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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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Volume 15, Issue 1

# Positive solutions for Hadamard differential systems with fractional integral conditions on an unbounded domain

• Corresponding author
• Nonlinear Dynamic Analysis Research Center. Department of Mathematics. Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand
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/ Sotiris K. Ntouyas
• Department of Mathematics, University of Ioannina, 451 10 Ioannina, Ioannina, Greece
• Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589 Saudi Arabia
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• Other articles by this author:
/ Suphawat Asawasamrit
• Nonlinear Dynamic Analysis Research Center. Department of Mathematics. Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand
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/ Chanon Promsakon
• Nonlinear Dynamic Analysis Research Center. Department of Mathematics. Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand
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Published Online: 2017-05-25 | DOI: https://doi.org/10.1515/math-2017-0057

## Abstract

In this paper, we investigate the existence of positive solutions for Hadamard type fractional differential system with coupled nonlocal fractional integral boundary conditions on an infinite domain. Our analysis relies on Guo-Krasnoselskii’s and Leggett-Williams fixed point theorems. The obtained results are well illustrated with the aid of examples.

MSC 2010: 26A33; 34A08; 34B18; 34B40

## 1 Introduction

Fractional calculus and fractional differential equations have been studied extensively during the last decades. Fractional derivatives provide a more excellent tool for the description of memory and hereditary properties of various materials and processes than integer derivatives. Engineers and scientists have developed new models that involve fractional differential equations. These models have been applied successfully in, e.g., physics, biomathematics, blood flow phenomena, ecology, environmental issues, viscoelasticity, aerodynamics, electrodynamics of complex medium, electrical circuits, electron-analytical chemistry, control theory, etc. For a systematic development of the topic, we refer to the books [1]-[7]. As an important issue for the theory of fractional differential equations, the existence of solutions to kinds of boundary value problems has attracted many scholars attention, and lots of excellent results have been obtained by means of fixed point theorems, upper and lower solutions technique, and so forth. A variety of results on initial and boundary value problems of fractional differential equations and inclusions can easily be found in the literature on the topic. For some recent results, we can refer to [8]-[18] and references cited therein.

Boundary value problems on infinite intervals appear often in applied mathematics and physics. Due to the fact that an infinite interval is noncompact, the discussion about boundary value problem on the infinite intervals is more complicated. Results on the existence of solutions of boundary value problems on infinite intervals for differential, difference and integral equations may be found in the monographs [19,20]. For boundary value problems of fractional order on infinite intervals we refer to [21]-[25].

Many researchers have shown their interest in the study of systems of fractional differential equations. The motivation for those works stems from both the intensive development of the theory of fractional calculus itself and the applications. See for example [26]-[30] where systems for fractional differential equations were studied by using Banach contraction mapping principle and Schaefer’s fixed point theorem.

Recently in [31] we investigated the existence of positive solutions for fractional differential equations of Hadamard type, with integral boundary condition on infinite intervals $Dαu(t)+a(t)f(u(t))=0,1<α≤2,t∈(1,∞),$(1) $u(1)=0,Dα−1u(∞)=∑i=1mλiIβiu(η),$(2)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 given constants. For some recent results on positive solutions of fractional differential equations we refer to [32]-[37] and references cited therein.

In [38] the existence of positive solutions were studied for the following fractional system of differential equations subject to the nonlocal Riemann-Liouville fractional integral boundary conditions of the form$Dpx(t)+f(t,x(t)),y(t))=0,1(3) where DΦ are Riemann-Liouville fractional derivatives of orders $\varphi \phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}\left\{p,q\right\},f,g\phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}C\left(\left[0,1\right]×{\mathbb{R}}_{+}^{2},{\mathbb{R}}_{+}\right),$Iϕ are Riemann-Liouville fractional integrals of order ϕ ∈ {γi, μj}, αi, βj > 0, i = 1,..., m, j = 1,..., n and the fixed constants 0 < η < ξ < 1.

In this paper we investigate the existence of positive solutions for the following fractional system of Hadamard differential equations subject to the fractional integral boundary conditions on an unbounded domain $Dpx(t)+a(t)f(x(t)),y(t))=0,1(4) where DΦ are Hadamard fractional derivatives of orders Φ ∈ {p, q} with lower limit $1,f,g\phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}C\left(\left[1,\mathrm{\infty }\right]×{\mathbb{R}}_{+}^{2},{\mathbb{R}}_{+}\right),$ IΦ are Hadamard fractional integrals of order Φ ∈ {αi, βj} with lower limit 1, λi, σj > 0, i,..., m, j = 1,..., n.

Applying first the well-known Guo-Krasnoselskii’s fixed point theorem we obtain the existence of at least one positive solution. Next we prove the existence of at least three distinct nonnegative solutions by using Leggett-Williams fixed point theorem.

The rest of the paper is organized as follows: In section 2, we present some preliminaries and lemmas that will be used to prove our main results. We also obtain the corresponding Green’s function and some of its properties. The main results are formulated and proved in Section 3. Especially in Subsection 3.1 we prove the existence of at least one positive solution while, in Subsection 3.2, we prove the existence of at least three distinct nonnegative solutions.

Examples illustrating our results are presented in Section 4.

## 2 Background materials and preliminaries

In this section, we present some notations and definitions of Hadamard fractional calculus (see [4]) and present preliminary results needed in our proofs later.

#### Definition 2.1

([14]). The Hadamard fractional integral of order Φ with lower limit 1 for a function g : [1, ∞) → is defined as$IΦg(t)=1Γ(q)∫1tlogtsΦ−1g(s)sds,Φ>0,$(5)provided the integral exists, where log(•) = loge(•).

#### Definition 2.2

([4]). The Hadamard derivative of fractional order Φ with lower limit 1 for a function g : [1, ∞) → ℝ ℝ is defined as$Dϕg(t)=1Γ(n−ϕ)tddtn∫1tlogtsn−ϕ−1g(s)sds,n−1<ϕ(6)where n = [Φ] + 1, [Φ] denotes the integer part of the real number Φ.

#### Lemma 2.3

([4, Property 2.24]). If a, α, β > 0 then$Daαlogtaβ−1(x)=Γ(β)Γ(β−α)logxaβ−α−1.$(7)and$Iαlogtaβ−1(x)=Γ(β)Γ(β+α)logxaβ+α−1.$(8)

#### Lemma 2.4

([4]). Let q < 0 and x ∈ C[1, ∞) ⋂ L1[1, ∞). Then the Hadamard fractional differential equation Dqx(t) = 0 has the solutions$x(t)=∑i=1nci(log⁡t)q−i,$and the following formula holds:$IqDqx(t)=x(t)+∑i=1nci(log⁡t)q−i,$where ci ∈ ℝ, i = 1, 2,...., n, and n — 1 < q < n.

#### Lemma 2.5

Suppose that the functions u,v ∈ C([1, ∞),ℝ+) and 1 < p, q ≤ 2. Then the following system$Dpx(t)+u(t)=0,t∈(1,∞),Dqy(t)+v(t)=0,t∈(1,∞),x(1)=0,Dp−1x(∞)=∑i=1mλiIαiy(η),y(1)=0,Dq−1y(∞)=∑j=1nσjIβjx(ξ),$(9)can be written in the equivalent integral equations of the form$x(t)=−1Γ(p)∫1tlogtsp−1u(s)dss+(log⁡t)p−1[Γ(q)Ω∫1∞u(s)dss−Γ(q)Ω∑i=1mλiΓ(q+αi)∫1ηlogηsq+αi−1u(s)dss+Λ1Ω∫1∞v(s)dss−Λ1Ω∑j=1nσjΓ(p+βj)∫1ξlogξsp+βj−1u(s)dss],$(10)and$y(t)=−1Γ(q)∫1t(log⁡ts)q−1v(s)dss+(log⁡t)q−1[Γ(p)Ω∫1∞v(s)dss−Γ(p)Ω∑j=1nσjΓ(p+βj)∫1ξ(log⁡ξs)p+βj−1u(s)dss+Λ2Ω∫1∞u(s)dss−Λ2Ω∑i=1mλiΓ(q+αi)∫1η(log⁡ηs)q+αi−1v(s)dss],$(11)where$Ω:=Γ(p)Γ(q)−Λ1Λ2>0,$with$Λ1:=∑i=1mλiΓ(q)Γ(q+αi)(log⁡η)q+αi−1andΛ2:=∑j=1nσjΓ(p)Γ(p+βj)(log⁡ξ)p+βj−1.$

#### Proof

By applying the Hadamard fractional integral of orders p and q to both sides of the first two equations of system (9), respectively, we obtain $x(t)=−1Γ(p)∫1tlog⁡tsp−1u(s)dss+c1(log⁡t)p−1+c2(log⁡t)p−2,y(t)=−1Γ(q)∫1tlog⁡tsq−1v(s)dss+k1(log⁡t)q−1+k2(log⁡t)q−2,$ where c1, c2, k1, k2 ∈ ℝ.

