# Multiple Positive Solutions for a System of Nonlinear Caputo-Type Fractional Differential Equations

Authors: Hongyu Li; Yang Chen
Journal: Journal of Function Spaces
 (2020)

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## Abstract

By using fixed-point index theory, we consider the existence of multiple positive solutions for a system of nonlinear Caputo-type fractional differential equations with the Riemann-Stieltjes boundary conditions.


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## Body

## 1. Introduction

Fractional order calculus is more widely used than integer order calculus (see [1]). Based on it, many researchers have been focused on the study of the existence of positive solutions for various fractional differential equations with some boundary conditions. We can refer to [2–35] for some recent works about fractional boundary value problems.In [14], Ma and Cui discussed the following fractional boundary value problem:
(1)cD0+θpy+µfy,py=0,y∈0,1,p0=p″0=0,p1=∫01pydAy,where θ∈2,3, Dc0+θ is the Caputo fractional derivative, f∈C0,1×0,+∞,0,+∞,µ>0 is a parameter. By using the Guo-Krasnosel’skii fixed-point theorem, they proved that the fractional differential equation has at least a positive solution when μ satisfies some conditions.A few researchers have studied the existence of solutions for the systems of nonlinear fractional differential equations (see [23–35]). For example, in [24], by monotone iterative technique and cone theory, Zhang et al. have studied the uniqueness of a solution for fractional systems with the Riemann-Stieltjes integral boundary condition. Hao et al. [32] have studied a system of fractional boundary value problems with two parameters; by using the Guo-Krasnosel’skii fixed-point theorem, they obtained the existence of positive solutions for the system in terms of different values of parameters. Meanwhile, a few researchers have been considering the systems for nonlinear integer order boundary value problems, such as in [33], where the researchers considered the existence of multiple positive solutions for a system of nonlinear third-order differential equations. In [34], by Banach’s contraction principle, the researchers considered the existence of a unique solution for a system of second order differential equations with coupled integral boundary conditions.Inspired by [14, 23–35], in this paper, we want to study the following system of fractional differential equations:
(2)−Dc0+θ1xt=f1t,xt,yt,t∈0,1,−Dc0+θ2yt=f2t,xt,yt,t∈0,1,x0=x″0=0,x1=∫01xtdA1t,y0=y″0=0,y1=∫01ytdA2t,where fi:0,1×0,+∞×0,+∞⟶0,+∞ is continuous; θi∈2,3;Dc0+θi is the Caputo fractional derivative; Ai is a bounded variation function with positive measure with Bi=∫01tdAit<1,i=1,2.The main purpose of this paper is that we prove that system (2) has one and multiple positive solutions by the fixed-point index theory.


## 2. Preliminaries

In this section, we only give the definition of Caputo’s fractional derivative. For some of the other definitions and properties of Caputo’s fractional derivative, we can refer to [2]. We mainly give the relevant Green functions and the used lemmas.Definition 1 (see [2, 14]).
For a functionx∈Cn0,+∞, we define Caputo’s fractional derivative of order θ>0 as follows:
(3)Dc0+θxt=1Γn−θ∫0tt−sn−θ−1xnsds,n−1<θ<n,where n is the smallest integer greater than or equal to θ.
From [14], we have the following lemmas.Lemma 2 (see [14]).
Letu∈C0,1 and θ1,θ2∈2,3. Then p is the solution of the linear Caputo fractional differential equation
(4)cD0+θipt+ut=0,t∈0,1,p0=p″0=0,p1=∫01ptdAit,if and only if pis the solution of the integral equation
(5)pt=∫01Kit,susds,where
(6)Kit,s=1Γθit1−Bi1−sθi−1−∫s1t−sθi−1dAit−t−sθi−1,0≤s,t≤1,t1−Bi1−sθi−1−∫s1t−sθi−1dAit,0≤t≤s≤1,and Bi=∫01tdAit<1,i=1,2.Lemma 3 (see [14]).
The above Green’s functionKit,si=1,2 has the following properties:
(i)
ΓθiKit,s≤1/1−Bi1−sθi−1,fort,s∈0,1(ii)
ΓθiKit,s≥Ni1−sθi−1,fort∈1/4,3/4,s∈0,1where
(7)Ni=min1−∫01tθi−1dAit41−Bi,mint∈1/4,3/4t1−tθi−2,i=1,2.Lemma 4 (see [36]).
LetE be a Banach space and P⊂E be a cone. Define Dr=u∈P:u<r, where r>0. Suppose that T:D¯r⟶P is completely continuous, and Tu≠u, for u∈∂Dr.
(i)
IfTu≤u,∀u∈∂Dr,theniT,Dr,P=1(ii)
IfTu≥u,∀u∈∂Dr,theniT,Dr,P=0


