# Existence of Solutions for a Class of Coupled Fractional Differential Systems with Nonlocal Boundary Conditions

Authors: Tingting Qi; Yansheng Liu; Yujun Cui
Journal: Journal of Function Spaces
 (2017)

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## Abstract

Applying Schauder fixed point theorem and Leray-Schauder nonlinear alternative theory, this paper is concerned with the existence of solutions to coupled fractional differential systems with fractional integral boundary value conditions. Meanwhile, two examples are worked out to illustrate the application of the main results.


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## Body

## 1. Introduction

Fractional differential equations have a wide range of applications in many science and engineering, such as in physics, chemistry, biology, and electrodynamics. We refer the reader to see [1–5]. The main reason is that fractional derivatives provide an excellent tool for the description of memory and hereditary properties of various materials and processes. Therefore, this topic has attracted much attention of scientists and engineers. More and more good results are obtained. See [6–10] and references therein.In recent years, fractional differential equations with the nonlinear terms involving fractional derivativeD0+α of unknown functions have been investigated by some authors. See [11–16] and references therein. For example, in [13], Su studied the following nonlinear coupled fractional differential systems: (1)D0+αut=ft,vt,D0+μvt,t∈0,1,D0+βvt=gt,ut,D0+δut,t∈0,1,u0=u1=v0=v1=0,where 1<α,β<2, μ,δ>0, α-δ≥1, β-μ≥1, and f,g:[0,1]×R×R→R are given continuous functions. D0+ρ(ρ=α,β,δ,μ) is the standard Riemann-Liouville fractional derivative. Applying Schauder fixed point theorem, the existence of solution was studied.On the other hand, integral boundary conditions have various applications in applied fields such as chemical engineering, underground water flow, blood flow problems, thermoelasticity, population dynamics, and finite element method approaches with the minimization of constitutive error. In consequence, the integral boundary value problem of fractional differential equations is gaining much importance and attention. See [17–21] and references therein. For instance, in [19], Zhao and Liu studied the following coupled fractional differential systems with integral boundary conditions:(2)D0+αCut+ft,vt=0,t∈0,1,D0+αCvt+gt,ut=0,t∈0,1,uj0=vj0=0,0≤j≤n-1,j≠1,u′1=λ∫01utdt,v′1=λ∫01vtdt,where n-1<α≤n, n≥3, 0≤λ<2. CD0+α is the Caputo fractional derivate and f,g:[0,1]×[0,+∞)→[0,+∞) are given continuous functions. By using the monotone method and the theory of fixed point index on cone, they investigated the existence and uniqueness of solution for this coupled system.In [21], Ahmad et al. investigated the following coupled fractional differential system with nonlocal and integral boundary value conditions: (3)D0+αCxt=ft,xt,yt,D0+γCyt,t∈0,T,D0+βCyt=gt,xt,D0+δCxt,yt,t∈0,T,x0=hy,∫0Tysds=μ1xη,y0=ϕx,∫0Txsds=μ2yξ,where 1<α,β≤2, 0<γ≤δ<1, η,ξ∈(0,T). f,g:[0,T]×R×R×R→R, h,ϕ:C([0,T],R)→R are given continuous functions and μ1,μ2 are real number. Applying Banach contraction mapping principle and Leray-Schauder nonlinear alternative theory, the existence and uniqueness of solution were studied.To the best of our knowledge, there are fewer results for coupled fractional differential systems with nonlocal and fractional integral boundary value conditions. Motivated by the above-mentioned references, we consider the existence of solutions of the following systems:(4)D0+αCxt=ft,yt,D0+γCyt,t∈0,1,D0+βCyt=gt,xt,D0+δCxt,t∈0,1,x0=hx,x1=aI0+θ1xη,y0=φy,y1=bI0+θ2yξ,where I0+θi(i=1,2) is Riemann-Liouville fractional integral, 1<α,β<2, 0<γ,δ<1, 0≤ξ,η≤1, θi>0(i=1,2), a,b∈R. f,g∈C([0,1]×R×R→R), and h,φ:C([0,1],R)→R are given continuous functionals. Applying Schauder fixed point theorem and Leray-Schauder nonlinear alternative theory, some existence results of solutions to this systems are obtained. Finally, two examples are worked out to illustrate the application of our results.The main features of this paper are as follows. (1) The coupled fractional differential systems with nonlocal and fractional integral boundary value conditions are first studied. (2) The nonlinear term here involves fractional derivative of unknown functions. (3) Nonlocal conditions such as x(0)=h(x) and y(0)=φ(y) and fractional integral boundary conditions such as x(1)=aI0+θ1x(η) and y(1)=bI0+θ2y(ξ) are more extensive and superior to local conditions.The rest of this paper is organized as follows. Section2 introduces some basic definitions and lemmas. In Section 3, the main results are presented. Finally, in Section 4, some examples are given to illustrate the effectiveness of the main results.