The conditions of (9) that x(1) = 0 and y(1) = 0 imply c2 = 0 and k2 = 0. Hence $x(t)=−1Γ(p)∫1tlog⁡tsp−1u(s)dss+c1(log⁡t)p−1,$(12) and $y(t)=−1Γ(q)∫1tlog⁡tsq−1v(s)dss+k1(log⁡t)q−1.$(13)

From Lemma 2.3, we obtain $Dp−1x(t)=−∫1tu(s)dss+c1Γ(p)andDq−1y(t)=−∫1tv(s)dss+k1Γ(q).$

By applying the Hadamard fractional integral of orders βj and αi to (12) and (13), and also substitution t = ξ and t = η, respectively, we get $Iβjx(ξ)=−1Γ(p+βj)∫1ξlog⁡ξsp+βj−1u(s)dss+c1Γ(p)Γ(p+βj)(log⁡ξ)p+βj−1,$ and $Iαiy(η)=−1Γ(q+αi)∫1ηlog⁡ηsq+αi−1v(s)dss+k1Γ(q)Γ(q+αi)(log⁡η)q+αi−1.$

Using the conditions of (9) and solving the system of linear equations we find the constants c1 and k1 as $c1=Γ(q)Ω∫1∞u(s)dss−Γ(q)Ω∑i=1mλiΓ(q+αi)∫1ηlog⁡ηsq+αi−1v(s)dss+Λ1Ω∫1∞v(s)dss−Λ1Ω∑j=1nσjΓ(p+βj)∫1ξlog⁡ξsp+βj−1u(s)dss,$ and $k1=Γ(p)Ω∫1∞v(s)dss−Γ(p)Ω∑j=1nσjΓ(p+βj)∫1ξlog⁡ξsp+βj−1u(s)dss+Λ2Ω∫1∞u(s)dss−Λ2Ω∑i=1mλiΓ(q+αi)∫1ηlog⁡ηsq+αi−1v(s)dss.$

Substituting the values of c1 and k1 in (12) and (13), we deduce the integral equations (10) and (11), respectively.

The converse follows by direct computation. This completes the proof.  □

#### Lemma 2.6

(Green’s function). The integral equations ((10)) and ((11)), in Lemma 2.5, can be expressed in the form of Green’s functions as$x(t)=∫1∞G1(t,s)u(s)dss+∫1∞G2(t,s)v(s)dss,$(14)$y(t)=∫1∞G3(t,s)v(s)dss+∫1∞G4(t,s)u(s)dss,$(15)where the Green’s functions Gi (t, s), i = 1, 2, 3, 4 are given by$G1(t,s)=gp(t,s)+Λ1Ω∑j=1nσj(log⁡t)p−1Γ(p+βj)gβjp(ξ,s),G2(t,s)=Γ(q)Ω∑i=1mλi(log⁡t)p−1Γ(q+αi)gαiq(η,s),G3(t,s)=gq(t,s)+Λ2Ω∑i=1mλi(log⁡t)q−1Γ(q+αi)gαiq(η,s),G4(t,s)=Γ(p)Ω∑j=1nσj(log⁡t)q−1Γ(p+βj)gβjp(ξ,s),$where$gϕ(t,s)=(log⁡t)ϕ−1−(log⁡(t/s))ϕ−1I∖(ϕ),1≤s≤t<∞,(log⁡t)ϕ−1Γ(ϕ),1≤t≤s<∞,$(16)and$gψϕ(ρ,s)=(log⁡ρ)ϕ+ψ−1−(log⁡(ρ/s))ϕ+ψ−1,1≤s≤ρ<∞,(log⁡ρ)ϕ+ψ−1,1≤ρ≤s<∞,$(17)with Φ ∈ {p, q}, ψ ∈ {αj, βi}, ρ ∈ {ξ, η}.

#### Proof

By Lemma 2.5, we have $x(t)=−1Γ(p)∫1t(log⁡ts)p−1u(s)dss+(log⁡t)p−1[Γ(q)Ω∫1∞u(s)dss−Γ(q)Ω∑i=1mλiΓ(q+αi)∫1η(log⁡ηs)q+αi−1v(s)dss+Λ1Ω∫1∞v(s)dss−Λ1Ω∑j=1nσjΓ(p+βj)∫1ξ(log⁡ξs)p+βj−1u(s)dss]+1Γ(p)∫1∞(log⁡t)p−1u(s)dss−(Γ(p)Γ(q)−Λ1Λ2)ΩΓ(p)∫1∞(log⁡t)p−1u(s)dss=∫1∞gp(t,s)u(s)dss+Λ1Ω∫1∞(log⁡t)p−1v(s)dss−Γ(q)Ω∑i=1mλi(log⁡t)p−1Γ(q+αi)∫1η(log⁡ηs)q+αj−1v(s)dss−Λ1Ω∑j=1nσj(log⁡t)p−1Γ(p+βj)∫1ξ(log⁡ξs)p+βj−1u(s)dss+Λ1Λ2ΩΓ(p)∫1∞(log⁡t)p−1u(s)dss=∫1∞gp(t,s)u(s)dss+Γ(q)Ω∑i=1mλi(log⁡t)p−1Γ(q+αi)∫1∞(log⁡η)q+αj−1v(s)dss−Γ(q)Ω∑i=1mλi(log⁡t)p−1Γ(q+αi)∫1η(log⁡ηs)q+αj−1v(s)dss−Λ1Ω∑j=1nσj(log⁡t)p−1Γ(p+βj)∫1ξ(log⁡ξs)p+βj−1u(s)dss+Λ1Ω∑j=1nσj(log⁡t)p−1Γ(p+βj)∫1∞(log⁡ξ)p+βj−1u(s)dss=∫1∞gp(t,s)u(s)dss+Λ1Ω∑j=1nσj(log⁡t)p−1Γ(p+βj)∫1∞gβjp(ξ,s)u(s)dss+Γ(q)Ω∑i=1mλi(log⁡t)p−1Γ(q+αi)∫1∞gαjq(η,s)v(s)dss=∫1∞G1(t,s)u(s)dss+∫1∞G2(t,s)v(s)dss,$ which yields that (14) is satisfied. In a similar way, we have $y(t)=−1Γ(q)∫1t(log⁡ts)q−1v(s)dss+(log⁡t)q−1[Γ(p)Ω∫1∞v(s)dss−Γ(p)Ω∑j=1nσjΓ(p+βj)∫1ξ(log⁡ξs)p+βj−1u(s)dss+Λ2Ω∫1∞u(s)dss−Λ2Ω∑i=1mλiΓ(q+αi)∫1η(log⁡ηs)q+αi−1v(s)dss]+1Γ(q)∫1∞(log⁡t)q−1v(s)dss−(Γ(p)Γ(q)−Λ1Λ2)ΩΓ(q)∫1∞(log⁡t)q−1v(s)dss=∫1∞gq(t,s)v(s)dss+Λ2Ω∫1∞(log⁡t)q−1u(s)dss−Γ(p)Ω∑j=1nσj(log⁡t)q−1Γ(p+βj)∫1ξ(log⁡ξs)p+βj−1u(s)dss−Λ2Ω∑i=1mλi(log⁡t)q−1Γ(q+αi)∫1η(log⁡ηs)q+αi−1v(s)dss+Λ1Λ2ΩΓ(q)∫1∞(log⁡t)q−1v(s)dss=∫1∞gq(t,s)v(s)dss+Γ(p)Ω∑j=1nσj(log⁡t)q−1Γ(p+βj)∫1∞(log⁡ξ)p+βj−1u(s)dss−Γ(p)Ω∑j=1nσj(log⁡t)q−1Γ(p+βj)∫1ξ(log⁡ξs)p+βj−1u(s)dss−Λ2Ω∑i=1mλi(log⁡t)q−1Γ(q+αi)∫1η(log⁡ηs)q+αi−1v(s)dss+Λ2Ω∑i=1mλi(log⁡t)q−1Γ(q+αi)∫1∞(log⁡η)q+αi−1v(s)dss=∫1∞gq(t,s)v(s)dss+Λ2Ω∑i=1mλi(log⁡t)q−1Γ(q+αi)∫1∞gαiq(η,s)v(s)dss+Γ(p)Ω∑j=1nσj(log⁡t)q−1Γ(p+βj)∫1ξgβjp(ξ,s)u(s)dss=∫1∞G3(t,s)v(s)dss+∫1∞G4(t,s)u(s)dss,$ which proves that (15) holds. This completes the proof.  □