## 3. Main Results

LetE=C0,1×C0,1 be a Banach space with the norm u,vE=u+v on E, where u=max0≤t≤1ut.Define the cone(8)P=u,v∈E:u≥0,v≥0,min1/4≤t≤3/4ut+vt≥Ku,vE,where
(9)K=minN11−B1,N21−B2<1.N1 and N2 are defined by (7), and the nonlinear operators T1, T2, and T are defined by
(10)T1x,yt=∫01K1t,sf1s,xs,ysds,t∈0,1,T2x,yt=∫01K2t,sf2s,xs,ysds,t∈0,1,Tx,y=T1x,y,T2x,y,x,y∈E,where Kit,si=1,2 is defined by (6).It is known that fixed points of the operatorT in P are positive solutions of the system (2).Lemma 5.
The operatorT:P⟶P is completely continuous.Proof.
Forx,y∈P, when t∈0,1, we have T1x,yt≥0,T2x,yt≥0.
When(11)t∈14,34,by Lemma 3, we obtain
(12)T1x,yt=∫01K1t,sf1s,xs,ysds≥N1Γθ1∫011−sθ1−1f1s,xs,ysds=N11−B1Γθ1∫011−sθ1−11−B1f1s,xs,ysds≥N11−B1maxt∈0,1∫01K1t,sf1s,xs,ysds=N11−B1T1x,y.
Similar to the proof of (12), for
(13)x,y∈P,t∈14,34,we have
(14)T2x,yt≥N21−B2T2x,y.
By (12) and (14), we get
(15)mint∈1/4,3/4T1x,yt+T2x,yt≥N11−B1T1x,y+N21−B2T2x,y≥KTx,yE.
By (15), TP⊂P. Similar to the proof of Lemma 6 in [14], we know that T1,T2:C0,1⟶C0,1 are completely continuous. So T:P⟶P is completely continuous.For convenience, some marks and assumptions are given.Some of the marks are as follows:(16)g0=limx,y→0+,0+f1t,x,yx+y,g¯0=limx,y→0+,0+f2t,x,yx+y,g∞=limx,y→+∞,+∞f1t,x,yx+y,g¯∞=limx,y→+∞,+∞f2t,x,yx+y,k1=N1KΓθ1∫1/43/41−sθ1−1ds,k3=N2KΓθ2∫1/43/41−sθ2−1ds,where K is defined by (9).
(17)k2=1Γθ11−B1∫011−sθ1−1ds,k4=1Γθ21−B2∫011−sθ2−1ds,k=mink1,k3,h=maxk2,k4.Some of the assumptions are as follows:S1. There exist a>0 and b2>b1>0 such that
(i)
f1t,x,y≥ax+y,f2t,x,y≥ax+y,∀t∈0,1,0≤x,y≤b1(ii)
f1t,x,y<b2/2h,f2t,x,y<b2/2h,∀t∈0,1,0≤x,y≤b2where a also satisfies ak>1/2, k and h are defined by (17).S2. There exists q>0 such that
(18)f1t,x,y>Kq2k,f2t,x,y>Kq2k,∀t∈0,1,x+y∈Kq,q,where K is defined by (9) and k is defined by (17).Theorem 6.
Suppose thatg0=g¯0=∞ and g∞=g¯∞=0 uniformly on t∈0,1. Then, system (2) has a positive solution.Proof.
