# Positive Solutions for Boundary Value Problems of Fractional Differential Equation with Integral Boundary Conditions

Authors: Qiao Sun; Hongwei Ji; Yujun Cui
Journal: Journal of Function Spaces
 (2018)

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## Abstract

By using two fixed-point theorems on cone, we discuss the existence results of positive solutions for the following boundary value problem of fractional differential equation with integral boundary conditions:D0+αx(t)+a(t)f(t,x(t))=0, t∈(0,1), x(0)=x′(0)=0, and x(1)=∫01x(t)dA(t).


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## Body

## 1. Introduction

Boundary value problem for fractional differential equation has aroused much attention in the past few years; many professors devoted themselves to the solvability of fractional differential equations, especially to the study of the existence of solutions for boundary value problems of fractional differential equation (see [1–28]). For example, Wang et al. [19] studied the existence of positive solutions for the following problem: (1)D0+αut+λft,ut=0,t∈0,1,with the boundary conditions (2)u0=u′0=u1=0,where D0+α is the Riemann-Liouville differential operator of 2<α≤3, λ is a positive parameter, and f may be singular at t=0,1 and may change sign. And Ma [14] discussed the positive solutions of (3)D0+αut+atft,ut=0,t∈0,1,u0=u′0=0,u1=∑i=1mβiuξi,where m≥1 is integer and ξi, βi>0.There have already been lots of books and papers involving the positive solutions for boundary value problems of fractional differential equation; however, only a few papers cover that for fractional differential equation boundary value problems with integral boundary conditions. Motivated by [14], we shall investigate the positive solutions of the following boundary value problem:(4)D0+αxt+atft,xt=0,t∈0,1,x0=x′0=0,x1=∫01xtdAt,where D0+α is the Riemann-Liouville differential operator of 2<α<3, A(t) is right continuous on 0,1, left continuous at t=1, and nondecreasing on 0,1 with A(0)=0, and ∫01u(t)dA(t) denotes the Riemann-Stieltjes integrals of u with respect to A. And a(t), f(t,x(t)) satisfies the following conditions:(H1)
a∈L0,1 is nonnegative and not identically zero on any compact subset of (0,1), σ=∫01tα-1dA(t)<1.(H2)
f:[0,1]×[0,+∞)→[0,+∞) is continuous.This paper consists of four sections. After the introduction, we recall some definitions, lemmas, and theorems in Section2. And the main results of this paper are stated in Section 3. In the last section, we give two examples of the main results.


