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

Authors: Dumitru Baleanu; Sayyedeh Zahra Nazemi; Shahram Rezapour
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We discuss the existence of positive solutions for the coupled system of multiterm singular fractional integrodifferential boundary value problemsD0+αu(t)+f1(t,u(t),v(t),(ϕ1u)(t),(ψ1v)(t),D0+pu(t),D0+μ1v(t),D0+μ2v(t),…,D0+μmv(t))=0,D0+βv(t)+f2(t,u(t),v(t),(ϕ2u)(t),(ψ2v)(t),D0+qv(t),D0+ν1u(t),D0+ν2u(t),…,D0+νmu(t))=0, u(i)(0)=0 and v(i)(0)=0 for all 0≤i≤n-2, [D0+δ1u(t)]t=1=0 for 2<δ1<n-1 and α-δ1≥1, [D0+δ2v(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<μi,νi<2(i=1,2,…,m), γj,λj:[0,1]×[0,1]→(0,∞) are continuous functions (j=1,2) and (ϕju)(t)=∫0t‍γj(t,s)u(s)ds,(ψjv)(t)=∫0t‍λj(t,s)v(s)ds. Here D is the standard Riemann-Liouville fractional derivative, fj(j=1,2) is a Caratheodory function, and fj(t,x,y,z,w,v,u1,u2,…,um) is singular at the value 0 of its variables.


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## Body

## 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:(1)D0+αu(t)+f1(D0+μmv(t)t,u(t),v(t),(ϕ1u)(t),(ψ1v)(t),D0+pu(t),D0+μ1v(t),D0+μ2v(t),…,D0+μmv(t))=0,D0+βv(t)+f2(D0+νmu(t)t,u(t),v(t),(ϕ2u)(t),(ψ2v)(t),D0+qv(t),D0+ν1u(t),D0+ν2u(t),…,D0+νmu(t))=0,(2)u(i)(0)=0∀0≤i≤n-2,[D0+δ1u(t)]t=1=0for2<δ1<n-1,α-δ1≥1,(3)v(i)(0)=0∀0≤i≤n-2,[D0+δ2v(t)]t=1=0for2<δ2<n-1,β-δ2≥1,
where n≥4, n-1<α, β<n, 0<p, q<1, 1<μi,νi<2 (i=1,2,…,m), γj and λj are positive-valued continuous functions on [0,1]×[0,1] (j=1,2),
(4)(ϕju)(t)=∫0t‍γj(t,s)u(s)ds,(ψjv)(t)=∫0t‍λj(t,s)v(s)ds,γ0j=supt∈[0,1]|∫0t‍γj(t,s)ds|<∞ and λ0j=supt∈[0,1]|∫0t‍λj(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 ℝm+5 and fj(t,x,y,z,w,v,u1,u2,…,um) 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,…,um):[0,1]→ℝ is measurable for all (x,y,z,w,v,u1,u2,…,um)∈𝔇,(ii)
f(t;·,·,…,·):𝔇→ℝ is continuous for almost all t∈[0,1],(iii)
for each compact subsetκ⊂𝔇, there is a function φκ∈L1[0,1] such that
(5)|f(t;x,y,z,w,v,u1,u2,…,um)|≤φκ(t),for almost all t∈[0,1] and all (x,y,z,w,v,u1,u2,…,um)∈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], D0+αu, D0+βv∈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
(6)D0+αu(t)+f(t,u(t),D0+μu(t))=0,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, Staněk reviewed the singular problem
(7)D0+αu(t)+f(t,u(t),u′(t),D0+μu(t))=0,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
(8)D0+αu(t)+f(t,u(t),D0+νu(t),D0+μu(t))=0,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