Byg0=g¯0=∞, we know that there exist r1>0 and m1>0 such that
(19)f1t,x,y≥m1x+y,f2t,x,y≥m1x+y,∀0≤x,y≤r1,t∈0,1,and m1k1>1/2, m1k3>1/2.
SetPr1=x,y∈P:x,yE<r1. By Lemma 3 and (19), for x,y∈∂Pr1,t∈1/4,3/4, we have
(20)T1x,yt=∫01K1t,sf1s,xs,ysds≥N1m1Γθ1∫1/43/41−sθ1−1xs+ysds≥m1N1KΓθ1∫1/43/41−sθ1−1dsx+y=m1k1x,yE>12x,yE.
Similar to the proof of (20), we have
(21)T2x,yt>12x,yE,∀x,y∈∂Pr1.
So by (20) and (21), we have
(22)Tx,yE=T1x,y+T2x,y>x,yE,∀x,y∈∂Pr1.
By Lemma4 and (22), we have
(23)iT,Pr1,P=0.
Fromg∞=g¯∞=0, we know that there exist r~>r1>0 and m2>0 such that
(24)f1t,x,y≤m2x+y,f2t,x,y≤m2x+y,∀x+y≥r~,t∈0,1,and m2h<1/4. In this part, we divide two cases. One case is that f1 and f2 are bounded. Namely, there exists W>0 such that
(25)f1t,x,y≤W,f2t,x,y≤W,∀x,y∈0,+∞,t∈0,1.
Taker2>max2Wh,r1. Set Pr2=x,y∈P:x,yE<r2. For x,y∈∂Pr2 and t∈0,1, we get
(26)T1x,yt=∫01K1t,sf1s,xs,ysds≤W∫01K1t,sds≤WΓθ11−B1∫011−sθ1−1ds≤Wh<r22=12x,yE.
The other case is thatf1 and f2 are unbounded. By the continuity of f1, there exists r2>maxr1,r~ such that
(27)f1t,x,y≤f1t,r2,r2,for0<x,y≤r2.
Hence, setPr2=x,y∈P:x,yE<r2. For x,y∈∂Pr2 and t∈0,1 we get
(28)T1x,yt=∫01K1t,sf1s,xs,ysds≤1Γθ11−B1∫011−sθ1−1f1s,r2,r2ds≤1Γθ11−B1∫011−sθ1−1m2r2+r2ds=2m2k2r2=2m2k2x,yE<2m2hx,yE<12x,yE.
So in either case, there always existsr2>r1 such that
(29)T1x,y<12x,yE,∀x,y∈∂Pr2.
Similarly, we have(30)T2x,y<12x,yE,∀x,y∈∂Pr2.
So by (29) and (30), we have
(31)Tx,yE=T1x,y+T2x,y<x,yE,∀x,y∈∂Pr2.
By Lemma4 and (31), we have
(32)iT,Pr2,P=1.
From (23) and (32), we obtain
(33)iT,Pr2\P¯r1,P=iT,Pr2,P−iT,Pr1,P=1.
SoT has a fixed point x,y in Pr2\P¯r1. It is obvious that x,y is a positive solution of system (2).Theorem 7.
Suppose that S1 holds and g∞=g¯∞=∞. Then, the system (2) has two positive solutions.Proof.
SetPb1=x,y∈P:x,yE<b1. By Lemma 3 and S1 (i), for x,y∈∂Pb1 and t∈1/4,3/4, we get
