# Existence and Uniqueness of Positive Solutions for a Coupled System of Fractional Differential Equations

Authors: Hanying Feng; Xiaofeng Zhang
Journal: Discrete Dynamics in Nature and Society
 (2022)

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## Abstract

In this article, we investigate a boundary value problem for a coupled differential system of fractional order that the nonlinear term depends on the unknown functions as well as their lower order fractional derivatives. Firstly, we give Green’s functions and prove their properties; secondly, the existence and uniqueness of positive solutions are obtained by using some fixed point theorems. In addition, two examples are presented to demonstrate the application of our main results.


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## 1. Introduction

Fractional differential equations have received considerable attention in recent years due to their wide applications in engineering, physics, economy, and control theory (see the monographs of Das [1], Kilbas et al. [2], and Podlubny [3]). There are a large number of papers dealing with the solvability of nonlinear fractional differential equations, such as [4–8] and references therein. The study of coupled differential systems of fractional order is very significant because this kind of system can often occur in applications. Recently, a series of investigations on boundary value problems for fractional differential equation systems with the nonlinearity depending on the fractional derivative have been presented. Most of them are devoted to the solvability of nonlinear fractional differential equation systems by using techniques of nonlinear analysis [9–15]. It is worth mentioning that the nonlinear terms in these papers are independent of the fractional derivative of the unknown functions. But the opposite case is more difficult and complicated, and this work attempts to deal exactly with this case.In [16], the authors investigated a three-point boundary value problem for a coupled system of nonlinear fractional differential equations given by(1)Dαut+ft,vt,Dpvt=0,0<t<1,Dβvt+gt,ut,Dqut=0,0<t<1,u0=0,u1=γuη,v0=0,v1=γvη,where 1<α,β<2,p,q,γ>0,0<η<1,α−q≥1,β−p≥1,γηα−1<1,γηβ−1<1,f,g:0,1×ℝ×ℝ⟶ℝ are given continuous functions. The nonlinear terms in the coupled system involve the fractional derivatives of the unknown functions.In [12], the author discussed the existence of solutions for a coupled differential system of fractional order:(2)Dαut+ft,vt,Dμvt=0,0<t<1,Dβvt+gt,ut,Dνut=0,0<t<1,u0=u1=v0=v1=0,where 1<α,β<2,μ,ν>0,α−ν≥1,β−μ≥1,f,g:0,1×ℝ×ℝ⟶ℝ are given functions.Motivated by the above work, in this paper, we consider the system of nonlinear fractional differential equations:(3)Dαut+ft,vt,Dγvt=0,0<t<1Dβvt+gt,ut,Dδut=0,0<t<1,u0=u′0=0,u′1=μu′ξ,v0=v′0=0,v′1=μv′ξ.where 2<α,β≤3,1<γ,δ≤2 and α−δ≥1,β−γ≥1,0<ξ<1,0<μξα−2<1,0<μξβ−2<1. Dα and Dβ are the standard Riemann–Liouville fractional derivatives and f,g:0,1×ℝ+×ℝ⟶ℝ+ are given continuous functions and depend on the unknown functions as well as their lower order fractional derivatives. The properties of Green’s function are investigated and the existence and uniqueness results of positive solutions are obtained by applying Schauder fixed point theorem and contraction mapping principle.


