# Multiplicity Solutions for Integral Boundary Value Problem of Fractional Differential Systems

Authors: Shiying Song; Yujun Cui
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

This paper deals with the existence and multiplicity of solutions for the integral boundary value problem of fractional differential systems:D0+α1u1t=f1t,u1t,u2t,D0+α2u2t=f2t,u1t,u2t,u10=0,D0+β1u10=0,D0+γ1u11=∫01D0+γ1u1ηdA1η,u20=0,D0+β2u20=0,D0+γ2u21=∫01D0+γ2u2ηdA2η,, where fi:0,1×0,∞×0,∞⟶0,∞ is continuous and αi−2<βi≤2,αi−γi≥1,2<αi≤3,γi≥1i=1,2.D0+α is the standard Riemann–Liouville’s fractional derivative of order α. Our result is based on an extension of the Krasnosel’skiĭ’s fixed-point theorem due to Radu Precup and Jorge Rodriguez-Lopez in 2019. The main results are explained by the help of an example in the end of the article.


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## Body

## 1. Introduction

With the deepening of people's understanding of mathematics, the knowledge of mathematics is more and more closely related to the way of production and life of human beings. In recent years, fractional calculus is very active in the field of applied mathematics. It can be applied not only in biochemistry, mathematical physics equation, physical science experiment, and other academic fields but also in precision production [1–3].In many recent papers are researched the fractional differential equations with the existence of the solutions [4–43]. For example, Zhang and Zhong [38] used the fixed-point theorem on cones to find the existence result of at least two positive solutions which are considered the nonlinear fractional differential equations of nonlocal boundary value problems as follows:(1)D0+αut+ft,ut=0,0<t≤1,D0+βu1=∑i=1∞ξiD0+βuηi,u0=0,D0+βu0=0,where 2<α≤3,1≤β≤2,α−β≥1,0<ξi,ηi<1 with ∑i=1∞ξiηiα−β−1<1.In [22], the authors obtained the uniqueness results of positive solution by the contraction map principle and obtained some existence results of positive solution through the fixed-point index theory, which is as follows:(2)D0+αut+ft,ut=0,0<t≤1,βuη=u1,u0=0,where 1<α≤2, 0<βηα−1<1,0<η<1, D0+α is the standard Riemann–Liouville differentiation, and the function f is continuous on 0,1×0,∞.However, in recent years, many scholars have begun to use the fixed-point index to study the existence and multiplicity of operator equation and operator systems [44–51]. For example, in [46], the authors use the fixed-point index in cones to study the existence, localization, and multiplicity of positive solutions to operator systems of the following form:(3)Liui=Fiu1,u2,ui∈DLi,i=1,2,where for each i,Li:DLi⊂X⟶Y is invertible, Fi:Y×Y⟶Y, and X,Y⊂C0,1. It should be noted that each component of the fixed-point operator systems may have the same or different behaviors.In particular, only few papers studied the existence and multiplicity of solutions to specific differential systems [52–58]. Therefore, in this paper, we will apply an extension of the Krasnosel’skiǐ’s fixed-point theorem to investigate the existence and multiplicity of solutions for a class of fractional differential systems. More precisely, the following fractional differential systems are studied:(4)D0+α1u1t=f1t,u1t,u2t,D0+α2u2t=f2t,u1t,u2t,u10=0,D0+β1u10=0,D0+γ1u11=∫01D0+γ1u1ηdA1η,u20=0,D0+β2u20=0,D0+γ2u21=∫01D0+γ2u2ηdA2η,where αi−2<βi≤2,αi−γi≥1,2<αi≤3,γi≥1, fi∈C0,1×0,∞2,0,∞, and Ai are nondecreasing on 0,1, left continuous at t=1.