(9)cD0+αu(t)+f(t,u(t),u′(t),cD0+μu(t))=0,u′(0)=0, u(1)=0, where 1<α<2, 0<μ<1 and f is a Lq-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=∫01‍|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 ACk[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 mj such that fj(t,x,y,z,w,v,u1,u2,…,um)≥mj for almost all t∈[0,1] and (x,y,z,w,v,u1,u2,…,um)∈𝔇;(H2)
there exist positive mappingsk1,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,
(10)M1=m1(α-δ1)Γ(α+1),M2=m2(β-δ2)Γ(β+1),limx→∞hj(x,x,…,x)x=0,(j=1,2),∫01‍P1((δ2-2)M2Γ(β-μm)sβ-μmM1sα-1,M2sβ-1,M1∫0s‍γ1(s,τ)τα-1dτ,M2∫0s‍λ1(s,τ)τβ-1dτ,(δ1-2)M1Γ(α-p)sα-p,(δ2-2)M2Γ(β-μ1)sβ-μ1,(δ2-2)M2Γ(β-μ2)sβ-μ2,…,(δ2-2)M2Γ(β-μm)sβ-μm)ds<∞,∫01‍P2((δ1-2)M1Γ(α-νm)M1sα-1,M2sβ-1,M1∫0s‍γ2(s,τ)τα-1dτ,M2∫0s‍λ2(s,τ)τβ-1dτ,(δ2-2)M2Γ(β-q)sβ-q,(δ1-2)M1Γ(α-ν1)sα-ν1,(δ1-2)M1Γ(α-ν2)sα-ν2,…,(δ1-2)M1Γ(α-νm)sα-νm)ds<∞,fj(t,x,y,z,w,v,u1,u2,…,um)≤Pj(x,y,z,w,v,u1,u2,…,um)+kj(t)hj(x,y,z,w,v,u1,u2,…,um)for almost all t∈[0,1] and (x,y,z,w,v,u1,u2,…,um)∈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 fnj (j=1,2) by
(11)fnj(t,x,y,z,w,v,u1,u2,…,um)=fj(t,χn(x),χn(y),χn(z),χn(w),χn(v),χn(u1),χn(u2),…,χn(um))
for almost all t∈[0,1] and (x,y,z,w,v,u1,u2,…,um)∈ℝm+5, where
(12)χn(x)={xifx≥1n,1nifx<1n.
Then condition (H1) implies that fnj∈Car([0,1]×ℝm+5) and fnj(t,x,y,z,w,v,u1,u2,…,um)≥mj for almost all t∈[0,1] and (x,y,z,w,v,u1,u2,…,um)∈ℝm+5, (j=1,2). Also, the condition (H2) implies
(13)fnj(t,x,y,z,w,v,u1,u2,…,um)≤Pj(1n,1n,…,1n)+kj(t)hj(1+x,1+y,1+z,1+w,1+v,1+u1,1+u2,…,1+um)
for almost all t∈[0,1] and (x,y,z,w,v,u1,u2,…,um)∈[0,∞)m+5 and
(14)fnj(t,x,y,z,w,v,u1,u2,…,um)≤Pj(x,y,z,w,v,u1,u2,…,um)+kj(t)hj(1+x,1+y,1+z,1+w,1+v,1+u1,1+u2,…,1+um)
for almost all t∈[0,1] and (x,y,z,w,v,u1,u2,…,um)∈𝔇. For obtaining solutions of the system (1), (2), and (3), for each natural number n, we will obtain solutions of the system
(15)D0+αu(t)+fn1(12t,u(t),v(t),(ϕ1u)(t),(ψ1v)(t),D0+pu(t),D0+μ1v(t),D0+μ2v(t),…,D0+μmv(t)12)=0,D0+βv(t)+fn2(12t,u(t),v(t),(ϕ2u)(t),(ψ2v)(t),D0+qv(t),D0+ν1u(t),D0+ν2u(t),…,D0+νmu(t)12)=0,
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 I0+α 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 D0+αx∈L1[0,1], then x(t)=I0+αD0+αx(t)+∑k=1n‍cktα-k for all t∈(0,1], where n=[α]+1 and ck∈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, D0+μx∈C[0,1], D0+νx∈C[0,1], D0+μx(t)=(1/Γ(2-μ))∫0t‍(t-s)1-μx′′(s)ds, and D0+νx(t)=(1/Γ(2-ν))∫0t‍(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.
Letn≥4 and n-1<α<n. For each g∈L1[0,1], u(t)=∫01‍Gα(t,s)g(s)ds is the unique solution of the equation D0+αu(t)+g(t)=0 in C2[0,1] which satisfies the boundary condition (2), where
(16)Gα(t,s)={tα-1(1-s)α-δ1-1-(t-s)α-1Γ(α)0≤s≤t≤1,tα-1(1-s)α-δ1-1Γ(α)0≤t≤s≤1.Proof.
