# Existence of Uniqueness and Nonexistence Results of Positive Solution for Fractional Differential Equations Integral Boundary Value Problems

Authors: Yongqing Wang
Journal: Journal of Function Spaces
 (2018)

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## Abstract

In this paper, we consider a class of fractional differential equations with conjugate type integral conditions. Both the existence of uniqueness and nonexistence of positive solution are obtained by means of the iterative technique. The interesting point lies in that the assumption on nonlinearity is closely associated with the spectral radius corresponding to the relevant linear operator.


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## Body

## 1. Introduction

In this paper, we consider the existence of uniqueness and nonexistence of positive solution for the following fractional differential equations:(1)D0+αut+ft,ut=0,0<t<1,u0=u′0=⋯=un-20=0,D0+βu1=∫0ηatD0+γutdt,where n-1<α≤n, n≥3, 0<β<1, Γ(α-β)∫0η‍a(t)tα-γ-1dt<Γ(α-γ), η∈(0,1], D0+α is the standard Riemann-Liouville derivative, f:(0,1)×[0,+∞)→[0,+∞) is continuous, and a(t)∈L1[0,1] is nonnegative.In the recent years, many results were obtained to deal with the existence of solutions for nonlinear differential equations by using nonlinear analysis methods; see [1–16] and references therein. The fractional nonlocal boundary value problems have particularly attracted a great deal of attention (see [17–27]). While there are a lot of works dealing with the existence and multiplicity of solutions for nonlinear fractional differential equations, the results dealing with the uniqueness of solution are relatively scarce (see [28–35]). The main tool used in most of the papers dealing with the uniqueness of solution is the Banach contraction map principle provided that the nonlinearity f is a Lipschitz continuous function. When 1≤β<α-1, and f is continuous on [0,1]×(-∞,+∞), Zhang and Zhong [34] established the uniqueness results of solution to problem (1) by using the Banach contraction map principle. It is worth mentioning that only positive solutions are meaningful in most practical problems. As far as we know, the nonexistence of positive solution has seldom been considered up to now.Motivated by the above work, the aim of this paper is to establish the existence of uniqueness and nonexistence of positive solution to problem (1). Our analysis relies on the iterative technique on the cone derived from the properties of the Green function. This article provides some new insights. Firstly, the uniqueness results are obtained under some conditions concerning the spectral radius with respect to the relevant linear operator. In addition, the error estimation of the iterative sequences is given. Secondly, we impose weaker positivity conditions on f; that is, the Lipschitz constant is generalized to a function and f(t,x) may be singular at t=0,1. Finally, the nonexistence results of positive solution are obtained under conditions concerning the spectral radius of the relevant linear operator.