## 2. Preliminary Results

In this section, we first introduce some definitions and lemmas for fractional calculus. For details, please refer to [1, 22].Definition 1.
The Caputo fractional derivative of orderα>0 of a function u:(0,∞)→R is given by (5)D0+αCut=1Γn-α∫0tunst-sα-n+1ds,where n=[α]+1,[α] denotes the integer part of the real number α.Definition 2.
The Riemann-Liouville fractional integral of orderα>0 of a function u:(0,∞)→R is given by (6)I0+αut=1Γα∫0tt-sα-1usds,where Γ is the gamma function.Lemma 3.
Letα>0. Then the fractional differential equation CD0+αu(t)=0 has solution (7)ut=c1+c2t+c3t2+···+cntn-1,where ci∈R,i=1,2,…,n,n=[α]+1.Lemma 4.
Letn-1<α≤n(n∈N). Then (8)I0+αD0+αCut=ut+c1+c2t+c3t2+···+cntn-1,where ci∈R,i=1,2,…,n,n=[α]+1.Lemma 5.
Letp>q≥0, and g∈L[a,b]. Then (9)I0+pI0+qgt=I0+p+qgt=I0+qI0+pgt,D0+pCI0+pgt=gt,D0+qCI0+pgt=I0+p-qgt,t∈a,b.Lemma 6.
Supposeh,φ:C[0,1]→R are given continuous functionals and ω,u∈L[0,1]. Then the boundary value problem (10)D0+αCxt=ωt,t∈0,1,D0+βCyt=ut,t∈0,1,x0=hx,x1=aI0+θ1xη,y0=φy,y1=bI0+θ2yξis equivalent to system(11)xt=1+τ1thx+at1-a▵1I0+α+θ1ωη-t1-a▵1I0+αω1+I0+αωt,(12)yt=1+τ2tφy+bt1-b▵2I0+β+θ2uξ-t1-b▵2I0+βu1+I0+βut,where 1<α,β<2,0≤ξ,η≤1,θi>0(i=1,2),a,b∈R,1-a▵1≠0,1-b▵2≠0, (13)▵1=ηθ1+1Γθ1+2,σ1=ηθ1Γθ1+1,τ1=aσ1-11-a▵1,▵2=ξθ2+1Γθ2+2,σ2=ξθ2Γθ2+1,τ2=bσ2-11-b▵2.Proof.
In view of Lemmas3 and 4, the solution of CD0+αx(t)=ω(t) is (14)xt=c1+c2t+I0+αωt,where ci∈R,i=1,2. From x(0)=h(x), one can get c1=h(x). By means of Lemma 5 and (14), we have (15)I0+θ1xt=c1tθ1Γθ1+1+c2tθ1+1Γθ1+2+I0+α+θ1ωt.Since x(1)=aI0+θ1x(η), one has (16)c1+c2+I0+αω1=ac1Γθ1+1ηθ1+ac2Γθ1+2ηθ1+1+aI0+α+θ1ωη,which implies (17)c21-a▵1=aσ1-1hx+aI0+α+θ1ωη-I0+αω1.Thus, (18)c1=hx,c2=aσ1-11-a▵1hx+a1-a▵1I0+α+θ1ωη-11-a▵1I0+αω1.Substituting c1 and c2 to (14), we obtain that (11) holds. Similarly, one can prove that (12) holds.LetX={x∈C[0,1]:CD0+δxt∈C[0,1]} be endowed with the norm xX=xc+CD0+δxc=maxt∈[0,1]|x(t)|+maxt∈[0,1]|CD0+δx(t)|. Then we have the following conclusions.Lemma 7.