It is easy to see that the functionsu(t)=-I0+αg(t)+c1tα-1+c2tα-2+⋯+cntα-n are solutions of D0+αu(t)+g(t)=0 in C(0,1] for all ci∈ℝ. Since [α]≥3, I0+αg∈AC[α]-1[0,1] and so u(t)=-I0+αg(t)+c1tα-1 are solutions of D0+αu(t)+g(t)=0 in C2[0,1], where c1∈ℝ. The boundary condition (2) implies that c2=⋯=cn-1=0 and
(17)0=c1Γ(α)Γ(α-δ1)(1)α-δ1-1-1Γ(α-δ1)∫01‍(1-s)α-δ1-1g(s)ds.
Thus, u(t)=(tα-1/Γ(α))∫01‍(1-s)α-δ1-1g(s)ds-(1/Γ(α))∫0t‍(t-s)α-1g(s)ds=∫01‍Gα(t,s)g(s)ds is the unique solution of the problem in C2[0,1].Note that the Green functionGα in Lemma 1 has some properties. For example, Gα, (∂/∂t)Gα and (∂2/∂t2)Gα are continuous functions on [0,1]×[0,1], Gα(t,s)>0 on (0,1)×(0,1), Gα(t,s)≤1/Γ(α) for all (t,s)∈[0,1]×[0,1], ∫01‍Gα(t,s)ds≥tα-1/(α-δ1)Γ(α+1) for all t∈[0,1], (∂/∂t)Gα(t,s)>0 on (0,1)×(0,1), (∂/∂t)Gα(t,s)≤1/Γ(α-1) for all (t,s)∈[0,1]×[0,1], ∫01‍(∂/∂t)Gα(t,s)ds≥tα-2/(α-δ1)Γ(α) for all t∈[0,1], (∂2/∂t2)Gα(t,s)>0 on (0,1)×(0,1), (∂2/∂t2)Gα(t,s)≤1/Γ(α-2) for all (t,s)∈[0,1]×[0,1], and ∫01‍(∂2/∂t2)Gα(t,s)ds≥(δ1-2)tα-3/(α-δ1)Γ(α-1) for all t∈[0,1]. Let X=C2[0,1]. Define the cone P on X×X by
(18)P={21(x,y)∈X×X:x(0)=x′(0)=y(0)=y′(0)=0,x(t),x′(t),x′′(t),y(t),y′(t),y′′(t)≥0,∀t∈[0,1]12}.
It is easy to see that for each (x,y)∈P, i=1,2,…,m and t∈[0,1] we have D0+px,D0+νix,D0+qy,D0+μiy∈C[0,1] and D0+px(t)≥0, D0+νix(t)≥0, D0+qy(t)≥0, D0+μiy(t)≥0. Now for each natural number n, define the operator Tn on P by
(19)Tn(x,y)(t)=(Tn1(x,y)(t)Tn2(x,y)(t))=(∫01‍Gα(t,s)fn1×(D0+μms,x(s),y(s),(ϕ1x)(s),(ψ1y)(s),D0+px(s),D0+μ1y(s),D0+μ2y(s),…,D0+μmy(s))ds∫01‍Gβ(t,s)fn2×(D0+μms,x(s),y(s),(ϕ2x)(s),(ψ2y)(s),D0+qy(s),D0+ν1x(s),D0+ν2x(s),…,D0+νmx(s))ds∫01‍Gα(t,s)fn1).Lemma 2.
For each natural numbern, Tn is a completely continuous operator on P.Proof.
