The Existence of Positive Solutions for a New Coupled System of Multiterm Singular Fractional Integrodifferential Boundary Value Problems

Abstract

We discuss the existence of positive solutions for the coupled system of multiterm singular fractional integrodifferential boundary value problems 𝐷^𝛼₀⁺𝑢(𝑡)+𝑓₁(𝑡,𝑢(𝑡),𝑣(𝑡),(𝜙₁𝑢)(𝑡),(𝜓₁𝑣)(𝑡),𝐷^𝑝₀₊𝑢(𝑡),𝐷^𝜇1₀₊𝑣(𝑡),𝐷^𝜇2₀₊𝑣(𝑡),…,𝐷^𝜇𝑚₀₊𝑣(𝑡))=0,𝐷^𝛽₀₊𝑓𝑣(𝑡)+₂(𝑡,𝑢(𝑡),𝑣(𝑡),(𝜙₂𝑢)(𝑡),(𝜓₂𝑣)(𝑡),𝐷^𝑞₀₊𝑣(𝑡),𝐷^𝜈1₀₊𝑢(𝑡),𝐷^𝜈2₀₊𝑢(𝑡),…,𝐷^𝜈𝑚₀₊𝑢(𝑡))=0, 𝑢^(𝑖)(0)=0 and 𝑣^(𝑖)(0)=0 for all 0≤𝑖≤𝑛−2, [𝐷^𝛿1₀₊𝑢(𝑡)]_𝑡=1=0 for 2<𝛿₁<𝑛−1 and 𝛼−𝛿₁≥1, [𝐷^𝛿2₀₊𝑣(𝑡)]_𝑡=1=0 for 2<𝛿₂<𝑛−1 and 𝛽−𝛿₂≥1, where 𝑛≥4, 𝑛−1<𝛼,𝛽<𝑛, 0<𝑝,𝑞<1, 1<𝜇_𝑖,𝜈_𝑖<2(𝑖=1,2,…,𝑚), 𝛾_𝑗,𝜆_𝑗∶[0,1]×[0,1]→(0,∞) are continuous functions (𝑗=1,2) and (𝜙_𝑗𝑢)(𝑡)=∫^𝑡₀𝛾_𝑗(𝑡,𝑠)𝑢(𝑠)𝑑𝑠,(𝜓_𝑗𝑣)(𝑡)=∫^𝑡₀𝜆_𝑗(𝑡,𝑠)𝑣(𝑠)𝑑𝑠. Here 𝐷 is the standard Riemann-Liouville fractional derivative, 𝑓_𝑗(𝑗=1,2) is a Caratheodory function, and 𝑓_𝑗(𝑡,𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚) is singular at the value 0 of its variables.

1. Introduction

During the last decade, there were a lot of manuscripts on fractional differential equations (see, e.g., [1–19] and the references therein). Fractional equations have been discussed extensively as valuable tools in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in fluid mechanics, viscoelasticity, edge detection, porous media, and electromagnetism, as well as in various other areas. For more examples and details, see [3, 13–15] and references therein. On the other hand, there are many works about the existence of positive solutions of fractional differential equations (see, e.g., [1, 2, 6–8, 10, 17, 19] and the references therein).

2. The Problem

In this paper, we investigate the following coupled system of multiterm singular fractional integrodifferential boundary value problem: 𝐷^𝛼₀⁺𝑢(𝑡)+𝑓₁𝑡,𝑢(𝑡),𝑣(𝑡),𝜙₁𝑢(𝑡),𝜓₁𝑣(𝑡),𝐷^𝑝₀₊𝑢(𝑡),𝐷^𝜇1₀₊𝑣(𝑡),𝐷^𝜇2₀₊𝑣(𝑡),…,𝐷^𝜇𝑚₀₊𝑣(𝑡)𝐷=0,_𝛽0+𝑣(𝑡)+𝑓₂𝑡,𝑢(𝑡),𝑣(𝑡),𝜙₂𝑢(𝑡),𝜓₂𝑣(𝑡),𝐷^𝑞₀₊𝑣(𝑡),𝐷^𝜈1₀₊𝑢(𝑡),𝐷^𝜈2₀₊𝑢(𝑡),…,𝐷^𝜈𝑚₀₊𝑢(𝑡)𝑢=0,_(𝑖)(0)𝐷=0∀0≤𝑖≤𝑛−2,_𝛿10+𝑢(𝑡)_𝑡=1=0for2<𝛿₁<𝑛−1,𝛼−𝛿₁𝑣≥1,_(𝑖)(0)𝐷=0∀0≤𝑖≤𝑛−2,_𝛿20+𝑣(𝑡)_𝑡=1=0for2<𝛿₂<𝑛−1,𝛽−𝛿₂≥1, (1)(2)(3) where 𝑛≥4, 𝑛−1<𝛼, 𝛽<𝑛, 0<𝑝, 𝑞<1, 1<𝜇_𝑖,𝜈_𝑖<2 (𝑖=1,2,…,𝑚), 𝛾_𝑗 and 𝜆_𝑗 are positive-valued continuous functions on [0,1]×[0,1] (𝑗=1,2), 𝜙_𝑗𝑢(𝑡)=^𝑡₀𝛾_𝑗()𝑢(𝑠)𝜓𝑡,𝑠𝑑𝑠,_𝑗𝑣(𝑡)=^𝑡₀𝜆_𝑗(𝑡,𝑠)𝑣(𝑠)𝑑𝑠, (4)𝛾^𝑗₀=sup_𝑡∈[0,1]|∫^𝑡₀𝛾_𝑗(𝑡,𝑠)𝑑𝑠|<∞ and 𝜆^𝑗₀=sup_𝑡∈[0,1]|∫^𝑡₀𝜆_𝑗(𝑡,𝑠)𝑑𝑠|<∞. Also, 𝑓₁ and 𝑓₂ satisfy the local Caratheodory condition on [0,1]×𝔇(𝑓₁,𝑓₂∈Car([0,1]×𝔇)), where 𝔇 is a subset of ℝ^𝑚+5 and 𝑓_𝑗(𝑡,𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚) may be singular at the value 0 of all its variables. Here, 𝐷 is the standard Riemann-Liouville fractional derivative. In fact, a function 𝑓 satisfies the local Caratheodory condition on [0,1]×𝔇 whenever(i)𝑓(⋅;𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚)∶[0,1]→ℝ is measurable for all (𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚)∈𝔇,(ii)𝑓(𝑡;⋅,⋅,…,⋅)∶𝔇→ℝ is continuous for almost all 𝑡∈[0,1],(iii)for each compact subset 𝜅⊂𝔇, there is a function 𝜑_𝜅∈𝐿¹[0,1] such that 𝑓𝑡;𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚≤𝜑_𝜅(𝑡), (5)for almost all 𝑡∈[0,1] and all (𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚)∈𝜅. The functions 𝑢,𝑣∈𝐶²[0,1] are called positive solutions of the problem (1), (2), and (3) whenever 𝑢>0 and 𝑣>0 on (0,1], 𝐷^𝛼₀⁺𝑢, 𝐷^𝛽₀₊𝑣∈𝐿¹[0,1], 𝑢, 𝑣 satisfy boundary conditions (2), (3), and (1) holds for almost all 𝑡∈[0,1]. In 2010, Agarwal et al. reviewed positive solutions of the singular Dirichlet problem 𝐷^𝛼₀⁺𝑢(𝑡)+𝑓𝑡,𝑢(𝑡),𝐷^𝜇₀₊𝑢(𝑡)=0, (6)𝑢(0)=𝑢(1)=0, where 1<𝛼<2, 0<𝜇≤𝛼−1 and 𝑓 satisfies the local Caratheodory condition on [0,1]×(0,∞)×ℝ [1]. In 2011, Stanek reviewed the singular problem 𝐷^𝛼₀⁺𝑢(𝑡)+𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),𝐷^𝜇₀₊𝑢(𝑡)=0, (7)𝑢(0)=0, 𝑢^′(0)=𝑢^′(1)=0, where 2<𝛼<3, 0<𝜇<1 and 𝑓 satisfies the local Caratheodory condition on [0,1]×𝔇 [19]. Here, 𝔇 is a subset of ℝ³. In 2012, Bai and Sun reviewed the singular problem 𝐷^𝛼₀⁺𝑢(𝑡)+𝑓𝑡,𝑢(𝑡),𝐷^𝜈₀⁺𝑢(𝑡),𝐷^𝜇₀₊𝑢(𝑡)=0, (8)𝑢(0)=𝑢^′(0)=𝑢^′′(0)=𝑢^′′(1)=0, where 3<𝛼≤4, 0<𝜈≤1, 1<𝜇≤2 and 𝑓 satisfies the local Caratheodory condition on [0,1]×𝔇 [8]. Again, 𝔇 is a subset of ℝ³. In this year, Agarwal et al. reviewed the singular problem ^𝑐𝐷^𝛼₀⁺𝑢(𝑡)+𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),^𝑐𝐷^𝜇₀₊𝑢(𝑡)=0, (9)𝑢^′(0)=0, 𝑢(1)=0, where 1<𝛼<2, 0<𝜇<1 and 𝑓 is a 𝐿^𝑞-Caratheodory function on [0,1]×𝔅 [2]. Here, 𝔅=(0,∞)×(−∞,0)×(−∞,0), 𝑞>1/(𝛼−1) and ^𝑐𝐷 is the Caputo fractional derivative. Many researchers have established the existence and uniqueness of solutions for some systems of nonlinear fractional differential equations, but there are few works about coupled system of multiterm nonlinear fractional differential equations (see, for example, [4, 5, 20, 21]). Also, there are a few papers discussing singular system of fractional differential equations (see, e.g., [6, 16]).

