# Positive Solutions for a System of Hadamard-Type Fractional Differential Equations with Semipositone Nonlinearities

Authors: Youzheng Ding; Jiqiang Jiang; Donal O’Regan; Jiafa Xu
Journal: Complexity
 (2020)

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## Abstract

In this paper, we use the fixed-point index and nonnegative matrices to study the existence of positive solutions for a system of Hadamard-type fractional differential equations with semipositone nonlinearities.


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## Body

## 1. Introduction

In this paper, we study the existence of positive solutions for the following system of Hadamard-type fractional boundary value problems:(1)DαDαξt=f1t,ξt,δξt,−Dαξt,ηt,δηt,−Dαηt,t∈1,e,DαDαηt=f2t,ξt,δξt,−Dαξt,ηt,δηt,−Dαηt,t∈1,e,ξ1=δξ1=δξe=0,Dαξ1=δDαξ1=δDαξe=0,η1=δη1=δηe=0,Dαη1=δDαη1=δDαηe=0,where α∈2,3, Dα is the Hadamard fractional derivative of order α, δ=td/dt (i.e., if u is ξ or η, then δut=td/dtut; δu1=limt⟶1+td/dtut=td/dtutt=1 etc.), and the nonlinearities fii=1,2 satisfy the semipositone condition:(H0)fi∈C1,e×ℝ+6,ℝ, and there exists M>0 such that(2)fit,x1,x2,x3,y1,y2,y3≥−M,fort,x1,x2,x3,y1,y2,y3∈1,e×ℝ+6,where ℝ+=0,+∞andℝ=−∞,+∞,i=1,2.Fractional problems arise in many applications in aerodynamics, signal and image processing, biophysics, blood flow phenomena, etc. (see, for example, [1–37] and the references therein). In [1], the authors used the Krasnoselskii–Zabreiko fixed-point theorem to study the existence of positive solutions for Riemann–Liouville-type fractional boundary value problems:(3)D0+αD0+αu=ft,u,u′,−D0+αu,t∈0,1,u0=u′0=u′1=D0+αu0=D0+α+1u0=D0+α+1u1=0,where f has a semipositone nonlinearity, i.e., it is bounded below and can be sign-changing, and in [2], the authors used the Guo–Krasnoselskii’s fixed-point theorem to investigate the existence of positive solutions for fractional multipoint boundary value problems:(4)D0+αut+ft,ut=0,0<t<1,u0=u′0=⋯=un−20=0,D0+pu1=∑i=1maiD0+quξi,where the nonlinear f may have singularities on the time and the space variables.Coupled systems of fractional-order differential equations have also been investigated by many authors (see [7–22, 25–36] and the references therein). In [7], the authors used coincidence degree theory to establish an existence result for a coupled system of nonlinear fractional differential equations:(5)D0+αut=ft,vt,D0+β−2vt,D0+β−1vt,0<t<1,D0+βvt=gt,ut,D0+α−2ut,D0+α−1ut,0<t<1,with the multipoint boundary conditions:(6)u0=0,D0+1α−1u0=D0+1α−1uη,u1=∑i=1m1αiuηi,v0=0,D0+β−1v0=D0+β−1uξ,v1=∑i=1m2βivξi.There are a few results in the literature on Hadamard-type fractional differential equations (see [23–36]). In [23], the authors used fixed-point methods to obtain some existence theorems for Hadamard-type fractional boundary value problems:(7)−Dαut=ft,ut,t∈1,e,u1=δu1=δue=0,where f grows superlinearly and sublinearly at infinity and can be sign-changing. The authors in [24] studied positive solutions for the p-Laplacian Hadamard fractional differential equations with integral boundary value problems:(8)DβφpDαut=ft,ut,1<t<e,u1=u′1=u′e=0,Dαu1=0,φpDαue=μ∫1eφpDαutdtt,where f has a p−1-superlinear and p−1-sublinear nonlinearity. In [25], the authors studied fractional impulsive Cauchy problems:(9)DH1+αut=ft,ut,α∈0,1,t∈1,e\t1,t2,…,tm,Δuti=HJ1+1−αuti+−HJ1+1−αuti−=pi,pi∈ℝ,i=1,2,…,m,JH1+1−αu1+=u0,u0∈ℝ,where DH1+α denotes the left-sided Hadamard fractional derivative and the nonlinearity f satisfies a Lipschitz condition, and in [26], the authors studied positive solutions for the system of Hadamard fractional differential equations involving coupled integral boundary conditions:(10)Dβut+f1t,ut,vt=0,1<t<e,Dβvt+f2t,ut,vt=0,1<t<e,u1=v1=u′1=v′1=0,ue=∫1ehsvsdss,ve=∫1egsusdss,where the nonlinearities fii=1,2 grow superlinearly and sublinearly.Motivated by the above, in this paper, we study the existence of positive solutions for the system of Hadamard-type fractional differential equation (1). The nonlinearities fii=1,2 can be sign-changing and can depend on the unknown functions and their derivatives. We use some appropriate nonnegative matrices to characterize the coupling behavior for the nonlinearities fii=1,2, and the nonlinearities fii=1,2 can grow superlinearly and sublinearly.