Letn be a natural number, (x,y)∈P,
(20)ρ1(t)=fn1(D0+μmt,x(t),y(t),(ϕ1x)(t),(ψ1y)(t),D0+px(t),D0+μ1y(t),D0+μ2y(t),…,D0+μmy(t))
and ρ2(t)=fn2(t,x(t), y(t), (ϕ2x)(t), (ψ2y)(t), D0+qy(t), D0+ν1x(t), D0+ν2x(t), …, D0+νmx(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 Gα, Gβ, (∂/∂t)Gα, (∂/∂t)Gβ, (∂2/∂t2)Gα, and (∂2/∂t2)Gβ are nonnegative and continuous functions on [0,1]×[0,1], Gα(0,s)=0, and Gβ(0,s)=0 for all s∈[0,1], we get Tn1(x,y),Tn2(x,y)∈C2[0,1], Tn1(x,y)(0)=Tn2(x,y)(0)=(Tn1(x,y))′(0)=(Tn2(x,y))′(0)=0, Tn1(x,y)≥0, Tn2(x,y)≥0, (Tn1(x,y))′≥0, (Tn2(x,y))′≥0, (Tn1(x,y))′′≥0, and (Tn2(x,y))′′≥0 on [0,1]. Thus, Tn maps P into P. Suppose that {(xk,yk)}k≥1 is a convergent sequence in P and limk→∞(xk,yk)=(x,y). Then, limk→∞xk(j)(t)=x(j)(t) and limk→∞yk(j)(t)=y(j)(t) uniformly on [0,1] for j=0,1,2. Since for p,q∈(0,1), μi,νi∈(1,2) for all i=1,2,…,m,
(21)|D0+pxk(t)-D0+px(t)|≤∥xk′′-x′′∥Γ(2-p)∫0t‍(t-s)1-pds≤∥xk′′-x′′∥Γ(3-p),|D0+qyk(t)-D0+qy(t)|≤∥yk′′-y′′∥Γ(2-q)∫0t‍(t-s)1-qds≤∥yk′′-y′′∥Γ(3-q),|D0+νixk(t)-D0+νix(t)|≤∥xk′′-x′′∥Γ(2-νi)∫0t‍(t-s)1-νids≤∥xk′′-x′′∥Γ(3-νi),|D0+μiyk(t)-D0+μiy(t)|≤∥yk′′-y′′∥Γ(2-μi)∫0t‍(t-s)1-μids≤∥yk′′-y′′∥Γ(3-μi),|(ϕjxk)(t)-(ϕjx)(t)|≤supt∈[0,1]|∫0t‍γj(t,s)ds|∥xk-x∥=γ0j∥xk-x∥,|(ψjyk)(t)-(ψjy)(t)|≤supt∈[0,1]|∫0t‍λj(t,s)ds|∥yk-y∥=λ0j∥yk-y∥,
for j=1,2 and k≥1, we have limk→∞D0+pxk(t)=D0+px(t), limk→∞D0+qyk(t)=D0+qy(t), limk→∞D0+νixk(t)=D0+νix(t), limk→∞D0+μiyk(t)=D0+μiy(t) for i=1,2,…,m, and limk→∞(ϕjxk)(t)=(ϕjx)(t), limk→∞(ψjyk)(t)=(ψjy)(t) uniformly on [0,1] for j=1,2. Also, we have
(22)∥D0+pxk∥≤∥xk′′∥Γ(2-p)∫0t‍(t-s)1-pds≤∥xk′′∥Γ(3-p),∥D0+qyk∥≤∥yk′′∥Γ(2-q)∫0t‍(t-s)1-qds≤∥yk′′∥Γ(3-q),∥D0+νixk∥≤∥xk′′∥Γ(2-νi)∫0t‍(t-s)1-νids≤∥xk′′∥Γ(3-νi),∥D0+μiyk∥≤∥yk′′∥Γ(2-μi)∫0t‍(t-s)1-μids≤∥yk′′∥Γ(3-μi),
and also ∥ϕjxk∥≤γ0j∥xk∥ and ∥ψjyk∥≤λ0j∥yk∥ for j=1,2 and k≥1. Now, put
(23)ρk1(t)=fn1(D0+μmt,xk(t),yk(t),(ϕ1xk)(t),ρk1(t)222=(ψ1yk)(t),D0+pxk(t),D0+μ1yk(t),D0+μ2yk(t),…,D0+μmyk(t))