## 2. Preliminaries

For the convenience of the readers, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literature [2–5, 7, 17].Definition 1 (see [2, 3]).
The Riemann–Liouville fractional integral of orderα>0 of a function f:0,+∞⟶ℝ is given by(4)Iαft=1Γα∫0tt−sα−1fsds,t>0,provided the integral exists.Definition 2 (see [2, 3]).
The Riemann–Liouville fractional derivative of orderα>0 for a function f:0,+∞⟶ℝ is given by(5)Dαft=ddtnIn−αft=1Γn−αddtn∫0tfst−sα−n+1ds,t>0,where n=α+1, in which α denotes the integer part of the number α.Lemma 1 (see [2, 4]).
Letα>0. If we assume u∈C0,1∩L0,1, then the fractional differential equation Dαut=0 has solutions(6)ut=C1tα−1+C2tα−2+⋯+Cntα−N,Ci∈ℝ,i=1,2,…,N,where N is the smallest integer greater than or equal to α.Lemma 2 (see [2, 4]).
Assume thatu∈C0,1∩L0,1 with a fractional derivative of order αα>0 that belongs to C0,1∩L0,1, then(7)IαDαut=ut+C1tα−1+C2tα−2+⋯+Cntα−N,for some Ci∈ℝ,i=1,2,…,N, where N is the smallest integer greater than or equal to α.Remark 1 (see [2, 3]).
The following properties are useful for our discussion:(8)IαIβft=Iα+βft,DαIαft=ft,α,β>0,a.e.t∈0,1,f∈L0,1;Dαtγ=Γγ+1Γγ+1−αtγ−α,α>0,γ>−1,t>0.
In the following, we present Green’s function of the fractional differential equation boundary value problem.Lemma 3.
Leth1∈C0,1 and 2<α≤3, the unique solution of problem(9)Dαut+h1t=0,0<t<1,(10)u0=u′0=0,u′1=μu′ξ,
isut=∫01G1t,sh1sds, where(11)G1t,s=1−sα−2tα−1−μξ−sα−2tα−1−t−sα−11−μξα−21−μξα−2Γα,0≤s≤t≤1,s≤ξ,1−sα−2tα−1−t−sα−11−μξα−21−μξα−2Γα,0<ξ≤s≤t≤1,1−sα−2tα−1−μξ−sα−2tα−11−μξα−2Γα,0≤t≤s≤ξ<1,1−sα−2tα−11−μξα−2Γα,0≤t≤s≤1,ξ≤s..
HereG1t,s is called Green’s function of BVP (9) and (10).Proof.
The functionut is said be a solution of BVP (9) and (10) if it satisfies the factional differential equation (9) and boundary conditions (10) in the classical sense. We may apply Lemma 2 to reduce equation (9) to an equivalent integral equation(12)ut=−Iαh1t+C1tα−1+C2tα−2+C3tα−3,for some C1,C2,C3∈ℝ. Consequently, the general solution of equation (9) is(13)ut=−1Γα∫0tt−sα−1h1sds+C1tα−1+C2tα−2+C3tα−3.
Byu0=u′0=0, there are C2=C3=0.
On the other hand,u′1=μu′ξ combining with(14)u′1=−∫011−sα−2Γα−1h1sds+α−1C1,u′ξ=−∫0ξξ−sα−2Γα−1h1sds+α−1C1ξα−2,one has C1=1/1−μξα−2∫011−sα−2h1s/Γαds−μ/1−μξα−2∫0ξξ−sα−2h1s/Γαds.
Therefore, the unique solution of problem (9) and (10) is(15)ut=−∫0tt−sα−1Γαh1sds+11−μξα−2∫011−sα−2tα−1Γαh1sds−μ1−μξα−2∫0ξξ−sα−2tα−1Γαh1sds.
Fort≤ξ<1, one has(16)ut=−∫0tt−sα−1Γαh1sds+11−μξα−2∫0t+∫tξ+∫ξ11−sα−2tα−1Γαh1sds−μ1−μξα−2∫0t+∫tξξ−sα−2tα−1Γαh1sds=∫0t1−sα−2tα−1−μξ−sα−2tα−1−t−sα−11−μξα−21−μξα−2Γαh1sds+∫tξ1−sα−2tα−1−μξ−sα−2tα−11−μξα−2Γαh1sds+∫ξ11−sα−2tα−11−μξα−2Γαh1sds=∫01G1t,sh1sds.
Fort≥ξ, one has(17)ut=−∫0ξ+∫ξtt−sα−1Γαh1sds+11−μξα−2∫0ξ+∫ξt+∫t11−sα−2tα−1Γαh1sds−μ1−μξα−2∫0ξξ−sα−2tα−1Γαh1sds=∫0ξ1−sα−2tα−1−μξ−sα−2tα−1−t−sα−11−μξα−21−μξα−2Γαh1sds+∫ξt1−sα−2tα−1−t−sα−11−μξα−21−μξα−2Γαh1sds+∫t11−sα−2tα−11−μξα−2Γαh1sds=∫01G1t,sh1sds.