## 2. Preliminaries

In this part, we first give the basic definitions, lemmas, and theorems related to fractional calculus.Definition 1 (see [2]).
Define the Riemann–Liouville fractional derivative of orderα>0 for function σ as(5)D0+ασt=1Γn−αdndtn∫0tσst−sα+1−nds,n=α+1,where Γ⋅ is the Euler gamma function.Definition 2 (see [2]).
Letσ define the Riemann–Liouville fractional integral of order α>0 for σ as(6)I0+ασt=1Γα∫0tσst−sα−1ds,where Γ⋅ is the Euler gamma function.Lemma 1 (see [2]).
Letn−1<α≤n and u∈C0,1∩L10,1; then,(7)I0+αD0+αut=ut+c1tα−1+c2tα−2+⋯+c1tα−n,where ci∈ℝ, i=1,2,….For convenience, we first consider the following linear fractional differential equation:(8)D0+α1u1t+σt=0,0<t≤1,D0+γ1u11=∫01D0+γ1u1ηdA1η,u10=0,D0+β1u10=0,where α1−2<β1≤2, α1−γ1≥1, 2<α1≤3, γ1≥1, and A1t is nondecreasing on 0,1, left continuous at t=1.Lemma 2.
Let1−∫01ηα1−γ1−1dA1η>0 and σ∈C0,1; then, boundary value problem (8) has an unique solution u1t=∫01G1t,sσsds, where(9)G1t,s=1Γα1p10tα1−1p1s1−sα1−γ1−1,0≤t≤s≤1;tα1−1p1s1−sα1−γ1−1−p10t−sα1−1,0≤s≤t≤1,p1s=1−∫s1η−sα1−γ1−11−sα1−γ1−1dA1η.Proof.
It follows from Lemma1 that u1t=−I0+α1σt+c1tα1−1+c2tα1−2+c3tα1−3. With consideration of the boundary value conditions u10=0, we can get c3=0. Consequently,(10)u1t=−I0+α1σ1t+c1tα1−1+c2tα1−2.
Notice thatD0+β1tα−i=Γα−i+1/Γα−i−β1+1tα−i−β1i=1,2; we get(11)D0+β1u1t=D0+β1−∫0t1Γ1α1t−sα1−1σsds+c1Γ1α1Γ1α1−β1tα1−β1−1+c2Γ1α1−1Γ1α1−β1−1tα1−β1−2.
Sinceα1−2<β1 and D0+β1u10=0, we conclude that c2=0. Therefore, (10) reduces to(12)u1t=−I0+α1σ1t+c1tα1−1.
Taking into account thatD0+γ1u11=∫01D0+γ1u1ηdA1η and D0+γ1I0+α1σ1t=I0+α1−γ1σ1t, we have(13)D0+γ1u11=−1Γα1−γ1∫011−sα1−γ1−1σsds+c1Γα1Γα1−γ1=∫01D0+γ1u1ηdA1η=∫01−1Γα1−γ1∫0ηη−sα1−γ1−1σsds+c1Γα1Γα1−γ1ηα1−γ1−1dA1η=−1Γα1−γ1∫01∫0ηη−sα1−γ1−1σsdsdA1η+c1Γα1Γα1−γ1∫01ηα1−γ1−1dA1η.
This yields(14)c1=∫011−sα1−γ1−1σsds−∫01∫0ηη−sα1−γ1−1σsdsdA1ηΓα11−∫01ηα1−γ1−1dA1η.
Taking the above equality into (12), we have(15)u1t=−I0+α1σt+c1tα1−1=−1Γα1∫0tt−sα1−1σsds+tα1−1Γα1p0∫011−sα1−γ1−1σsds−∫01∫0ηη−sα1−γ1−1σsdsdA1η=−1Γα1∫0tt−sα1−1σsds+tα1−1Γα1p0∫011−sα1−γ1−1σsds∫01∫s1η−sα1−γ1−1dA1ησsds=−1Γα1∫0tt−sα1−1σsds+tα1−1Γα1p0∫011−sα1−γ1−1psσsds=∫01G1t,sσsds,where G1t,s and ps are given in Lemma 2.The following proposition of Green’s functionG1t,s will be used throughout the paper.Lemma 3.
The functionp1s>0,s∈0,1, and p1 is nondecreasing on 0,1.Proof.
