# Positive Solutions for(n-1,1)-Type Singular Fractional Differential System with Coupled Integral Boundary Conditions

Authors: Ying Wang; Lishan Liu; Xinguang Zhang; Yonghong Wu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We study the positive solutions of the(n-1,1)-type fractional differential system with coupled integral boundary conditions. The conditions for the existence of positive solutions to the system are established. In addition, we derive explicit formulae for the estimation of the positive solutions and obtain the unique positive solution when certain additional conditions hold. An example is then given to demonstrate the validity of our main results.


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## Body

## 1. Introduction

This paper is motivated by the boundary value problem(1)D0+5/2ut+sint+cos⁡tut1-tv3=0,D0+5/2vt+v3ett1-tu=0,0<t<1,u0=u′0=0,u1=12∫01vtdt,v0=v′0=0,v1=∫01utdt2,
which arises in a variety of disciplinary areas such as mechanics, chemical physics, mathematical biology, flows, fluid electrical networks, and viscoelasticity (see [1–6] and the references cited therein). In problem (1), the nonlinearity f(t,x,y) may be singular at t=0,1 and y=0; g(t,x,y) may be singular at t=0,1 and x=0.Research on fractional order integrodifferential operators dates back to the end of the 19th century, when Riemann and Liouville introduced the first definition of the fractional derivative. However, this field of study started to become attractive to engineers only in the late 1960s, when fractional derivative description of some real systems was observed. It was found that fractional operators are nonlocal and are more suitable for constructing models possessing memory effect in a long time period, and hence fractional differential equations possess many advantages.In this paper, we consider the existence of positive solutions for a nonlinear singular fractional differential system with coupled boundary conditions:(2)D0+α1ut+ft,ut,vt=0,D0+α2vt+gt,ut,vt=0,0<t<1,u0=u′0=⋯=un-2=0,u1=μ1∫01vsdA1s,v0=v′0=⋯=vn-2=0,v1=μ2∫01usdA2s,
where n-1<αi≤n, n≥2, and D0+αi is the standard Riemann-Liouville derivative. μi>0, Ai is right continuous on [0,1), left continuous at t=1, and nondecreasing on [0,1], Ai(0)=0, and ∫01x(s)dAi(s) denotes the Riemann-Stieltjes integrals of x with respect to Ai(i=1,2). f:(0,1)×[0,+∞)×(0,+∞)→[0,+∞) and g:(0,1)×(0,+∞)×[0,+∞)→[0,+∞) are two continuous functions, and f(t,x,y) may be singular at t=0,1 and y=0, while g(t,x,y) may be singular at t=0,1 and x=0.Coupled boundary value problem arises naturally in the research of Sturm-Liouville problems, reaction-diffusion equations, mathematical biology, and so on. In recent years, there has been a significant development in ordinary and partial differential equations involving fractional derivatives with coupled boundary conditions, as shown by [7–16] and the references therein. By using the nonlinear alternative of the Leray and Schauder theorem and the Krasnoselskii fixed point theorem in a cone, Bai and Fang in [17] obtained some results of existence of positive solutions by considering the singular coupled system of nonlinear fractional differential equations:
(3)Dsu+ft,v=0,Dpv+gt,u=0,0<t<1,
where 0<s, p<1, Ds, Dp are two standard Riemann-Liouville fractional derivative, and f,g:(0,1]×[0,+∞)→[0,+∞) are two given continuous functions and are singular at t=0.Wang et al. [18] study the following system of nonlinear fractional differential equations:
(4)D0+αut+ft,vt=0,D0+βvt+gt,ut=0,0<t<1,u0=v0=0,u1=auξ,v1=bvξ,
where 1<α, β<2, 0≤a, b<1, 0<ξ<1, f,g:[0,1]×[0,+∞)→[0,+∞) are continuous functions, and D0+α, D0+β are two standard Riemann-Liouville fractional derivatives. By using the Banach fixed point theorem and the nonlinear alternative of Leray-Schauder type, the existence and uniqueness of a positive solution are obtained.In this paper, we consider the existence and uniqueness of positive solutions for the singular system (2). The work presented in this paper has the following new features. Firstly, until now, coupled integral boundary value problems for fractional differential system as system (2) have seldom been considered when f(t,x,y) may be singular at t=0,1 and y=0, and g(t,x,y) may be singular at t=0,1 and x=0. Also ∫01x(s)dAi(s) denotes the Riemann-Stieltjes integral, and thus system (2) includes the multipoint problems and integral problems as special cases. Secondly, by using the well-known fixed point theorem due to Guo-Krasnoselskii, we not only obtain the existence of positive solutions for system (2), but also obtain the uniqueness of system (2).A vectoru,v is said to be a positive solution of system (2) if and only if u,v satisfies (2) and ut>0, v(t)>0 for any t∈(0,1].