andρk2(t)=fn2(t,xk(t),yk(t),(ϕ2xk)(t),(ψ2yk)(t),D0+qyk(t), D0+ν1xk(t), D0+ν2xk(t),…,D0+νmxk(t)) for all k≥1. Then, limk→∞ρkj(t)=ρj(t) for almost all t∈[0,1] and j=1,2. Since fnj∈Car([0,1]×ℝm+5) and {xk}k≥1, {yk}k≥1 are bounded sequences in C2[0,1] and also by using the above inequalities, {D0+pxk}k≥1, {D0+qyk}k≥1, {D0+νixk}k≥1, {D0+μiyk}k≥1 for i=1,2,…,m and {ϕjxk}k≥1, {ψjxk}k≥1 for j=1,2 are bounded sequences in C[0,1], there exist g1,g2∈L1[0,1] such that mj≤ρkj(t)≤gj(t) for almost all t∈[0,1] and k≥1. Thus, by using the Lebesgue dominated convergence theorem and the following relations:
(24)|Tn1(xk,yk)(t)-Tn1(x,y)(t)|≤1Γ(α)∫01‍|ρk1(s)-ρ1(s)|ds,|(Tn1(xk,yk))′(t)-(Tn1(x,y))′(t)|≤1Γ(α-1)∫01‍|ρk1(s)-ρ1(s)|ds,|(Tn1(xk,yk))′′(t)-(Tn1(x,y))′′(t)|≤1Γ(α-2)∫01‍|ρk1(s)-ρ1(s)|ds,|Tn2(xk,yk)(t)-Tn2(x,y)(t)|≤1Γ(β)∫01‍|ρk2(s)-ρ2(s)|ds,|(Tn2(xk,yk))′(t)-(Tn2(x,y))′(t)|≤1Γ(β-1)∫01‍|ρk2(s)-ρ2(s)|ds,|(Tn2(xk,yk))′′(t)-(Tn2(x,y))′′(t)|≤1Γ(β-2)∫01‍|ρk2(s)-ρ2(s)|ds,
we get limk→∞(Tn1(xk,yk))(j)(t)=(Tn1(x,y))(j)(t) and limk→∞(Tn2(xk,yk))(j)(t)=(Tn2(x,y))(j)(t) uniformly on [0,1] for j=0,1,2. This implies that Tn1 and Tn2 are continuous operators and so Tn is a continuous operator. Let {(xk,yk)}k≥1 be a bounded sequence in P. We show that the sequence {Tn(xk,yk)}k≥1 is relatively compact in X×X. By using the Arzela-Ascoli theorem, it is sufficient to prove that {Tn(xk,yk)}k≥1 is bounded in X×X and {(Tn(xk,yk))′′} is equicontinuous on [0,1]. Choose a positive number S such that ∥xk∥≤S, ∥yk∥≤S, ∥xk′∥≤S, ∥yk′∥≤S, ∥xk′′∥≤S, ∥yk′′∥≤S for all k≥1. By using the above inequalities, one can see that ∥D0+pxk∥≤S/Γ(3-p), ∥D0+qyk∥≤S/Γ(3-q), ∥D0+νixk∥≤S/Γ(3-νi), and ∥D0+μiyk∥≤S/Γ(3-μi) for i=1,2,…,m and ∥ϕjxk∥≤γ0jS and ∥ψjxk∥≤λ0jS for j=1,2. Since
(25)0≤Tn1(xk,yk)(t)=∫01‍Gα(t,s)ρk1(s)ds≤1Γ(α)∫01‍g1(s)ds=∥g1∥1Γ(α),0≤(Tn1(xk,yk))′(t)=∫01‍∂∂tGα(t,s)ρk1(s)ds≤1Γ(α-1)∫01‍g1(s)ds=∥g1∥1Γ(α-1),0≤(Tn1(xk,yk))′′(t)=∫01‍∂2∂t2Gα(t,s)ρk1(s)ds≤1Γ(α-2)∫01‍g1(s)ds=∥g1∥1Γ(α-2),∥Tn1(xk,yk)∥*≤∥g1∥1/Γ(α-2). Similarly, we can verify that ∥Tn2(xk,yk)∥*≤∥g2∥1/Γ(β-2). Thus,
(26)∥Tn(xk,yk)∥**≤max{∥g1∥1Γ(α-2),∥g2∥1Γ(β-2)}
for all k. Hence, {Tn(xk,yk)}k≥1 is bounded in X×X. Also, for 0≤t1<t2≤1 we have
(27)|(Tn1(xk,yk))′′(t2)-(Tn1(xk,yk))′′(t1)|≤t2α-3-t1α-3Γ(α-2)∫01‍(1-s)α-δ1-1ρk1(s)ds+1Γ(α-2)×|∫0t2‍(t2-s)α-3ρk1(s)ds-∫0t1‍(t1-s)α-3ρk1(s)ds|≤∥ρk1∥1Γ(α-2)(t2α-3-t1α-3)+1Γ(α-2)×[∫t1t2‍(t2-s)α-3ρk1(s)ds+∫0t1‍((t2-s)α-3-(t1-s)α-3)ρk1(s)ds]≤∥g1∥1Γ(α-2)(t2α-3-t1α-3)+1Γ(α-2)×[∫0t1‍(t2-t1)α-3∥g1∥1+∫0t1‍((t2-s)α-3-(t1-s)α-3)g1(s)ds].
