# Positive Solutions for a System of Fractional Integral Boundary Value Problems Involving Hadamard-Type Fractional Derivatives

Authors: Haiyan Zhang; Yaohong Li; Jiafa Xu
Journal: Complexity
 (2019)

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## Abstract

In this paper, we use fixed-point index to study the existence of positive solutions for a system of Hadamard fractional integral boundary value problems involving nonnegative nonlinearities. By virtue of integral-type Jensen inequalities, some appropriate concave and convex functions are used to depict the coupling behaviors for our nonlinearitiesfii=1,2.


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## Body

## 1. Introduction

In this paper, we study the existence of positive solutions for the system of Hadamard fractional integral boundary value problems:(1)−DαHut=f1t,ut,vt,t∈1,e,−DαHvt=f2t,ut,vt,t∈1,e,uj1=vj1=0,ue=∫1ehtutdtt,ve=∫1ehtvtdtt,where α∈n−1,n is a real number with n≥3, j=0,1,2,…,n−2, and DαH is the Hadamard fractional derivative. The nonlinearities fi∈C1,e×ℝ+×ℝ+,ℝ+,ℝ+=0,+∞. Moreover, the function h on 1,e satisfies the condition:(H0)h≥0 with ∫1ehtlogtα−1dt/t∈0,1.In recent years, the fractional calculus and fractional differential equations are of importance in mathematics, physics, electroanalytical chemistry, capacitor theory, electrical circuits, biology, control theory, and fluid dynamics [1–20]. For example, in [1], the author considered the fractional n−1,1-type conjugate boundary value problems:(2)D0+αut+λft,ut=0,0<t<1,uj0=0,u1=0,0≤j≤n−2,where α∈n−1,n, n≥3, and D0+α is the Riemann–Liouville’s fractional derivative. By means of Leray–Schauder type and Krasnosel’skii’s fixed-point theorems, the author derived an interval of parameter λ such that (2) has multiple positive solutions when any λ lies in the interval.On the other hand, we note that coupled systems of fractional differential equations have also been investigated by many authors, see [21–32]. For example, in [21], the authors used a fixed-point theorem of increasing φ-h,r-concave operators to establish the existence and uniqueness of solutions for a system of four-point boundary value problems involving Hadamard fractional derivatives:(3)DαHut+ft,vt=lf,t∈1,e,DβHvt+gt,ut=lg,t∈1,e,uj1=vj1=0,0≤j≤n−2,ue=avξ,ve=buη,ξ,η∈1,e,where f,g∈C1,e×ℝ,ℝ and lfandlg are two positive parameters. In [22], the authors established positive solutions for the coupled Hadamard fractional integral boundary value problems:(4)DαHut+λft,ut,vt=0,t∈1,e,λ>0,DβHvt+λgt,ut,vt=0,t∈1,e,λ>0,uj1=vj1=0,0≤j≤n−2,ue=μ∫1evsdss,ve=ν∫1eusdss,where the nonlinearities fandg satisfy either of the following conditions:HYang1: there exists θ1,θ2⊂1,e such that liminfu⟶+∞mint∈θ1,θ2ft,u,v/u=+∞ and liminfv⟶+∞mint∈θ1,θ2gt,u,v/v=+∞.HYang2: there exists θ1,θ2⊂1,e such that liminfv⟶+∞mint∈θ1,θ2ft,u,v/v=+∞ and liminfu⟶+∞mint∈θ1,θ2gt,u,v/u=+∞.Inspired by the aforementioned works, in this paper, we use the fixed-point index to consider the existence of positive solutions for system (1) of fractional integral boundary value problems involving Hadamard-type fractional derivatives. Based on integral-type Jensen inequalities, some appropriate concave and convex functions are used to depict the coupling behaviors for the nonlinearities fii=1,2. Moreover, our a priori estimates for positive solutions are derived by developing some appropriate nonnegative matrices when fii=1,2 grow sublinearly at ∞. These conditions here are different from that in HYang1 and HYang2.