## 2. Preliminaries and Lemmas

In what follows, we present some necessary definitions about fractional calculus theory.Definition 1 (see [2, 19]).
Letα>0 and let u be piecewise continuous on (0,+∞) and integrable on any finite subinterval of [0,+∞). Then, for t>0, we call
(5)I0+αut=1Γα∫0tt-sα-1usds
the Riemann-Liouville fractional integral of u of order α.Definition 2 (see [2, 19]).
The Riemann-Liouville fractional derivative of orderα>0, n-1≤α<n, n∈N, is defined as
(6)D0+αut=1Γn-αddtn∫0tt-sn-α-1usds,
where N denotes the natural number set and the function u(t) is n times continuously differentiable on [0,+∞).Lemma 3 (see [2]).
Letα>0; then,
(7)I0+αD0+αutM=ut+c1tα-1+c2tα-2+⋯+cntα-n,
where c1,c2,…,cn∈(-∞,+∞) and n is the smallest integer greater than or equal to α.Lemma 4.
Lethi∈C(0,1)∩L1(0,1)(i=1,2), and the following condition (H0) holds:H0
(8)k1=∫01tα2-1dA1t>0,k2=∫01tα1-1dA2t>0,1-μ1μ2k1k2>0.Then the system subjected to the coupled boundary conditions
(9)D0+α1ut+h1t=0,D0+α2vt+h2t=0,0<t<1,u0=u′0=⋯=un-2=0,u1=μ1∫01vsdA1s,v0=v′0=⋯=vn-2=0,v1=μ2∫01usdA2s
has an integral representation
(10)ut=∫01K1t,sh1sds+∫01H1t,sh2sds,vt=∫01K2t,sh2sds+∫01H2t,sh1sds,
where
(11)K1t,s=μ1μ2k1tα1-11-μ1μ2k1k2∫01G1t,sdA2t+G1t,s,H1t,s=μ1tα1-11-μ1μ2k1k2∫01G2t,sdA1t,K2t,s=μ2μ1k2tα2-11-μ1μ2k1k2∫01G2t,sdA1t+G2t,s,H2t,s=μ2tα2-11-μ1μ2k1k2∫01G1t,sdA2t,(12)Git,s=1Γαit1-sαi-1-t-sαi-1,0≤s≤t≤1,t1-sαi-1,0≤t≤s≤1,i=1,2.Proof.
By Lemma3, the system (9) is equivalent to the following integral equations system:
(13)ut=u1tα1-1+∫01G1t,sh1sds,(14)vt=v1tα2-1+∫01G2t,sh2sds.
Integrating (13) and (14) with respect to dA2(t) and dA1(t), respectively, we have
(15)∫01utdA2tM=u1∫01tα1-1dA2t+∬01G1t,sh1sdsdA2t,∫01vtdA1tM=v1∫01tα2-1dA1t+∬01G2t,sh2sdsdA1t,
which yield
(16)1μ1u1-k1v1=∬01G2t,sh2sdsdA1t,-k2u1+1μ2v1=∬01G1t,sh1sdsdA2t.
It follows from
(17)1μ1-k1-k21μ2=1-μ1μ2k1k2μ1μ2≠0
that the system of (16) has a unique solution, which can be represented as
(18)u1=μ11-μ1μ2k1k2M·∬01G2t,sh2sdsdA1tMMM+μ2k1∬01G1t,sh1sdsdA2t,v1=μ21-μ1μ2k1k2M·∬01G1t,sh1sdsdA2tMMM+μ1k2∬01G2t,sh2sdsdA1t.
Substituting (18) into (13) and (14), we have
(19)ut=μ1tα1-11-μ1μ2k1k2MM·∬01G2t,sh2sdsdA1tMMMM+μ2k1∬01G1t,sh1sdsdA2tMM+∫01G1t,sh1sdsM=∫01K1t,sh1sds+∫01H1t,sh2sds,vt=μ2tα2-11-μ1μ2k1k2MM·∬01G1t,sh1sdsdA2tMMM+μ1k2∬01G2t,sh2sdsdA1tMM+∫01G2t,sh2sdsM=∫01K2t,sh2sds+∫01H2t,sh1sds.
So (10) holds. This completes the proof of the lemma.Lemma 5.
Fort,s∈[0,1], the functions Ki(t,s) and Hi(t,s)  (i=1,2) defined by (11) possess the following properties:
(20)K1t,s,H2t,s≤ρs1-sα1-1,K2t,s,H1t,s≤ρs1-sα2-1,(21)K1t,s,H1t,s≤ρtα1-1,K2t,s,H2t,s≤ρtα2-1,(22)K1t,s≥ϱtα1-1s1-sα1-1,H2t,s≥ϱtα2-1s1-sα1-1,(23)K2t,s≥ϱtα2-1s1-sα2-1,H1t,s≥ϱtα1-1s1-sα2-1,
where
(24)ρ=max⁡1Γα1-1μ1μ2k11-μ1μ2k1k2∫01dA2t+1,μ1Γα2-11-μ1μ2k1k2∫01A1t,1Γα2-1μ2μ1k21-μ1μ2k1k2∫01dA1t+1,μ2Γα1-11-μ1μ2k1k2∫01dA2t,ϱ=min⁡μ1μ2k1Γα11-μ1μ2k1k2∫011-ttα1-1dA2t,μ1Γα21-μ1μ2k1k2∫011-ttα2-1dA1t,μ2μ1k2Γα21-μ1μ2k1k2∫011-ttα2-1dA1t,μ2Γα11-μ1μ2k1k2∫011-ttα1-1dA2t.Proof.
