is a set-valued map, then a point x∈X is called a fixed point for T if x∈T(x). T is said to be bounded on bounded sets if T(B):=∪x∈BT(x) is a bounded subset of X for all bounded sets B in X. T is said to be compact if T(B) is relatively compact for any bounded sets B in X. T is said to be totally compact if T(X)- is a compact subset of X. T is said to be upper semicontinuous if for any open set D⊂X, the set {x∈X:T(x)⊂D} is open in X. T is called completely continuous if it is upper semicontinuous and, for every bounded subset A⊂X, T(A) is relatively compact. It is well known that a compact set-valued map T with nonempty compact values is upper semicontinuous if and only if T has a closed graph.We define the graph ofT to be the set Gr(T)={(x,y)∈X×Y,y=T(x)} and recall a useful result regarding connection between closed graphs and upper semicontinuity.Lemma 1 (see [24, Proposition 1.2]).
IfT:X→𝒫cl(Y) is upper semicontinuous, then Gr(T) is a closed subset of X×Y, that is, for every sequence {xn}n∈ℕ⊂X and {yn}n∈ℕ⊂Y, if when n→∞, xn→x*, yn→y*, and yn∈T(xn), then y*∈T(x*). Conversely, if T is completely continuous and has a closed graph, then it is upper semicontinuous.For the convenience of the reader, we present here the following nonlinear alternative of the Leray-Schauder type and its consequences.Theorem 2 (nonlinear alternative for Kakutani maps [25]).
LetX be a Banach space, C a closed convex subset of X, U an open subset of C, and 0∈U. Suppose that T:U-→𝒫cl,cv(C) is an upper semicontinuous compact map; here 𝒫cl,cv(C) denotes the family of nonempty, compact convex subsets of C. Then, either(i)
Thas a fixed point in U or(ii)
there is au∈∂U and λ∈(0,1) with u∈λT(u).Definition 3.
The multifunctionT:X→𝒫(X) is said to be lower semicontinuous if for any closed subset C⊂X, the subset {s∈X:T(s)⊂C} is closed.IfF:[0,+∞)×ℝ×ℝ→𝒫(ℝ) is a set-valued map with compact values and x∈C([0,+∞),ℝ), we define
(4)SF(x):={f∈L1([0,+∞),ℝ):f(t)∈F(t,x(t),x′(t))a.e.[0,+∞)}.
Then, F is of a lower semicontinuous type if SF(·) is a lower semicontinuous with closed and decomposable values.Theorem 4 (see [26]).
LetS be a separable metric space and G:S→𝒫(L1([0,+∞),ℝ)) be a lower semicontinuous set-valued map with closed decomposable values. Then G has a continuous selection (i.e., there exists a continuous mapping g:S→L1([0,+∞),ℝ) such that g(s)∈G(s) for all s∈S).Definition 5.
Consider the following.(i)
A set-valued mapG:[0,+∞)→𝒫(ℝ) with nonempty compact convex values is said to be measurable if for any x∈ℝ the function t→d(x,G(t)) is measurable.(ii)
A set-valued mapF:[0,+∞)×ℝ×ℝ→𝒫(ℝ) is said to be Carathéodory if t→F(t,x,y) is measurable for all x,y∈ℝ and (x,y)→F(t,x,y) is upper semicontinuous for almost all t∈[0,+∞).(iii)
F is said to be L1-Carathéodory if for any l>0 there exists hl∈L1([0,+∞),ℝ) such that sup{|v|:v∈F(t,x,y)}≤hl(t)a.e.[0,+∞);∀x,y∈ℝ.Finally, the following results are easily deduced from the theoretical limit set properties.Lemma 6 (see [27, Lemma 1.1.9]).
Let{Kn}n∈ℕ⊂K⊂X be a sequence of subsets where K is a compact subset of a separable Banach space X. Then,
(5)co-(limsupn→∞Kn)=⋂N>0‍co-(⋃n≥N‍Kn),
where co-(A) refers to the closure of the convex hull of A.Lemma 7 (see [27, Lemma 1.4.13]).
LetX and Y be two metric spaces. If G:X→𝒫cp(Y) is an upper semicontinuous, then, for each x0∈X,
(6)limsupx→x0G(x)=G(x0).Definition 8.
LetX be a Banach space. A sequence {xn}n∈ℕ⊂L1([a,b],X) is said to be semicompact if(a)
it is integrably bounded; that is, there existsq∈L1([a,b],ℝ+) such that
(7)|xn(t)|E≤q(t),fora.e.t∈[a,b]andeveryn∈ℕ,(b)
the image sequence{xn(t)}n∈ℕ is relatively compact in E for a.e. t∈[a,b].The following important result follows from the Dunford-Pettis theorem (see [28, Proposition 4.2.1]).Lemma 9.
Every semicompact sequenceL1([a,b],X) is weakly compact in L1([a,b],X).When the nonlinearity takes convex values, Mazur’s Lemma, 1933, may be useful.Lemma 10 (see [29, Theorem 21.4]).
LetE be a normed space and {xk}k∈ℕ⊂E a sequence weakly converging to a limit x∈E. Then, there exists a sequence of convex combinations ym=∑k=1m‍αmkxk with αmk>0 for k=1,2,…,m and ∑k=1m‍αmk=1 which converges strongly to x.Lemma 11 (see [30]).
LetX be defined as before and M⊂X. Then M is relatively compact in X if the following conditions hold:(a)
M is uniformly bounded in X;(b)
the functions fromM are equicontinuous on any compact interval of [0,+∞);(c)
the functions fromM are equiconvergent; that is, for any given ϵ>0, there exists a T=T(ϵ)>0 such that |f(t)-f(+∞)|<ϵ, for any t>T, f∈M.Definition 12 (see [6]).