## 2. Preliminaries

In this paper, we only provide some necessary definitions and lemmas for the Hadamard fractional derivative. For more details about Hadamard fractional calculus, see the book [33].Definition 1.
The Hadamard derivative of fractional orderq for a function g:1,∞⟶ℝ is defined as(5)DqHgt=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⋅.Definition 2.
The Hadamard fractional integral of orderq for a function g is defined as(6)IqHgt=1Γq∫1tlogt−logsq−1gsdss,q>0.Lemma 1.
Letq>0 and u∈C1,∞∩​L11,∞. Then, the Hadamard fractional differential equation DqHut=0 has the solution(7)ut=c1logtq−1+c2logtq−2+⋯+cnlogtq−n,where ci∈ℝ, n−1<q<n,n=q+1, and i=1,2,…,n.Lemma 2.
Letq>0 and u∈C1,∞∩​L11,∞. Then, we have the following formula:(8)IqHDqHut=ut+c1logtq−1+c2logtq−2+⋯+cnlogtq−n,where ci and n are as in Lemma 1 and i=1,2,…,n.Lemma 3.
Suppose that (H0) holds. Letf∈C1,e. Then, the boundary value problems(9)−DαHut=ft,t∈1,e,uj1=0,ue=∫1ehtutdtt,has a unique solution(10)ut=∫1eGt,sfsdss,where(11)Gt,s=G1t,s+logtα−11−∫1ehtlogtα−1dt/t∫1ehtG1t,sdtt,G1t,s=1Γαlogtα−11−logsα−1−logt−logsα−1,1≤s≤t≤e,logtα−11−logsα−1,1≤t≤s≤e.Proof.
Using Lemma 2, we have(12)ut=c1logtα−1+c2logtα−2+⋯+cnlogtα−n−1Γα∫1tlogt−logsα−1fsdss,where ci∈ℝ, i=1,2,…,n. By uj1=0,j=0,1,…,n−2, we have ci=0,i=2,3,…,n. Hence,(13)ut=c1logtα−1−1Γα∫1tlogt−logsα−1fsdss.
Then, we knowue=c1−1/Γα∫1e1−logsα−1fsds/s. Using the condition ue=∫​1ehtutdt/t, we have(14)c1−1Γα∫1e1−logsα−1fsdss=c1∫1ehtlogtα−1dtt−1Γα∫1eht∫1tlogt−logsα−1fsdssdtt.
Then, (H0) implies that(15)c1=1Γα1−∫1ehtlogtα−1dt/t∫1e1−logsα−1fsdss−1Γα1−∫1ehtlogtα−1dt/t∫1eht∫1tlogt−logsα−1fsdssdtt=1Γα1−∫1ehtlogtα−1dt/t∫1e1−logsα−1fsdss−1Γα1−∫1ehtlogtα−1dt/t∫1e1−logsα−1fsdss.
As a result, we have(16)ut=logtα−1Γα1−∫1ehtlogtα−1dt/t∫1e1−logsα−1fsdss−logtα−1Γα1−∫1ehtlogtα−1dt/t∫1eht∫selogt−logsα−1fsdttdss−1Γα∫1tlogt−logsα−1fsdss=logtα−1Γα1−∫1ehtlogtα−1dt/t∫1e1−logsα−1fsdss+1Γα∫1elogtα−11−logsα−1fsdss−logtα−1Γα1−∫1ehtlogtα−1dt/t∫1eht∫1elogt−logsα−1fsdttdss−1Γα∫1tlogt−logsα−1fsdss−1Γα∫1elogtα−11−logsα−1fsdss=∫1eG1t,sfsdss+∫1ehtlogtα−1dt/tΓα1−∫1ehtlogtα−1dt/t∫1elogtα−11−logsα−1fsdss−logtα−1Γα1−∫1ehtlogtα−1dt/t∫1eht∫selogt−logsα−1fsdttdss=∫1eG1t,sfsdss+logtα−1∫1ehtlogtα−1dt/t∫1e1−logsα−1fsds/s−∫1eht∫selogt−logsα−1fsdt/tds/sΓα1−∫1ehtlogtα−1dt/t=∫1eG1t,sfsdss+logtα−11−∫1ehtlogtα−1dt/t∫1e∫1ehtG1t,sdttfsdss=∫1eGt,sfsdss.
