# Existence Results for Langevin Fractional Differential Inclusions Involving Two Fractional Orders with Four-Point Multiterm Fractional Integral Boundary Conditions

Authors: Ahmed Alsaedi; Sotiris K. Ntouyas; Bashir Ahmad
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We discuss the existence of solutions for Langevin fractional differential inclusions involving two fractional orders with four-point multiterm fractional integral boundary conditions. Our study relies on standard fixed point theorems for multivalued maps and covers the cases when the right-hand side of the inclusion has convex as well as nonconvex values. Illustrative examples are also presented.


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## Body

## 1. Introduction

We consider a boundary value problem of nonlinear Langevin fractional differential inclusions involving two fractional orders with four-point multiterm fractional integral boundary conditions given by(1)cDp(cDq+λ)x(t)∈F(t,x(t)),0<t<1,x(0)=∑i=1n‍βi(Iμix)(ζ),x(1)=∑i=1n‍αi(Iνix)(η),0<ζ<η<1,
where 0<p,q≤1, cDq denotes the Caputo fractional derivative of order q, λ is a real number, Iκ is the Riemann-Liouville fractional integral of order κ>0  (κ=νi,μi;i=1,2,…,n,n∈ℕ={1,2,…}), and αi,βi∈ℝ are suitably chosen constants.In recent years, the boundary value problems of fractional order differential equations have emerged as an important area of research, since these problems have applications in various disciplines of science and engineering such as mechanics, electricity, chemistry, biology, economics, control theory, signal and image processing, polymer rheology, regular variation in thermodynamics, biophysics, aerodynamics, viscoelasticity and damping, electrodynamics of complex medium, wave propagation, and blood flow phenomena [1–5]. Many researchers have studied the existence theory for nonlinear fractional differential equations with a variety of boundary conditions; for instance, see the papers [6–17] and the references therein.The Langevin equation (first formulated by Langevin in 1908) is found to be an effective tool to describe the evolution of physical phenomena in fluctuating environments [18]. For some new developments on the fractional Langevin equation, see, for example, [19–26].The main objective of this paper is to develop the existence theory for a class of problems of the type (1), when the right-hand side is convex as well as nonconvex valued. We establish three existence results: the first result is obtained by means of the nonlinear alternative of Leray-Schauder type; the second one relies on the nonlinear alternative of Leray-Schauder type for single-valued maps together with a selection theorem due to Bressan and Colombo for lower semicontinuous multivalued maps with nonempty closed and decomposable values; and a fixed point theorem due to Covitz and Nadler for contraction multivalued maps is applied to get the third result. The methods used are well known; however their exposition in the framework of problem (1) is new.The paper is organized as follows: in Section2, we recall some preliminary facts that we need in the sequel, and Section 3 contains our main results.


## 2. Preliminaries

### 2.1. Fractional Calculus

Let us recall some basic definitions of fractional calculus [1–3].Definition 1.
For at leastn-times differentiable function g:[0,∞)→ℝ, the Caputo derivative of fractional order q is defined as
(2)cDqg(t)=1Γ(n-q)∫0t‍(t-s)n-q-1g(n)(s)ds,n-1<q<n,n=[q]+1,
where [q] denotes the integer part of the real number q.Definition 2.
The Riemann-Liouville fractional integral of orderq is defined as
(3)Iqg(t)=1Γ(q)∫0t‍g(s)(t-s)1-qds,q>0,
provided the integral exists.Lemma 3.
Forq>0, the general solution of the fractional differential equation Dqx(t)=0 is given by
(4)x(t)=c0+c1t+⋯+cn-1tn-1,
where ci∈ℝ,i=1,2,…,n-1  (n=[q]+1).In view of Lemma3, it follows that
(5)IqDqx(t)=x(t)+c0+c1t+⋯+cn-1tn-1,
for some ci∈ℝ,i=1,2,…,n-1  (n=[q]+1).In the following,AC1([0,T],ℝ) will denote the space of functions x:[0,T]→ℝ that are absolutely continuous and whose first derivative is absolutely continuous.Definition 4.
A functionx∈AC1([0,T],ℝ) is called a solution of problem (1) if there exists a function v∈L1([0,T],ℝ) with v(t)∈F(t,x(t)), a.e. [0,T] such that cDp(cDq+λ)x(t)=v(t), a.e. [0,T], and x(0)=∑i=1n‍βi(Iμix)(ζ),x(1)=∑i=1n‍αi(Iνix)(η).Lemma 5.
Leth∈C[0,1]. Then the boundary value problem
(6)cDp(cDq+λ)x(t)=h(t),0<t<1,x(0)=∑i=1n‍βi(Iμix)(ζ),x(1)=∑i=1n‍αi(Iνix)(η),0<ζ<η<1,
has a unique solution(7)x(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s)∫0s‍(s-r)p-1Γ(p)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s)}ds)du,where
(8)Δ=Δ1Δ2-Δ3Δ4≠0,Δ1=∑i=1n‍βiζμi+qΓ(μi+q+1),Δ2=(1-∑i=1n‍αiηνiΓ(νi+1)),Δ3=(1-∑i=1n‍βiζμiΓ(μi+1)),Δ4=(∑i=1n‍αiηνi+qΓ(νi+q+1)-1Γ(q+1)).Proof.
As argued in [23], the solution of cDp(cDq+λ)x(t)=h(t) can be written as
(9)x(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s))ds-c0Γ(q+1)tq-c1.
Using the given conditions in (9) together with (8), we find that
(10)Δ3c1-Δ1c0=-𝒥1,Δ2c1-Δ4c0=𝒥2,
where(11)𝒥1=∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s)}ds)du,𝒥2=∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s)}ds)du.Solving (10) for c0 and c1, we find that
(12)c0=1Δ(Δ2𝒥1+Δ3𝒥2),c1=1Δ(Δ4𝒥1+Δ1𝒥2).
Substituting these values in (9), we find the desired solution.In order to simplify the computations in the main results, we present a technical lemma, concerning the bounds of the operators𝒥1 and 𝒥2 defined in the proof of the above lemma.Lemma 6.
One has(13)|𝒥1|≤∑i=1n‍|βi|ζμi+p+q∥h∥Γ(μi+p+q+1)+∑i=1n‍|βi|ζμi+q|λ|∥x∥Γ(μi+q+1),|𝒥2|≤∥h∥Γ(p+q+1)+|λ|∥x∥Γ(q+1)+∑i=1n‍|αi|ηνi+p+q∥h∥Γ(νi+p+q+1)+∑i=1n‍|αi|ηνi+q|λ|∥x∥Γ(νi+q+1).Proof.
By using the following property of beta function(14)B(b+1,a)=∫01‍(1-s)a-1sbds=Γ(a)Γ(b+1)Γ(a+b+1),
we have(15)|𝒥1|≤∑i=1n‍|βi|×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)|h(r)|dr+|λ||x(s)|}ds)du≤∑i=1n‍|βi|∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{spΓ(p+1)∥h∥+|λ|∥x∥}ds)du≤∑i=1n‍|βi|∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1sp∥h∥Γ(q)Γ(p+1)ds+∫0u‍(u-s)q-1Γ(q)|λ|∥x∥ds)du≤∑i=1n‍|βi|∫0ζ‍(ζ-u)μi-1Γ(μi)up+q∥h∥Γ(q)Γ(p+1)×B(p+1,q)du+∑i=1n‍|βi|∫0ζ‍(ζ-u)μi-1Γ(μi)uqΓ(q+1)|λ|∥x∥du≤∑i=1n‍|βi|ζμi+p+q∥h∥Γ(μi)Γ(q)Γ(p+1)×B(p+1,q)B(p+q+1,μi)+∑i=1n‍|βi|ζμi+qΓ(μi)Γ(q+1)B(q+1,μi)|λ|∥x∥=∑i=1n‍|βi|ζμi+p+q∥h∥Γ(μi+p+q+1)+∑i=1n‍|βi|ζμi+q|λ|∥x∥Γ(μi+q+1).By a similar way, we have
(16)|𝒥2|≤|v(u)|Γ(p+q+1)+|λ||x(u)|Γ(q+1)+∑i=1n‍|αi|ηνi+p+q|v(u)|Γ(νi+p+q+1)+∑i=1n‍|αi|ηνi+q|λ||x(u)|Γ(νi+q+1),
which completes the proof.In the following, for convenience, we put(17)A1=|Δ3|+|Δ1|Γ(q+1)|Δ|Γ(q+1),A2=|Δ2|+|Δ4|Γ(q+1)|Δ|Γ(q+1),Ω=1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1),Ψ=|λ|(1+A1)Γ(q+1)+|λ|A1∑i=1n‍|αi|ηνi+qΓ(νi+q+1)+A2∑i=1n‍|βi|ζμi+qΓ(μi+q+1).


