# The Existence Results of Solutions for System of Fractional Differential Equations with Integral Boundary Conditions

Authors: Dongxia Zan; Run Xu
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

In this paper, we investigated the system of fractional differential equations with integral boundary conditions. By using a fixed point theorem in the Banach spaces, we get the existence of solutions for the fractional differential system. By constructing iterative sequences for any given initial point in space, we can approximate this solution. As an application, an example is presented to illustrate our main results.


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## Body

## 1. Introduction

The fractional differential equations attracted many people’s attention, and they have many applications in different fields of science and biology. In recent years, the research on the qualitative properties of fractional differential equations has become a hot topic, and many results have been obtained; see [1–29]. From the literature, there are many articles investing the existence of solutions or the solutions for some nonlinear systems; see [4–10, 21].In [6], Su studied the following system of nonlinear fractional differential equations:(1)Dαut=ft,vt,Dμvt,Dβvt=gt,ut,Dνut,u0=v0=u1=v1=0,where 0<t<1,1<α,β<2,μ,ν>0,α-ν,β-μ≥1,f,g are given functions and Dα,Dβ are Riemann-Liouville fractional derivatives of order α,β. The author gave sufficient conditions for the existence of solutions of system (1).In [7], Wang et al. studied the existence of solutions for the following system of nonlinear fractional differential equations:(2)Dαut=ft,vt,Dβvt=gt,ut,u0=v0=0,u1=auξ,v1=bvξ,where 0<t<1,1<α,β<2,0<ξ<1,0≤a,b≤1,f,g are given functions and Dα,Dβ are Riemann-Liouville fractional derivatives of order α,β.In [8], Yang studied the existence of solutions for the following system of nonlinear fractional differential equations:(3)Dαut=atft,vt,Dβvt=btgt,ut,u0=v0=0,u1=∫01ϕtutdt,v1=∫01ψtvtdt,where 0<t<1,1<α,β<2,a,b∈C((0,1),[0,∞)),ϕ,ψ∈L1[0,1] are nonnegative and f,g∈C([0,1]×[0,+∞),[0,+∞)), and Dα,Dβ are also the standard Riemann-Liouville fractional derivatives of order α,β.In [9], Alsulami et al. studied the existence of solutions for the following system of nonlinear Caputo fractional integrodifferential equations:(4)Dαcut=ft,ut,vt,Dβcvt=gt,ut,vt,subject to the nonseparated coupled boundary conditions:(5)u0=λ1vT,u′0=λ2v′T,v0=μ1uT,v′0=μ2u′T,where t∈[0,T],1<α,β≤2,Dαc,Dβc are Caputo fractional derivatives of order α,β,f,g are functions, and λi,μi are real constants with λiμi≠1(i=1,2).In [10], Yang et al. studied the existence of solutions for the following system of fractional differential equations with integral boundary conditions:(6)Dαut=ft,vt,Dβvt=gt,ut,u0=v0=0,u1=∫01ϕtutdt,v1=∫01ψtvtdt,where 0<t<1,1<α,β<2,ϕ,ψ∈L1[0,1] are nonnegative and f,g∈C([0,1]×[0,+∞),[0,+∞)), and Dα,Dβ are also the standard Riemann-Liouville fractional derivatives of order α,β.In this paper, we investigate the following system of fractional differential equations:(7)Dαut+ft,vt=0,Dβvt+gt,wt=0,Dγwt+rt,ut=0,u0=v0=w0=0,u1=∫01ϕtutdt,v1=∫01ψtvtdt,w1=∫01ηtwtdt,where 0<t<1,1<α,β,γ<2, and Dα,Dβ,Dγ are also the standard Riemann-Liouville fractional derivatives of order α,β,γ. The system of (7) satisfies the following conditions:(1°)f,g,r∈C([0,1]×[0,+∞),[0,+∞)).(2°)ϕ,ψ,η:[0,1]→[0,∞) with ϕ,ψ,η∈L1[0,1] and θi∈(0,1)(i=1,2,3),θj>0(j=4,5,6),where(8)θ1=∫01tα-1ϕtdt,θ2=∫01tβ-1ψtdt,θ3=∫01tγ-1ηtdt;θ4=∫01tα-1ϕt1-tdt,θ5=∫01tβ-1ψt1-tdt,θ6=∫01tγ-1ηt1-tdt.We know 0<θ4≤θ1<1,0<θ5≤θ2<1,0<θ6≤θ3<1.