Let ‖𝑥‖₁=∫¹₀|𝑥(𝑡)|𝑑𝑡 be the norm of 𝐿¹[0,1], ‖𝑥‖=max{|𝑥(𝑡)|∶𝑡∈[0,1]} the norm of 𝐶[0,1], and ‖𝑢‖_∗=max{‖𝑢‖,‖𝑢^′‖,‖𝑢^′′‖} the norm of 𝑋=𝐶²[0,1]. As you know, (𝑋,‖⋅‖_∗) and (𝑋×𝑋,‖⋅‖_∗∗) are Banach spaces, where ‖(𝑢,𝑣)‖_∗∗=max{‖𝑢‖_∗,‖𝑣‖_∗}. Suppose that 𝐴𝐶[0,1] and 𝐴𝐶^𝑘[0,1] are the spaces of absolutely continuous functions and functions having absolutely continuous 𝑘th derivatives on [0,1], respectively. In this paper, we suppose that the functions 𝑓₁ and 𝑓₂ in (1) satisfy the following conditions:(H₁)𝑓₁,𝑓₂∈Car([0,1]×𝔇), where 𝔇=(0,∞)^𝑚+5. Also, there exists a positive constant 𝑚_𝑗 such that 𝑓_𝑗(𝑡,𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚)≥𝑚_𝑗 for almost all 𝑡∈[0,1] and (𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚)∈𝔇;(H₂) there exist positive mappings 𝑘₁,𝑘₂∈𝐿¹[0,1], 𝑃₁,𝑃₂∈𝐶(𝔇), and ℎ₁,ℎ₂∈𝐶([0,∞)^𝑚+5) such that 𝑃₁ and 𝑃₂ are nonincreasing and ℎ₁ and ℎ₂ are nondecreasing in all their variables, 𝑀₁=𝑚₁𝛼−𝛿₁Γ(𝛼+1),𝑀₂=𝑚₂𝛽−𝛿₂Γ,𝛽+1lim_𝑥→∞ℎ_𝑗(𝑥,𝑥,…,𝑥)𝑥=0,𝑗=1,2₁₀,𝑃₁𝑀₁𝑠^𝛼−1,𝑀₂𝑠^𝛽−1,𝑀₁^𝑠₀𝛾₁(𝑠,𝜏)𝜏^𝛼−1𝑑𝜏,𝑀₂^𝑠₀𝜆₁(𝑠,𝜏)𝜏^𝛽−1𝛿𝑑𝜏,₁−2𝑀₁Γ𝑠𝛼−𝑝^𝛼−𝑝,𝛿₂−2𝑀₂Γ𝛽−𝜇₁𝑠^𝛽−𝜇1,𝛿₂−2𝑀₂Γ𝛽−𝜇₂𝑠^𝛽−𝜇2𝛿,…,²−2𝑀₂Γ𝛽−𝜇_𝑚𝑠^{𝛽−𝜇𝑚}𝑑𝑠<∞,₁₀𝑃₂𝑀₁𝑠^𝛼−1,𝑀₂𝑠^𝛽−1,𝑀₁^𝑠₀𝛾₂(𝑠,𝜏)𝜏^𝛼−1𝑑𝜏,𝑀₂^𝑠₀𝜆₂(𝑠,𝜏)𝜏^𝛽−1𝛿𝑑𝜏,₂−2𝑀₂Γ𝑠𝛽−𝑞^𝛽−𝑞,𝛿₁−2𝑀₁Γ𝛼−𝜈₁𝑠^𝛼−𝜈1,𝛿₁−2𝑀₁Γ𝛼−𝜈₂𝑠^𝛼−𝜈2𝛿,…,¹−2𝑀₁Γ𝛼−𝜈_𝑚𝑠^{𝛼−𝜈𝑚}𝑓𝑑𝑠<∞,_𝑗𝑡,𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚≤𝑃_𝑗𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚+𝑘_𝑗(𝑡)ℎ_𝑗𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚 (10)for almost all 𝑡∈[0,1] and (𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚)∈𝔇. Since we suppose that the problem (1) is singular, we use regularization and sequential techniques for the existence of positive solutions of the problem. In this way, for each natural number 𝑛 define the function 𝑓^𝑗_𝑛 (𝑗=1,2) by 𝑓^𝑗_𝑛𝑡,𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚=𝑓_𝑗𝑡,𝜒_𝑛(𝑥),𝜒_𝑛𝑦,𝜒_𝑛(𝑧),𝜒_𝑛(𝑤),𝜒_𝑛(𝑣),𝜒_𝑛𝑢₁,𝜒_𝑛𝑢₂,…,𝜒_𝑛𝑢_𝑚 (11) for almost all 𝑡∈[0,1] and (𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚)∈ℝ^𝑚+5, where 𝜒_𝑛(𝑥)=1𝑥if𝑥≥𝑛,1𝑛1if𝑥<𝑛. (12) Then condition (𝐻₁) implies that 𝑓^𝑗_𝑛∈Car([0,1]×ℝ^𝑚+5) and 𝑓^𝑗_𝑛(𝑡,𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚)≥𝑚_𝑗 for almost all 𝑡∈[0,1] and (𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚)∈ℝ^𝑚+5, (𝑗=1,2). Also, the condition (𝐻₂) implies 𝑓^𝑗_𝑛𝑡,𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚≤𝑃_𝑗1𝑛,1𝑛1,…,𝑛+𝑘_𝑗(𝑡)ℎ_𝑗1+𝑥,1+𝑦,1+𝑧,1+𝑤,1+𝑣,1+𝑢₁,1+𝑢₂,…,1+𝑢_𝑚 (13) for almost all 𝑡∈[0,1] and (𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚)∈[0,∞)^𝑚+5 and 𝑓^𝑗_𝑛𝑡,𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚≤𝑃_𝑗𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚+𝑘_𝑗(𝑡)ℎ_𝑗1+𝑥,1+𝑦,1+𝑧,1+𝑤,1+𝑣,1+𝑢₁,1+𝑢₂,…,1+𝑢_𝑚 (14) for almost all 𝑡∈[0,1] and (𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚)∈𝔇. For obtaining solutions of the system (1), (2), and (3), for each natural number 𝑛, we will obtain solutions of the system 𝐷^𝛼₀⁺𝑢(𝑡)+𝑓¹_𝑛𝑡,𝑢(𝑡),𝑣(𝑡),𝜙₁𝑢(𝑡),𝜓₁𝑣(𝑡),𝐷^𝑝₀₊𝑢(𝑡),𝐷^𝜇1₀₊𝑣(𝑡),𝐷^𝜇2₀₊𝑣(𝑡),…,𝐷^𝜇𝑚₀₊𝑣(𝑡)𝐷=0,_𝛽0+𝑣(𝑡)+𝑓²_𝑛𝑡,𝑢(𝑡),𝑣(𝑡),𝜙₂𝑢(𝑡),𝜓₂𝑣(𝑡),𝐷^𝑞₀₊𝑣(𝑡),𝐷^𝜈1₀₊𝑢(𝑡),𝐷^𝜈2₀₊𝑢(𝑡),…,𝐷^𝜈𝑚₀₊𝑢(𝑡)=0, (15) via the boundary conditions (2) and (3) and by using solutions of this system, we will obtain solution of the system (1), (2), and (3). It has been proved that the fractional integral 𝐼^𝛼₀⁺ maps 𝐿¹[0,1] into 𝐿¹[0,1] whenever 𝛼∈(0,1) and maps 𝐿¹[0,1] into 𝐴𝐶^[𝛼]−1[0,1] whenever 𝛼≥1 [19]. Here, [𝛼] means the integral part of 𝛼 and 𝐴𝐶⁰[0,1]=𝐴𝐶[0,1]. Suppose that 𝛼>0 and 𝛼 is not a natural number. If 𝑥∈𝐶(0,1] and 𝐷^𝛼₀⁺𝑥∈𝐿¹[0,1], then 𝑥(𝑡)=𝐼^𝛼₀⁺𝐷^𝛼₀⁺𝑥(𝑡)+∑^𝑛_𝑘=1𝑐_𝑘𝑡^𝛼−𝑘 for all 𝑡∈(0,1], where 𝑛=[𝛼]+1 and 𝑐_𝑘∈ℝ for 𝑘=1,2,…,𝑛 [7]. Suppose that 𝜈∈(0,1), 𝜇∈(1,2), 𝑥∈𝐶²[0,1], and 𝑥(0)=𝑥^′(0)=0. Then, 𝐷^𝜇₀₊𝑥∈𝐶[0,1], 𝐷^𝜈₀⁺𝑥∈𝐶[0,1], 𝐷^𝜇₀₊𝑥(𝑡)=(1/Γ(2−𝜇))∫^𝑡₀(𝑡−𝑠)^1−𝜇𝑥^′′(𝑠)𝑑𝑠, and 𝐷^𝜈₀⁺𝑥(𝑡)=(1/Γ(2−𝜈))∫^𝑡₀(𝑡−𝑠)^1−𝜈𝑥^′′(𝑠)𝑑𝑠 [2].

3. Main Results

Now, we are ready to state and prove our main results. One can find main idea of next result in [17].

Lemma 1. Let 𝑛≥4 and 𝑛−1<𝛼<𝑛. For each 𝑔∈𝐿¹[0,1], 𝑢(𝑡)=∫¹₀𝐺_𝛼(𝑡,𝑠)𝑔(𝑠)𝑑𝑠 is the unique solution of the equation 𝐷^𝛼₀⁺𝑢(𝑡)+𝑔(𝑡)=0 in 𝐶²[0,1] which satisfies the boundary condition (2), where 𝐺_𝛼()=𝑡𝑡,𝑠_𝛼−1(1−𝑠)^{𝛼−𝛿1−1}−(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑡0≤𝑠≤𝑡≤1,_𝛼−1(1−𝑠)^{𝛼−𝛿1−1}Γ(𝛼)0≤𝑡≤𝑠≤1. (16)

Proof. It is easy to see that the functions 𝑢(𝑡)=−𝐼^𝛼₀⁺𝑔(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛 are solutions of 𝐷^𝛼₀⁺𝑢(𝑡)+𝑔(𝑡)=0 in 𝐶(0,1] for all 𝑐_𝑖∈ℝ. Since [𝛼]≥3, 𝐼^𝛼₀⁺𝑔∈𝐴𝐶^[𝛼]−1[0,1] and so 𝑢(𝑡)=−𝐼^𝛼₀⁺𝑔(𝑡)+𝑐₁𝑡^𝛼−1 are solutions of 𝐷^𝛼₀⁺𝑢(𝑡)+𝑔(𝑡)=0 in 𝐶²[0,1], where 𝑐₁∈ℝ. The boundary condition (2) implies that 𝑐₂=⋯=𝑐_𝑛−1=0 and 0=𝑐₁Γ(𝛼)Γ𝛼−𝛿₁(1)^{𝛼−𝛿1−1}−1Γ𝛼−𝛿₁¹₀(1−𝑠)^{𝛼−𝛿1−1}𝑔(𝑠)𝑑𝑠. (17) Thus, 𝑢(𝑡)=(𝑡^𝛼−1/Γ(𝛼))∫¹₀(1−𝑠)^{𝛼−𝛿1−1}𝑔(𝑠)𝑑𝑠−(1/Γ(𝛼))∫^𝑡₀(𝑡−𝑠)^𝛼−1𝑔(𝑠)𝑑𝑠=∫¹₀𝐺_𝛼(𝑡,𝑠)𝑔(𝑠)𝑑𝑠 is the unique solution of the problem in 𝐶²[0,1].

Note that the Green function 𝐺_𝛼 in Lemma 1 has some properties. For example, 𝐺_𝛼, (𝜕/𝜕𝑡)𝐺_𝛼 and (𝜕²/𝜕𝑡²)𝐺_𝛼 are continuous functions on [0,1]×[0,1], 𝐺_𝛼(𝑡,𝑠)>0 on (0,1)×(0,1), 𝐺_𝛼(𝑡,𝑠)≤1/Γ(𝛼) for all (𝑡,𝑠)∈[0,1]×[0,1], ∫¹₀𝐺_𝛼(𝑡,𝑠)𝑑𝑠≥𝑡^𝛼−1/(𝛼−𝛿₁)Γ(𝛼+1) for all 𝑡∈[0,1], (𝜕/𝜕𝑡)𝐺_𝛼(𝑡,𝑠)>0 on (0,1)×(0,1), (𝜕/𝜕𝑡)𝐺_𝛼(𝑡,𝑠)≤1/Γ(𝛼−1) for all (𝑡,𝑠)∈[0,1]×[0,1], ∫¹₀(𝜕/𝜕𝑡)𝐺_𝛼(𝑡,𝑠)𝑑𝑠≥𝑡^𝛼−2/(𝛼−𝛿₁)Γ(𝛼) for all 𝑡∈[0,1], (𝜕²/𝜕𝑡²)𝐺_𝛼(𝑡,𝑠)>0 on (0,1)×(0,1), (𝜕²/𝜕𝑡²)𝐺_𝛼(𝑡,𝑠)≤1/Γ(𝛼−2) for all (𝑡,𝑠)∈[0,1]×[0,1], and ∫¹₀(𝜕²/𝜕𝑡²)𝐺_𝛼(𝑡,𝑠)𝑑𝑠≥(𝛿₁−2)𝑡^𝛼−3/(𝛼−𝛿₁)Γ(𝛼−1) for all 𝑡∈[0,1]. Let 𝑋=𝐶²[0,1]. Define the cone 𝑃 on 𝑋×𝑋 by 𝑥(0)𝑃=𝑥,𝑦∈𝑋×𝑋∶=𝑥^′(0)=𝑦(0)=𝑦^′(0)𝑥(𝑡)=0,,𝑥^′(𝑡),𝑥^′′(𝑡),𝑦(𝑡),𝑦^′(𝑡),𝑦^′′(𝑡)[.≥0,∀𝑡∈0,1] (18) It is easy to see that for each (𝑥,𝑦)∈𝑃, 𝑖=1,2,…,𝑚 and 𝑡∈[0,1] we have 𝐷^𝑝₀₊𝑥,𝐷^𝜈𝑖₀₊𝑥,𝐷^𝑞₀₊𝑦,𝐷^𝜇𝑖₀₊𝑦∈𝐶[0,1] and 𝐷^𝑝₀₊𝑥(𝑡)≥0, 𝐷^𝜈𝑖₀₊𝑥(𝑡)≥0, 𝐷^𝑞₀₊𝑦(𝑡)≥0, 𝐷^𝜇𝑖₀₊𝑦(𝑡)≥0. Now for each natural number 𝑛, define the operator 𝑇_𝑛 on 𝑃 by 𝑇_𝑛(𝑡)=𝑇𝑥,𝑦_1𝑛(𝑡)𝑇𝑥,𝑦_2𝑛(𝑡)=𝑥,𝑦₁₀𝐺_𝛼(𝑡,𝑠)𝑓¹_𝑛×𝑠,𝑥(𝑠),𝑦(𝑠),𝜙₁𝑥(𝑠),𝜓₁𝑦(𝑠),𝐷^𝑝₀₊𝑥(𝑠),𝐷^𝜇1₀₊𝑦(𝑠),𝐷^𝜇2₀₊𝑦(𝑠),…,𝐷^𝜇𝑚₀₊𝑦(𝑠)𝑑𝑠₁₀𝐺_𝛽(𝑡,𝑠)𝑓²_𝑛×𝑠,𝑥(𝑠),𝑦(𝑠),𝜙₂𝑥(𝑠),𝜓₂𝑦(𝑠),𝐷^𝑞₀₊𝑦(𝑠),𝐷^𝜈1₀₊𝑥(𝑠),𝐷^𝜈2₀₊𝑥(𝑠),…,𝐷^𝜈𝑚₀₊𝑥(𝑠)𝑑𝑠. (19)

Lemma 2. For each natural number 𝑛, 𝑇_𝑛 is a completely continuous operator on 𝑃.