(34)T1x,yt=∫01K1t,sf1s,xs,ysds≥N1Γθ1∫1/43/41−sθ1−1f1s,xs,ysds≥N1aΓθ1∫1/43/41−sθ1−1xs+ysds≥N1aKΓθ1∫1/43/41−sθ1−1dsx+y=ak1x,yE≥akx,yE≥12x,yE.
So(35)T1x,y>12x,yE,∀x,y∈∂Pb1.
Similar to the proof of (35), we have
(36)T2x,y>12x,yE,∀x,y∈∂Pb1.
So by (35) and (36), we have
(37)Tx,yE=T1x,y+T2x,y>x,yE,∀x,y∈∂Pb1.
By Lemma4 and (37), we have
(38)iT,Pb1,P=0.
SetPb2=x,y∈P:x,yE<b2. By Lemma 3 and S1 (ii), for x,y∈∂Pb2 and t∈0,1, we get
(39)T1x,yt=∫01K1t,sf1s,xs,ysds≤1Γθ11−B1∫011−sθ1−1f1s,xs,ysds<b22hΓθ11−B1∫011−sθ1−1ds=b2k22h≤b22=12x,yE.
So(40)T1x,y<12x,yE,∀x,y∈∂Pb2.
Similarly,(41)T2x,y<12x,yE,∀x,y∈∂Pb2.
So by (40) and (41), we have
(42)Tx,yE=T1x,y+T2x,y<x,yE,∀x,y∈∂Pb2.
By Lemma4 and (42), we have
(43)iT,Pb2,P=1.
Byg∞=g¯∞=∞, there exist m3>0 and b>b2 such that
(44)f1t,x,y≥m3x+y,f2t,x,y≥m3x+y,∀x+y≥b,t∈0,1,and m3k>1/2.
Chooseb3>maxb2,b/K. Set Pb3=x,y∈P:x,yE<b3. For x,y∈∂Pb3(45)mint∈1/4,3/4xt+yt≥Kx+y=Kb3>b.
Then, by (44) and Lemma 3, we get
(46)T1x,yt=∫01K1t,sf1s,xs,ysds≥N1Γθ1∫1/43/41−sθ1−1f1s,xs,ysds≥N1m3Γθ1∫1/43/41−sθ1−1xs+ysds≥N1m3KΓθ1∫1/43/41−sθ1−1dsx+y=m3k1x,yE≥m3kx,yE>12x,yE.
So(47)T1x,y>12x,yE,∀x,y∈∂Pb3.
Similarly,(48)T2x,y>12x,yE,∀x,y∈∂Pb3.
So by (47) and (48), we have
(49)Tx,yE=T1x,y+T2x,y>x,yE,∀x,y∈∂Pb3.
By Lemma4 and (49), we have
(50)iT,Pb3,P=0.
Since0<b1<b2<b3, by (38), (43), and (50), we get
(51)iT,Pb3\P¯b2,P=iT,Pb3,P−iT,Pb2,P=−1,(52)iT,Pb2\P¯b1,P=iT,Pb2,P−iT,Pb1,P=1.
From (51), we know that T has a fixed point x1,y1∈Pb3\P¯b2. From (52), we know that T has another fixed pointx2,y2∈Pb2\Pb1. So the system (2) has two positive solutionsx1,y1 and x2,y2, with 0<x2,y2E<b2<x1,y1E.Theorem 8.
Letg0=g¯0=0 and g∞=g¯∞=0. In addition, suppose that S2 holds. Then, system (2) has two positive solutions.Proof.
Byg0=g¯0=0, we know that there exist c1>0 and h1∈0,q such that
(53)f1t,x,y≤c1x+y,f2t,x,y≤c1x+y,∀0≤x,y≤h1,t∈0,1,and c1h<1/2. Set Ph1=x,y∈P:x,yE<h1.
By Lemma3 and (53), for x,y∈∂Ph1 and t∈0,1, we get
