# Iterative Positive Solutions to a Coupled Hadamard-Type Fractional Differential System on Infinite Domain with the Multistrip and Multipoint Mixed Boundary Conditions

Authors: Xinran Du; Yuan Meng; Huihui Pang
Journal: Journal of Function Spaces
 (2020)

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## Abstract

This paper is devoted to the existence of positive solutions for a nonlinear coupled Hadamard fractional differential system, with multistrip and multipoint mixed boundary conditions on an infinite interval. Based on the Arzelá-Ascoli theorem, we establish an important lemma to prove the complete continuity of operators on the infinite interval. Using the monotone iterative technique, the existence criteria for positive extremal solutions can be acquired, and an example is given to illustrate the feasibility of the above study as well.


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## Body

## 1. Introduction

Fractional calculus, which is organically united with integral calculus, extends the concept of classical integral calculus to the whole real number line and even the complex plane. In latest researches, it has been fully shown that fractional calculus definitely has some excellent properties. This kind of calculus is nonlocal in nature, which can accurately describe several materials and processes with traits of heredity and memory [1, 2]. There are numerous applications in a variety of disciplinary fields such as characterization of anomalous diffusion [3], random processes [4], viscoelasticity [5], non-Newtonian fluid mechanics [6], and biomathematics [7]. Moreover, differential equations can succinctly establish the relationship between variables and their derivatives. Therefore, the study of differential equations with fractional calculus is of significance theoretically and practically. In fact, fractional differential equations have been widely focused on and studied in depth. For the recent development of the topic, many researchers studied the existence [8–10], uniqueness [11, 12], and multiplicity of the solutions to fractional differential problems [13, 14].Meanwhile, it is observed that there have been many studies based on Riemann-Liouville- and Caputo-type fractional differential equations. However, there is another kind of fractional differential equation, which is based on the Hadamard-type fractional calculus definition, found in the literature due to Hadamard (see [15]). It is worth mentioning that Hadamard’s construction of fractional integrodifferential, including a logarithmic function of arbitrary exponent, is a fractional power of the type xd/dxa. Therefore, it is well suited to the case of the half-axis and is invariant relative to dilation. As a result, the Hadamard fractional definition will be a strong tool to expand the interval further. From a finite interval to an infinite semi-infinite interval, this is theoretically a high generalization of many existing models, more general and innovative. This model has already been used in applied mathematics and physics. A detailed description about the boundary value problems of the Hadamard fractional derivative and integral on infinite intervals can be seen in [16–20].In [16], Thiramanus et al. considered the positive solution for Hadamard fractional differential equations on infinite intervals by applying the idea of Leggett-Williams and Guo-Krasnoselskii fixed point theorems
(1)Daut+atfut=0,1<a≤2,t∈1,∞,u1=0,Da−1u∞=∑i=1mλiIβiuη,where Dα denotes the Hadamard fractional derivative of order α,η∈1,∞, and Iβi is the Hadamard fractional integral of order βi>0,i=1,2,⋯,m, and λi≥0,i=1,2,⋯,m are the given constants.In [19], Pei et al. not only established the existence of positive solutions but also sought the positive minimal and maximal solutions and got two explicit monotone iterative sequences which converge to the extremal solutions
(2)HDaut+ft,ut,HIγut,HDa−1ut=0,1<a<2,t∈1,+∞,u1=0,HDau∞=∑i=1mλiHIβiun,where HDα denotes Hadamard fractional derivative, η∈1,∞ and HI. is the Hadamard fractional integral. γ,βi,λi≥0i=1,2,⋯,m are the given constants and α,η,βi,λi satisfy Γa>∑i=1mλiΓa/Γa+βilogηa+βi−1.Furthermore, the system of fractional differential equations boundary value problems has also received much attention. More and more researchers invest effort in this direction; see [20, 21] and the references therein.More recently in [20], Yang investigated the extremal iterative solutions for a coupled system of nonlinear Hadamard fractional differential equations with Cauchy initial value conditions by using the comparison principle and the monotone iterative technique combined with the method of upper and lower solutions
(3)Da+axt=ft,xt,yt,Ja+1−axa+=x0∗,a∈0,1,t∈a,b,Da+ayt=gt,xt,yt,Ja+1−aya+=y0∗,a∈0,1,t∈a,b,where f,g∈Ca,b×ℝ×ℝ,ℝ,x0∗,y0∗∈ℝ,x0∗≤y0∗,Da+a, and Ja+a are the left-sided Hadamard fractional derivative and integral of order α, respectively.In [21], Ahmad and Ntouyas proved the existence and uniqueness of positive solutions of a class of boundary value problems of fractional differential equations by utilizing the Leray-Schauder alternative principle and the Banach contraction mapping principle
(4)Daut=ft,ut,vt,1<t<e,1<a≤2,Dβvt=gt,ut,vt,1<t<e,1<β≤2,u1=0,ue=Iγuσ1=1Γγ∫1σ1logσsγ−1ussds,u1=0,ve=Iγvσ2=1Γγ∫1σ2logσ2sγ−1vssds,where D. and I., respectively, denote the Hadamard fractional derivative and Hadamard fractional integral, and f,g:1,e×ℝ3⟶ℝ are given continuous functions.Inspired by the aforementioned work, to get more extensive results, we investigate the existence of iterative positive solutions to the following Hadamard fractional differential systems(5)Dta1ut+f1t,ut,vt,Dta1−1ut,Dta2−1vt=0,t∈1,∞,Dta2vt+f2t,ut,vt,Dta1−1ut,Dta2−1vt=0,t∈1,∞,subject to the coupled fractional-order integral and discrete mixed boundary conditions
(6)u1=0,Dta1−1u∞=∑i=1mλ1iItβ1ivξi+∑j=1nbjvηj,v1=0,Dta2−1v∞=∑i=1mλ2iItβ2iuξi+∑j=1nbjuηj,where Dtαk denotes the Hadamard fractional derivative of order αk and Itβki is the Hadamard fractional integral of order βki, for k=1,2, and i=1,2,⋯,m.β1i>0,β2i>0,1<α1,α2<2,λ1i≥0,λ2i≥0,1<ξi<∞,fori=1,2,⋯,m;bj≥0,1<ηj<∞, for j=1,2,⋯,n.The main aim of this paper is to investigate the coupled nonlinear fractional differential system of the Hadamard type on infinite intervals subject to the coupled multistrip and multipoint mixed boundary conditions. This kind of condition is a linear combination of values at the multiple band integrals and the different discrete points, which highly summarizes the characteristics of boundary conditions in the existing study. There are two unknown functionsut,vt which influence each other in this system. And the nonlinear part, f1 and f2, contain lower order derivative operators Dtα1−1ut,Dtα2−1vt as well. Based on the above model, the present research results are not abundant or even almost blank; therefore, we fill the gap in this paper.The proofs of our main result are derived by using the monotone iterative method, which gets two significant benefits. Firstly, by selecting the appropriate initial function vector of concise form, the existence of solution can be guaranteed with the effective and succinct process. Moreover, we can seek the approximate positive solution under a different level of precision, which has more practical application value. Through comprehensive consideration, we construct the initial iterative function vector which satisfies the multiconstraints from the cone and the monotonicity of the complete continuous operatorT and its respective derivatives. More researches on the details of the method can be found in these references [22–27].This article is structured as follows: Section2 contains numerous relative definitions and lemmas. Also, there are some main results and their proofs. In Section 3, we present the existence results of monotone iterative positive solutions. Besides, the main results are illustrated by a practical example.


