# An Integral Boundary Value Problem of Fractional Differential Equations with a Sign-Changed Parameter in Banach Spaces

Authors: Chen Yang; Yaru Guo; Chengbo Zhai
Journal: Complexity
 (2021)

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## Abstract

This paper is to investigate the existence and uniqueness of solutions for an integral boundary value problem of new fractional differential equations with a sign-changed parameter in Banach spaces. The main used approach is a recent fixed point theorem of increasingΨ−h,r-concave operators defined on ordered sets. In addition, we can present a monotone iterative scheme to approximate the unique solution. In the end, two simple examples are given to illustrate our main results.


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## Body

## 1. Introduction

With the intensive development of theory and applications of fractional calculus, fractional differential equations have been paid great interest in many fields and thus different boundary conditions of fractional differential equations have also attracted much attention, see [1–42]. However, we can see that there are very few results reported on solutions for boundary value problems of fractional differential equations in Banach spaces.In 2010, by using contraction mapping principle, fixed point index theory of completely continuous operator, and so on, Bai [2] obtained the existence results of positive solutions for the following three-point problem involving with fractional derivative:(1)D0+βut+ft,ut=0,0<t<1,u0=0,γuη=u1,where 1<β≤2,0<η<1,0<γηβ−1<1, D0+β is the Riemann–Liouville (R-L for short) fractional derivative, and the function ft,u is continuous on 0,1×0,+∞.In 2014, by using Guo–Krasnoselskii fixed point theorem, Cabada and Hamdi [4] obtained the existence of positive solutions for the following fractional differential equations with an integral condition:(2)D0+βut+ft,ut=0,0<t<1,u0=u′0=0,u1=ρ∫01utdt,where 2<β≤3,0<ρ<β, D0+β is the R-L fractional derivative, and ft,u is a continuous function.As we know, most of the existing results of solutions for fractional differential equation boundary value problems have been obtained in real spaceR. There exists very few papers studied in abstract spaces, and we can find [3, 9] and others.In 2018, Chen and Gao [3] discussed the following problem of fractional differential equations in a Banach space E:(3)−D0+βut=ft,ut,t∈J,u0=u′0=θ,u1=ρ∫01utdt,where 2<β≤3,0<ρ<β,J=0,1, D0+β is the R-L fractional derivative, f:J×K⟶K is continuous, here K is a normal cone in E, and θ denotes the zero element of E. They obtained the existence results of positive solutions by using fixed point index theory of condensing mapping.Motivated by these works, we consider the following new form of fractional differential equation with an integral boundary condition:(4)−D0+βut=ft,ut+γgt,t∈J,u0=u′0=θ,u1=ρ∫01utdt,where 2<β≤3, 0<ρ<β, and βandρ are real numbers, γ∈R is a sign-changed parameter, J=0,1, D0+β is the R-L fractional derivative, f:J×K⟶K is continuous, g:J⟶K is continuous, here K is a normal cone in E, and θ is the zero element of E. From the literature, we know that problem (4) is a new form of fractional differential equations. Different from the previous works, we use a recent fixed point theorem for Ψ−h,r-concave operators defined in ordered set Ph,r (see [43]) to study (4), and we obtain the existence and uniqueness of solution, while the uniqueness is not treated in the literature [3]. It should be pointed out that this method is interesting for solving nonlinear differential boundary value problems. In particular, we do not need the existence of upper-lower solutions, which is a critical condition in many articles.


## 2. Preliminaries

In this section, we state some known facts on concave operators; one can see [43–45].LetE be a Banach space, which has a partial order induced by a cone K⊂E, i.e., x≤y if and only if y−x∈K. For x,y∈E, the notation x∼y means that there are μ>0 and ν>0 such that μx≤y≤νx. Clearly, ∼ is an equivalence relation. For h>θ (i.e., h≥θ and h≠θ), define Ph=x∈E|x∼h. Clearly, Ph⊂K is convex. Let r∈K with θ≤r≤h, and define Ph,r=x∈E|x+r∈Ph, that is,(5)Ph,r=x∈E|there existμ=μh,r,x>0,ν=νh,r,x>0such thatμh≤x+r≤νh.Evidently,Ph=Ph,θ.Next, we list the definition ofΨ−h,r-concave operator and fixed point theorems that will play a key role in this paper.Definition 1.
(see [43]). Let T:Ph,r⟶E be a given operator which satisfies, for x∈Ph,r,λ∈0,1, and there is Ψλ>λ such that Tλx+λ−1r≥ΨλTx+Ψλ−1r. Then, we call T a Ψ−h,r-concave operator.Lemma 1.
(see [43]). Let T be an increasing Ψ−h,r-concave operator and P be normal, Th∈Ph,r. Then, T has a unique fixed point x∗ in Ph,r. Take any v0∈Ph,r, putting a sequence vn=Tvn−1, n=1,2,…; then, vn−x∗⟶0 as n⟶∞.Lemma 2.
(see [44]). Let T be an increasing Ψ−h,θ-concave operator and P be normal, Th∈Ph. Then, T has a unique fixed point x∗ in Ph. Take any v0∈Ph, making a sequence vn=Tvn−1, n=1,2,…; then, vn−x∗⟶0 as n⟶∞.