By [20], for any t,s∈[0,1], we have
(25)1-ttαi-1s1-sαi-1ΓαiM≤Git,s≤s1-sαi-1Γαi-1,i=1,2.
It follows from (11) and (25) that
(26)K1t,s=μ1μ2k1tα1-11-μ1μ2k1k2∫01G1t,sdA2t+G1t,s≤μ1μ2k1tα1-11-μ1μ2k1k2∫01s1-sα1-1Γα1-1dA2t+s1-sα1-1Γα1-1≤1Γα1-1μ1μ2k1tα1-11-μ1μ2k1k2∫01dA2t+1s1-sα1-1≤1Γα1-1μ1μ2k11-μ1μ2k1k2∫01dA2t+1s1-sα1-1≤ρs1-sα1-1,H2t,s=μ2tα2-11-μ1μ2k1k2∫01G1t,sdA2t≤s1-sα1-1Γα1-1μ2tα2-11-μ1μ2k1k2∫01dA2t≤μ2Γα1-11-μ1μ2k1k2∫01dA2ts1-sα1-1≤ρs1-sα1-1.
As for the proof of (26), we have
(27)K2t,s,H1t,s≤ρs1-sα2-1,t,s∈0,1;
that is, (20) holds.
By [20], for any t,s∈[0,1], we have
(28)1-ttαi-1s1-sαi-1ΓαiM≤Git,s≤tαi-11-tΓαi-1,i=1,2.
So, by (11) and (28), we have
(29)K1t,s=μ1μ2k1tα1-11-μ1μ2k1k2∫01G1t,sdA2t+G1t,s≤μ1μ2k1tα1-11-μ1μ2k1k2∫01tα1-11-tΓα1-1dA2tM+tα1-11-tΓα1-1≤μ1μ2k1tα1-11-μ1μ2k1k2∫011Γα1-1dA2t+tα1-1Γα1-1≤1Γα1-1μ1μ2k11-μ1μ2k1k2∫01dA2t+1tα1-1≤ρtα1-1,H2t,s=μ2tα2-11-μ1μ2k1k2∫01G1t,sdA2t≤μ2tα2-11-μ1μ2k1k2∫01tα1-11-tΓα1-1dA2t≤μ2tα2-11-μ1μ2k1k2∫011Γα1-1dA2t≤ρtα2-1.
Proceeding as for the proof of (29), we have
(30)K2t,s≤ρtα2-1,H1t,s≤ρtα1-1,t∈0,1;
thus (21) holds.
On the other hand, it follows from (11) and (25) that
(31)K1t,s=μ1μ2k1tα1-11-μ1μ2k1k2∫01G1t,sdA2t+G1t,s≥μ1μ2k1tα1-11-μ1μ2k1k2∫011-ttα1-1s1-sα1-1Γα1dA2t≥μ1μ2k1Γα11-μ1μ2k1k2∫011-ttα1-1dA2t·tα1-1s1-sα1-1≥ϱtα1-1s1-sα1-1,H2t,s=μ2tα2-11-μ1μ2k1k2∫01G1t,sdA2t≥μ2tα2-11-μ1μ2k1k2∫011-ttα1-1s1-sα1-1Γα1dA2t≥μ2Γα11-μ1μ2k1k2∫011-ttα1-1dA2t·tα2-1s1-sα1-1≥ϱtα2-1s1-sα1-1,
which implies that (22) holds. Similarly, we also have
(32)K2t,s≥ϱtα2-1s1-sα2-1,H1t,s≥ϱtα1-1s1-sα2-1,t∈0,1.
This completes the proof of the lemma.From Lemma5, we have the following conclusion.Remark 6.
Fort,τ,s∈[0,1], we have
(33)Kit,s≥ωtαi-1Kiτ,s,Hit,s≥ωtαi-1Hiτ,s,i=1,2,K1t,s≥ωtα1-1H2τ,s,H2t,s≥ωtα2-1K1τ,s,K2t,s≥ωtα2-1H1τ,s,H1t,s≥ωtα1-1K2τ,s,
where ω=ϱ/ρ, 0<ω<1.Throughout this paper, we will work in the spaceX=C[0,1]×C[0,1], which is a Banach space if it is endowed with the norm
(34)u,v=max⁡u,v,u=max⁡0≤t≤1ut,v=max⁡0≤t≤1vt.