Before establishing some properties of the Green’s functions, we set $M1=1Γ(p)+Λ1Ω∑j=1nσj(log⁡ξ)p+Bj⋅−1Γ(p+βj),M2=Γ(q)Ω∑i=1mλi(log⁡η)q+αi−1Γ(q+αi),M3=1Γ(q)+Λ2Ω∑i=1mλi(log⁡η)q+αi−1Γ(q+αi),M4=Γ(p)Ω∑j=1nσj(log⁡ξ)p+Bj˙−1Γ(p+βj).$

#### Lemma 2.7

The Green’s functions Gi(t, s), i = 1, 2, 3, 4, satisfy the following properties:

• (P1)Gi (t, s), i = 1, 2, 3, 4 are continuous for (t, s) ∈ [1, ∞) × [1,∞);

• (P2)Gi(t, s) ≥ 0, i = 1, 2, 3, 4 for all (t, s) ∈ [1, ∞) × [1,∞);

• $(P3)G1(t,s)1+(log⁡t)p−1+(log⁡t)q−1≤M1,G2(t,s)1+(log⁡t)p−1+(log⁡t)q−1≤M2,G3(t,s)1+(log⁡t)p−1+(log⁡t)q−1≤M3,G4(t,s)1+(log⁡t)p−1+(log⁡t)q−1≤M4;$

• $(P4)infξ≤t≤ξk1G1(t,s)1+(log⁡t)p−1+(log⁡t)q−1≥Λ1Ω∑j=1nσj(log⁡ξ)p−1gβjp.(ξ,s)Γ(p+βj)(1+(log⁡ξ)p−1+(log⁡ξ)q−1),infξ≤t≤ξk1G2(t,s)1+(log⁡t)p−1+(log⁡t)q−1≥Γ(q)Ω∑j=1mλi(log⁡η)p−1gαiq(η,s)Γ(q+αi)(1+(log⁡η)p−1+(log⁡η)q−1),infξ≤t≤ξk1G3(t,s)1+(log⁡t)p−1+(log⁡t)q−1≥Λ2Ω∑j=1mλi(log⁡η)q−1gαiq(η,s)Γ(q+αi)(1+(log⁡η)p−1+(log⁡η)q−1),infξ≤t≤ξk1G4(t,s)1+(log⁡t)p−1+(log⁡t)q−1≥Γ(p)Ω∑j=1nσj(log⁡ξ)q−1gβjp.(ξ,s)Γ(p+βj)(1+(log⁡ξ)p−1+(log⁡ξ)q−1),foranyκ1,κ2>1.$

#### Proof

It is easy to prove that (P1) and (P2) hold.

To prove (P3), for (s, t) ∈ [1, ∞) × [1, ∞), we have $G1(t,s)1+(log⁡t)p−1+(log⁡t)q−1=gp(t,s)1+(log⁡t)p−1+(log⁡t)q−1+Λ1Ω∑j=1nσj(log⁡t)p−1gβjp(ξ,s)Γ(p+βj)(1+(log⁡t)p−1+(log⁡t)q−1)≤(log⁡t)p−1Γ(p)(1+(log⁡t)p−1+(log⁡t)q−1)+Λ1Ω∑j=1nσj(log⁡t)p−1gβjp(ξ,s)Γ(p+βj)(1+(log⁡t)p−1+(log⁡t)q−1)≤1Γ(p)+Λ1Ω∑j=1nσjgβjp(ξ,s)Γ(p+βj).$

In similar manner, we can prove that $G2(t,s)1+(log⁡t)p−1+(log⁡t)q−1<_Γ(q)Ω∑i=1nλigαiq(η,s)Γ(q+αi),G3(t,s)1+(log⁡t)p−1+(log⁡t)q−1<_1Γ(q)+Λ2Ω∑i=1mλigαiq(η,s)Γ(q+αi),G4(t,s)1+(log⁡t)p−1+(log⁡t)q−1<_Γ(p)Ω∑j=1nσjgβjp(ξ,s)Γ(p+βj).$

From (17), it follows that (P3) holds.

To prove (P4), from the positivity of gp(t, s) and ${g}_{{\beta }_{j}}^{p}\left(\xi ,s\right)$, j = 1,..., n, we have for any κ1 > 1, $minξ≤t≤ξκ1G1(t,s)1+(log⁡t)p−1+(log⁡t)q−1=minξ≤t≤ξκ1⁡[gp(t,s)1+(log⁡t)p−1+(log⁡t)q−1+Λ1Ω∑j=1nσj(log⁡t)p−1gβjp.(ξ,s)Γ(p+βj)(1+(log⁡t)p−1+(log⁡t)q−1)]≥minξ≤t≤ξκ1[Λ1Ω∑j=1nσj(log⁡t)p−1gβjp.(ξ,s)Γ(p+βj)(1+(log⁡t)p−1+(log⁡t)q−1)]≥Λ1Ω∑j=1nσj(log⁡ξ)p−1gβjp.(ξ,s)Γ(p+βj)(1+(log⁡ξ)p−1+(log⁡ξ)q−1).$

In the same way, for any κ2 > 1, we have $minη≤t≤ηκ2G2(t,s)1+(log⁡t)p−1+(log⁡t)q−1≥Γ(q)Ω∑i=1mλi(log⁡η)p−1gαiq(η,s)Γ(q+αi)(1+(log⁡η)p−1+(log⁡η)q−1),minη≤t≤ηκ2G3(t,s)1+(log⁡t)p−1+(log⁡t)q−1≥Λ2Ω∑i=1mλi(log⁡η)q−1gαiq(η,s)Γ(q+αi)(1+(log⁡η)p−1+(log⁡η)q−1),minη≤t≤ηκ1G4(t,s)1+(log⁡t)p−1+(log⁡t)q−1≥Γ(p)Ω∑j=1nσj(log⁡ξ)q−1gβjp.(ξ,s)Γ(p+βj)(1+(log⁡ξ)p−1+(log⁡ξ)q−1).$

Therefore, (P4) is proved.  □

## 3 Main results

Define the set $E={(x,y)∈C([1,∞),R)×C([1,∞),R):supt∈[1,∞)|x(t)|+|y(t)|1+(log⁡t)p−1+(log⁡t)q−1<∞},$ and the norm $∥(x,y)∥=supt∈[1,∞)|x(t)|+|y(t)|1+(log⁡t)p−1+(log⁡t)q−1.$

It is clear that (E, || • ||) is a Banach space.

#### Lemma 3.1

Let U ⊂ E be a bounded set. If the following conditions hold:

1. for any $\left(u,v\right)\left(t\right)\in U,\frac{u\left(t\right)+v\left(t\right)}{1+\left(\mathrm{log}t{\right)}^{p-1}+\left(\mathrm{log}t{\right)}^{q-1}}$is equicontinuous on any compact interval of [1, ∞);

2. for any ε > 0, there exists a constant T = T(ε) > 0 such that$u(t1)+v(t1)1+(log⁡t1)p−1+(log⁡t1)q−1−u(t2)+v(t2)1+(log⁡t2)p−1+(log⁡t2)q−1<ϵ$(18)for any t1,t2 ≥ T and (u,v) ∈ U,

then U is relatively compact in E.

