# Positive Solutions for a Coupled System of Nonlinear Semipositone Fractional Boundary Value Problems

Authors: S. Nageswara Rao; M. Zico Meetei
Journal: International Journal of Differential Equations
 (2019)

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## Abstract

In this paper, we consider a four-point coupled boundary value problem for system of the nonlinear semipositone fractional differential equationD0+αu(t)+λf(t,u(t),v(t))=0,0<t<1, D0+αv(t)+μg(t,u(t),v(t))=0,0<t<1, u(0)=v(0)=0,a1D0+βu(1)=b1D0+βv(ξ), a2D0+βv(1)=b2D0+βu(η),η,ξ∈(0,1), where the coefficients ai,bi,i=1,2 are real positive constants, α∈(1,2],β∈(0,1],D0+α, D0+β are the standard Riemann-Liouville derivatives. Values of the parameters λ and μ are determined for which boundary value problem has positive solution by utilizing a fixed point theorem on cone.


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## Body

## 1. Introduction

In recent years, fractional-order calculus has been one of the most rapidly developing areas of mathematical analysis. In fact, a natural phenomenon may depend not only on the time instant but also on the previous time history, which can be successfully modeled by fractional calculus. Fractional-order differential equations are naturally related to systems with memory, as fractional derivatives are usually nonlocal operators. Thus, fractional differential equations (FDEs) play an important role because of their applications in various fields of science, such as mathematics, physics, chemistry, optimal control theory, finance, biology, and engineering [1–6]. In particular, a great interest has been shown by many authors in the subject of fractional-order boundary value problems (BVPs), and a variety of results for BVPs equipped with different kinds of boundary conditions have been obtained; for instance, see [7–18] and the references cited therein.We consider the four-point coupled system of nonlinear fractional differential equations:(1)D0+αut+λft,ut,vt=0,0<t<1,D0+αvt+μgt,ut,vt=0,0<t<1,with the coupled boundary conditions(2)u0=v0=0,a1D0+βu1=b1D0+βvξ,a2D0+βv1=b2D0+βuη,η,ξ∈0,1,where α∈(1,2],β∈(0,1], D0+α and D0+β are the standard Riemann-Liouville derivatives, f,g∈C([0,1]×[0,+∞)×[0,+∞),[0,+∞)) and ai,bi,i=1,2 are real positive constants.Here we emphasize that our problem is new in the sense of nonseparated coupled boundary conditions introduced here. To the best of our knowledge, fractional-order coupled system (1) has yet to be studied with the boundary conditions (2). In consequence, our findings of the present work will be a useful contribution to the existing literature on the topic. The existence of positive solution results for the given problem is new, though they are proved by applying the well-known fixed point theorem.We present intervals for parametersλ,μ,f, and g such that the above problem (1)-(2) has at least one positive solution. By a positive solution (1)-(2), we mean a pair of functions (u,v)∈C[0,1]×C[0,1] satisfying (1) and (2) with u(t)≥0,v(t)≥0 for all t∈[0,1] and u(t)>0,v(t)>0.We use the following notations for our convenience:(3)Ki=∫01Gi1,sdsandLi=∫01Hi1,sdsfori=1,2.Ai=∫s∈IGi1,sdsandBi=∫s∈IHi1,sdsfori=1,2.Before stating our results, we make precise assumptions throughout the paper:(H1)
The functionsf,g∈C((0,1)×[0,∞)×[0,∞),(-∞,∞)) and there exist functions p1,p2∈C([0,1]×[0,∞)) such that f(t,u,v)≥-p1(t) and g(t,u,v)≥-p2(t) for any t∈[0,1] and (u,v)∈[0,∞).(H2)
a1,a2,b1,b2 are positive constants such that a1a2≥b1b2/ξ1-α+βη1-α+β.(H3)
f(t,0,0)>0,g(t,0,0)>0 for all t∈[0,1].(H4)
The functionsf,g∈C((0,1)×[0,∞)×[0,∞),(-∞,∞)), f,g may be singular at t=0 and/or t=1, and there exist functions p1,p2∈C((0,1), [0,∞)),α1,α2∈C((0,1),(0,∞)),β1,β2∈C([0,1]×[0,∞),[0,∞)) such that -p1(t)≤f(t,u,v)≤α1(t)β1(t,u,v),-p2(t)≤g(t,u,v)≤α2(t)β2(t,u,v) for all t∈(0,1),u,v∈[0,∞), with 0<∫01pi(s)ds<∞,0<∫01αi(s)ds<∞,i=1,2.(H5)
There existst∈I=[1/4,3/4]⊂(0,1) such that(4)f∞=limu+v→∞⁡mint∈I⁡ft,u,vu+v=∞org∞=limu+v→∞⁡mint∈I⁡gt,u,vu+v=∞.The rest of the paper is organized as follows. In Section2, we construct the Green functions for the associated linear fractional-order boundary value problems and estimate the bounds for these Green functions. In Section 3, we establish the existence of at least one positive solution of the boundary value problem (1)-(2) by applying fixed point theorem. Finally, as an application, we give an example to illustrate our result.


