# Existence of Positive Solution for Fractional Differential Systems with Multipoint Boundary Value Conditions

Authors: Zhi-Wei Lv
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we apply the fixed-point theorems ofγ concave and −γ convex operators to establish the existence of positive solutions for fractional differential systems with multipoint boundary conditions. Two examples are given to support our results.


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## Body

## 1. Introduction

Consider the following system of nonlinear fractional differential equations(1)D0+αut+μ1ft,vt+μ2gt,vt=0,0<t<1,D0+αvt+μ1ft,ut+μ2gt,ut=0,0<t<1,(2)D0+αut+μ1ft,ut+μ2gt,vt=0,0<t<1,D0+αvt+μ1ft,vt+μ2gt,ut=0,0<t<1,with the multipoint boundary conditions with the multipoint boundary conditions(3)D0+βu1=∑i=1m−2ξiD0+βuηi,u0=0,D0+βv1=∑i=1m−2ξiD0+βvηi,v0=0,where D0+α is the standard Riemann–Liouville fractional derivative, f:J×0,∞⟶0,∞is  continuous,1<α≤2, 0≤β≤1, 0≤α−β−1, 0<ξi,ηi<1, i=1,2,…,m−2, ∑i=1m−2ξiηiα−β−1<1,J=0,1, and μ1,μ2∈0,+∞, μ1≥μ2.Differential equations with fractional order have been applied in various areas of science and engineering. For their applications, there has been a sharp increase in studying fractional differential equations (see [1–18] and references therein). Meanwhile, the theory of boundary value problems with multipoint boundary conditions has various applications in applied fields, which have been studied by many authors (cf., e.g., [19–26]). Many authors have studied these problems by using different methods, such as monotone iterative technique, the method of upper and lower solutions, fixed-point theorems in cones, nonlinear alternatives of Leray–Schauder, and coincidence degree theory. However, concave (convex) operators are a class of important operators, which can be used in nonlinear differential and integral equations (cf., e.g., [27–31]). Moreover, few papers can be reported on the existence of solutions for coupled systems of fractional differential equations with multipoint boundary conditions by using fixed-point theorems of γ concave and −γ convex operators.In [25], we considered the following m-point boundary value problem for fractional differential equation(4)D0+αut+ft,ut=0,0<t<1,with the multipoint boundary conditions u0=0,D0+βu1=∑i=1m−2ξiD0+βuηi, where D0+α is the standard Riemann–Liouville fractional derivative, n=α+1, f:0,1×0,∞⟶0,∞ is continuous, 1<α≤2, 0≤β≤1, 0≤α−β−1, 0<ξi,ηi<1, i=1,2,…,m−2, and ∑i=1m−2ξiηiα−β−1<1.In [26], Henderson and Luca studied the following system of nonlinear second-order ordinary