Proof. Let 𝑛 be a natural number, (𝑥,𝑦)∈𝑃, 𝜌¹(𝑡)=𝑓¹_𝑛𝑡,𝑥(𝑡),𝑦(𝑡),𝜙₁𝑥(𝑡),𝜓₁𝑦(𝑡),𝐷^𝑝₀₊𝑥(𝑡),𝐷^𝜇1₀₊𝑦(𝑡),𝐷^𝜇2₀₊𝑦(𝑡),…,𝐷^𝜇𝑚₀₊𝑦(𝑡) (20) and 𝜌²(𝑡)=𝑓²_𝑛(𝑡,𝑥(𝑡), 𝑦(𝑡), (𝜙₂𝑥)(𝑡), (𝜓₂𝑦)(𝑡), 𝐷^𝑞₀₊𝑦(𝑡), 𝐷^𝜈1₀₊𝑥(𝑡), 𝐷^𝜈2₀₊𝑥(𝑡), …, 𝐷^𝜈𝑚₀₊𝑥(𝑡)). Then, 𝜌¹,𝜌²∈𝐿¹[0,1] and there exist positive constants 𝑚₁ and 𝑚₂ such that 𝜌¹(𝑡)≥𝑚₁ and 𝜌²(𝑡)≥𝑚₂ for almost all 𝑡∈[0,1]. Since 𝐺_𝛼, 𝐺_𝛽, (𝜕/𝜕𝑡)𝐺_𝛼, (𝜕/𝜕𝑡)𝐺_𝛽, (𝜕²/𝜕𝑡²)𝐺_𝛼, and (𝜕²/𝜕𝑡²)𝐺_𝛽 are nonnegative and continuous functions on [0,1]×[0,1], 𝐺_𝛼(0,𝑠)=0, and 𝐺_𝛽(0,𝑠)=0 for all 𝑠∈[0,1], we get 𝑇¹_𝑛(𝑥,𝑦),𝑇²_𝑛(𝑥,𝑦)∈𝐶²[0,1], 𝑇¹_𝑛(𝑥,𝑦)(0)=𝑇²_𝑛(𝑥,𝑦)(0)=(𝑇¹_𝑛(𝑥,𝑦))^′(0)=(𝑇²_𝑛(𝑥,𝑦))^′(0)=0, 𝑇¹_𝑛(𝑥,𝑦)≥0, 𝑇²_𝑛(𝑥,𝑦)≥0, (𝑇¹_𝑛(𝑥,𝑦))^′≥0, (𝑇²_𝑛(𝑥,𝑦))^′≥0, (𝑇¹_𝑛(𝑥,𝑦))^′′≥0, and (𝑇²_𝑛(𝑥,𝑦))^′′≥0 on [0,1]. Thus, 𝑇_𝑛 maps 𝑃 into 𝑃. Suppose that {(𝑥_𝑘,𝑦_𝑘)}_𝑘≥1 is a convergent sequence in 𝑃 and lim_𝑘→∞(𝑥_𝑘,𝑦_𝑘)=(𝑥,𝑦). Then, lim_𝑘→∞𝑥_𝑘^(𝑗)(𝑡)=𝑥^(𝑗)(𝑡) and lim_𝑘→∞𝑦_𝑘^(𝑗)(𝑡)=𝑦^(𝑗)(𝑡) uniformly on [0,1] for 𝑗=0,1,2. Since for 𝑝,𝑞∈(0,1), 𝜇_𝑖,𝜈_𝑖∈(1,2) for all 𝑖=1,2,…,𝑚, 𝐷^𝑝₀₊𝑥_𝑘(𝑡)−𝐷^𝑝₀₊𝑥(𝑡≤𝑥_𝑘′′)−𝑥^′′Γ2−𝑝^𝑡₀(𝑡−𝑠)^1−𝑝𝑥𝑑𝑠≤^𝑘′′−𝑥^′′Γ,𝐷3−𝑝_𝑞0+𝑦_𝑘(𝑡)−𝐷^𝑞₀₊𝑦(𝑡≤𝑦_𝑘′′)−𝑦^′′Γ2−𝑞^𝑡₀(𝑡−𝑠)^1−𝑞𝑦𝑑𝑠≤^𝑘′′−𝑦^′′Γ,𝐷3−𝑞_𝜈𝑖0+𝑥_𝑘(𝑡)−𝐷^𝜈𝑖₀₊𝑥(𝑡≤𝑥_𝑘′′)−𝑥^′′Γ2−𝜈_𝑖^𝑡₀(𝑡−𝑠)^1−𝜈𝑖𝑥𝑑𝑠≤^𝑘′′−𝑥^′′Γ3−𝜈_𝑖,𝐷^𝜇𝑖₀₊𝑦_𝑘(𝑡)−𝐷^𝜇𝑖₀₊𝑦(𝑡≤𝑦_𝑘′′)−𝑦^′′Γ2−𝜇_𝑖^𝑡₀(𝑡−𝑠)^1−𝜇𝑖𝑦𝑑𝑠≤^𝑘′′−𝑦^′′Γ3−𝜇_𝑖,𝜙_𝑗𝑥_𝑘(𝑡)−𝜙_𝑗𝑥(𝑡)≤sup_𝑡∈[0,1]^𝑡₀𝛾_𝑗(𝑡,𝑠)𝑑𝑠𝑥_𝑘−𝑥=𝛾^𝑗₀𝑥_𝑘,𝜓−𝑥_𝑗𝑦_𝑘(𝑡)−𝜓_𝑗𝑦(𝑡)≤sup_𝑡∈[0,1]^𝑡₀𝜆_𝑗(𝑡,𝑠)𝑑𝑠𝑦_𝑘−𝑦=𝜆^𝑗₀𝑦_𝑘−𝑦, (21) for 𝑗=1,2 and 𝑘≥1, we have lim_𝑘→∞𝐷^𝑝₀₊𝑥_𝑘(𝑡)=𝐷^𝑝₀₊𝑥(𝑡), lim_𝑘→∞𝐷^𝑞₀₊𝑦_𝑘(𝑡)=𝐷^𝑞₀₊𝑦(𝑡), lim_𝑘→∞𝐷^𝜈𝑖₀₊𝑥_𝑘(𝑡)=𝐷^𝜈𝑖₀₊𝑥(𝑡), lim_𝑘→∞𝐷^𝜇𝑖₀₊𝑦_𝑘(𝑡)=𝐷^𝜇𝑖₀₊𝑦(𝑡) for 𝑖=1,2,…,𝑚, and lim_𝑘→∞(𝜙_𝑗𝑥_𝑘)(𝑡)=(𝜙_𝑗𝑥)(𝑡), lim_𝑘→∞(𝜓_𝑗𝑦_𝑘)(𝑡)=(𝜓_𝑗𝑦)(𝑡) uniformly on [0,1] for 𝑗=1,2. Also, we have 𝐷^𝑝₀₊𝑥_𝑘≤𝑥_𝑘^′′Γ2−𝑝^𝑡₀(𝑡−𝑠)^1−𝑝𝑥𝑑𝑠≤^𝑘′′Γ,𝐷3−𝑝_𝑞0+𝑦_𝑘≤𝑦_𝑘^′′Γ2−𝑞^𝑡₀(𝑡−𝑠)^1−𝑞𝑦𝑑𝑠≤^𝑘′′Γ,𝐷3−𝑞_𝜈𝑖0+𝑥_𝑘≤𝑥_𝑘^′′Γ2−𝜈_𝑖^𝑡₀(𝑡−𝑠)^1−𝜈𝑖𝑥𝑑𝑠≤^𝑘′′Γ3−𝜈_𝑖,𝐷^𝜇𝑖₀₊𝑦_𝑘≤𝑦_𝑘^′′Γ2−𝜇_𝑖^𝑡₀(𝑡−𝑠)^1−𝜇𝑖𝑦𝑑𝑠≤^𝑘′′Γ3−𝜇_𝑖, (22) and also ‖𝜙_𝑗𝑥_𝑘‖≤𝛾^𝑗₀‖𝑥_𝑘‖ and ‖𝜓_𝑗𝑦_𝑘‖≤𝜆^𝑗₀‖𝑦_𝑘‖ for 𝑗=1,2 and 𝑘≥1. Now, put 𝜌¹_𝑘(𝑡)=𝑓¹_𝑛𝑡,𝑥_𝑘(𝑡),𝑦_𝑘(𝑡),𝜙₁𝑥_𝑘(𝑡),𝜓₁𝑦_𝑘(𝑡),𝐷^𝑝₀₊𝑥_𝑘(𝑡),𝐷^𝜇1₀₊𝑦_𝑘(𝑡),𝐷^𝜇2₀₊𝑦_𝑘(𝑡),…,𝐷^𝜇𝑚₀₊𝑦_𝑘(𝑡) (23) and𝜌²_𝑘(𝑡)=𝑓²_𝑛(𝑡,𝑥_𝑘(𝑡),𝑦_𝑘(𝑡),(𝜙₂𝑥_𝑘)(𝑡),(𝜓₂𝑦_𝑘)(𝑡),𝐷^𝑞₀₊𝑦_𝑘(𝑡), 𝐷^𝜈1₀₊𝑥_𝑘(𝑡), 𝐷^𝜈2₀₊𝑥_𝑘(𝑡),…,𝐷^𝜈𝑚₀₊𝑥_𝑘(𝑡)) for all 𝑘≥1. Then, lim_𝑘→∞𝜌^𝑗_𝑘(𝑡)=𝜌^𝑗(𝑡) for almost all 𝑡∈[0,1] and 𝑗=1,2. Since 𝑓^𝑗_𝑛∈Car([0,1]×ℝ^𝑚+5) and {𝑥_𝑘}_𝑘≥1, {𝑦_𝑘}_𝑘≥1 are bounded sequences in 𝐶²[0,1] and also by using the above inequalities, {𝐷^𝑝₀₊𝑥_𝑘}_𝑘≥1, {𝐷^𝑞₀₊𝑦_𝑘}_𝑘≥1, {𝐷^𝜈𝑖₀₊𝑥_𝑘}_𝑘≥1, {𝐷^𝜇𝑖₀₊𝑦_𝑘}_𝑘≥1 for 𝑖=1,2,…,𝑚 and {𝜙_𝑗𝑥_𝑘}_𝑘≥1, {𝜓_𝑗𝑥_𝑘}_𝑘≥1 for 𝑗=1,2 are bounded sequences in 𝐶[0,1], there exist 𝑔₁,𝑔₂∈𝐿¹[0,1] such that 𝑚_𝑗≤𝜌^𝑗_𝑘(𝑡)≤𝑔_𝑗(𝑡) for almost all 𝑡∈[0,1] and 𝑘≥1. Thus, by using the Lebesgue dominated convergence theorem and the following relations: 𝑇¹_𝑛𝑥_𝑘,𝑦_𝑘(𝑡)−𝑇¹_𝑛(𝑡≤1𝑥,𝑦)Γ(𝛼)¹₀𝜌¹_𝑘(𝑠)−𝜌¹(𝑠𝑇)𝑑𝑠,_1𝑛𝑥_𝑘,𝑦_𝑘^′(𝑡)−𝑇¹_𝑛𝑥,𝑦^′(𝑡≤1)Γ(𝛼−1)¹₀𝜌¹_𝑘(𝑠)−𝜌¹(𝑠𝑇)𝑑𝑠,_1𝑛𝑥_𝑘,𝑦_𝑘^′′(𝑡)−𝑇¹_𝑛𝑥,𝑦^′′(𝑡≤1)Γ(𝛼−2)¹₀𝜌¹_𝑘(𝑠)−𝜌¹(𝑠𝑇)𝑑𝑠,_2𝑛𝑥_𝑘,𝑦_𝑘(𝑡)−𝑇²_𝑛(𝑡≤1𝑥,𝑦)Γ𝛽¹₀𝜌²_𝑘(𝑠)−𝜌²(𝑠𝑇)𝑑𝑠,_2𝑛𝑥_𝑘,𝑦_𝑘^′(𝑡)−𝑇²_𝑛𝑥,𝑦^′(𝑡≤1)Γ𝛽−1¹₀𝜌²_𝑘(𝑠)−𝜌²(𝑠𝑇)𝑑𝑠,_2𝑛𝑥_𝑘,𝑦_𝑘^′′(𝑡)−𝑇²_𝑛𝑥,𝑦^′′(𝑡≤1)Γ𝛽−2¹₀𝜌²_𝑘(𝑠)−𝜌²(𝑠)𝑑𝑠, (24) we get lim_𝑘→∞(𝑇¹_𝑛(𝑥_𝑘,𝑦_𝑘))^(𝑗)(𝑡)=(𝑇¹_𝑛(𝑥,𝑦))^(𝑗)(𝑡) and lim_𝑘→∞(𝑇²_𝑛(𝑥_𝑘,𝑦_𝑘))^(𝑗)(𝑡)=(𝑇²_𝑛(𝑥,𝑦))^(𝑗)(𝑡) uniformly on [0,1] for 𝑗=0,1,2. This implies that 𝑇¹_𝑛 and 𝑇²_𝑛 are continuous operators and so 𝑇_𝑛 is a continuous operator. Let {(𝑥_𝑘,𝑦_𝑘)}_𝑘≥1 be a bounded sequence in 𝑃. We show that the sequence {𝑇_𝑛(𝑥_𝑘,𝑦_𝑘)}_𝑘≥1 is relatively compact in 𝑋×𝑋. By using the Arzela-Ascoli theorem, it is sufficient to prove that {𝑇_𝑛(𝑥_𝑘,𝑦_𝑘)}_𝑘≥1 is bounded in 𝑋×𝑋 and {(𝑇_𝑛(𝑥_𝑘,𝑦_𝑘))^′′} is equicontinuous on [0,1]. Choose a positive number 𝑆 such that ‖𝑥_𝑘‖≤𝑆, ‖𝑦_𝑘‖≤𝑆, ‖𝑥^′_𝑘‖≤𝑆, ‖𝑦^′_𝑘‖≤𝑆, ‖𝑥_𝑘^′′‖≤𝑆, ‖𝑦_𝑘^′′‖≤𝑆 for all 𝑘≥1. By using the above inequalities, one can see that ‖𝐷^𝑝₀₊𝑥_𝑘‖≤𝑆/Γ(3−𝑝), ‖𝐷^𝑞₀₊𝑦_𝑘‖≤𝑆/Γ(3−𝑞), ‖𝐷^𝜈𝑖₀₊𝑥_𝑘‖≤𝑆/Γ(3−𝜈_𝑖), and ‖𝐷^𝜇𝑖₀₊𝑦_𝑘‖≤𝑆/Γ(3−𝜇_𝑖) for 𝑖=1,2,…,𝑚 and ‖𝜙_𝑗𝑥_𝑘‖≤𝛾^𝑗₀𝑆 and ‖𝜓_𝑗𝑥_𝑘‖≤𝜆^𝑗₀𝑆 for 𝑗=1,2. Since 0≤𝑇¹_𝑛𝑥_𝑘,𝑦_𝑘(𝑡)=¹₀𝐺_𝛼(𝑡,𝑠)𝜌¹_𝑘(𝑠)≤1𝑑𝑠Γ(𝛼)¹₀𝑔₁(𝑠)𝑔𝑑𝑠=¹¹Γ(𝛼),0≤𝑇¹_𝑛𝑥_𝑘,𝑦_𝑘^′(𝑡)=¹₀𝜕𝐺𝜕𝑡^𝛼(𝑡,𝑠)𝜌¹_𝑘(𝑠)≤1𝑑𝑠Γ(𝛼−1)¹₀𝑔₁(𝑠)𝑔𝑑𝑠=¹¹Γ(),𝑇𝛼−10≤¹_𝑛𝑥_𝑘,𝑦_𝑘^′′(𝑡)=¹₀𝜕²𝜕𝑡²𝐺_𝛼(𝑡,𝑠)𝜌¹_𝑘(𝑠)≤1𝑑𝑠Γ(𝛼−2)¹₀𝑔₁(𝑠)𝑔𝑑𝑠=¹¹Γ(),𝛼−2 (25)‖𝑇¹_𝑛(𝑥_𝑘,𝑦_𝑘)‖_∗≤‖𝑔₁‖₁/Γ(𝛼−2). Similarly, we can verify that ‖𝑇²_𝑛(𝑥_𝑘,𝑦_𝑘)‖_∗≤‖𝑔₂‖₁/Γ(𝛽−2). Thus, 𝑇_𝑛𝑥_𝑘,𝑦_𝑘∗∗𝑔≤max¹¹Γ(),𝑔𝛼−2Γ𝛽−²¹2 (26) for all 𝑘. Hence, {𝑇_𝑛(𝑥_𝑘,𝑦_𝑘)}_𝑘≥1 is bounded in 𝑋×𝑋. Also, for 0≤𝑡₁<𝑡₂≤1 we have 𝑇¹_𝑛𝑥_𝑘,𝑦_𝑘^′′𝑡₂−𝑇¹_𝑛𝑥_𝑘,𝑦_𝑘^′′𝑡₁≤𝑡₂^𝛼−3−𝑡₁^𝛼−3Γ(𝛼−2)¹₀(1−𝑠)^{𝛼−𝛿1−1}𝜌¹_𝑘(𝑠)1𝑑𝑠+Γ()×𝛼−2_𝑡20𝑡₂−𝑠^𝛼−3𝜌¹_𝑘(𝑠)𝑑𝑠−^𝑡1₀𝑡₁−𝑠^𝛼−3𝜌¹_𝑘(𝑠)≤𝜌𝑑𝑠_1𝑘1Γ()𝑡𝛼−2^2𝛼−3−𝑡₁^𝛼−3+1Γ()×𝛼−2_𝑡2𝑡1𝑡₂−𝑠^𝛼−3𝜌¹_𝑘(𝑠)+𝑑𝑠_𝑡10𝑡₂−𝑠^𝛼−3−𝑡₁−𝑠^𝛼−3𝜌¹_𝑘(𝑠)≤𝑔𝑑𝑠₁₁Γ()𝑡𝛼−2^2𝛼−3−𝑡₁^𝛼−3+1Γ()×𝑡𝛼−2₂−𝑡₁^𝛼−3𝑔₁₁+^𝑡1₀𝑡₂−𝑠^𝛼−3−𝑡₁−𝑠^𝛼−3𝑔₁(𝑠)𝑑𝑠. (27) Similarly, we have 𝑇²_𝑛𝑥_𝑘,𝑦_𝑘^′′𝑡₂−𝑇²_𝑛𝑥_𝑘,𝑦_𝑘^′′𝑡₁≤𝑔₂₁Γ𝑡𝛽−2^2𝛽−3−𝑡₁^𝛽−3+1Γ×𝑡𝛽−2₂−𝑡₁^𝛽−3𝑔₂₁+^𝑡1₀𝑡₂−𝑠^𝛽−3−𝑡₁−𝑠^𝛽−3𝑔₂(𝑠)𝑑𝑠. (28) Let 𝜖>0 be given. Since the functions 𝑡^𝛼−3 and 𝑡^𝛽−3 are uniformly continuous on [0,1] and the functions |𝑡−𝑠|^𝛼−3 and |𝑡−𝑠|^𝛽−3 are uniformly continuous on [0,1]×[0,1], there exists 𝛿>0 such that 0<𝑡₂^𝛼−3−𝑡₁^𝛼−3<𝜖, 0<𝑡₂^𝛽−3−𝑡₁^𝛽−3<𝜖, 0<(𝑡₂−𝑠)^𝛼−3−(𝑡₁−𝑠)^𝛼−3<𝜖, and 0<(𝑡₂−𝑠)^𝛽−3−(𝑡₁−𝑠)^𝛽−3<𝜖 for all 0≤𝑡₁<𝑡₂≤1, 𝑡₂−𝑡₁<𝛿 and 0≤𝑠≤𝑡₁. If 0≤𝑡₁<𝑡₂≤1 and 𝑡₂−𝑡₁<min{𝛿,√^𝛼−3𝜖,√^𝛽−3𝜖}, then 𝑇¹_𝑛𝑥_𝑘,𝑦_𝑘^′′𝑡₂−𝑇¹_𝑛𝑥_𝑘,𝑦_𝑘^′′𝑡₁≤3𝜖Γ()𝑔𝛼−2¹¹ (29) and |(𝑇²_𝑛(𝑥_𝑘,𝑦_𝑘))^′′(𝑡₂)−(𝑇²_𝑛(𝑥_𝑘,𝑦_𝑘))^′′(𝑡₁)|≤(3𝜖/Γ(𝛽−2))‖𝑔₂‖₁. Therefore, {(𝑇_𝑛(𝑥_𝑘,𝑦_𝑘))^′′} is equicontinuous on [0,1] and so 𝑇_𝑛 is a completely continuous operator on 𝑃.