This completes the proof.In what follows, we study some useful inequalities for Green’s functions in (11). We first provide a result in [1].Letht∈C0,1, and then the Riemann–Liouville boundary-value problem(17)D0+αut+ht=0,0<t<1,2≤n−1<α≤n,uj0=u1=0,0≤j≤n−2,has a unique solution ut=∫01Ht,shsds, where(18)Ht,s=1Γαtα−11−sα−1−t−sα−1,0≤s≤t≤1,tα−11−sα−1,0≤t≤s≤1.Moreover, Green’s functionH satisfies the inequalities:(19)Γαktqs≤Ht,s≤α−1qs,fort,s∈0,1,where kt=tα−11−t/Γαandqs=s1−sα−1/Γα.ComparingG1 with H, using logtand logs to replace tands, from (11) and (19), we obtain the function G1 satisfies the inequalities:(20)logtα−11−logtlogs1−logsα−1Γα≤G1t,s≤α−1logs1−logsα−1Γα,t,s∈1,e.This, for allt,s∈1,e, implies that(21)logtα−11−∫1ehtlogtα−1dt/t∫1ehtG1t,sdtt≤α−1∫1ehtdt/tΓα1−∫1ehtlogtα−1dt/tlogs1−logsα−1,and(22)logtα−11−∫1ehtlogtα−1dt/t∫1ehtG1t,sdtt≥logtα−1∫1ehtlogtα−11−logtdt/tΓα1−∫1ehtlogtα−1dt/tlogs1−logsα−1.Lemma 4.
Letϕt=logt1−logtα−1,wheret∈1,e. Then there exist(23)κ1=α2ΓαΓ2α+2+Γα2Γ2α∫1ehtlogtα−11−logtdt/t1−∫1ehtlogtα−1dt/t,κ2=α−1Γα+21+∫1ehtdt/t1−∫1ehtlogtα−1dt/t,such that(24)κ1ϕs≤∫1eGt,sϕtdtt≤κ2ϕs,fors∈1,e.Proof.
Using (20)–(22), for all s∈1,e, we have(25)∫1eGt,sϕtdtt≤∫1eα−1ϕsΓαϕtdtt+∫1eα−1∫1ehtdt/tΓα1−∫1ehtlogtα−1dt/tϕsϕtdtt=κ2ϕs,∫1eGt,sϕtdtt≥∫1elogtα−11−logtϕsΓαϕtdtt+∫1elogtα−1∫1ehtlogtα−11−logtdt/tΓα1−∫1ehtlogtα−1dt/tϕsϕtdtt=κ1ϕs.
This completes the proof.From Lemma 3, we know (1) is equivalent to the following Hammerstein-type integral equations:(26)utvt=∫1eGt,sf1s,us,vsdss∫1eGt,sf2s,us,vsdss.LetE≔C1,e,u≔maxt∈1,eut,andP≔u∈E:ut≥0,∀t∈1,e. Then E,⋅ becomes a real Banach space and P a cone on E. Moreover, E×E is a Banach space with the norm x,y=x+y, and P×P is a cone on E×E. Therefore, we define operators Aii=1,2 and A as follows:(27)A1u,vt=∫1eGt,sf1s,us,vsdss,A2u,vt=∫1eGt,sf2s,us,vsdss,Au,vt=A1,A2u,vt,foru,v∈P,t∈1,e.Note thatGandfii=1,2 are nonnegative continuous functions, so the operators Ai:P×P⟶Pi=1,2 and A:P×P⟶P×P are three completely continuous operators. Moreover, if u,v∈P×P\0 is a fixed point of A, then u,v is a positive solution for (1). Therefore, in what follows, we turn to study the existence of fixed points of the operator A.Lemma 5.