## 2. Green Functions and Bounds

In this section, we construct the Green functions for the associated linear fractional-order boundary value problems and estimate the bounds for these Green functions, which are needed to establish the main results.Lemma 1.
Letα>0. Then, the differential equation D0+αu(t)=0 has a solution (5)ut=c1tα-1+c2tα-2+⋯+cntα-nfor some ci∈R,i=1,2,…,n, where n is the smallest integer greater than or equal to α.Lemma 2.
Letα>0. Then, I0+αD0+αu(t)=u(t)+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.Lemma 3.
LetΔ=Γ(α)N≠0 and N=a1a2-b1b2ξα-β-1ηα-β-1. Let x,y∈C[0,1] be given functions. Then, the boundary value problem,(6)D0+αut+xt=0,0<t<1,D0+αvt+yt=0,0<t<1,u0=v0=0,a1D0+βu1=b1D0+βvξ,a2D0+βv1=b2D0+βuη,ξ,η∈0,1,has an integral representation(7)ut=∫01G1t,sxsds+∫01H1t,sysds,vt=∫01G2t,sysds+∫01H2t,sxsds,where(8)G1t,s=1Δa1a2tα-11-sα-β-1-Nt-sα-1-b1b2tα-1ξα-β-1η-sα-β-1,0≤s≤t≤1,s≤η,a1a2tα-11-sα-β-1-Nt-sα-1,0≤s≤t≤1,s≥η,a1a2tα-11-sα-β-1-b1b2tα-1ξα-β-1η-sα-β-1,0≤t≤s≤1,s≤η,a1a2tα-11-sα-β-1,0≤t≤s≤1,s≥η,(9)G2t,s=1Δa1a2tα-11-sα-β-1-Nt-sα-1-b1b2tα-1ηα-β-1ξ-sα-β-1,0≤s≤t≤1,s≤ξ,a1a2tα-11-sα-β-1-Nt-sα-1,0≤s≤t≤1,s≥ξ,a1a2tα-11-sα-β-1-b1b2tα-1ηα-β-1ξ-sα-β-1,0≤t≤s≤1,s≤ξ,a1a2tα-11-sα-β-1,0≤t≤s≤1,s≥ξ,(10)H1t,s=1Δa2b1tα-1ξα-β-11-sα-β-1-a2b1tα-1ξ-sα-β-1,s≤ξ,a2b1tα-1ξα-β-11-sα-β-1,s≥ξ,(11)H2t,s=1Δa1b2tα-1ηα-β-11-sα-β-1-a1b2tα-1η-sα-β-1,s≤η,a1b2tα-1ηα-β-11-sα-β-1,s≥η.Lemma 4.
Assume that condition(H2) is satisfied. Then, the Green functions G1(t,s) and H1(t,s) defined, respectively, by (8) and (10) are nonnegative, for all t,s∈[0,1].Lemma 5.
Assume that condition(H2) is satisfied. Then, the Green functions G2(t,s) and H2(t,s) defined, respectively, by (9) and (11) are nonnegative, for all t,s∈[0,1].Lemma 6.
Assume that condition(H2) is satified. Then, the Green functions G1(t,s) and H1(t,s) defined, respectively, by (8) and (10) have the following properties: (C1)
G1(t,s)≤G1(1,s), H1(t,s)≤H1(1,s) for all (t,s)∈[0,1]×[0,1],(C2)
G1(t,s)≥(1/4)α-1G1(1,s), H1(t,s)≥(1/4)α-1H1(1,s), for all (t,s)∈I×(0,1), where I=[1/4,3/4].Lemma 7.
Assume that condition(H2) is satified. Then, the Green functions G2(t,s) and H2(t,s) defined, respectively, by (9) and (11) have the following properties: (C3)
G2(t,s)≤G2(1,s) and H2(t,s)≤H2(1,s) for all (t,s)∈[0,1]×[0,1],(C4)
G2(t,s)≥(1/4)α-1G2(1,s) and H2(t,s)≥(1/4)α-1H2(1,s), for all (t,s)∈I×(0,1), where I=[1/4,3/4].In the proof of our main results, we shall use the nonlinear alternative of Leray-Schauder type and the Guo-Krasnosel’skii fixed point theorem presented below [19, 20].Theorem 8.
LetX be a Banach space with Ω⊂X closed and convex. Assume U is a relatively open subset of Ω with 0∈U, and let S:U¯→Ω be a completely continuous operator (continuous and compact). Then, either (i)
S has a fixed point in U¯, or(ii)
there existu∈∂U and v∈(0,1) such that u=vSu.Theorem 9 (Krasnosel’skii).
LetB be a Banach space, and let P⊂B be a cone in B. Assume that Ω1 and Ω2 are two bounded open subsets of B with 0∈Ω1⊂Ω¯1⊂Ω2, and let T:P∩(Ω¯2∖Ω1)→P be a completely continuous operator such that either (i)
Tu≤u,u∈P∩∂Ω1, and Tu≥u,u∈P∩∂Ω2, or(ii)
Tu≥u,u∈P∩∂Ω1, and Tu≤u,u∈P∩∂Ω2.
Then,T has a fixed point in P∩(Ω¯2∖Ω1).