## 2. Preliminaries

In this section, we will present here the definitions, some lemmas from the theory of fractional calculus, and some auxiliary results for the proof of our main results.Definition 1 [1].
Ifgt∈Xcpℝ+, then the Hadamard fractional integral of order γ for a function g is defined as
(7)Itγgt=1Γγ∫1tlogtsγ−1gssds,γ>0,provided the integral exists. Such a space, which we denote by Xcpℝ+c∈ℝ;1≤p≤∞, consists of those complex-valued Lebesgue measurable functions g on ℝ for which ∥g∥Xcp<∞, with
(8)gxcp<∫0∞tcgtpdtt1/p1≤p≤∞,gxc∞<esssupx∈ℝ+xcgx.
In particular, whenc=1/p the space x1/ppℝ+ coincides with the Lp-space:
(9)X1/ppℝ+=Lpℝ+.Definition 2 [1].
The Hadamard derivative of fractional orderγ for a function g:1,∞⟶R is defined as
(10)Dtγgt=1Γn−γtddtn∫1tlogtsn−γ−1gssds,n−1<γ<n,n=γ+1,where γ denotes the integer part of the real number γ and log·=loge·.Lemma 3 [2].
Ifa,γ,β>0, then
(11)Daγlogtaβ−1x=ΓβΓβ−γlogxaβ−γ−1,Iaγlogtaβ−1x=ΓβΓβ−γlogxaβ−γ−1.Lemma 4 [2].
Letγ>0 and x∈C1,∞∩L11,∞. Then, the Hadamard fractional differential equation Dtγxt=0 has the solutions
(12)xt=∑i=1ncilogtγ−i,and the following formula holds:
(13)DtγItγut=ut,where γ>0, and
(14)ItγDtγxt=xt+∑i=1ncilogtγ−i,where ci∈R,i=1,2,⋯,n, and n−1<γ<n.For convenience, we denote(15)l1=∑i=1mλ1iΓβ1i∫1ξilogξisβ1i−1logsa2−1dss+∑j=1nbjlogηja2−1,l2=∑i=1mλ2iΓβ2i∫1ξilogξisβ2i−1logsa1−1dss+∑j=1nbjlogηja1−1.In the forthcoming analysis, we always need the following assumptions:(F1)
1<αk≤2,βki>0,fork=1,2andi=1,2,⋯,m(F2)
ηj,ξi>1,bj≥0,λ1i,λ2i≥0,fori=1,2,⋯,m,j=1,2,⋯,n(F3)
1−l1l2/Γα1Γα2>0, where l1,l2 are defined by (15)(F4)
fk:1,∞×0,+∞4⟶0,+∞ is continuous k=1,2Subject to BVP (5) and (6), we consider a corresponding linear boundary value problem as follows and establish the expression of the corresponding Green’s functions.Lemma 5.
Assume that (F1)-(F3) hold. Forh1,h2∈C1,∞ with 0<∫1∞hs1+1/sds<∞, the fractional differential system
(16)Dta1ut+h1t=0,Dta2vt+h2t=0,t∈1,∞,with boundary condition (6) has an integral representation
(17)ut=∫1∞K1t,sh1sds+∫1∞H1t,sh2sds,vt=∫1∞K2t,sh2sds+∫1∞H2t,sh1sds,where
(18)K1t,s=g1t,s+l1logtα1−1Γα1Γα2−l1l2∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−1g1τ,sdττ+∑j=1nbjg1ηj,s,H1t,s=Γα2logtα1−1Γα1Γα2−l1l2∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−1g2τ,sdττ+∑j=1nbjg2ηj,s,(19)K2t,s=g2t,s+l2logtα2−1Γα1Γα2−l1l2∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−1g2τ,sdττ+∑j=1nbjg2ηj,s,H2t,s=Γα1logtα2−1Γα1Γa2−l1l2∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−1g1τ,sdττ+∑j=1nbjg1ηj,s,and for k=1,2,
(20)gkt,s=1Γαklogtαk−1−logtsαk−11s,1≤s≤t<∞,logtαk−118,1≤t≤s<∞.Proof.