## 3. Main Results

LetE be an ordered Banach space, positive cone K⊂E is normal, θ is the zero element of E, and E has an identity element e, i.e., ex=x,∀x∈E. In this section, let J=0,1, and we apply Lemmas 1 and 2 to study problem (4) in abstract Banach space ℬ=CJ,E; the space of all continuous E-value functions on interval J with the norm uc=maxut:t∈J. Then, CJ,E is an ordered Banach space induced by the cone P=u∈CJ,E|ut∈K,t∈J, and P is also a normal cone in CJ,E.Lemma 3.
(see [3, 4]). Assume 2<β≤3 and 0<ρ<β. Then, for every y∈CJ,E, the problem of fractional linear differential equation in Banach spaces(6)−D0+βut=yt,0<t<1,u0=u′0=θ,u1=ρ∫01utdt,has a unique solution ut=∫01Gt,sysds, where(7)Gt,s=tβ−11−sβ−1β−ρ+ρs−β−ρt−sβ−1β−ρΓβ,0≤s≤t≤1,tβ−11−sβ−1β−ρ+ρsβ−ρΓβ,0≤t≤s≤1.Lemma 4.
(see [3, 4]). Let 2<β≤3 and ρ≥0. The function Gt,s given by (7) has the following properties:(i)
Gt,s>0 for t,s∈0,1 and ρ∈0,β(ii)
Gt,s≤β/β−ρΓβ for t,s∈0,1 and ρ∈0,β(iii)
1−sβ−1ρs/β−ρΓβtβ−1≤Gt,s≤1−sβ−1β−ρ+ρs/β−ρΓβtβ−1, for t,s∈0,1 and ρ∈0,βNext, we intend to give some existence and uniqueness results for problem (4). We break our discussion up into two sections. Two cases of interest are given in Section 3.1 for γ<0 and in Section 3.2 for γ≥0, respectively. In Section 3.3, we discuss the existence and uniqueness corollaries for problem (4) in real space CJ,R.