Let
(35)K=u,v∈X:ut≥ωtα1-1u,v,vt≥ωtα2-1u,v,t∈0,1;
then K is a cone in X. For 0<r<R, denote
(36)Kr,R=u,v∈K:r≤u,v≤R,Kr=u,v∈K:u,v<r.
In what follows, we list some conditions to be used later:(H1)
f:(0,1)×[0,+∞)×(0,+∞)→[0,+∞) is continuous, f(t,x,y) is nondecreasing in x and nonincreasing in y, and there exist λ1,δ1∈(0,1) such that
(37)cλ1ft,x,y≤ft,cx,y,x,y>0,c∈0,1,ft,x,cy≤c-δ1ft,x,y,x,y>0,c∈0,1;g:(0,1)×(0,+∞)×[0,+∞)→[0,+∞) is continuous, g(t,x,y) is nonincreasing in x and nondecreasing in y, and there exist λ2,δ2∈(0,1) such that
(38)cλ2gt,x,y≤gt,x,cy,x,y>0,c∈0,1,gt,cx,y≤c-δ2gt,x,y,x,y>0,c∈0,1;(H2)
(39)∫01s1-sα1-1fs,1,sα2-1ds<+∞,∫01s1-sα2-1gs,sα1-1,1ds<+∞.Remark 7.
By(H1), we have
(40)fs,sα2-1,1≤fs,1,sα2-1,gs,1,sα1-1≤gs,sα1-1,1.
This together with (H2) yields
(41)∫01s1-sα1-1fs,sα2-1,1dsM≤∫01s1-sα1-1fs,1,sα2-1ds<+∞,∫01s1-sα2-1gs,1,sα1-1dsM≤∫01s1-sα1-1gs,sα1-1,1ds<+∞.From the above assumptions(H0)–(H2), for any (u,v)∈K∖{(0,0)}, we define an integral operator T:K∖{(0,0)}→X by
(42)Tu,vtM=T1u,vt,T2u,vt,0≤t≤1,
where
(43)T1u,vtM=∫01K1t,sfs,us,vsdsMM+∫01H1t,sgs,us,vsds,T2u,vtM=∫01K2t,sgs,us,vsdsMM+∫01H2t,sfs,us,vsds.
Now we claim that T is well defined for (u,v)∈K∖{(0,0)}. In fact, for any (u,v)∈K∖{(0,0)}, we have
(44)ωtα1-1u,v≤ut≤u,v,ωtα2-1u,v≤vt≤u,v,t∈0,1.
Let c be a positive number such that (u,v)/c<1, c>1. From (H1) and (44), we have
(45)ft,ut,vt≤ft,c,ωtα2-1u,v≤cλ1ft,1,ωu,vctα2-1≤cλ1+δ1ωu,v-δ1ft,1,tα2-1,gt,ut,vt≤gt,ωtα1-1u,v,c≤cλ2gt,ωu,vctα1-1,1≤cλ2+δ2ωu,v-δ2gt,tα1-1,1.
Hence, for any t∈[0,1], by Lemma 5 and (45), we have
(46)T1u,vt=∫01K1t,sfs,us,vsds+∫01H1t,sgs,us,vsds≤ρcλ1+δ1ωu,v-δ1∫01s1-sα1-1fs,1,sα2-1ds+cλ2+δ2ωu,v-δ2·∫01s1-sα2-1gs,sα1-1,1ds<+∞.
Similarly, for any t∈[0,1], we have
(47)T2u,vt≤ρcλ1+δ1ωu,v-δ1∫01s1-sα1-1fs,1,sα2-1ds+cλ2+δ2ωu,v-δ2·∫01s1-sα2-1gs,sα1-1,1ds<+∞.Together with the continuity ofKi(t,s) and Hi(t,s)  (i=1,2), it is easy to see that Tiu,v∈C[0,1], for u,v∈K∖{(0,0)}. Therefore T:K∖0,0→X is well defined.Obviously,(u,v) is a positive solution of system (2) if and only if (u,v) is a fixed point of T in K∖{(0,0)}.Lemma 8.
Assume that(H0)–(H2) hold. Then T:K[r1,r2]→K is a completely continuous operator.Proof.
First, we showTK[r1,r2]⊆K.
For any(u,v)∈K[r1,r2], 0≤t, τ≤1, by Remark 6, we have
(48)T1u,vt=∫01K1t,sfs,us,vsds+∫01H1t,sgs,us,vsds≥∫01ωtα1-1K1τ,sfs,us,vsds+∫01ωtα1-1H1τ,sgs,us,vsds≥ωtα1-1∫01K1τ,sfs,us,vsds+∫01H1τ,sgs,us,vsds≥ωtα1-1T1u,vτ,T1u,vt=∫01K1t,sfs,us,vsds+∫01H1t,sgs,us,vsds≥∫01ωtα1-1H2τ,sfs,us,vsds+∫01ωtα1-1K2τ,sgs,us,vsds≥ωtα1-1∫01H2τ,sfs,us,vsds+∫01K2τ,sgs,us,vsds≥ωtα1-1T2u,vτ.