The Riemann-Liouville fractional integral operator of orderα>0, of function f∈L1(ℝ+), is defined as
(8)I0+αf(t)=1Γ(α)∫0t‍(t-s)α-1f(s)ds,
where Γ(·) is the Euler gamma function.Definition 13 (see [31]).
The Riemann-Liouville fractional derivative of orderα>0, n-1<α<n, n∈ℕ is defined as
(9)D0+αf(t)=1Γ(n-α)(ddt)n∫0t‍(t-s)n-α-1f(s)ds,
where the function f(t) has absolutely continuous derivatives up to order (n-1).Lemma 14 (see [31]).
The equalityD0+γI0+γf(t)=f(t), γ>0 holds for f∈L1(0,1).Lemma 15 (see [31]).
Letα>0 and u∈C(0,1)∩L1(0,1). Then, the differential equation
(10)D0+αu(t)=0
has a unique solution u(t)=c1tα-1+c2tα-2+⋯+cntα-n, ci∈ℝ, i=1,…,n, where n-1<α<n.Lemma 16 (see [31]).
Letα>0. Then, the following equality holds for u∈L1(0,1), D0+αu∈L1(0,1):
(11)I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2+⋯+cntα-n,ci∈ℝ, i=1,…,n, where n-1<α≤n.ByAC1([0,+∞),ℝ) we denote the space of continuous real-valued functions whose first derivative exists and it is absolutely continuous on [0,+∞). In this paper, we will use the following space E to the study (1) which is denoted by
(12)E={u∈AC1([0,+∞),ℝ):sup0≤t<+∞|u(t)|1+tα-1,E=sup0≤t<+∞|u′(t)|1+tα-1<+∞}.From [32], we know that E is a Banach space equipped with the norm
(13)∥u∥=max{sup0≤t<+∞|u(t)|1+tα-1,sup0≤t<+∞|u′(t)|1+tα-1}.In what follows,I=[0,+∞), α∈(2,3), and Δ=∑i=1m-2‍βiξiα-1. Next, we need the following technical result proved in [1].Lemma 17 (see [1]).
For anyh∈L1([0,+∞),ℝ), the problem
(14)D0+αu(t)+h(t)=0,0<t<∞,2<α<3,u(0)=u′(0)=0,Dα-1u(+∞)=∑i=1m-2‍βiu(ξi),
has a unique solution u(t) that
(15)u(t)=∫0+∞‍G(t,s)h(s)ds,
where
(16)G(t,s)=G1(t,s)+G2(t,s),(17)G1(t,s)=1Γ(α){tα-1-(t-s)α-1,0≤s≤t<+∞,tα-1,0≤t≤s<+∞,(18)G2(t,s)=∑i=1m-2‍βitα-1Γ(α)-ΔG1(ξi,s).Note thatG(t,s)>0,∀t,s∈[0,+∞), (e.g., Lemma  3.2 in [1]) and from the definition of G1(t,s), we have the following  (e.g., Remark  3.1 in [1]):
(19)G1(t,s)1+tα-1≤1Γ(α),G(t,s)1+tα-1≤L1for(t,s)∈[0,+∞)×[0,+∞),
where
(20)L1=1Γ(α)(1+∑i=1m-2‍βiξm-2α-1(Γ(α)-Δ)).
Also, one can get
(21)∂G(t,s)/∂t1+tα-1≤L2for(t,s)∈[0,+∞)×[0,+∞),
where
(22)L2=2(α-1)Γ(α)(1+∑i=1m-2‍βiξm-2α-1(Γ(α)-Δ)).Lemma 18.
The functionG(t,s) defined by (16) satisfies
(23)limt→+∞G(t,s)1+tα-1=∑i=1m-2‍βiξiα-1Γ(α)(Γ(α)-Δ).By calculation, it is easy to prove that Lemma18 holds. So, we omit its proof here.


## 3. Main Results

Now we are able to present the existence results for problem (1).


### 3.1. The Upper Semicontinuous Case

To obtain the complete continuity of existence solutions operator, the following lemma is still needed.Lemma 19 (see [32]).
LetV={u∈E∣∥u∥<l}(l>0),V1={u(t)/(1+tα-1):u∈V}. If   V1 is equicontinuous on any compact interval of [0,+∞) and equiconvergent at infinity, then V is relatively compact on E.Definition 20.
V
1 is called equiconvergent at infinity if and only if for all ϵ>0, there exists v(ϵ)>0 such that for all u∈V1,t1,t2≥v, it holds
(24)|u(t1)1+t1α-1-u(t2)1+t2α-1|<ϵ.Theorem 21.
The Carathéodory multivalued mapF:I×ℝ×ℝ→𝒫(ℝ) has nonempty, compact, convex values and satisfies the following.(H1)
There exists a continuous nondecreasing functionψ:[0,+∞)→(0,+∞) and φ∈L1([0,+∞), ℝ+) such that∥F(t,x,y)∥𝒫≔sup{|v(t)/(1+tα-1)|:v∈F(t,x,y)}≤φ(t)ψ(∥x∥),fora.e.t∈I  and  each  x,y∈ℝ.(H2)
There exists a constantM>0 such that
(25)Mmax{L1,L2}ψ(M)∫0+∞‍φ(s)ds>1.
Then, problem (1) has at least one solution.Proof.