### 3.1. Investigation for the Caseγ<0

Let(8)rt=−γ∫01Gt,sgsds,t∈0,1,r^=−γββ−ρΓβ∫01gsds,(9)ht=Htβ−1∫01gsds,withH≥−γββ−ρΓβ.To prove the main results, we need the following assumptions:(H1)
f:0,1×−r^,θ∪K⟶E is continuous and g:0,1⟶K is continuous, and ft,u≤ft,v for −r^≤u≤v.(H2)
Forλ∈0,1 and y∈θ,r^, there is 1>Ψλ>λ such that ft,λx+λ−1y≥Ψλft,x,t∈0,1,x∈E.(H3)
ft,θ≥θ with ft,θ≡θ for t∈0,1, and there are two constants a2≥a1>0 such that(10)∫01t1−tβ−1ft,θdt≥a1∫01gtdt,ft,H∫01gsds≤a2gt,t∈0,1.Theorem 1.
Assumee∈K and (H1)–(H3) are satisfied; then, problem (4) has a unique solution u∗ in Ph,r, where randh are given in (8) and (9). Furthermore, set a sequence by(11)vnt=∫01Gt,sfs,vn−1sds+γ∫01Gt,sgsds,n=1,2,…,for any taken v0∈Ph,r, and one has vnt⟶u∗t as n⟶∞.Proof.
Fort∈0,1, by Lemma 4, one has(12)rt=−γ∫01Gt,sgsds≤−γ∫011−sβ−1β−ρ+ρsβ−ρΓβtβ−1gsds,≤−γββ−ρΓβ∫01gsds⋅tβ−1≤Htβ−1∫01gsds=ht.
Obviously, it followsrt≥θ, from γ<0 and Lemma 4.
Asθ≤rt≤ht, we have r∈P. In the following, we will consider a set Ph,r=u∈CJ,E|u+r∈Ph. By means of Lemma 3, if problem (4) has a solution u, then(13)ut=∫01Gt,sfs,us+γgsds=∫01Gt,sfs,usds+γ∫01Gt,sgsds=∫01Gt,sfs,usds−rt.
So, for anyu∈Ph,r and t∈0,1, we can define an operator by Tut=∫01Gt,sfs,usds−rt. Hence, ut is a solution of problem (4) if and only if u is a fixed point of T.
Now, we show thatT:Ph,r⟶E satisfied the definition of Ψ−h,r-concave operator. For every λ∈0,1, u∈Ph,r, and from condition (H2), one has(14)Tλu+λ−1rt=∫01Gt,sfs,λus+λ−1rsds−rt≥Ψλ∫01Gt,sfs,usds−Ψλrt+Ψλ−1rt=ΨλTut+Ψλ−1rt.
Therefore, we obtain(15)Tλu+λ−1r≥ΨλTu+Ψλ−1r,λ∈0,1,u∈Ph,r.
In view of Definition1, T is a Ψ−h,r-concave operator.
Next, we prove thatT:Ph,r⟶E is increasing. Since u∈Ph,r, we get u+r∈Ph, so there is τ>0 such that ut+rt≥τht; thus, we obtain(16)ut≥τht−rt≥−rt≥−r^.
By condition (H1),T:Ph,r⟶E is increasing.
The main step is to prove thatTh∈Ph,r; in other words, we prove Th+r∈Ph. By (H1), (H3), and Lemma 4, one has(17)Tht+rt=∫01Gt,sfs,hsds−rt+rt=∫01Gt,sfs,hsds=∫01Gt,sfs,Hsβ−1∫01gzdzds≤∫011−sβ−1β−ρ+ρsβ−ρΓβtβ−1fs,H∫01gzdzds≤tβ−1β−ρΓβ∫011−sβ−1β−ρ+ρsa2gsds≤a2ββ−ρΓβtβ−1∫01gsds=a2βHβ−ρΓβHtβ−1∫01gsds=a2βHβ−ρΓβht,Tht+rt=∫01Gt,sfs,hsds=∫01Gt,sfs,Hsβ−1∫01gzdzds≥∫011−sβ−1ρsβ−ρΓβtβ−1fs,θds=ρβ−ρΓβ∫01s1−sβ−1fs,θds⋅tβ−1≥a1ρβ−ρΓβ∫01gsds⋅tβ−1=a1ρHβ−ρΓβH∫01gsds⋅tβ−1=a1ρHβ−ρΓβht.
Let(18)l1=a1ρHβ−ρΓβ,l2=a2βHβ−ρΓβ.
SinceΓβ>0, H>0, and from (H1) and (H3), we can easily get l2≥l1>0. It follows that Th+r∈Ph.
Finally, by using Lemma1, operator T has a unique fixed point u∗ in Ph,r, and then,(19)u∗t=∫01Gt,sfs,u∗sds−rt,t∈0,1.
Consequently,u∗t is the unique solution of problem (4) in Ph,r. Taking any v0∈Ph,r, the sequence vn=Tvn−1, n=1,2,…, satisfies vn⟶u∗ as n⟶∞. That is,(20)vnt=∫01Gt,sfs,vn−1sds+γ∫01Gt,sgsds,n=1,2,…,and vnt⟶u∗t as n⟶∞.