## 2. Preliminaries

For details about Hadamard fractional calculus, see the book [37].Definition 1.
The Hadamard derivative of fractional orderq for a function g:1,∞⟶ℝ is defined as(11)Dqgt=1Γn−qtddtn∫1tlogt−logsn−q−1gsdss,n−1<q<n,where n=q+1, q denotes the integer part of the real number q, and log⋅=loge⋅.
Now, we establish Green’s functions associated with the system of (1). In order to do this, we first consider the following auxiliary problem:(12)DαDαxt=ft,xt,δxt,−Dαxt,x1=δx1=δxe=0,Dαx1=δDαx1=δDαxe=0,where αandδ are as in (1), and f∈C1,e×ℝ+3,ℝ satisfies the following semipositone condition:
(H0)′ There existsM>0 such that ft,x1,x2,x3≥−M, for t,x1,x2,x3∈1,e×ℝ+3.
Let−Dαxt=yt for t∈1,e. Then, from Lemma 2.3 of [36], we have(13)xt=c1logtα−1+c2logtα−2+c3logtα−3−1Γα∫1tlogt−logsα−1ysdss,where ci∈ℝ,i=1,2,3. This, together with the boundary conditions x1=δx1=δxe=0, implies that(14)xt=1Γα∫1elogtα−11−logsα−2ysdss−1Γα∫1tlogt−logsα−1ysdss=∫1eH1t,sysdss,where(15)H1t,s=1Γαlogtα−11−logsα−2−logt−logsα−1,1≤s≤t≤e,logtα−11−logsα−2,1≤t≤s≤e.
Let(16)H2t,s=δH1t,s=α−1Γαlogtα−21−logsα−2−logt−logsα−2,1≤s≤t≤e,logtα−21−logsα−2,1≤t≤s≤e.
Then, we have(17)δxt=δ∫1eH1t,sysdss=∫1eH2t,sysdss.
Therefore, substitutingxandδx into problem (12), we have(18)−Dαyt=ft,∫1eH1t,sysdss,∫1eH2t,sysdss,yt,y1=δy1=δye=0.
From (13) and (14), we see that problem (18) is equivalent to the following Hammerstein-type integral equation:(19)yt=∫1eH1t,sfs,∫1eH1s,τyτdττ,∫1eH2s,τyτdττ,ysdss.Lemma 1.
Letφt=logt1−logtα−2, t∈1,e. Then, the functions Hii=1,2 have the following properties:(i)
Hi∈C1,e2,ℝ+, i=1,2,(ii)
For allt,s∈1,e, the following inequalities hold:(20)logtα−1φs≤ΓαH1t,s≤φs,(21)α−1α−2logtα−21−logtφs≤ΓαH2t,s≤α−1logtα−3φs.Lemma1 (ii) is a direct result of [1, Lemma 4], so we omit the proof.Lemma 2.