From Lemma4, we can reduce (16) and (6) to the following equivalent integral equations:
(21)ut=−1Γα1∫1tlogtsα1−1h1sdss+c11logtα1−1+c12logtα1−2,vt=−1Γα2∫1tlogtsα2−1h2sdss+c21logtα2−1+c22logtα2−2.where c11,c12,c21,c22 are constants.Fromu1=v1=0, we have c12=c22=0. Further, we use the second condition of (6) to reduce (21) to
(22)ut=logtα1−1Γα1∑i=1mλ1iΓβ1i∫1ξilogξisβ1i−1vsdss+∑j=1nbjvηj+∫1∞g1t,sh1sds,vt=logtα2−1Γα2∑i=1mλ2iΓβ2i∫1ξilogξisβ2i−1usdss+∑j=1nbjuηj+∫1∞g2t,sh2sds.Then, we can get(23)∑i=1mλ1iΓβ1i∫1ξilogξisβ1i−1vsdss+∑j=1nbjvηj=1Γα2∑i=1mλ1iΓβ1i∫1ξilogξisβ1i−1logsα2−1dss+∑j=1nbjlogηjα2−1⋅∑i=1mλ2iΓβ2i∫1ξilogξisβ2i−1usdss+∑j=1nbjuηj+∑j=1nbj∫1∞g2ηj,sh2sds+∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−11τ∫1∞g2τ,sh2sdsdτ,(24)∑i=1mλ2iΓβ2i∫1ξilogξisβ2i−1usdss+∑j=1nbjuηj=1Γα1∑i=1mλ2iΓβ2i∫1ξilogξisβ2i−1logsα1−1dss+∑j=1nbjlogηjα1−1⋅∑i=1mλ1iΓβ1i∫1ξilogξisβ1i−1vsdss+∑j=1nbjuηj+∑j=1nbj∫1∞g1ηj,sh1sds+∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−11τ∫1∞g1τ,sh1sdsdτ.Combining (15), (23), and (24), it can be seen that
(25)∑i=1mλ1iΓβ1i∫1ξilogξisβ1i−1vsdss+∑j=1nbjvηj=11−l1l2/Γα1Γα2l1Γα2∑j=1nbj∫1∞g1ηj,sh1sds+∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−11τ∫1∞g1τ,sh1sdsdτ+∑j=1nbj∫1∞g2ηj,sh2sds+∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−11τ∫1∞g2τ,sh2sdsdτ,where lkk=1,2 is defined by (15). From (22) and (25), we have
(26)ut=∫1∞g1τ,sh1sds+1Γα1logtα1−11−l1l2/Γα1Γα2∫1∞l1Γα2∑j=1nbjg1ηj,s+∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−1g1τ,sdττh1sds+∫1∞∑j=1nbjg2ηj,s+∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−1g2τ,sdττh2sds=∫1∞K1t,sh1sds+∫1∞H1t,sh2sds,where K1t,s and H1t,s are introduced by (19). Similarly, we also have
(27)vt∫1∞K2t,sh2sds+∫1∞H2t,sh1sds,where K2t,s and H2t,s are also given by (19).This completes the proof of the lemma.Moreover, according to (17) and Lemma 4, the fractional order derivative of solution (17) can be expressed as
(28)Dtα1−1ut=∫1∞K3t,sh1sds+∫1∞H3t,sh2sds,Dtα2−1vt=∫1∞K4t,sh2sds+∫1∞H4t,sh1sds,where
(29)K3t,s=Dtα1−1g1t,s+l1Γα1Γα1Γα2−l1l2∑j=1nbjg1ηj,s+∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−1g1τ,sdττ,(30)H3t,s=11−l1l2/Γα1Γα2∑j=1nbjg2ηj,s+∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−1g2τ,sdττ,(31)K4t,s=Dtα2−1g2t,s+l2Γα2Γα1Γα2−l1l2∑j=1nbjg2ηj,s+∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−1g2τ,sdττ,(32)H4t,s=11−l1l2/Γα1Γα2∑j=1nbjg1ηj,s+∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−1g1τ,sdττ,and for k=1,2,
(33)Dtαk−1gkt,s=0,1≤s≤t<∞,1s,1≤t≤s<∞.Lemma 6.
Assume that (F1) holds. Then, the functionsgkt,s and Dtakgkt,s, for k=1,2, defined by (20) and (33) have the following properties:
(1)
0≤gkt,s≤logtαk−1/Γαk1/s+1,fort,s∈1,∞(2)
0≤Dtαk−1gkt,s≤1/s+1,fort,s∈1,∞Proof.
(1)
For1≤s≤t≤∞, we have(34)gkt,s=1Γαklogtαk−1−logtsαk−11s≥0,gkt,s=1Γαklogtαk−1−logtsαk−11s≤logtαk−1Γαk1s≤logtαk−1Γαk1s+1.
For1≤t≤s≤∞, we have
(35)0≤gkt,s=1Γαklogtαk−11s≤logtαk−1Γαk1s+1.(2)
According to (33), we can easily get(36)0≤Dtαk−1gkt,s≤1s+1,fort,s∈1,∞.
This completes the proof of the lemma.For convenience, we denote(37)ϱ1=1Γα11+l1Γα1Γα2−l1l2∑j=1nbjlogηjα1−1+∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−1logτα1−1dττ,(38)ϱ2=1Γα21+l2Γα1Γα2−l1l2∑j=1nbjlogηjα2−1+∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−1logτα2−1dττ,(39)ρ1=1Γα1Γα2−l1l2∑j=1nbjlogηjα2−1+∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−1logτα2−1dττ,(40)ρ2=1Γα1Γα2−l1l2∑j=1nbjlogηjα1−1+∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−1logτα1−1dττ.Lemma 7.
Assume that (F1)-(F3) hold. Then, fort,s∈1,∞×1,∞, the functions Kit,s and Hit,s, for i=1,2,3,4, defined by (18), (19), and (29), satisfy the following results:
(1)
0≤K1t,s/1+logtα1−1≤1/s+1ϱ1,0≤K2t,s/1+logtα2−1≤1/s+1ϱ2,0≤H1t,s/1+logtα1−1≤1/s+1ρ1,0≤H2t,s/1+logtα2−1≤1/s+1ρ2(2)
0≤K3t,s≤Γα11/s+1ϱ1,0≤K4t,s≤Γα21/s+1ϱ2,0≤H3t,s≤Γα11/s+1ρ1,0≤H4t,s≤Γα11/s+1ρ2.Proof.
(1)
According to (F3), Lemma6 and the definition of Kit,s, we obtain(41)0≤K1t,s1+logtα1−1=g1t,s1+logtα1−1+l1Γα1Γα2−l1l2logtα1−11+logtα1−1∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−1g1τ,sdττ+∑j=1nbjg1ηj,s≤1Γα11s+1+l1Γα1Γα2−l1l2∑i=1mλ2iΓβ2iΓα1∫1ξilogξiτβ2i−1logτα1−11s+1dττ+1Γα1∑j=1nbjlogηjα1−11s+1≤1s+11Γα11+l1Γα1Γα2−l1l2∑j=1nbjlogηjα1−1+∑i=1mλ2iβ2i∫1ξilogξiτβ2i−1logτα1−1dττ=1s+1ϱ1,(42)0≤K3t,s=Dtα1−1g1t,s+l1Γα1Γα1Γα2−l1l2∑j=1nbjg1ηj,s+∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−1g1τ,sdττ≤1s+1+l1Γα1Γα1Γα2−l1l2∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−1logτα1−11s+1dττ+1Γα1∑j=1nbjlogηjα1−11s+1=1s+11+l1Γα1Γα2−l1l2∑j=1nbjlogηjα1−1+∑i=1mλ2iΓβ2i∫1ξilogξiτβ2i−1logτα1−1dττ=Γα11s+1ϱ1,
whereϱ1 is defined by (37). Similarly, we get
(43)0≤K2t,s1+logtα2−1≤1s+1ϱ2,0≤K4t,s≤Γα21s+1ϱ2,where ϱ2 is defined by (38).