Similarly, we have
(28)|(Tn2(xk,yk))′′(t2)-(Tn2(xk,yk))′′(t1)|≤∥g2∥1Γ(β-2)(t2β-3-t1β-3)+1Γ(β-2)×[∫0t1‍(t2-t1)β-3∥g2∥1+∫0t1‍((t2-s)β-3-(t1-s)β-3)g2(s)ds].
Let ϵ>0 be given. Since the functions tα-3 and tβ-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<t2α-3-t1α-3<ϵ, 0<t2β-3-t1β-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,ϵα-3,ϵβ-3}, then
(29)|(Tn1(xk,yk))′′(t2)-(Tn1(xk,yk))′′(t1)|≤3ϵΓ(α-2)∥g1∥1
and |(Tn2(xk,yk))′′(t2)-(Tn2(xk,yk))′′(t1)|≤(3ϵ/Γ(β-2))∥g2∥1. Therefore, {(Tn(xk,yk))′′} is equicontinuous on [0,1] and so Tn is a completely continuous operator on P.We need the following result (see [10]).Lemma 3.
LetY be a Banach space, P a cone in Y, and Ω1 and Ω2 bounded open balls in Y centered at the origin with Ω-1⊂Ω2. Suppose that T:P∩(Ω-2∖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∩(Ω-2∖Ω1).Theorem 4.
For each natural numbern, the system (15), (2), and (3) has a solution (un,vn)∈P such that un≥M1tα-1 and vn(t)≥M2tβ-1 for all t∈[0,1].Proof.
Let(x,y)∈P and n≥1 be given. Then, Tn1(x,y)(t)≥m1∫01‍Gα(t,s)ds≥M1tα-1 and Tn2(x,y)(t)≥m2∫01‍Gβ(t,s)ds≥M2tβ-1 for all t∈[0,1]. Hence, ∥Tn1(x,y)∥≥M1 and ∥Tn2(x,y)∥≥M2 and so ∥Tn1(x,y)∥*≥M1 and ∥Tn2(x,y)∥*≥M2. Thus, ∥Tn(x,y)∥**≥max{M1,M2}=M. Now, put
(30)Ω1={(x,y)∈X×X:∥(x,y)∥**<M}.
Then, ∥Tn(x,y)∥**≥∥(x,y)∥** for all (x,y)∈P∩∂Ω1. Let Wnj=Pj(1/n,1/n,…,1/n) for j=1,2. Then, we have
(31)0≤(Tn1(x,y))′′(t)≤1Γ(α-2)×∫01‍(D0+μ2y(s)Wn1+k1(s)h1(D0+μmy1+x(s),1+y(s),1+(ϕ1x)(s),1+(ψ1y)(s),1+D0+px(s),1+D0+μ1y(s),1+D0+μ2y(s),…,1+D0+μmy(s)))ds≤1Γ(α-2)×(∥D0+μmy∥Wn1+h1(∥D0+μmy∥1+∥x∥,1+∥y∥,1+∥ϕ1x∥,1+∥ψ1y∥,1+∥D0+px∥,1+∥D0+μ1y∥,1+∥D0+μ2y∥,…,1+∥D0+μmy∥)∥k1∥1),0≤(Tn1(x,y))′(t)=∫0t‍(Tn1(x,y))′′(s)ds≤1Γ(α-2)×(∥D0+μmy∥Wn1+h1(∥D0+μmy∥1+∥x∥,1+∥y∥,1+∥ϕ1x∥,1+∥ψ1y∥,1+∥D0+px∥,1+∥D0+μ1y∥,1+∥D0+μ2y∥,…,1+∥D0+μmy∥)∥k1∥1),0≤Tn1(x,y)(t)=∫0t‍(Tn1(x,y))′(s)ds≤1Γ(α-2)(D0+μ1yWn1+h1(∥D0+μmy∥1+∥x∥,1+∥y∥,1+∥ϕ1x∥,1+∥ψ1y∥,1+∥D0+px∥,1+∥D0+μ1y∥,1+∥D0+μ2y∥,…,1+∥D0+μmy∥)∥k1∥1)
for all (x,y)∈P and t∈[0,1]. Hence,
(32)∥Tn1(x,y)∥*≤1Γ(α-2)(∥y∥*Γ(3-μm)Wn1+h1(∥y∥*Γ(3-μm)1+∥x∥*,1+∥y∥*,1+γ01∥x∥*,1+λ01∥y∥*,1+∥x∥*Γ(3-p),1+∥y∥*Γ(3-μ1),1+∥y∥*Γ(3-μ2),…,1+∥y∥*Γ(3-μm))∥k1∥1)≤1Γ(α-2)(Wn1+h1(∥(x,y)∥**Γ(3-μm)1+∥(x,y)∥**,1+∥(x,y)∥**,1+γ01∥(x,y)∥**,1+λ01∥(x,y)∥**,1+∥(x,y)∥**Γ(3-p),1+∥(x,y)∥**Γ(3-μ1),1+∥(x,y)∥**Γ(3-μ2),…,1+∥(x,y)∥**Γ(3-μm))∥k1∥1).