### 2.2. Background Material for Multivalued Analysis

Now we recall some basic definitions on multivalued maps [27–29].LetC([0,1]) denote a Banach space of continuous functions from [0,1] into ℝ with the norm ∥x∥=supt∈[0,1]|x(t)|. Let L1([0,1],ℝ) be the Banach space of measurable functions x:[0,1]→ℝ which are Lebesgue integrable and normed by ∥x∥L1=∫01‍|x(t)|dt.For a normed space(X,∥·∥), let
(18)𝒫cl(X)={Y∈𝒫(X):Yisclosed},𝒫b(X)={Y∈𝒫(X):Yisbounded},𝒫cp(X)={Y∈𝒫(X):Yiscompact},𝒫cp,c(X)={Y∈𝒫(X):Yiscompactandconvex}.
A multivalued map G:X→𝒫(X)(i)
isconvex (closed) valued if G(x) is convex (closed) for all x∈X;(ii)
isbounded on bounded sets if G(𝔹)=∪x∈𝔹G(x) is bounded in X for all 𝔹∈𝒫b(X) (i.e., supx∈𝔹{sup{|y|:y∈G(x)}}<∞);(iii)
is calledupper semicontinuous (u.s.c.) on X if, for each x0∈X, the set G(x0) is a nonempty closed subset of X and if, for each open set N of X containing G(x0), there exists an open neighborhood 𝒩0 of x0 such that G(𝒩0)⊆N;(iv)
G is lower semicontinuous (l.s.c.) if the set {y∈X:G(y)∩B≠∅} is open for any open set B in E;(v)
is said to becompletely continuous if G(𝔹) is relatively compact for every 𝔹∈𝒫b(X);(vi)
is said to bemeasurable if, for every y∈ℝ, the function
(19)t↦d(y,G(t))=inf{|y-z|:z∈G(t)}is measurable;(vii)
has a fixed point if there is x∈X such that x∈G(x). The fixed point set of the multivalued operator G will be denoted by FixG.Definition 7.
A multivalued mapF:[0,1]×ℝ→𝒫(ℝ) is said to be Carathéodory if(i)
t↦F(t,x) is measurable for each x∈ℝ;(ii)
x↦F(t,x) is upper semicontinuous for almost all t∈[0,1];
Further a Carathéodory functionF is called L1-Carathéodory if (iii)
for eachα>0, there exists φα∈L1([0,1],ℝ+) such that
(20)∥F(t,x)∥=sup{|v|:v∈F(t,x)}≤φα(t)for all∥x∥∞≤α and for a. e. t∈[0,1].For eachx∈C([0,1],ℝ), define the set of selections of F by
(21)SF,x:={v∈L1([0,1],ℝ):v(t)∈F(t,x(t))fora.e.  t∈[0,1]}.We define the graph ofG to be the set Gr(G)={(x,y)∈X×Y,y∈G(x)} and recall two useful results regarding closed graphs and upper semicontinuity.Lemma 8 (see [27, Proposition 1.2]).
IfG:X→𝒫cl(Y) is u.s.c., then Gr(G) 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∈G(xn), then y*∈G(x*). Conversely, if G is completely continuous and has a closed graph, then it is upper semicontinuous.Lemma 9 (see [30]).
LetX be a Banach space. Let F:[0,T]×ℝ→𝒫cp,c(X) be an L1-Carathéodory multivalued map, and let Θ be a linear continuous mapping from L1([0,1],X) to C([0,1],X). Then the operator
(22)Θ∘SF:C([0,1],X)→𝒫cp,c(C([0,1],X)),x↦(Θ∘SF)(x)=Θ(SF,x,y)
is a closed graph operator in C([0,1],X)×C([0,1],X).We recall the well-known nonlinear alternative of Leray-Schauder for multivalued maps.Lemma 10 (nonlinear alternative for Kakutani maps [31]).
LetE be a Banach space, C a closed convex subset of E,U an open subset of C, and 0∈U. Suppose that F:U¯→𝒫c,cv(C) is an upper semicontinuous compact map; here 𝒫c,cv(C) denotes the family of nonempty, compact convex subsets of C. Then either(i)
F has a fixed point in U¯, or(ii)
there is au∈∂U and λ∈(0,1) with u∈λF(u).Definition 11.
LetA be a subset of [0,1]×ℝ. A is ℒ⊗ℬ measurable if A belongs to the σ-algebra generated by all sets of the form 𝒥×𝒟, where 𝒥 is Lebesgue measurable in [0,1] and 𝒟 is Borel measurable in ℝ.Definition 12.
A subset𝒜 of L1([0,1],ℝ) is decomposable if, for all u,v∈𝒜, and measurable 𝒥⊂[0,1]=J, the function uχ𝒥+vχJ-𝒥∈𝒜, where χ𝒥 stands for the characteristic function of 𝒥.Lemma 13 (see [32]).
LetY be a separable metric space, and let N:Y→𝒫(L1([0,1],ℝ)) be a lower semicontinuous (l.s.c.) multivalued operator with nonempty closed and decomposable values. Then N has a continuous selection; that is, there exists a continuous function (single-valued) g:Y→L1([0,1],ℝ) such that g(x)∈N(x) for every x∈Y.Let(X,d) be a metric space induced from the normed space (X;∥·∥). Consider Hd:𝒫(X)×𝒫(X)→ℝ∪{∞} given by
(23)Hd(A,B)=max{supa∈Ad(a,B),supb∈Bd(A,b)},
where d(A,b)=infa∈Ad(a;b) and d(a,B)=infb∈Bd(a;b). Then (𝒫b,cl(X),Hd) is a metric space (see [33]).Definition 14.
A multivalued operatorN:X→𝒫cl(X) is called(a)
γ-Lipschitz if and only if there exists γ>0 such that
(24)Hd(N(x),N(y))≤γd(x,y)foreachx,y∈X;(b)
a contraction if and only if it isγ-Lipschitz with γ<1.Lemma 15 (see [34]).
Let(X,d) be a complete metric space. If N:X→𝒫cl(X) is a contraction, then FixN≠∅.