(2)
According to (F3), Lemma6, and the definition of Hkt,sk=1,2,3,4, we infer that(44)0≤H1t,s1+logtα1−1=Γα2Γα1Γα2−l1l2logtα1−11+logtα1−1∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−1g2τ,sdττ+∑j=1nbjg2ηj,s≤Γα2Γα1Γα2−l1l2logtα1−11+logtα1−1∑i=1mλ1iΓβ1iΓα2∫1ξilogξiτβ1i−1logα2−11s+1dττ+1Γα2∑j=1nbjlogηjα2−11s+1≤1s+11Γα1Γα2−l1l2∑j=1nbjlogηjα2−1+∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−1logτα2−1dττ=1s+1ρ1,(45)0≤H3t,s=Γα1Γα2Γα1Γα2−l1l2∑i=1mλ1iΓβ1i∫1ξilogξiτβ1i−1g2τ,sdττ+∑j=1nbjg2ηj,s≤Γα1Γα2Γα1Γα2−l1l2∑i=1mλ1iΓβ1iΓα2∫1ξilogξiτβ1i−1logτα2−11s+1dττ+1Γα2∑j=1nbjlogηjα2−11s+1=Γα11s+1ρ1,where ρ1 is defined by (39). Analogously, we get
(46)0≤H2t,s1+logtα2−1≤1s+1ρ1,0≤H4t,s≤Γα21s+1ρ2, where ρ2 is defined by (40).
This completes the proof of the lemma.LetX=uu∈C1,∞, and Dtα1−1u∈C1,∞, supt∈1,∞ut/1+logtα1−1 and supt∈1,∞Dtα1−1t<∞ is endowed with the norm
(47)uX=maxu11,∣Dtα1−1u12, where u11=supt∈1,∞ut/1+logtα1−1,Dtα1−1u12=supt∈1,∞Dtα1−1t.Also, letY=vv∈C1,∞, Dtα2−1v∈C1,∞, supt∈1,∞vt/1+logtα2−1<∞ and supt∈1,∞Dtα2−1vt<∞ is endowed with the norm
(48)vY=maxv21,Dtα2−1v22,where u21=supt∈1,∞ut/1+logtα2−1,Dtα2−1v22=supt∈1,∞Dtα2−1vt. Then, we introduce the product space X×Y,u,v endowed with the norm u,v=maxuX,vY and define a partial order over the product space:
(49)u1v1≥u2v2,while u1t≥u2t, v1t≥v2t, Dtα1−1u1t≥Dtα1−1u2t, Dtα2−1v1t≥Dtα2−1v2t,t∈1,∞.Lemma 8.
X×Y,u,v is a Banach space.Proof.
Letunn=1∞ be a Cauchy sequence in the space X,·X. Then, ∀ε>0, ∃N>0 such that
(50)un−un11=supt∈1,∞un−um1+logtα1−1<ε,Dtα1−1un−um12=supt∈1,∞Dtα1−1un−Dtα1−1um<ε,for n, m>N. Therefore, for any fixed t0∈1,∞, we have unt0n=1∞, Dtα1−1unt0n=1∞ which are Cauchy sequences in ℝ. So we can associate to each t∈1,∞ for a unique ut, wt. Letting n⟶∞, we obtain un−um/1+logtα1−1<ε and wt−Dtα1−1umt<ε for all m>N and t∈1,∞. It easily shows that um⟶u and Dtα1−1um⟶w in X as m⟶∞. In other words,unn=1∞ uniformly converge to u, also Dtα1−1unn=1∞ uniformly converge to w on 1,∞.Now we should prove thatw=Dtα1−1u, while u, w∈C1,∞.Fort∈1,∞, we have
(51)Itα1−1Dtα1−1unt−Itα1−1t1+logtα1−1≤11+logtα1−11Γa1−1∫1tlogtsa1−2Dtα1−1uns−wsdss≤1Γa1logta1−11+logta1−1supt∈1,∞Dtα1−1uns−ws.In view of the convergence ofunn=1∞, we obtain that limn→∞Itα1−1Dtα1−1unt=Itα1−1wt uniformly for t∈1,∞. Otherwise, by Lemma 4, we have Itα1−1Dtα1−1unt=unt+c1logta1−2, for t∈1,∞. The two facts imply that
(52)limn→∞Itα1−1Dtα1−1unt=limn→∞unt+c1logta1−2=Itα1−1wt,fort∈1,∞.Combining withlimn→∞unt=ut for t∈1,∞, we have
(53)ut+c1logta1−2=Itα1−1wt,fort∈1,∞.Take the derivative ofα1−1-order on both sides of equation (53). In consequence, we have
(54)Dtα1−1Itα1−1wt=Dtα1−1ut,fort∈1,∞.From Lemma4, it leads to
(55)wt=Dtα1−1ut,fort∈1,∞,which proves that X,.X is a Banach space.Also,Y,.Y is a Banach space which can be proved in the same way. Moreover, the product space X×Y,u,v is also a Banach space with the norm u,v=maxuX,vY.The proof of this lemma is completed.Further, we define a coneP⊂X×YbyP=u,v∈X×Y:ut≥0,vt≥0,Dtα1−1ut≥0,Dtα2−1vt≥0,t∈1,∞. For all u,v∈P, in view of Lemma 5 and (F4), let T:P⟶P be the operator defined by
(56)Tu,vt=T1u,vtT2u,vt,where
(57)T1u,vt=∫1∞K1t,sf1u,vsds+∫1∞H1t,sf2u,vsds,T2u,vt=∫1∞K2t,sf2u,vsds+∫1∞H2t,sf1u,vsds,and for convenience, we set
(58)f1u,vs≜f1s,us,vs,Dtα1−1us,Dtα2−1vs,f2u,vs≜f2s,us,vs,Dtα1−1us,Dtα2−1vs.Also, according to (28), it is easy to see that
(59)Dtα1−1T1u,vt=∫1∞K3t,sf1u,vsds+∫1∞H3t,sf2u,vsds,Dtα2−1T2u,vt=∫1∞K4t,sf2u,vsds+∫1∞H4t,sf1u,vsds.Now, we list the following assumptions:(H1) For k=1,2, there exist nonnegative functions akt,bkt, ckt, dkt, and ekt defined on 1,∞ and constants 0≤pk,qk,mk,lk<1 such that
(60)fkt,u1,u2,u3,u4≤aKt+bktu1pk+cktu2qk+dktu3mk+ektu3lk,t∈1,∞,and denote that
(61)ak∗=∫1∞akt1t+1dt<,bk∗=∫1∞bkt1+logtα1−1pk1t+1dt<∞,ck∗=∫1∞ckt1+logtα2−1qk1t+1dt<∞,dk∗=∫1∞dkt1t+1dt<∞,ek∗=∫1∞ekt1t+1dt<∞.(H2) fk is increasing with respect to the second, third, fourth, and last variables, and fkt,0,0,0,0≢0,t∈1,∞, for k=1,2.Lemma 9.