(
X
,
·
X
) is a Banach space.
LetY={y∈C[0,1]:CD0+γy(t)∈C[0,1]} be endowed with yY=yc+CD0+γyc=maxt∈[0,1]|y(t)|+maxt∈[0,1]|CD0+γy(t)|. Then (Y,·Y) is a Banach space. Let (x,y)X×Y=xX+yY. Clearly (X×Y,·X×Y) is also a Banach space.
Defining an operatorT:X×Y→X×Y by (19)Tx,yt=T1x,yt,T2x,yt,∀x∈X,y∈Y,where(20)T1x,yt=1+τ1thx+at1-a▵1I0+α+θ1ft,yt,D0+γCytt=η-t1-a▵1I0+αft,yt,D0+γCytt=1+I0+αft,yt,D0+γCyt,T2x,yt=1+τ2tφy+bt1-b▵2I0+β+θ2gt,xt,D0+δCxtt=ξ-t1-b▵2I0+βgt,xt,D0+δCxtt=1+I0+βgt,xt,D0+δCxt,▵i,σi,τi(i=1,2) are described as in Lemma 6.Lemma 8.
Supposef,g∈C([0,1]×R2,R). Then (x,y)∈X×Y is a solution of problem (4) if and only if (x,y)∈X×Y is a fixed point of the operator T.Proof.
By Lemma6, the necessity is obvious. Now we show sufficiency.
Suppose(x,y)∈X×Y is a fixed point of the operator T. This together with (19) indicates(21)xt=1+τ1thx+at1-a▵1I0+α+θ1ft,yt,D0+γCytt=η-t1-a▵1I0+αft,yt,D0+γCytt=1+I0+αft,yt,D0+γCyt,yt=1+τ2tφy+bt1-b▵2I0+β+θ2gt,xt,D0+δCxtt=ξ-t1-b▵2I0+βgt,xt,D0+δCxtt=1+I0+βgt,xt,D0+δCxt.
Notice thatCD0+ρtρ-m=0, m=1,2,…,N, where N is the smallest integer greater than or equal to ρ. Therefore, (22)D0+αCxt=D0+αCI0+αft,yt,D0+γCyt=ft,yt,D0+γCyt.Similarly, one has (23)D0+βCyt=D0+βCI0+βgt,xt,D0+δCxt=gt,xt,D0+δCxt.By direct computation, we can easily get x(0)=h(x),y(0)=φ(y),x(1)=aI0+θ1x(η),y(1)=bI0+θ2y(ξ). Therefore, (x,y)∈X×Y is a solution of (4).Lemma 9 ([23] (Leray-Schauder nonlinear alternative)).
LetF be a Banach space, Ω be a bounded open subset of F, and 0∈Ω,T:Ω¯→F be a completely continuous operator. Then, either there exists x∈∂Ω,λ>1 such that Tx=λx or there exists a fixed point x∗∈Ω¯.