## 2. Preliminaries and Basic Lemmas

In this section, we will present some definitions and lemmas, which will be used throughout this paper.From [1], the definitions of Riemann-Liouville fractional integral and derivative are given as follows:(9)Ia+qft=1Γq∫att-sq-1fsds,where t>a,q>0,  f∈C[a,b] and Γ is the gamma function.(10)Da+qft=1Γn-qdndtn∫att-sn-q-1fsds,where t>a,n-1<q<n and f:[a,∞)→R.Lemma 1 (see [2]).
If∫01ϕ(t)tα-1dt≠1,  1<α<2, then, for any σ∈C[0,1], the unique solution of the boundary value problem,(11)Dαut+σt=0,0<t<1,u0=0,u1=∫01ϕtutdt,is given by(12)ut=∫01Gαt,sσsds,where(13)Gαt,s=G1αt,s+G2αt,s,(14)G1αt,s=tα-11-sα-1-t-sα-1Γα,0≤s≤t≤1,tα-11-sα-1Γα,0≤t≤s≤1,(15)G2αt,s=tα-11-θ1∫01ϕtG1αt,sdt.Then G(t,s)=(Gα(t,s),Gβ(t,s),Gγ(t,s)) is Green’s function of (7).Lemma 2 (see [2]).
If∫01ϕ(t)tα-1dt∈[0,1), then Gα(t,s)≥0 is continuous for all t,s∈[0,1],Gα(t,s)>0 for all t,s∈(0,1), where Gα(t,s) is given in (13).Lemma 3 (see [3]).
The functionG1α(t,s) has the following properties:(16)α-1Γαtα-11-t1-sα-1s≤G1αt,s≤1Γαtα-11-t1-sα-2,where t,s∈[0,1] and  G1α(t,s) is given in (14).Lemma 4.
Assume condition (2°) holds; thenGα(t,s) satisfies the following property:(17)α-1θ4s1-sα-1tα-11-θ1Γα≤Gαt,s≤tα-11-sα-2Γα1-t+θ41-θ1,where Gα(t,s) is given in (13) and θ1,θ4 are defined in (8).Proof.
From (13)–(15), we have(18)Gαt,s=G1αt,s+G2αt,s=G1αt,s+tα-11-θ1∫01ϕtG1αt,sdt≤1Γαtα-11-t1-sα-2+tα-11-θ1∫01ϕt1Γαtα-11-t1-sα-2dt=tα-11-sα-2Γα1-t+θ41-θ1.From [10], we have(19)α-1θ4s1-sα-1tα-11-θ1Γα≤Gαt,s.Throughout this paper, we list the following definitions and marks for our convenience. For details, see [11–14].Let(E,∥·∥) be a real Banach space, for any μ,ν∈E.(1)
IfP⊂E, then cone P is called partially ordered, i.e., μ≤ν⇔ν-μ∈P.(2)
o is the zero element of E.(3)
There is a constantM>0; if P is called normal, then ‖μ‖≤M‖ν‖ when o≤μ≤ν.(4)
Let the mappingA:E→E, if A is an increasing operator, then Aμ≤Aν when μ≤ν.(5)
The notation~ is an equivalence relation. μ~ν; i.e., there exist λ>0 and γ>0 such that λμ≤ν≤γμ.(6)
SetPh={μ∈E∣μ~h}, where h>o.(7)
Φ is the class of function φ:(0,1)→(0,1), which satisfies the condition φ(t)>t for t∈(0,1).Lemma 5 (see [14]).
LetP be a normal cone in a real Banach space E,  h>o. The operator T satisfies the following conditions:
(i)T is an increasing
(ii) There ish0∈Ph such that Th0∈Ph
(iii) For anyx∈P and t∈(0,1), there exists φ∈Φ such that T(tμ)≥φ(t)T(μ)
Then,
(1)T has a unique fixed point μ∗∈Ph
(2) for any initial valueμ0∈Ph, constructing successively the sequence μn+1=Tμn,  n=0,1,…, we have μn→μ∗ as n→∞.