### 3.2. Investigation for the Caseγ≥0

In this section, we can get the unique result by using Lemma2. And, some other assumptions are listed in the following:(H4)
f:0,1×K⟶K is continuous with ft,θ≡θ for t∈0,1, g:0,1⟶K is continuous.(H5)
Fort∈0,1, ft,x is increasing with respect to the second variable, and there are l1,l2>0 such that(21)l1e=ρβ−ρΓβ∫01s1−sβ−1fs,θ+γgsds,l2e=1β−ρΓβ∫011−sβ−1β−ρ+ρsfs,e+γgsds.(H6)
Forλ∈0,1, there exists 1>Ψλ>λ such that ft,λx≥Ψλft,x,∀t∈0,1,x∈K.Theorem 2.
Assume (H4)–(H6) are satisfied; then, for any takenγ≥0, problem (4) has a unique positive solution u∗ in Ph, where ht=tβ−1e,t∈0,1. Moreover, taking any initial value v0∈Ph, the sequence(22)vnt=∫01Gt,sfs,vn−1s+γgsds,n=1,2,…,satisfies vnt⟶u∗t as n⟶∞.Proof.
Foru∈Ph, we define an operator T by(23)Tut=∫01Gt,sfs,us+γgsds,t∈0,1,with Gt,s given as in Lemma 3.
From Lemma3, ut is the solution of problem (4) if and only if u is the fixed point of T. Noting that ft,x≥θ and gt≥θ, it is easy to check that T:P⟶P. From conditions (H4) and (H5), it is easy to prove that T:P⟶P is increasing.
Next, we show thatT:P⟶P satisfied the definition of Ψ−h,θ-concave operator. From (H6), for λ∈0,1 and u∈P, we obtain(24)Tλut=∫01Gt,sfs,λus+γgsds≥Ψλ∫01Gt,sfs,usds+γ∫01Gt,sgsds≥ΨλTut,t∈0,1.
Hence,Tλu≥ΨλTu,∀u∈P,andλ∈0,1.
In the following, we prove thatTh∈Ph. Firstly, note that ht=tβ−1e, and from (H4), (H5), and Lemma 4, it follows that(25)Tht=∫01Gt,sfs,hs+γgsds=∫01Gt,sfs,sβ−1e+γgsds≥∫011−sβ−1ρsβ−ρΓβtβ−1fs,θ+γgsds=ρβ−ρΓβ∫01s1−sβ−1fs,θ+γgsds⋅tβ−1=l1e⋅tβ−1=l1ht.
Secondly, fort∈0,1, one has(26)Tht=∫01Gt,sfs,sβ−1e+γgsds≤∫011−sβ−1β−ρ+ρsβ−ρΓβtβ−1fs,e+γgsds=tβ−1β−ρΓβ∫011−sβ−1β−ρ+ρsfs,e+γgsds=l2e⋅tβ−1=l2ht.
Sincee∈P, from (H5), it is easy to prove that(27)l2−l1e=l2e−l1e,=1β−ρΓβ∫011−sβ−1β−ρ+ρsfs,e+γgsds−ρβ−ρΓβ∫01s1−sβ−1fs,θ+γgsds=1Γβ∫011−sβ−1fs,e+γgsds+ρβ−ρΓβ∫01s1−sβ−1fs,e+γgsds−ρβ−ρΓβ∫01s1−sβ−1fs,θ+γgsds=1Γβ∫011−sβ−1fs,e+γgsds+ρβ−ρΓβ∫01s1−sβ−1fs,e−fs,θds≥θ.
Hencel2≥l1>0. It follows that Th∈Ph. Finally, by using Lemma 2, the operator T has a unique fixed point u∗ in Ph, and(28)u∗t=∫01Gt,sfs,u∗sds+γgsds,t∈0,1.
Consequently,u∗t is the unique solution of problem (4) in Ph. For v0∈Ph, the sequence vn=Tvn−1, n=1,2,…, satisfies vn⟶u∗ as n⟶∞. That is,(29)vnt=∫01Gt,sfs,vn−1s+γgsds,n=1,2,…,satisfies vnt⟶u∗t as n⟶∞.Remark 1.
The form of problem (4) is more general. Our method is new to study fractional differential equations in Banach spaces, which guarantees the existence and uniqueness of solutions or positive solutions. We also can make an iteration to approximate the unique solution.