## 3. Main Results

For convenience, some notations and assumptions are stated as follows:(24)A1=a1-a▵11Γθ1+α+1+11-a▵11Γα+1+1Γα+1,A2=b1-b▵21Γθ2+β+1+11-b▵21Γβ+1+1Γβ+1,B1=1+τ1+τ1Γ2-δ,B2=1+τ2+τ2Γ2-γ,C1=a1-a▵11Γθ1+α+1+11-a▵11Γα+1+1Γα,C2=b1-b▵21Γθ2+β+1+11-b▵21Γβ+1+1Γβ,D1=A1+C1Γ2-δ,D2=A2+C2Γ2-γ,P1=a1+b1ϕ1Q+c1ψ1Q,P2=a2+b2ϕ2Q+c2ψ2Q.(H1)
f,g:[0,1]×R2→R are continuous functions and there exist constants ki>0,li>0, and mi,ni∈(0,1),i=1,2, such that(25)ft,u,v≤k0+k1um1+k2vm2,∀t∈0,1,u,v∈R;gt,u,v≤l0+l1un1+l2vn2,∀t∈0,1,u,v∈R.(H2)
h,φ:C[0,1]→R are continuous functionals, h(0)=φ(0)=0, and there exist constants K,L>0,μ,λ∈(0,1) such that, for all x,y∈C[0,1],(26)hx≤Kxcμ,φy≤Lycλ.H3
f,g:[0,1]×R2→R are continuous functions, and there exist ai,bi,ci,di≥0(i=1,2) and nondecreasing functions ϕi,ψi∈C([0,+∞),[0,+∞))(i=1,2), such that(27)ft,u,v≤a1+b1ϕ1u+c1ψ1v,∀t∈0,1,u,v∈R;(28)gt,u,v≤a2+b2ϕ2u+c2ψ2v,∀t∈0,1,u,v∈R.(H4)
There existsQ>0 such that B1KQμ+D1P1+B2LQλ+D2P2<Q, where K,L,μ,λ are described as in (H2).Theorem 10.
Assume that(H1)-(H2) hold. Then FDE (4) has at least one solution.Proof.
Let(29)BR=x,y∈X×Y∣x,yX×Y≤R,where(30)R≥max8B1K1/1-μ,8B2L1/1-λ,8k0D1,8l0D2,8k1D11/1-m1,8l1D21/1-n1,8k2D11/1-m2,8l2D21/1-n2.
Now we prove thatT:BR→BR. In fact, for any (x,y)∈BR, it follows from Definition 2 and (H1) that(31)T1x,yt≤1+τ1hx+a1-a▵1I0+α+θ1ft,yt,D0+γCytt=η+11-a▵1I0+αft,yt,D0+γCytt=1+I0+αft,yt,D0+γCyt≤1+τ1KRμ+a1-a▵11Γθ1+α+1k0+k1Rm1+k2Rm2+11-a▵11Γα+1k0+k1Rm1+k2Rm2+1Γα+1k0+k1Rm1+k2Rm2=1+τ1KRμ+k0+k1Rm1+k2Rm2a1-a▵11Γθ1+α+1+11-a▵11Γα+1+1Γα+1≤1+τ1KRμ+A1k0+k1Rm1+k2Rm2.
On the other hand, by Definition1, we can obtain (32)D0+δCT1x,yt=1Γ1-δ∫0tT1x,y′st-sδds.Repeating a process similar to that of (31), it is easy to see (33)T1x,y′t≤τ1KRμ+a1-a▵11Γα+θ1+1k0+k1Rm1+k2Rm2+11-a▵11Γα+1k0+k1Rm1+k2Rm2+1Γαk0+k1Rm1+k2Rm2=τ1KRμ+k0+k1Rm1+k2Rm2a1-a▵11Γα+θ1+1+11-a▵11Γα+1+1Γα≤τ1KRμ+C1k0+k1Rm1+k2Rm2.It follows from (32) and (33) that (34)D0+δCT1x,yt≤1Γ1-δ∫0tT1x,y′st-sδds≤1Γ1-δ∫0tτ1KRμ+C1k0+k1Rm1+k2Rm2t-sδds≤1Γ2-δτ1KRμ+C1k0+k1Rm1+k2Rm2.By virtue of (31) and (34), we have (35)T1x,yX≤B1KRμ+k0+k1Rm1+k2Rm2D1≤R8+R8+R8+R8=R2.Similarly, we can get (36)T2x,yY≤B2LRλ+l0+l1Rn1+l2Rn2D2≤R8+R8+R8+R8=R2.Therefore, (37)Tx,yX×Y=T1x,yX+T2x,yY≤R2+R2≤R,which means that T:BR→BR.
From the continuity of the functionsf,g,h, and φ, it is not difficult to see that T is continuous.