## 3. Main Results

SetX={u(t)∣u(t)∈C[0,1]}, the norm ‖u‖X=maxt∈[0,1]|u(t)|. For (u,v,w)∈X×X×X, we define the norm ‖u,v,w‖X×X×X=maxt∈[0,1]{‖u‖X,‖v‖X,‖w‖X}.We defineS={(u,v,w)∈X×X×X∣u(t)≥0,v(t)≥0,w(t)≥0},  P={u∈X∣u(t)≥0}.We give the definition of the partially ordered in the spaceX×X×X; let (u1,v1,w1),(u2,v2,w2)∈S, then (u1,v1,w1)≤(u2,v2,w2)⇔u1≤u2,v1≤v2,w1≤w2.We also defineλ(h1,h2,h3)=(λh1,λh2,λh3), where λ>0,h1,h2,h3>o.Lemma 6.
DefineSh={(μ,ν,ω); there exist λ(μ,ν,ω),γ(μ,ν,ω)>0 such that λ(h1,h2,h3)≤(μ,ν,ω)≤γ(h1,h2,h3)},(20)Ph1=u∈X∣ut≥0,Ph2=v∈X∣vt≥0,Ph3=w∈X∣wt≥0;And then Sh=Ph1×Ph2×Ph3, where h=(h1,h2,h3).Proof.
On one hand, for any(μ,ν,ω)∈Sh, there exists λ,γ>0 such that λh≤(μ,ν,ω)≤γh. Then,(21)λh1,λh2,λh3=λh1,h2,h3≤μ,ν,ω≤γh1,h2,h3=γh1,γh2,γh3.From the definition of the partial order, we can get λh1≤μ≤γh1,λh2≤ν≤γh2,λh3≤ω≤γh3. So μ∈Ph1,ν∈Ph2,ω∈Ph3.
That is to say,(μ,ν,ω)∈Ph1×Ph2×Ph3.
Hence,Sh⊂Ph1×Ph2×Ph3.
On the other hand, for any(μ,ν,ω)∈Ph1×Ph2×Ph3, we have(22)μ∈Ph1,ν∈Ph2,ω∈Ph3.There exist λ1,λ2,λ3,γ1,γ2,γ3>0, such that(23)λ1h1≤μ≤γ1h1,λ2h2≤ν≤γ2h2,λ3h3≤ω≤γ3h3.Let λ=min⁡{λ1,λ2,λ3},γ=max⁡γ1,γ2,γ3.
From the definition of the partial order, we have(24)λh1,h2,h3=λh1,λh2,λh3≤λ1h1,λ2h2,λ3h3≤μ,ν,ω≤γ1h1,γ2h2,γ3h3≤γh1,γh2,γh3=γh1,h2,h3.That is, (μ,ν,ω)∈Sh, and thus Ph1×Ph2×Ph3⊂Sh.
HenceSh=Ph1×Ph2×Ph3 holds.Lemma 7.
Assume conditions (1°) and (2°) hold; then(u,v,w)∈S is a solution of system (7) if and only if(25)ut=∫01Gαt,sfs,vsds,vt=∫01Gβt,sgs,wsds,wt=∫01Gγt,srs,usds.From Lemma 1, we can get that (25) holds.Theorem 8.
Assume condition (2°) and the following conditions hold:
(1)f(t,x),g(t,x),r(t,x) are continuous and f(t,0),g(t,0),r(t,0)≢0
(2)f(t,x),g(t,x),r(t,x) are increasing functions in x, for any t∈[0,1]
(3) There existsφ1,φ2,φ3∈Φ such that(26)ft,λu≥φ1λft,u,gt,λu≥φ2λgt,u,rt,λu≥φ3λrt,ufor any t∈[0,1],u∈[0,∞) and λ∈(0,1).
Then system (7) has a unique positive solution (u∗,v∗,w∗)∈Sh, where h(t)=(tα-1,tβ-1,tγ-1),t∈[0,1].Proof.
For(u,v,w)∈S, we define operators T1,T2,T3, and T by(27)T1ut=∫01Gαt,sfs,vsds,T2vt=∫01Gβt,sgs,wsds,T3wt=∫01Gγt,srs,usds,Tu,v,wt=∫01Gαt,sfs,vsds,∫01Gβt,sgs,wsds,∫01Gγt,srs,usds.If T1,T2,T3:P→P then T:S→S, that is,(28)Tu,v,wt=T1ut,T2vt,T3wt=ut,vt,wt;
To begin with, we prove that condition (i) of Lemma5 holds. For any (u1,v1,w1),(u2,v2,w2)∈S with (u1,v1,w1)≤(u2,v2,w2), according to the definition of the partially order, we have u1(t)≤u2(t),v1(t)≤v2(t),w1(t)≤w2(t).