### 3.3. Direct Corollaries

LetE=R, ℬ = C0,1, and P=u∈C0,1,ut≥0,t∈0,1. We can get some obvious results from Theorems 1 and 2. Set(30)rt=−γ∫01Gt,sgsds,∀t∈0,1,(31)r^=−γββ−ρΓβ∫01gsds,ht=Htβ−1∫01gsds,withH≥−γββ−ρΓβ.(H1’)
f:0,1×−r^,+∞⟶−∞,+∞ is continuous, ft,u≤ft,v for −r^≤u≤v<+∞, and g:0,1⟶0,+∞ is continuous.(H2’)
Forλ∈0,1 and y∈0,r^, there is Ψλ>λ such that ft,λx+λ−1y≥Ψλft,x,∀t∈0,1,x∈−∞,+∞.(H3’)
ft,0≥0,ft,0≡0 for t∈0,1, and there are two constants a2≥a1>0 such that(32)∫01t1−tβ−1ft,0dt≥a1∫01gtdt,ft,H∫01gsds≤a2gt,∀t∈0,1.Corollary 1.
Letβ∈2,3 and ρ∈0,β. Assume that (H1’)–(H3’) hold; then, for γ<0, the boundary value problem(33)−D0+βut=ft,ut+γgt,0≤t≤1,u0=u′0=0,u1=ρ∫01utdt,has a unique solutionu∗ in Ph,r, where randh are given in (30 and 31). Furthermore, set a sequence(34)vnt=∫01Gt,sfs,vn−1sds+γ∫01Gt,sgsds,n=1,2,…,for any takenv0∈Ph,r, and one has vnt⟶u∗t as n⟶∞.Corollary 2.
Letβ∈2,3 and ρ∈0,β. Assume that(H4’)
f:0,1×0,+∞⟶0,+∞ is continuous with ft,0≡0 for t∈0,1 and g:0,1⟶0,+∞ is continuous(H5’)
Fort∈0,1, ft,x is increasing in the second variable(H6’)
Forλ∈0,1, there exists Ψλ>λ such that ft,λx≥Ψλft,x,∀t∈0,1,x∈0,+∞; then, for γ≥0, the boundary value problem(35)−D0+βut=ft,ut+γgt,0<t<1,u0=u′0=0,u1=ρ∫01utdt,has a unique solution u∗ in Ph, where ht=tβ−1. Furthermore, put a sequence by(36)vnt=∫01Gt,sfs,vn−1s+γgsds,n=1,2,…,for any taken v0∈Ph, and one has vnt⟶u∗t as n⟶∞.


## 3.1. Investigation for the Caseγ<0

Let(8)rt=−γ∫01Gt,sgsds,t∈0,1,r^=−γββ−ρΓβ∫01gsds,(9)ht=Htβ−1∫01gsds,withH≥−γββ−ρΓβ.To prove the main results, we need the following assumptions:(H1)
f:0,1×−r^,θ∪K⟶E is continuous and g:0,1⟶K is continuous, and ft,u≤ft,v for −r^≤u≤v.(H2)
Forλ∈0,1 and y∈θ,r^, there is 1>Ψλ>λ such that ft,λx+λ−1y≥Ψλft,x,t∈0,1,x∈E.(H3)
ft,θ≥θ with ft,θ≡θ for t∈0,1, and there are two constants a2≥a1>0 such that(10)∫01t1−tβ−1ft,θdt≥a1∫01gtdt,ft,H∫01gsds≤a2gt,t∈0,1.Theorem 1.
Assumee∈K and (H1)–(H3) are satisfied; then, problem (4) has a unique solution u∗ in Ph,r, where randh are given in (8) and (9). Furthermore, set a sequence by(11)vnt=∫01Gt,sfs,vn−1sds+γ∫01Gt,sgsds,n=1,2,…,for any taken v0∈Ph,r, and one has vnt⟶u∗t as n⟶∞.Proof.