LetX=E and consider M>0 as in (25). It is obvious that the existence of solutions to problem (1) is reduced to the existence of the solutions of the integral inclusion
(26)u(t)∈∫0+∞‍G(t,s)F(s,u(s),u′(s))ds,t∈I,
where G(t,s) is defined by (16) and (17). Consider the set-valued map, T:E→𝒫(X) is defined by
(27)T(u):={v∈X;v(t)=∫0+∞‍G(t,s)f(s)ds,f∈SF(u)-}.
We show thatT satisfies the hypotheses of Theorem 2.
Claim
1.We show that T(u)⊂X is convex for any u∈X. If v1,v2∈T(u), then, there exist f1,f2∈SF(u) such that for any t∈I one has
(28)vi(t)=∫0+∞‍G(t,s)fi(s)ds,i=1,2.
Let 0≤λ≤1. Then, for any t∈I, we have
(29)(λv1+(1-λ)v2)(t)=∫0+∞‍G(t,s)[λf1(s)+(1-λ)f2(s)]ds.
The values of F are convex; thus, SF(u) is a convex set and hence λv1+(1-λ)v2∈T(u).
Claim
2. We show that T is bounded on bounded sets of X. Let B be any bounded subset of X. Then, there exists m>0 such that ∥u∥≤m for all u∈B. If v∈T(u), then there exists f∈SF(u) such that v(t)=∫0+∞‍G(t,s)f(s)ds. One may write the following for any t∈I:
(30)|v(t)1+tα-1|≤∫0+∞‍|G(t,s)1+tα-1||f(s)|ds≤L1∫0+∞‍φ(s)ψ(∥u∥)ds≤L1|φ|1ψ(m).
On the other hand,
(31)|v′(t)1+tα-1|≤∫0+∞‍∂G(t,s)/∂t1+tα-1f(s)ds≤L2∫0+∞‍φ(s)ψ(∥u∥)ds≤L2|φ|1ψ(m),
and therefore
(32)∥v∥=maxt∈I{|v(t)1+tα-1|,|v′(t)1+tα-1|}≤max{L1,L2}|φ|1ψ(m),
for all v∈T(u); that is, T(B) is bounded.
Claim
3.We show that T maps bounded the sets into equi-continuous sets. Let B be any bounded subset of X as before and v∈T(u) for some u∈B. Then, there exists f∈SF(u) such that v(t)=∫0+∞‍G(t,s)f(s)ds. So, for any T0∈(0,+∞) and t1,t2∈[0,T0], without loss of generality, we may assume that t2>t1 and one can get the following:
(33)|v(t1)1+t1α-1-v(t2)1+t2α-1|≤∫0+∞‍|G1(t1,s)1+t1α-1-G1(t2,s)1+t2α-1|f(s)ds+∑i=1m-2‍βiΓ(α)-Δ|t1α-11+t1α-1-t2α-11+t2α-1|×∫0+∞‍G1(ξi,s)f(s)ds≤∫0+∞‍|G1(t1,s)1+t1α-1-G1(t2,s)1+t1α-1|φ(s)ψ(∥u∥)ds+∫0+∞‍|G1(t2,s)1+t1α-1-G1(t2,s)1+t2α-1|φ(s)ψ(∥u∥)ds+∑i=1m-2‍βiξm-2α-1Γ(Γ(α)-Δ|t1α-11+t1α-1-t2α-11+t2α-1|×∫0+∞‍φ(s)ψ(∥u∥)ds.
On the other hand, we get(34)∫0+∞‍|G1(t1,s)1+t1α-1-G1(t2,s)1+t1α-1|φ(s)ψ(∥u∥)ds≤(∫0t1‍+∫t1t2‍+∫t2+∞‍)|G1(t1,s)1+t1α-1-G1(t2,s)1+t1α-1|×φ(s)ψ(∥u∥)ds≤ψ(m)∫0t1‍(t2α-1-t1α-1)+((t2-s)α-1-(t1-s)α-1)1+t1α-1≤ψ(m)∫0t1‍×φ(s)ds+ψ(m)∫t1t2‍(t2α-1-t1α-1)+(t2-s)α-11+t1α-1φ(s)ds+ψ(m)∫t2+∞‍(t2α-1-t1α-1)1+t1α-1φ(s)ds→0uniformlyast1→t2.
Similar to (34), we have
(35)∫0+∞‍|G1(t2,s)1+t1α-1-G1(t2,s)1+t2α-1|φ(s)ψ(∥u∥)ds→0uniformlyast1→t2.
From (34) and (35), we have
(36)|v(t1)1+t1α-1-v(t2)1+t2α-1|→0uniformlyast1→t2.
Similar to (36), one can get
(37)|v′(t1)1+t1α-1-v′(t2)1+t2α-1|→0uniformlyast1→t2.
Therefore, T(B) is an equicontinuous set in X.
Claim
4. We show that T is equiconvergent at ∞. Let v∈T(u) for some u∈B. Then, there exists f∈SF(u) such that v(t)=∫0+∞‍G(t,s)f(s)ds. So, we have the following:
(38)∫0+∞‍f(s)ds≤ψ(m)∫0+∞‍φ(s)ds<+∞,limt→+∞|v(t)1+tα-1|=limt→+∞11+tα-1∫0+∞‍G(t,s)f(s)ds=∑i=1m-2‍βiξiα-1Γ(α)(Γ(α)-Δ)∫0+∞‍f(s)ds-∑i=1m-2‍βiΓ(α)(Γ(α)-Δ)×∫0ξi‍(ξi-s)α-1f(s)ds<∞,
and, similarly, one has
(39)limt→+∞|v′(t)1+tα-1|<∞.
Therefore, T(B) is equiconvergent at infinity.
Therefore, with Lemma11, Lemma 19 and Claims 2–4, we conclude that T is completely continuous.
Claim
5.