From Lemma6 and condition (2), we have(29)T1u1t=∫01Gαt,sfs,v1sds≤∫01Gαt,sfs,v2sds=T1u2t,T2v1t=∫01Gβt,sgs,w1sds≤∫01Gβt,sgs,w2sds=T2v2t,T3w1t=∫01Gγt,srs,u1sds≤∫01Gγt,srs,u2sds=T3w2t.So, we can get the following result:(30)Tu1,v1,w1t=T1u1t,T2v1t,T3w1t≤T1u2t,T2v2t,T3w2t=Tu2,v2,w2t.Hence, T is an increasing operator.
Then, we prove that condition (ii) of Lemma5 holds. Let h0(t)=h(t)=(h1(t),h2(t),h3(t))=(tα-1,tβ-1,tγ-1),t∈[0,1].
According to Lemma7, we only need to prove T1h1∈Ph1,T2h2∈Ph2,T3h3∈Ph3.
Applying (17) and Theorem 3.1 of [10], we have(31)T1h1t=∫01Gαt,sfs,h2sds≥α-1θ41-θ1Γαh1t∫01s1-sα-1fs,0ds,T2h2t=∫01Gβt,sgs,h3sds≥β-1θ51-θ2Γβh2t∫01s1-sβ-1gs,0ds,T3h3t=∫01Gγt,srs,h1sds≥γ-1θ61-θ3Γγh3t∫01s1-sγ-1rs,0ds.By computation we have(32)T1h1t=∫01Gαt,sfs,h2sds≤∫011-sα-2Γα1-t+θ41-θ1tα-1fs,sβ-1ds≤1-t+θ4/1-θ1Γαh1t∫011-sα-2fs,1ds,T2h2t=∫01Gβt,sgs,h3sds≤∫011-sβ-2Γβ1-t+θ51-θ2tβ-1gs,sγ-1ds≤1-t+θ5/1-θ2Γβh2t∫011-sβ-2gs,1ds,T3h3t=∫01Gγt,srs,h1sds≤∫011-sγ-2Γγ1-t+θ61-θ3tγ-1rs,sα-1ds≤1-t+θ6/1-θ3Γγh3t∫011-sγ-2rs,1ds.From [10], we know f(s,1)>f(s,0)≥0,s∈[0,1]. Then we have(33)0<∫01s1-sα-1fs,0ds≤∫011-sα-2fs,1ds,0<∫01s1-sβ-1gs,0ds≤∫011-sβ-2gs,1ds,0<∫01s1-sγ-1rs,0ds≤∫011-sγ-2rs,1ds.Let(34)m1=α-1θ41-θ1Γα∫01s1-sα-1fs,0ds>0,m2=1-t+θ4/1-θ1Γα∫011-sα-2fs,1ds>0,m3=β-1θ51-θ2Γβ∫01s1-sβ-1gs,0ds>0,m4=1-t+θ5/1-θ2Γβ∫011-sβ-2gs,1ds>0,m5=γ-1θ61-θ3Γγ∫01s1-sγ-1rs,0ds>0,m6=1-t+θ6/1-θ3Γγ∫011-sγ-2rs,1ds>0.Since (1-t)+θ4/1-θ1-(α-1)θ4/1-θ1=(1-t)(1-θ1)+(2-α)θ4/1-θ1>0,
we can obtainm1≤m2,m3≤m4,m5≤m6.
Hence,m1h1(t)≤T1h1(t)≤m2h1(t),m3h2(t)≤T2h2(t)≤m4h2(t),m5h3(t)≤T3h3(t)≤m6h3(t).
That is,T1h1∈Ph1,T2h2∈Ph2,T3h3∈Ph3.
Therefore,Th0∈Sh.
Last, we prove that condition (iii) of Lemma5 holds. For any λ∈(0,1) and (u,v,w)∈S, by condition (3), we can get(35)T1λut=∫01Gαt,sfs,λvsds≥φ1λ∫01Gαt,sfs,vsds=φ1λT1ut,T2λvt=∫01Gβt,sgs,λwsds≥φ2λ∫01Gβt,sgs,wsds=φ2λT2vt,T3λwt=∫01Gγt,srs,λusds≥φ3λ∫01Gγt,srs,usds=φ3λT3wt.Thus,(36)Tλu,v,wt=T1λut,T2λvt,T3λwt≥φ1λT1ut,φ2λT2vt,φ3λT3wt.Let φ(λ)=min⁡{φ1(λ),φ2(λ),φ3(λ)}. Hence, T(λ(u,v,w))(t)≥φ(λ)T(u,v,w)(t).
We have proved that the operatorT satisfies all the conditions of Lemma 5. Hence T has a unique fixed point (u∗,v∗,w∗)∈Sh.
Moreover, we give the sequences, whose limit is the solution of system (7).
For any(u0,v0,w0)∈Sh, we can construct the sequences as follows:(37)unt=∫01Gαt,sfs,vn-1sds,vnt=∫01Gβt,sgs,wn-1sds,wnt=∫01Gγt,srs,un-1sds.The limit of un(t),vn(t),wn(t) exists as u∗,v∗,w∗, respectively.
From Lemma (16), we have(38)unt→u∗n→∞,vnt→v∗n→∞,wnt→w∗n→∞.Hence, (u∗,v∗,w∗) is a positive solution of system (7).