We need the following result (see [10]).

Lemma 3. Let 𝑌 be a Banach space, 𝑃 a cone in 𝑌, and Ω₁ and Ω₂ bounded open balls in 𝑌 centered at the origin with Ω₁⊂Ω₂. Suppose that 𝑇∶𝑃∩(Ω₂⧵Ω₁)→𝑃 is a completely continuous operator such that ‖𝑇𝑥‖≥‖𝑥‖ for all 𝑥∈𝑃∩𝜕Ω₁ and ‖𝑇𝑥‖≤‖𝑥‖ for all 𝑥∈𝑃∩𝜕Ω₂. Then 𝑇 has a fixed point in 𝑃∩(Ω₂⧵Ω₁).

Theorem 4. For each natural number 𝑛, the system (15), (2), and (3) has a solution (𝑢_𝑛,𝑣_𝑛)∈𝑃 such that 𝑢_𝑛≥𝑀₁𝑡^𝛼−1 and 𝑣_𝑛(𝑡)≥𝑀₂𝑡^𝛽−1 for all 𝑡∈[0,1].

Proof. Let (𝑥,𝑦)∈𝑃 and 𝑛≥1 be given. Then, 𝑇¹_𝑛(𝑥,𝑦)(𝑡)≥𝑚₁∫¹₀𝐺_𝛼(𝑡,𝑠)𝑑𝑠≥𝑀₁𝑡^𝛼−1 and 𝑇²_𝑛(𝑥,𝑦)(𝑡)≥𝑚₂∫¹₀𝐺_𝛽(𝑡,𝑠)𝑑𝑠≥𝑀₂𝑡^𝛽−1 for all 𝑡∈[0,1]. Hence, ‖𝑇¹_𝑛(𝑥,𝑦)‖≥𝑀₁ and ‖𝑇²_𝑛(𝑥,𝑦)‖≥𝑀₂ and so ‖𝑇¹_𝑛(𝑥,𝑦)‖_∗≥𝑀₁ and ‖𝑇²_𝑛(𝑥,𝑦)‖_∗≥𝑀₂. Thus, ‖𝑇_𝑛(𝑥,𝑦)‖_∗∗≥max{𝑀₁,𝑀₂}=𝑀. Now, put Ω₁=𝑥,𝑦∈𝑋×𝑋∶𝑥,𝑦_∗∗<𝑀. (30) Then, ‖𝑇_𝑛(𝑥,𝑦)‖_∗∗≥‖(𝑥,𝑦)‖_∗∗ for all (𝑥,𝑦)∈𝑃∩𝜕Ω₁. Let 𝑊^𝑗_𝑛=𝑃_𝑗(1/𝑛,1/𝑛,…,1/𝑛) for 𝑗=1,2. Then, we have 0≤𝑇¹_𝑛𝑥,𝑦^′′(𝑡)≤1Γ()×𝛼−2₁₀𝑊¹_𝑛+𝑘₁(𝑠)ℎ₁(𝑠),(𝑠)𝜙1+𝑥1+𝑦,1+₁𝑥(𝑠),1+𝜓₁𝑦(𝑠),1+𝐷^𝑝₀₊𝑥(𝑠),1+𝐷^𝜇1₀₊𝑦(𝑠),1+𝐷^𝜇2₀₊𝑦(𝑠),…,1+𝐷^𝜇𝑚₀₊𝑦(𝑠)≤1𝑑𝑠Γ()×𝑊𝛼−2_1𝑛+ℎ₁‖𝑥‖𝑦,𝜙1+,1+1+₁𝑥,1+𝜓₁𝑦,1+𝐷^𝑝₀₊𝑥,1+𝐷^𝜇1₀₊𝑦,1+𝐷^𝜇2₀₊𝑦𝐷,…,1+^𝜇𝑚₀₊𝑦𝑘₁₁,0≤𝑇¹_𝑛𝑥,𝑦^′(𝑡)=^𝑡₀𝑇¹_𝑛𝑥,𝑦^′′(𝑠)≤1𝑑𝑠Γ()×𝑊𝛼−2_1𝑛+ℎ₁‖𝑥‖,𝑦𝜙1+1+,1+₁𝑥,1+𝜓₁𝑦,1+𝐷^𝑝₀₊𝑥,1+𝐷^𝜇1₀₊𝑦,1+𝐷^𝜇2₀₊𝑦,…,1+𝐷^𝜇𝑚₀₊𝑦𝑘₁₁,0≤𝑇¹_𝑛𝑥,𝑦(𝑡)=^𝑡₀𝑇¹_𝑛𝑥,𝑦^′(𝑠)≤1𝑑𝑠Γ()𝑊𝛼−2^1𝑛+ℎ₁‖𝑥‖𝑦,𝜙1+,1+1+₁𝑥,1+𝜓₁𝑦,1+𝐷^𝑝₀₊𝑥,1+𝐷^𝜇1₀₊𝑦,1+𝐷^𝜇2₀₊𝑦,…,1+𝐷^𝜇𝑚₀₊𝑦𝑘₁₁ (31) for all (𝑥,𝑦)∈𝑃 and 𝑡∈[0,1]. Hence, 𝑇¹_𝑛𝑥,𝑦_∗≤1Γ()𝑊𝛼−2^1𝑛+ℎ₁1+‖𝑥‖_∗,1+𝑦_∗,1+𝛾¹₀‖𝑥‖_∗,1+𝜆¹₀𝑦_∗‖𝑥‖,1+^∗Γ,𝑦3−𝑝1+^∗Γ3−𝜇₁𝑦,1+^∗Γ3−𝜇₂𝑦,…,1+^∗Γ3−𝜇_𝑚𝑘₁₁≤1Γ()𝑊𝛼−2^1𝑛+ℎ₁1+𝑥,𝑦_∗∗,1+𝑥,𝑦_∗∗,1+𝛾¹₀𝑥,𝑦_∗∗,1+𝜆¹₀𝑥,𝑦_∗∗,1+𝑥,𝑦_∗∗Γ,3−𝑝1+𝑥,𝑦_∗∗Γ3−𝜇₁,1+𝑥,𝑦_∗∗Γ3−𝜇₂,…,1+𝑥,𝑦_∗∗Γ3−𝜇_𝑚𝑘₁₁. (32) Similarly, we have 𝑇²_𝑛𝑥,𝑦_∗≤1Γ𝑊𝛽−2^2𝑛+ℎ₂1+𝑥,𝑦_∗∗,1+𝑥,𝑦_∗∗,1+𝛾²₀𝑥,𝑦_∗∗,1+𝜆²₀𝑥,𝑦_∗∗,1+𝑥,𝑦_∗∗Γ,3−𝑞1+𝑥,𝑦_∗∗Γ3−𝜈₁,1+𝑥,𝑦_∗∗Γ3−𝜈₂,…,1+𝑥,𝑦_∗∗Γ3−𝜈_𝑚𝑘₂₁ (33) for all (𝑥,𝑦)∈𝑃. Since lim_𝑥→∞ℎ_𝑗(𝑥,𝑥,…,𝑥)/𝑥=0 for 𝑗=1,2, there exists 𝑆>𝑀>0 such that 1Γ()𝑊𝛼−2^1𝑛+ℎ₁1+𝑆,1+𝑆,1+𝛾¹₀𝑆,1+𝜆¹₀𝑆𝑆,1+Γ𝑆3−𝑝,1+Γ3−𝜇₁,𝑆1+Γ3−𝜇₂𝑆,…,1+Γ3−𝜇_𝑚𝑘₁₁1≤𝑆,Γ𝑊𝛽−2^2𝑛+ℎ₂1+𝑆,1+𝑆,1+𝛾²₀𝑆,1+𝜆²₀𝑆𝑆,1+Γ𝑆3−𝑞,1+Γ3−𝜈₁,𝑆1+Γ3−𝜈₂𝑆,…,1+Γ3−𝜈_𝑚𝑘₂₁≤𝑆. (34) Put Ω₂={(𝑥,𝑦)∈𝑋×𝑋s.t‖(𝑥,𝑦)‖_∗∗<𝑆}. Then, the above inequalities imply that ‖𝑇_𝑛(𝑥,𝑦)‖_∗∗≤‖(𝑥,𝑦)‖_∗∗ for all (𝑥,𝑦)∈𝑃∩𝜕Ω₂. Now by using Lemma 3, we conclude that the operator 𝑇_𝑛 has a fixed point (𝑢_𝑛,𝑣_𝑛) in 𝑃∩(Ω₂⧵Ω₁). It is easy to see that (𝑢_𝑛,𝑣_𝑛) is a desired solution of the system, that is, 𝑢_𝑛≥𝑀₁𝑡^𝛼−1 and 𝑣_𝑛(𝑡)≥𝑀₂𝑡^𝛽−1 for all 𝑡∈[0,1].

We need the following lemma.

Lemma 5. The set of solutions {(𝑢_𝑛,𝑣_𝑛)}_𝑛≥1 of the system (15), (2), and (3) is a relatively compact subset of 𝑋×𝑋.