Letp be a continuous concave function. Then, if φ is an integrable function on 0,1, we have(28)p∫01φtdt≥∫01pφtdt.Proof.
Let0=t0<t1<t2<⋯<tn−1<tn=1, for all n∈ℕ+, and Δti=ti−ti−1,d=maxΔti,i=1,2,…,n. Then, note that ∑i=1nΔti=1, for all ξi∈ti−1,ti,wherei=1,2,…,n, we have(29)p∫01φtdt=plimd⟶0∑i=1nφξiΔti=limd⟶0p∑i=1nφξiΔti≥limd⟶0∑i=1npφξiΔti=∫01pφtdt.
This completes the proof.Remark 1.
Ifp is a continuous convex function in Lemma 5, then (28) can be changed into the inverse inequality:(30)p∫01φtdt≤∫01pφtdt.Lemma 6 (see [34]).
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(31)ω−Aω≠λω0,∀λ≥0,ω∈∂Ω∩​P,then iA,Ω∩​P,P=0, where i denotes the fixed-point index on P.Lemma 7 (see [34]).
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(32)ω−λAω≠0,∀λ∈0,1,ω∈∂Ω∩​P,then iA,Ω∩​P,P=1.


## 3. Main Results

Lemma 8.
LetP0=u∈P:∫1eutϕtdt/t≥ω0u. Then Bu∈P0, where(33)But=∫1eGt,susdss,u∈P,where(34)ω0=α2Γ2αα−1Γ2α+21+∫1ehtlogtα−11−logtdt/t1−∫1ehtlogtα−1dt/t1+∫1ehtdt/t1−∫1ehtlogtα−1dt/t−1.Proof.
From the definition ofG, for all t,τ,s∈1,e, we have(35)Gt,s≥logtα−11−logtϕsΓα+logtα−1∫1ehtlogtα−11−logtdt/tΓα1−∫1ehtlogtα−1dt/tϕs≥logtα−11−logtΓα1+∫1ehtlogtα−11−logtdt/t1−∫1ehtlogtα−1dt/tϕs=logtα−11−logtΓα1+∫1ehtlogtα−11−logtdt/t1−∫1ehtlogtα−1dt/t⋅α−1Γα1+∫1ehtdt/t1−∫1ehtlogtα−1dt/tϕs⋅Γαα−11+∫1ehtdt/t1−∫1ehtlogtα−1dt/t−1≥logtα−11−logtα−11+∫1ehtlogtα−11−logtdt/t1−∫1ehtlogtα−1dt/t1+∫1ehtdt/t1−∫1ehtlogtα−1dt/t−1Gτ,s.
Then ifu∈P, we have(36)∫1eButϕtdtt=∫1eϕt∫1eGt,susdssdtt≥∫1eϕt∫1elogtα−11−logtα−11+∫1ehtlogtα−11−logtdt/t1−∫1ehtlogtα−1dt/t1+∫1ehtdt/t1−∫1ehtlogtα−1dt/t−1Gτ,susdssdtt=α2Γ2αα−1Γ2α+21+∫1ehtlogtα−11−logtdt/t1−∫1ehtlogtα−1dt/t1+∫1ehtdt/t1−∫1ehtlogtα−1dt/t−1Buτ.
Note that the arbitrariness ofτ∈0,1, we have(37)∫1eButϕtdtt≥ω0Bu.