The proof is finished.Lemma 4.
The functionG1t,s defined by equation (11) possesses the following properties:(1)
G1t,s>0, for t,s∈0,1;(2)
tα−1G11,s≤G1t,s<G11,s, for t,s∈0,1.Proof.
Let(18)g1t,s=1−sα−2tα−1−μξ−sα−2tα−1−t−sα−11−μξα−2,g2t,s=1−sα−2tα−1−t−sα−11−μξα−2,g3t,s=1−sα−2tα−1−μξ−sα−2tα−1,g4t,s=1−sα−2tα−1.(1)
Fors≤t,s≤ξ, let Δ=1−s/ξα−2tα−1−t−sα−1, If Δ≥0, by 2<α≤3,0<μξα−2<1,0<ξ<1, there is(19)g1t,s=1−sα−2tα−1−μξ−sα−2tα−1−t−sα−11−μξα−2=1−sα−2tα−1−t−sα−1−μξα−21−sξα−2tα−1−t−sα−1>1−sα−2tα−1−1−sξα−2tα−1=tα−11−sα−2−1−sξα−2≥0, If Δ<0, by 2<α≤3,0<μξα−2<1,0<ξ<1, there is(20)g1t,s=1−sα−2tα−1−t−sα−1−μξα−21−sξα−2tα−1−t−sα−1=tα−11−sα−2−1−stα−1−μξα−21−sξα−2−1−stα−1>0. For ξ≤s≤t, by 2<α≤3,0<μξα−2<1,0<1−μξα−2<1, there is(21)g2t,s=1−sα−2tα−1−t−sα−11−μξα−2>1−sα−1tα−1−t−sα−1=t−tsα−1−t−sα−1≥0. For t≤s≤ξ, by 0<μξα−2<1,0<ξ<1, there is(22)g3t,s=1−sα−2tα−1−μξ−sα−2tα−1=1−sα−2tα−1−μξα−21−sξα−2tα−1>1−sα−2tα−1−1−sξα−2tα−1=tα−11−sα−2−1−sξα−2≥0. For t≤s,ξ≤s, g4t,s>0 holds clearly. Therefore, G1t,s>0, for t,s∈0,1.(2)
In the following, we provegit,si=1,2,3,4 are monotone increasing functions for t. For s≤t,s≤ξ, if t≤ξ, by 0<μξα−2<1,0<ξ<1, there is(23)∂g1t,s∂t=α−11−sα−2tα−2−μα−1ξ−sα−2tα−2−α−1t−sα−21−μξα−2=α−11−sα−2tα−2−t−sα−2−μξα−21−sξα−2tα−2−t−sα−2>α−11−sα−2tα−2−1−sξα−2tα−2=α−1tα−21−sα−2−1−sξα−2≥0, if t>ξ, by 0<μξα−2<1,0<ξ<1, there is(24)∂g1t,s∂t=α−11−sα−2tα−2−μα−1ξ−sα−2tα−2−α−1t−sα−21−μξα−2=α−11−sα−2tα−2−t−sα−2−μξα−21−sξα−2tα−2−t−sα−2=α−1tα−21−sα−2−1−stα−2+μξα−21−stα−2−1−sξα−2>0. For ξ≤s≤t, by 2<α≤3,0<μξα−2<1,0<1−μξα−2<1, there is(25)∂g2t,s∂t=α−11−sα−2tα−2−α−1t−sα−21−μξα−2>α−11−sα−2tα−2−t−sα−2>α−1t−tsα−1−t−sα−1≥0. For t≤s≤ξ, by 0<μξα−2<1,0<ξ<1, there is(26)∂g3t,s∂t=α−11−sα−2tα−2−μα−1ξ−sα−2tα−2=α−11−sα−2tα−2−μξα−21−sξα−2tα−2>α−1tα−21−sα−2−1−sξα−2≥0. For t≤s,ξ≤s, ∂g4t,s/∂t=α−11−sα−2tα−2>0. So G1t,s is monotone increasing function for t. Therefore, G1t,s<G11,s, for t,s∈0,1.
On the other hand, when0<s≤t<1,s≤ξ, we have(27)G1t,s=tα−11−μξα−2Γα1−sα−2−μξ−sα−2−1−stα−11−μξα−2≥tα−11−μξα−2Γα1−sα−2−μξ−sα−2−1−sα−11−μξα−2=tα−1G11,s.
When0<ξ≤s≤t<1, we obtain(28)G1t,s=tα−11−μξα−2Γα1−sα−2−1−stα−11−μξα−2≥tα−11−μξα−2Γα1−sα−2−1−sα−11−μξα−2=tα−1G11,s.
In the similar discussion, we can deduce(29)G1t,s≥tα−1G11,s,where(30)G11,s=1−sα−2−μξ−sα−2−1−sα−11−μξα−21−μξα−2Γα,0≤s≤ξ,1−sα−2−1−sα−11−μξα−21−μξα−2Γα,ξ≤s≤1.
The proof is finished.
Similarly, we can obtainG2t,s if α is replaced by β,(31)G2t,s=1−sβ−2tβ−1−μξ−sβ−2tβ−1−t−sβ−11−μξβ−21−μξβ−2Γβ,0≤s≤t≤1,s≤ξ,1−sβ−2tβ−1−t−sβ−11−μξβ−21−μξβ−2Γβ,0<ξ≤s≤t≤1,1−sβ−2tβ−1−μξ−sβ−2tβ−11−μξβ−2Γβ,0≤t≤s≤ξ<1,1−sβ−2tβ−11−μξβ−2Γβ,0≤t≤s≤1,ξ≤s.
The functionG2t,s defined by (31) has the same properties as G1t,s.Lemma 5 (see [17]).
IfS is a closed, bounded and convex subset of a Banach space X and T:S⟶S is completely continuous, then T has a fixed point in S.