Then, we have
(49)T1u,vt≥ωtα1-1T1u,v,T1u,vt≥ωtα1-1T2u,v;
that is,
(50)T1u,vt≥ωtα1-1T1u,v,T2u,v.
In the same way as (48), we can prove that
(51)T2u,vt≥ωtα2-1T1u,v,T2u,v.
Therefore, we have TK[r1,r2]⊆K.
Next, we showT:K[r1,r2]→K is continuous.
Let(un,vn),(u,v)∈K[r1,r2], such that (un,vn)-(u,v)→0(n→+∞). Obviously, r1≤(u,v), (un,vn)≤r2 for all n; choose c such that (u,v)/c<1, (un,vn)/c<1, c>1. So, by Lemma 5, (H1), and (43), we have
(52)T1un,vnt-T1u,vt=∫01K1t,sfs,uns,vnsds+∫01H1t,sgs,uns,vnsds-∫01K1t,sfs,us,vsds+∫01H1t,sgs,us,vsds≤2ρ∫01s1-sα1-1·fs,uns,vns+fs,us,vsds+2ρ∫01s1-sα2-1·gs,uns,vns+gs,us,vsds≤ρ∫01scλ1+δ1ωu,v-δ1+ωun,vn-δ1·∫01s1-sα1-1fs,1,sα2-1ds+cλ2+δ2ωu,v-δ2+ωun,vn-δ2·∫01s1-sα2-1gs,sα1-1,1ds≤ρ2cλ1+δ1ωr2-δ1∫01s1-sα1-1fs,1,sα2-1ds+2cλ2+δ2ωr2-δ2∫01s1-sα2-1gs,sα1-1,1ds<+∞,t∈0,1.
By (52), for any ε>0, we can find a sufficiently large natural number m>0, for all n, such that
(53)ρ∫Hms1-sα1-1·fs,uns,vns+fs,us,vsds+ρ∫Hms1-sα2-1·gs,uns,vns+gs,us,vsds<ε2,
where H(m)=0,1/m∪1-1/m,1. On the other hand, for each (u¯,v¯)∈K[r1,r2] and t∈1/m,1-1/m, we have
(54)ωϖr1≤u¯t,v¯t≤r2,
where ϖ=min⁡tαi-1:t∈1/m,1-1/m,i=1,2. Since f(t,x,y) and g(t,x,y) are uniformly continuous in 1/m,1-1/m×ωϖr1,r2×ωϖr1,r2, we have
(55)lim⁡n→+∞fs,uns,vns-fs,us,vs=0,lim⁡n→+∞gs,uns,vns-gs,us,vs=0
hold uniformly on s∈1/m,1-1/m. Then the Lebesgue dominated convergence theorem yields that
(56)ρ∫1/m1-1/ms1-sα1-1·fs,uns,vns-fs,us,vsds⟶0,asn⟶+∞,ρ∫1/m1-1/ms1-sα2-1·gs,uns,vns-gs,us,vsds⟶0,asn⟶+∞.
So, for the above ε>0, there exists a sufficiently large natural number N0 such that when n>N0, we have
(57)ρ∫1/m1-1/ms1-sα1-1·fs,uns,vns-fs,us,vsds<ε4,ρ∫1/m1-1/ms1-sα2-1·gs,uns,vns-gs,us,vsds<ε4.
It follows from (53) and (57) that
(58)T1un,vn-T1u,v≤∫01K1t,sfs,uns,vns-fs,us,vsds+∫01H1t,sgs,uns,vns-gs,us,vsds≤ρ∫1/m1-1/ms1-sα1-1fs,uns,vns-fs,us,vsds+ρ∫1/m1-1/ms1-sα2-1gs,uns,vns-gs,us,vsds+ρ∫Hms1-sα1-1fs,uns,vns+fs,us,vsds+ρ∫Hms1-sα2-1gs,uns,vns+gs,us,vsds<ε,n>N0.
This implies that the operator T1:K[r1,r2]→C[0,1] is continuous. Similarly, we can prove T2:K[r1,r2]→C[0,1] is continuous. So T:K[r1,r2]→K is continuous.
Finally, we showT:K[r1,r2]→K is compact.
LetD⊂K[r1,r2] be any bounded set; then, for any (u,v)∈D, we have r1≤(u,v)≤r2. Choose c, such that (u,v)/c<1, c>1. By (45), for any (u,v)∈D, t∈[0,1], we have
(59)Tiu,vt≤ρ∫01s1-sα1-1fs,us,vsds+∫01s1-sα2-1gs,us,vsds≤ρcλ1+δ1ωr1-δ1∫01s1-sα1-1fs,1,sα2-1ds+cλ2+δ2ωr1-δ2∫01s1-sα2-1gs,sα1-1,1ds<+∞,i=1,2.