T is upper semicontinuous. To this end, it is sufficient to show that T has a closed graph. Let vn∈T(un)  such that vn→v and un→u, as n→+∞. Then, there exists m>0 such that ∥un∥≤m. We will prove that v∈T(u) means that there exists fn∈SF(un) such that, for a.e. t∈I, we have vn(t)=∫0+∞‍G(t,s)fn(s)ds. Then, we need to show that v∈T(u).
Condition (H1) implies thatfn(t)∈φ(t)ψ(m)B1(0). Then, {fn}n∈ℕ is integrably bounded in L1(I,ℝ). Since F has compact values, we deduce that {fn}n is semicompact. By Lemma 9, there exists a subsequence, still denoted as {fn}n∈ℕ, which converges weakly to some limit f∈L1(I,ℝ). Moreover, the mapping Γ:L1(I,ℝ)→X=E defined by
(40)Γ(g)(t)=∫0+∞‍G(t,s)g(s)ds
is a continuous linear operator. Then, it remains continuous if these spaces are endowed with their weak topologies [29, 33]. Moreover, for a.e. t∈I, un(t) converges to u(t). Then, we have
(41)v(t)=∫0+∞‍G(t,s)f(s)ds.
It remains to prove that f∈F(t,u(t),u′(t)), a.e. t∈I. Mazur’s Lemma (see Lemma 10) yields the existence of αin≥0, i=n,…,k(n) such that ∑i=1k(n)‍αin=1 and the sequence of convex combinations gn(·)=∑i=1k(n)‍αinfi(·) converges strongly to f in L1. Using Lemma 6, we obtain that
(42)v(t)∈⋂n≥1‍{gn(t)}-,a.e.t∈I⊂⋂n≥1‍co-{fk(t),k≥n}⊂⋂n≥1‍co-{⋃n≥1‍F(t,uk(t),uk′(t))}=co-(limsupk→+∞F(t,uk(t),uk′(t))).
However, the fact that the multivalued x→F(·,x,x′) is upper semicontinuous and has compact values, together with Lemma 7, implies that
(43)limsupn→+∞F(t,un(t),un′(t))=F(t,u(t),u′(t)),a.e.t∈I.
This along with (42) yields that f(t)∈co-F(t,u(t),u′(t)). Finally, F(·,·,·) has closed, convex values; hence, f(t)∈F(t,u(t),u′(t)), a.e. t∈I. Thus, v∈T(u), proving that T has a closed graph. Finally, with Lemma 1 and the compactness of T, we conclude that T is upper semicontinuous.
Claim
6. A priori bounds on solutions. Let u be a solution of (1). Then, there exists f∈L1([0,+∞),ℝ) with f∈SF(u) such that u(t)=∫0+∞‍G(t,s)f(s)ds. In view of (H1), and using the computations in Claim 2 above, for each t∈[0,+∞), we obtain
(44){|u(t)1+tα-1|,|u′(t)1+tα-1|}≤max{L1,L2}∫0+∞‍f(s)ds≤max{L1,L2}ψ(∥u∥)×∫0+∞‍φ(s)ds.
Consequently,
(45)∥u∥max{L1,L2}ψ(∥u∥)∫0+∞‍φ(s)ds≤1.
In view of (H2), there existsM such that ∥u∥≠M. Let us  consider the following:
(46)U:={u∈AC1([0,+∞),ℝ):∥u∥<M}.
Note that the operatorT:U-→𝒫(AC1([0,+∞)) is upper semicontinuous and completely continuous. From the choice of U, there is no u∈∂U such that u=λT(u) for some λ∈(0,1). Consequently, by the nonlinear alternative of the Leray-Schauder type (Theorem 2), we deduce that T has a fixed point u∈U which is a solution of the problem (1). This completes the proof.


### 3.2. The Lipschitz Case

Now we prove the existence of solutions for the problem (1) with a nonconvex-valued right hand side by applying a fixed point theorem for multivalued maps according to Covitz and Nadler [34].Definition 22.
A multivalued operatorN:X→𝒫cl(X) is called the following:(a)
γ-Lipschitz if and only if there exists γ>0 such that dH(N(x),N(y))≤γd(x,y) for each x,y∈X;(b)
a contraction if and only if it isγ-Lipschitz with γ<1.Lemma 23 (Covitz-Nadler, [34]).
Let(X,d) be a complete metric space. If N:X→𝒫cl(X) is a contraction, then FixN≠∅.Definition 24.
A measurable multivalued functionF:[0,+∞)→𝒫(X) is said to be integrably bounded if there exists a function f∈L1([0,+∞),X) such that for all v∈F(t), ∥v∥≤f(t) for a.e. t∈[0,+∞).Theorem 25.
Assume that the following condition holds:(H4)
F:I×ℝ×ℝ→𝒫cp(ℝ)  such that F(·,x,y):[0,+∞)→𝒫cp(ℝ) is measurable for each x,y∈ℝ;(H5)
There existl1,l2:[0,+∞)→[0,+∞) which are not identical zero on any closed subinterval of [0,+∞), and
(47)∫0+∞‍(1+sα-1)li(s)ds<+∞,i=1,2,such that for almost all t∈[0,+∞),
(48)dH(F(t,x1,y1),F(t,x2,y2))≤l1(t)|x1-x2|+l2(t)|y1-y2|
for all x1,x2,y1, and y2∈ℝ with d(0,F(t,0,0))≤l1(t)+l2(t) for almost all t∈[0,+∞).
Then, the boundary value problem (1) has at least one solution on I=[0,+∞) if
(49)max{L1,L2}(∫0+∞‍(1+sα-1)(l1(s)+l2(s))ds)<1.Proof.