differential equation(5)u″t+ctfvt=0,t∈0,T,v″t+dtgut=0,t∈0,T,with the multipoint boundary conditions αu0−βu′0=0,uT=∑i=1m−2iuξi+a0a,m∈ℕ,m≥3 and γv0−δv′0=0,vT=∑i=1m−2bivηi+b0,n∈ℕ,m≥3. By using the Schauder fixed-point theorem, the existence of positive solutions was investigated.Motivated by above papers, in this paper, we investigate the existence of positive solutions for systems (1)–(3).In this paper, we need the following assumptions that we shall use in the sequel:H1 f,g∈CJ×ℝ+,ℝ+, ft,x and gt,x are increasing in x for x∈ℝ+, gt,0≠0H2 There exists a constant γ∈0,1 such that ft,λx≥λγft,x and gt,λx≥λγgt,x, ∀t∈J,λ∈0,1,x∈ℝ+H3 There exists a constant δ0>0 such that ft,x≥δ0gt,x, t∈J,x∈ℝ+H4 f∈CJ×ℝ+,ℝ+, ft,x is increasing in x for x∈ℝ+, ft,0≠0H5 There exists a constant γ∈0,1 such that ft,λx≥λγft,x, ∀t∈J,λ∈0,1,x∈ℝ+H6 f,g∈CJ×ℝ+,ℝ+, ft,x is nondecreasing in x, and gt,y is nonincreasing in yH7 ft,x and gt,y are bounded in J×ℝ+H8 There exists 0≤γ1<1 such that ft,kx≥kγ1ft,x, and there exists 0≤γ2<1 such that gt,kx≤k−γ2gt,x, where k∈0,1, 0≤γ1+γ2<1Here are our main results.Theorem 1.
Suppose thatH1−H3 hold. Then, equations (1)–(3) have a unique positive solution u∗,v∗ in Ph×Ph, where ht=tα−1,t∈J. Moreover, for any initial value u0∈Ph and v0∈Ph, constructing successively the sequence(6)unt=∫01Gt,sμ1fs,vn−1s+μ2gs,vn−1sds,n=0,1,2,…,vnt=∫01Gt,sμ1fs,un−1s+μ2gs,un−1sds,n=0,1,2,…,we have unt,vnt⟶u∗t,v∗t as n⟶∞.Corollary 1.
Suppose thatH4−H5 holds. Then, system(7)D0+αut+μ1ft,vt=0,0<t<1,D0+αvt+μ1ft,ut=0,0<t<1,with the multipoint boundary conditions(8)D0+βu1=∑i=1m−2ξiD0+βuηi,u0=0,D0+βv1=∑i=1m−2ξiD0+βvηi,v0=0,has a unique positive solution u∗,v∗ in Ph×Ph, where ht=tα−1,t∈J, D0+α is the standard Riemann–Liouville fractional derivative, f:J×0,∞⟶0,∞ is continuous, 1<α≤2, 0≤β≤1, 0≤α−β−1, 0<ξi,ηi<1, i=1,2,…,m−2, ∑i=1m−2ξiηiα−β−1<1,J=0,1, and μ1,μ2∈0,+∞, μ1≥μ2. Moreover, for any initial value u0∈Ph and v0∈Ph, constructing successively the sequence(9)unt=μ1∫01Gt,sfs,vn−1sds,n=0,1,2,…,vnt=μ1∫01Gt,sfs,un−1sds,n=0,1,2,…,we have unt,vnt⟶u∗t,v∗t as n⟶∞.Theorem 2.
Suppose thatH6−H8 hold. Then, equations (1)–(3) have exactly one positive solution u∗,v∗∈u0,v0×u0,v0, where u0,v0∈P with u0≤v0, and constructing successively the sequence(10)unt=∫01Gt,sμ1fs,un−1s+μ2gs,vn−1sds,n=0,1,2,…,vnt=∫01Gt,sμ1fs,vn−1s+μ2gs,un−1sds,n=0,1,2,…,we have unt,vnt⟶u∗t,v∗t as n⟶∞.
The rest of this paper is organized as follows. In Section2, we present some background materials and preliminaries. Section 3 deals with the existence results. In Section 4, two examples are given to illustrate the result.