So T(D) is bounded in X.
In what follows, we show thatTi(D) is equicontinuous. In fact, by (59), for any ε>0, there exists a sufficiently large natural number m0, for all (u,v)∈D, such that
(60)ρ∫Hm0s1-sα1-1fs,us,vsds+∫Hm0s1-sα2-1gs,us,vsds<ε4.
Let
(61)M0=max⁡ft,x,y:1m0≤t≤1-1m0,ωϖ0r1≤x≤r2,ωϖ0r1≤y≤r21m0,M0′=max⁡gt,x,y:1m0≤t≤1-1m0,ωϖ0r1≤x≤r2,ωϖ0r1≤y≤r21m0,
where ϖ0=min⁡tαi-1:t∈1/m0,1-1/m0,i=1,2>0.
By the uniform continuity ofK1(t,s), H1(t,s) on [0,1]×[0,1], for the above ε>0, there exists δ0>0 such that, for any t1,t2∈[0,1], |t1-t2|<δ0, s∈1/m0,1-1/m0, we have
(62)K1t1,s-K1t2,s<ε4M01-2m0-1,H1t1,s-H1t2,s<ε4M0′1-2m0-1.
Thus, when t1,t2∈[0,1], |t1-t2|<δ0, s∈1/l0,l0-1/l0, for any (u,v)∈D, we get
(63)T1u,vt1-T1u,vt2M=∫01K1t1,sfs,us,vsdsMMM+∫01H1t1,sgs,us,vsdsMMM-∫01K1t2,sfs,us,vsdsMMMMM+∫01H1t2,sgs,us,vsdsM≤∫1/m01-1/m0K1t1,s-K1t2,sMMMMM·fs,us,vsdsMM+∫1/m01-1/m0H1t1,s-H1t2,sMMMMMM·gs,us,vsdsMM+∫Hm0K1t1,s-K1t2,sMMMMMM·fs,us,vsdsMM+∫Hm0H1t1,s-H1t2,sMMMMMM·gs,us,vsdsM≤ε2+2ρ∫Hm0s1-sα1-1MMMMMMMM·fs,us,vsdsMM+2ρ∫Hm0s1-sα2-1MMMMMMM·gs,us,vsds<ε.
This means that T1(D) is equicontinuous. By the Arzela-Ascoli theorem, T1(D) is a relatively compact set. In the same way, we can show that T2(D) is a relatively compact set. So T:K[r1,r2]→K is compact.
From the above discussion, together with the fact thatT:K[r1,r2]→K is continuous, we get that T:K[r1,r2]→K is completely continuous. This completes the proof of the lemma.In order to obtain the existence of positive solutions of system (2), we will use the following cone compression and expansion fixed point theorem.Lemma 9 (see [21]).
LetP be a positive cone in a Banach space E, Ω1 and Ω2 are bounded open sets in E, θ∈Ω1, Ω¯1⊂Ω2, and A:P∩Ω¯2∖Ω1→P is a completely continuous operator. If the following conditions are satisfied:
(64)Ax≤x,∀x∈P∩∂Ω1,Ax≥x,∀x∈P∩∂Ω2,
or
(65)Ax≥x,∀x∈P∩∂Ω1,∥Ax∥≤∥x∥,∀x∈P∩∂Ω2,
then A has at least one fixed point in P∩(Ω¯2∖Ω1).


## 3. Main Results

Theorem 10.
Assume that(H0)–(H2) hold; then system (2) has at least one positive solution (u*,v*), and there exists a real number 0<l<1 satisfying
(66)ltα¯-1≤u*t≤l-1tα*-1,ltα¯-1≤v*t≤l-1tα*-1,t∈0,1,
where α¯=max⁡α1,α2, α*=min⁡α1,α2.Proof.
First, we show that system (2) has at least one positive solution.
Chooser,R, such that
(67)0<r≤min⁡∫01s1-sα1-1fs,sα1-1,1ds1/1-λ1,1213α1-1ϱωλ1·∫01s1-sα1-1fs,sα1-1,1ds1/1-λ1,12,R≥max⁡·gs,sα1-1,1ds∫01s1/1-max⁡λ1,λ2,1ω,2ρ∫01s1-sα1-1fs,1,sα2-1ds+ρ∫01s1-sα2-1·gs,sα1-1,1ds∫01s1/1-max⁡λ1,λ2,1ω,2.
For any(u,v)∈∂Kr, we have
(68)rωtα1-1≤ut≤r,rωtα2-1≤vt≤r,t∈0,1.