## 2. Preliminaries

Firstly, for convenience we recall some definitions, lemmas, and theorems.Definition 1 (see [29, 30]).
Letf∈L1(R+) define the Riemann-Liouville fractional integral of order α>0 for f as (5)I0+αft=1Γα∫0tfst-sα-1ds,where Γ(α) is Euler gamma function.Definition 2 (see [29, 30]).
Define the Riemann-Liouville fractional derivative of orderα>0 for f as (6)D0+αft=1Γn-αddtn∫0tfst-sα+1-nds,n=α+1,where f has absolutely continuous derivatives up to order (n-1) on R+.Lemma 3.
Lety∈C0,1, 2<α≤3; then the boundary value problem (7)D0+αxt+yt=0,t∈0,1,x0=x′0=0,x1=∫01xtdAthas the unique solution x(t)=∫01G1(t,s)y(s)ds, where(8)G1t,s=Gt,s+tα-11-σ∫01Gτ,sdAτ,(9)Gt,s=tα-11-sα-1-t-sα-1Γα,0≤s≤t≤1;tα-11-sα-1Γα,0≤t≤s≤1.Proof.
The boundary value problem can be converted to an equivalent integral equation:(10)xt=-I0+αyt+c1tα-1+c2tα-2+c3tα-3,c1,c2,c3∈R.Then the solution is (11)xt=-1Γα∫0tyst-sα-1ds+c1tα-1+c2tα-2+c3tα-3.It follows from the boundary conditions x(0)=x′(0)=0 that c3=c2=0 and (12)c1=1Γα∫01ys1-sα-1ds+∫01xsdAs.Thus we get (13)xt=-1Γα∫0tyst-sα-1ds+1Γα∫01ys1-sα-1ds·tα-1+∫01xsdAs·tα-1=∫01Gt,sysds+∫01xsdAs·tα-1.Then we can obtain (14)∫01xsdAs=∫01∫01Gs,τyτdτ+∫01xτdAτsα-1dAs=∫01∫01Gs,τyτdτdAs+∫01xsdAs·∫01tα-1dAt,which means (15)∫01xsdAs=∫01∫01Gs,τyτdτdAs1-∫01tα-1dAt.So(16)xt=∫01Gt,sysds+∫01∫01Gs,τyτdτdAs1-∫01tα-1dAstα-1=∫01Gt,s+tα-11-σ∫01Gτ,sdAτysds=∫01G1t,sysds.Lemma 4 (see [20]).
G ( t , s ) defined in (9) has the following properties:(i)
G(t,s)>0, t,s∈(0,1).(ii)
G(t,s)=G(1-s,1-t), t,s∈[0,1].(iii)
k(1-t)k(s)≤Γ(α)G(t,s)≤(α-1)k(s), t,s∈[0,1] where k(t)=t1-tα-1.Lemma 5.
If (H2) is satisfied, thenG1(t,s) defined in (8) has the following properties: (17)k1-tks≤ΓαG1t,s≤Lks,t,s∈0,1,where L=(α-1)(1+∫01dA(s)/1-σ).Proof.
The proof can be easily accomplished by Lemma4, so we omitted it.Theorem 6 (see [31]).
LetE be a Banach space and P⊆E be a cone in E. Suppose that Ω1,Ω2 are two bounded open sets of E with Ω1¯⊂Ω2. Assume that T:P∩(Ω2¯∖Ω1)→P is a completely continuous operator such that either(i)
Tx≤x for any x∈P∩∂Ω1 and Tx≥x for any x∈P∩∂Ω2 or(ii)
Tx≥x for any x∈P∩∂Ω1 and Tx≤x for any x∈P∩∂Ω2.
ThenT has a fixed point in P∩(Ω2¯∖Ω1).Theorem 7 (see [32]).
LetP be a cone of a Banach space E. Pc={x∈P:x<c}. θ is a nonnegative continuous concave function on P, such that, for any x∈Pc¯, θ(x)≤x, and P(θ,b,d)=x∈P:θx≥b,x≤d. Assume that T:Pc¯→Pc¯ is completely continuous, and there exist constants a<b<d≤c such that(c1)
{x∈P(θ,b,d):θ(x)>b}≠∅ and θ(Tx)>b for x∈P(θ,b,d);(c2)
Tx<a for x∈Pa¯;(c3)
θ(Tx)>b for any x∈P(θ,b,c) with Tx>d.
ThenT has at least three fixed points x1, x2, x3 with x1<a, b<θ(x2), x3>a, and θ(x3)<b.LetE=C[0,1] be a Banach space with the maximum norm x=max0≤t≤1xt; define the cone P⊆E as (18)P=x∈E:xt≥0,t∈0,1.Define a continuous operatorT:P→E as (19)Txt=∫01G1t,sasfs,xsds.Lemma 8.
Assume that (H1) and (H2) hold; thenT:P→P is a completely continuous operator.Proof.
The lemma can be easily proven, so we omitted it.