(54)T1x,yt=∫01K1t,sf1s,xs,ysds≤c1Γθ11−B1∫011−sθ1−1xs+ysds≤c1x+yΓθ11−B1∫011−sθ1−1ds=c1k2x,yE<c1hx,yE<12x,yE.
So(55)T1x,y<12x,yE,∀x,y∈∂Ph1.
Similarly,(56)T2x,y<12x,yE,∀x,y∈∂Ph1.
So by (55) and (56), we have
(57)Tx,yE=T1x,y+T2x,y<x,yE,∀x,y∈∂Ph1.
By Lemma4 and (57), we have
(58)iT,Ph1,P=1.
Fromg∞=g¯∞=0, we know that there exist c2>0 and H>h1 such that
(59)f1t,x,y≤c2x+y,f2t,x,y≤c2x+y,∀x+y≥H,t∈0,1,and c2h<1/4. In this part, we divide two cases. One case is that f1 and f2 are bounded. Namely, there exists W>0 such that
(60)f1t,x,y≤W,f2t,x,y≤W,∀x,y∈0,+∞,t∈0,1.
Takeh2>max2Wh,q. Set Ph2=x,y∈P:x,yE<h2. For x,y∈∂Ph2 and t∈0,1, we get
(61)T1x,yt=∫01K1t,sf1s,xs,ysds≤W∫01K1t,sds≤WΓθ11−B1∫011−sθ1−1ds≤Wh<h22=12x,yE.
Another case is thatf1 and f2 are unbounded. By the continuity of f1, there exists h2>maxh1,q,H such that
(62)f1t,x,y≤f1t,h2,h2,for0<x,y≤h2.
Hence, setPh2=x,y∈P:x,yE<h2. For x,y∈∂Ph2, we get
(63)T1x,yt=∫01K1t,sf1s,xs,ysds≤1Γθ11−B1∫011−sθ1−1f1s,h2,h2ds≤1Γθ11−B1∫011−sθ1−1c2h2+h2ds=2c2k2h2=2c2k2x,yE<2c2hx,yE<12x,yE.
So in either case, there always existsh2>maxh1,q such that
(64)T1x,y<12x,yE,∀x,y∈∂Ph2.
Similarly, we have(65)T2x,y<12x,yE,∀x,y∈∂Ph2.
Thus(66)Tx,yE=T1x,y+T2x,y<x,yE,∀x,y∈∂Ph2.
So by Lemma4 and (66), we get
(67)iT,Ph2,P=1.
SetPq=x,y∈P:x,yE<q. For x,y∈∂Pq(68)mint∈1/4,3/4xt+yt≥Kx+y=Kq.
Then, by S2 and Lemma 3, we get
(69)T1x,yt=∫01K1t,sf1s,xs,ysds≥N1Γθ1∫1/43/41−sθ1−1f1s,xs,ysds>N1Kq2kΓθ1∫1/43/41−sθ1−1ds>q2=12x,yE.
So(70)T1x,y>12x,yE,∀x,y∈∂Pq.
Similarly,(71)T2x,y>12x,yE,∀x,y∈∂Pq.
So by (70) and (71), we have
(72)Tx,yE=T1x,y+T2x,y>x,yE,∀x,y∈∂Pq.
By Lemma4 and (72), we have
(73)iT,Pq,q=0.
Since0<h1<q<h2, by (58), (67), and (73), we get iT,Pq\P¯h1,P=iT,Pq,P−iT,Ph1,P=−1. So T has a fixed point x1,y1∈Pq\P¯h1.iT,Ph2\P¯q,P=iT,Ph2,P−iT,Pq,P=1. So T has another fixed point x2,y2∈Ph2\P¯q. Therefore, system (2) has at least two positive solutions.