Letκ1=αΓα−1/Γ2α, κ2=1/αα−1Γα, κ3=α−1α−2Γα/Γ2α, and κ4=Γα−1/Γ2α−2. Then,(22)κ1φs≤∫1eH1t,sφtdtt≤κ2φs,fors∈1,e,κ3φs≤∫1eH2t,sφtdtt≤κ4φs,fors∈1,e.These inequalities can be deduced from Lemma1 (ii), so we omit the proof.Letwt=M∫1eH1t,sds/s, where M is defined in (H0)′. Then, w satisfies the following boundary value problem:(23)−Dαwt=M,w1=δw1=δwe=0.In what follows, we consider the following boundary value problems:(24)−Dαyt=Ft,∫1eH1t,sys−wsdss,∫1eH2t,sys−wsdss,yt−wt,y1=δy1=δye=0,where(25)Ft,x1,x2,x3=ft,x1,x2,x3+M,t∈1,e,forxi≥0,i=1,2,3,ft,0,0,0+M,t∈1,e, for else cases.Then,F∈C1,e×ℝ+3,ℝ+, and we have the following lemma.Lemma 3.
(i) Ify∗ is a solution of (24) with y∗t≥wt,t∈1,e, then y∗−w is a positive solution of (18)
(ii)
Ify∗ is a solution of (18), then y∗+w is a solution of (24)Proof.
Lety∗ be a solution of (24) with y∗t≥wt,t∈1,e. Then, we substitute y∗−w into problem (18) and obtain(26)−Dαy∗−wt=ft,∫1eH1t,sy∗−wsdss,∫1eH2t,sy∗−wsdss,y∗−wt,y∗−w1=δy∗−w1=δy∗−we=0.
Considering (23), we have(27)−Dαy∗t+Dαwt+M=ft,∫1eH1t,sy∗−wsdss,∫1eH2t,sy∗−wsdss,y∗−wt+M,−Dαy∗t=Ft,∫1eH1t,sy∗−wsdss,∫1eH2t,sy∗−wsdss,y∗−wt.
This, together withy∗1=δy∗1=δy∗e=0, implies that (26) holds, and y∗−w is a positive solution of (18).
Substitutey∗+w into (24), and we get(28)−Dαy∗+wt=Ft,∫1eH1t,sy∗sdss,∫1eH2t,sy∗sdss,y∗t,y∗+w1=δy∗+w1=δy∗+we=0.
Considering (23), we have(29)−Dαy∗t−Dαwt=ft,∫1eH1t,sy∗sdss,∫1eH2t,sy∗sdss,y∗t+M,−Dαy∗t=ft,∫1eH1t,sy∗sdss,∫1eH2t,sy∗sdss,y∗t.
This, together withy∗1=δy∗1=δy∗e=0, implies y∗+w is a solution of (24). This completes the proof.
From (13) and (14), we also find that problem (24) is equivalent to the following Hammerstein-type integral equation:(30)yt=∫1eH1t,sFs,∫1eH1s,τyτ−wτdττ,∫1eH2s,τyτ−wτdττ,ys−wsdss.
LetE≔C1,e,y≔maxt∈1,eyt,andP≔y∈E:yt≥0,∀t∈1,e. Then, E,⋅ is a real Banach space and P a cone on E. Then, E×E is also a Banach space, equipped with the norm: u,v=maxu,v, where uandv are norms in E. From (30), we define an operator A:P⟶P as follows:(31)Ayt=∫1eH1t,sFs,∫1eH1s,τyτ−wτdττ,∫1eH2s,τyτ−wτdττ,ys−wsdss,y,w∈P.
Note that our functionsH1,H2,F,andw are continuous, so the operator A is a completely continuous operator. Moreover, if there is a y∈P, a fixed point of A, and yt≥wt,t∈1,e, then we have that yt−wt is a positive solution of (18). From (12) and (18), we have xt=∫1eH1t,sy−wsds/s is a positive solution of (12). Therefore, in what follows, we study the existence of fixed points of the operator A, which are greater than w.Lemma 4.
LetP0=y∈P:yt≥logtα−1y,∀t∈1,e. Then, AP⊂P0.This is a direct result from (20) in Lemma 1, so we omit its proof.Note that our aim is to find a fixed point ofA, which is greater than w. If there is a y∈P such that Ay=y, and yt≥wt,t∈1,e, then if y≥M/α−1Γα, Lemma 4 enables us to obtain(32)yt−wt≥logtα−1y−MΓα∫1elogtα−11−logsα−2dss≥0.Consequently, ifA has fixed point y∈P with y≥M/α−1Γα, then yt≥wt,t∈1,e, and xt=∫1eH1t,sy−wsds/s is a positive solution of (12).Lemma 5 (see [38]).