## 2. Preliminaries

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory and lemmas.Definition 1.
The fractional integral of orderα>0 of a function u:(0,+∞)→R is given by(2)I0+αut=1Γα∫0tt-sα-1usdsprovided that the right-hand side is point-wise defined on (0,+∞).Definition 2.
The Riemann-Liouville fractional derivative of orderα>0 of a function u:(0,+∞)→R is given by (3)D0+αut=1Γn-αddtn∫0tt-sn-α-1usds,where n=[α]+1; [α] denotes the integer part of number α, provided that the right-hand side is point-wise defined on (0,+∞).For convenience, we here list the assumptions to be used throughout the paper.( A 1 )  a(t)∈L1[0,1] is nonnegative, and(4)Δ≔Γα-γ-Γα-β∫0ηattα-γ-1dt>0;( A 2 )  f:(0,1)×[0,+∞)→[0,+∞) is continuous.Lemma 3 ([34]).
For anyy∈L[0,1]∩C(0,1), the unique solution of the boundary value problem(5)D0+αut+yt=0,0<t<1,u0=u′0=⋯=un-20=0,D0+βu1=∫0ηatD0+γut,is(6)ut=∫01Gt,sysds,where(7)Gt,s=G1t,s+hstα-1,G1t,s=1Γαtα-11-sα-β-1,0≤t≤s≤1,tα-11-sα-β-1-t-sα-1,0≤s≤t≤1,G2t,s=1Γαtα-γ-11-sα-β-1,0≤t≤s≤1,tα-γ-11-sα-β-1-t-sα-γ-1,0≤s≤t≤1,hs=Γα-γΔ∫0ηatG2t,sdt.Lemma 4.
The functionG1(t,s) has the following properties:(1)
G1(t,s)>0,∀t,s∈(0,1);(2)
ΓαG1t,s≤tα-11-sα-β-1,∀t,s∈[0,1];(3)
βs1-sα-β-1tα-1≤ΓαG1t,s≤s1-sα-β-1,∀t,s∈[0,1].Proof.
It is obvious that (1) and (2) hold. In the following, we will prove (3).
Case (i) (0<s≤t<1). Noticing α>2, we have (8)∂∂ttα-1-t-sα-11-sα-2=α-1tα-21-t-st1-sα-2≥0,which implies(9)tα-1-t-sα-11-sα-2≤1-1-s=s.Noticing 0<β<1, we have (10)tα-11-sα-β-1-t-sα-1=1-sα-β-1tα-1-t-sα-11-sα-β-1≤1-sα-β-1tα-1-t-sα-11-sα-2≤s1-sα-β-1.
On the other hand, it follows from(11)∂∂sβs+1-sβ≤0,∀s∈0,1that(12)βs+1-sβ≤1,∀s∈0,1.Therefore,(13)tα-11-sα-β-1-t-sα-1≥tα-11-sα-β-1-t-sβt-tsα-β-1=tα-11-1-stβ1-sα-β-1≥tα-11-1-sβ1-sα-β-1≥βs1-sα-β-1tα-1.
Case (ii) (0≤t≤s≤1). It is easy to see that (14)tα-11-sα-β-1≤sα-11-sα-β-1≤s1-sα-β-1.
On the other hand, we have(15)tα-11-sα-β-1≥stα-11-sα-β-1≥βs1-sα-β-1tα-1.
It follows from (10), (13), (14), and (15) that (3) holds.Lemma 5.
The functionG(t,s) has the following properties:(1)
G(t,s)>0,∀t,s∈(0,1);(2)
Gt,s≤tα-1Φ1s,∀t,s∈[0,1];(3)
βtα-1Φ2s≤Gt,s≤Φ2s,∀t,s∈[0,1],
where(16)Φ1s=1-sα-β-1Γα+hs,Φ2s=s1-sα-β-1Γα+hs.Proof.
It can be directly deduced from Lemma4 and the definition of G(t,s), so we omit the proof.LetE=C[0,1] be endowed with the maximum norm u=max0≤t≤1u(t), Br={u∈E:u<r}. Define cones P, Q by(17)P=u∈E:ut≥0,Q=u∈P:thereexistslu>0suchthatβutα-1≤ut≤lutα-1.It is clear that Q is nonempty set since tα-1∈Q.For convenience, we list here one more assumption to be used later:( A 3 ) There exists λ∈C(0,1)∩L[0,1] such that(18)ft,x-ft,y≤λtx-y,t∈0,1,x,y∈0,∞.Moreover,(19)0<∫01Φ1sλsds<+∞;0<∫01Φ1sfs,0ds<+∞.Define operatorsA and T as follows:(20)Aut=∫01Gt,sfs,usdsTλut=∫01Gt,sλsusds.Lemma 6.
Assume that(A3) holds; then A:P→Q.Proof.
It is clear thatf(t,x)≤λ(t)x+f(t,0),  ∀x≥0. For any u∈P, we have (21)Aut=∫01Gt,sfs,usds≤∫01Φ1sλsuds+∫01Φ1sfs,0ds.Then A is well-defined on P. It follows from Lemma 5 that A:P→Q.By virtue of the Krein-Rutmann theorem and Lemma5, we have the following lemma.Lemma 7.
Assume that(A3) holds. Then Tλ:P→Q is a completely continuous linear operator. Moreover, the spectral radius r(Tλ)>0 and Tλ has a positive eigenfunction φ1 corresponding to its first eigenvalue rTλ-1; that is, Tλφ1=r(Tλ)φ1.