## 3. Main Results and Proof

In this section, we establish the existence and uniqueness of positive solutions for (3).The vectoru,v∈X×Y is said to be a positive solution of BVP (3) if u,v satisfies (3) and ut>0,vt>0 for t∈0,1, where X=ut|ut∈C10,1∩Dδut∈C0,1,1<δ≤2 and Y=vt|vt∈C10,1∩Dγvt∈C0,1,1<γ≤2.By using Green’s functionsGit,si=1,2 from Section 2, problem (3) can be written equivalently as the following nonlinear system of integral equations:(32)ut=∫01G1t,sfs,vs,Dγvsds,0≤t≤1,vt=∫01G2t,sgs,us,Dδusds,0≤t≤1..We define the spaceX=ut|∈C10,1 and Dδut∈C0,1,1<δ≤2 and the norm(33)uX=u+Dδu=max0≤t≤1ut+max0≤t≤1Dδut.Lemma 6.
X,⋅X is a Banach space.Proof.
Letutn=1∞ be a Cauchy sequence in the space X,⋅X; then clearly unn=1∞ and Dδunn=1∞ are Cauchy sequences in the space C0,1. Therefore, unn=1∞ and Dδunn=1∞ converge to some v and w on 0,1 uniformly and v,w∈C0,1. We need to prove that w=Dδv.
Note that(34)IδDδunt−Iδwt≤1Γδ∫0tt−sδ−1Dδuns−wsds≤1Γδ+1maxt∈0,1Dδunt−wt.
By the convergence ofDδunn=1∞, we have limn⟶∞IδDδunt=Iδwt uniformly for t∈0,1. On the other hand, by Remark 1 one has IδDδunt=unt. Hence, vt=Iδwt, Remark 1 implies that it is equivalent to the relation w=Dδv. This completes the proof.
Also we define the spaceY=vt|vt∈C10,1 and Dγvt∈C0,1,1<γ≤2 and the norm(35)vY=v+Dγv=max0≤t≤1vt+max0≤t≤1Dγvt.
AgainY,⋅Y is a Banach space.
Obviously the product spaceX×Y,⋅X×Y is a Banach space with norm(36)u,vX×Y=maxuX,vY,for u,v∈X×Y.
We define the operatorsT1:Y⟶X,T2:X⟶Y and T:X×Y⟶X×Y by(37)T1vt=∫01G1t,sfs,vs,Dγvsds,0≤t≤1,T2ut=∫01G2t,sgs,us,Dδusds,0≤t≤1,and Tu,v=T1v,T2u,u,v∈X×Y. Thus, the solutions of our problem (3) are the fixed points of the operator T.
Now, we give the main result of this work. For convenience, we introduce the following notations:(38)p=1−μξα−2−1,q=1−μξβ−2−1,A=pα−1Γα+1+p+μpΓα−δ+1,B=qβ−1Γβ+1+q+μqΓβ−γ+1.Theorem 1.