Next we show thatT(BR) are equicontinuous. For this sake, let (38)M1=maxt∈0,1,x≤R,y≤R⁡ft,x,y,M2=maxt∈0,1,x≤R,y≤R⁡gt,x,y.For any t1,t2∈[0,1](t1<t2), one has (39)T1x,yt2-T1x,yt1≤τ1KRμt2-t1+a1-a▵1M1Γα+θ1+1t2-t1+M11-a▵11Γα+1t2-t1+1Γα∫0t2t2-sα-1fs,ys,D0+γCysds-1Γα∫0t1t1-sα-1fs,ys,D0+γCysds≤τ1KRμt2-t1+a1-a▵1M1Γα+θ1+1t2-t1+M11-a▵11Γα+1t2-t1+M1Γα+1t2α-t1α.On the other hand, we have (40)D0+δCT1x,yt2-D0+δCT1x,yt1=1Γ1-δ∫0t2T1x,y′st2-sδds-1Γ1-δ∫0t1T1x,y′st1-sδds≤1Γ1-δ∫0t1t2-sδ-t1-sδt1-sδt2-sδT1x,y′sds+∫t1t21t2-sδT1x,y′sds.Notice that (41)T1x,y′s≤τ1KRμ+M1C1.This together with (39)-(40) guarantees that (42)D0+δCT1x,yt2-D0+δCT1x,yt1≤1Γ1-δτ1KRμ+M1C1∫0t1t2-sδ-t1-sδt1-sδt2-sδds+∫t1t21t2-sδds≤1Γ2-δτ1KRμ+M1C1t21-δ-t11-δ+2t2-t11-δ.Similarly, we can get(43)T2x,yt2-T2x,yt1≤τ2LRλt2-t1+b1-b▵2M2Γβ+θ2+1t2-t1+M21-b▵21Γβ+1t2-t1+M2Γβ+1t2β-t1β.D0+γCT2x,yt2-D0+γCT2x,yt1≤1Γ2-γτ2LRλ+M2C2t21-γ-t11-γ+2t2-t11-γ.This together with (39), (42), and (43) implies that T(BR) are equicontinuous. By virtue of the Arzela-Ascoli theorem, we can infer that the operator T is completely continuous. Hence, applying Schauder fixed theorem, FDE (4) has at least one solution (x,y) in BR.Theorem 11.
Assume that(H2)–(H4) hold. Then FDE (4) has at least one solution.Proof.
Firstly, we show thatT is completely continuous operator. From the continuity of the functions f,g,h, and φ, it follows that the operator T is continuous. Let BQ={(x,y)∈X×Y∣(x,y)X×Y≤Q}, where Q is described as in (H4); we now prove that T(BQ) is relatively compact.
For any(x,y)∈BQ, by similar computation as (31) and (34), we obtain that(44)T1x,yt≤1+τ1KQμ+A1a1+b1ϕ1Q+c1ψ1Q,D0+δCT1x,yt≤1Γ2-δτ1KQμ+C1a1+b1ϕ1Q+c1ψ1Q,which imply (45)T1x,ytX≤1+τ1+τ1Γ2-δKQμ+D1a1+b1ϕ1Q+c1ψ1Q.Similarly, we have (46)T2x,ytY≤1+τ2+τ2Γ2-γLQλ+D2a2+b2ϕ2Q+c2ψ2Q.Combining (45) and (46), one can get (47)Tx,yX×Y≤1+τ1+τ1Γ2-δKQμ+D1a1+b1ϕ1Q+c1ψ1Q+1+τ2+τ2Γ2-γLQλ+D2a2+b2ϕ2Q+c2ψ2Q=B1KQμ+D1P1+B2LQλ+D2P2.By (47) and (H4) we know T(BQ)⊂BQ. Using similar computation as in (39), (40), (42), and (43), we can obtain T(BQ) are equicontinuous. Thus, T is completely continuous by Arzela-Ascoli theorem.
We now claim thatT has at least one solution in X×Y.
Suppose there exists(x,y)∈∂BQ such that (x,y)=λT(x,y) for some λ∈(0,1). This together with (44) implies that(48)xt=λT1x,yt≤T1x,yt,D0+δCxt=λD0+δCT1x,yt≤D0+δCT1x,yt.By virtue of (44)–(47), it is easy to see (49)x,yX×Y≤B1KQμ+D1P1+B2LQλ+D2P2<Q.This is a contradiction with (x,y)∈∂BQ. Therefore, it follows from Lemma 9 that the operator T has a fixed point (x,y)∈B¯Q. Thus FDE (4) has at least one solution in X×Y.Remark 12.
Comparing(H1) with (H3), we know that condition (H3) is more extensive than (H1). However, the assumptions of Theorem 10 are easier to verify than that of Theorem 11.