## 2.1. Fractional Calculus

Let us recall some basic definitions of fractional calculus [1–3].Definition 1.
For at leastn-times differentiable function g:[0,∞)→ℝ, the Caputo derivative of fractional order q is defined as
(2)cDqg(t)=1Γ(n-q)∫0t‍(t-s)n-q-1g(n)(s)ds,n-1<q<n,n=[q]+1,
where [q] denotes the integer part of the real number q.Definition 2.
The Riemann-Liouville fractional integral of orderq is defined as
(3)Iqg(t)=1Γ(q)∫0t‍g(s)(t-s)1-qds,q>0,
provided the integral exists.Lemma 3.
Forq>0, the general solution of the fractional differential equation Dqx(t)=0 is given by
(4)x(t)=c0+c1t+⋯+cn-1tn-1,
where ci∈ℝ,i=1,2,…,n-1  (n=[q]+1).In view of Lemma3, it follows that
(5)IqDqx(t)=x(t)+c0+c1t+⋯+cn-1tn-1,
for some ci∈ℝ,i=1,2,…,n-1  (n=[q]+1).In the following,AC1([0,T],ℝ) will denote the space of functions x:[0,T]→ℝ that are absolutely continuous and whose first derivative is absolutely continuous.Definition 4.
A functionx∈AC1([0,T],ℝ) is called a solution of problem (1) if there exists a function v∈L1([0,T],ℝ) with v(t)∈F(t,x(t)), a.e. [0,T] such that cDp(cDq+λ)x(t)=v(t), a.e. [0,T], and x(0)=∑i=1n‍βi(Iμix)(ζ),x(1)=∑i=1n‍αi(Iνix)(η).Lemma 5.
Leth∈C[0,1]. Then the boundary value problem
(6)cDp(cDq+λ)x(t)=h(t),0<t<1,x(0)=∑i=1n‍βi(Iμix)(ζ),x(1)=∑i=1n‍αi(Iνix)(η),0<ζ<η<1,
has a unique solution(7)x(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s)∫0s‍(s-r)p-1Γ(p)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s)}ds)du,where
(8)Δ=Δ1Δ2-Δ3Δ4≠0,Δ1=∑i=1n‍βiζμi+qΓ(μi+q+1),Δ2=(1-∑i=1n‍αiηνiΓ(νi+1)),Δ3=(1-∑i=1n‍βiζμiΓ(μi+1)),Δ4=(∑i=1n‍αiηνi+qΓ(νi+q+1)-1Γ(q+1)).Proof.
As argued in [23], the solution of cDp(cDq+λ)x(t)=h(t) can be written as
(9)x(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s))ds-c0Γ(q+1)tq-c1.
Using the given conditions in (9) together with (8), we find that
(10)Δ3c1-Δ1c0=-𝒥1,Δ2c1-Δ4c0=𝒥2,
where(11)𝒥1=∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s)}ds)du,𝒥2=∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)h(r)dr-λx(s)}ds)du.Solving (10) for c0 and c1, we find that
(12)c0=1Δ(Δ2𝒥1+Δ3𝒥2),c1=1Δ(Δ4𝒥1+Δ1𝒥2).
Substituting these values in (9), we find the desired solution.In order to simplify the computations in the main results, we present a technical lemma, concerning the bounds of the operators𝒥1 and 𝒥2 defined in the proof of the above lemma.Lemma 6.
One has(13)|𝒥1|≤∑i=1n‍|βi|ζμi+p+q∥h∥Γ(μi+p+q+1)+∑i=1n‍|βi|ζμi+q|λ|∥x∥Γ(μi+q+1),|𝒥2|≤∥h∥Γ(p+q+1)+|λ|∥x∥Γ(q+1)+∑i=1n‍|αi|ηνi+p+q∥h∥Γ(νi+p+q+1)+∑i=1n‍|αi|ηνi+q|λ|∥x∥Γ(νi+q+1).Proof.
By using the following property of beta function(14)B(b+1,a)=∫01‍(1-s)a-1sbds=Γ(a)Γ(b+1)Γ(a+b+1),
we have(15)|𝒥1|≤∑i=1n‍|βi|×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)|h(r)|dr+|λ||x(s)|}ds)du≤∑i=1n‍|βi|∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{spΓ(p+1)∥h∥+|λ|∥x∥}ds)du≤∑i=1n‍|βi|∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1sp∥h∥Γ(q)Γ(p+1)ds+∫0u‍(u-s)q-1Γ(q)|λ|∥x∥ds)du≤∑i=1n‍|βi|∫0ζ‍(ζ-u)μi-1Γ(μi)up+q∥h∥Γ(q)Γ(p+1)×B(p+1,q)du+∑i=1n‍|βi|∫0ζ‍(ζ-u)μi-1Γ(μi)uqΓ(q+1)|λ|∥x∥du≤∑i=1n‍|βi|ζμi+p+q∥h∥Γ(μi)Γ(q)Γ(p+1)×B(p+1,q)B(p+q+1,μi)+∑i=1n‍|βi|ζμi+qΓ(μi)Γ(q+1)B(q+1,μi)|λ|∥x∥=∑i=1n‍|βi|ζμi+p+q∥h∥Γ(μi+p+q+1)+∑i=1n‍|βi|ζμi+q|λ|∥x∥Γ(μi+q+1).By a similar way, we have
(16)|𝒥2|≤|v(u)|Γ(p+q+1)+|λ||x(u)|Γ(q+1)+∑i=1n‍|αi|ηνi+p+q|v(u)|Γ(νi+p+q+1)+∑i=1n‍|αi|ηνi+q|λ||x(u)|Γ(νi+q+1),
which completes the proof.In the following, for convenience, we put(17)A1=|Δ3|+|Δ1|Γ(q+1)|Δ|Γ(q+1),A2=|Δ2|+|Δ4|Γ(q+1)|Δ|Γ(q+1),Ω=1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1),Ψ=|λ|(1+A1)Γ(q+1)+|λ|A1∑i=1n‍|αi|ηνi+qΓ(νi+q+1)+A2∑i=1n‍|βi|ζμi+qΓ(μi+q+1).