Proof. It is easy to check that 𝑢_𝑛(𝑡)=¹₀𝐺_𝛼(𝑡,𝑠)𝑓¹_𝑛𝑠,𝑢_𝑛(𝑠),𝑣_𝑛(𝑠),𝜙₁𝑢_𝑛(𝑠),𝜓₁𝑣_𝑛(𝑠),𝐷^𝑝₀₊𝑢_𝑛(𝑠),𝐷^𝜇1₀₊𝑣_𝑛(𝑠),𝐷^𝜇2₀₊𝑣_𝑛(𝑠)𝐷,…,_𝜇𝑚0+𝑣_𝑛(𝑠)𝑣𝑑𝑠,_𝑛(𝑡)=¹₀𝐺_𝛽(𝑡,𝑠)𝑓²_𝑛𝑠,𝑢_𝑛(𝑠),𝑣_𝑛(𝑠),𝜙₂𝑢_𝑛(𝑠),𝜓₂𝑣_𝑛(𝑠),𝐷^𝑞₀₊𝑣_𝑛(𝑠),𝐷^𝜈1₀₊𝑢_𝑛(𝑠),𝐷^𝜈2₀₊𝑢_𝑛(𝑠),…,𝐷^𝜈𝑚₀₊𝑢_𝑛(𝑠)𝑑𝑠 (35) for 𝑡∈[0,1] and 𝑛≥1, satisfy in Theorem 4. Also, 𝑢^′_𝑛(𝑡)≥𝑚₁∫¹₀(𝜕/𝜕𝑡)𝐺_𝛼(𝑡,𝑠)𝑑𝑠≥𝑚₁𝑡^𝛼−2/Γ(𝛼)(𝛼−𝛿₁), 𝑢_𝑛^′′(𝑡)≥𝑚₁∫¹₀(𝜕²/𝜕𝑡²)𝐺_𝛼(𝑡,𝑠)𝑑𝑠≥𝑚₁(𝛿₁−2)𝑡^𝛼−3/Γ(𝛼−1)(𝛼−𝛿₁), 𝑣^′_𝑛(𝑡)≥𝑚₂∫¹₀(𝜕/𝜕𝑡)𝐺_𝛽(𝑡,𝑠)𝑑𝑠≥𝑚₂𝑡^𝛽−2/Γ(𝛽)(𝛽−𝛿₂), and 𝑣_𝑛^′′(𝑡)≥𝑚₂∫¹₀(𝜕²/𝜕𝑡²)𝐺_𝛽(𝑡,𝑠)𝑑𝑠≥𝑚₂(𝛿₂−2)𝑡^𝛽−3/Γ(𝛽−1)(𝛽−𝛿₂) for 𝑡∈[0,1] and 𝑛≥1. Moreover, 𝐷^𝑝₀₊𝑢_𝑛(𝑡)=1Γ2−𝑝^𝑡₀(𝑡−𝑠)^1−𝑝𝑢_𝑛^′′(𝑠)≥𝑚𝑑𝑠₁𝛿₁−2Γ2−𝑝Γ(𝛼−1)𝛼−𝛿₁×^𝑡₀(𝑡−𝑠)^1−𝑝𝑠^𝛼−3𝐷𝑑𝑠,_𝑞0+𝑣_𝑛(𝑡)=1Γ2−𝑞^𝑡₀(𝑡−𝑠)^1−𝑞𝑣_𝑛^′′(𝑠)≥𝑚𝑑𝑠₂𝛿₂−2Γ2−𝑞Γ𝛽−1𝛽−𝛿₂×^𝑡₀(𝑡−𝑠)^1−𝑞𝑠^𝛽−3𝐷𝑑𝑠,_𝜈𝑖0+𝑢_𝑛(𝑡)=1Γ2−𝜈_𝑖^𝑡₀(𝑡−𝑠)^1−𝜈𝑖𝑢_𝑛^′′(𝑠)≥𝑚𝑑𝑠₁𝛿₁−2Γ2−𝜈_𝑖Γ(𝛼−1)𝛼−𝛿₁×^𝑡₀(𝑡−𝑠)^1−𝜈𝑖𝑠^𝛼−3𝐷𝑑𝑠,_𝜇𝑖0+𝑣_𝑛(𝑡)=1Γ2−𝜇_𝑖^𝑡₀(𝑡−𝑠)^1−𝜇𝑖𝑣_𝑛^′′(𝑠)≥𝑚𝑑𝑠₂𝛿₂−2Γ2−𝜇_𝑖Γ𝛽−1𝛽−𝛿₂×^𝑡₀(𝑡−𝑠)^1−𝜇𝑖𝑠^𝛽−3𝑑𝑠 (36) for 𝑖=1,2,…,𝑚. Since ^𝑡₀(𝑡−𝑠)^1−𝑝𝑠^𝛼−3𝑑𝑠=𝑡^{𝛼−𝑝−1}×¹₀(1−𝜉)^1−𝑝𝜉^𝛼−3=ΓΓ()𝑑𝜉2−𝑝𝛼−2Γ𝑡𝛼−𝑝^{𝛼−𝑝−1}≥Γ2−𝑝Γ(𝛼−2)Γ𝑡𝛼−𝑝^𝛼−𝑝_𝑡0,(𝑡−𝑠)^1−𝑞𝑠^𝛽−3ΓΓ𝑑𝑠≥2−𝑞𝛽−2Γ𝑡𝛽−𝑞^𝛽−𝑞_𝑡0,(𝑡−𝑠)^1−𝜈𝑖𝑠^𝛼−3Γ𝑑𝑠≥2−𝜈_𝑖Γ(𝛼−2)Γ𝛼−𝜈_𝑖𝑡^{𝛼−𝜈𝑖},_𝑡0(𝑡−𝑠)^1−𝜇𝑖𝑠^𝛽−3Γ𝑑𝑠≥2−𝜇_𝑖Γ𝛽−2Γ𝛽−𝜇_𝑖𝑡^{𝛽−𝜇𝑖} (37) for 𝑖=1,2,…,𝑚, we get 𝐷^𝑝₀₊𝑢_𝑛(𝑡)≥(𝑚₁(𝛿₁−2)/(𝛼−2)(𝛼−𝛿₁)Γ(𝛼−𝑝))𝑡^𝛼−𝑝, 𝐷^𝑞₀₊𝑣_𝑛(𝑡)≥(𝑚₂(𝛿₂−2)/(𝛽−2)(𝛽−𝛿₂)Γ(𝛽−𝑞))𝑡^𝛽−𝑞, 𝐷^𝜈𝑖₀₊𝑢_𝑛(𝑡)≥(𝑚₁(𝛿₁−2)/(𝛼−2)(𝛼−𝛿₁)Γ(𝛼−𝜈_𝑖))𝑡^{𝛼−𝜈𝑖}, and 𝐷^𝜇𝑖₀₊𝑣_𝑛(𝑡)≥(𝑚₂(𝛿₂−2)/(𝛽−2)(𝛽−𝛿₂)Γ(𝛽−𝜇_𝑖))𝑡^{𝛽−𝜇𝑖} for all 𝑡∈[0,1] and 𝑛≥1. Since 𝑚₁1⋅min𝛼−𝛿₁Γ(),1𝛼+1(𝛼−2)𝛼−𝛿₁=𝑀₁ (38) and 𝑚₂⋅min{1/(𝛽−𝛿₂)Γ(𝛽+1),1/(𝛽−2)(𝛽−𝛿₂)}=𝑀₂, for each 𝑡∈[0,1] and 𝑛≥1, we get 𝑢_𝑛(𝑡)≥𝑀₁𝑡^𝛼−1, 𝑣_𝑛(𝑡)≥𝑀₂𝑡^𝛽−1, 𝐷^𝑝₀₊𝑢_𝑛(𝑡)≥((𝛿₁−2)𝑀₁/Γ(𝛼−𝑝))𝑡^𝛼−𝑝, 𝐷^𝑞₀₊𝑣_𝑛(𝑡)≥((𝛿₂−2)𝑀₂/Γ(𝛽−𝑞))𝑡^𝛽−𝑞, 𝐷^𝜈𝑖₀₊𝑢_𝑛(𝑡)≥((𝛿₁−2)𝑀₁/Γ(𝛼−𝜈_𝑖))𝑡^{𝛼−𝜈𝑖}, and 𝐷^𝜇𝑖₀₊𝑣_𝑛(𝑡)≥((𝛿₂−2)𝑀₂/Γ(𝛽−𝜇_𝑖))𝑡^{𝛽−𝜇𝑖} for 𝑖=1,2,…,𝑚 and (𝜙_𝑗𝑢_𝑛)(𝑡)≥𝑀₁∫^𝑡₀𝛾_𝑗(𝑡,𝑠)𝑠^𝛼−1𝑑𝑠 and (𝜓_𝑗𝑣_𝑛)(𝑡)≥𝑀₂∫^𝑡₀𝜆_𝑗(𝑡,𝑠)𝑠^𝛽−1𝑑𝑠 for 𝑗=1,2. Thus, 𝑃₁𝑢_𝑛(𝑡),𝑣_𝑛(𝑡),𝜙₁𝑢_𝑛(𝑡),𝜓₁𝑣_𝑛(𝑡),𝐷^𝑝₀₊𝑢_𝑛(𝑡),𝐷^𝜇1₀₊𝑣_𝑛(𝑡),𝐷^𝜇2₀₊𝑣_𝑛(𝑡),…,𝐷^𝜇𝑚₀₊𝑣_𝑛(𝑡)≤𝑃₁𝑀₁𝑡^𝛼−1,𝑀₂𝑡^𝛽−1,𝑀₁^𝑡₀𝛾₁(𝑡,𝑠)𝑠^𝛼−1𝑀𝑑𝑠,_2𝑡0𝜆₁(𝑡,𝑠)𝑠^𝛽−1𝛿𝑑𝑠,¹−2𝑀₁Γ𝑡𝛼−𝑝^𝛼−𝑝,𝛿₂−2𝑀₂Γ𝛽−𝜇₁𝑡^𝛽−𝜇1,𝛿₂−2𝑀₂Γ𝛽−𝜇₂𝑡^𝛽−𝜇2𝛿,…,₂−2𝑀₂Γ𝛽−𝜇_𝑚𝑡^{𝛽−𝜇𝑚},𝑃₂𝑢_𝑛(𝑡),𝑣_𝑛(𝑡),𝜙₂𝑢_𝑛(𝑡),𝜓₂𝑣_𝑛(𝑡),𝐷^𝑞₀₊𝑣_𝑛(𝑡),𝐷^𝜈1₀₊𝑢_𝑛(𝑡),𝐷^𝜈2₀₊𝑢_𝑛(𝑡),…,𝐷^𝜈𝑚₀₊𝑢_𝑛(𝑡)≤𝑃₂𝑀₁𝑡^𝛼−1,𝑀₂𝑡^𝛽−1,𝑀₁^𝑡₀𝛾₂(𝑡,𝑠)𝑠^𝛼−1𝑀𝑑𝑠,_2𝑡0𝜆₂(𝑡,𝑠)𝑠^𝛽−1𝛿𝑑𝑠,²−2𝑀₂Γ𝑡𝛽−𝑞^𝛽−𝑞,𝛿₁−2𝑀₁Γ𝛼−𝜈₁𝑡^𝛼−𝜈1,𝛿₁−2𝑀₁Γ𝛼−𝜈₂𝑡^𝛼−𝜈2𝛿,…,₁−2𝑀₁Γ𝛼−𝜈_𝑚𝑡^{𝛼−𝜈𝑚} (39) for all 𝑡∈[0,1] and 𝑛≥1 and so 0≤𝑢_𝑛^′′(𝑡)=¹₀𝜕²𝜕𝑡²𝐺_𝛼(𝑡,𝑠)𝑓¹_𝑛𝑠,𝑢_𝑛(𝑠),𝑣_𝑛(𝑠),𝜙₁𝑢_𝑛(𝑠),𝜓₁𝑣_𝑛(𝑠),𝐷^𝑝₀₊𝑢_𝑛(𝑠),𝐷^𝜇1₀₊𝑣_𝑛(𝑠),𝐷^𝜇2₀₊𝑣_𝑛(𝑠),…,𝐷^𝜇𝑚₀₊𝑣_𝑛(𝑠)≤1𝑑𝑠Γ(𝛼−2)¹₀𝑃₁𝑀₁𝑠^𝛼−1,𝑀₂𝑠^𝛽−1,𝑀₁^𝑠₀𝛾₁(𝑠,𝜏)𝜏^𝛼−1𝑀𝑑𝜏,_2𝑠0𝜆₁(𝑠,𝜏)𝜏^𝛽−1𝛿𝑑𝜏,₁−2𝑀₁Γ𝑠𝛼−𝑝^𝛼−𝑝,𝛿₂−2𝑀₂Γ𝛽−𝜇₁𝑠^𝛽−𝜇1,𝛿₂−2𝑀₂Γ𝛽−𝜇₂𝑠^𝛽−𝜇2𝛿,…,₂−2𝑀₂Γ𝛽−𝜇_𝑚𝑠^{𝛽−𝜇𝑚}𝑑𝑠+ℎ₁1+𝑢_𝑛∗,1+𝑣_𝑛∗,1+𝛾¹₀𝑢_𝑛∗,1+𝜆¹₀𝑣_𝑛∗,𝑢1+^𝑛∗Γ𝑣3−𝑝,1+^𝑛∗Γ3−𝜇₁,𝑣1+^𝑛∗Γ3−𝜇₂𝑣,…,1+^𝑛∗Γ3−𝜇_𝑚×¹₀𝑘₁(𝑠)=1𝑑𝑠Γ()×Λ𝛼−2₁+ℎ₁1+𝑢_𝑛∗,1+𝑣_𝑛∗,1+𝛾¹₀𝑢_𝑛∗,1+𝜆¹₀𝑣_𝑛∗,𝑢1+^𝑛∗Γ𝑣3−𝑝,1+^𝑛∗Γ3−𝜇₁,𝑣1+^𝑛∗Γ3−𝜇₂𝑣,…,1+^𝑛∗Γ3−𝜇_𝑚𝑘₁₁,0≤𝑢^′_𝑛(𝑡)=^𝑡₀𝑢_𝑛^′′(𝑠)≤1𝑑𝑠Γ()×Λ𝛼−2₁+ℎ₁1+𝑢_𝑛∗,1+𝑣_𝑛∗,1+𝛾¹₀𝑢_𝑛∗,1+𝜆¹₀𝑣_𝑛∗,𝑢1+^𝑛∗Γ𝑣3−𝑝,1+^𝑛∗Γ3−𝜇₁,𝑣1+^𝑛∗Γ3−𝜇₂𝑣,…,1+^𝑛∗Γ3−𝜇_𝑚𝑘₁₁,0≤𝑢_𝑛(𝑡)=^𝑡₀𝑢^′_𝑛(𝑠)≤1𝑑𝑠Γ()×Λ𝛼−2₁+ℎ₁1+𝑢_𝑛∗,1+𝑣_𝑛∗,1+𝛾¹₀𝑢_𝑛∗,1+𝜆¹₀𝑣_𝑛∗,𝑢1+^𝑛∗Γ𝑣3−𝑝,1+^𝑛∗Γ3−𝜇₁,𝑣1+^𝑛∗Γ3−𝜇₂𝑣,…,1+^𝑛∗Γ3−𝜇_𝑚𝑘₁₁ (40) for all 𝑡∈[0,1] and 𝑛≥1 and also Λ₁<∞, where Λ₁=¹₀𝑃₁𝑀₁𝑠^𝛼−1,𝑀₂𝑠^𝛽−1,𝑀₁^𝑠₀𝛾₁(𝑠,𝜏)𝜏^𝛼−1𝑀𝑑𝜏,_2𝑠0𝜆₁(𝑠,𝜏)𝜏^𝛽−1𝛿𝑑𝜏,₁−2𝑀₁Γ𝑠𝛼−𝑝^𝛼−𝑝,𝛿₂−2𝑀₂Γ𝛽−𝜇₁𝑠^𝛽−𝜇1,𝛿₂−2𝑀₂Γ𝛽−𝜇₂𝑠^𝛽−𝜇2𝛿,…,₂−2𝑀₂Γ𝛽−𝜇_𝑚𝑠^{𝛽−𝜇𝑚}𝑑𝑠. (41) Hence, 𝑢_𝑛∗≤1Γ()Λ𝛼−2¹+ℎ₁1+𝑢_𝑛∗,1+𝑣_𝑛∗,1+𝛾¹₀𝑢_𝑛∗,1+𝜆¹₀𝑣_𝑛∗,𝑢1+^𝑛∗Γ𝑣3−𝑝,1+^𝑛∗Γ3−𝜇₁,𝑣1+^𝑛∗Γ3−𝜇₂𝑣,…,1+^𝑛∗Γ3−𝜇_𝑚𝑘₁₁≤1Γ()Λ𝛼−2¹+ℎ₁1+𝑢_𝑛,𝑣_𝑛∗∗,1+𝑢_𝑛,𝑣_𝑛∗∗,1+𝛾¹₀𝑢_𝑛,𝑣_𝑛∗∗,1+𝜆¹₀𝑢_𝑛,𝑣_𝑛∗∗,𝑢1+^𝑛,𝑣_𝑛∗∗Γ𝑢3−𝑝,1+^𝑛,𝑣_𝑛∗∗Γ3−𝜇₁,𝑢1+^𝑛,𝑣_𝑛∗∗Γ3−𝜇₂𝑢,…,1+^𝑛,𝑣_𝑛∗∗Γ3−𝜇_𝑚𝑘₁₁=𝐺¹_𝑛 (42) for all 𝑛≥1. Similarly, we have 𝑣_𝑛∗≤1Γ×Λ𝛽−2₂+ℎ₂1+𝑢_𝑛,𝑣_𝑛∗∗,1+𝑢_𝑛,𝑣_𝑛∗∗,1+𝛾²₀𝑢_𝑛,𝑣_𝑛∗∗,1+𝜆²₀𝑢_𝑛,𝑣_𝑛∗∗,𝑢1+^𝑛,𝑣_𝑛∗∗Γ𝑢3−𝑞,1+^𝑛,𝑣_𝑛∗∗Γ3−𝜈₁,𝑢1+^𝑛,𝑣_𝑛∗∗Γ3−𝜈₂𝑢,…,1+^𝑛,𝑣_𝑛∗∗Γ3−𝜈_𝑚𝑘₂₁=𝐺²_𝑛 (43) for all 𝑛≥1, where Λ₂=¹₀𝑃₂𝑀₁𝑠^𝛼−1,𝑀₂𝑠^𝛽−1,𝑀₁^𝑠₀𝛾₂(𝑠,𝜏)𝜏^𝛼−1𝑀𝑑𝜏,_2𝑠0𝜆₂(𝑠,𝜏)𝜏^𝛽−1𝛿𝑑𝜏,₂−2𝑀₂Γ𝑠𝛽−𝑞^𝛽−𝑞,𝛿₁−2𝑀₁Γ𝛼−𝜈₁𝑠^𝛼−𝜈1,𝛿₁−2𝑀₁Γ𝛼−𝜈₂𝑠^𝛼−𝜈2𝛿,…,₁−2𝑀₁Γ𝛼−𝜈_𝑚𝑠^{𝛼−𝜈𝑚}𝑑𝑠. (44) Thus, ‖(𝑢_𝑛,𝑣_𝑛)‖_∗∗≤max{𝐺¹_𝑛,𝐺²_𝑛} for all 𝑛≥1. Since lim_𝑥→∞ℎ_𝑗(𝑥,𝑥,…,𝑥)/𝑥=0 for 𝑗=1,2, there exists 𝐿>0 such that 1Γ()Λ𝛼−2¹+ℎ₁1+𝜐,1+𝜐,1+𝛾¹₀𝜐,1+𝜆¹₀𝜐𝜐,1+Γ,𝜐3−𝑝1+Γ3−𝜇₁,𝜐1+Γ3−𝜇₂𝜐,…,1+Γ3−𝜇_𝑚𝑘₁₁1<𝜐,ΓΛ𝛽−2²+ℎ₂1+𝜐,1+𝜐,1+𝛾²₀𝜐,1+𝜆²₀𝜐𝜐,1+Γ,𝜐3−𝑞1+Γ3−𝜈₁,𝜐1+Γ3−𝜈₂𝜐,…,1+Γ3−𝜈_𝑚𝑘₂₁<𝜐 (45) for all 𝜐≥𝐿. Consequently, ‖(𝑢_𝑛,𝑣_𝑛)‖_∗∗<𝐿 for all 𝑛≥1 and so {(𝑢_𝑛,𝑣_𝑛)}_𝑛≥1 is a bounded sequence in 𝑋×𝑋. It remains to prove that {(𝑢_𝑛,𝑣_𝑛)^′′} is equicontinuous on [0,1]. Put 𝑉₁=ℎ₁1+𝐿,1+𝐿,1+𝛾¹₀𝐿,1+𝜆¹₀𝐿𝐿,1+Γ𝐿3−𝑝,1+Γ3−𝜇₁,𝐿1+Γ3−𝜇₂𝐿,…,1+Γ3−𝜇_𝑚,𝑉₂=ℎ₂1+𝐿,1+𝐿,1+𝛾²₀𝐿,1+𝜆²₀𝐿𝐿,1+Γ𝐿3−𝑞,1+Γ3−𝜈₁,𝐿1+Γ3−𝜈₂𝐿,…,1+Γ3−𝜈_𝑚,Φ₁(𝑡)=𝑃₁𝑀₁𝑡^𝛼−1,𝑀₂𝑡^𝛽−1,𝑀₁^𝑡₀𝛾₁(𝑡,𝑠)𝑠^𝛼−1𝑀𝑑𝑠,_2𝑡0𝜆₁(𝑡,𝑠)𝑠^𝛽−1𝛿𝑑𝑠₁−2𝑀₁Γ𝑡𝛼−𝑝^𝛼−𝑝,𝛿₂−2𝑀₂Γ𝛽−𝜇₁𝑡^𝛽−𝜇1,𝛿₂−2𝑀₂Γ𝛽−𝜇₂𝑡^𝛽−𝜇2𝛿,…,₂−2𝑀₂Γ𝛽−𝜇_𝑚𝑡^{𝛽−𝜇𝑚},Φ₂(𝑡)=𝑃₂𝑀₁𝑡^𝛼−1,𝑀₂𝑡^𝛽−1,𝑀₁^𝑡₀𝛾₂(𝑡,𝑠)𝑠^𝛼−1𝑀𝑑𝑠,_2𝑡0𝜆₂(𝑡,𝑠)𝑠^𝛽−1𝛿𝑑𝑠,₂−2𝑀₂Γ𝑡𝛽−𝑞^𝛽−𝑞,𝛿₁−2𝑀₁Γ𝛼−𝜈₁𝑡^𝛼−𝜈1,𝛿₁−2𝑀₁Γ𝛼−𝜈₂𝑡^𝛼−𝜈2𝛿,…,¹−2𝑀₁Γ𝛼−𝜈_𝑚𝑡^{𝛼−𝜈𝑚} (46) for all 𝑡∈(0,1]. Note that, Λ_𝑗=∫¹₀Φ_𝑗(𝑡)𝑑𝑡 for 𝑗=1,2 and 𝑓¹_𝑛𝑡,𝑢_𝑛(𝑡),𝑣_𝑛(𝑡),𝜙₁𝑢_𝑛(𝑡),𝜓₁𝑣_𝑛(𝑡),𝐷^𝑝₀₊𝑢_𝑛(𝑡),𝐷^𝜇1₀₊𝑣_𝑛(𝑡),𝐷^𝜇2₀₊𝑣_𝑛(𝑡),…,𝐷^𝜇𝑚₀₊𝑣_𝑛(𝑡)≤Φ₁(𝑡)+𝑉₁𝑘₁(𝑡),𝑓²_𝑛𝑡,𝑢_𝑛(𝑡),𝑣_𝑛(𝑡),𝜙₂𝑢_𝑛(𝑡),𝜓₂𝑣_𝑛(𝑡),𝐷^𝑞₀₊𝑣_𝑛(𝑡),𝐷^𝜈1₀₊𝑢_𝑛(𝑡),𝐷^𝜈2₀₊𝑢_𝑛(𝑡),…,𝐷^𝜈𝑚₀₊𝑢_𝑛(𝑡)≤Φ₂(𝑡)+𝑉₂𝑘₂(𝑡) (47) for almost all 𝑡∈[0,1] and 𝑛≥1. If 0≤𝑡₁<𝑡₂≤1, then 𝑢_𝑛^′′𝑡₂−𝑢_𝑛^′′𝑡₁=¹₀𝜕²𝜕𝑡²𝐺_𝛼𝑡₂−𝜕,𝑠²𝜕𝑡²𝐺_𝛼𝑡₁𝑓,𝑠_1𝑛𝑠,𝑢_𝑛(𝑠),𝑣_𝑛(𝑠),𝜙₁𝑢_𝑛(𝑠),𝜓₁𝑣_𝑛(𝑠),𝐷^𝑝₀₊𝑢_𝑛(𝑠),𝐷^𝜇1₀₊𝑣_𝑛(𝑠),𝐷^𝜇2₀₊𝑣_𝑛(𝑠),…,𝐷^𝜇𝑚₀₊𝑣_𝑛(𝑠)≤1𝑑𝑠Γ()𝑡𝛼−2^2𝛼−3−𝑡₁^𝛼−31₀Φ₁(𝑠)+𝑉₁𝑘₁(𝑠)+𝑑𝑠_𝑡2𝑡1𝑡₂−𝑠^𝛼−3Φ₁(𝑠)+𝑉₁𝑘₁(𝑠)+𝑑𝑠_𝑡10𝑡₂−𝑠^𝛼−3−𝑡₁−𝑠^𝛼−3×Φ₁(𝑠)+𝑉₁𝑘₁(𝑠)≤1𝑑𝑠Γ()𝑡𝛼−2^2𝛼−3−𝑡₁^𝛼−3Λ₁+𝑉₁𝑘₁₁+𝑡₂−𝑡₁^𝛼−3Λ₁+𝑉₁𝑘₁₁+^𝑡1₀𝑡₂−𝑠^𝛼−3−𝑡₁−𝑠^𝛼−3×Φ₁(𝑠)+𝑉₁𝑘₁(𝑠)𝑑𝑠. (48) Similarly, we have 𝑣_𝑛^′′𝑡₂−𝑣_𝑛^′′𝑡₁≤1Γ𝑡𝛽−2^2𝛽−3−𝑡₁^𝛽−3Λ₂+𝑉₂𝑘₂₁+𝑡₂−𝑡₁^𝛽−3Λ₂+𝑉₂𝑘₂₁+^𝑡1₀𝑡₂−𝑠^𝛽−3−𝑡₁−𝑠^𝛽−3×Φ₂(𝑠)+𝑉₂𝑘₂(𝑠)𝑑𝑠. (49) Let 𝜖>0 be given. Choose 𝛿₀>0 such that 0<𝑡₂^𝛼−3−𝑡₁^𝛼−3<𝜖, 0<𝑡₂^𝛽−3−𝑡₁^𝛽−3<𝜖, 0<𝑡₂−𝑠^𝛼−3−𝑡₁−𝑠^𝛼−3<𝜖 (50) and 0<(𝑡₂−𝑠)^𝛽−3−(𝑡₁−𝑠)^𝛽−3<𝜖 for all 0≤𝑡₁<𝑡₂≤1, 𝑡₂−𝑡₁<𝛿₀ and 0≤𝑠≤𝑡₁. Thus, for each 0≤𝑡₁<𝑡₂≤1 and 𝑡₂−𝑡₁<min{𝛿₀,√^𝛼−3𝜖,√^𝛽−3𝜖}, we have |𝑢_𝑛^′′(𝑡₂)−𝑢_𝑛^′′(𝑡₁)|≤(3𝜖/Γ(𝛼−2))(Λ₁+𝑉₁‖𝑘₁‖₁) and |𝑣_𝑛^′′(𝑡₂)−𝑣_𝑛^′′(𝑡₁)|≤(3𝜖/Γ(𝛽−2))(Λ₂+𝑉₂‖𝑘₂‖₁). Therefore, {(𝑢_𝑛,𝑣_𝑛)^′′}_𝑛≥1 is equicontinuous on [0,1]. This completes the proof.