## 3. Main Results

In this section, we investigate the existence of positive solutions for our problem (1)-(2).We consider the system of nonlinear fractional differential equations(12)D0+αxt+λft,xt-q1t⋆,yt-q2t⋆+p1t=0,0<t<1,D0+αyt+μgt,xt-q1t⋆,yt-q2t⋆+p2t=0,0<t<1,with the boundary conditions(13)x0=y0=0,a1D0+βx1=b1D0+βyξ,a2D0+βy1=b2D0+βxη,η,ξ∈0,1,where a modified function [z(t)]⋆ for any z∈C[0,1] by (14)zt⋆=zt,ifzt≥0,andzt⋆=0,ifzt=0.Here (q1,q2) with(15)q1t=λ∫01G1t,sp1sds+μ∫01H1t,sp2sds,t∈0,1,q2t=μ∫01G2t,sp2sds+λ∫01H2t,sp1sds,t∈0,1,is solution of the system of fractional differential equations(16)D0+αq1t+λp1t=0,0<t<1,D0+αq2t+μp2t=0,0<t<1,with the boundary conditions(17)q10=q20=0,a1D0+βq11=b1D0+βq2ξ,a2D0+βq21=b2D0+βq1η,η,ξ∈0,1.Under the assumptions (H1) and (H2) or (H2) and (H4), we have q1(t)≥0,q2(t)≥0 for all t∈[0,1].We shall prove that there exists a solution(x,y) for the boundary value problem (12)-(13) with x(t)≥q1(t) and y(t)≥q2(t) on [0,1],x(t)>q1(t),y(t)>q2(t) on (0,1). In this case, (u,v) with u(t)=x(t)-q1(t) and v(t)=y(t)-q2(t),t∈[0,1] represents a positive solution of boundary value problem (1)-(2).By using Lemma3, a solution of the system(18)xt=λ∫01G1t,sfs,xs-q1s⋆,ys-q2s⋆+p1sds+μ∫01H1t,sgs,xs-q1s⋆,ys-q2s⋆+p2sds,t∈0,1,yt=μ∫01G2t,sgs,xs-q1s⋆,ys-q2s⋆+p2sds+λ∫01H2t,sfs,xs-q1s⋆,ys-q2s⋆+p1sds,t∈0,1,is a solution for problem (12)-(13).We consider the Banach spaceX=C[0,1] with supremum norm · and the Banach space Y=X×X with the norm (u,v)=u+v. We define the cone P⊂Y(19)P=x,y∈Y:xt≥0,yt≥0∀t∈0,1andmint∈I⁡xt+yt≥14α-1x,y,where I=[1/4,3/4].Forλ,μ>0, we define the operators Q1,Q2:Y→X and Q:Y→Y defined by Q(x,y)=(Q1(x,y),Q2(x,y)),(x,y)∈Y with (20)Q1x,y=λ∫01G1t,sfs,xs-q1s⋆,ys-q2s⋆+p1sds+μ∫01H1t,sgs,xs-q1s⋆,ys-q2s⋆+p2sds,t∈0,1,Q2x,y=μ∫01G2t,sgs,xs-q1s⋆,ys-q2s⋆+p2sds+λ∫01H2t,sfs,xs-q1s⋆,ys-q2s⋆+p1sds,t∈0,1.It is clear that if (x,y) is a fixed point of operator Q, then (x,y) is a solution of problem (12)-(13).Lemma 10.
If(H1) and (H2) or (H2) and (H4) hold, then operator Q:P→P is a completely continuous operator.Proof.