## 2. Background Materials and Preliminaries

Definition 1 (see [6]).
The fractional integral of orderα with the lower limit t0 for a function f is defined as(11)Iαft=1Γα∫t0tt−sα−1fsds,t>t0,α>0,where Γ is the gamma function.Definition 2 (see [6]).
For a functionf:0,∞⟶ℝ, the Riemann–Liouville derivative of fractional order is defined as(12)Dαft=1Γn−αddtn∫0tt−sn−α−1fsds,α>0,n=α+1.Definition 3 (see [28]).
LetE be a real Banach space and P be a cone in E which defined a partial ordering in E by x≤y if and only if y−x∈P. P is said to be normal if there exists a positive constant N such that θ≤x≤y implies x≤Ny. P is called solid if its interior P∘ is nonempty.Definition 4 (see [28]).
An operatorA:D×D⟶E is said to be mixed monotone if Ax,y is nondecreasing in x and nonincreasing in y, i.e., ∀xi,yi∈Di=1,2,x1≤x2, and y2≤y1 imply Ax1,y1≤Ax2,y2.Definition 5 (see [28]).
For allx,y∈E, the notation x∼y means that there exists λ>0 and μ>0 such that λx≤y≤μx. Clearly, ∼ is an equivalence relation. Given h>θ (i.e., h≥θ and h≠θ), we denote by Ph the set Ph=x∈Ex∼h. It is easy to see that Ph⊂P.Definition 6 (see [30]).
LetD=P or D=P∘ and γ be a real number with 0≤γ<1. An operator A:P⟶P is said to be γconcave−γconvex if it satisfies(13)Aλx≥λγAx,Aλx≤λ−γAx,∀λ∈0,1,x∈D.Definition 7 (see [31]).
An operatorA:E⟶E is said to be homogeneous if it satisfies(14)Aλx=λAx,∀λ>0,x∈E.
An operatorA:P⟶P is said to be subhomogeneous if it satisfies(15)Aλx≥λAx,∀λ∈0,1,x∈P.Lemma 1 (see [25]).
Lety∈C0,1. Then, the fractional differential equation(16)D0+αut+yt=0,0<t<1,1<α≤2,D0+βu1=∑i=1m−2ξiD0+βuηi,u0=0,has a unique solution which is given by(17)ut=∫01Gt,sysds,where(18)Gt,s=G1t,s+G2t,s,in which(19)G1t,s=1Γαtα−11−sα−β−1−t−sα−1,0≤s≤t≤1,1Γαtα−11−sα−β−1,0≤t≤s≤1,G2t,s=1AΓα∑0≤s≤ηiξiηiα−β−1tα−11−sα−β−1−ξitα−1ηi−sα−β−1,t∈0,1,1AΓα∑ηi≤s≤1ξiηiα−β−1tα−11−sα−β−1,t∈0,1,where(20)A=1−∑i=1m−2ξiηiα−β−1.Lemma 2.
Letht=tα−1, then Gt,s in Lemma 1 has the following property:(i)
Gt,s≥ht1/Γα1−sα−β−1−1−sα−1+1/AΓα∑i=1m−2ξiηiα−β−11−sα−β−1−ξiηi−sα−β−1(ii)
Gt,s≤ht1/Γα1−sα−β−1+1/AΓα∑i=1m−2ξiηiα−β−11−sα−β−1Proof.
For0≤s≤t≤1,1<α≤2, we have(21)tα−11−sα−β−1−t−sα−1=tα−11−sα−β−1−1−stα−1≥tα−11−sα−β−1−1−sα−1≥0.
Thus,(22)G1t,s≥tα−1Γα1−sα−β−1−1−sα−1≥0.
From [25], we have(23)G2t,s≥tα−1AΓα∑i=1m−2ξiηiα−β−11−sα−β−1−ξiηi−sα−β−1≥0.
This means that (i) holds. From Lemma1, we know that (ii) is obvious.Theorem 3 (see [31]).
LetP be a normal cone in a real Banach space E and A:P⟶P be an increasing γ−concave operator and B:P⟶P be an increasing subhomogeneous operator. Assume that(i)
There ish>θ such that Ah∈Ph and Bh∈Ph(ii)
There exists a constantδ0>0 such that Ax≥δ0Bx, ∀x∈P
Then, operator equationAx+Bx=x has a unique solution x∗ in Ph. Moreover, constructing successively the sequence yn=Ayn−1+Byn−1,n=1,2,… for any initial value y0∈Ph, we have yn⟶x∗ as n⟶∞.Theorem 4 (see [30]).
LetP be a normal cone of the real Banach space E and A:P×P⟶P be a mixed monotone operator. Suppose that(i)
For fixedy, A⋅,y:P⟶P is γ1 concave; for fixed x, Ax:P⟶P is −γ2 convex, where 0≤γ1+γ2<1(ii)
There exist elementsu0,v0∈P with u0≤v0 and a real number r0>0 such that(24)u0≥r0v0,u0≤Au0,v0,Av0,u0≤v0.
Then,A has exactly one fixed point x∗ in u0,v0, and constructing successively the sequence(25)xn=Axn−1,yn−1,yn=Ayn−1,xn−1,n=1,2,…,for any initial value x0,y0∈u0,v0×u0,v0, we have xn⟶x∗,yn⟶y∗n⟶∞.