We transform problem (1) into a fixed point problem. Consider the set-valued map T:AC1[0,+∞)→𝒫(AC1[0,+∞)) defined at the beginning of the proof of Theorem 21. It is clear that the fixed point of T are solutions of the problem (1).
Note that since the set-valued mapF(·,u(·)) is measurable with the measurable selection theorem (e.g., Theorem III.6 in [35]) it admits a measurable selection f:I→ℝ. Moreover, since F is integrably bounded, f∈L1([0,+∞),ℝ). Therefore, SF(u)≠∅.
We will prove thatT fulfills the assumptions of Covitz-Nadler contraction principle (Lemma 23).
First, we note that sinceSF(u)≠∅, T(u)≠∅ for any u∈AC1([0,+∞),ℝ).
Second, we prove thatT(u) is closed for any u∈AC1([0,+∞),ℝ). Let {un}n≥0∈T(u) such that un→u0 in AC1([0,+∞),ℝ). Then u0∈AC1([0,+∞),ℝ) and there exists fn∈SF(u) such that
(50)un(t)=∫0+∞‍G(t,s)fn(s)ds.
Since F has compact values, we may pass onto a subsequence (if necessary) to obtain that fn converges to f∈L1(([0,+∞),ℝ)) in L1(([0,+∞),ℝ)). In particular, f∈SF(u), and for any t∈[0,+∞) we have
(51)un(t)→u0(t)=∫0+∞‍G(t,s)f(s)ds,
that is, u0∈T(u) and T(u) is closed.
Next we show thatT is a contraction on AC1([0,+∞),ℝ). Let u1,u2∈AC1([0,+∞),ℝ) and v1∈T(u1). Then there exist f1∈SF(u1) such that
(52)v1(t)=∫0+∞‍G(t,s)f1(s)ds,t∈[0,+∞).
Consider the set-valued map
(53)H(t):=F(t,u2(t),u2′(t))∩{u∈ℝ;|f1(t)-u|≤l1(t)|x1-x2|+l2(t)|x1′-x2′|},t∈[0,+∞).
By (H5), we have
(54)dH(F(t,x1,x1′),F(t,x2,x2′))≤l1(t)|x1-x2|+l2(t)|x1′-x2′|,
hence H has nonempty closed values. Moreover, since H is measurable (e.g., Proposition III.4 in [35]), there exists f2 which is a measurable selection of H. It follows that f2∈SF(u2) and for any t∈[0,+∞),
(55)|f1(t)-f2(t)|≤l1(t)|x1-x2|+l2(t)|x1′-x2′|.
Define
(56)v2(t)=∫0+∞‍G(t,s)f2(s)ds,t∈[0,+∞),
and one can get
(57)|v1(t)1+tα-1-v2(t)1+tα-1|≤∫0+∞‍G(t,s)1+tα-1|f1(s)-f1(s)|ds≤L1∫0+∞‍[l1(s)|x1(s)-x2(s)|+l2(s)|x1′(s)-x2′(s)|]ds≤L1∫0+∞‍(1+sα-1)[l1(s)|x1(s)-x2(s)1+sα-1|≤L1∫0+∞‍(1+sα-1)+l2(s)|x1′(s)-x2′(s)1+sα-1|]ds≤max{L1,L2}∥x1-x2∥∫0+∞‍(1+sα-1)(l1(s)+l2(s))ds.
Similarly, we have
(58)|v1′(t)1+tα-1-v2′(t)1+tα-1|≤max{L1,L2}∥x1-x2∥×∫0+∞‍(1+sα-1)(l1(s)+l2(s))ds.
Therefore,
(59)∥v1-v2∥≤max{L1,L2}∥x1-x2∥×∫0+∞‍(1+sα-1)(l1(s)+l2(s))ds.
From an analogous reasoning by interchanging the roles of u1 and u2, it follows that
(60)dH(T(u1),T(u2))≤max{L1,L2}∥x1-x2∥×∫0+∞‍(1+sα-1)(l1(s)+l2(s))ds.
Since T is a contraction, it follows by the Lemma 23 that T admits a fixed point which is a solution to problem (1).


## 3.1. The Upper Semicontinuous Case

To obtain the complete continuity of existence solutions operator, the following lemma is still needed.Lemma 19 (see [32]).
LetV={u∈E∣∥u∥<l}(l>0),V1={u(t)/(1+tα-1):u∈V}. If   V1 is equicontinuous on any compact interval of [0,+∞) and equiconvergent at infinity, then V is relatively compact on E.Definition 20.
V
1 is called equiconvergent at infinity if and only if for all ϵ>0, there exists v(ϵ)>0 such that for all u∈V1,t1,t2≥v, it holds
(24)|u(t1)1+t1α-1-u(t2)1+t2α-1|<ϵ.Theorem 21.
The Carathéodory multivalued mapF:I×ℝ×ℝ→𝒫(ℝ) has nonempty, compact, convex values and satisfies the following.(H1)
There exists a continuous nondecreasing functionψ:[0,+∞)→(0,+∞) and φ∈L1([0,+∞), ℝ+) such that∥F(t,x,y)∥𝒫≔sup{|v(t)/(1+tα-1)|:v∈F(t,x,y)}≤φ(t)ψ(∥x∥),fora.e.t∈I  and  each  x,y∈ℝ.(H2)
There exists a constantM>0 such that
(25)Mmax{L1,L2}ψ(M)∫0+∞‍φ(s)ds>1.
Then, problem (1) has at least one solution.Proof.