## 2.2. Background Material for Multivalued Analysis

Now we recall some basic definitions on multivalued maps [27–29].LetC([0,1]) denote a Banach space of continuous functions from [0,1] into ℝ with the norm ∥x∥=supt∈[0,1]|x(t)|. Let L1([0,1],ℝ) be the Banach space of measurable functions x:[0,1]→ℝ which are Lebesgue integrable and normed by ∥x∥L1=∫01‍|x(t)|dt.For a normed space(X,∥·∥), let
(18)𝒫cl(X)={Y∈𝒫(X):Yisclosed},𝒫b(X)={Y∈𝒫(X):Yisbounded},𝒫cp(X)={Y∈𝒫(X):Yiscompact},𝒫cp,c(X)={Y∈𝒫(X):Yiscompactandconvex}.
A multivalued map G:X→𝒫(X)(i)
isconvex (closed) valued if G(x) is convex (closed) for all x∈X;(ii)
isbounded on bounded sets if G(𝔹)=∪x∈𝔹G(x) is bounded in X for all 𝔹∈𝒫b(X) (i.e., supx∈𝔹{sup{|y|:y∈G(x)}}<∞);(iii)
is calledupper semicontinuous (u.s.c.) on X if, for each x0∈X, the set G(x0) is a nonempty closed subset of X and if, for each open set N of X containing G(x0), there exists an open neighborhood 𝒩0 of x0 such that G(𝒩0)⊆N;(iv)
G is lower semicontinuous (l.s.c.) if the set {y∈X:G(y)∩B≠∅} is open for any open set B in E;(v)
is said to becompletely continuous if G(𝔹) is relatively compact for every 𝔹∈𝒫b(X);(vi)
is said to bemeasurable if, for every y∈ℝ, the function
(19)t↦d(y,G(t))=inf{|y-z|:z∈G(t)}is measurable;(vii)
has a fixed point if there is x∈X such that x∈G(x). The fixed point set of the multivalued operator G will be denoted by FixG.Definition 7.
A multivalued mapF:[0,1]×ℝ→𝒫(ℝ) is said to be Carathéodory if(i)
t↦F(t,x) is measurable for each x∈ℝ;(ii)
x↦F(t,x) is upper semicontinuous for almost all t∈[0,1];
Further a Carathéodory functionF is called L1-Carathéodory if (iii)
for eachα>0, there exists φα∈L1([0,1],ℝ+) such that
(20)∥F(t,x)∥=sup{|v|:v∈F(t,x)}≤φα(t)for all∥x∥∞≤α and for a. e. t∈[0,1].For eachx∈C([0,1],ℝ), define the set of selections of F by
(21)SF,x:={v∈L1([0,1],ℝ):v(t)∈F(t,x(t))fora.e.  t∈[0,1]}.We define the graph ofG to be the set Gr(G)={(x,y)∈X×Y,y∈G(x)} and recall two useful results regarding closed graphs and upper semicontinuity.Lemma 8 (see [27, Proposition 1.2]).
IfG:X→𝒫cl(Y) is u.s.c., then Gr(G) 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∈G(xn), then y*∈G(x*). Conversely, if G is completely continuous and has a closed graph, then it is upper semicontinuous.Lemma 9 (see [30]).
LetX be a Banach space. Let F:[0,T]×ℝ→𝒫cp,c(X) be an L1-Carathéodory multivalued map, and let Θ be a linear continuous mapping from L1([0,1],X) to C([0,1],X). Then the operator
(22)Θ∘SF:C([0,1],X)→𝒫cp,c(C([0,1],X)),x↦(Θ∘SF)(x)=Θ(SF,x,y)
is a closed graph operator in C([0,1],X)×C([0,1],X).We recall the well-known nonlinear alternative of Leray-Schauder for multivalued maps.Lemma 10 (nonlinear alternative for Kakutani maps [31]).
LetE be a Banach space, C a closed convex subset of E,U an open subset of C, and 0∈U. Suppose that F:U¯→𝒫c,cv(C) is an upper semicontinuous compact map; here 𝒫c,cv(C) denotes the family of nonempty, compact convex subsets of C. Then either(i)
F has a fixed point in U¯, or(ii)
there is au∈∂U and λ∈(0,1) with u∈λF(u).Definition 11.
LetA be a subset of [0,1]×ℝ. A is ℒ⊗ℬ measurable if A belongs to the σ-algebra generated by all sets of the form 𝒥×𝒟, where 𝒥 is Lebesgue measurable in [0,1] and 𝒟 is Borel measurable in ℝ.Definition 12.
A subset𝒜 of L1([0,1],ℝ) is decomposable if, for all u,v∈𝒜, and measurable 𝒥⊂[0,1]=J, the function uχ𝒥+vχJ-𝒥∈𝒜, where χ𝒥 stands for the characteristic function of 𝒥.Lemma 13 (see [32]).
LetY be a separable metric space, and let N:Y→𝒫(L1([0,1],ℝ)) be a lower semicontinuous (l.s.c.) multivalued operator with nonempty closed and decomposable values. Then N has a continuous selection; that is, there exists a continuous function (single-valued) g:Y→L1([0,1],ℝ) such that g(x)∈N(x) for every x∈Y.Let(X,d) be a metric space induced from the normed space (X;∥·∥). Consider Hd:𝒫(X)×𝒫(X)→ℝ∪{∞} given by
(23)Hd(A,B)=max{supa∈Ad(a,B),supb∈Bd(A,b)},
where d(A,b)=infa∈Ad(a;b) and d(a,B)=infb∈Bd(a;b). Then (𝒫b,cl(X),Hd) is a metric space (see [33]).Definition 14.
A multivalued operatorN:X→𝒫cl(X) is called(a)
γ-Lipschitz if and only if there exists γ>0 such that
(24)Hd(N(x),N(y))≤γd(x,y)foreachx,y∈X;(b)
a contraction if and only if it isγ-Lipschitz with γ<1.Lemma 15 (see [34]).
Let(X,d) be a complete metric space. If N:X→𝒫cl(X) is a contraction, then FixN≠∅.