IfH1 holds, for any u,v∈X×Y, we have
(62)∫1∞fkt,ut,vt,Dtα1−1ut,Dtα2−1vt1t+1dt≤ak∗+bk∗u,vpk+ck∗u,vqk+dk∗u,vmk+ek∗u,vlk,fork=1,2.Proof.
For anyu,v∈X×Y, we have
(63)fkt,ut,vt,Dtα1−1ut,Dtα2−1vt≤akt+bktutpk+cktvtqk+dktDtα1−1utmk+ektDtα2−1vtlk=akt+dktDtα1−1utmkektDtα2−1vtlk+bkt1+logtα1−1pkutpk1+logtα1−1pk+ckt1+logtα2−1qkvtqk1+logtα2−1qk≤akt+bkt1+logtα1−1pkuXpk+ckt1+logtα2−1qkvYqk+dktuYmk+ektuXlk≤akt+bk1+logtα1−1pku,vpk+ckt1+logtα2−1qku,vqk+dktu,vmk+ektu,vlk.
Hence,(64)∫1∞fkt,ut,vt,Dtα1−1ut,Dtα2−1vt1t+1dt≤∫1∞akt1t+1dt+∫1∞bkt1=logtα1−1pku,vpk1t+1dt+∫1∞ckt1+logtα2−1qk1t+1dt+∫1∞dktu,vmk1t+1dt+∫1∞ektu,vlk1t+1dt=ak∗+bk∗u,vpk+ck∗u,vqk+dk∗u,vmk+ek∗u,vlk.
The proof is completed.Lemma 10.
LetU⊂X be a bounded set. Then U is relatively compact in X if the following conditions hold:
(i)
For anyut∈U,ut/1+logtα1−1 and Dtα1−1ut are equicontinuous on any compact interval of 1,∞(ii)
for anyε>0 there exists a constant T=Tε>1 such that(65)ut11+logt1α1−1−ut21+logt2α1−1<∈andDtα1−1ut1−Dtα1−1ut2<∈,for any t1,t2≥T and u∈U.Proof.
According to the conditions, we only need to prove thatU is totally bounded. The proof consists of the following two parts.
(a)
Consider the case thatt∈1,T, we can define a Banach space. U1,T=ut:ut∈U,t∈1,T, with the norm u∞=supt∈1,Tut/1+logtα1−1. Arzelá-Ascoli theorem and (i) can ensure that U1,T is relatively a compact set, which means U1,T is a totally bounded set. In other words, for ∀ε>0, there exist finitely many balls Bεui such that(66)U1,T⊂∪i=1nBεui,where
(67)Bεui=ut∈U1,T:u−ui∞=supt∈1,Tut1+logtα1−1−uit1+logtα1−1<ε.
Similarly, define a Banach spaceU1,Tα1−1=Dα1−1ut:ut∈U1,T equipped with norm Dtα1−1ut∞=supt∈1,TDtα1−1ut. That is to say, for ∀∈>0, there exist finitely many balls BεDtα1−1uj such that
(68)U1,Tα1−1⊂∪j=1nBεDtα1−1uj,uj∈U1,T,where
(69)BεDtα1−1uj=Dtα1−1ut∈D1,Tα1−1:Dtα1−1u−Dtα1−1uj∞=supt∈1,TDtα1−1u−Dtα1−1ujt<ε.(b)
DefineUij=ut∈U:u1,T∈Bεui,Dtα1−1u1,T∈BεDtα1−1uj.
Evidently,U1,T⊂∪1≤i≤n1≤j≤mUij1,T. Let us consider uij∈Uij. Then, the set U can be covered by the balls B4εuij,i=1,2,⋯,n,j=1,2,⋯,m, where B4εuij=ut∈U:u−uijx<4ε.
From (a), we can get that there existi and j such that u1,T∈Bεui,Dtα1−1u1,T∈BεDtα1−1uj. Thus, for any t∈1,T, we get
(70)ut1+logtα1−1−uijt1+logtα1−1≤ut1+logtα1−1−uit1+logtα1−1+uit1+logtα1−1−uijt1+logtα1−1<2ε,(71)Dtα1−1ut−Dtα1−1uijt≤Dtα1−1ut−Dtα1−1ujt+Dtα1−1ujt−Dtα1−1uijt<2ε.
Hence, fort∈T,∞, inequalities in (i) and (70) indicate that
(72)ut1+logtα1−1−uijt1+logtα1−1≤ut1+logtα1−1−uT1+logtα1−1+uT1+logtα1−1−uijT1+logtα1−1+uijT1+logtα1−1−uijt1+logtα1−1<ε+2ε+ε=4ε.
Likely, by inequalities in (i) and (71) we know that
(73)Dtα1−1ut−Dtα1−1uijt≤Dtα1−1ut−Dtα1−1uT+Dtα1−1uT−Dtα1−1uijT+Dtα1−1uijT−Dtα1−1uijt<ε+2ε+ε=4ε.
It follows from all the above thatu−uijX<4ε, which implies that U is a totally bounded set.
The proof is finished.
There are similar conclusions for spaceY; we omit the details here.Lemma 11 [21].
The operatorT:P⟶P is completely continuous.Proof.