LetE be a real Banach space and P a cone on E. Suppose that Ω⊂E is a bounded open set and that A:Ω¯∩P⟶P is a continuous compact operator. If there exists a ω0∈P\0 such that(33)ω−Aω≠λω0,∀λ≥0,ω∈∂Ω∩P,then iA,Ω∩P,P=0, where i denotes the fixed-point index on P.Lemma 6 (see [38]).
LetE be a real Banach space and P a cone on E. Suppose that Ω⊂E is a bounded open set with 0∈Ω and that A:Ω¯∩P⟶P is a continuous compact operator. If(34)ω−λAω≠0,∀λ∈0,1,ω∈∂Ω∩P,then iA,Ω∩P,P=1.


## 3. Positive Solutions for (1)

From Section2, let −Dαξt=ut,−Dαηt=vt, and then, we should consider the following system of Hadamard-type fractional boundary value problems:(35)−Dαut=F1t,∫1eH1t,sus−wsdss,∫1eH2t,sus−wsdss,ut−wt,∫1eH1t,svs−wsdss,∫1eH2t,svs−wsdss,vt−wt−Dαvt=F2t,∫1eH1t,sus−wsdss,∫1eH2t,sus−wsdss,ut−wt,∫1eH1t,svs−wsdss,∫1eH2t,svs−wsdss,vt−wtu1=δu1=δue=v1=δv1=δve=0,where(36)Fit,x1,x2,x3,y1,y2,y3=fit,x1,x2,x3,y1,y2,y3+M,t∈1,e,forxj,yj≥0,i=1,2,j=1,2,3,fit,0,0,0,0,0,0+M,t∈1,e,i=1,2,for  else cases.We define some operators as follows:(37)B1u,vt=∫1eH1t,sF1s,∫1eH1s,τuτ−wτdττ,∫1eH2s,τuτ−wτdττ,us−ws,∫1eH1s,τvτ−wτdττ,∫1eH2s,τvτ−wτdττ,vs−wsdss,B2u,vt=∫1eH1t,sF2s,∫1eH1s,τuτ−wτdττ,∫1eH2s,τuτ−wτdττ,us−ws,∫1eH1s,τvτ−wτdττ,∫1eH2s,τvτ−wτdττ,vs−wsdss,Bu,vt=B1,B2u,vt,foru,v∈P,t∈1,e.Then, if there is au,v∈P such that Bu,v=u,v with u,v≥M/α−1Γα, i.e., (35) has a solution u,v with ut,vt≥wt,t∈1,e, and thus, ξt,ηt=∫1eH1t,sus−wsds/s,∫1eH1t,svs−wsds/s is a positive solution for (1) and t∈1,e.Now, we list our assumptions for the functionsFii=1,2 as follows:(H1) There existaji,bji≥0 and lj>0j=1,2,i=1,2,3 such that(38)F1t,x1,x2,x3,y1,y2,y3F2t,x1,x2,x3,y1,y2,y3≥a11x1+a12x2+a13x3+b11y1+b12y2+b13y3−l1a21x1+a22x2+a23x3+b21y1+b22y2+b23y3−l2,∀t,x1,x2,x3,y1,y2,y3∈1,e×ℝ+6,and the coefficientsaji,bjij=1,2,i=1,2,3 satisfy(39)κ1a11κ1+a12κ3+a13<1,κ1b21κ1+b22κ3+b23<1,detκ1b11κ1+b12κ3+b13κ1a11κ1+a12κ3+a13−1κ1b21κ1+b22κ3+b23−1κ1a21κ1+a22κ3+a23>0.