The operatorsQ1 and Q2 are well defined. To prove this, let (x,y)∈P be fixed with (x,y)=L~. Then we have (21)xs-q1s⋆≤xs≤x≤x,y=L~,∀s∈0,1,ys-q2s⋆≤ys≤y≤x,y=L~,∀s∈0,1.If (H1) and (H2) hold, then we deduce easily that Q1(x,y)(t)<∞ and Q2(x,y)(t)<∞ for all t∈[0,1]. If (H2) and (H4) hold, we deduce, for all t∈[0,1]:(22)Q1x,y≤λ∫01G11,sα1sβ1s,xs-q1s⋆,ys-q2s⋆+p1sds+μ∫01H11,sα2sβ2s,xs-q1s⋆,ys-q2s⋆+p2sds,≤Mλ∫01G11,sα1s+p1sds+μ∫01H11,sα2s+p2sds<∞,Q2x,y≤μ∫01G21,sα2sβ2s,xs-q1s⋆,ys-q2s⋆+p2sds+λ∫01H21,sα1sβ1s,xs-q1s⋆,ys-q2s⋆+p1sds,≤Mμ∫01G21,sα2s+p2sds+λ∫01H21,sα1s+p1sds<∞,where M=max⁡{maxt∈[0,1],u,v∈[0,L~]⁡β1(t,u,v), maxt∈[0,1],u,v∈[0,L~]⁡β2(t,u,v),1}.
Thus,Q:P→Y is well defined.
Next, we show thatT:P→P. For any fixed (x,y)∈P, by Lemmas 6 and 7, we have (23)mint∈I⁡Q1x,yt=mint∈I⁡λ∫01G1t,sfs,xs-q1s⋆,ys-q2s⋆+p1sds+μ∫01H1t,sgs,xs-q1s⋆,ys-q2s⋆+p2sds≥14α-1λ∫01G11,sfs,xs-q1s⋆,ys-q2s⋆+p1sds+μ∫01H11,sgs,xs-q1s⋆,ys-q2s⋆+p2sds≥14α-1Q1x,y.Similarly, mint∈I⁡Q2(x,y)(t)≥(1/4)α-1Q2(x,y). Therefore, (24)mint∈I⁡Q1x,yt+Q2x,yt≥14α-1Q1x,y+14α-1Q2x,y=14α-1Q1x,y,Q2x,y=14α-1Qx,y.Hence, Q(x,y)∈P. This implies that Q(P)⊂P. According to the Ascoli-Arzela theorem, we can easily get that Q:P→P is completely continuous.Theorem 11.
Assume that(H1)-(H3) hold. Then, there exist constants λ0>0 and μ0>0 such that, for any λ∈(0,λ0] and μ∈(0,μ0], the boundary value problem (1)-(2) has at least one positive solution.Proof.
Letδ∈(0,1) be fixed. From (H1) and (H3), there exist R0∈(0,1] such that(25)ft,u,v≥δft,0,0,gt,u,v≥δgt,0,0,∀t∈0,1,u,v∈0,R0.We define (26)f¯R0=maxt∈0,1,u,v∈0,R0⁡ft,u,v+p1t≥maxt∈0,1⁡δft,0,0+p1t>0,g¯R0=maxt∈0,1,u,v∈0,R0⁡gt,u,v+p2t≥maxt∈0,1⁡δgt,0,0+p2t>0,λ0=max⁡R08K1f¯R0,R08L2f¯R0,μ0=max⁡R08L1g¯R0,R08K2g¯R0.We will show that, for any λ∈(0,λ0] and μ∈(0,μ0], problem (12)-(13) has at least one positive solution.
So, letλ∈(0,λ0] and μ∈(0,μ0] be arbitrary but fixed for the moment. We define the set U={(x,y)∈P,(x,y)<R0}. We suppose that there exist (x,y)∈∂U((x,y)=R0 or x+y=R0) and θ∈(0,1) such that (x,y)=θQ(x,y) or x=θQ1(x,y),y=θQ2(x,y).