After direct computation, we will get(16)p1′s=∫s1α1−γ1−1η−sα1−γ1−21−sα1−γ1−11−s2α1−γ1−1dA1η−∫s1η−sα1−γ1−11−sα1−γ1−2α1−γ1−11−s2α1−γ1−1dA1η=α1−γ1−1∫s1η−sα1−γ1−2−η−sα1−γ1−11−s−11−sα1−γ1−1dA1η=α1−γ1−1∫s1η−sα1−γ1−21−η−s/1−s1−sα1−γ1−1dA1η=α1−γ1−1∫s11−η/1−sη−sα1−γ1−21−sα1−γ1−1dA1η≥0,s∈0,1.
Then, we conclude thatp1s is a nondecreasing function on 0,1, which implies that p1s≥p10=1−∫01ηα1−γ1−1dA1η>0,s∈0,1. This proves the content of lemma.Lemma 4.
The functionG1t,s has the following properties:(i)
G1t,s>0,∂/∂tG1t,s>0,0<t,s<1(ii)
maxt∈0,1G1t,s=G11,s,0≤s≤1(iii)
G1t,s≥tα1−1G11,s,0≤t,s≤1Proof.
According to Lemma2, we have learned that Green’s function G1 is divided into two cases, and next, we will prove three properties of G1, respectively.(i)
When0≤t≤s≤1,(17)G1t,s=tα1−1Γα1p101−sα1−γ1−1p1s>0,then by a direct calculation, it is easy to get(18)∂∂tG1t,s=p1s1−sα1−γ1−1Γα1p10α1−1tα1−2≥0.When0≤s≤t≤1,(19)G1t,s=1p10Γα1tα1−1p1s1−sα1−γ1−1−p10t−sα1−1≥1p10Γα1tα1−1p101−sα1−γ1−1−p10t−sα1−1=1p10Γα1tα1−1p101−sα1−γ1−1−p101−stα1−1=tα1−1Γα11−sα1−γ1−1−1−stα1−1≥0,∂∂tG1t,s=1p10Γα1α1−1tα1−2p1s1−sα1−γ1−1−α1−1t−sα1−2p10=α1−1p10Γα1tα1−2p1s1−sα1−γ1−1−p10t−sα1−2=α1−1p10Γα1tα1−2p1s1−sα1−γ1−1−p101−stα1−2≥α1−1Γα1tα1−21−sα1−γ1−1−1−stα1−2≥0.(ii)
Based on the property (i), it follows thatG1t,s is increasing with respect to t. Obviously, we have(20)maxt∈0,1G1t,s=G11,s=1p10Γα1p1s1−sα1−γ1−1−p101−sα1−1.(iii)
Fort,s∈0,1×0,1, we discuss two cases.
When0≤t≤s≤1, G1t,s=tα1−1/p10Γα11−sα1−γ1−1p1s, it is easy to get that G1t,s≥tα1−1G11,s.
When0≤s≤t≤1,(21)G1t,s=1p10Γα1tα1−1p1s1−sα1−γ1−1−p10t−sα1−1=tα1−1p10Γα1p1s1−sα1−γ1−1−p101−stα1−1≥tα1−1G11,s.
This yields the desired result.The main proof of this research uses the following theorem in [46].Theorem 1 (see [46]).
LetX,⋅ be a Banat space, K1,K2⊂X two cones, and K=K1×K2 the corresponding cone of X2=X×X. Let pi,qi>0 with pi≠qi, Upi=u∈Ki:u<pi. Assume that N:W1×W2¯⟶K, T=T1,T2 is a compact map (where Wi=Upi∪Uqi for i=1,2) and there exist φi∈Ki\0,i=1,2, such that for each i∈1,2, the following condition is satisfied in W1×W2¯:(i)
λxi≠Tix for xi=pi and λ≥1(ii)
xi≠Tix+μφi for xi=qi and μ≥0
Then,(1)
T has at least one fixed point in K such that xi∈Upi\Uqi¯ for i=1,2 if pi>qi for i=1,2(2)
T has at least two fixed points located in Up1\Uq1¯×Up2 and Up1\Uq1¯×Uq2\Up2¯ if q1<p1 and q2>p2(3)
T has at least two fixed points located in Up1×Up2Uq2¯ and Uq1\Up1¯×Up2\Uq2¯ if q1>p1 and q2<p2(4)
T has at least four fixed points located in Up1×Up2,Up1×Uq2\Up2¯,Uq1\Up1¯×Up2,Uq2\Up1¯×Uq2\Up2¯ if pi<qi for i=1,2