There are four steps to complete the whole proof.(a)
T:P⟶P is uniformly bounded
LetΩ be any bounded subset of P, then there exists ϕ>0 such that ∥u,v∥X×Y≤ϕ for all u,v∈Ω; it follows that
(74)T1u,v=supt∈1,∞∫1∞K1t,s1+logtα1−1f1u,vsds+∫1∞H1t,s1+logtα1−11+logtα1−1f2u,vsds≤ϱ1∫1∞f1u,vs1s+1ds+ρ1∫1∞f2u,vs1s+1ds≤ϱ1a1∗+b1∗u,vp1+c1∗u,vq1+d1∗u,vm1+e1∗u,vl1+ρ1a2∗+b2∗u,vp2+c2∗u,vq2+d2∗u,vm2+e2∗u,vl2≤ϱ1a1∗+b1∗ϕp1+c1∗ϕq1+d1∗ϕm1+e1∗ϕl1+ρ1a2∗+b2∗ϕp2+c2∗ϕq2+d2∗ϕm2+e1∗ϕl2<∞.Dtα1−1T1u,v=supt∈1,∞K3t,sf1u,vsds+∫1∞H3t,sf2u,vsds≤Γα1ϱ1∫1∞f1u,vs1s+1ds+Γα1ρ1∫1∞f2u,vs1s+1ds≤Γα1ϱ1a1∗+b1∗u,vp1+c1∗u,vq1+d1∗u,vm1+e1∗u,vl1+Γα1ρ1a2∗+b2∗u,vp2+c2∗u,vq2+d2∗u,vm2+e2∗u,vl2≤Γα1ϱ1a1∗+b1∗ϕp1+c1∗ϕq1+d1∗ϕm1+e1∗ϕl1+Γα1ρ1a2∗+b2∗ϕp2+c2∗ϕq2+d2∗ϕm2+e2∗ϕl2<∞.
So, foru,v∈Ω, T1Ω is uniformly bounded. Furthermore, we get T2Ω is also uniformly bounded.
Therefore, it follows from the above inequalities that the operatorT is uniformly bounded.
(b)
The operatorT:P⟶P is equicontinuous on any compact interval of 1,∞×1,∞.
LetI⊂1,∞×1,∞ be any compact interval. We assume that ∀t1,t2∈I, t2>t1, then
(75)T1u,vt11+logt1α1−1−T1u,vt21+logt2α1−1≤∫1∞K1t2,slogt2α1−1−K1t1,s1+logt1α1−1sf1u,vsdss+∫1∞H1t2,s1+logt2α1−1−H1t1,s1+logt1α1−1sf2u,vsdss.
For any compact setI×I, H1t,s/1+logtα1−1 and K1t,s/1+logtα1−1 are uniformly continuous as H1t,s,K1t,s∈C1,∞×1,∞. Notice that this function is only related to t when s≥t, so it is uniformly continuous on I×1,∞\I. Above all, for all s∈1,∞ and t1,t2∈I, ∀ε>0,∃δε>0 such that if ∣t2−t1∣<δ, then
(76)K1t2,s1+logt2α1−1−K1t1,s1+logt1α1−1<ε,(77)H1t2,s1+logt2α1−1−H1t1,s1+logt1α1−1<ε.
Combined with Lemma9, we get for ∀u,v∈Ω,k=1,2,
(78)∫1∞fks,us,vs,Dtα1−1us,Dtα2−1vsdss<∫1∞fks,us,vs,Dtα1−1us,Dtα2−1vs1s+1ds<∞.
Together with (75) and (76), the above inequality implies that T1u,vt/1+logtα1−1 is equicontinuous on I. Similar conclusion can be drawn in T2u,vt/1+logtα2−1.
Note that(79)Dtα1−1T1u,vt=∫1∞K3t,sf1u,vsds+∫1∞H3t,sf2u,vsds.
SinceK3t,s and H3t,s∈C1,∞×1,∞ do not depend on t, we can obtain that Dtα2−1T1u,vt is equicontinuous on I. Moreover, Dtα2−1T2u,vt is also equicontinuous on I.
Hence,T:P⟶P is equicontinuous on any compact interval of 1,∞×1,∞.
(c)
The operatorT and Dtα1−1T are equiconvergent at ∞
For anyu,v∈Ω, we have
(80)limt→∞T1u,vt1+logtα1−1≤limt→∞∫1∞K1t,s1+logtα1−1f1u,vsds+∫1∞H1t,s1+logtα1−1f2u,vsds≤limt→∞ϱ1+ρ1a1∗+b1∗u,vp1+c1∗u,vq1+d1∗u,vm1+e1∗u,vl1≤ϱ1+ρ1a1∗+b1∗ϕp1+c1∗ϕq1+d1∗ϕm1+e1∗ϕl1<∞,then combining with the conclusion of the T2, we get TΩ which is equiconvergent at ∞.For any u,v∈Ω, we have
(81)limt→∞Dtα1−1T1u,vt≤limt→∞∫1∞K3t,sf1u,vsds+∫1∞H3t,sf2u,vsds≤Γα1ϱ1+ρ1a1∗+b1∗u,vp1+c1∗u,vq1+d1∗u,vm1+e1∗u,vl1≤Γα1ϱ1+ρ1a1∗+b1∗ϕp1+c1∗ϕq1+d1∗ϕm1+e1∗ϕl1<∞,then combining with the conclusion of the Dtα1−1T2u,vt, which implies that Dtα1−1Tu,vt is equiconvergent at ∞.
(d)
The operatorT:P⟶P is continuous
Letu,vn⟶u,vn⟶∞, then ∥u,vn∥<∞,∥u,v∥<∞, we also let
(82)f1un,vns=f1s,uns,vns,Dsα1−1uns,Dsα2−1vns,f2un,vns=f2s,uns,vns,Dsα1−1uns,Dsα2−1vns.
We have,(83)T1un,vnt1+logtα1−1=∫1∞K1t,s1+logtα1−1f1un,vnsds+∫1∞H1t,s1+logtα1−1f2un,vnsds≤ϱ1+ρ1a1∗+b1∗u,vp1+c1∗u,vq1+d1∗u,vm1+e1∗u,vl1<∞,Dtα1−1T1un,vnt=∫1∞K3t,sf1un,vnsds+∫1∞H3t,sf2un,vnsds≤Γα1ϱ1+ρ1a1∗+b1∗u,vp1+c1∗u,vq1+d1∗u,vm1+e1∗u,vl1<∞.
With the Lebesgue-dominated convergence theorem and continuity off1,f2, we guarantee that
(84)limn→∞∫1∞K1t,s1+logtα1−1f1un,vnsds+∫1∞H1t,s1+logtα1−1f2un,vnsds=∫1∞K1t,s1+logtα1−1f1s,us,vs,Dsα1−1us,Dsα2−1vsds+∫1∞H1t,s1+logtα1−1f2s,us,vs,Dsα1−1us,Dsα2−1vsds,limn→∞∫1∞K3t,sf1un,vnsds+∫1∞H3t,sf2un,vnsds=∫1∞K3t,sf1u,vsds+∫1∞H3t,sf2u,vsds.