The proof is similar to that of Lemma 2.8 in [31], and is omitted.

Now, we define the positive cone PE by $P={(x,y)∈E:x(t)≥0,y(t)≥0on[1,∞)},$ and the operator 𝒯 : PE by 𝒯(x, y)(t) = (A(x, y)(t), B(x, y)(t)) for all t ∈ [1, ∞), where the operators A : PE and B : PE are defined by $A(x,y)(t):=∫1∞G1(t,s)a(s)f(x(s),y(s))dss+∫1∞G2(t,s)b(s)g(x(s),y(s))dss,B(x,y)(t):=∫1∞G3(t,s)b(s)g(x(s),y(s))dss+∫1∞G4(t,s)a(s)f(x(s),y(s))dss.$(19)

Throughout this paper we assume that the following conditions hold:

• (H1)The functions f, gC([0,∞) × [0, ∞), [0, ∞)), f(x, y), g(x, y) ≠ 0 on any subinterval of (0, ∞) × (0, ∞) and f, g are bounded on [0, ∞) × [0, ∞);

• (H2)a, b : [1, ∞) → [1, ∞) are not identical zero on any closed subinterval of [1, ∞) and $0<∫1∞a(s)dss<∞,0<∫1∞b(s)dss<∞.$

Next, we are going to prove that the operator 𝒯 is completely continuous.

#### Lemma 3.2

Let (H1) and (H2) hold. Then 𝒯 : P → P is completely continuous.

#### Proof

Firstly, we will show that 𝒯 is uniformly bounded on P. Let Ω* = {(x, y) ∈ P : ||(x, y)|| ≤ r} ≤ P be a bounded set. From (H1), there exist two positive constants L1 and L2 such that $|f(x(t),y(t))|≤L1,|g(x(t),y(t))|≤L2,∀(x,y)∈Ω∗.$

Then, from (H2), for any (x, y) ∈ Ω*, we have $∥T(x,y)∥=supt∈[1,∞)11+(log⁡t)p−1+(log⁡t)q−1[∫1∞G1(t,s)a(s)f(x(s),y(s))dss+∫1∞G2(t,s)b(s)g(x(s),y(s))dss+∫1∞G3(t,s)b(s)g(x(s),y(s))dss+∫1∞G4(t,s)a(s)f(x(s),y(s))dss]≤M1∫1∞a(s)f(x(s),y(s))dss+M2∫1∞b(s)g(x(s),y(s))dss+M3∫1∞b(s)g(x(s),y(s))dss+M4∫1∞a(s)f(x(s),y(s))dss≤(M1+M4)L1∫1∞a(s)dss+(M2+M3)L2∫1∞b(s)dss<∞.$

This means that the operator 𝒯 is uniformly bounded.

Secondly, we will show that 𝒯 is equicontinuous on any compact interval of [1, ∞). For any S > 1, t1, t2 ∈ [1, S], and (x, y) ∈ Ω*, without loss of generality, we assume that t1 < t2. Then, we have $|A(x,y)(t2)1+(log⁡t2)p−1+(log⁡t2)q−1−A(x,y)(t1)1+(log⁡t1)p−1+(log⁡t1)q−1|=|∫1∞G1(t2,s)1+(log⁡t2)p−1+(log⁡t2)q−1a(s)f(x(s),y(s))dss+∫1∞G2(t2,s)1+(log⁡t2)p−1+(log⁡t2)q−1b(s)g(x(s),y(s))dss−∫1∞G1(t1,s)1+(log⁡t1)p−1+(log⁡t1)q−1a(s)f(x(s),y(s))dss−∫1∞G2(t1,s)1+(log⁡t1)p−1+(log⁡t1)q−1b(s)g(x(s),y(s))dss|≤|∫1∞(G1(t2,s)1+(log⁡t2)p−1+(log⁡t2)q−1−G1(t1,s)1+(log⁡t1)p−1+(log⁡t1)q−1)a(s)f(x(s),y(s))dss|+|∫1∞(G2(t2,s)1+(log⁡t2)p−1+(log⁡t2)q−1−G2(t1,s)1+(log⁡t1)p−1+(log⁡t1)q−1)b(s)g(x(s),y(s))dss|.$

Since $|G1(t2,s)1+(log⁡t2)p−1+(log⁡t2)q−1−G1(t1,s)1+(log⁡t1)p−1+(log⁡t1)q−1|≤|gp(t2,s)−gp(t1,s)1+(log⁡t2)p−1+(log⁡t2)q−1+|(log⁡t2)p−1−(log⁡t1)p−1+(log⁡t2)q−l−(log⁡t1)q−l|gp(t1,s)(1+(log⁡t2)p−1+(log⁡t2)q−1)(1+(log⁡t1)p−1+(log⁡t1)q−1)+|(log⁡t2)p−1−(log⁡t1)p−1+(log⁡t2)p−1(log⁡t1)q−1−(log⁡t1)p−1(log⁡t2)q−1|(1+(log⁡t2)p−1+(log⁡t2)q−1)(1+(log⁡t1)p−1+(log⁡t1)q−1)Λ1Ω∑j=1nσjgβjp(ξ,s)Γ(p+βj),$ and $∫1∞|gp(t2,s)−gp(tl,s)|1+(log⁡t2)p−1+(log⁡t2)q−1a(s)f(x(s),y(s))dss≤∫1t1|(log⁡t2)p−1−(log⁡t1)p−1|+|(log⁡(t2/s))p−1−(log⁡(t1/s))p−1|Γ)(1+(log⁡t2)p−1+(log⁡t2)q−1)a(s)f(x(s),y(s))dss+∫t1t2|(log⁡t2)p−1−(log⁡t1)p−1|+|(log⁡(t2/s))p−1|Γ(p)(1+(log⁡t2)p−1+(log⁡t2)q−1)a(s)f(x(s),y(s))dss+∫t2∞|(log⁡t2)p−1−(log⁡t1)p−1|Γ(p)(1+(log⁡t2)p−1+(log⁡t2)q−1)a(s)f(x(s),y(s))dss,$ we have that $∫1∞(G1(t2,s)1+(log⁡t2)p−1+(log⁡t2)q−1−G1(t1,s)1+(log⁡t1)p−1+(log⁡t1)q−1)a(s)f(x(s),y(s))dss→0,$ uniformly as t1t2. In addition, we can find that $G2(t2,s)1+(log⁡t2)p−1+(log⁡t2)q−1−G2(t1,s)1+(log⁡t1)p−1+(log⁡t1)q−1≤(|(log⁡t2)p−1−(log⁡t1)p−1|+|(log⁡t2)p−1(log⁡t1)q−1−(log⁡t1)p−1(log⁡t2)q−1|)(1+(log⁡t2)p−1+(log⁡t2)q−1)(1+(log⁡t1)p−1+(log⁡t1)q−1)×Γ(q)Ω∑i=1mλigαiq(η,s)Γ(q+αi),$ which leads to $∫1∞(G2(t2,s)1+(log⁡t2)p−1+(log⁡t2)q−1−G2(t1,s)1+(log⁡t1)p−1+(log⁡t1)q−1)b(s)g(x(s),y(s))dss→0,$ uniformly as t1t2. Hence $A(x,y)(t2)1+(log⁡t2)p−1+(log⁡t2)q−1−A(x,y)(t1)1+(log⁡t1)p−1+(log⁡t1)q−1→0,uniformlyast1,→t2.$

In the similar way, we can prove that $B(x,y)(t2)1+(log⁡t2)p−1+(log⁡t2)q−1−B(x,y)(t1)1+(log⁡t1)p−1+(log⁡t1)q−1→0,$ as t1t2 independently of (x, y) ∈ Ω*. Therefore 𝒯Ω* is equicontinuous on [1, ∞).