## 3. Main Results

In this section, we shall investigate the existence of positive solutions for systems (1)–(3). We consider the space E=C0,1,ℝ equipped with the norm u=sup0≤t≤1ut. Let P=u∈Eut≥0, then P is a cone in E. Let ℝ+=0,+∞.From Lemma1, we know that (1)–(3) can be translated into the following equation(26)ut=∫01Gt,sμ1fs,vs+μ2gs,vsds,vt=∫01Gt,sμ1fs,us+μ2gs,usds,and (2)-(3) can be translated into the following equation(27)ut=∫01Gt,sμ1fs,us+μ2gs,vsds,vt=∫01Gt,sμ1fs,vs+μ2gs,usds.Thus,u,v is a solution of (1)–(3) if and only if u,v is a solution of system (26), and u,v is a solution of (2)-(3) if and only if u,v is a solution of system (27).For convenience, we denote(28)p1s=1Γα1−sα−β−1−1−sα−1+1AΓα∑i=1m−2ξiηiα−β−11−sα−β−1−ξiηi−sα−β−1,p2s=1Γα1−sα−β−1+1AΓα∑i=1m−2ξiηiα−β−11−sα−β−1.Now, we prove Theorem1, Corollary 1, and Theorem 2.Proof of Theorem 1.
Define two operators(29)Aut=μ1∫01Gt,sfs,vsds,But=μ2∫01Gt,sgs,vsds.
Thus,(30)Aut=μ1∫01Gt,sfs,∫01Gs,τμ1fτ,uτ+μ2gτ,uτdτds,But=μ2∫01Gt,sgs,∫01Gs,τμ1fτ,uτ+μ2gτ,uτdτds.
Owing to [25], we know that Gt,s>0,t,s∈0,1. By H1, we have A:P⟶P and B:P⟶P. It is obvious that u,v is the solution of problem (26) if and only if u,v is the solution of u=Au+Bu,v=Av+Bv.Step 1:A and B are two increasing operators.Setu,u¯∈P,u≤u¯. It follows from (26) that(31)vt=∫01Gt,sμ1fs,us+μ2gs,usds,v¯t=∫01Gt,sμ1fs,u¯s+μ2gs,u¯sds.FromH1, we have(32)vt≤v¯t,t∈J.According toH1 and (29), we obtain(33)Aut=μ1∫01Gt,sfs,vsds≤Au¯t=μ1∫01Gt,sfs,v¯sds.Thus,A is an increasing operator. Similarly, we can see that B is an increasing operator.Step 2:A is a γconcave operator, and B is a subhomogeneous operator.In fact, forλ∈0,1,γ∈0,1,u∈P,t∈J, from (30) and H2, we have(34)Aλut=μ1∫01Gt,sfs,∫01Gs,τμ1fτ,λuτ+μ2gτ,λuτdτds≥μ1∫01Gt,sfs,∫01Gs,τμ1λγfτ,uτ+μ2λγgτ,uτdτds=μ1∫01Gt,sfs,λγ∫01Gs,τμ1fτ,uτ+μ2gτ,uτdτds≥μ1λγγ∫01Gt,sfs,∫01Gs,τμ1fτ,uτ+μ2gτ,uτdτds≥μ1λγ∫01Gt,sfs,∫01Gs,τμ1fτ,uτ+μ2gτ,uτdτds=λγAut.Similarly, we can get(35)Bλut≥λγBut.Thus,(36)Bλut≥λBut.Therefore, we can see thatA is a γconcave operator, and B is a subhomogeneous operator.Step 3:Ah∈Ph and Bh∈Ph.Combining (29), (30), H1, and Lemma 2, one has, for t∈J,(37)Aht=μ1∫01Gt,sfs,∫01Gs,τμ1fτ,hτ+μ2gτ,hτdτds≥μ1∫01htp1sfs,∫01hsp1τμ1fτ,0+μ2gτ,0dτds.Aht=μ1∫01Gt,sfs,∫01Gs,τμ1fτ,hτ+μ2gτ,hτdτds≤μ1∫01htp2sfs,∫01hsp2τμ1fτ,1+μ2gτ,1dτds.Let(38)m=μ1∫01p1sfs,∫01hsp1τμ1fτ,0+μ2gτ,0dτds,m¯=μ1∫01p2sfs,∫01hsp2τμ1fτ,1+μ2gτ,1dτds.Noting thatH3 and gt,0≠0, we obtain(39)m>0,m¯>0.Therefore,(40)mht≤Aht≤m¯ht,which implies thatAh∈Ph. By a closely similar way, we have Bh∈Ph.Step 4: There exists a constantδ0>0 such that Au≥δ0Bu and Av≥δ0Bv, ∀u∈P.