Similarly, we have
(33)∥Tn2(x,y)∥*≤1Γ(β-2)(∥(x,y)∥**Γ(3-νm)Wn2+h2(∥(x,y)∥**Γ(3-νm)1+∥(x,y)∥**,1+∥(x,y)∥**,1+γ02∥(x,y)∥**,1+λ02∥(x,y)∥**,1+∥(x,y)∥**Γ(3-q),1+∥(x,y)∥**Γ(3-ν1),1+∥(x,y)∥**Γ(3-ν2),…,1+∥(x,y)∥**Γ(3-νm))∥k2∥1)
for all (x,y)∈P. Since limx→∞hj(x,x,…,x)/x=0 for j=1,2, there exists S>M>0 such that
(34)1Γ(α-2)(SΓ(3-μm)Wn1+h1(SΓ(3-μm)1+S,1+S,1+γ01S,1+λ01S,1+SΓ(3-p),1+SΓ(3-μ1),1+SΓ(3-μ2),…,1+SΓ(3-μm))∥k1∥1)≤S,1Γ(β-2)(Wn2+h2(SΓ(3-μm)1+S,1+S,1+γ02S,1+λ02S,1+SΓ(3-q),1+SΓ(3-ν1),1+SΓ(3-ν2),…,1+SΓ(3-νm))∥k2∥1)≤S.
Put Ω2={(x,y)∈X×Xs.t∥(x,y)∥**<S}. Then, the above inequalities imply that ∥Tn(x,y)∥**≤∥(x,y)∥** for all (x,y)∈P∩∂Ω2. Now by using Lemma 3, we conclude that the operator Tn has a fixed point (un,vn) in P∩(Ω¯2∖Ω1). It is easy to see that (un,vn) is a desired solution of the system, that is, un≥M1tα-1 and vn(t)≥M2tβ-1 for all t∈[0,1].We need the following lemma.Lemma 5.
The set of solutions{(un,vn)}n≥1 of the system (15), (2), and (3) is a relatively compact subset of X×X.Proof.