Iff,g:0,1×ℝ+×ℝ⟶ℝ+ are continuous functions and ft,0,0,gt,0,0 are not identically zero. Suppose that one of the following conditions is satisfied.(H1) There are two nonnegative functions at,bt∈C0,1 such that. ft,x,y≤at+c1|x|ρ1+c2|y|ρ2 and gt,x,y≤bt+d1|x|θ1+d2|y|θ2, where ci,di≥0,0<ρi,θi<1 for i=1,2(H2) ft,x,y≤c1|x|ρ1+c2|y|ρ2 and gt,x,y≤d1|x|θ1+d2|y|θ2, where ci,di≥0,ρi,θi>1 for i=1,2
Then boundary value problem (3) has a positive solution ut,vt,t∈0,1.Proof.
First, let condition (H1) be valid. Define(39)S=u,v|u,v∈X×Y,u≥0,v≥0,u,vX×Y≤R,t∈0,1,where R≥max3Ac11/1−ρ1,3Ac21/1−ρ2,3Bd11/1−θ1,3Bd21/1−θ2,3k,3l., and(40)k=∫01G11,sasds+1+pΓα−δ∫011−sα−δ−1a1ds+μpΓα−δ∫0ξξ−sα−2aξds,l=∫01G21,sbsds+1+qΓβ−γ∫011−sβ−γ−1b1ds+μqΓβ−γ∫0ξξ−sβ−2bξds.
Observe thatS is a closed convex set of the Banach space X×Y.
Now we prove thatT:S⟶S. For any u,v∈S, from the nonnegativeness of f,g and Lemma 4, it is easy to know that T1vt≥0,T2ut≥0.
Again by Remark1 and Lemma 4, we have(41)T1vt=∫01G1t,sfs,vs,Dγvsds≤∫01G1t,sasds+c1Rρ1+c2Rρ2∫01G1t,sds≤∫01G11,sasds+c1Rρ1+c2Rρ2∫01G11,sds<∫01G11,sasds+c1Rρ1+c2Rρ21α−1Γα1−μξα−2,(42)DδT1vt=|−Iα−δft,vt,Dγvt+Γα−1tα−δ−1Γα−δ1−μξα−2Iα−1f⋅,v⋅,Dγv⋅1−μIα−1f⋅,v⋅,Dγv⋅ξ|≤Iα−δat+Γα−1pΓα−δIα−1a1+Γα−1μpΓα−δIα−1aξ+c1Rρ1+c2Rρ2Γα−δ∫0tt−sα−δ−1ds+p∫011−sα−2ds+μp∫0ξξ−sα−2ds≤1+pΓα−δ∫011−sα−δ−1a1ds+μpΓα−δ∫0ξξ−sα−2aξds+c1Rρ1+c2Rρ21+p+μpΓα−δ+1.
Hence,(43)T1vX≤k+c1Rρ1+c2Rρ2A≤R3+R3+R3=R.
Similarly, one has(44)T2uY≤l+d1Rθ1+d2Rθ2B≤R3+R3+R3=R.Therefore, we get Tu,vX×Y≤R. Notice that T1vt,T2ut,DδT1vt and DγT2ut are continuous on 0,1. Thus, we have T:S⟶S.
Now we are in the position to let (H2) be satisfied. Choose(45)0<R≤min2Ac111−ρ1,2Ac211−ρ2,2Bd111−θ1,2Bd211−θ2.
Repeating arguments similar to that above we can obtain(46)T1vX≤c1Rρ1+c2Rρ2A=Ac1Rρ1−1R+Ac2Rρ2−1R≤Ac12Ac1R+Ac22Ac2R=R2+R2=R,(47)T2uY≤d1Rθ1+d2Rθ2B≤R2+R2=R.
Consequently, we haveT:S⟶S.
In view of the continuity ofG1,G2,f, and g, it is easy to see that the operator T is continuous.