Now, we give our main result.

Theorem 6. The system (1), (2), and (3) has a positive solution (𝑢,𝑣) such that 𝑢(𝑡)≥𝑀₁𝑡^𝛼−1, 𝑣(𝑡)≥𝑀₂𝑡^𝛽−1, 𝐷^𝑝₀₊𝑢(𝑡)≥((𝛿₁−2)𝑀₁/Γ(𝛼−𝑝))𝑡^𝛼−𝑝, 𝐷^𝑞₀₊𝑣(𝑡)≥((𝛿₂−2)𝑀₂/Γ(𝛽−𝑞))𝑡^𝛽−𝑞, 𝐷^𝜈𝑖₀₊𝑢(𝑡)≥((𝛿₁−2)𝑀₁/Γ(𝛼−𝜈_𝑖))𝑡^{𝛼−𝜈𝑖}, and 𝐷^𝜇𝑖₀₊𝑣(𝑡)≥((𝛿₂−2)𝑀₂/Γ(𝛽−𝜇_𝑖))𝑡^{𝛽−𝜇𝑖} for 𝑖=1,2,…,𝑚 and 𝑡∈[0,1] and (𝜙_𝑗𝑢)(𝑡)≥𝑀₁∫^𝑡₀𝛾_𝑗(𝑡,𝑠)𝑠^𝛼−1𝑑𝑠 and 𝜓_𝑗𝑣(𝑡)≥𝑀₂^𝑡₀𝜆_𝑗(𝑡,𝑠)𝑠^𝛽−1𝑑𝑠 (51) for 𝑗=1,2 and 𝑡∈[0,1].