## 4. An Example

Example 1.
Consider the system of fractional differential equations:(39)D5/4ut+vtτ1+b1t=0,D6/5vt+wtτ2+b2t=0,D7/6wt+utτ3+b3t=0u0=v0=w0=0,u1=∫01t3utdt,v1=∫01t2vtdt,w1=∫01twtdt,where 0<t<1,τ1,τ2,τ3∈(0,1),b1,b2,b3:[0,1]→[0,∞) are continuous with bi≢0(i=1,2,3), then system (39) has a unique positive solution.Proof.
Compared with (7), we know α=5/4,β=6/5,γ=7/6 and(40)ft,x=xτ1+b1t,gt,x=xτ2+b2t,rt,x=xτ3+b3t,ϕt=t3,ψt=t2,ηt=t;After calculation, we get(41)θ1=∫01tα-1ϕtdt=417,θ2=∫01tβ-1ψtdt=516,θ3=∫01tγ-1ηtdt=613,θ4=∫01tα-1ϕt1-tdt=16357,θ5=∫01tβ-1ψt1-tdt=25336,θ6=∫01tγ-1ηt1-tdt=36247.We find θ1,θ2,θ3∈(0,1),θ4,θ5,θ6>0. The functions f,g,r are continuous and f(t,0)=b1(t)≢0,g(t,0)=b2(t)≢0,r(t,0)=b3(t)≢0. In addition, f(t,x),g(t,x),r(t,x) are increasing in x.
Letφ1(t)=tτ1,φ2(t)=tτ2,φ3(t)=tτ3, then φ1,φ2,φ3∈Φ.
For anyλ∈(0,1), we have(42)ft,λx=λτ1xτ1+b1t≥λτ1xτ1+b1t=φ1λft,x,gt,λx=λτ2xτ2+b2t≥λτ2xτ2+b2t=φ2λgt,x,rt,λx=λτ3xτ3+b3t≥λτ3xτ3+b3t=φ3λrt,x;So, system (39) satisfies all conditions of Theorem 8.
Hence, (39) has a unique positive solution (u∗,v∗,w∗)∈Sh,
whereh(t)=(h1(t),h2(t),h3(t))=(t1/4,t1/5,t1/6), and we can get this positive solution by the following way.
For any(u0,v0,w0)∈Sh, we can construct the sequences as follows:(43)unt=∫01Gαt,svn-1sτ1+b1sds,vnt=∫01Gβt,sgwn-1sτ2+b2sds,wnt=∫01Gγt,srun-1sτ3+b3sds,n=1,2,…,where(44)G1αt,s=t1/41-s1/4-t-s1/4Γ5/4,0≤s≤t≤1,t1/41-s1/4Γ5/4,0≤t≤s≤1,G1βt,s=t1/51-s1/5-1-s1/5Γ6/5,0≤s≤t≤1,t1/51-s1/5Γ6/5,0≤t≤s≤1,G1γt,s=t1/61-s1/6-t-s1/6Γ7/6,0≤s≤t≤1,t1/61-s1/6Γ7/6,0≤t≤s≤1,G2αt,s=t1/41-θ1∫01t3G1αt,sds=17t1/413Γ5/44171-s1/4-451-s5/4+48451-s9/4-3845851-s13/4+153699451-s17/4,G2βt,s=t1/51-θ2∫01t2G1βt,sds=16t1/511Γ6/55161-s1/5-561-s6/5+25331-s11/5-1255281-s16/5,G2γt,s=t1/61-θ3∫01tG1γt,sds=13t1/67Γ7/66131-s1/6-671-s7/6+36911-s13/6,Gαt,s=G1αt,s+G2αt,s,Gβt,s=G1βt,s+G2βt,s,Gγt,s=G1γt,s+G2γt,s.Then, the limit of un(t),vn(t),wn(t) is the solution of system (39).
That is,(45)unt→u∗n→∞,vnt→v∗n→∞,wnt→w∗n→∞.From Theorem 8, system (39) has a unique positive solution.



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