(H2) There existsQit in 1,e such that(40)∫1eφtQitdtt<Mα−1,Fit,x1,x2,x3,y1,y2,y3≤Qit,∀t,x1,x2,x3,y1,y2,y3∈1,e×0,Mα−1Γα6,i=1,2.(H3) There exista˜ji,b˜ji≥0 and l˜j>0j=1,2,i=1,2,3 such that(41)F1t,x1,x2,x3,y1,y2,y3F2t,x1,x2,x3,y1,y2,y3≤a˜11x1+a˜12x2+a˜13x3+b˜11y1+b˜12y2+b˜13y3+l˜1a˜21x1+a˜22x2+a˜23x3+b˜21y1+b˜22y2+b˜23y3+l˜2,∀t,x1,x2,x3,y1,y2,y3∈1,e×ℝ+6,and the coefficientsa˜ji,b˜jij=1,2,i=1,2,3 satisfy(42)κ2κ2a˜11+κ4a˜12+a˜13<1,κ2κ2b˜21+κ4b˜22+b˜23<1,det1−κ2κ2a˜11+κ4a˜12+a˜13−κ2κ2b˜11+κ4b˜12+b˜13−κ2κ2a˜21+κ4a˜22+a˜231−κ2κ2b˜21+κ4b˜22+b˜23>0.(H4) There existsQ˜it in 1,e and t0∈1,e such that(43)∫1eφtQ˜itdtt>Mα−1logt0α−1,Fit,x1,x2,x3,y1,y2,y3≥Q˜it,∀t,x1,x2,x3,y1,y2,y3∈1,e×0,Mα−1Γα6,i=1,2.LetBρ=u∈P:u<ρ for ρ>0. Then, we have B¯ρ=u∈P:u≤ρ and ∂Bρ=u∈P:u=ρ.Theorem 1.
Suppose that (H0) to (H2) hold. Then, (1) has at least one positive solution.Proof.
For convenience, letℳ=M/α−1Γα. We next show that there is a R1>ℳ such that(44)u,v≠Bu,v+λϕ0,ϕ0,foru,v∈∂BR1∩P×P,∀λ≥0,where ϕ0∈P0 is a fixed element. Suppose not, i.e., suppose there exist u,v∈∂BR1∩P×P,λ0≥0 such that(45)u,v=Bu,v+λ0ϕ0,ϕ0.
Combining this and Lemma4, we have(46)u,v∈P0.
Using (45), we obtain(47)ut=B1u,vt+λ0ϕt≥B1u,vt,vt=B2u,vt+λ0ϕt≥B2u,vt,t∈1,e.
From (H1), we have(48)utvt≥∫1eH1t,sa11∫1eH1s,τuτ−wτdττ+a12∫1eH2s,τuτ−wτdττ+a13us−ws+b11∫1eH1s,τvτ−wτdττ+b12∫1eH2s,τvτ−wτdττ+b13vs−ws−l1dss∫1eH1t,sa21∫1eH1s,τuτ−wτdττ+a22∫1eH2s,τuτ−wτdττ+a23us−ws+b21∫1eH1s,τvτ−wτdττ+b22∫1eH2s,τvτ−wτdττ+b23vs−ws−l2dss.
Multiplying byφt and integrating from 1 to e, Lemma 2 enables us to obtain(49)∫1eutφtdtt∫1evtφtdtt≥κ1a11κ1+a12κ3+a13∫1eut−wtφtdtt+κ1b11κ1+b12κ3+b13∫1evt−wtφtdtt−l1Γα−1α2α−1Γ2ακ1a21κ1+a22κ3+a23∫1eut−wtφtdtt+κ1b21κ1+b22κ3+b23∫1evt−wtφtdtt−l2Γα−1α2α−1Γ2α.
This implies that(50)κ1b11κ1+b12κ3+b13κ1a11κ1+a12κ3+a13−1κ1b21κ1+b22κ3+b23−1κ1a21κ1+a22κ3+a23∫1evtφtdtt∫1eutφtdtt≤κ1κ1a11+b11+κ3a12+b12+a13+b13∫1ewtφtdtt+l1Γα−1α2α−1Γ2ακ1κ1a21+b21+κ3a22+b22+a23+b23∫1ewtφtdtt+l2Γα−1α2α−1Γ2α≔N1N2.