Fort∈0,1, by Lemma 4, one has(12)rt=−γ∫01Gt,sgsds≤−γ∫011−sβ−1β−ρ+ρsβ−ρΓβtβ−1gsds,≤−γββ−ρΓβ∫01gsds⋅tβ−1≤Htβ−1∫01gsds=ht.
Obviously, it followsrt≥θ, from γ<0 and Lemma 4.
Asθ≤rt≤ht, we have r∈P. In the following, we will consider a set Ph,r=u∈CJ,E|u+r∈Ph. By means of Lemma 3, if problem (4) has a solution u, then(13)ut=∫01Gt,sfs,us+γgsds=∫01Gt,sfs,usds+γ∫01Gt,sgsds=∫01Gt,sfs,usds−rt.
So, for anyu∈Ph,r and t∈0,1, we can define an operator by Tut=∫01Gt,sfs,usds−rt. Hence, ut is a solution of problem (4) if and only if u is a fixed point of T.
Now, we show thatT:Ph,r⟶E satisfied the definition of Ψ−h,r-concave operator. For every λ∈0,1, u∈Ph,r, and from condition (H2), one has(14)Tλu+λ−1rt=∫01Gt,sfs,λus+λ−1rsds−rt≥Ψλ∫01Gt,sfs,usds−Ψλrt+Ψλ−1rt=ΨλTut+Ψλ−1rt.
Therefore, we obtain(15)Tλu+λ−1r≥ΨλTu+Ψλ−1r,λ∈0,1,u∈Ph,r.
In view of Definition1, T is a Ψ−h,r-concave operator.
Next, we prove thatT:Ph,r⟶E is increasing. Since u∈Ph,r, we get u+r∈Ph, so there is τ>0 such that ut+rt≥τht; thus, we obtain(16)ut≥τht−rt≥−rt≥−r^.
By condition (H1),T:Ph,r⟶E is increasing.
The main step is to prove thatTh∈Ph,r; in other words, we prove Th+r∈Ph. By (H1), (H3), and Lemma 4, one has(17)Tht+rt=∫01Gt,sfs,hsds−rt+rt=∫01Gt,sfs,hsds=∫01Gt,sfs,Hsβ−1∫01gzdzds≤∫011−sβ−1β−ρ+ρsβ−ρΓβtβ−1fs,H∫01gzdzds≤tβ−1β−ρΓβ∫011−sβ−1β−ρ+ρsa2gsds≤a2ββ−ρΓβtβ−1∫01gsds=a2βHβ−ρΓβHtβ−1∫01gsds=a2βHβ−ρΓβht,Tht+rt=∫01Gt,sfs,hsds=∫01Gt,sfs,Hsβ−1∫01gzdzds≥∫011−sβ−1ρsβ−ρΓβtβ−1fs,θds=ρβ−ρΓβ∫01s1−sβ−1fs,θds⋅tβ−1≥a1ρβ−ρΓβ∫01gsds⋅tβ−1=a1ρHβ−ρΓβH∫01gsds⋅tβ−1=a1ρHβ−ρΓβht.
Let(18)l1=a1ρHβ−ρΓβ,l2=a2βHβ−ρΓβ.
SinceΓβ>0, H>0, and from (H1) and (H3), we can easily get l2≥l1>0. It follows that Th+r∈Ph.
Finally, by using Lemma1, operator T has a unique fixed point u∗ in Ph,r, and then,(19)u∗t=∫01Gt,sfs,u∗sds−rt,t∈0,1.
Consequently,u∗t is the unique solution of problem (4) in Ph,r. Taking any v0∈Ph,r, the sequence vn=Tvn−1, n=1,2,…, satisfies vn⟶u∗ as n⟶∞. That is,(20)vnt=∫01Gt,sfs,vn−1sds+γ∫01Gt,sgsds,n=1,2,…,and vnt⟶u∗t as n⟶∞.