This completes the proof.LetK=α−1/Γα1+∫1ehtdt/t/1−∫1ehtlogtα−1dt/t. Then, maxt,s∈1,eGt,s≤K. Now, we list our assumptions for fii=1,2:(H1)fi∈C1,e×ℝ+×ℝ+,ℝ+,i=1,2.(H2) There existp1,q1∈Cℝ+,ℝ+ and c1>0 such that(i)
p1 is a strictly increasing concave function on ℝ+ and limz⟶+∞p1z=+∞(ii)
f1t,u,vf2t,u,v≥p1v−c1q1u−c1,∀t,u,v∈1,e×ℝ+×ℝ+(iii)
∃γ1∈κ1−2,+∞ such that p1Kq1z≥γ1Kz−c1,∀z∈ℝ+(H3) There existp2,q2∈Cℝ+,ℝ+ and r1>0 such that(i)
p2 is a strictly increasing convex function on ℝ+ and p20=0(ii)
f1t,u,vf2t,u,v≤p2vq2u,∀t,u,v∈1,e×0,r1×0,r1(iii)
∃γ2∈0,κ2−2 such that p2Kq2z≤γ2Kz,∀z∈0,r1(H4) There existp3,q3∈Cℝ+,ℝ+ and r2>0 such that(i)
p3 is a strictly increasing concave function on ℝ+(ii)
f1t,u,vf2t,u,v≥p3vq3u,∀t,u,v∈1,e×0,r2×0,r2(iii)
∃γ3∈κ1−2,+∞ such that p3Kq3z≥γ3Kz,∀z∈0,r2(H5) There exista11,b11,a12,b12≥0 and l1,l2>0 such that(38)a11κ2<1,b12κ2<1,det1−a11κ2−a12κ2−b11κ21−b12κ2>0,f1t,u,vf2t,u,v≤a11u+a12v+l1b11u+b12v+l2,∀t,u,v∈1,e×ℝ+×ℝ+.DefineBρ=z∈E:z<ρ for ρ>0. We adopt the convention in the sequel that c1,c2,… stand for different positive constants.Theorem 1.
Suppose that (H1)–(H3) hold. Then, (1) has at least one positive solution.Proof.
LetM1=u,v∈P×P:u,v=Au,v+μu∗,v∗,μ≥0, where u∗,v∗∈P0 are two given elements. Then, we claim that M1 is a bounded set in P×P. We define operators fi:P×P⟶Pi=1,2 as follows:(39)fiu,vt=fit,ut,vt,foru,v∈P,t∈1,e,i=1,2.
Now, if there existsu,v∈M1, then we have u=A1u,v+μu∗=Bf1u,v+μu∗ and v=A2u,v+μv∗=Bf2u,v+μv∗. From Lemma 8, we have(40)u,v∈P0.
Moreover, together with (H2) (ii), we can obtain that(41)ut≥A1u,vt≥∫1eGt,sp1vs−c1dss,vt≥A2u,vt≥∫1eGt,sq1us−c1dss,≥∫1eGt,sq1usdss−c2.
Using (H2) (i) and (iii), we have(42)p1vt≥p1vt+c2−p1c2≥p1∫1eGt,sq1usdss−p1c2=p1∫01Gt,exq1uexdx−p1c2≥∫01p1Gt,exq1uexdx−p1c2=∫01p1Gt,exKKq1uex+1−Gt,exK⋅0dx−p1c2≥∫01Gt,exKp1Kq1uexdx−p1c2≥∫01Gt,exKγ1Kuex−c1dx−p1c2≥γ1∫1eGt,susdss−c3.
Therefore, we have(43)ut≥∫1eGt,sγ1∫1eGs,τuτdττ−c3dss−c2≥γ1∫1e∫1eGt,sGs,τuτdssdττ−c4.
Recall thatϕt=logt1−logtα−1,wheret∈1,e. Therefore, we multiply both sides of the above by ϕt, integrate over 1,e, and use Lemma 4 to obtain(44)∫1eutϕtdtt≥γ1∫1eϕt∫1e∫1eGt,sGs,τuτdssdττdtt−c4∫1eϕtdtt≥γ1κ12∫1eutϕtdtt−c4ΓαΓα+2.