## 3. Main Results

### 3.1. The Carathéodory Case

In this section, we are concerned with the existence of solutions for the problem (1) when the right-hand side has convex as well as nonconvex values. Initially, we assume that F is a compact and convex valued multivalued map.Theorem 16.
Suppose that(H1)
the mapF:[0,1]×ℝ→𝒫(ℝ) is Carathéodory and has nonempty compact and convex values;(H2)
there exist a continuous nondecreasing functionψ:[0,∞)→(0,∞) and function p∈L1([0,1],ℝ+) such that
(25)∥F(t,x)∥𝒫:=sup{|v|:v∈F(t,x)}≤p(t)ψ(∥x∥)for each(t,u)∈[0,1]×ℝ;(H3)
there exists a numberM>0 such that
(26)M(1-Ψ)ψ(M)∥p∥L1Ω>1,with Ψ<1, where Ω,Ψ are defined in (17).
Then BVP (1) has at least one solution.Proof.
Let us introduce the operatorN:C([0,1],ℝ)→𝒫(C([0,1],ℝ)) as(27)N(x)={h∈C([0,1],ℝ):h(t)={∫0t‍(t-s)q-1Γ(q)(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q){∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du}for v∈SF,x. We will show that the operator N satisfies the assumptions of the nonlinear alternative of Leray-Schauder type. The proof consists of several steps. As a first step, we show that N(x)  is convex for eachx∈C([0,1],ℝ). For that, let h1,h2∈N(x). Then there exist v1,v2∈SF,x such that, for each t∈[0,1], we have(28)hi(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vi(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vi(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vi(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vi(r)dr-λx(s)}ds)du,for i=1,2. Let 0≤ω≤1. Then, for each t∈[0,1] and putting z(r)=ωv1(r)+(1-ω)v2(r), we have
(29)[ωh1+(1-ω)h2](t)=∫0t‍(t-s)q-1Γ(q)(∫0s‍(s-r)p-1Γ(p)z(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)(∫0s‍(s-r)p-1Γ(p)z(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)z(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)z(r)dr-λx(s)}ds)du.
SinceSF,x is convex (F has convex values), therefore it follows that ωh1+(1-ω)h2∈N(x).
Next, we show thatN(x)  maps bounded sets into bounded sets inC([0,1],ℝ). For a positive number r, let Bρ={x∈C([0,1],ℝ):∥x∥≤ρ} be a bounded set in C([0,1],ℝ). Then, for each h∈N(x),x∈Bρ, there exists v∈SF,x such that
(30)h(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))𝒥2-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))𝒥1,|h(t)|≤∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v(r)|dr+|λ||x(s)|)ds+A1|𝒥2|+A2|𝒥1|≤ψ(∥x∥)∥p∥L1Γ(p+q+1)+|λ|∥x∥Γ(q+1)+A1(ψ(∥x∥)∥p∥L1Γ(p+q+1)+|λ|∥x∥Γ(q+1)+∑i=1n‍|αi|[ηνi+p+qψ(∥x∥)∥p∥L1Γ(νi+p+q+1)+ηνi+q|λ|∥x∥Γ(νi+q+1)])+A2∑i=1n‍|βi|(ζμi+p+qψ(∥x∥)∥p∥L1Γ(μi+p+q+1)+ζμi+q|λ|∥x∥Γ(μi+q+1))=ψ(∥x∥)∥p∥L1×{∑i=1n‍1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)}+∥x∥{∑i=1n‍|λ|(1+A1)Γ(q+1)+|λ|A1∑i=1n‍|αi|ηνi+qΓ(νi+q+1)+A2∑i=1n‍|βi|ζμi+qΓ(μi+q+1)}=Ωψ(∥x∥)∥p∥L1+Ψ∥x∥.
Then
(31)∥h∥≤Ωψ(ρ)∥p∥L1+Ψρ.
Now we show thatNmaps bounded sets into equicontinuous sets ofC([0,1],ℝ). Let t′,t′′∈[0,1] with t′<t′′ and x∈Bρ, where Bρ, as above, is a bounded set of C([0,1],ℝ). For each h∈N(x), we obtain
(32)|h(t′′)-h(t′)|≤|∫0t′′‍(t′′-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v(r)|dr+|λ||x(s)|)ds-∫0t′‍(t′-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v(r)|dr+|λ||x(s)|)ds|+|Δ3||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥2|+|Δ2||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥1|≤|∫0t′‍[(t′′-s)q-1-(t′-s)q-1]Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v(r)|dr+|λ||x(s)|)ds+∫t′t′′‍(t′′-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v(r)|dr+|λ||x(s)|)ds∫0t′‍[(t′′-s)q-1-(t′-s)q-1]Γ(q)|+|Δ3||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥2|+|Δ2||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥1|≤|∫0t′‍[(t′′-s)q-1-(t′-s)q-1]Γ(q)×(∥p∥L1ψ(ρ)∫0s‍(s-r)p-1Γ(p)dr+|λ|ρ)ds+∫t′t′′‍(t′′-s)q-1Γ(q)×(∥p∥L1ψ(ρ)∫0s‍(s-r)p-1Γ(p)dr+|λ|ρ)ds∫0t′‍[(t′′-s)q-1-(t′-s)q-1]Γ(q)|+|Δ3||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥2|+|Δ2||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥1|.
Obviously the right-hand side of the above inequality tends to zero independently ofx∈Bρ as t′′-t′→0. As N satisfies the above three assumptions, therefore it follows by Ascoli-Arzelá theorem that N:C([0,1],ℝ)→𝒫(C([0,1],ℝ)) is completely continuous.
In our next step, we show thatNhas a closed graph. Let xn→x*,hn∈N(xn), and hn→h*. Then we need to show that h*∈N(x*). Associated with hn∈N(xn), there exists vn∈SF,xn such that, for each t∈[0,1],(33)hn(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s)}ds)du.Thus we have to show that there exists v*∈SF,x* such that, for each t∈[0,1],(34)h*(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s)}ds)du.
Let us consider the continuous linear operatorΘ:L1([0,1],ℝ)→C([0,1],ℝ) so that
(35)v↦Θ(v)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du.
Observe that(36)∥hn(t)-h*(t)∥=∥∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)(vn(r)-v*(s))dr)ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)(vn(r)-v*(s))dr)ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)(vn(r)-v*(s))dr}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)(vn(r)-v*(s))dr}ds)du∥,which tends to zero as n→∞.
Thus, it follows from Lemma9 that Θ∘SF is a closed graph operator. Further, we have hn(t)∈Θ(SF,xn). Since xn→x*, it follows that(37)h*(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s)}ds)du,for some v*∈SF,x*.
Finally,we discuss a priori bounds on solutions. Let x be a solution of (1). Then there exists v∈L1([0,1],ℝ) with v∈SF,x such that, for t∈[0,1], we have(38)h(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du.Using the computations proving that N(x) maps bounded sets into bounded sets and the notations (17), we have
(39)∥x∥≤ψ(∥x∥)∥p∥L1×{∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)}+∥x∥{∑i=1n‍|βi|ζμi+qΓ(μi+q+1)|λ|(1+A1)Γ(q+1)+|λ|A1∑i=1n‍|αi|ηνi+qΓ(νi+q+1)+A2∑i=1n‍|βi|ζμi+qΓ(μi+q+1)}=Ωψ(∥x∥)∥p∥L1+Ψ∥x∥.
Consequently(40)∥x∥(1-Ψ)ψ(∥x∥)∥p∥L1Ω≤1.
In view of (H3), there exists M such that ∥x∥≠M. Let us set
(41)U={x∈C([0,1],ℝ):∥x∥<M+1}.
Note that the operator N:U¯→𝒫(C([0,1],ℝ)) is upper semicontinuous and completely continuous. From the choice of U, there is no x∈∂U such that x∈μN(x) for some μ∈(0,1). Consequently, by the nonlinear alternative of Leray-Schauder type [31], we deduce that N has a fixed point x∈U¯ which is a solution of the problem (1). This completes the proof.


### 3.2. The Lower Semicontinuous Case

Next, we study the case whereF is not necessarily convex valued. Our approach here is based on the nonlinear alternative of Leray-Schauder type combined with the selection theorem of Bressan and Colombo for lower semicontinuous maps with decomposable values.Theorem 17.
Assume that(H2)-(H3) and the following conditions hold:(H4)
F:[0,1]×ℝ→𝒫(ℝ) is a nonempty compact-valued multivalued map such that
(a)
(t,x)↦F(t,x) is ℒ⊗ℬ measurable,(b)
x↦F(t,x) is lower semicontinuous for each t∈[0,1];(H5)
for eachσ>0, there exists φσ∈L1([0,1],ℝ+) such that
(42)∥F(t,x)∥=sup{|y|:y∈F(t,x)}≤φσ(t)∀∥x∥≤σandfora.e.t∈[0,1].Then the boundary value problem (1) has at least one solution on [0,1].Proof.
It follows from(H4) and (H5) that F is of l.s.c. type [35]. Then from Lemma 13, there exists a continuous function f:C([0,1],ℝ)→L1([0,1],ℝ) such that f(x)∈ℱ(x) for all x∈C([0,1],ℝ).
Consider the problem(43)cDp(cDq+λ)x(t)=F(x(t)),0<t<1,x(0)=∑i=1n‍βi(Iμix)(ζ),x(1)=∑i=1n‍αi(Iνix)(η),0<ζ<η<1.
Observe that, ifx∈AC1([0,1]) is a solution of (43), then x is a solution to the problem (1). In order to transform the problem (43) into a fixed point problem, we define the operator N¯ as
(44)(N¯x)(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)f(x(r))dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)f(x(r))dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)f(x(r))dr-λx(s)∫0s‍}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)f(x(r))dr-λx(s)}ds)du.
It can easily be shown that N¯ is continuous and completely continuous. The remaining part of the proof is similar to that of Theorem 16. So we omit it. This completes the proof.