## 3.2. Investigation for the Caseγ≥0

In this section, we can get the unique result by using Lemma2. And, some other assumptions are listed in the following:(H4)
f:0,1×K⟶K is continuous with ft,θ≡θ for t∈0,1, g:0,1⟶K is continuous.(H5)
Fort∈0,1, ft,x is increasing with respect to the second variable, and there are l1,l2>0 such that(21)l1e=ρβ−ρΓβ∫01s1−sβ−1fs,θ+γgsds,l2e=1β−ρΓβ∫011−sβ−1β−ρ+ρsfs,e+γgsds.(H6)
Forλ∈0,1, there exists 1>Ψλ>λ such that ft,λx≥Ψλft,x,∀t∈0,1,x∈K.Theorem 2.
Assume (H4)–(H6) are satisfied; then, for any takenγ≥0, problem (4) has a unique positive solution u∗ in Ph, where ht=tβ−1e,t∈0,1. Moreover, taking any initial value v0∈Ph, the sequence(22)vnt=∫01Gt,sfs,vn−1s+γgsds,n=1,2,…,satisfies vnt⟶u∗t as n⟶∞.Proof.
Foru∈Ph, we define an operator T by(23)Tut=∫01Gt,sfs,us+γgsds,t∈0,1,with Gt,s given as in Lemma 3.
From Lemma3, ut is the solution of problem (4) if and only if u is the fixed point of T. Noting that ft,x≥θ and gt≥θ, it is easy to check that T:P⟶P. From conditions (H4) and (H5), it is easy to prove that T:P⟶P is increasing.
Next, we show thatT:P⟶P satisfied the definition of Ψ−h,θ-concave operator. From (H6), for λ∈0,1 and u∈P, we obtain(24)Tλut=∫01Gt,sfs,λus+γgsds≥Ψλ∫01Gt,sfs,usds+γ∫01Gt,sgsds≥ΨλTut,t∈0,1.
Hence,Tλu≥ΨλTu,∀u∈P,andλ∈0,1.
In the following, we prove thatTh∈Ph. Firstly, note that ht=tβ−1e, and from (H4), (H5), and Lemma 4, it follows that(25)Tht=∫01Gt,sfs,hs+γgsds=∫01Gt,sfs,sβ−1e+γgsds≥∫011−sβ−1ρsβ−ρΓβtβ−1fs,θ+γgsds=ρβ−ρΓβ∫01s1−sβ−1fs,θ+γgsds⋅tβ−1=l1e⋅tβ−1=l1ht.
Secondly, fort∈0,1, one has(26)Tht=∫01Gt,sfs,sβ−1e+γgsds≤∫011−sβ−1β−ρ+ρsβ−ρΓβtβ−1fs,e+γgsds=tβ−1β−ρΓβ∫011−sβ−1β−ρ+ρsfs,e+γgsds=l2e⋅tβ−1=l2ht.
Sincee∈P, from (H5), it is easy to prove that(27)l2−l1e=l2e−l1e,=1β−ρΓβ∫011−sβ−1β−ρ+ρsfs,e+γgsds−ρβ−ρΓβ∫01s1−sβ−1fs,θ+γgsds=1Γβ∫011−sβ−1fs,e+γgsds+ρβ−ρΓβ∫01s1−sβ−1fs,e+γgsds−ρβ−ρΓβ∫01s1−sβ−1fs,θ+γgsds=1Γβ∫011−sβ−1fs,e+γgsds+ρβ−ρΓβ∫01s1−sβ−1fs,e−fs,θds≥θ.
Hencel2≥l1>0. It follows that Th∈Ph. Finally, by using Lemma 2, the operator T has a unique fixed point u∗ in Ph, and(28)u∗t=∫01Gt,sfs,u∗sds+γgsds,t∈0,1.
Consequently,u∗t is the unique solution of problem (4) in Ph. For v0∈Ph, the sequence vn=Tvn−1, n=1,2,…, satisfies vn⟶u∗ as n⟶∞. That is,(29)vnt=∫01Gt,sfs,vn−1s+γgsds,n=1,2,…,satisfies vnt⟶u∗t as n⟶∞.Remark 1.
The form of problem (4) is more general. Our method is new to study fractional differential equations in Banach spaces, which guarantees the existence and uniqueness of solutions or positive solutions. We also can make an iteration to approximate the unique solution.


## 3.3. Direct Corollaries

LetE=R, ℬ = C0,1, and P=u∈C0,1,ut≥0,t∈0,1. We can get some obvious results from Theorems 1 and 2. Set(30)rt=−γ∫01Gt,sgsds,∀t∈0,1,(31)r^=−γββ−ρΓβ∫01gsds,ht=Htβ−1∫01gsds,withH≥−γββ−ρΓβ.(H1’)
f:0,1×−r^,+∞⟶−∞,+∞ is continuous, ft,u≤ft,v for −r^≤u≤v<+∞, and g:0,1⟶0,+∞ is continuous.(H2’)
Forλ∈0,1 and y∈0,r^, there is Ψλ>λ such that ft,λx+λ−1y≥Ψλft,x,∀t∈0,1,x∈−∞,+∞.(H3’)
ft,0≥0,ft,0≡0 for t∈0,1, and there are two constants a2≥a1>0 such that(32)∫01t1−tβ−1ft,0dt≥a1∫01gtdt,ft,H∫01gsds≤a2gt,∀t∈0,1.Corollary 1.
Letβ∈2,3 and ρ∈0,β. Assume that (H1’)–(H3’) hold; then, for γ<0, the boundary value problem(33)−D0+βut=ft,ut+γgt,0≤t≤1,u0=u′0=0,u1=ρ∫01utdt,has a unique solutionu∗ in Ph,r, where randh are given in (30 and 31). Furthermore, set a sequence(34)vnt=∫01Gt,sfs,vn−1sds+γ∫01Gt,sgsds,n=1,2,…,for any takenv0∈Ph,r, and one has vnt⟶u∗t as n⟶∞.Corollary 2.
Letβ∈2,3 and ρ∈0,β. Assume that(H4’)
f:0,1×0,+∞⟶0,+∞ is continuous with ft,0≡0 for t∈0,1 and g:0,1⟶0,+∞ is continuous(H5’)
Fort∈0,1, ft,x is increasing in the second variable(H6’)
Forλ∈0,1, there exists Ψλ>λ such that ft,λx≥Ψλft,x,∀t∈0,1,x∈0,+∞; then, for γ≥0, the boundary value problem(35)−D0+βut=ft,ut+γgt,0<t<1,u0=u′0=0,u1=ρ∫01utdt,has a unique solution u∗ in Ph, where ht=tβ−1. Furthermore, put a sequence by(36)vnt=∫01Gt,sfs,vn−1s+γgsds,n=1,2,…,for any taken v0∈Ph, and one has vnt⟶u∗t as n⟶∞.