Solving this inequality, from (40), we have(45)u≤ω0−1∫1eutϕtdtt≤ω0−1c4Γαγ1κ12−1Γα+2.
On the other hand, we estimate the norm ofv. Multiplying both sides of the first inequality of (41) by ϕt, integrating over 1,e, and using Lemma 4, we obtain(46)∫1eutϕtdtt≥κ1∫1eϕtp1vt−c1dtt.
This implies that(47)∫1eϕtp1vtdtt≤κ1−1c4Γαγ1κ12−1Γα+2+c1ΓαΓα+2.
Without loss of generality, we may assumevt≢0, then v>0. Note that v∈P0, we have(48)v≤1ω0∫1evtϕtdtt=vω0p1v∫01vexvp1vϕexdx≤vω0p1v∫01p1vexϕexdx,p1v≤1ω0∫01p1vexϕexdx=1ω0∫1ep1vtϕtdtt≤1ω0κ1−1c4Γαγ1κ12−1Γα+2+c1ΓαΓα+2.
Combining (H2) (i) (limz⟶+∞p1z=+∞), there exists N1 such that v≤N1.
Up to now, we have proved the boundedness ofM1. Taking R1>N1+ω0−1c4Γα/γ1κ12−1Γα+2 and R1>r1 (r1 is defined by (H3)), we have(49)u,v≠Au,v+μu∗,v∗,foru,v∈∂BR1∩​P×P,μ≥0.
Then, Lemma 6 enables us to obtain(50)iA,BR1∩​P×P,P×P=0.
Next, we show that(51)u,v≠μAu,v,foru,v∈∂Br1∩​P×P,μ∈0,1.
If this claim is not true, then there existu,v∈∂Br1∩​P×P,μ∈0,1 such that(52)u,v=μAu,v.
Combining (H3) (ii), we obtain(53)ut≤A1u,vt≤∫1eGt,sp2vsdss,vt≤A2u,vt≤∫1eGt,sq2usdss.
From (H3) (i) and (iii), we have(54)p2vt≤p2∫1eGt,sq2usdss≤∫01p2Gt,exq2uexdx=∫01p2Gt,exKKq2uex+1−Gt,exK⋅0dx≤∫01Gt,exKp2Kq2uexdx≤γ2∫1eGt,susdss.
Consequently, we have(55)ut≤γ2∫1e∫1eGt,sGs,τuτdττdss.
Multiplying both sides of the above byϕt, integrating over 1,e, and using Lemma 4, we obtain(56)∫1eutϕtdtt≤γ2κ22∫1eutϕtdtt.
Note thatγ2∈0,κ2−2, we have ∫1eutϕtdt/t=0 and ut≡0 for t∈1,e. Moreover, using (54), we have p2vt≡0 for t∈1,e. From (H3) (i), we have vt≡0 for t∈1,e. Therefore, this contradicts to u,v∈∂Br1∩​P×P, r1>0. This also implies that (51) holds. Then, Lemma 7 enables us to obtain(57)iA,Br1∩​P×P,P×P=1.
From (50) and (57), we have(58)iA,BR1\B¯r1∩​P×P,P×P=iA,BR1∩​P×P,P×P−iA,Br1∩​P×P,P×P=0−1=−1.
Therefore, the operator A has at least one fixed point onBR1\B¯r1∩​P×P. Equivalently, (1) has at least one positive solution. This completes the proof.Theorem 2.
Suppose that (H1) and (H4)-(H5) hold. Then, (1) has at least one positive solution.Proof.
Forr2 in (H4), we first show that(59)u,v≠Au,v+μu∗,v∗,foru,v∈∂Br2∩​P×P,μ≥0,where u∗,v∗∈P are two given elements. Indeed, if this claim is false, there exist u,v∈∂Br2∩​P×P,μ≥0 such that(60)u,v=Au,v+μu∗,v∗.