Next, we proofTS is equicontinuous. We take M=max0≤t≤1ft,vt,Dγvt,N=max0≤t≤1gt,ut,Dδut, for any u,v∈S, let t1,t2∈0,1 with t1<t2, then we have(48)T1vt2−T1vt1=∫01G1t2,s−G1t1,sfs,vs,Dγvsds≤MΓα∫0t2t2−sα−1ds−∫0t1t1−sα−1ds+MΓα1−μξα−2∫011−sα−2dst2α−1−t1α−1+μMΓα1−μξα−2∫0ξξ−sα−2dst2α−1−t1α−1≤MΓα+1t2α−t1α+Mp1+μα−1Γαt2α−1−t1α−1,(49)DδT1vt2−DδT1vt1≤Iα−δft2,vt2,Dγvt2−Iα−δft1,vt1,Dγvt1+Γα−1pΓα−δIα−1f1,v1,Dγv1t2α−δ−1−t1α−δ−1+Γα−1μpΓα−δIα−1fξ,vξ,Dγvξt2α−δ−1−t1α−δ−1≤MΓα−δ∫0t2t2−sα−δ−1ds−∫0t1t1−sα−δ−1ds+MpΓα−δ∫011−sα−2dst2α−δ−1−t1α−δ−1+μMpΓα−δ∫0ξξ−sα−2dst2α−δ−1−t1α−δ−1≤MΓα−δ+1t2α−δ−t1α−δ+Mp1+μα−1Γα−δt2α−δ−1−t1α−δ−1.
Similarly,(50)T2ut2−T2ut1≤NΓβ+1t2β−t1β+Nq1+μβ−1Γβt2β−1−t1β−1,DγT2ut2−DγT2ut1≤NΓβ−γ+1t2β−γ−t1β−γ+Nq1+μβ−1Γβ−γt2β−γ−1−t1β−γ−1.
Now, using the fact that the functionstα,tα−1,tα−δ,tα−δ−1,tβ,tβ−1,tβ−γ,tβ−γ−1 are uniformly continuous on the interval 0,1, we conclude that TS is equicontinuous. Obviously, it is uniformly bounded since TS⊆S. Thus, T is completely continuous. By using Lemma 5, we conclude that the boundary value problem (3) has at least one positive solution.Theorem 2.
Assume thatf,g satisfy(51)|ft,x,y−ft,x¯,y¯|<mx−x¯|+|y−y¯,|gt,x,y−gt,x¯,y¯|<nx−x¯|+|y−y¯,for t∈0,1,x,x¯∈ℝ+,y,y¯∈ℝ. Then, the boundary value problem (3) has a unique positive solution if(52)m<pα−1Γα+2pΓα−δ+1−1,n<qβ−1Γβ+2qΓβ−γ+1−1.Proof.
We consider the operatorT:S⟶S, by the definition of Git,si=1,2, for ui,vi∈Si=1,2, we have the estimate(53)T1v1t−T1v2t=∫01G1t,sfs,v1s,Dγv1s−fs,v2s,Dγv2sds<∫01G11,smv1s−v2s+Dγv1s−Dγv2sds≤m∫01G11,sdsv1−v2<mpα−1Γαv1−v2,sDδT1v1t−DδT1v2t≤Iα−δft,v1t,Dγv1t−ft,v2t,Dγv2t+Γα−1tα−δ−1Γα−δ1−μξα−2I0+α−1f1,v11,Dγv11−f1,v21,Dγv21+μΓα−1tα−δ−1Γα−δ1−μξα−2I0+α−1fξ,v1ξ,Dγv1ξ−fξ,v2ξ,Dγv2ξ<mv1−v2Γα−δ∫0tt−sα−δ−1ds+11−μξα−2∫011−sα−2ds+μ1−μξα−2∫0ξξ−sα−2d=mv1−v2Γα−δtα−δα−δ+1α−11−μξα−2+μξα−1α−11−μξα−2≤mv1−v2Γα−δ1−μξα−2+1+μξα−1α−δ1−μξα−2≤2mpΓα−δ+1v1−v2.
So,T1v1t−T1v2t<η1v1−v2, where(54)η1=mpα−1Γα+2pΓα−δ+1<1.
Similarly,T2u1t−T2u2t<η2u1−u2, where(55)η2=nqβ−1Γβ+2qΓβ−γ+1<1.
Therefore,(56)Tu1,v1−Tu2,v2<maxη1,η2u1−u2,v1−v2.
By using contraction mapping principle, we conclude that problem (3) has a unique positive solution.