Thirdly, we will show that the operator 𝒯 is equiconvergent at ∞. From the first step, for any (x, y) ∈ Ω*, we have $∫1∞a(s)f(x(s),y(s))dss≤L1∫1∞a(s)dss<∞,$ and $∫1∞b(s)g(x(s),y(s))dss≤L2∫1∞b(s)dss<∞.$

Then $limt→∞|A(x,y)(t)1+(log⁡t)p−1+(log⁡t)q−1|=limt→∞|∫1∞G1(t,s)1+(log⁡t)p−1+(log⁡t)q−1a(s)f(x(s),y(s))dss+∫1∞G2(t,s)1+(log⁡t)p−1+(log⁡t)q−1b(s)g(x(s),y(s))dss|≤limt→∞∫1∞((log⁡t)p−1Γ(p)(1+(log⁡t)p−1+(log⁡t)q−1)+Λ1Ω∑j=1nσj(log⁡t)p−1gβjp.(ξ,s)Γ(p+βj)(1+(log⁡t)p−1+(log⁡t)q−1))a(s)f(x(s),y(s))dss+limt→∞∫1∞(Γ(q)Ω∑i=1mλi(log⁡t)p−1gαiq(η,s)Γ(q+αi)(1+(log⁡t)p−1+(log⁡t)q−1))b(s)g(x(s),y(s))dss≤M1L1∫1∞a(s)dss+M2L2∫1∞b(s)dss<∞,$ and $limt→∞|B(x,y)(t)1+(log⁡t)p−1+(log⁡t)q−1|=limt→∞|∫1∞G3(t,s)1+(log⁡t)p−1+(log⁡t)q−1b(s)g(x(s),y(s))dss+∫1∞G4(t,s)1+(log⁡t)p−1+(log⁡t)q−1a(s)f(x(s),y(s))dss|≤limt→∞∫1∞((log⁡t)q−1Γ(q)(1+(log⁡t)p−1+(log⁡t)q−1)+Λ2Ω∑i=1mλi(log⁡t)q−1gαiq(η,s)Γ(q+αi)(1+(log⁡t)p−1+(log⁡t)q−1))b(s)g(x(s),y(s))dss+limt→∞∫1∞(Γ(p)Ω∑j=1nσj(log⁡t)q−1gβjp.(ξ,s)Γ(p+βj)(1+(log⁡t)p−1+(log⁡t)q−1))a(s)f(x(s),y(s))dss≤M3L2∫1∞b(s)dss+M4L1∫1∞a(s)dss<∞,$ which imply $limt→∞T(x,y)(t)1+(log⁡t)p−1+(log⁡t)q−1<∞.$ Hence 𝒯Ω* is equiconvergent at infinity.

Finally, we will prove that the operator 𝒯 is continuous. Let (xn, yn)→ ∞ as n → ∞ in P. By applying the Lebesgue dominated convergence and the continuity of f and g guarantee that $∫1∞a(s)f(xn(s),yn(s))dss→∫1∞a(s)f(x(s),y(s))dssasn→∞,$ and $∫1∞b(s)g(xn(s),yn(s))dss→∫1∞b(s)g(x(s),y(s))dssasn→∞.$ Therefore, we get $∥T(xn,yn)−T(x,y)∥=∥(A(xn,yn)(t)−A(x,y)(t),B(xn,yn)(t)−B(x,y)(t))∥=supt∈[1,∞)|A(xn,yn)(t)−A(x,y)(t)|+|B(xnyn)(t)−B(x,y)(t)|1+(log⁡t)p−1+(1og′t)q−1≤M1∫1∞a(s)|f(xn(s),yn(s))−f(x(s),y(s))|dss+M2∫1∞b(s)|g(xn(s),yn(s))−g(x(s),y(s))|dss+M3∫1∞b(s)|g(xn(s),yn(s))−g(x(s),y(s))|dss+M4∫1∞a(s)|f(xn(s),yn(s))−f(x(s),y(s))|dss→0asn→∞.$ Thus, the operator 𝒯 is continuous.

By applying Lemma 3.1, we deduce that the operator 𝒯 : PP is completely continuous. This completes the proof.  □

To prove our main results, we set the following constants $m1=Λ1Ω∑j=1nσj(log⁡ξ)2p+Bj⋅−2Γ(p+βj)(1+(log⁡ξ)p−1+(log⁡ξ)q−1),m2=Γ(q)Ω∑i=1mλi(log⁡η)p+q+αi−2Γ(q+αi)(1+(log⁡η)p−1+(log⁡η)q−1),m3=Λ2Ω∑i=1mλi(log⁡η)2q+αi−2Γ(q+αi)(1+(log⁡η)p−1+(log⁡η)q−1),m4=Γ(p)Ω∑j=1nσj(log⁡ξ)p+q+Bj−2Γ(p+βj)(1+(log⁡ξ)p−1+(log⁡ξ)q−1),n1=∫1∞a(s)dss,n2=∫1∞b(s)dss,n3=∫ξk1ξa(s)dss,n4=∫ηk2ηb(s)dss,Λ3=n3(m1+m4),Λ4=n4(m2+m3),Λ5=n1(M1+M4),Λ6=n2(M2+M3).$

## 3.1 Existence result via Guo-Krasnoselskii’s fixed point theorem

In this subsection, the existence theorems of at least one positive solution will be established using the Guo-Krasnoselskii fixed point theorem.

#### Theorem 3.3

(Guo-Krasnoselskii fixed point theorem) [39] Let E be a Banach space, and let 𝒫E be a cone. Assume that Φ1, Φ2 are bounded open subsets of E with 0 ∈ ϕ1, ̄ϕ1 ⊂ Φ2, and let Q : 𝒫 ∩ (̄Φ2 \ Φ1) → 𝒫 be a completely continuous operator such that:

1. ||Qx||≥||x||, x𝒫 ∩ ∂ Φ1, and ||Qx||≤||x||, x𝒫Φ2; or

2. ||Qx||≤||x||, x𝒫 ∩ ∂ Φ1, and ||Qx||≥||x||, x𝒫Φ2.

Then Q has a fixed point in 𝒫 ∩ (̄Φ21).

#### Theorem 3.4

let f, g : + × ++ be continuous functions. Suppose that there exist positive constants ${\rho }_{1}<{\rho }_{2},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\theta }_{1}\in \left({\mathrm{\Lambda }}_{3}^{-1},\mathrm{\infty }\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\theta }_{2}\in \left({\mathrm{\Lambda }}_{4}^{-1},\mathrm{\infty }\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\theta }_{3}\in \left(0,{\mathrm{\Lambda }}_{5}^{-1}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\theta }_{4}\in \left(0,{\mathrm{\Lambda }}_{6}^{-1}\right)$. In addition, assume that the following conditions hold: $\left({H}_{3}\right)f\left(x,y\right)\underset{}{\ge }\frac{{\theta }_{1}{\rho }_{1}}{2}\phantom{\rule{thinmathspace}{0ex}}\mathit{f}\mathit{o}\mathit{r}\phantom{\rule{thinmathspace}{0ex}}\left(x,y\right)\in \left[0,{\rho }_{1}\right]×\left[0,{\rho }_{1}\right]\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}g\left(x,y\right)\underset{}{\ge }\frac{{\theta }_{2}{\rho }_{1}}{2}\phantom{\rule{thinmathspace}{0ex}}\mathit{f}\mathit{o}\mathit{r}\phantom{\rule{thinmathspace}{0ex}}\left(x,y\right)\in \left[0,{\rho }_{1}\right]×\left[0,{\rho }_{1}\right]\phantom{\rule{thinmathspace}{0ex}};$$\left({H}_{4}\right)f\left(x,y\right)\le \frac{{\theta }_{3}{\rho }_{2}}{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{f}\mathit{o}\mathit{r}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(x,y\right)\in \left[0,{\rho }_{2}\right]×\left[0,{\rho }_{2}\right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g\left(x,y\right)\le \frac{{\theta }_{4}{\rho }_{2}}{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(x,y\right)\in \left[0,{\rho }_{2}\right]×\left[0,{\rho }_{2}\right].$Then, the problem (4) has at least one positive solution (x, y) such that$ρ1<∥(x,y)∥<ρ2.$