We deduce that(27)xt-q1t⋆=xt-q1t≤xt≤R0,ifxt-q1t≥0,xt-q1t⋆=0,forxt-q1t<0,∀t∈0,1,yt-q2t⋆=yt-q2t≤yt≤R0,ifyt-q2t≥0,yt-q2t⋆=0,foryt-q2t<0,∀t∈0,1.Then by Lemma 3, for all t∈[0,1], we obtain(28)xt=θQ1x,yt<Q1x,yt≤λ∫01G11,sf¯R0ds+μ∫01H11,sg¯R0ds≤λ0K1f¯R0+μ0L1g¯R0≤R08+R08=R04,yt=θQ2x,yt<Q2x,yt≤μ∫01G21,sg¯R0ds+λ∫01H21,sf¯R0ds≤μ0K2g¯R0+λ0L2f¯R0≤R08+R08=R04.Hence, x≤R0/4 and y≤R0/4. Then, R0=(x,y)=x+y≤R0/4+R0/4=R0/2, which is contradiction.
Therefore, by Theorem8 (with Ω=P), we deduce that Q has a fixed point (x0,y0)∈U¯∩P. That is, (x0,y0)=Q(x0,y0) or x0=Q1(x0,y0),y0=Q2(x0,y0), and x0+y0≤R0 with x0≥(1/4)α-1x0 and y0(t)≥(1/4)α-1y0 for all t∈[0,1]. Moreover, by (25), we conclude(29)x0t=Q1x0,y0t≥λ∫01G1t,sδft,0,0+p1sds+μ∫01H1t,sδgt,0,0+p2sds≥λ∫01G1t,sp1sds+μ∫01H1t,sp2sds=q1t,∀t∈0,1,x0t>λ∫01G1t,sp1sds+μ∫01H1t,sp2sds=q1t,∀t∈0,1,y0t=Q2x0,y0t≥μ∫01H2t,sδgt,0,0+p2sds+λ∫01G2t,sδft,0,0+p1sds≥μ∫01H2t,sp2sds+λ∫01G2t,sp1sds=q2t,∀t∈0,1,y0t>μ∫01H2t,sp2sds+λ∫01G2t,sp1sds=q2t,∀t∈0,1.Therefore, x0(t)≥q1(t),y0(t)≥q2(t) for all t∈[0,1], and x0(t)>q1(t),y0(t)>q2(t) for all t∈(0,1). Let u0(t)=x0(t)-q1(t) and v0(t)=y0(t)-q2(t) for all t∈[0,1]. Then, u0(t)≥0,v0(t)≥0 for all t∈[0,1],u0(t)>0,v0(t)>0 for all t∈(0,1). Therefore, (u0,v0) is a positive solution of (1)-(2).Theorem 12.
Assume that(H1),(H4), and (H5) hold. Then, there exist λ⋆>0 and μ⋆>0 such that, for any λ∈(0,λ⋆] and μ∈(0,μ⋆], the boundary value problem (1)-(2) has at least one positive solution.Proof.
We choose a positive number(30)R1>max⁡1,2∫01G11,sp1s+G21,sp2sdsand we define the set Ω1={(x,y)∈P,(x,y)<R1}.
We introduce(31)λ⋆=min1,R14M1∫01G11,sα1s+p1sds-1,R14M1∫01H21,sα1s+p1sds-1,μ⋆=min⁡1,R14M2∫01H11,sα2s+p2sds-1,R14M2∫01G21,sα2s+p2sds-1,with (32)M1=max⁡maxt∈0,1,u,v≥0,u+v≤R1⁡β1t,u,v,1,M2=max⁡maxt∈0,1,u,v≥0,u+v≤R1⁡β2t,u,v,1.Let λ∈(0,λ⋆] and μ∈(0,μ⋆]. Then, for any (x,y)∈P∩∂Ω1 and s∈[0,1], we have(33)xs-q1s⋆≤xs≤x≤R1,ys-q2s⋆≤ys≤y≤R1.Then, for any (x,y)∈P∩∂Ω1, we obtain (34)Q1x,y≤λ∫01G11,sα1sβ1s,xs-q1s⋆,ys-q2s⋆+p1sds+μ∫01H11,sα2sβ2s,xs-q1s⋆,ys-q2s⋆+p2sds,≤λ⋆M1∫01G11,sα1s+p1sds+μ⋆M2∫01H11,sα2s+p2sds≤R14+R14=R12=x,y2,Q2x,y≤μ∫01G21,sα2sβ2s,xs-q1s⋆,ys-q2s⋆+p2sds+μ∫01H21,sα1sβ1s,xs-q1s⋆,ys-q2s⋆+p1sds,≤μ⋆M2∫01G21,sα2s+p2sds+λ⋆M1∫01H21,sα1s+p1sds≤R14+R14=R12=x,y2.Therefore,(35)Qx,y=Q1x,y+Q2x,y≤x,y,∀x,y∈P∩∂Ω1.On the other hand, we choose a constant L>0 such that (36)λ142α-1A1L≥4,μ142α-1A2L≥4.From (H5), we deduce that there exists a constant M0>0 such that(37)ft,u,v≥Lu+vorgt,u,v≥Lu+v,∀t∈I,u,v≥0,u+v≥M0.Now we define (38)R2=max⁡2R1,4αM0,4∫01G11,sp1s+H11,sp2sds>0,and let Ω2={(x,y)∈P,(x,y)<R2}.
We suppose thatf∞=∞, that is, f(t,u,v)≥L(u+v) for all t∈I and u,v≥0,u+v≥M0. Then, for any (x,y)∈P∩∂Ω2, we have (x,y)=R2 or x+y=R2. We deduce that x≥R2/2 or y≥R2/2.
We suppose thatx≥R2/2. Then, for any (x,y)∈P∩∂Ω2, we obtain (39)xt-q1t=xt-λ∫01G1t,sp1sds-μ∫01H1t,sp2sds≥xt-14α-1∫01G11,sp1sds+∫01H11,sp2sds≥xt-xtx∫01G11,sp1s+H11,sp2sds=xt1-1x∫01G11,sp1s+H11,sp2sds≥1-2R2∫01G11,sp1s+H11,sp2sds≥12xt≥0.Therefore, we conclude (40)xt-q1t⋆=xt-q1t≥12xt≥1214α-1x≥1414α-1R2=14αR2≥M0,∀t∈I.Hence,(41)xt-q1t⋆+yt-q2t⋆≥xt-q1t⋆=xt-q1t≥M0,∀t∈I.Then, for any (x,y)∈P∩∂Ω2 and t∈I, by (37) and (41), we deduce (42)ft,xt-q1t⋆,yt-q2t⋆≥Lxt-q1t⋆+yt-q2t⋆≥Lxt-q1t⋆≥L2xt,∀t∈I.It follows that, for any (x,y)∈P∩∂Ω2,t∈I, we obtain (43)Q1x,yt≥λ∫01G1t,sfs,xs-q1s⋆,ys-q2s⋆+p1sds≥λ∫s∈IG1t,sfs,xs-q1s⋆,ys-q2s⋆+p1sds≥14α-1λ∫s∈IG11,sLxs-q1s⋆ds≥λ14α-1A114α-1L4R2=λL4142α-1A1R2≥R2.Then, Q1(x,y)≥(x,y) and(44)Qx,y≥x,y,∀x,y∈P∩∂Ω2.If y≥R2/2, then by a similar approach, we obtain again relation (44).
We suppose thatg∞=∞, that is, g(t,u,v)≥L(u+v), for all t∈I and u,v≥0,u+v≥M0. Then, for any (x,y)∈P∩∂Ω2, we have (x,y)=R2. Hence, x≥R2/2 or y≥R2/2.
Ifx≥R2/2, then for any (x,y)∈P∩∂Ω2 we deduce in a similar manner as above that x(t)-q1(t)≥1/2x(t) for all t∈[0,1] and (45)Q1x,yt≥μ∫01G2t,sgs,xs-q1s⋆,ys-q2s⋆+p2sds≥μ∫s∈IG2t,sgs,xs-q1s⋆,ys-q2s⋆+p2sds≥14α-1μ∫s∈IG21,sLxs-q1s⋆ds≥μ14α-1A214α-1L4R2ds=μL4142α-1A2R2≥R2,∀t∈I.Hence, we obtain relation (44). If y≥R2/2, then in a similar way as above, we deduce again relation (44). Therefore, by Theorem 9, relation (35), and (44), we conclude that Q has a fixed point (x,y)∈P∩(Ω2¯∖Ω1).