## 4. Examples

To illustrate our main results, we present two examples.Example 1.
Consider the following boundary value problem:(37)−D0+5/2ut=ft,ut−2t,0<t<1,u0=u′0=0,u1=12∫01utdt,where(38)ft,u=835u+157Γ7/2t5/2−1625u+2Γ7/2t7/21/3,and β=5/2,ρ=1/2,γ=−2,gt=t,andt∈0,1. After a simple calculation,(39)rt=−γ∫01Gt,sgsds=2t5/21−tΓ7/2+t5/22Γ9/2=2Γ7/2+17Γ7/2t5/2−2Γ7/2t7/2,t∈0,1,where Gt,s is given as Lemma 1. Moreover, ht=Ht3/2 with H≥5/4Γ5/2,∀t∈0,1. Then,(40)rt=2t5/21−tΓ7/2+t5/22Γ9/2≥0,rt=2Γ7/2+17Γ7/2t5/2−2Γ7/2t7/2≤2+171Γ7/2t5/2<54Γ5/2t3/2≤Ht3/2=ht.
Furthermore,r^=5/4Γ5/2. We can see that f:0,1×−5/4Γ5/2,+∞⟶−∞,+∞ is continuous and increasing in the second variable, and(41)ft,0=157Γ7/2t5/2−2Γ7/2t7/21/3=rt1/3≥0,t∈0,1,with ft,0≡0, t∈0,1. Thus, conditions (H1’) and (H3’) are satisfied. Obviously, it can be expressed as ft,ut=rt/r^ut+rt1/3 and(42)rtr^=835t5/2−1625t7/2,t∈0,1.
From the Remark 4 in [43], we obtain(43)ft,λx+λ−1y≥Ψλft,x,∀t∈0,1,x∈−∞,+∞,y∈0,r^,where Ψλ=λ1/3>λ,λ∈0,1; then, condition (H1’) is satisfied. In view of Corollary 1, problem (37) has a unique solution u∗ in Ph,r. For v0∈Ph,r, let(44)vnt=∫01Gt,sfs,vn−1sds+−2∫01Gt,stds=∫01Gt,s835vn−1s+157Γ7/2s5/2−1625vn−1s+2Γ7/2s7/21/3ds−2Γ7/2+17Γ7/2t5/2+2Γ7/2t7/2,n=1,2,…
We havevnt⟶u∗t as n⟶∞, t∈0,1.Example 2.
Consider the following boundary value problem:(45)−D0+5/2ut=53t2+tu+25/2+γgt,0<t<1,u0=u′0=0,u1=12∫01utdt,where ft,u=5/3t2+tu+22/5,t∈0,1, and β=5/2,ρ=1/2,γ=0,andgt≥0,t∈0,1 are continuous. Take ht=tβ−1,t∈0,1. It is clear that f:0,1×0,+∞⟶0,+∞ is continuous and increasing in the second variable, ft,0=10/3t2+2t2/5≥0 with ft,0≡0, t∈0,1, and thus, conditions (H4’) and (H5’) are satisfied. Next, for λ∈0,1,(46)ft,λu=53t2+tλu+22/5≥53t2+tλu+2λ2/5,=λ2/553t2+tu+22/5,t∈0,1,where Ψλ=λ2/5>λ,λ∈0,1; then, condition (H6’) is satisfied. In view of Corollary 2, problem (45) has a unique solution u∗ in Ph. For v0∈Ph, set(47)vnt=∫01Gt,s53t2+tvn−1t+22/5,n=1,2,…,And we have vnt⟶u∗t as n⟶∞, t∈0,1.


## 5. Conclusion

In this paper, we study an integral boundary value problem (4) with sign-changed parameter in Banach spaces. By taking two functions r and h and using fixed point theorems of increasing Ψ−h,r-concave operators defined on ordered set Ph,r, we establish some new existence and uniqueness criteria for problem (4) dependent on different parameters, which also give some different answers to the same type of fractional differential equations’ boundary value problem in the literature [3]. Our results can guarantee the existence of a unique solution with not supposing the existence of upper-lower solutions. As applications, two good examples are presented to illustrate the main results.



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