### 3.3. 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 map due to Covitz and Nadler [34].Theorem 18.
Assume that the following conditions hold:(H6)
F:[0,1]×ℝ→𝒫cp(ℝ) is such that F(·,x):[0,1]→𝒫cp(ℝ) is measurable for each x∈ℝ;(H7)
Hd(F(t,x),F(t,x-))≤m(t)|x-x-| for almost all t∈[0,1] and x,x-∈ℝ with m∈C([0,1],ℝ+) and d(0,F(t,0))≤m(t) for almost all t∈[0,1].
Then the boundary value problem (1) has at least one solution on [0,1] if Ω∥m∥L1<1, that is,
(45){1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)}∥m∥L1<1.Proof.
Observe that the setSF,x is nonempty for each x∈C([0,1],ℝ) by the assumption (H6), so F has a measurable selection (see Theorem III.6 [36]). Now we show that the operator N satisfies the assumptions of Lemma 15. To show that N(x)∈𝒫cl((C[0,1],ℝ)) for each x∈C([0,1],ℝ), let {un}n≥0∈N(x) be such that un→u(n→∞) in C([0,1],ℝ). Then u∈C([0,1],ℝ), and there exists vn∈SF,x such that, for each t∈[0,1],(46)un(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s)}ds)du.
AsF has compact values, we pass onto a subsequence to obtain that vn converges to v in L1([0,1],ℝ). Thus, v∈SF,x and, for each t∈[0,1], (47)un(t)→u(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du.Hence, u∈N(x).
Next we show that there existsγ<1 such that
(48)Hd(N(x),N(x-))≤γ∥x-x-∥foreachx,x-∈C([0,1],ℝ).
Let x,x-∈C([0,1],ℝ) and h1∈N(x). Then there exists v1(t)∈F(t,x(t)) such that, for each t∈[0,1], (49)h1(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v1(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v1(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v1(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v1(r)dr-λx(s)}ds)du.
By(H7), we have
(50)Hd(F(t,x),F(t,x-))≤m(t)|x(t)-x-(t)|.
So, there exists w(t)∈F(t,x-(t)) such that
(51)|v1(t)-w(t)|≤m(t)|x(t)-x-(t)|,t∈[0,1].
DefineU:[0,1]→𝒫(ℝ) by
(52)U(t)={w∈ℝ:|v1(t)-w|≤m(t)|x(t)-x-(t)|}.
Since the multivalued operator U(t)∩F(t,x-(t)) is measurable (Proposition III.4 [36]), there exists a function v2(t) which is a measurable selection for V. So v2(t)∈F(t,x-(t)), and for each t∈[0,1], we have |v1(t)-v2(t)|≤m(t)|x(t)-x-(t)|.
For eacht∈[0,1], let us define
(53)h2(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v2(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v2(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v2(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v2(r)dr-λx(s)}ds)du.
Thus, (54)|h1(t)-h2(t)|=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v1(r)-v2(r)|dr)ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v1(r)-v2(r)|dr)ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)|v1(r)-v2(r)|dr}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)|v1(r)-v2(r)|dr}ds)du.Hence,
(55)∥h1-h2∥≤{∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)}×∥m∥L1∥x-x¯∥.
Analogously, interchanging the roles of x and x¯, we obtain
(56)Hd(N(x),N(x-))≤γ∥x-x¯∥≤{∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)}×∥m∥L1∥x-x¯∥.
SinceN is a contraction, it follows from Lemma 15 that N has a fixed point x which is a solution of (1). This completes the proof.Remark 19.
It is important to note that several new interesting special results of the present work can be obtained by fixing the parameters involved in the given problem. Some of these results are listed in the following.(i)
Our results correspond to the multivalued extension of the Dirichlet problem considered in [23] for βi=0=αi,i=1,2,…,n.(ii)
In case we takeβi=0,i=1,2,…,n, we obtain the results for Langevin fractional differential inclusions with the three-point integral boundary conditions of the following form:
(57)x(0)=0,x(1)=∑i=1n‍αi(Iνix)(η).(iii)
By takingβi=0=αi,i=2,3,…,n, and ν1=1=μ1, we get the results for Langevin fractional differential inclusions with four-point nonlocal integral boundary conditions of the following type:
(58)x(0)=β1∫0ζ‍x(s)ds,x(1)=α1∫0η‍x(s)ds.Note that we obtain the typical integral boundary condition in the limitζ,η→1-.Example 20.
Consider the problem(59)cD2/3(cD4/5+110)x(t)∈F(t,x(t)),0<t<1,x(0)=12(I1/2x)(14)-13(I3/4x)(14)+14(I3/2x)(14)-15(I4/3x)(14),x(1)=12(I1/4x)(13)-23(I3/4x)(13)+34(I5/4x)(13)-45(I7/4x)(13),
where F:[0,1]×ℝ→𝒫(ℝ) is a multivalued map given by
(60)x→F(t,x)=[14x5x5+3+18(t+1),14sinx+14(t+1)].
For f∈F, we have
(61)|f|≤max[14x5x5+3+18(t+1),14sinx+14(t+1)]≤34.
Here p=2/3,  q=4/5,  λ=1/10,  η=1/3,  n=4,  α1=1/2,  α2=-2/3,  α3=3/4,  α4=-4/5,  ν1=1/4,  ν2=3/4,  ν3=5/4,  ν4=7/4,  ζ=1/4,  β1=1/2,  β2=-1/3,  β3=1/4,  β4=-1/5,  μ1=1/2,  μ2=3/4,  μ3=3/2,  and μ4=4/3. Clearly,
(62)∥F(t,x)∥𝒫:=sup{|y|:y∈F(t,x)}≤p(t)ψ(∥x∥),x∈ℝ,
with p(t)=1,ψ(∥x∥)=3/4. Using the given data, it is found that
(63)Δ1=0.126506,Δ2=0.804143,Δ3=0.849082,Δ4=-0.983656,Δ=0.936934,A1=1.108019,A2=1.971367,Ω=1.764724,Ψ=0.47047.
Thus,
(64)M>ψ(M)∥p∥L1Ω1-Ψ≈2.499467.
Clearly, all the conditions of Theorem 16 are satisfied. So there exists at least one solution of the problem (59) on [0,1].Example 21.
Consider the fractional inclusion boundary value problem (59) with F:[0,1]×ℝ→𝒫(ℝ) given by
(65)F(t,x)=[0,sinx(2+t)2+112].
Then, we have
(66)sup{|u|:u∈F(t,x)}≤112+1(2+t)2,Hd(F(t,x),F(t,x-))≤m(t)|x-x-|,
where m(t)=1/(2+t)2. With ∥m∥L1=1/6 and Ω=1.764724 (from Example 20), it is found that
(67)γ=Ω∥m∥L1≈0.294121<1.
Since all the conditions of Theorem 18 are satisfied, therefore the problem (59) with F given by (65) has at least one solution on [0,1].