Solving this matrix inequality, we have(51)∫1evtφtdtt∫1eutφtdtt≤κ1a21κ1+a22κ3+a231−κ1a11κ1+a12κ3+a131−κ1b21κ1+b22κ3+b23κ1b11κ1+b12κ3+b13detκ1b11κ1+b12κ3+b13κ1a11κ1+a12κ3+a13−1κ1b21κ1+b22κ3+b23−1κ1a21κ1+a22κ3+a23N1N2.
Consequently, there existN3,N4>0 such that(52)∫1evtφtdtt∫1eutφtdtt≤N3N4.
Note (46) and we have(53)vu≤N3Γα+1Γα−1Γ2αN4Γα+1Γα−1Γ2α.
ChooseR1>ℳ,N3Γα+1Γα−1/Γ2α,N4Γα+1Γα−1/Γ2α. Then, (44) holds. Lemma 5 gives(54)iB,BR1∩P×P,P×P=0.
Next, we prove that(55)u,v≠λBu,v,foru,v∈∂Bℳ∩P×P,∀λ∈0,1.
Suppose not. Then, there existu,v∈∂Bℳ∩P×P,λ1∈0,1 such that(56)u,v=λ1Bu,v.
This implies that(57)u≤B1u,v,v≤B2u,v.
However, from (H2), we have(58)B1u,vt=∫1eH1t,sF1s,∫1eH1s,τuτ−wτdττ,∫1eH2s,τuτ−wτdττ,us−ws,∫1eH1s,τvτ−wτdττ,∫1eH2s,τvτ−wτdττ,vs−wsdss≤1Γα∫1eφsQ1sdss<Mα−1Γα.
Note that from (H2), we haveu=M/α−1Γα. Hence, we obtain B1u,v<u. Similarly, B2u,v<v. This is a contradiction. As a result, (55) holds. From Lemma 6, we have(59)iB,Bℳ∩P×P,P×P=1.
From (54) and (59), we have(60)iB,BR1\B¯ℳ∩P×P,P×P=iB,BR1∩P×P,P×P−iB,Bℳ∩P×P,P×P=0−1=−1.
Therefore, the operatorB has at least one fixed point on BR1\B¯ℳ∩P×P. Thus, (1) has at least one positive solution. This completes the proof.Theorem 2.
Suppose that (H0), (H3) and (H4) hold. Then, (1) has at least one positive solution.Proof.
We first show that there is aR2>M/α−1Γα such that(61)u,v≠λBu,v,foru,v∈∂BR2∩P×P,∀λ∈0,1.
If this claim is false, there existu,v∈∂BR2∩P×P,λ2∈0,1 such that(62)u,v=λ2Bu,v.
Lemma4 implies that(63)u,v∈P0.
Moreover, from (H3), we have(64)utvt≤∫1eH1t,sa˜11∫1eH1s,τuτ−wτdττ+a˜12∫1eH2s,τuτ−wτdττ+a˜13us−ws+b˜11∫1eH1s,τvτ−wτdττ+b˜12∫1eH2s,τvτ−wτdττ+b˜13vs−ws+l˜1dss∫1eH1t,sa˜21∫1eH1s,τuτ−wτdττ+a˜22∫1eH2s,τuτ−wτdττ+a˜23us−ws+b˜21∫1eH1s,τvτ−wτdττ+b˜22∫1eH2s,τvτ−wτdττ+b˜23vs−ws+l˜2dss.
Multiply byφt on both sides of the above and integrate over 1,e and use Lemma 2 to obtain(65)∫1eutφtdtt∫1evtφtdtt≤κ2κ2a˜11+κ4a˜12+a˜13∫1eut−wtφtdtt+κ2κ2b˜11+κ4b˜12+b˜13∫1evt−wtφtdtt+l˜1Γα−1α2α−1Γ2ακ2κ2a˜21+κ4a˜22+a˜23∫1eut−wtφtdtt+κ2κ2b˜21+κ4b˜22+b˜23∫1evt−wtφtdtt+l˜2Γα−1α2α−1Γ2α.