## 4. Examples

Example 1.
Consider the following coupled system:(50)D0+5/4Cxt=125+t2cos⁡t+12yt1/5+141+tD0+1/4Cyt1/3,t∈0,1,D0+3/2Cyt=e-t3+t2+14xt1/7+142+tsin⁡D0+1/2Cxt1/5,t∈0,1,x0=14∫01xtdt1/7,x1=3I0+1/4x13,y0=12∫01ytdt1/5,y1=2I0+1/2y12,where α=5/4, β=3/2, γ=1/4, δ=1/2, a=3, b=2, θ1=1/4, θ2=1/2, η=1/3, ξ=1/2.
(50) can be regarded as the form of (4), where (51)ft,u,v=125+t2cos⁡t+12ut1/5+141+tvt1/3,gt,u,v=e-t3+t2+14ut1/7+142+tsin⁡vt1/5,hx=14∫01xtdt1/7,φy=12∫01ytdt1/5.It is not difficult to see (52)ft,u,v≤15+12u1/5+14v1/3,gt,u,v≤13+14u1/7+18v1/5,hx≤14xc1/7,φy≤12yc1/5.Choose k0=1/5, k1=1/2, k2=1/4, l0=1/3, l1=1/4, l2=1/8, m1=1/5, m2=1/3, n1=1/7, n2=1/5, K=1/4, L=1/2, μ=1/7, λ=1/5. Obviously, h(0)=φ(0)=0. Therefore, (H1) and (H2) are satisfied. Immediately, we can conclude that FDE (4) has at least one solution by Theorem 10.Example 2.
Consider the following problem:(53)D0+3/2Cxt=1-t23000+y2t70001+y+t5000D0+1/2Cyt2,t∈0,1,D0+5/4Cyt=1+t24000+1-tx4t40001+x2+t21000D0+1/2Cxt3,t∈0,1,x0=1200maxt∈0,1⁡xt1/3,x1=4I0+1/4x14,y0=1400maxt∈0,1⁡yt1/2,y1=2I0+1/2y12,where α=3/2, β=5/4, γ=1/2, δ=1/2, a=4, b=2, θ1=1/4, θ2=1/2, η=1/4, ξ=1/2.
(53) can be regarded as the form of (4), where (54)ft,u,v=1-t23000+u270001+u+t5000v2,gt,u,v=1+t24000+1-tu440001+u2+t21000v3,hx=1200maxt∈0,1⁡xt1/3,φy=1400maxt∈0,1yt1/2.It is not difficult to see (55)ft,u,v≤13000+u27000+v25000,gt,u,v≤11000+14000u2+11000v3,hx≤1200xc1/3,φy≤1400yc1/2.Choose a1=1/3000, b1=1/7000, c1=1/5000, ϕ1(|u|)=|u|2, ψ1(|v|)=|v|2, a2=1/1000, b2=1/4000, c2=1/1000, ϕ2(|u|)=|u|2, ψ2(|v|)=|v|3, K=1/200, L=1/400, μ=1/3, λ=1/2. Now we need to verify (H4). By direct calculation, it is easy to see (56)▵1=0.0781,σ1=0.7801,τ1=3.0838,▵2=0.2660,σ2=0.1995,τ2=-1.2842,A1=11.2022,B1=4.9537,C1=10.7321,D1=14.2296,P1=0.0051,A2=5.9608,B2=0.6447,C2=5.9846,D2=7.6490,P2=0.0690.Thus, we can get B1KQμ+D1P1+B2LQλ+D2P2=0.0394+0.0726+0.0032+0.5278=0.7582<4 for Q=4, which means that (H4) holds. Consequently, FDE (4) has at least one solution by Theorem 11.



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