## 4. Example

In this section, we give an example to illustrating our result. Let(46)α=32,β=14,η=23,ξ=13,a1=a2=1,b1=b2=1.Consider the system of fractional differential equations,(47)D0+3/2ut+λft,ut,vt=0,t∈0,1,D0+3/2vt+μgt,ut,vt=0,t∈0,1,u0=v0=0,D0+1/4u1=D0+1/4v13,D0+1/4v1=D0+1/4u23,where f(t,u,v)=(u+v)3+cos⁡u,g(t,u,v)=(u+v)1/3+cos⁡v. We have p1(t)=p2(t)=1 for all t∈[0,1], and then assumption (H1) is satisfied. Besides, assumption (H3) is also satisfied, because f(t,0,0)=1 and g(t,0,0)=1 for all t∈[0,1]. Let δ=1/3<1 and R0=1. Then f(t,u,v)≥δf(t,0,0)=1/3, g(t,u,v)≥δg(t,0,0)=1/3,∀t∈[0,1],u,v∈[0,1]. In addition, (48)f¯R0=f¯1=maxt∈0,1,u,v∈0,1⁡ft,u,v+p1t≈9.999848,g¯R0=g¯1=maxt∈0,1,u,v∈0,1⁡gt,u,v+p2t≈3.259769.We also obtain Δ=(0.8865)(0.3133)≈0.2778>0, M1=992, M2=1280, K1=0.1488, K2=0.01598, L1=0.0536, L2=0.1268, and then λ0=max⁡{R0/8K1f¯(R0),R0/8K2f¯(R0)}≈0.782239674, μ0=max⁡{R0/8L1g¯(R0),R0/8L2g¯(R0)}≈0.7154155. We can apply Theorem 11. So we conclude that there exist λ0,μ0>0 such that, for every λ∈(0,λ0] and μ∈(0,μ0], the boundary value problem (47) has at least one positive solution.


## 5. Conclusions

This paper studies the existence of positive solution of a four-point coupled system of nonlinear fractional differential equations. We give sufficient conditions onλ,μ,f, and g such that the system has at least one positive solution. The existence of positive solution is discussed by using Guo-Krasnosel’skii fixed point theorem. Also, an example which illustrates the obtained result is presented.



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*Source: 2893857-2019-02-03.xml*