This implies that(66)1−κ2κ2a˜11+κ4a˜12+a˜13−κ2κ2b˜11+κ4b˜12+b˜13−κ2κ2a˜21+κ4a˜22+a˜231−κ2κ2b˜21+κ4b˜22+b˜23∫1eutφtdtt∫1evtφtdtt≤l˜1Γα−1α2α−1Γ2αl˜2Γα−1α2α−1Γ2α.
Solving this matrix inequality, we obtain(67)∫1eutφtdtt∫1evtφtdtt≤1−κ2κ2b˜21+κ4b˜22+b˜23κ2κ2b˜11+κ4b˜12+b˜13κ2κ2a˜21+κ4a˜22+a˜231−κ2κ2a˜11+κ4a˜12+a˜13l˜1Γα−1/α2α−1Γ2αl˜2Γα−1/α2α−1Γ2αdet1−κ2κ2a˜11+κ4a˜12+a˜13−κ2κ2b˜11+κ4b˜12+b˜13−κ2κ2a˜21+κ4a˜22+a˜231−κ2κ2b˜21+κ4b˜22+b˜23.
Note thatu,v∈P0, there exist N5,N6>0 such that(68)uv≤N5Γα+1Γα−1Γ2αN6Γα+1Γα−1Γ2α.
ChooseR2>maxℳ,N5Γα+1Γα−1/Γ2α,N6Γα+1Γα−1/Γ2α. Then, (61) holds. From Lemma 6, we have(69)iB,BR2∩P×P,P×P=1.
Next, we prove that(70)u,v≠Bu,v+λϕ1,ϕ1,foru,v∈∂Bℳ∩P×P,∀λ≥0,where ϕ1∈P is a fixed element. Suppose not. Then, there exist u,v∈∂Bℳ∩P×P,λ3≥0 such that(71)u,v=Bu,v+λ3ϕ1,ϕ1.
This implies that(72)u≥B1u,v,v≥B2u,v.
However, from (H4), we have(73)B1u,vt0=∫1eH1t0,sF1s,∫1eH1s,τuτ−wτdττ,∫1eH2s,τuτ−wτdττ,us−ws,∫1eH1s,τvτ−wτdττ,∫1eH2s,τvτ−wτdττ,vs−wsdss≥logt0α−1Γα∫1eφsQ˜1sdss>logt0α−1ΓαMα−1logt0α−1.
Note that from (H2), we haveu=M/α−1Γα. Hence, we obtain B1u,v≥B1u,vt0>u. Similarly, B2u,v>v. This has a contradiction, and thus, (70) holds. From Lemma 5, we obtain(74)iB,Bℳ∩P×P,P×P=0.
From (69) and (74), we have(75)iB,BR2\B¯ℳ∩P×P,P×P=iB,BR2∩P×P,P×P−iB,Bℳ∩P×P,P×P=1−0=1.
Therefore, the operatorB has at least one fixed point on BR2\B¯ℳ∩P×P. Thus, (1) has at least one positive solution. This completes the proof.Letα=2.5. Then, we have κ1=0.09,κ2=0.2,κ3=0.004,κ4=0.44, M/α−1Γα=0.502M, and ∫1eφtdt/t=0.27.Example 1.
Leta11=50,a12=1000,a13=2, b21=52,b22=1002,b23=2.2, a21=140,a22=2900,a23=15, b11=130,b12=2800,b13=12, and(76)F1t,x1,x2,x3,y1,y2,y3=2005−γ1M1−γ150x1+1000x2+2x3+130y1+2800y2+12y3γ1,F2t,x1,x2,x3,y1,y2,y3=2064−γ1M1−γ1140x1+2900x2+15x3+52y1+1002y2+2.2y3γ2,for t,x1,x2,x3,y1,y2,y3∈1,e×ℝ+6, where γ1,γ2>1. Then, we have(77)κ1a11κ1+a12κ3+a13=0.945<1,κ1b21κ1+b22κ3+b23=0.98<1,κ1b11κ1+b12κ3+b13κ1a11κ1+a12κ3+a13−1κ1b21κ1+b22κ3+b23−1κ1a21κ1+a22κ3+a23=3.141−0.055−0.023.528>0.