#### Proof

It follows from Lemma 3.2 that the operator 𝒯 : 𝒫𝒫 is completely continuous. Define Φ1 = {(x, y) ∈ E : ||(x, y)|| < ρ1}. Hence, for any (x, y) ∈ 𝒫Φ1, we have 0 ≤ x(t) ≤ ρ1 and 0 ≤ y(t) ≤ ρ1 for all t ∈ [1, ∞). From assumption (H3) and Lemma 2.7, we get $∥T(x,y)∥=supt∈[1,∞)|A(x,y)(t)|+|B(x,y)(t)|1+(log⁡t)p−1+(log⁡t)q−1=supt∈[1,∞)11+(log⁡t)p−1+(log⁡t)q−1[∫1∞G1(t,s)a(s)f(x(s),y(s))dss+∫1∞G2(t,s)b(s)g(x(s),y(s))dss+∫1∞G3(t,s)b(s)g(x(s),y(s))dss+∫1∞G4(t,s)a(s)f(x(s),y(s))dss]≥∫1∞inft∈[ξ,k1ξ]G1(t,s)a(s)f(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss+∫1∞inft∈[η,k2η]G2(t,s)b(s)g(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss+∫1∞inft∈[η,k2η]G3(t,s)b(s)g(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss+∫1∞inft∈[ξ,k1ξ]G4(t,s)a(s)f(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss≥Λ1Ω∑j=1nσj(log⁡ξ)p−1Γ(p+βj)(1+(log⁡ξ)p−1+(log⁡ξ)q−1)∫ξk1ξgβjp(ξ,s)a(s)f(x(s),y(s))dss+Γ(q)Ω∑i=1mλi(log⁡η)p−1Γ(q+αi)(1+(log⁡η)p−1+(log⁡η)q−1)∫ηk2ηgαiq(η,s)b(s)g(x(s),y(s))dss+Λ2Ω∑i=1λi(log⁡η)q−1Γ(q+αi)(1+(log⁡η)p−1+(log⁡η)q−1)m∫ηk2ηgαiq(η,s)b(s)g(x(s),y(s))dss+rτ(p)Ω∑j=1nσj(log⁡ξ)q−1Γ(p+βj)(1+(log⁡ξ)p−1+(log⁡ξ)q−1)∫ξk1ξgβjp(ξ,s)a(s)f(x(s),y(s))dss≥12m1ρ1θ1∫ξk1ξa(s)dss+12m2ρ1θ2∫ηk2ηb(s)dss+12m3ρ1θ1∫ηk2ηb(s)dss+12m4ρ1θ2∫ξk1ξa(s)dss=12Λ3θ1ρ1+12Λ4θ2ρ1≥ρ1,$ which implies that ||𝒯(x,y)|| ≥ ||(x,y)|| for (x,y) ∈ 𝒫 ∩ ∂Φ1.

Next, we define Φ2 = {(x,y) ∈ E : ||(x,y)|| < ρ2}. Hence, for any (x,y) ∈ 𝒫 ∩∂Φ2, we have 0 ≤ x(t) ≤ ρ2 and 0 ≤ y(t) ≤ ρ2 for all t ∈ [1,∞). Using the condition (H4), we obtain $∥T(x,y)∥=supt∈[1,∞)11+(log⁡t)p−1+(log⁡t)q−1[∫1∞G1(t,s)a(s)f(x(s),y(s))dss+∫1∞G2(t,s)b(s)g(x(s),y(s))dss+∫1∞G3(t,s)b(s)g(x(s),y(s))dss+∫1∞G4(t,s)a(s)f(x(s),y(s))dss]≤12M1θ3ρ2∫1∞a(s)dss+12M2θ4ρ2∫1∞b(s)dss+12M3θ4ρ2∫1∞b(s)dss+12M4θ3ρ2∫1∞a(s)dss=12Λ5θ3ρ2+12Λ6θ4ρ2≤ρ2,$ which leads to ||𝒯(x,y)|| ≤ ||(x,y)|| for (x,y) ∈ 𝒫 ∩ ∂Φ2.

Therefore, by the first part of Theorem 3.3, we deduce that the operator 𝒯 has a fixed point in 𝒫 ∩ (̄Φ21) which is a positive solution of problem (4). Thus the problem (4) has at least one positive solution (x, y) such that $ρ1<∥(x,y)∥<ρ2.$ This completes the proof. □

Similarly to the previous theorem, we can prove the following result.

#### Theorem 3.5

Let f, g : ℝ+ × ℝ+ → ℝ+ be continuous functions. Assume that there exist positive constants ${\rho }_{1}<{\rho }_{2},\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\theta }_{1}\in \left({\mathrm{\Lambda }}_{3}^{-1},\mathrm{\infty }\right),{\theta }_{2}\in \left({\mathrm{\Lambda }}_{4}^{-1},\mathrm{\infty }\right),{\theta }_{3}\in \left(0,{\mathrm{\Lambda }}_{5}^{-1}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\theta }_{4}\in \left(0,{\mathrm{\Lambda }}_{6}^{-1}\right)$. In addition, assume that the following conditions hold:

$\left({H}_{5}\right)f\left(x,y\right)\le \frac{{\theta }_{3}{\rho }_{1}}{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{f}\mathit{o}\mathit{r}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(x,y\right)\in \left[0,{\rho }_{1}\right]×\left[0,{\rho }_{1}\right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g\left(x,y\right)\le \frac{{\theta }_{4}{\rho }_{1}}{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{f}\mathit{o}\mathit{r}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(x,y\right)\in \left[0,{\rho }_{1}\right]×\left[0,{\rho }_{1}\right];$

$\left({H}_{6}\right)f\left(x,y\right)\underset{}{\ge }\frac{{\theta }_{1}{\rho }_{2}}{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{f}\mathit{o}\mathit{r}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(x,y\right)\in \left[0,{\rho }_{2}\right]×\left[0,{\rho }_{2}\right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g\left(x,y\right)\underset{}{\ge }\frac{{\theta }_{2}{\rho }_{2}}{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{f}\mathit{o}\mathit{r}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(x,y\right)\in \left[0,{\rho }_{2}\right]×\left[0,{\rho }_{2}\right].$

Then, the problem (4) has at least one positive solution (x, y) such that$ρ1<∥(x,y)∥<ρ2.$

## 3.2 Existence result via Leggett-Williams fixed point theorem

In this subsection, the existence of at least three positive solutions will be proved using the Leggett-Williams fixed point theorem.

#### Definition 3.6

A continuous mapping ω : 𝒫 → [0,∞) is said to be a nonnegative continuous concave functional on the cone 𝒫 of a real Banach space E provided that$ω(λx+(1−λ)y)≥λω(x)+(1−λ)ω(y)$for all x,y𝒫 and λ ∈ [0,1].

Let a, b,d > 0 be given constants and define ${\mathcal{P}}_{d}=\left\{\left(x,y\right)\in \mathcal{P}:\parallel \left(x,y\right)\parallel

#### Theorem 3.7

(Leggett-Williams fixed point theorem) [40] Let 𝒫 be a cone in the real Banach space E and c>0 be a constant. Assume that there exists a concave nonnegative continuous functional ω on 𝒫 with ω (x) ≤ ||x|| for all x ∈ ̄𝒫c. Let Q : ̄𝒫c → ̄𝒫c be a completely continuous operator. Suppose that there exist constants 0 < a < b < dc such that the following conditions hold:

1. $\left\{x\in \mathcal{P}\left(\omega ,b,d\right):\omega \left(x\right)>b\right\}\ne \mathrm{\varnothing }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\omega \left(Qx\right)>b\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{f}\mathit{o}\mathit{r}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\in \mathcal{P}\left(\omega ,b,d\right)$;

2. $\parallel Qx\parallel ;

3. $\omega \left(Qx\right)>b\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{f}\mathit{o}\mathit{r}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\in \mathcal{P}\left(\omega ,b,c\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{w}\mathit{i}\mathit{t}\mathit{h}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\parallel Qx\parallel >d$.

Then Q has at least three fixed points x1, x2 and x3 in ̄𝒫c. In addition, ||x1|| < a, ω(x2) > b, ||x3|| > a with ω(x3) < b.