Foru∈P, t∈J, by H3, we have(41)∫01Gt,sfs,vsds≥δ0∫01Gt,sgs,vsds.
This means thatAu≥δ0Bu,u∈P. Similarly, we have Av≥δ0Bv,v∈P.
Therefore, by simple computation, the conditions in Theorem3 are satisfied. This implies that the operator equation Au+Bu=u has a unique solution u∗ in Ph, and the operator equation Av+Bv=v has a unique solution v∗ in Ph. Thus, u=Au+Bu,v=Av+Bv has a unique solution u∗,v∗ in Ph×Ph. For any initial value u0∈Ph and v0∈Ph, we can construct the following sequence:(42)un+1t=∫01Gt,sμ1fs,vns+μ2gs,vnsds,n=0,1,2,…,vn+1t=∫01Gt,sμ1fs,uns+μ2gs,unsds,n=0,1,2,…
This follows thatunt,vnt⟶u∗t,v∗t as n⟶∞.Proof of Corollary 1.
In Theorem3, we let B be a null operator, Theorem 3 also holds. By Theorem 1, we conclude that Corollary 1 holds.Proof of Theorem 2.
Define the following operator:(43)unt=Aun−1t,vn−1t=∫01Gt,sμ1fs,un−1s+μ2gs,vn−1sds,n=0,1,2,…,vnt=Avn−1t,un−1t=∫01Gt,sμ1fs,vn−1s+μ2gs,un−1sds,n=0,1,2,…
FromH6, we know that A:P×P⟶P is a mixed monotone operator.Step 1: we will prove that the condition (i) of Theorem4 holdsFromH8, we know that, for γ1,γ2∈0,1,0<γ1+γ2<1,k1,k2∈0,1,(44)Aku,v=∫01Gt,sμ1fs,kus+μ2gs,vsds≥∫01Gt,skγ1μ1fs,us+μ2gs,vsds≥kγ1∫01Gt,sμ1fs,us+μ2gs,vsds=kγ1Au,v,Au,kv=∫01Gt,sμ1fs,us+μ2gs,kvsds≤∫01Gt,sμ1fs,us+k−γ2μ2gs,vsds≤k−γ2∫01Gt,sμ1fs,us+μ2gs,vsds=k−γ2Au,v.Step 2: we will verify that the condition (ii) of Theorem4 holds
It follows fromH7 that there exists M1>0 and M2>0 such that(45)ft,x≤M1,gt,x≤M2,t,x∈J×ℝ+.
Let(46)v0t=∫01Gt,sμ1M1+μ2M2ds,u0t=∫01Gt,sμ1fs,0ds,r0=μ1mins∈Jfs,0μ1M1+μ2M2.
Obviously,(47)Av0,u0=∫01Gt,sμ1fs,v0s+μ2gs,u0sds≤∫01Gt,sμ1M1+μ2M2ds=v0t,Au0,v0=∫01Gt,sμ1fs,u0s+μ2gs,v0sds≥∫01Gt,sμ1fs,0ds=u0t,u0t=∫01Gt,sμ1fs,0ds≥μ1mins∈Jfs,0μ1M1+μ2M2∫01Gt,sμ1M1+μ2M2ds=r0v0t.
Therefore, the conditions of Theorem4 are satisfied. This means that (2)-(3) has exactly one positive solution u∗,v∗∈u0,v0×u0,v0, where u0,v0∈P with u0≤v0. By constructing successively the sequence(48)unt=∫01Gt,sμ1fs,un−1s+μ2gs,vn−1sds,n=0,1,2,…,vnt=∫01Gt,sμ1fs,vn−1s+μ2gs,un−1sds,n=0,1,2,…,we obtain unt,vnt⟶u∗t,v∗t as n⟶∞.