## 3. Main Results

Theorem 8.
Assume that(A3) holds. Then (1) has a unique positive solution if the spectral radius r(Tλ)∈(0,1).Proof.
It follows from0<∫01‍Φ1(s)f(s,0)ds<+∞ that θ is not a fixed point of A. Then we only need to prove that A has a unique fixed point in Q.
Firstly, we will proveA has a fixed point in Q.
For anyu∈Q, let (22)lu=∫01Φ1sλsusds.Then (23)βutα-1≤Tλut≤lutα-1.By Lemma 7, we have that Tλφ1=r(Tλ)φ1. It is easy to see that (24)βφ1rTλtα-1≤φ1t≤lφ1rTλtα-1.
For anyu0∈Q, set (25)un=Aun-1,n=1,2,⋯.We may suppose that u1-u0≠θ (otherwise, the proof is finished). It follows from (23) and (24) that (26)Tλu1-u0≤rTλlu1-u0βφ1φ1.Then, we have (27)u2-u1=∫01Gt,sfs,u1s-fs,u0sds≤∫01Gt,sλsu1s-u0sds≤rTλlu1-u0βφ1φ1.By induction, we can get (28)un+1-un≤rTλnlu1-u0βφ1φ1,n=1,2,⋯.Then, for any n,m∈N, one has (29)un+m-un≤un+m-un+m-1+⋯+un+1-un≤rTλn+m-1+⋯+rTλnlu1-u0βφ1φ1≤rTλnlu1-u01-rLλβφ1φ1.It follows from r(Tλ)<1 that(30)un+m-um→0n→∞,which implies {un} is a Cauchy sequence. Therefore, there exists u∗∈Q such that {un} converges to u∗. Clearly u∗ is a fixed point of A.
In the following, we will prove the fixed point ofA is unique.
Supposev≠u∗ is a positive fixed point of A. Then there exists l|u∗-v|>0 such that (31)Tλu∗-v≤rTλlu∗-vβφ1φ1.Therefore,(32)Au∗-Av=∫01Gt,sfs,u∗s-fs,vsds≤∫01Gt,sλsu∗s-vsds≤rTλlu∗-vβφ1φ1.By induction, we can get (33)Anu∗-Anv≤rTλnlu∗-vβφ1φ1.It follows from r(Tλ)<1 that (34)u∗-v=Anu∗-Anv→0n→∞,which implies the positive fixed point of A is unique.Remark 9.
The unique positive solutionu∗ of (1) can be approximated by the iterative schemes: for any u0∈Q, let (35)un=Aun-1,n=1,2,…,and then un→u∗. Furthermore, we have error estimation(36)un-u∗≤rTλnlu1-u01-rTλβφ1φ1,and with the rate of convergence (37)un-u∗=OrTλn.Remark 10.
The spectral radius satisfiesr(Tλ)=limn→∞Tλn1/n and r(Tλ)≤Tλn1/n. Particularly, (38)rTλ≤Tλ=sup0≤t≤1⁡∫01Gt,sλsds.Theorem 11.
Assume that the following condition holds:
( A 4 ) There exists λ1∈C(0,1)∩L[0,1] satisfying (39)0<∫01Φ1sλ1sds<+∞,such that(40)ft,x≤λ1tx,t∈0,1,x∈0,∞.Then (1) has no positive solution if the spectral radius r(Tλ1)∈(0,1), where (41)Tλ1ut=∫01Gt,sλ1susds.Proof.
We only need to prove thatA has no fixed point in Q∖{θ}. Otherwise, there exists v∈Q∖{θ}, such that Av=v.
By Lemma7, we have that the spectral radius r(Tλ1)>0 and Tλ1 has a positive eigenfunction ψ1 satisfying (42)Tλ1ψ1=rTλ1ψ1.It is clear that ψ1∈Q∖{θ}. Therefore, there exists c1>0 such that (43)v≤c1ψ1.It follows from f(t,x)≤b1(t)x that v=Av≤Tλ1v. It is obvious that Tλ1 is increasing on Q. By induction, we can get v≤Tλ1nv, ∀n=1,2,3,⋯. Thus, (44)v≤Tλ1nv≤Tλ1nc1ψ1=c1rTλ1nψ1,∀n=1,2,3,⋯.Noticing r(Tλ1)<1, we have v=θ, which contradicts with v∈Q∖{θ}.Theorem 12.
Assume that the following condition holds:
( A 5 ) There exists λ2∈C(0,1)∩L[0,1] satisfying (45)0<∫01Φ1sλ2sds<+∞,such that(46)ft,x≥λ2tx,t∈0,1,x∈0,∞.Then (1) has no positive solution if the spectral radius r(Tλ2)>1, where (47)Tλ2ut=∫01Gt,sλ2susds.Proof.
Suppose that there existsv∈Q∖{θ}, such that Av=v.
By Lemma7, we have that the spectral radius r(Tλ2)>0 and Tλ2 has a positive eigenfunction ψ2 satisfying (48)Tλ2ψ2=rTλ2ψ2.It is clear that ψ2∈Q∖{θ}. Therefore, there exists c2>0 such that (49)v≥c2ψ1.
It follows fromf(t,x)≥λ2(t)x that v=Av≥Tλ2v. Noticing that Tλ2 is increasing on Q, by induction, we have v≥Tλ2nv, ∀n=1,2,3,⋯. Thus, (50)v≥Tλ2nv≥Tλ2nc2ψ2=c2rTλ2nψ2,∀n=1,2,3,⋯.It follows from r(Tλ1)>1 that v=∞, which contradicts with v∈Q.