So we have(85)T1un,vnt−T1u,vt=supt∈1,∞∫1∞K1t,s1+logtα1−1f1un,vns−f1u,vsds+∫1∞H1t,s1+logtα1−1f2un,vns−f2u,vsds⟶0,n⟶∞,Dtα1−1T1un,vn−Dtα1−1T1u,v=supt∈1,∞Dtα1−1T1un,vn−Dtα1−1T1u,v⟶0,n⟶∞.
Above all,T1 is continuous. Meanwhile, we can obtain that T2 is continuous. In consequence, T is continuous.
From the above steps, we get the operatorT:P⟶P is completely continuous. The proof is completed.


## 3. Existence Results of Monotone Iterative Positive Solutions

Now, based on Lemma7, Lemma 8, and Lemma 11, in what follows, we show that there exist positive extremal solutions for BVP (5) and (6) by the monotone iterative method.Theorem 12.
Assume that (F1)-(F4) hold, for0≤pk,qk,mk,lk<1, k=1,2, choose
(86)A1=maxϱ1,ϱ2,ρ1,ρ2,(87)A2≥maxak∗,bk∗,ck∗,dk∗,ek∗,andA2<∞,fork=1,2,(88)p0=maxpk,qk,mk,lk,fork=1,2,(89)Λ=max10∗A1∗A21/1−p0,1.
Under the conditions (H1) and (H2), BVP (5) and (6) have positive solutions u∗,v∗ and w∗,z∗ satisfying 0≤u∗,v∗X×Y≤∧, which limn→∞un,vn=u∗,v∗,
(90)un,vn=Tun−1,vn−1=T1un−1,vn−1T2un−1,vn−1,n=1,2,⋯,(91)u0tv0t=Λlogtα1−1Λlogtα2−1,
0≤w∗,z∗X×Y≤Λ, which limn→∞wn,zn=w∗,z∗,
(92)wn,zn=Twn−1,zn−1=T1wn−1,zn−1T2wn−1,zn−1,n=1,2,⋯,(93)w0tz0t=00.Proof.
DenoteB=u,v∈X×Y:∥u,v∥X×Y≤Λ, where Λ is introduced by (86). In the following, we first prove that T:B⟶B. For any u,v∈B, in view of Lemma 7 and Lemma 9, we have
(94)T1u,v=supt∈1,∞∫1∞K1t,s1+logtα1−1f1u,vsds+∫1∞H1t,s1+logtα1−1f2u,vsds≤ϱ1∫1∞f1u,vs1s+1ds+ρ1∫1∞f2u,vs1s+1ds≤ϱ1a1∗+b1∗u,vp1+c1∗u,vq1+d1∗u,vm1+e1∗u,vl1+ρ1a2∗+b2∗u,vp2+c2∗u,vq2+d2∗u,vm2+e2∗u,vl2≤ϱ1a1∗+b1∗Λp1+c1∗Λq1+d1∗Λm1+e1∗Λl1+ρ1a2∗+b2∗Λp2+c2∗Λq2+d2∗Λm2+e2∗Λl2≤Λ,Dtα1−1T1u,v=supt∈1,∞∫1∞K3t,sf1u,vsds+∫1∞H3t,sf2u,vsds≤Γα1ϱ1∫1∞f1u,vs1s+1ds+Γα1ρ1∫1∞f2u,vs1s+1ds≤Γα1ϱ1a1∗+b1∗u,vp1+c1∗u,vq1+d1∗u,vm1+e1∗u,vl1+Γα1ρ1a2∗+b2∗u,vp2+c2∗u,vq2+d2∗u,vm2+e2∗u,vl2≤ϱ1a1∗+b1∗Λp1+c1∗Λq1+d1∗Λm1+e1∗Λl1+ρ1a2∗+b2∗Λp2+c2∗Λq2+d2∗Λm2+e2∗Λl2≤Λ,further, we can also get T2u,v≤Λ,Dtα2−1T2u,v≤Λ.
As a result, we obtain(95)Tu,v≤Λ,
andT:B⟶B.According to (91) and (93), it is obvious that u0,v0,w0,z0∈B. By using the completely continuous operator T, we define the sequences un,vn and wn,zn as un,vn=Tun−1,vn−1,wn,zn=Twn−1,zn−1, for n=1,2,⋯ Since T:B⟶B, we get that un,vn,wn,zn∈B, for n=1,2,⋯Fort∈1,∞, according to the definition of the iterative scheme, we have
(96)T1u0,v0t=∫1∞K1t,sf1u0,v0sds+∫1∞H1t,sf2u0,v0sds≤∫1∞logtα1−1ϱ1f1u0,v0s1s+1ds+∫1∞logtα1−1ρ1f2u0,v0s1s+1ds≤ϱ1logtα1−1a1∗+b1∗u,vp1+c1∗u,vq1+d1∗u,vm1+e1∗u,vl1+ρ1logtα1−1a2∗+b2∗u,vp2+c2∗u,vq2+d2∗u,vm2+e2∗u,vl2≤Λlogtα1−1=u0t.Similarly,(97)T2u0,v0t≤Λlogtα2−1=v0t.From (96) and (97), we can get
(98)u1tv1t=T1u0,v0tT2u0,v0t≤u0tv0t.Next, we discuss the monotonicity of fractional derivative ofu,v. From (91), we obtain
(99)Dtα1−1u0t=ΛΓα1≥0,Dtα2−1u0t=ΛΓα2≥0.Hence, one has(100)Dtα1−1u1t=Dtα1−1T1u0,v0t=∫1∞K3t,sf1u0,v0sds+∫1∞H3t,sf2u0,v0sds≤ΛΓα1=Dtα1−1u0t,(101)Dtα2−1v1t≤ΛΓα2=Dtα2−1v0t.Combining (100) and (101), it is easy to see that
(102)Dtα1−1u1tDtα2−1v1t=Dtα1−1T1u0,v0tDtα2−1T2u0,v0t≤Dtα1−1u0tDtα2−1v0t.Thus, fort≥1, from (98), (102), and H2, we do the second iteration
(103)u2tv2t=T1u1,v1tT2u1,v1t≤T1u0,v0tT2u0,v0t=u1tv1t,Dtα1−1u2tDtα2−1v2t=Dtα1−1T1u1,v1tDtα2−1T2u1,v1t≤Dtα1−1T1u0,v0tDtα2−1T2u0,v0t=Dtα1−1u1tDtα2−1v1t.By induction, forn=0,1,2,⋯, one has