This, together with (H4) (ii), implies that(61)ut≥A1u,vt≥∫1eGt,sp3vsdss,vt≥A2u,vt≥∫1eGt,sq3usdss.
Similar to (42), we have(62)p3vt≥p3∫1eGt,sq3usdss≥∫1eGt,sKp3Kq3usdss.
From (H4) (iii), we have(63)ut≥∫1eGt,sp3vsdss≥γ3∫1e∫1eGt,sGs,τuτdssdττ.
Multiplying both sides of the above byϕt, integrating over 1,e, and using Lemma 4, we obtain(64)∫1eutϕtdtt≥γ3κ12∫1eutϕtdtt,where ϕt=logt1−logtα−1,t∈1,e. Consequently, γ3κ12>1 implies that ∫1eutϕtdt/t=0 and ut≡0 for t∈1,e. Note that (65), should be(65)∫1eGt,sp3vsdss≤ut≡0,∀t∈1,e.
From (H4) (i), this indicates thatp3vs≡0 and vs≡0 for s∈1,e. Therefore, u=v=0 contradicts to u,v∈∂Br2∩​P×P and (59) holds. Then, Lemma 6 enables us to obtain(66)iA,Br2∩​P×P,P×P=0.
LetM2=u,v∈P×P:u,v=μAu,v,μ∈0,1. Then, we prove that M2 is a bounded set in P×P. If u,v∈M2, then we have(67)u=μA1u,v,v=μA2u,v,foru,v∈P×P.
From Lemma 8, we have(68)u,v∈P0.
Moreover, by (H5), we have(69)utvt≤∫1eGt,sa11us+a12vs+l1dss∫1eGt,sb11us+b12vs+l2dss.
Multiplying both sides of the above byϕt, integrating over 1,e, and using Lemma 4, we obtain(70)∫1eutϕtdtt∫1evtϕtdtt≤κ2∫1eϕta11ut+a12vt+l1dttκ2∫1eϕtb11ut+b12vt+l2dtt.
Consequently, we have(71)1−a11κ2−a12κ2−b11κ21−b12κ2∫1eutϕtdtt∫1evtϕtdtt≤κ2l1ΓαΓα+2κ2l2ΓαΓα+2.
Solving this matrix inequality, we have(72)∫1eutϕtdtt∫1evtϕtdtt≤1−b12κ2a12κ2b11κ21−a11κ2κ2l1Γα/Γα+2κ2l2Γα/Γα+21−a11κ21−b12κ2−a12b11κ22.
This implies that(73)∫1eutϕtdtt≤κ2Γα/Γα+21−b12κ2l1+a12κ2l21−a11κ21−b12κ2−a12b11κ22,∫1evtϕtdtt≤κ2Γα/Γα+2b11κ2l1+1−a11κ2l21−a11κ21−b12κ2−a12b11κ22.
Note thatu,v∈P0, we have(74)u≤ω0−1κ2Γα/Γα+21−b12κ2l1+a12κ2l21−a11κ21−b12κ2−a12b11κ22,v≤ω0−1κ2Γα/Γα+2b11κ2l1+1−a11κ2l21−a11κ21−b12κ2−a12b11κ22.
TakingR2>ω0−1κ2Γα/Γα+21−b12κ2l1+a12κ2l2+b11κ2l1+1−a11κ2l2/1−a11κ21−b12κ2−a12b11κ22 and R2>r2 (r2 is defined by (H4)), we have(75)u,v≠μAu,v,foru,v∈∂BR2∩​P×P,μ∈0,1.
Then, Lemma 7 enables us to obtain(76)iA,BR2∩​P×P,P×P=1.
From (66) and (76), we have(77)iA,BR2\B¯r2∩​P×P,P×P=iA,BR2∩​P×P,P×P−iA,Br2∩​P×P,P×P=1−0=1.