#### Theorem 3.8

Let f, g : ℝ+ × ℝ+ → ℝ+ be continuous functions. Suppose that there exist two constants k1, k2 > 1 such that [ξ, k1 ξ] ∩ [η, k2η] ≠ θ. In addition, assume that there exist positive constants a < b < c satisfying

$\left({H}_{7}\right)f\left(x,y\right)<\frac{a}{2{\mathrm{\Lambda }}_{5}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}g\left(x,y\right)>\frac{a}{2{\mathrm{\Lambda }}_{6}}\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}\left(x,y\right)\in \left[0,a\right]×\left[0,a\right];$

$\left({H}_{8}\right)f\left(x,y\right)<\frac{b}{2{\mathrm{\Lambda }}_{3}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}g\left(x,y\right)>\frac{b}{2{\mathrm{\Lambda }}_{4}}\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}\left(x,y\right)\in \left[b,c\right]×\left[b,c\right];$

$\left({H}_{9}\right)f\left(x,y\right)<\frac{c}{2{\mathrm{\Lambda }}_{5}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}g\left(x,y\right)>\frac{b}{2{\mathrm{\Lambda }}_{6}}\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}\left(x,y\right)\in \left[0,c\right]×\left[0,c\right].$

Then, the problem (4) has at least three positive solutions (x1, y1), (x2, y2) and (x3, y3) such that $\parallel \left({x}_{1},{y}_{1}\right)\parallel b\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\parallel \left({x}_{3},{y}_{3}\right)\parallel >a\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{w}\mathit{i}\mathit{t}\mathit{h}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{i}\mathfrak{n}{\mathrm{f}}_{{\tau }_{1}\le t\le {\tau }_{2}}\left({x}_{3},{y}_{3}\right)\left(t\right)

#### Proof

At first, we will show that the operator 𝒯 : ̄𝒫c → ̄𝒫c. For any (x, y) ∈ ̄𝒫c, we have ||(x, y)|| ≤ c. Using the condition (H9) and Lemma 2.7, we obtain $∥T(x,y)∥=supt∈[1,∞)11+(log⁡t)p−1+(log⁡t)q−1[∫1∞G1(t,s)a(s)f(x(s),y(s))dss+∫1∞G2(t,s)b(s)g(x(s),y(s))dss+∫1∞G3(t,s)b(s)g(x(s),y(s))dss+∫1∞G4(t,s)a(s)f(x(s),y(s))dss]≤∫1∞supt∈[1,∞)G1(t,s)1+(log⁡t)p−1+(log⁡t)q−1a(s)f(x(s),y(s))dss+∫1∞supt∈[1,∞)G2(t,s)1+(log⁡t)p−1+(log⁡t)q−1b(s)g(x(s),y(s))dss+∫1∞supt∈[1,∞)G3(t,s)1+(log⁡t)p−1+(log⁡t)q−1b(s)g(x(s),y(s))dss+∫1∞supt∈[1,∞)G4(t,s)1+(log⁡t)p−1+(log⁡t)q−1a(s)f(x(s),y(s))dss≤c2Λ5M1n1+c2Λ6M2n2+c2Λ6M3n2+c2Λ5M4n1=c,$ which yields 𝒯 : ̄𝒫c → ̄𝒫c. Secondly, we let (x, y) ∈ ̄Pa. By applying condition (H1 ), we have $∥T(x,y)∥≤∫1∞supt∈[1,∞)G1(t,s)1+(log⁡t)p−1+(log⁡t)q−1a(s)f(x(s),y(s))dss+∫1∞supt∈[1,∞)G2(t,s)1+(log⁡t)p−1+(log⁡t)q−1b(s)g(x(s),y(s))dss+∫1∞supt∈[1,∞)G3(t,s)1+(log⁡t)p−1+(log⁡t)q−1b(s)g(x(s),y(s))dss+∫1∞supt∈[1,∞)G4(t,s)1+(log⁡t)p−1+(log⁡t)q−1a(s)f(x(s),y(s))dss This means that the condition (i i) of Theorem 3.7 is satisfied.

Thirdly, we define ${\tau }_{1}=max\left\{\xi ,\eta \right\},{\tau }_{2}=min\left\{{k}_{1}\xi ,{k}_{2}\eta \right\}$ and also a concave nonnegative continuous functional ω on E by $ω((x,y))=infτ1≤t≤τ2⁡|x(t)|+|y(t)|1+(log⁡t)p−1+(log⁡t)q−1.$ By choosing $(x,y)(t)=((b+c)2(1+(log⁡t)p−1+(log⁡t)q−1),(b+c)2(1+(log⁡t)p−1+(log⁡t)q−1)),$ we deduce that (x, y)(t) ∈ ̄𝒫(ω, b, c) and ω((x, y)) > b. Hence it follows that {(x, y) ∈ ̄𝒫(ω, b, c) : ω((x, y)) > b} ≠ Ø. Therefore, if (x, y) ∈ ̄𝒫(ω, b, c), then we have bx(t) ≤ c and by(t) ≤ c for t ∈ [τ1, τ2]. Using condition (H8) and Lemma 2.7, we obtain $ω(T(x,y)(t))=infτ1≤t≤τ2|A(x,y)(t)|+|B(x,y)(t)|1+(log⁡t)p−1+(log⁡t)q−1=infτ1≤t≤τ211+(log⁡t)p−1+(log⁡t)q−1[∫1∞G1(t,s)a(s)f(x(s),y(s))dss+∫1∞G2(t,s)b(s)g(x(s),y(s))dss+∫1∞G3(t,s)b(s)g(x(s),y(s))dss+∫1∞G4(t,s)a(s)f(x(s),y(s))dss]≥∫1∞infτ1≤t≤τ2G1(t,s)a(s)f(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss+∫1∞infτ1≤t≤τ2G2(t,s)b(s)g(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss+∫1∞infτ1≤t≤τ2G3(t,s)b(s)g(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss+∫1∞infτ1≤t≤τ2G4(t,s)a(s)f(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss>b2Λ3m1n3+b2Λ4m2n4+b2Λ4m3n4+b2Λ3m4n3=b.$ Therefore, ω(𝒯(x,y)) > b for all (x,y) ∈ P(ω,b,c). This means that the condition (i) of Theorem 3.7 is fulfilled.

Finally, we assume that (x,y)(t) ∈ 𝒫(ω,b,c) with ||𝒯 (x,y)|| > d, where b < dc. Then, we have bx(t) ≤ c and by(t) ≤ c for t ∈ [τ1, τ2]. From condition (H8) and Lemma 2.7, we get $ω(T(x,y)(t))=infτ1≤t≤τ2⁡11+(log⁡t)p−1+(log⁡t)q−1[∫1∞G1(t,s)a(s)f(x(s),y(s))dss+∫1∞G2(t,s)b(s)g(x(s),y(s))dss+∫1∞G3(t,s)b(s)g(x(s),y(s))dss+∫1∞G4(t,s)a(s)f(x(s),y(s))dss]≥∫1∞infτ1≤t≤τ2⁡G1(t,s)a(s)f(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss+∫1∞infτ1≤t≤τ2⁡G2(t,s)b(s)g(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss+∫1∞infτ1≤t≤τ2⁡G3(t,s)b(s)g(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss+∫1∞infτ1≤t≤τ2⁡G4(t,s)a(s)f(x(s),y(s))1+(log⁡t)p−1+(log⁡t)q−1dss>b2Λ3n3(m1+m4)+b2Λ4n4(m2+m3)=b.$ It follows that the condition (iii) of Theorem 3.7 is satisfied. Hence, by applying Theorem 3.7, we deduce that the problem (4) has at least three positive solutions (x1, y1), (x2, y2) and (x3, y3) such that ||(x1, y1|| < a, $\mathrm{i}\mathfrak{n}{\mathrm{f}}_{{\tau }_{1}\le t\le {\tau }_{2}}\left({x}_{2},{y}_{2}\right)\left(t\right)>b\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\parallel \left({x}_{3},{y}_{3}\right)\parallel >a\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}with\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{i}\mathfrak{n}{\mathrm{f}}_{{\tau }_{1}\le t\le {\tau }_{2}}\left({x}_{3},{y}_{3}\right)\left(t\right). This completes the proof.  □

## 4 Examples

In this section, we present two examples to illustrate our results.