(104)un+1tvn+1t≤untvnt,Dtα1−1un+1tDtα2−1vn+1t≤Dtα1−1untDtα2−1vnt,fort≥1.Due to the complete continuity of the operatorT, there exist u∗,v∗ such that unk,vnk⟶u∗,v∗ as k⟶∞, which can be obtained through that un,vnn=1∞ has a convergent subsequence unk,vnkk=1∞. This together with (104) implies that limn→∞un,vn=u∗,v∗. Additionally, T is continuous and un+1,vn+1=Tun,vn; it implies Tu∗,v∗=u∗,v∗.For the sequencewn,znn=1∞, we apply a similar argument. For t∈1,∞, we have
(105)w1tz1t=T1w0,z0tT2w0,z0t=∫1∞K1t,sf1w0,z0sds+∫1∞H1t,sf2w0,z0sds,∫1∞K2t,sf2w0,z0sds+∫1∞H2t,sf1w0,z0sds≥00=w0tz0t,Dtα1−1w1tDtα2−1z1t=Dtα1−1T1w0,z0tDtα2−1T2w0,z0t=∫1∞K3t,sf1w0,z0sds+∫1∞H3t,sf2w0,z0sds∫1∞K4t,sf2w0,z0sds+∫1∞H4t,sf1w0,z0sds≥00=Dtα1−1w0tDtα2−1z0t.To sum up, fort≥1, we have
(106)w2tz2t=T1w1,z1tT2w1,z1t≥T1w0,z0tT2w0,z0t=w1tz1t,Dtα1−1w2tDtα2−1z2t=Dtα1−1T1w1,z1tDtα2−1T2w1,z1t≥Dtα1−1T1w0,z0tDtα2−1T2w0,z0t=Dtα1−1w1tDtα2−1z1t.Analogously, forn=1,2,⋯, one has
(107)wn+1tzn+1t≥wntznt,Dtα1−1wn+1tDtα2−1zn+1t≥Dtα1−1wntDtα2−1znt,fort≥1.Similarly, we can also get thatlimn→∞wn,zn=w∗,z∗. Additionally, T is continuous and wn+1,zn+1=Twn,zn; it implies that Tw∗,z∗=w∗,z∗.Consequently, there existu∗,v∗ and w∗,z∗ in B which are nonnegative extremal solutions of BVP (5) and (6). Since fkt,0,0,0,0≢0,∀t∈1,∞, for k=1,2, then zero is not a solution of problem (5). It is obvious that u∗,v∗t>0,w∗,z∗t>0. The proof is completed.Example 13.
Fort∈1,∞, consider the following fractional differential system:
(108)Dt1.8ut+117e−t+ut0.5e−t101+logt0.80.5+vt0.3e−t321+logt0.80.3+logt+1t+13Dt0.8ut0.6+logt+1t+32t+1Dt0.6vt0.1=0,Dt1.6vt+14πe−t+ut0.7e−t501+logt0.60.7+vt0.4e−t181+logt0.60.4+logt+1t+13Dt0.8ut0.2+logt+1t+42t+1Dt0.6vt0.8=0,with the coupled integral and discrete mixed boundary conditions:
(109)u1=0,Dt0.8u∞=∑i=12λ1iItβ1ivξi+∑j=12bjvηj,v1=0,Dt0.6v∞=∑i=12λ2iItβ2iuξi+∑j=12bjuηj.
In this model, we set(110)λ11=12,λ21=27,β11=1.5,β21=1.2,ξ1=1.5,b1=13,η1=1.2,λ12=14,λ22=37,β12=2.5,β22=2.2,ξ2=1.7,b2=23,η2=2.5.
It is obvious that (F1) and (F2) hold. By calculation, we get(111)l1=0.7878,l2=0.7384,ϱ1=3.5674,ϱ2=3.7186,ρ1=2.4778,ρ2=2.3226.
According to the forms off1u,vt and f2u,vt shown in (108), we have
(112)a1t=117e−t,a2t=14πe−t,b1t=e−t101+logt0.80.5,b2t=e−t501+logt0.60.7,c1t=e−t311+logt0.80.3,c2t=e−t181+logt0.60.4,d1t=tt+13,d2t=tt+13,e1t=tt+32t+1,e2t=tt+42t+1.
We can show that
f1u,vt≤11/7e−1+ut0.5e−t/101+logt0.80.5+vt0.3e−t/321+logt0.80.3+t/t+13Dt0.8ut0.6+t/t+32t+1Dt0.6vt0.1=a1t+b1tut0.5+c1tvt0.3+d1tDt0.8ut0.6+e1tDt0.6vt0.1, similarly, we get
(113)f2u,vt≤a2t+b2tut0.7+c2tvt0.4+d2tDt0.8ut0.2+e2tDt0.6vt0.8.
And, we have(114)a1∗=∫1∞a1t1t+1dt<2∫1∞117e−tdt=227e<∞,b1∗=∫1∞b1t1+logtα1−10.51t+1dt=∫1∞110e−t1t+1dt<15e<∞,c1∗=∫1∞c1t1+logtα2−10.31t+1dt=∫1∞132e−t1t+1dt<116e<∞,d1∗=∫1∞d1t1t+1dt=∫1∞1t+12dt=12<∞,e1∗=∫1∞e1t1t+1dt=∫1∞1t+32dt=14<∞,also, a2∗,b2∗,c2∗,d2∗,e2∗<∞, which imply that (H1) holds.From the expression of functionf, we can know that f is increasing with respect to the second, third, fourth and last variables, so H2 holds.SetA1=3.7186, A2=1/2, and p0=0.8, we get Λ=10.363. Thus, by Theorem 12, BVP (108) and (109) have monotone positive solutions u∗,v∗ and w∗,z∗, respectively, in 0,Λlogtα−1, which can be approximated by the following iterative sequences:
(115)un+1tvn+1t=T1un,vntT2un,vnt=∫01K1t,sf1un,vnsds+∫01H1t,sf2un,vnsds∫01K2t,sf2un,vnsds+∫01H2t,sf1un,vnsds,wn+1tzn+1t=∫01K1t,sf1wn,znsds+∫01H1t,sf2wn,znsds∫01K2t,sf2wn,znsds+∫01H2t,sf1wn,znsds,and the initial values are
(116)u0tv0t=Λlogtα1−1Λlogtα2−1=10.363logt0.810.363logt0.6,w0tz0t=00.



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*Source: 6508075-2020-07-03.xml*