Therefore, the operatorA has at least one fixed point on BR2\B¯r2∩​P×P. Equivalently, (1) has at least one positive solution. This completes the proof.
In (1), let n=3,α=2.5,andht=logt, t∈1,e. Then, ∫1ehtlogtα−1dt/t=∫1elogtαdt/t=2/7∈0,1 and (H0) holds. Moreover, we can calculate K,κ1,andκ2 as follows:(78)K=α−1Γα1+∫1ehtdt/t1−∫1ehtlogtα−1dt/t=1.5Γ2.51+∫1elogtdt/t1−∫1elogt2.5dt/t≈1.92,κ1=α2ΓαΓ2α+2+Γα2Γ2α∫1ehtlogtα−11−logtdt/t1−∫1ehtlogtα−1dt/t=2.52Γ2.5Γ7+Γ2.52Γ5Γ3.5Γ5.575≈0.014,κ2=α−1Γα+21+∫1ehtdt/t1−∫1ehtlogtα−1dt/t=1.5Γ4.51+∫1elogtdt/t1−∫1elogt2.5dt/t≈0.22.Example 1.
Letf1t,u,v=u+vγ1,f2t,u,v=u+vγ2, p1v=v1/3,q1u=u4, p2v=v2,andq2u=u, for t,u,v∈1,e×ℝ+×ℝ+, where γ1>2andγ2>4. Then, we have(i)
liminfv⟶+∞f1t,u,v/p1v=liminfv⟶+∞u+vγ1/v1/3≥liminfv⟶+∞vγ1/v1/3=+∞, for all t,u∈1,e×ℝ+(ii)
liminfu⟶+∞f2t,u,v/q1u=liminfu⟶+∞u+vγ2/u4≥liminfu⟶+∞uγ2/u4=+∞, for all t,v∈1,e×ℝ+(iii)
limsupu+v⟶0+f1t,u,v/p2v=limsupu+v⟶0+u+vγ1/v2=0, for all t∈1,e(iv)
limsupu+v⟶0+f2t,u,v/q2u=limsupu+v⟶0+u+vγ2/u=0, for all t∈1,e(v)
liminfz⟶+∞p1Kq1z/z=liminfz⟶+∞K3z4/3/z=+∞(vi)
limsupz⟶0+p2Kq2z/z=limsupz⟶0+K2z2/z=0
Therefore, (H2)-(H3) hold.Example 2.
Leta11=0.05,a12=0.6,b11=0.4,andb12=0.08, then we calculate a11κ2=0.011<1,b12κ2=0.0176<1, and(79)1−a11κ2−a12κ2−b11κ21−b12κ2=0.989−0.132−0.0880.9824≈0.96.
Letf1t,u,v=a11u+a12vγ3,f2t,u,v=b11u+b12vγ4, p3v=v, and q3u=u3/4, for t,u,v∈1,e×ℝ+×ℝ+, where γ3∈0,1/2andγ4∈0,3/4. Then for all t∈1,e, we have(80)liminfa11u+a12v⟶0+f1t,u,vp3v=liminfa11u+a12v⟶0+a11u+a12vγ3v1/2≥liminfa11u+a12v⟶0+a12vγ3v1/2=+∞,liminfb11u+b12v⟶0+f2t,u,vq3u=liminfb11u+b12v⟶0+b11u+b12vγ4u3/4≥liminfb11u+b12v⟶0+b11uγ4u3/4=+∞,limsupa11u+a12v⟶+∞f1t,u,va11u+a12v=limsupa11u+a12v⟶+∞a11u+a12vγ3a11u+a12v=0,limsupb11u+b12v⟶+∞f2t,u,vb11u+b12v=limsupb11u+b12v⟶+∞b11u+b12vγ4b11u+b12v=0,liminfz⟶0+p3Kq3zz=liminfz⟶0+Kz3/8z=+∞.
As a result, (H4)-(H5) hold.



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*Source: 2671539-2019-10-13.xml*