Moreover, note that whent,x1,x2,x3,y1,y2,y3∈1,e×0,M/α−1Γα6, we have Fi≤M, i=1,2. Consequently, if we choose Qit≡M,t∈1,e,i=1,2, then ∫1eφtQitdt/t<M/α−1.Note that we have(78)liminfa11x1+a12x2+a13x3+b11y1+b12y2+b13y3⟶+∞F1t,x1,x2,x3,y1,y2,y3a11x1+a12x2+a13x3+b11y1+b12y2+b13y3=liminfa11x1+a12x2+a13x3+b11y1+b12y2+b13y3⟶+∞2005−γ1M1−γ150x1+1000x2+2x3+130y1+2800y2+12y3γ1a11x1+a12x2+a13x3+b11y1+b12y2+b13y3=+∞,liminfa21x1+a22x2+a23x3+b21y1+b22y2+b23y3⟶+∞F2t,x1,x2,x3,y1,y2,y3a21x1+a22x2+a23x3+b21y1+b22y2+b23y3=liminfa21x1+a22x2+a23x3+b21y1+b22y2+b23y3⟶+∞2064−γ1M1−γ1140x1+2900x2+15x3+52y1+1002y2+2.2y3γ2a21x1+a22x2+a23x3+b21y1+b22y2+b23y3=+∞.As a result, (H1) and (H2) hold.Example 2.
Lett0=e,a˜11=0.075,a˜12=0.034,a˜13=0.015, b˜21=0.08,b˜22=0.035,b˜23=0.016, b˜11=0.09,b˜12=0.04,b˜13=0.02, a˜21=0.08,a˜22=0.05,a˜23=0.018, and(79)F1t,x1,x2,x3,y1,y2,y3=2.7Me0.14Mexp−0.075x1+0.034x2+0.015x3+0.09y1+0.04y2+0.02y3,F2t,x1,x2,x3,y1,y2,y3=3Me0.14Mexp−0.08x1+0.05x2+0.018x3+0.08y1+0.035y2+0.016y3,for t,x1,x2,x3,y1,y2,y3∈1,e×ℝ+6. Then, we have logt0α−1=1, and(80)κ2κ2a˜11+κ4a˜12+a˜13=0.009<1,κ2κ2b˜21+κ4b˜22+b˜23=0.009<1,1−κ2κ2a˜11+κ4a˜12+a˜13−κ2κ2b˜11+κ4b˜12+b˜13−κ2κ2a˜21+κ4a˜22+a˜231−κ2κ2b˜21+κ4b˜22+b˜23=0.991−0.011−0.01120.991>0.
Moreover, whent,x1,x2,x3,y1,y2,y3∈1,e×0,M/α−1Γα6, we obtain F1≥2.7MandF2≥3M, and if we choose Q˜1t≡2.7M and Q˜2t≡3M, t∈1,e, we also have ∫1eφtQ˜itdt/t>M/α−1logt0α−1.Note that we have(81)limsupa˜11x1+a˜12x2+a˜13x3+b˜11y1+b˜12y2+b˜13y3⟶+∞F1t,x1,x2,x3,y1,y2,y3a˜11x1+a˜12x2+a˜13x3+b˜11y1+b˜12y2+b˜13y3=limsupa˜11x1+a˜12x2+a˜13x3+a˜11y1+b˜12y2+b˜13y3⟶+∞2.7Me0.14Mexp−0.075x1+0.034x2+0.015x3+0.09y1+0.04y2+0.02y3a˜11x1+a˜12x2+a˜13x3+b˜11y1+b˜12y2+b˜13y3=0,limsupa˜21x1+a˜22x2+a˜23x3+b˜21y1+b˜22y2+b˜23y3⟶+∞F2t,x1,x2,x3,y1,y2,y3a˜21x1+a˜22x2+a˜23x3+b˜21y1+b˜22y2+b˜23y3=limsupa˜21x1+a˜22x2+a˜23x3+b˜21y1+b˜22y2+b˜23y3⟶+∞3Me0.14Mexp−0.08x1+0.05x2+0.018x3+0.08y1+0.035y2+0.016y3a˜21x1+a˜22x2+a˜23x3+b˜21y1+b˜22y2+b˜23y3=0.As a result, (H3) and (H4) hold.



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*Source: 9742418-2020-01-08.xml*