## 3.1. The Carathéodory Case

In this section, we are concerned with the existence of solutions for the problem (1) when the right-hand side has convex as well as nonconvex values. Initially, we assume that F is a compact and convex valued multivalued map.Theorem 16.
Suppose that(H1)
the mapF:[0,1]×ℝ→𝒫(ℝ) is Carathéodory and has nonempty compact and convex values;(H2)
there exist a continuous nondecreasing functionψ:[0,∞)→(0,∞) and function p∈L1([0,1],ℝ+) such that
(25)∥F(t,x)∥𝒫:=sup{|v|:v∈F(t,x)}≤p(t)ψ(∥x∥)for each(t,u)∈[0,1]×ℝ;(H3)
there exists a numberM>0 such that
(26)M(1-Ψ)ψ(M)∥p∥L1Ω>1,with Ψ<1, where Ω,Ψ are defined in (17).
Then BVP (1) has at least one solution.Proof.
Let us introduce the operatorN:C([0,1],ℝ)→𝒫(C([0,1],ℝ)) as(27)N(x)={h∈C([0,1],ℝ):h(t)={∫0t‍(t-s)q-1Γ(q)(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q){∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du}for v∈SF,x. We will show that the operator N satisfies the assumptions of the nonlinear alternative of Leray-Schauder type. The proof consists of several steps. As a first step, we show that N(x)  is convex for eachx∈C([0,1],ℝ). For that, let h1,h2∈N(x). Then there exist v1,v2∈SF,x such that, for each t∈[0,1], we have(28)hi(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vi(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vi(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vi(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vi(r)dr-λx(s)}ds)du,for i=1,2. Let 0≤ω≤1. Then, for each t∈[0,1] and putting z(r)=ωv1(r)+(1-ω)v2(r), we have
(29)[ωh1+(1-ω)h2](t)=∫0t‍(t-s)q-1Γ(q)(∫0s‍(s-r)p-1Γ(p)z(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)(∫0s‍(s-r)p-1Γ(p)z(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)z(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)z(r)dr-λx(s)}ds)du.
SinceSF,x is convex (F has convex values), therefore it follows that ωh1+(1-ω)h2∈N(x).
Next, we show thatN(x)  maps bounded sets into bounded sets inC([0,1],ℝ). For a positive number r, let Bρ={x∈C([0,1],ℝ):∥x∥≤ρ} be a bounded set in C([0,1],ℝ). Then, for each h∈N(x),x∈Bρ, there exists v∈SF,x such that
(30)h(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))𝒥2-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))𝒥1,|h(t)|≤∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v(r)|dr+|λ||x(s)|)ds+A1|𝒥2|+A2|𝒥1|≤ψ(∥x∥)∥p∥L1Γ(p+q+1)+|λ|∥x∥Γ(q+1)+A1(ψ(∥x∥)∥p∥L1Γ(p+q+1)+|λ|∥x∥Γ(q+1)+∑i=1n‍|αi|[ηνi+p+qψ(∥x∥)∥p∥L1Γ(νi+p+q+1)+ηνi+q|λ|∥x∥Γ(νi+q+1)])+A2∑i=1n‍|βi|(ζμi+p+qψ(∥x∥)∥p∥L1Γ(μi+p+q+1)+ζμi+q|λ|∥x∥Γ(μi+q+1))=ψ(∥x∥)∥p∥L1×{∑i=1n‍1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)}+∥x∥{∑i=1n‍|λ|(1+A1)Γ(q+1)+|λ|A1∑i=1n‍|αi|ηνi+qΓ(νi+q+1)+A2∑i=1n‍|βi|ζμi+qΓ(μi+q+1)}=Ωψ(∥x∥)∥p∥L1+Ψ∥x∥.
Then
(31)∥h∥≤Ωψ(ρ)∥p∥L1+Ψρ.
Now we show thatNmaps bounded sets into equicontinuous sets ofC([0,1],ℝ). Let t′,t′′∈[0,1] with t′<t′′ and x∈Bρ, where Bρ, as above, is a bounded set of C([0,1],ℝ). For each h∈N(x), we obtain
(32)|h(t′′)-h(t′)|≤|∫0t′′‍(t′′-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v(r)|dr+|λ||x(s)|)ds-∫0t′‍(t′-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v(r)|dr+|λ||x(s)|)ds|+|Δ3||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥2|+|Δ2||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥1|≤|∫0t′‍[(t′′-s)q-1-(t′-s)q-1]Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v(r)|dr+|λ||x(s)|)ds+∫t′t′′‍(t′′-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v(r)|dr+|λ||x(s)|)ds∫0t′‍[(t′′-s)q-1-(t′-s)q-1]Γ(q)|+|Δ3||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥2|+|Δ2||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥1|≤|∫0t′‍[(t′′-s)q-1-(t′-s)q-1]Γ(q)×(∥p∥L1ψ(ρ)∫0s‍(s-r)p-1Γ(p)dr+|λ|ρ)ds+∫t′t′′‍(t′′-s)q-1Γ(q)×(∥p∥L1ψ(ρ)∫0s‍(s-r)p-1Γ(p)dr+|λ|ρ)ds∫0t′‍[(t′′-s)q-1-(t′-s)q-1]Γ(q)|+|Δ3||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥2|+|Δ2||Δ|Γ(q+1)|(t′′)q-(t′)q||𝒥1|.
Obviously the right-hand side of the above inequality tends to zero independently ofx∈Bρ as t′′-t′→0. As N satisfies the above three assumptions, therefore it follows by Ascoli-Arzelá theorem that N:C([0,1],ℝ)→𝒫(C([0,1],ℝ)) is completely continuous.
In our next step, we show thatNhas a closed graph. Let xn→x*,hn∈N(xn), and hn→h*. Then we need to show that h*∈N(x*). Associated with hn∈N(xn), there exists vn∈SF,xn such that, for each t∈[0,1],(33)hn(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s)}ds)du.Thus we have to show that there exists v*∈SF,x* such that, for each t∈[0,1],(34)h*(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s))ds-∑i=1n‍αi∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s)}ds)du.
Let us consider the continuous linear operatorΘ:L1([0,1],ℝ)→C([0,1],ℝ) so that
(35)v↦Θ(v)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du.
Observe that(36)∥hn(t)-h*(t)∥=∥∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)(vn(r)-v*(s))dr)ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)(vn(r)-v*(s))dr)ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)(vn(r)-v*(s))dr}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)(vn(r)-v*(s))dr}ds)du∥,which tends to zero as n→∞.
Thus, it follows from Lemma9 that Θ∘SF is a closed graph operator. Further, we have hn(t)∈Θ(SF,xn). Since xn→x*, it follows that(37)h*(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v*(r)dr-λx(s)}ds)du,for some v*∈SF,x*.
Finally,we discuss a priori bounds on solutions. Let x be a solution of (1). Then there exists v∈L1([0,1],ℝ) with v∈SF,x such that, for t∈[0,1], we have(38)h(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du.Using the computations proving that N(x) maps bounded sets into bounded sets and the notations (17), we have
(39)∥x∥≤ψ(∥x∥)∥p∥L1×{∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)}+∥x∥{∑i=1n‍|βi|ζμi+qΓ(μi+q+1)|λ|(1+A1)Γ(q+1)+|λ|A1∑i=1n‍|αi|ηνi+qΓ(νi+q+1)+A2∑i=1n‍|βi|ζμi+qΓ(μi+q+1)}=Ωψ(∥x∥)∥p∥L1+Ψ∥x∥.
Consequently(40)∥x∥(1-Ψ)ψ(∥x∥)∥p∥L1Ω≤1.
In view of (H3), there exists M such that ∥x∥≠M. Let us set
(41)U={x∈C([0,1],ℝ):∥x∥<M+1}.
Note that the operator N:U¯→𝒫(C([0,1],ℝ)) is upper semicontinuous and completely continuous. From the choice of U, there is no x∈∂U such that x∈μN(x) for some μ∈(0,1). Consequently, by the nonlinear alternative of Leray-Schauder type [31], we deduce that N has a fixed point x∈U¯ which is a solution of the problem (1). This completes the proof.


## 3.2. The Lower Semicontinuous Case

Next, we study the case whereF is not necessarily convex valued. Our approach here is based on the nonlinear alternative of Leray-Schauder type combined with the selection theorem of Bressan and Colombo for lower semicontinuous maps with decomposable values.Theorem 17.
Assume that(H2)-(H3) and the following conditions hold:(H4)
F:[0,1]×ℝ→𝒫(ℝ) is a nonempty compact-valued multivalued map such that
(a)
(t,x)↦F(t,x) is ℒ⊗ℬ measurable,(b)
x↦F(t,x) is lower semicontinuous for each t∈[0,1];(H5)
for eachσ>0, there exists φσ∈L1([0,1],ℝ+) such that
(42)∥F(t,x)∥=sup{|y|:y∈F(t,x)}≤φσ(t)∀∥x∥≤σandfora.e.t∈[0,1].Then the boundary value problem (1) has at least one solution on [0,1].Proof.
It follows from(H4) and (H5) that F is of l.s.c. type [35]. Then from Lemma 13, there exists a continuous function f:C([0,1],ℝ)→L1([0,1],ℝ) such that f(x)∈ℱ(x) for all x∈C([0,1],ℝ).
Consider the problem(43)cDp(cDq+λ)x(t)=F(x(t)),0<t<1,x(0)=∑i=1n‍βi(Iμix)(ζ),x(1)=∑i=1n‍αi(Iνix)(η),0<ζ<η<1.
Observe that, ifx∈AC1([0,1]) is a solution of (43), then x is a solution to the problem (1). In order to transform the problem (43) into a fixed point problem, we define the operator N¯ as
(44)(N¯x)(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)f(x(r))dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)f(x(r))dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)f(x(r))dr-λx(s)∫0s‍}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)f(x(r))dr-λx(s)}ds)du.
It can easily be shown that N¯ is continuous and completely continuous. The remaining part of the proof is similar to that of Theorem 16. So we omit it. This completes the proof.