LetX=E and consider M>0 as in (25). It is obvious that the existence of solutions to problem (1) is reduced to the existence of the solutions of the integral inclusion
(26)u(t)∈∫0+∞‍G(t,s)F(s,u(s),u′(s))ds,t∈I,
where G(t,s) is defined by (16) and (17). Consider the set-valued map, T:E→𝒫(X) is defined by
(27)T(u):={v∈X;v(t)=∫0+∞‍G(t,s)f(s)ds,f∈SF(u)-}.
We show thatT satisfies the hypotheses of Theorem 2.
Claim
1.We show that T(u)⊂X is convex for any u∈X. If v1,v2∈T(u), then, there exist f1,f2∈SF(u) such that for any t∈I one has
(28)vi(t)=∫0+∞‍G(t,s)fi(s)ds,i=1,2.
Let 0≤λ≤1. Then, for any t∈I, we have
(29)(λv1+(1-λ)v2)(t)=∫0+∞‍G(t,s)[λf1(s)+(1-λ)f2(s)]ds.
The values of F are convex; thus, SF(u) is a convex set and hence λv1+(1-λ)v2∈T(u).
Claim
2. We show that T is bounded on bounded sets of X. Let B be any bounded subset of X. Then, there exists m>0 such that ∥u∥≤m for all u∈B. If v∈T(u), then there exists f∈SF(u) such that v(t)=∫0+∞‍G(t,s)f(s)ds. One may write the following for any t∈I:
(30)|v(t)1+tα-1|≤∫0+∞‍|G(t,s)1+tα-1||f(s)|ds≤L1∫0+∞‍φ(s)ψ(∥u∥)ds≤L1|φ|1ψ(m).
On the other hand,
(31)|v′(t)1+tα-1|≤∫0+∞‍∂G(t,s)/∂t1+tα-1f(s)ds≤L2∫0+∞‍φ(s)ψ(∥u∥)ds≤L2|φ|1ψ(m),
and therefore
(32)∥v∥=maxt∈I{|v(t)1+tα-1|,|v′(t)1+tα-1|}≤max{L1,L2}|φ|1ψ(m),
for all v∈T(u); that is, T(B) is bounded.
Claim
3.We show that T maps bounded the sets into equi-continuous sets. Let B be any bounded subset of X as before and v∈T(u) for some u∈B. Then, there exists f∈SF(u) such that v(t)=∫0+∞‍G(t,s)f(s)ds. So, for any T0∈(0,+∞) and t1,t2∈[0,T0], without loss of generality, we may assume that t2>t1 and one can get the following:
(33)|v(t1)1+t1α-1-v(t2)1+t2α-1|≤∫0+∞‍|G1(t1,s)1+t1α-1-G1(t2,s)1+t2α-1|f(s)ds+∑i=1m-2‍βiΓ(α)-Δ|t1α-11+t1α-1-t2α-11+t2α-1|×∫0+∞‍G1(ξi,s)f(s)ds≤∫0+∞‍|G1(t1,s)1+t1α-1-G1(t2,s)1+t1α-1|φ(s)ψ(∥u∥)ds+∫0+∞‍|G1(t2,s)1+t1α-1-G1(t2,s)1+t2α-1|φ(s)ψ(∥u∥)ds+∑i=1m-2‍βiξm-2α-1Γ(Γ(α)-Δ|t1α-11+t1α-1-t2α-11+t2α-1|×∫0+∞‍φ(s)ψ(∥u∥)ds.
On the other hand, we get(34)∫0+∞‍|G1(t1,s)1+t1α-1-G1(t2,s)1+t1α-1|φ(s)ψ(∥u∥)ds≤(∫0t1‍+∫t1t2‍+∫t2+∞‍)|G1(t1,s)1+t1α-1-G1(t2,s)1+t1α-1|×φ(s)ψ(∥u∥)ds≤ψ(m)∫0t1‍(t2α-1-t1α-1)+((t2-s)α-1-(t1-s)α-1)1+t1α-1≤ψ(m)∫0t1‍×φ(s)ds+ψ(m)∫t1t2‍(t2α-1-t1α-1)+(t2-s)α-11+t1α-1φ(s)ds+ψ(m)∫t2+∞‍(t2α-1-t1α-1)1+t1α-1φ(s)ds→0uniformlyast1→t2.
Similar to (34), we have
(35)∫0+∞‍|G1(t2,s)1+t1α-1-G1(t2,s)1+t2α-1|φ(s)ψ(∥u∥)ds→0uniformlyast1→t2.
From (34) and (35), we have
(36)|v(t1)1+t1α-1-v(t2)1+t2α-1|→0uniformlyast1→t2.
Similar to (36), one can get
(37)|v′(t1)1+t1α-1-v′(t2)1+t2α-1|→0uniformlyast1→t2.
Therefore, T(B) is an equicontinuous set in X.
Claim
4. We show that T is equiconvergent at ∞. Let v∈T(u) for some u∈B. Then, there exists f∈SF(u) such that v(t)=∫0+∞‍G(t,s)f(s)ds. So, we have the following:
(38)∫0+∞‍f(s)ds≤ψ(m)∫0+∞‍φ(s)ds<+∞,limt→+∞|v(t)1+tα-1|=limt→+∞11+tα-1∫0+∞‍G(t,s)f(s)ds=∑i=1m-2‍βiξiα-1Γ(α)(Γ(α)-Δ)∫0+∞‍f(s)ds-∑i=1m-2‍βiΓ(α)(Γ(α)-Δ)×∫0ξi‍(ξi-s)α-1f(s)ds<∞,
and, similarly, one has
(39)limt→+∞|v′(t)1+tα-1|<∞.
Therefore, T(B) is equiconvergent at infinity.
Therefore, with Lemma11, Lemma 19 and Claims 2–4, we conclude that T is completely continuous.
Claim
5.