## 3. Main Results

LetX=C0,1, x∞=maxt∈0,1xt, K=x∈C0,1:xt≥0, and Pi=x∈C0,1:xt≥tαi−1x∞i=1,2. Then, X becomes a real Banach space with the norm ⋅∞ and K,andPi are cones on X. Also the product space X×X is a Banach space endowed with norm x,y=maxx∞,y∞ and P1×P2 is a cone in X×X.For convenience, we present some basic conditions as follows which we will be used later:(H1)fi∈C0,1×0,+∞×0,+∞,0,+∞i=1,2.(H2) There existri,βi∈0,+∞ and δi∈0,1i=1,2 such thatf1t,u1,u2>N1r1 for t,u1,u2∈δ1,1×h1r1,r1×0,R2f1t,u1,u2<M1β1 for t,u1,u2∈0,1×0,β1×0,R2f2t,u1,u2>N2r2 for t,u1,u2∈δ2,1×0,R1×h2r2,r2f2t,u1,u2<M2β2 for t,u1,u2∈0,1×0,R1×0,β2where(22)Ri=maxri,βi,hi=δiαi−1,Ni=1/hi∫δi1Gi1,sds,Mi=1/∫01Gi1,sds,Git,s=1pi0Γαitαi−11−sαi−γi−1pis,0≤t≤s≤1,tαi−1pis1−sαi−γi−1−pi0t−sαi−1,0≤s≤t≤1,pis=1−∫s1η−sαi−γi−11−sαi−γi−1dAiη.(H3)1−∫01ηαi−γi−1dAiη>0, i=1,2.Employing Lemma2 and the condition of (H1), system (4) has the following integral representation:(23)u1t=∫01G1t,sf1s,u1s,u2sds,t∈0,1,u2t=∫01G2t,sf2s,u1s,u2sds,t∈0,1.Let us define two operatorsTi:K×K⟶Xi=1,2 as follows:(24)T1u1,u2t=∫01G1t,sf1s,u1s,u2sds,t∈0,1,T2u1,u2t==∫01G2t,sf2s,u1s,u2sds,t∈0,1.Then, we can define an operatorT:K×K⟶X×X as follows:(25)Tu1,u2=T1u1,u2,T2u1,u2,u1,u2∈K×K.Lemma 5.
Assume that (H1), (H2), and (H3) hold. Then,T:K×K⟶P1×P2 is completely continuous.Proof.
Firstly, we prove thatT:K×K⟶P1×P2. In fact, for u1,u2∈K×K, by (H1), it is obvious that Tiu1,u2t≥0 for i=1,2 and t∈0,1. In addition, if u1,u2∈K×K, then(26)Tiu1,u2t=∫01Git,sf1s,u1s,u2sds≤∫01Gi1,sf1s,u1s,u2sds,t∈0,1.
So,(27)Tiu1,u2∞≤∫01Gi1,sf1s,u1s,u2sds.
On the other hand, for anyu1,u2∈K×K and any t∈0,1, it follows from Lemma 4 that(28)Tiu1,u2t=∫01G1t,sf1s,u1s,u2sds≥tαi−1∫01G11,sf1s,u1s,u2sds≥tαi−1T1u1,u2∞.
Thus, from the above discussion, we conclude thatT:K×K⟶P1×P2, and then, it obviously shows that T is well defined. The complete continuity of operator T can be given by a standard argument with the help of the Arzela–Arscoli Theorem. We omit the details.Theorem 2.
Assume that (H1), (H2), and (H3) hold. Then, we have(i)
Ifr1<β1 and r2<β2, then (23) has at least a positive solution located in Uβ1\Ur1¯×Uβ2\Ur2¯, where Uri=u∈Pi:u∞<ri(ii)
Ifr1<β1 and r2>β2, then (23) has at least two positive solutions located in Uβ1\Ur1¯×Uβ2 and Uβ1\Ur1¯×Ur2\Uβ2¯(iii)
Ifr1>β1 and r2<β2, then (23) has at least two positive solutions located in Uβ1×Uβ2\Ur2¯ and Ur1\Uβ1¯×Uβ2\Ur2¯(iv)
Ifr1>β1 and r2>β2, then (23) has at least three positive solutions located in Uβ1×Uβ2,Uβ1×Ur2\Uβ2¯,Ur1\Uβ1¯×Uβ2,Ur2Uβ1¯×Ur2\Uβ2¯Proof.
It follows from Lemma5 that the existence of a positive solution of problem (23) is equivalent to the existence of a nontrivial fixed point of T in P1×P2. Let Wi=u∈Pi:u∞<Ri.
First, note that ifu=u1,u2∈W1×W2¯ with u1∞=r1, then u2∞≤R2 and by the definition of P1,(29)r1tα1−1≤u1t≤r1,0≤u2t≤R2,t∈0,1.