By Lemma 5, Remark 7, and (H1), for any (u,v)∈∂Kr, we get
(69)T1u,vt=∫01K1t,sfs,us,vsds+∫01H1t,sgs,us,vsds≥∫01K1t,sfs,us,vsds≥ϱ∫01tα1-1s1-sα1-1fs,rωsα1-1,rds≥ϱ∫01tα1-1s1-sα1-1fs,rωsα1-1,1ds≥ϱtα1-1rλ1ωλ1∫01s1-sα1-1fs,sα1-1,1ds≥13α1-1ϱrλ1ωλ1∫01s1-sα1-1fs,sα1-1,1ds≥r=u,v,t∈13,23.
This guarantees
(70)Tu,v≥u,v,u,v∈∂Kr.
On the other hand, for any(u,v)∈∂KR, we have
(71)Rωtα1-1≤ut≤R,Rωtα2-1≤vt≤R,t∈0,1.
By Lemma 5, (H1), and (H2), for any (u,v)∈∂KR, t∈[0,1], we get
(72)T1u,vt=∫01K1t,sfs,us,vsds+∫01H1t,sgs,us,vsds≤ρ∫01s1-sα1-1fs,R,Rωsα2-1ds+ρ∫01s1-sα2-1gs,Rωsα1-1,Rds≤ρ∫01s1-sα1-1fs,R,sα2-1ds+ρ∫01s1-sα2-1gs,sα1-1,Rds≤ρRλ1∫01s1-sα1-1fs,1,sα2-1ds+ρRλ2∫01s1-sα2-1gs,sα1-1,1ds≤ρRmax⁡⁡λ1,λ2∫01s1-sα1-1fs,1,sα2-1ds+∫01s1-sα2-1gs,sα1-1,1ds≤R=u,v.
In the same way as (72), we have
(73)T2u,vt≤R=u,v,u,v∈∂KR.
So,
(74)Tu,v≤u,v,u,v∈∂KR.
It follows from (70), (74), and Lemmas 8 and 9 that T has a fixed point (u*,v*) with 0<r≤(u*,v*)≤R. It is easy to see that (u*,v*) is a positive solution of system (2).
Next, we show there exists a real number0<l<1 such that the positive solution (u*,v*) in system (2) satisfies
(75)ltα¯-1≤u*t≤l-1tα*-1,ltα¯-1≤v*t≤l-1tα*-1,t∈0,1,
where α¯=max⁡α1,α2, α*=min⁡α1,α2.
From Lemma8, we know (u*,v*)∈K∖{(0,0)}. So, we have
(76)ωtα1-1u*,v*≤u*t≤u*,v*,ωtα2-1u*,v*≤v*t≤u*,v*,t∈0,1.
Choose c, such that (u*,v*)/c<1, c>1/ω; by Lemma 5 and (H1), for t∈[0,1], we have
(77)u*t=∫01K1t,sfs,u*s,v*sds+∫01H1t,sgs,u*s,v*sds≤∫01ρtα1-1fs,c,ωsα2-1u*,v*ds+∫01ρtα1-1gs,ωtα1-1u*,v*,cds≤ρtα1-1∫01fs,c,ωu*,v*csα2-1ds+ρtα1-1∫01gs,ωu*,v*csα1-1,cds≤cλ1+δ1ωu*,v*-δ1ρtα1-1∫01fs,1,sα2-1ds+cλ2+δ2ωu*,v*-δ2ρtα1-1∫01gs,sα1-1,1ds≤cλ1+δ1ωR-δ1ρtα1-1∫01fs,1,sα2-1ds+cλ2+δ2ωR-δ2ρtα1-1∫01gs,sα1-1,1ds≤Mtα*-1,
where
(78)M=cλ1+δ1ωR-δ1ρ∫01fs,1,sα2-1dsM+cλ2+δ2ωR-δ2ρ∫01gs,sα1-1,1ds.
In the same way as (77), we also have v(t)≤Mtα*-1, t∈[0,1]. Choose
(79)l=min⁡ωr,1M,12;
then we have
(80)ltα¯-1≤u*t≤l-1tα*-1,ltα¯-1≤v*t≤l-1tα*-1,t∈0,1.
This completes the proof of Theorem 10.Theorem 11.
Assume that(H0)–(H2) hold. If λ1+δ1<1 and λ2+δ2<1, then system (2) has a unique positive solution for α1=α2.Proof.
Assume that system (2) has two different positive solutions (u1,v1) and (u2,v2). Denote α=α1=α2; by Theorem 10, there exists 0<l1<1, 0<l2<1, such that
(81)l1tα-1≤u1t≤1l1tα-1,l1tα-1≤v1t≤1l1tα-1,t∈0,1,l2tα-1≤u2t≤1l2tα-1,l2tα-1≤v2t≤1l2tα-1,t∈0,1.
Thus, we have
(82)l1l2u2t≤u1t≤1l1l2u2t,l1l2v2t≤v1t≤1l1l2v2t,t∈0,1.
Obviously, l1l2≠1. Let
(83)L=sup⁡l:lu2t≤u1t≤l-1u2t,lv2t≤v1t≤l-1v2t,t∈0,1.