T is upper semicontinuous. To this end, it is sufficient to show that T has a closed graph. Let vn∈T(un)  such that vn→v and un→u, as n→+∞. Then, there exists m>0 such that ∥un∥≤m. We will prove that v∈T(u) means that there exists fn∈SF(un) such that, for a.e. t∈I, we have vn(t)=∫0+∞‍G(t,s)fn(s)ds. Then, we need to show that v∈T(u).
Condition (H1) implies thatfn(t)∈φ(t)ψ(m)B1(0). Then, {fn}n∈ℕ is integrably bounded in L1(I,ℝ). Since F has compact values, we deduce that {fn}n is semicompact. By Lemma 9, there exists a subsequence, still denoted as {fn}n∈ℕ, which converges weakly to some limit f∈L1(I,ℝ). Moreover, the mapping Γ:L1(I,ℝ)→X=E defined by
(40)Γ(g)(t)=∫0+∞‍G(t,s)g(s)ds
is a continuous linear operator. Then, it remains continuous if these spaces are endowed with their weak topologies [29, 33]. Moreover, for a.e. t∈I, un(t) converges to u(t). Then, we have
(41)v(t)=∫0+∞‍G(t,s)f(s)ds.
It remains to prove that f∈F(t,u(t),u′(t)), a.e. t∈I. Mazur’s Lemma (see Lemma 10) yields the existence of αin≥0, i=n,…,k(n) such that ∑i=1k(n)‍αin=1 and the sequence of convex combinations gn(·)=∑i=1k(n)‍αinfi(·) converges strongly to f in L1. Using Lemma 6, we obtain that
(42)v(t)∈⋂n≥1‍{gn(t)}-,a.e.t∈I⊂⋂n≥1‍co-{fk(t),k≥n}⊂⋂n≥1‍co-{⋃n≥1‍F(t,uk(t),uk′(t))}=co-(limsupk→+∞F(t,uk(t),uk′(t))).
However, the fact that the multivalued x→F(·,x,x′) is upper semicontinuous and has compact values, together with Lemma 7, implies that
(43)limsupn→+∞F(t,un(t),un′(t))=F(t,u(t),u′(t)),a.e.t∈I.
This along with (42) yields that f(t)∈co-F(t,u(t),u′(t)). Finally, F(·,·,·) has closed, convex values; hence, f(t)∈F(t,u(t),u′(t)), a.e. t∈I. Thus, v∈T(u), proving that T has a closed graph. Finally, with Lemma 1 and the compactness of T, we conclude that T is upper semicontinuous.
Claim
6. A priori bounds on solutions. Let u be a solution of (1). Then, there exists f∈L1([0,+∞),ℝ) with f∈SF(u) such that u(t)=∫0+∞‍G(t,s)f(s)ds. In view of (H1), and using the computations in Claim 2 above, for each t∈[0,+∞), we obtain
(44){|u(t)1+tα-1|,|u′(t)1+tα-1|}≤max{L1,L2}∫0+∞‍f(s)ds≤max{L1,L2}ψ(∥u∥)×∫0+∞‍φ(s)ds.
Consequently,
(45)∥u∥max{L1,L2}ψ(∥u∥)∫0+∞‍φ(s)ds≤1.
In view of (H2), there existsM such that ∥u∥≠M. Let us  consider the following:
(46)U:={u∈AC1([0,+∞),ℝ):∥u∥<M}.
Note that the operatorT:U-→𝒫(AC1([0,+∞)) is upper semicontinuous and completely continuous. From the choice of U, there is no u∈∂U such that u=λT(u) for some λ∈(0,1). Consequently, by the nonlinear alternative of the Leray-Schauder type (Theorem 2), we deduce that T has a fixed point u∈U which is a solution of the problem (1). This completes the proof.


## 3.2. The Lipschitz Case

Now we prove the existence of solutions for the problem (1) with a nonconvex-valued right hand side by applying a fixed point theorem for multivalued maps according to Covitz and Nadler [34].Definition 22.
A multivalued operatorN:X→𝒫cl(X) is called the following:(a)
γ-Lipschitz if and only if there exists γ>0 such that dH(N(x),N(y))≤γd(x,y) for each x,y∈X;(b)
a contraction if and only if it isγ-Lipschitz with γ<1.Lemma 23 (Covitz-Nadler, [34]).
Let(X,d) be a complete metric space. If N:X→𝒫cl(X) is a contraction, then FixN≠∅.Definition 24.
A measurable multivalued functionF:[0,+∞)→𝒫(X) is said to be integrably bounded if there exists a function f∈L1([0,+∞),X) such that for all v∈F(t), ∥v∥≤f(t) for a.e. t∈[0,+∞).Theorem 25.
Assume that the following condition holds:(H4)
F:I×ℝ×ℝ→𝒫cp(ℝ)  such that F(·,x,y):[0,+∞)→𝒫cp(ℝ) is measurable for each x,y∈ℝ;(H5)
There existl1,l2:[0,+∞)→[0,+∞) which are not identical zero on any closed subinterval of [0,+∞), and
(47)∫0+∞‍(1+sα-1)li(s)ds<+∞,i=1,2,such that for almost all t∈[0,+∞),
(48)dH(F(t,x1,y1),F(t,x2,y2))≤l1(t)|x1-x2|+l2(t)|y1-y2|
for all x1,x2,y1, and y2∈ℝ with d(0,F(t,0,0))≤l1(t)+l2(t) for almost all t∈[0,+∞).
Then, the boundary value problem (1) has at least one solution on I=[0,+∞) if
(49)max{L1,L2}(∫0+∞‍(1+sα-1)(l1(s)+l2(s))ds)<1.Proof.