## 3. Main Results

We define the following notation: givenδ∈(0,1/2), take(20)h=minδ≤t≤1-δk1-t=minδ≤t≤1-δ1-ttα-1,M=ΓαL∫01ksasds,N=Γαh∫δ1-δksasds.Now we can obtain the following theorems.Theorem 9.
Suppose that (H1) and (H2) are satisfied; there exist two positive constantsr2>r1>0 such that(H3)
f(t,x)≤Mr2, (t,x)∈[0,1]×[0,r2];(H4)
f(t,x)≥Nr1, (t,x)∈[0,1]×[0,r1].
Then boundary value problem (4) has at least one positive solution x∈P such that r1≤x≤r2.Proof.
The solution of boundary value problem (4) is equivalent to the fixed point of operator T. Let Ω2={x∈P:x<r2}; when x∈∂Ω2, for any t∈[0,1] we have 0≤x(t)≤r2. By Lemma 5 and (H3) we get (21)Txt=∫01G1t,sasfs,xsds≤LMr2Γα∫01ksasds=r2=x,which means when x∈∂Ω2, Tx≤x.
LetΩ1={x∈P:x<r1}; when x∈∂Ω1, for any t∈[0,1] we have 0≤x(t)≤r1. By Lemma 5 and (H4) we get (22)Txt=∫01G1t,sasfs,xsds≥k1-tNr1Γα∫01ksasds≥minδ≤t≤1-δk1-tNr1Γα∫δ1-δksasds=r1=x,which means when x∈∂Ω1, Tx≥x.
It follows from Theorem6 that we know that T has at least one fixed point in (Ω2¯∖Ω1), which means that the boundary value problem (4) has at least one solution.Theorem 10.
Suppose that (H1) and (H2) are satisfied; there exist four positive constantsa, b, c, d with 0<a<b<h/Ld<d<c, such that(H5)
f(t,x)<Ma, (t,x)∈[0,1]×[0,a];(H6)
f(t,x)>Nb, (t,x)∈[δ,1-δ]×[b,d];(H7)
f(t,x)≤Mc, (t,x)∈[0,1]×[0,c].
Then boundary value problem (4) has at least three positive solutions x1, x2, x3, such that (23)max0≤t≤1x1t<a,b<minδ≤t≤1-δx2t<max0≤t≤1x2t≤c,a<max0≤t≤1x3t≤c,minδ≤t≤1-δx3t<b.Proof.
Define a nonnegative continuous concave functionθ on P as (24)θx=minδ≤t≤1-δxt.If x∈Pc¯={x∈P:x≤c}, then x≤c; it follows from (H7) that f(t,x)≤Mc; hence (25)Tx≤LΓα∫01ksasfs,xsds≤LMcΓα∫01ksasds=c.Thus, T(Pc¯)⊆Pc¯. It follows from Lemma 8 that T is completely continuous. In the same way, let x∈Pa¯; it follows from (H5) that f(t,x)<Ma for any t∈[0,1], which shows that condition (c2) of Theorem 7 is fulfilled.
Letx(t)=b+d/2; it is easy to know that x∈P(θ,b,d) and {x∈P(θ,b,d):θ(x)>b}≠∅. If x∈P(θ,b,d), we have b≤x(t)≤d for any t∈[δ,1-δ]. We know f(t,x(t))>Nb for δ≤t≤1-δ by (H6). So we get (26)θTx=minδ≤t≤1-δTxt=minδ≤t≤1-δ∫01G1t,sasfs,xsds≥minδ≤t≤1-δ∫01k1-tksΓαasfs,xsds>NhbΓα∫δ1-δksasds=b.So condition (c1) of Theorem 7 holds.
Whenx∈P(θ,b,c) with Tx>d, noting that (27)Tx≤LΓα∫01ksasfs,xsds,thus (28)Txt≥k1-tΓα∫01ksasfs,xsds≥k1-tLTx,so we obtain(29)θTx=minδ≤t≤1-δTxt≥minδ≤t≤1-δ⁡k1-tLTx>hLd>b.That is to say, (c3) is satisfied.
All conditions of Theorem7 are satisfied, so T has at least three fixed points x1, x2, x3, which means that the boundary value problem (4) has at least three positive solutions x1, x2, x3, such that (30)max0≤t≤1x1t<a,b<minδ≤t≤1-δx2t<max0≤t≤1x2t≤c,a<max0≤t≤1x3t≤c,minδ≤t≤1-δx3t<b.The proof of this theorem is finished.


## 4. Some Examples

Now we present two examples to illustrate our main results.Example 1.
Let us see the following problem:(31)D0+5/2xt+14sin⁡t+x+3=0,t∈0,1,x0=x′0=0,x1=∫01xtdt.Chooseδ=1/3; we obtain that L=4, h=2/243, M=105π/64≈2.91, and N=76545π/7042-256≈183.48. Then for any (t,x)∈[0,1]×[0,2], we have f(t,x)=1/4sin⁡t+x+3≤5.25≤Mr2≈5.82, and for any (t,x)∈[0,1]×[0,0.01], we get f(t,x)=1/4sin⁡t+x+3≥3≥Nr1≈1.83.Then the boundary value problem has at least one positive solutionx∈P such that 0.01≤x≤2.Example 2.
We now study the following problem:(32)D0+5/2xt+ft,x=0,t∈0,1,x0=x′0=0,x1=∫01xtdt,where (33)ft,x=t3+x2,x≤1;183+t3+x,x>1.Chooseδ=1/3; we have M≈2.91, N≈183.48, L=4, and h≈0.1283. Let a=1/2, b=1, c=100, and d=35; then for any (t,x)∈[0,1]×[0,1/2], we have f(t,x)=t/3+x2≤0.59<Ma≈1.46, for any (t,x)∈[1/3,2/3]×[1,35], we have f(t,x)=183+t/3+x≥184.11>Nb≈183.48, and for any (t,x)∈[0,1]×[0,100], we have f(t,x)=183+t/3+x≤283.34<Mc≈291. Then by Theorem 10, we conclude that this boundary value problem has at least three positive solutions x1, x2, x3, such that (34)max0≤t≤1x1t<12,1<minδ≤t≤1-δx2t<max0≤t≤1x2t≤100,12<max0≤t≤1x3t≤100,minδ≤t≤1-δx3t<1.



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