## 4. Example

Example 1.
Consider the following integral boundary value problem:(51)D0+5/2ut+ft,ut=0,0<t<1,u0=u′0=0,D0+1/2u1=∫011tD0+1/2utdt,with(52)ft,x=1+x+sin⁡x2Ct,where(53)C=83π-π2.By direct calculations, we have(54)Δ=Γ2-Γ2∫01ttdt=13.It is clear that (A1) and (A2) hold. Clearly, we have (55)G1t,s=43πt3/21-s,0≤t≤s≤1,t3/21-s-t-s3/2,0≤s≤t≤1,G2t,s=43πt1-s,0≤t≤s≤1,s1-t,0≤s≤t≤1,hs=163πs1-s,Φ1s=41-s3π+hs,Gt,s=G1t,s+hstα-1.Let λ(t)=1/Ct; then we have (56)ft,x-ft,y≤λtx-y,t∈0,1,x,y∈0,∞.It is easy to get that (A3) holds.Denote(57)et≡1,t∈0,1.By direct calculations, we have(58)Tλet=∫01Gt,sλsds=8t3/23πC-πt22C,(59)Tλ2et=∫01Gt,sλsTλesds=43πC257-27π280t3/2-835t7/2+9π21024t4≤43πC257-27π280t3/2-835t7/2+9π21024t7/2≈43πC20.41135t3/2-0.14183t7/2.∀t∈[0,1], we have (60)8t3/23πC-πt22C′=4t1/2-πtπC≥0,0.41135t3/2-0.14183t7/2′=0.617025t1/2-0.496405t5/2≥0.Therefore,(61)max0≤t≤1⁡Tλet=Tλe1=1,max0≤t≤1⁡Tλ2et≤43πC20.41135-0.14183≈0.39898.It is obvious that(62)Lλ=max0≤t≤1⁡Tλet=1,and(63)Lλ2=max0≤t≤1⁡Tλ2et<1.which implies that (64)Lλ21/2<1,Lemma 7 and Remark 10 can guarantee that (65)0<rLλ≤Lλ21/2<1<Lλ=1.So all of the assumptions of Theorem 8 are satisfied. As a result, BVP (51) has a unique positive solution.Example 2.
Consider BVP (51) with(66)ft,x=x+sin⁡x2Ct.Clearly, we have(67)ft,x≤λtx,t∈0,1,x∈0,∞.It is not difficult to check that (A4) holds.It follows from Example1 that (68)0<rLλ<1<Lλ=1.So all of the assumptions of Theorem 11 are satisfied. As a result, BVP (51) has no positive solution.


## 5. Conclusions

In this paper, we consider the existence of positive solution for fractional differential equations with conjugate type integral conditions. Both the existence of uniqueness and nonexistence of positive solution are established under conditions closely associated with the spectral radius with respect to the relevant linear operator. In addition, the unique positive solution can be approximated by an iterative scheme, and the error estimation of the iterative sequences is also given.



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