We transform problem (1) into a fixed point problem. Consider the set-valued map T:AC1[0,+∞)→𝒫(AC1[0,+∞)) defined at the beginning of the proof of Theorem 21. It is clear that the fixed point of T are solutions of the problem (1).
Note that since the set-valued mapF(·,u(·)) is measurable with the measurable selection theorem (e.g., Theorem III.6 in [35]) it admits a measurable selection f:I→ℝ. Moreover, since F is integrably bounded, f∈L1([0,+∞),ℝ). Therefore, SF(u)≠∅.
We will prove thatT fulfills the assumptions of Covitz-Nadler contraction principle (Lemma 23).
First, we note that sinceSF(u)≠∅, T(u)≠∅ for any u∈AC1([0,+∞),ℝ).
Second, we prove thatT(u) is closed for any u∈AC1([0,+∞),ℝ). Let {un}n≥0∈T(u) such that un→u0 in AC1([0,+∞),ℝ). Then u0∈AC1([0,+∞),ℝ) and there exists fn∈SF(u) such that
(50)un(t)=∫0+∞‍G(t,s)fn(s)ds.
Since F has compact values, we may pass onto a subsequence (if necessary) to obtain that fn converges to f∈L1(([0,+∞),ℝ)) in L1(([0,+∞),ℝ)). In particular, f∈SF(u), and for any t∈[0,+∞) we have
(51)un(t)→u0(t)=∫0+∞‍G(t,s)f(s)ds,
that is, u0∈T(u) and T(u) is closed.
Next we show thatT is a contraction on AC1([0,+∞),ℝ). Let u1,u2∈AC1([0,+∞),ℝ) and v1∈T(u1). Then there exist f1∈SF(u1) such that
(52)v1(t)=∫0+∞‍G(t,s)f1(s)ds,t∈[0,+∞).
Consider the set-valued map
(53)H(t):=F(t,u2(t),u2′(t))∩{u∈ℝ;|f1(t)-u|≤l1(t)|x1-x2|+l2(t)|x1′-x2′|},t∈[0,+∞).
By (H5), we have
(54)dH(F(t,x1,x1′),F(t,x2,x2′))≤l1(t)|x1-x2|+l2(t)|x1′-x2′|,
hence H has nonempty closed values. Moreover, since H is measurable (e.g., Proposition III.4 in [35]), there exists f2 which is a measurable selection of H. It follows that f2∈SF(u2) and for any t∈[0,+∞),
(55)|f1(t)-f2(t)|≤l1(t)|x1-x2|+l2(t)|x1′-x2′|.
Define
(56)v2(t)=∫0+∞‍G(t,s)f2(s)ds,t∈[0,+∞),
and one can get
(57)|v1(t)1+tα-1-v2(t)1+tα-1|≤∫0+∞‍G(t,s)1+tα-1|f1(s)-f1(s)|ds≤L1∫0+∞‍[l1(s)|x1(s)-x2(s)|+l2(s)|x1′(s)-x2′(s)|]ds≤L1∫0+∞‍(1+sα-1)[l1(s)|x1(s)-x2(s)1+sα-1|≤L1∫0+∞‍(1+sα-1)+l2(s)|x1′(s)-x2′(s)1+sα-1|]ds≤max{L1,L2}∥x1-x2∥∫0+∞‍(1+sα-1)(l1(s)+l2(s))ds.
Similarly, we have
(58)|v1′(t)1+tα-1-v2′(t)1+tα-1|≤max{L1,L2}∥x1-x2∥×∫0+∞‍(1+sα-1)(l1(s)+l2(s))ds.
Therefore,
(59)∥v1-v2∥≤max{L1,L2}∥x1-x2∥×∫0+∞‍(1+sα-1)(l1(s)+l2(s))ds.
From an analogous reasoning by interchanging the roles of u1 and u2, it follows that
(60)dH(T(u1),T(u2))≤max{L1,L2}∥x1-x2∥×∫0+∞‍(1+sα-1)(l1(s)+l2(s))ds.
Since T is a contraction, it follows by the Lemma 23 that T admits a fixed point which is a solution to problem (1).


## 4. Application

Consider the fractional boundary value problem,(61)D0+5/2u(t)+F(t,u(t),u′(t))∋0,0<t<+∞,u(0)=u′(0)=0,D3/2u(+∞)-18u(18)-14u(1)=0.
Here m=4,α=5/2, β1=1/8,β2=1/4,ξ1=1/8,  and ξ2=1, and F:I×ℝ×ℝ→𝒫(ℝ) is a multivalued map given by
(62)F(t,x,y)=[e-t(1+t3/2)(|x+y|5|x+y|5+5+9),2e-t(1+t3/2)(|x+y|3|x+y|3+3+1)].
For v∈F, we have
(63)|v(t)1+t3/2|≤max(e-t(|x+y|5|x+y|5+5+9),2e-t(|x+y|3|x+y|3+3+1))≤10e-t,x,y∈ℝ.
Thus,
(64)∥F(t,x,y)∥𝒫:=sup{|v(t)1+t3/2|:v∈F(t,x,y)}≤10e-t=φ(t)ψ(∥x∥),x,y∈ℝ,
with φ(t)=e-t, ψ(∥x∥)=10.Also, by direct calculation, we can obtain thatL1=1.01529 and L2=3.045869. Further, by using the following condition:
(65)Mmax{L1,L2}ψ(M)∫0+∞‍φ(s)ds>1,
we find that M>30.45869. Clearly, all the conditions of Theorem 21 are satisfied. So, there exists at least one solution of problem (1) on I.



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*Source: 823961-2013-03-28.xml*