It is easy to see that 0<l1l2≤L<1, and
(84)Lu2t≤u1t≤1Lu2t,lv2t≤v1t≤1Lv2t,t∈0,1.
By (H1), we have
(85)ft,u1t,v1tM≥ft,Lu2t,1Lv2t≥Lλ1+δ1ft,u2t,v2tM≥Lσft,u2t,v2t,gt,u1t,v1tM≥gt,Lu2t,1Lv2t≥Lλ2+δ2gt,u2t,v2tM≥Lσgt,u2t,v2t,
where σ=max⁡{λ1+δ1,λ2+δ2}<1. Consider the following:
(86)ft,u2t,v2tM≥ft,Lu1t,1Lv1t≥Lλ1+δ1ft,u1t,v1tM≥Lσft,u1t,v1t,gt,u2t,v2tM≥gt,Lu1t,1Lv1t≥Lλ2+δ2gt,u1t,v1tM≥Lσgt,u1t,v1t.
From (85), for t∈[0,1], we have
(87)u1tM=T1u1,v1t=∫01K1t,sfs,u1s,v1sdsMM+∫01H1t,sgs,u1s,v1sdsM≥Lσ∫01K1t,sfs,u2s,v2sdsMM+Lσ∫01H1t,sgs,u2s,v2sdsM≥LσT1u2,v2t=Lσu2t,(88)u2tM=T1u2,v2t=∫01K1t,sfs,u2s,v2sdsMM+∫01H1t,sgs,u2s,v2sdsM≥Lσ∫01K1t,sfs,u1s,v1sdsMM+Lσ∫01H1t,sgs,u1s,v1sdsM≥LσT1u1,v1t=Lσu1t.
Similarly, from (86), we also have
(89)v1t≥Lσv2t,v2t≥Lσv1t,t∈0,1.
Therefore, we obtain
(90)Lσu2t≤u1t≤1Lσu2t,Lσv2t≤v1t≤1Lσv2t,t∈0,1.
Notice that 0<L<1, 0<σ<1; we have Lσ>L, so it is a contradiction with the maximality of L. Therefore, system (2) has a unique positive solution (u*,v*). This completes the proof of Theorem 11.Remark 12.
By the proof of Theorems10 and 11, we obtained the positive solutions of system (2) suppose (H0)-(H2) hold; and the uniqueness of the solution to the system is established provided that system (2) satisfies the additional conditions (λ1+δ1<1 and λ2+δ2<1) in theorem.


## 4. Example

Now we consider the existence and uniqueness of positive solutions for the fractional differential system (1). Obviously,
(91)α1=α2=52,μ1=12,μ2=1,A1t=t,A2t=t1/2.
We also have
(92)k1=∫01tα2-1dA1t=∫01t3/2dt=25>0,k2=∫01tα1-1dA2t=∫01t3/2dt1/2=12∫01tdt=14>0,1-μ1μ2k1k2=1920>0.
So, the condition (H0) holds. For
(93)ft,x,y=sint+cos⁡txt1-ty3,gt,x,y=y3ett1-tx,
it is easy to see that f:(0,1)×[0,+∞)×(0,+∞)→(0,+∞) is continuous, f(t,x,y) is nondecreasing in x and nonincreasing in y, g:(0,1)×(0,+∞)×[0,+∞)→[0,+∞) is continuous, and g(t,x,y) is nonincreasing in x and nondecreasing in y. Take
(94)λ1=1120,δ1=25,λ2=35,δ2=15.
Then, we know that the condition (H1) holds. As
(95)∫01s1-sα1-1fs,1,sα2-1ds≤2∫01s1/21-sds=815<+∞,∫01s1-sα2-1gs,sα1-1,1ds≤∫01s-1/41-sds=1621<+∞,
the condition (H2) also holds.Therefore, by Theorem10, we get that system (1) has at least one positive solution (u*,v*)∈K[r1,r2]. For
(96)α1=α2,λ1+δ1=1120+25=1920<1,λ2+δ2=35+15=45<1,
by Theorem 11, we get that (u*,v*) is the unique positive solution of system (1).Remark 13.
The example not only implies thatf(t,x,y) can be singular at t=0,1, y=0, and g(t,x,y) can be singular at t=0,1, x=0, but also indicates that there is a large number of functions that satisfy the conditions of the theorems which we discuss in this paper. Also, the conditions in our theorems are easy to check.


## 5. Conclusions

In this paper, the(n-1,1)-type singular fractional differential system with coupled boundary conditions has been investigated. Based on the well-known Guo-Krasnoselskii fixed point theorem, the existence of solutions for the (n-1,1)-type fractional differential system with coupled integral boundary conditions is presented; also the uniqueness of positive solution is established when certain additional conditions are satisfied. The example given demonstrates the effectiveness and feasibility of our results.



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*Source: 142391-2014-11-03.xml*