## 4. Applications

Example 9.
We study the following Caputo-type fractional system:(74)−Dc0+5/2xt=2+3txt+yt,t∈0,1,−Dc0+5/2yt=1+t2+t4xt+yt5,t∈0,1,x0=x″0=0,x1=12∫01xtdt,y0=y″0=0,y1=12∫01ytdt,where θ1=θ2=5/2, A1t=A2t=1/2t, B1=B2=1/4, f1t,x,y=2+3tx+y, and f2t,x,y=1+t2+t4x+y5. Obviously, g0=g¯0=∞ and g∞=g¯∞=0. From Theorem 6, system (74) has a positive solution.Example 10.
We study the following Caputo-type fractional system:(75)−Dc0+5/2xt=f1t,xt,yt,t∈0,1,−Dc0+5/2yt=f2t,xt,yt,t∈0,1,x0=x″0=0,x1=12∫01xtdt,y0=y″0=0,y1=12∫01ytdt,where θ1=θ2=5/2, A1t=A2t=1/2t, and B1=B2=1/4.
Take(76)f1t,x,y=104t+1,t∈0,1,0≤x,y≤12,104t+1x+y,t∈0,1,12≤x,y≤1,2×104t+1,t∈0,1,1≤x,y≤106,10−8t+12x+y2,t∈0,1,x,y≥106,and
(77)f2t,x,y=104t+1,t∈0,1,0≤x,y≤12,104t+1x+y,t∈0,1,12≤x,y≤1,2×104t+1,t∈0,1,1≤x,y≤106,10−14t+14x+y3,t∈0,1,x,y≥106.
By simple computation, we getK=9/256, k=993−1/327680π, and h=32/45π.
Choosea=7000, b2=106, and b1=1, then b2/2h=45π/64⋅106.
When0≤x, y≤b1=1,
(1)
0≤x,y≤1/2,f1t,x,y=104t+1≥104≥7000x+y; f2t,x,y=104t+1≥104≥7000x+y(2)
1/2≤x,y≤1,f1t,x,y=104t+1x+y≥104x+y≥7000x+y; f2t,x,y=104t+1x+y≥104x+y≥7000x+y
When0≤x, y≤b2=106,
(1)
0≤x,y≤1/2,f1t,x,y=104t+1≤2×104<45π/64⋅106; f2t,x,y=104t+1≤2×104<45π/64⋅106(2)
1/2≤x,y≤1,f1t,x,y=104t+1x+y≤4×104<45π/64⋅106; f2t,x,y=104t+1x+y≤4×104<45π/64⋅106(3)
1≤x,y≤106,f1t,x,y=2×104t+1≤4×104<45π/64⋅106; f2t,x,y=2×104t+1≤4×104<45π/64⋅106
Obviously,g∞=g¯∞=∞. So by Theorem 7, system (74) has at least two positive solutions.Example 11.
We study the following Caputo-type fractional system:(78)−Dc0+5/2xt=f1t,xt,yt,t∈0,1,−Dc0+5/2yt=f2t,xt,yt,t∈0,1,x0=x″0=0,x1=12∫01xtdt,y0=y″0=0,y1=12∫01ytdt,where θ1=θ2=5/2, A1t=A2t=1/2t, and B1=B2=1/4.
Take(79)f1t,x,y=12×108t+1x+y2,t∈0,1,0≤x,y≤10−4,104t+1x+y,t∈0,1,10−4≤x,y≤10−2,4t+1x+y,t∈0,1,x,y≥10−2,and
(80)f2t,x,y=12×108t+1x+y2,t∈0,1,0≤x,y≤10−4,104t+1x+y,t∈0,1,10−4≤x,y≤10−2,8×10−2t+1x+y2,t∈0,1,x,y≥10−2.Take q=0.001, then Kq=9/25600, Kq/2k≈0.7778. When x+y∈Kq,q, f1t,x,y=f2t,x,y=104t+1x+y≥2≥Kq/2k. Obviously, g0=g¯0=0 and g∞=g¯∞=0. So, by Theorem 8, system (78) has at least two positive solutions.



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