## 3.3. 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 map due to Covitz and Nadler [34].Theorem 18.
Assume that the following conditions hold:(H6)
F:[0,1]×ℝ→𝒫cp(ℝ) is such that F(·,x):[0,1]→𝒫cp(ℝ) is measurable for each x∈ℝ;(H7)
Hd(F(t,x),F(t,x-))≤m(t)|x-x-| for almost all t∈[0,1] and x,x-∈ℝ with m∈C([0,1],ℝ+) and d(0,F(t,0))≤m(t) for almost all t∈[0,1].
Then the boundary value problem (1) has at least one solution on [0,1] if Ω∥m∥L1<1, that is,
(45){1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)}∥m∥L1<1.Proof.
Observe that the setSF,x is nonempty for each x∈C([0,1],ℝ) by the assumption (H6), so F has a measurable selection (see Theorem III.6 [36]). Now we show that the operator N satisfies the assumptions of Lemma 15. To show that N(x)∈𝒫cl((C[0,1],ℝ)) for each x∈C([0,1],ℝ), let {un}n≥0∈N(x) be such that un→u(n→∞) in C([0,1],ℝ). Then u∈C([0,1],ℝ), and there exists vn∈SF,x such that, for each t∈[0,1],(46)un(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)vn(r)dr-λx(s)}ds)du.
AsF has compact values, we pass onto a subsequence to obtain that vn converges to v in L1([0,1],ℝ). Thus, v∈SF,x and, for each t∈[0,1], (47)un(t)→u(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v(r)dr-λx(s)}ds)du.Hence, u∈N(x).
Next we show that there existsγ<1 such that
(48)Hd(N(x),N(x-))≤γ∥x-x-∥foreachx,x-∈C([0,1],ℝ).
Let x,x-∈C([0,1],ℝ) and h1∈N(x). Then there exists v1(t)∈F(t,x(t)) such that, for each t∈[0,1], (49)h1(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v1(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v1(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v1(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v1(r)dr-λx(s)}ds)du.
By(H7), we have
(50)Hd(F(t,x),F(t,x-))≤m(t)|x(t)-x-(t)|.
So, there exists w(t)∈F(t,x-(t)) such that
(51)|v1(t)-w(t)|≤m(t)|x(t)-x-(t)|,t∈[0,1].
DefineU:[0,1]→𝒫(ℝ) by
(52)U(t)={w∈ℝ:|v1(t)-w|≤m(t)|x(t)-x-(t)|}.
Since the multivalued operator U(t)∩F(t,x-(t)) is measurable (Proposition III.4 [36]), there exists a function v2(t) which is a measurable selection for V. So v2(t)∈F(t,x-(t)), and for each t∈[0,1], we have |v1(t)-v2(t)|≤m(t)|x(t)-x-(t)|.
For eacht∈[0,1], let us define
(53)h2(t)=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v2(r)dr-λx(s))ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)v2(r)dr-λx(s))ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v2(r)dr-λx(s)}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)v2(r)dr-λx(s)}ds)du.
Thus, (54)|h1(t)-h2(t)|=∫0t‍(t-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v1(r)-v2(r)|dr)ds-1Δ(tqΔ3+Δ1Γ(q+1)Γ(q+1))×[∫01‍(1-s)q-1Γ(q)×(∫0s‍(s-r)p-1Γ(p)|v1(r)-v2(r)|dr)ds-∑i=1n‍αi×∫0η‍(η-u)νi-1Γ(νi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)|v1(r)-v2(r)|dr}ds)du]-1Δ(tqΔ2+Δ4Γ(q+1)Γ(q+1))×∑i=1n‍βi×∫0ζ‍(ζ-u)μi-1Γ(μi)×(∫0u‍(u-s)q-1Γ(q)×{∫0s‍(s-r)p-1Γ(p)|v1(r)-v2(r)|dr}ds)du.Hence,
(55)∥h1-h2∥≤{∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)}×∥m∥L1∥x-x¯∥.
Analogously, interchanging the roles of x and x¯, we obtain
(56)Hd(N(x),N(x-))≤γ∥x-x¯∥≤{∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)1+A1Γ(p+q+1)+A1∑i=1n‍|αi|ηνi+p+qΓ(νi+p+q+1)+A2∑i=1n‍|βi|ζμi+p+qΓ(μi+p+q+1)}×∥m∥L1∥x-x¯∥.
SinceN is a contraction, it follows from Lemma 15 that N has a fixed point x which is a solution of (1). This completes the proof.Remark 19.
It is important to note that several new interesting special results of the present work can be obtained by fixing the parameters involved in the given problem. Some of these results are listed in the following.(i)
Our results correspond to the multivalued extension of the Dirichlet problem considered in [23] for βi=0=αi,i=1,2,…,n.(ii)
In case we takeβi=0,i=1,2,…,n, we obtain the results for Langevin fractional differential inclusions with the three-point integral boundary conditions of the following form:
(57)x(0)=0,x(1)=∑i=1n‍αi(Iνix)(η).(iii)
By takingβi=0=αi,i=2,3,…,n, and ν1=1=μ1, we get the results for Langevin fractional differential inclusions with four-point nonlocal integral boundary conditions of the following type:
(58)x(0)=β1∫0ζ‍x(s)ds,x(1)=α1∫0η‍x(s)ds.Note that we obtain the typical integral boundary condition in the limitζ,η→1-.Example 20.
Consider the problem(59)cD2/3(cD4/5+110)x(t)∈F(t,x(t)),0<t<1,x(0)=12(I1/2x)(14)-13(I3/4x)(14)+14(I3/2x)(14)-15(I4/3x)(14),x(1)=12(I1/4x)(13)-23(I3/4x)(13)+34(I5/4x)(13)-45(I7/4x)(13),
where F:[0,1]×ℝ→𝒫(ℝ) is a multivalued map given by
(60)x→F(t,x)=[14x5x5+3+18(t+1),14sinx+14(t+1)].
For f∈F, we have
(61)|f|≤max[14x5x5+3+18(t+1),14sinx+14(t+1)]≤34.
Here p=2/3,  q=4/5,  λ=1/10,  η=1/3,  n=4,  α1=1/2,  α2=-2/3,  α3=3/4,  α4=-4/5,  ν1=1/4,  ν2=3/4,  ν3=5/4,  ν4=7/4,  ζ=1/4,  β1=1/2,  β2=-1/3,  β3=1/4,  β4=-1/5,  μ1=1/2,  μ2=3/4,  μ3=3/2,  and μ4=4/3. Clearly,
(62)∥F(t,x)∥𝒫:=sup{|y|:y∈F(t,x)}≤p(t)ψ(∥x∥),x∈ℝ,
with p(t)=1,ψ(∥x∥)=3/4. Using the given data, it is found that
(63)Δ1=0.126506,Δ2=0.804143,Δ3=0.849082,Δ4=-0.983656,Δ=0.936934,A1=1.108019,A2=1.971367,Ω=1.764724,Ψ=0.47047.
Thus,
(64)M>ψ(M)∥p∥L1Ω1-Ψ≈2.499467.
Clearly, all the conditions of Theorem 16 are satisfied. So there exists at least one solution of the problem (59) on [0,1].Example 21.
Consider the fractional inclusion boundary value problem (59) with F:[0,1]×ℝ→𝒫(ℝ) given by
(65)F(t,x)=[0,sinx(2+t)2+112].
Then, we have
(66)sup{|u|:u∈F(t,x)}≤112+1(2+t)2,Hd(F(t,x),F(t,x-))≤m(t)|x-x-|,
where m(t)=1/(2+t)2. With ∥m∥L1=1/6 and Ω=1.764724 (from Example 20), it is found that
(67)γ=Ω∥m∥L1≈0.294121<1.
Since all the conditions of Theorem 18 are satisfied, therefore the problem (59) with F given by (65) has at least one solution on [0,1].



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*Source: 869837-2013-05-02.xml*