## 4. Examples

Now, we present two examples to illustrate the main results.Example 1.
Consider the following system of fractional differential equations:(57)D5/2ut+et1+et+sin2t2|vt|ρ1+|D7/6vt|ρ2=0,0<t<1,D7/3vt+t1+t+cos2t4|ut|θ1+|D5/4ut|θ2=0,0<t<1,u0=u′0=0,u′1=μu′ξ,v0=v′0=0,v′1=μv′ξ.
In system (57), α=5/2,β=7/3,γ=7/6,δ=5/4,μ=3/4,ξ=1/4. Let(58)ft,v,D7/6vt=et1+et+sin2t2|vt|ρ1+|D5/4vt|ρ2,gt,u,D5/4ut=t1+t+cos2t4|ut|θ1+|D7/6ut|θ2,for t,u,v∈0,1×ℝ+×ℝ.
We deduce(59)ft,x,y≤at+c1xρ1+c2yρ2,0<ρ1,ρ2<1,gt,x,y≤bt+d1xθ1+d2yθ2,0<θ1,θ2<1.where at=et/1+et,bt=t/1+t,c1=c2=1/2,d1=d2=1/4.
Therefore, by Theorem1, there exists one positive solution.Example 2.
Consider the following system of fractional differential equations:(60)D5/2ut+e−t6t2+|vt|+|D7/6vt|=0,0<t<1,D7/3vt+sin2t8t+|ut|+|D5/4ut|=0,0<t<1,u0=u′0=0,u′1=μu′ξ,v0=v′0=0,v′1=μv′ξ.
In system (60), α=5/2,β=7/3,γ=7/6,δ=5/4,μ=3/4,ξ=1/4. We can easily get(61)pα−1Γα+2pΓα−δ+1−1≈0.2741,qβ−1Γβ+2qΓβ−γ+1−1≈0.2522.
We deduce(62)ft,x,y−ft,x¯,y¯<16x−x¯+y−y¯,|gt,x,y−gt,x¯,y¯|<18x−x¯+y−y¯.where m=1/6<0.2741,n=1/8<0.2522. Therefore, by Theorem 2, there exists a unique positive solution.


## 5. Conclusion

The existence and uniqueness of positive solutions for a coupled differential system of fractional order with three-point boundary conditions are obtained. The main tools are some fixed point theorems in cones. It is worth mentioning that the nonlinear terms in the system are dependent on the fractional derivative of the unknown functions.



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