Proof. By using Theorem 4, for each natural number 𝑛, the system (15), (2), and (3) has a solution (𝑢_𝑛,𝑣_𝑛) in 𝑃. Also by using Lemma 5, the set {(𝑢_𝑛,𝑣_𝑛)}_𝑛≥1 is a relatively compact subset of 𝑋×𝑋. By using the Arzela-Ascoli theorem, without loss of generality we can assume that {(𝑢_𝑛,𝑣_𝑛)}_𝑛≥1 is convergent in 𝑋×𝑋 to some element (𝑢,𝑣) of 𝑃. It is easy to check that (𝑢,𝑣) satisfy the boundary conditions (2) and (3) and also lim_𝑛→∞𝐷^𝑝₀₊𝑢_𝑛=𝐷^𝑝₀₊𝑢, lim_𝑛→∞𝐷^𝑞₀₊𝑣_𝑛=𝐷^𝑞₀₊𝑣, lim_𝑛→∞𝐷^𝜈𝑖₀₊𝑢_𝑛=𝐷^𝜈𝑖₀₊𝑢, and lim_𝑛→∞𝐷^𝜇𝑖₀₊𝑣_𝑛=𝐷^𝜇𝑖₀₊𝑣 for 𝑖=1,2,…,𝑚 and lim_𝑛→∞𝜙_𝑗𝑢_𝑛=𝜙_𝑗𝑢 and lim_𝑛→∞𝜓_𝑗𝑣_𝑛=𝜓_𝑗𝑣 for 𝑗=1,2. Thus, it is easy to see that (𝑢,𝑣) satisfy the desired conditions. Also, lim_𝑛→∞𝑓¹_𝑛𝑡,𝑢_𝑛(𝑡),𝑣_𝑛(𝑡),𝜙₁𝑢_𝑛(𝑡),𝜓₁𝑣_𝑛(𝑡),𝐷^𝑝₀₊𝑢_𝑛(𝑡),𝐷^𝜇1₀₊𝑣_𝑛(𝑡),𝐷^𝜇2₀₊𝑣_𝑛(𝑡),…,𝐷^𝜇𝑚₀₊𝑣_𝑛(𝑡)=𝑓₁𝑡,𝑢(𝑡),𝑣(𝑡),𝜙₁𝑢(𝑡),𝜓₁𝑣(𝑡),𝐷^𝑝₀₊𝑢(𝑡),𝐷^𝜇1₀₊𝑣(𝑡),𝐷^𝜇2₀₊𝑣(𝑡),…,𝐷^𝜇𝑚₀₊𝑣(𝑡),lim_𝑛→∞𝑓²_𝑛𝑡,𝑢_𝑛(𝑡),𝑣_𝑛(𝑡),𝜙₂𝑢_𝑛(𝑡),𝜓₂𝑣_𝑛(𝑡),𝐷^𝑞₀₊𝑣_𝑛(𝑡),𝐷^𝜈1₀₊𝑢_𝑛(𝑡),𝐷^𝜈2₀₊𝑢_𝑛(𝑡),…,𝐷^𝜈𝑚₀₊𝑢_𝑛(𝑡)=𝑓₂𝑡,𝑢(𝑡),𝑣(𝑡),𝜙₂𝑢𝑡),𝜓₂(𝑣(𝑡),𝐷^𝑞₀₊𝑣(𝑡),𝐷^𝜈1₀₊𝑢(𝑡),𝐷^𝜈2₀₊𝑢(𝑡),…,𝐷^𝜈𝑚₀₊𝑢(𝑡) (52) for almost all 𝑡∈[0,1]. If 𝜉=sup{‖(𝑢_𝑛,𝑣_𝑛)‖_∗∗∶𝑛≥1}, then ‖𝐷^𝑝₀₊𝑢_𝑛‖≤𝜉/Γ(3−𝑝), ‖𝐷^𝑞₀₊𝑣_𝑛‖≤𝜉/Γ(3−𝑞), ‖𝐷^𝜈𝑖₀₊𝑢_𝑛‖≤𝜉/Γ(3−𝜈_𝑖), and ‖𝐷^𝜇𝑖₀₊𝑣_𝑛‖≤𝜉/Γ(3−𝜇_𝑖) for 𝑖=1,2,…,𝑚 and 𝑛≥1‖𝜙_𝑗𝑢_𝑛‖≤𝜉𝛾^𝑗₀ and ‖𝜓_𝑗𝑣_𝑛‖≤𝜉𝜆^𝑗₀ for 𝑗=1,2 and 𝑛≥1. Thus, 0≤𝐺_𝛼(𝑡,𝑠)𝑓¹_𝑛𝑠,𝑢_𝑛(𝑠),𝑣_𝑛(𝑠),𝜙₁𝑢_𝑛(𝑠),𝜓₁𝑣_𝑛(𝑠),𝐷^𝑝₀₊𝑢_𝑛(𝑠),𝐷^𝜇1₀₊𝑣_𝑛(𝑠),𝐷^𝜇2₀₊𝑣_𝑛(𝑠),…,𝐷^𝜇𝑚₀₊𝑣_𝑛(𝑠)≤1Γ(𝛼)Φ₁(𝑠)+ℎ₁1+𝜉,1+𝜉,1+𝛾¹₀𝜉,1+𝜆¹₀𝜉𝜉,1+Γ𝜉3−𝑝,1+Γ3−𝜇₁,𝜉1+Γ3−𝜇₂𝜉,…,1+Γ3−𝜇_𝑚𝑘₁(𝑠),0≤𝐺_𝛽(𝑡,𝑠)𝑓²_𝑛𝑠,𝑢_𝑛(𝑠),𝑣_𝑛(𝑠),𝜙₂𝑢_𝑛(𝑠),𝜓₂𝑣_𝑛(𝑠),𝐷^𝑞₀₊𝑣_𝑛(𝑠),𝐷^𝜈1₀₊𝑢_𝑛(𝑠),𝐷^𝜈2₀₊𝑢_𝑛(𝑠),…,𝐷^𝜈𝑚₀₊𝑢_𝑛(𝑠)≤1Γ𝛽Φ₂(𝑠)+ℎ₂1+𝜉,1+𝜉,1+𝛾²₀𝜉,1+𝜆²₀𝜉𝜉,1+Γ𝜉3−𝑞,1+Γ3−𝜈₁,𝜉1+Γ3−𝜈₂𝜉,…,1+Γ3−𝜈_𝑚𝑘₂(𝑠) (53) for almost all (𝑡,𝑠)∈[0,1]×[0,1] and 𝑛≥1. Hence, by using the Lebesgue dominated convergence theorem we get 𝑢(𝑡)=¹₀𝐺_𝛼(𝑡,𝑠)𝑓₁𝑠,𝑢(𝑠),𝑣(𝑠),𝜙₁𝑢(𝑠),𝜓₁𝑣(𝑠),𝐷^𝑝₀₊𝑢(𝑠),𝐷^𝜇1₀₊𝑣(𝑠),𝐷^𝜇2₀₊𝑣(𝑠)𝐷,…,_𝜇𝑚0+𝑣(𝑠)𝑣(𝑡)=𝑑𝑠,₁₀𝐺_𝛽(𝑡,𝑠)𝑓₂𝑠,𝑢(𝑠),𝑣(𝑠),𝜙₂𝑢(𝑠),𝜓₂𝑣(𝑠),𝐷^𝑞₀₊𝑣(𝑠),𝐷^𝜈1₀₊𝑢(𝑠),𝐷^𝜈2₀₊𝑢(𝑠),…,𝐷^𝜈𝑚₀₊𝑢(𝑠)𝑑𝑠 (54) for all 𝑡∈[0,1]. Therefore, (𝑢,𝑣) is a positive solution of the system (1), (2), and (3).

4. Example

Here, we give an example to illustrate our last result.

Example 1. Let 𝑚₁>0, 𝑚₂>0, 𝑛≥4, 𝑛−1<𝛼,𝛽<𝑛, 0<𝑝,𝑞<1, 1<𝜇_𝑖,𝜈_𝑖<2 for 𝑖=1,2,…,𝑚, 𝜌₁,𝜌₂,𝜎₁,𝜎₂∈𝐿¹[0,1], 𝜎₁(𝑡)≥𝑚₁ and 𝜎₂(𝑡)≥𝑚₂ for almost all 𝑡∈[0,1]. Suppose that 𝑎₁,𝑐₁∈(0,1/(𝛼−1)), 𝑎₂,𝑐₂∈(0,1/(𝛽−1)), 𝑎₅∈(0,1/(𝛼−𝑝)), 𝑐₅∈(0,1/(𝛽−𝑞)), 𝑎^′_𝑖∈(0,1/(𝛽−𝜇_𝑖)), and 𝑐^′_𝑖∈(0,1/(𝛼−𝜈_𝑖)) for 𝑖=1,2,…,𝑚 and also 𝑏₁,𝑏₂,𝑏₃,𝑏₄,𝑏₅,𝑏^′₁,𝑏^′₂,…,𝑏^′_𝑚,𝑑₁,𝑑₂,𝑑₃,𝑑₄,𝑑₅,𝑑^′₁,𝑑^′₂,…, 𝑑^′_𝑚∈(0,1). Assume that 𝑎₃, 𝑎₄, 𝑐₃ and 𝑐₄ are positive real numbers such that ∫¹₀(∫^𝑠₀𝛾₁(𝑠,𝜏)𝜏^𝛼−1𝑑𝜏)^−𝑎3𝑑𝑠<∞, ∫¹₀(∫^𝑠₀𝛾₂(𝑠,𝜏)𝜏^𝛼−1𝑑𝜏)^−𝑐3𝑑𝑠<∞, ∫¹₀(∫^𝑠₀𝜆₁(𝑠,𝜏)𝜏^𝛽−1𝑑𝜏)^−𝑎4𝑑𝑠<∞, and ∫¹₀(∫^𝑠₀𝜆₂(𝑠,𝜏)𝜏^𝛽−1𝑑𝜏)^−𝑐4𝑑𝑠<∞. Define the functions 𝑓₁ and 𝑓₂ on [0,1]×𝔇 by 𝑓₁𝑡,𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚=1𝑥^𝑎1+1𝑦^𝑎2+1𝑧^𝑎3+1𝑤^𝑎4+1𝑣^𝑎5+1𝑢^𝑎′1₁+1𝑢^𝑎′2₂1+⋯+𝑢^𝑎′𝑚_𝑚+𝜌₁(𝑡)𝑥^𝑏1+𝑦^𝑏2+𝑧^𝑏3+𝑤^𝑏4+𝑣^𝑏5+𝑢^𝑏′1₁+𝑢^𝑏′2₂+⋯+𝑢^𝑏′𝑚_𝑚+𝜎₁(𝑡),𝑓₂𝑡,𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚=1𝑥^𝑐1+1𝑦^𝑐2+1𝑧^𝑐3+1𝑤^𝑐4+1𝑣^𝑐5+1𝑢^𝑐′1₁+1𝑢^𝑐′2₂1+⋯+𝑢^𝑐′𝑚_𝑚+𝜌₂(𝑡)𝑥^𝑑1+𝑦^𝑑2+𝑧^𝑑3+𝑤^𝑑4+𝑣^𝑑5+𝑢^𝑑′1₁+𝑢^𝑑′2₂+⋯+𝑢^𝑑′𝑚_𝑚+𝜎₂(𝑡). (55) Note that the functions 𝑓₁ and 𝑓₂ satisfy the conditions (𝐻₁) and (𝐻₂), where 𝑃₁𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚=1𝑥^𝑎1+1𝑦^𝑎2+1𝑧^𝑎3+1𝑤^𝑎4+1𝑣^𝑎5+1𝑢^𝑎′1₁+1𝑢^𝑎′2₂1+⋯+𝑢^𝑎′𝑚_𝑚,𝑃₂𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚=1𝑥^𝑐1+1𝑦^𝑐2+1𝑧^𝑐3+1𝑤^𝑐4+1𝑣^𝑐5+1𝑢^𝑐′1₁+1𝑢^𝑐′2₂1+⋯+𝑢^𝑐′𝑚_𝑚,ℎ₁𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚=1+𝑥^𝑏1+𝑦^𝑏2+𝑧^𝑏3+𝑤^𝑏4+𝑣^𝑏5+𝑢^𝑏′1₁+𝑢^𝑏′2₂+⋯+𝑢^𝑏′𝑚_𝑚,ℎ₂𝑥,𝑦,𝑧,𝑤,𝑣,𝑢₁,𝑢₂,…,𝑢_𝑚=1+𝑥^𝑑1+𝑦^𝑑2+𝑧^𝑑3+𝑤^𝑑4+𝑣^𝑑5+𝑢^𝑑′1₁+𝑢^𝑑′2₂+⋯+𝑢^𝑑′𝑚_𝑚, (56)𝑘₁(𝑡)=|𝜌₁(𝑡)|+𝜎₁(𝑡) and 𝑘₂(𝑡)=|𝜌₂(𝑡)|+𝜎₂(𝑡). Theorem 6 guarantees that the system (1), (2), and (3) via these functions has a positive solution (𝑢,𝑣) satisfying the desired inequalities in Theorem 6 whenever 𝑀₁=𝑚₁/(𝛼−𝛿₁)Γ(𝛼+1) and 𝑀₂=𝑚₂/(𝛽−𝛿₂)Γ(𝛽+1).

5. Conclusions

One of the most interesting branches is obtaining solutions of singular fractional differential equations via boundary value problems. Having these thought in mind we discuss the existence of positive solutions for a coupled system of multiterm singular fractional integrodifferential boundary value problems. An illustrative example illustrates the applicability of the proposed method.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The research of the second and third authors was supported by Azarbaijan Shahid Madani University. Also, the authors express their gratitude to the referees for their helpful suggestions which improved the final version of this paper.