## 4. Examples

Example 1.
Letα=3/2,β=1/2,m=4,ξ1=η1=1/4,ξ2=η2=1/2,μ1,μ2∈0,+∞,μ1≥μ2. Consider the following boundary value problem(49)D0+3/2ut+μ1ft,vt+μ2gt,vt=0,0<t<1,D0+3/2vt+μ1ft,ut+μ2gt,ut=0,0<t<1,with the multipoint boundary conditions(50)D0+1/2u1=∑i=1m−2ξiD0+1/2uηi,u0=0,D0+1/2v1=∑i=1m−2ξiD0+1/2vηi,v0=0.
Here,(51)ft,x=x1/3+t2+2,gt,x=x1/31+t21+x1/3+t2+1.
Thus,(52)α−β−1=0,∑i=1m−2ξiηiα−β−1=ξ1+ξ2=14+12<1.
Setγ=1/3. Obviously, f,g∈CJ×ℝ+,ℝ+ and are increasing with respect to the second argument, gt,0≥1>0. For λ∈0,1,t∈J,x∈ℝ+, we can notice that(53)gt,λx=λ1/3x1/31+t21+λ1/3x1/3+t2+1≥λ1/3x1/31+t21+x1/3+λ1/3t2+1=λγgt,x,ft,λx=λ1/3x1/3+t2+2≥λ1/3x1/3+t2+2=λγft,x.
Fort∈J,x∈ℝ+, we deduce that (54)ft,x=x1/3+t2+2≥x1/31+t21+x1/3+t2+1=δ0gt,x,where(55)δ0=1.
Thus, the assumptions ofH1−H3 are satisfied. By Theorem 1, system (49)-(50) has a unique positive solution in Ph×Ph, where ht=tα−1, t∈0,1.Example 2.
Lettingα=3/2,β=1/2,m=4,ξ1=η1=1/4,ξ2=η2=1/2,μ1,μ2∈0,+∞,μ1≥μ2, we consider the following problem(56)D0+3/2ut+μ1ft,ut+μ2gt,vt=0,0<t<1,D0+3/2vt+μ1ft,vt+μ2gt,ut=0,0<t<1,with the multipoint boundary conditions(57)D0+1/2u1=∑i=1m−2ξiD0+1/2uηi,u0=0,D0+1/2v1=∑i=1m−2ξiD0+1/2vηi,v0=0.
Here,(58)ft,x=1+x1/32+x1/3+t+1,gt,y=1+y−1/2+t2+1.
We deduce that(59)α−β−1=0,∑i=1m−2ξiηiα−β−1=ξ1+ξ2=14+12<1.
Setγ=1/3,γ1=1/3,γ2=1/2. It is clear that f,g∈CJ×ℝ+,ℝ+ and ft,x is nondecreasing in x, and gt,y is nonincreasing in y, ft,x≤1+1+1=3, gt,y≤1+0−1/2+1+1=3, and 0<γ1+γ2=5/6. Moreover, for k∈0,1, we can find that(60)ft,kx=1+kx1/32+kx1/3+t+1≥1+kx1/32+x1/3+t+1≥k1/31+x1/32+x1/3+t+1=k1/3ft,x,gt,ky=1+ky−1/2+t2+1=k−1/21k+y−1/2+t2+1≤k−1/21+y−1/2+t2+1≤k−1/21+y−1/2+t2+1=k−1/2gt,y.
Then, all the conditions of Theorem2 are fulfilled. Consequently, there exist u0,v0∈P, and system (56)-(57) has exactly one positive solution in u0,v0×u0,v0.



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