In the following, we conclude that fori∈1,2, the following properties hold:(30)λui≠Tiu,forui∞=βi,λ≥1;ui≠Tiu+μtαi−1,forui∞=ri,μ≥0,guaranteeing the validity of Theorem 1.
In fact, ifu1∞=β1 and λu1=T1u for a λ1≥1, then by (H2),(31)u1t≤λu1t=T1u1t≤∫01G11,sf1s,u1s,u2sds<M1β1∫01G11,sds=β1,t∈0,1,whence, in particular, we conclude β1<β1, a contradiction. Now, if u1=T1u+μ1tα1−1 for u1∞=r1 and μ1≥0, then by (H2), we obtain(32)u1t=T1ut+μ1tα1−1≥∫δ11G1t,sf1s,u1s,u2sds≥δ1α1−1∫δ11G11,sf1s,u1s,u2sds>δ1α1−1∫δ11G11,sN1r1ds=r1,for all t∈0,1. This yields the contradiction r1<r1. Hence, (30) holds for i=1. Similarly, (30) is true for i=2.Example 1.
Consider the following integral boundary value problem of fractional differential systems:(33)D0+11/4u1t=u1tarctanu1t+11u2t+12+100,D0+5/2u2t=12sint+12cosu1tu2t+u2t+3,u10=0,D0+3/4u10=0,D0+5/4u11=∫01D0+5/4u1τdτ,u20=0,D0+1/2u20=0,D0+3/2u21=12D0+3/2u212.
Then, (33) has at least two positive solutions u1,v1andu2,v2 with u1,v1∈U100\U0.72¯×U5 and u2,v2∈U100\U0.72¯×U7100\U5¯.
To see this, we will apply Theorem2 with(34)α1=114,γ1=54,α2=52,γ2=32,f1t,u1,u2=u1arctanu1+11u2+12+100,f2t,u1,u2=12sint+12cosu1u2+u2210+2,A1t=t,A2t=12,t≤12,1,12<t≤1.
Clearly,(35)1−∫01tα1−γ1−1dA1t=1−∫01t1/2dt=13>0,1−∫01tα2−γ2−1dA2t=1−12=12>0.
Thus, (H3) holds.
Take(36)G1t,s=1Γ11/4t7/41−s1/21+2s,0≤t≤s≤1,t7/41−s1/21+2s−t−s7/4,0≤s≤t≤1,G2t,s=1Γ5/22t3/2p2s,0≤t≤s≤1,2t3/2p2s−t−s3/2,0≤s≤t≤1,p1s=1+2s3,p2s=12,s≤12,1,12<s≤1.
Letδ1=δ2=1/4. By simple computation, we have(37)h1≈0.0884,h2=0.125,M1≈1.9231,M2≈1.2085,N1≈24.4377,N2≈10.0794.
We chooser1=0.72, β1=100, r2=7100, and β2=5. Then, R1=100, R2=7100,(38)f1t,u1,u2>25πh1r1>N1r1,t,u1,u2∈δ1,1×h1r1,r1×0,R2,f1t,u1,u2<101π2β1<M1β1,t,u1,u2∈0,1×0,β1×0,R2,f2t,u1,u2>32+h22r2210>N2r2,t,u1,u2∈δ2,1×0,R1×h2r2,r2,f2t,u1,u2<3+β2210<M2β2,t,u1,u2∈0,1×0,R1×0,β2.
Consequently,H2 holds with r1<β1 and r2>β2, and our conclusion follows from Theorem 2.


## 4. Conclusions

In this paper, we investigate the existence and multiplicity of positive solutions for the integral boundary value problem of higher-order fractional differential systems. This result is based on an extension of the Krasnosel’skiǐ’s fixed-point theorem due to Radu Precup and Jorge Rodriguez-Lopez in [46]. We rewrite the original fractional differential systems as equivalent fractional integral systems. With the help of properties of Green’s function, we obtain some sufficient conditions of existence and multiplicity of positive solutions. Finally, an example is presented to illustrate the effectiveness of the main result. The interesting point is that the integral boundary condition is dependent on the lower-order fractional derivative.



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