#### Example 4.1

Consider the following Hadamard fractional differential system subject to boundary conditions on an unbounded domain$D3/2x(t)+e−rf(x(t),y(t))=0,t∈(1,∞),D5/3y(t)+t−2g(x(t),y(t))=0,t∈(1,∞),x(1)=0,D1/2x(∞)=23I1/2y(74)+π5I3/2y(74),y(1)=0,D2/3y(∞)=25I1/3x(83)+17I2/3x(83)+2eI4/3x(83),$(20)where$f(x,y)=x1+y(14−x)+y1+x(14−y)+7;0≤x,y≤1/4,6+e116−xy+sin2⁡((14−x)(14−y));1/4≤x,y<∞,$and$g(x,y)={x21+x2(14−y)+y41+y4(14−x)+2;0≤x,y≤1/4,2+e−xysin4((x−14)(y−14)); 1/4≤x,y<∞.$ Here $p=3/2,q=5/3,a\left(t\right)={e}^{-t},b\left(t\right)={t}^{-2},m=2,\eta =7/4,{\lambda }_{1}=2/3,{\alpha }_{1}=1/2,{\lambda }_{2}=\pi /5,{\alpha }_{2}=3/2,n=3,\xi =8/3,{\sigma }_{1}=\sqrt{2}/5,{\beta }_{1}=1/3,{\sigma }_{2}=1/7,{\beta }_{2}=2/3,{\sigma }_{3}=2/\sqrt{e}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\beta }_{3}=4/3$. We find that Λ1 = 0.3512388401 and Λ2 = 0.9782224108 which leads to Ω = 0.4564474804 > 0. In addition, we can compute that M1 = 1.977763776, M2 = 0.7695054857, M3 = 1.941574800, M4 = 2.143121504, m1 = 0.2825156640, m2 = 0.2371688476, m3 = 0.2332983559, m4 = 0.7105323246, n1 = 0.2193839344 and n2 = 0.5000000000. By choosing k1 = 2 and k2 = 3, we also obtain n3 = 0.01925300492, n4 = 0.1451247166, Λ3 = 0.01911915781, Λ4 = 0.06827641958, Λ5 = 0.9040560259 and Λ6 = 1.355540143. Observe that the functions f, g, a and b satisfy the conditions .(H1)-(H2).

Choosing ρ1 = 1/4, ρ2 = 100, θ1 = 54 ∈ (Λ-13, ∞) = (52.30355908, ∞), θ2 = 16 ∈ (Λ-14, ∞) = (14.64634505, ∞) θ3 = 1 ∈ (0, Λ-15) = (0, 1.106126137) and θ4 = (0.5 ∈ (0, Λ-16) = (0, 0.7377133058), we obtain $f(x,y)≥7≥θ1ρ12andg(x,y)≥2≥θ2ρ12,$ for 0 ≤ x,y ≤ 1/4. Also we have $f(x,y)≤50≤θ3ρ22andg(x,y)≤25≤θ4ρ22,$ for 1/4 ≤ x, y < ∞.

Hence the conditions (H3)-(H4) hold. By Theorem 3.4, we conclude that the problem (20) has at least one positive solution (x,y) such that 1/4 < ||(x,y)|| < 100.

#### Example 4.2

Consider the following Hadamard fractional differential system subject to boundary conditions on an unbounded domain$D7/4x(t)+t−3/4f(x(t),y(t))=0,t∈(1,∞),D9/5y(t)+e−2tg(x(t),y(t))=0,t∈(1,∞),x(1)=0,D3/4x(∞)=π13I1/4y(95)+712I1/2y(95)+215I3/4y(95)y(1)=0,D4/5y(∞)=316I1/5x(73)+25I2/5x(73)+13e2I3/5x(73)+38πI4/5X(73),$(21) where$f(x,y)=x5(1+x)(35−x)+y5(1+y)(35−y)+3100;0≤x,y≤3/5,3100e−|x−y|+3(x−35)+2(y−35);3/5≤x,y≤7/5,403100+e−((7/5)−y)sin2⁡(75−x);7/5≤x,y<∞,$ and $g(x,y)=xy(35−x)(35−y)+2;0≤x,y≤3/5,2e−|x2−y2|+230(x−35)+220(y−35);3/5≤x,y≤7/5,362+e−2((7/5)−x)sin4⁡(75−y);7/5≤x,y<∞.$ Here $p=7/4,q=9/5,a\left(t\right)={t}^{-3/4},b\left(t\right)={e}^{-2t},m=3,\eta =9/5,{\lambda }_{1}=\sqrt{\pi }/13,{\alpha }_{1}=1/4,{\lambda }_{2}=7/12,{\alpha }_{2}=1/2,{\lambda }_{3}=\sqrt{2}/15,{\alpha }_{2}=3/4,n=4,\xi =7/3,{\sigma }_{1}=3/16,{\beta }_{1}=1/5,{\sigma }_{2}=2/\sqrt{5},{\beta }_{2}=2/5,{\sigma }_{3}=1/{e}^{2}$ and β3 = 3/5, σ4 = 3/8ￗ and β4 = 4/5. We find that Λ1 = 0.3324581038 and Λ2 = 0.8725814055 which leads to Ω = 0.5659031627 > 0. In addition, we can compute that M1 = 1.645835954, M2 = 0.5874823216, M3 = 1.624063246, M4 = 1.541927070, m1 = 0.1785392532, m2 = 0.1696242483, m3 = 0.1547484152, m4 = 0.4894898091, n1 = 1.333333333 and n2 = 0.04890051071. By choosing k1 = 4 and k2 = 5, it is easy to see that [ξ, k1ξ] ∩ [η, k2η] ≠θ. Then we also obtain n3 = 0.4565504861, n4 = 0.006160413528, Λ3 =0.3049889931, Λ4 = 0.001998269744, Λ5 = 4.250350698 and Λ6 = 0.1081457077. From above information, the conditions (H1)-(H2) are fulfilled.

Choosing a = 3/5, b = 7/5, c = 80, we get $f(x,y)≤0.06600000000andg(x,y)≤2.008100000000,$ which yields for 0 ≤ x, y ≤ 3/5, $f(x,y)<0.07058241103=a2Λ5andg(t,x,y)<2.774035201=a2Λ6.$ In addition, we obtain $f(x,y)≥4.030000000andg(x,y)≥362.000000,$ which leads to $f(x,y)>2.295164796=b2Λ3andg(x,y)>350.3030570=b2Λ4,$ for 7/5 ≤ x, y ≤ 80. Also we have for 0 ≤ x, y ≤ 80. $f(x,y)≤9.410988138=c2Λ5andg(x,y)≤369.8713601=c2Λ6.$ It is easy to see that ${\tau }_{1}=max\left\{\xi ,\eta \right\}=7/3,{\tau }_{2}=min\left\{{k}_{1}\xi ,{k}_{2}\eta \right\}=9$.

Therefore, the conditions (H7)-(H9) of Theorem 3.8 hold. Applying Theorem 3.8, we deduce that the problem (21) has at least three positive solutions (x1, y1), (x2, y2) and (x3, y3) such that ||(x1, y1|| < 3/5, $\mathrm{i}\mathfrak{n}{\mathrm{f}}_{7/3\le t\le 9}\left({x}_{2},{y}_{2}\right)\left(t\right)>7/5\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\parallel \left({x}_{3},{y}_{3}\right)\parallel >3/5\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}with\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{i}\mathfrak{n}{\mathrm{f}}_{7/3\le t\le 9}\left({x}_{3},{y}_{3}\right)\left(t\right)<7/5.$

## Acknowledgement

This research was funded by King Mongkut’s University of Technology North Bangkok. Contract no. KMUTNB-GEN-59-18.

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Accepted: 2017-03-21

Published Online: 2017-05-25

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 645–666, ISSN (Online) 2391-5455,

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