It is easy to check that(35)un(t)=∫01‍Gα(t,s)fn1(D0+qs,un(s),vn(s),(ϕ1un)(s),(ψ1vn)(s),D0+pun(s),D0+μ1vn(s),D0+μ2vn(s),…,D0+μmvn(s))ds,vn(t)=∫01‍Gβ(t,s)fn2(D0+qs,un(s),vn(s),(ϕ2un)(s),(ψ2vn)(s),D0+qvn(s),D0+ν1un(s),D0+ν2un(s),…,D0+νmun(s))ds
for t∈[0,1] and n≥1, satisfy in Theorem 4. Also, un′(t)≥m1∫01‍(∂/∂t)Gα(t,s)ds≥m1tα-2/Γ(α)(α-δ1), un′′(t)≥m1∫01‍(∂2/∂t2)Gα(t,s)ds≥m1(δ1-2)tα-3/Γ(α-1)(α-δ1), vn′(t)≥m2∫01‍(∂/∂t)Gβ(t,s)ds≥m2tβ-2/Γ(β)(β-δ2), and vn′′(t)≥m2∫01‍(∂2/∂t2)Gβ(t,s)ds≥m2(δ2-2)tβ-3/Γ(β-1)(β-δ2) for t∈[0,1] and n≥1. Moreover,
(36)D0+pun(t)=1Γ(2-p)∫0t‍(t-s)1-pun′′(s)ds≥m1(δ1-2)Γ(2-p)Γ(α-1)(α-δ1)×∫0t‍(t-s)1-psα-3ds,D0+qvn(t)=1Γ(2-q)∫0t(t-s)1-qvn′′(s)ds≥m2(δ2-2)Γ(2-q)Γ(β-1)(β-δ2)×∫0t(t-s)1-qsβ-3ds,D0+νiun(t)=1Γ(2-νi)∫0t(t-s)1-νiun′′(s)ds≥m1(δ1-2)Γ(2-νi)Γ(α-1)(α-δ1)×∫0t(t-s)1-νisα-3ds,D0+μivn(t)=1Γ(2-μi)∫0t(t-s)1-μivn′′(s)ds≥m2(δ2-2)Γ(2-μi)Γ(β-1)(β-δ2)×∫0t‍(t-s)1-μisβ-3ds
for i=1,2,…,m. Since
(37)∫0t‍(t-s)1-psα-3ds=tα-p-1×∫01‍(1-ξ)1-pξα-3dξ=Γ(2-p)Γ(α-2)Γ(α-p)tα-p-1≥Γ(2-p)Γ(α-2)Γ(α-p)tα-p,∫0t(t-s)1-qsβ-3ds≥Γ(2-q)Γ(β-2)Γ(β-q)tβ-q,∫0t(t-s)1-νisα-3ds≥Γ(2-νi)Γ(α-2)Γ(α-νi)tα-νi,∫0t(t-s)1-μisβ-3ds≥Γ(2-μi)Γ(β-2)Γ(β-μi)tβ-μi
for i=1,2,…,m, we get D0+pun(t)≥(m1(δ1-2)/(α-2)(α-δ1)Γ(α-p))tα-p, D0+qvn(t)≥(m2(δ2-2)/(β-2)(β-δ2)Γ(β-q))tβ-q, D0+νiun(t)≥(m1(δ1-2)/(α-2)(α-δ1)Γ(α-νi))tα-νi, and D0+μivn(t)≥(m2(δ2-2)/(β-2)(β-δ2)Γ(β-μi))tβ-μi for all t∈[0,1] and n≥1. Since
(38)m1·min{1(α-δ1)Γ(α+1),1(α-2)(α-δ1)}=M1
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 un(t)≥M1tα-1, vn(t)≥M2tβ-1, D0+pun(t)≥((δ1-2)M1/Γ(α-p))tα-p, D0+qvn(t)≥((δ2-2)M2/Γ(β-q))tβ-q, D0+νiun(t)≥((δ1-2)M1/Γ(α-νi))tα-νi, and D0+μivn(t)≥((δ2-2)M2/Γ(β-μi))tβ-μi for i=1,2,…,m and (ϕjun)(t)≥M1∫0t‍γj(t,s)sα-1ds and (ψjvn)(t)≥M2∫0t‍λj(t,s)sβ-1ds for j=1,2. Thus,
(39)P1(D0+μ2vnun(t),vn(t),(ϕ1un)(t),(ψ1vn)(t),D0+pun(t),D0+μ1vn(t),D0+μ2vn(t),…,D0+μmvn(t))≤P1((δ2-2)M2Γ(β-μm)tβ-μmM1tα-1,M2tβ-1,M1∫0tγ1(t,s)sα-1ds,M2∫0tλ1(t,s)sβ-1ds,(δ1-2)M1Γ(α-p)tα-p,(δ2-2)M2Γ(β-μ1)tβ-μ1,(δ2-2)M2Γ(β-μ2)tβ-μ2,…,(δ2-2)M2Γ(β-μm)tβ-μm),P2(un(t),vn(t),(ϕ2un)(t),(ψ2vn)(t),D0+qvn(t),D0+ν1un(t),D0+ν2un(t),…,D0+νmun(t))≤P2(M1tα-1,M2tβ-1,M1∫0tγ2(t,s)sα-1ds,M2∫0t‍λ2(t,s)sβ-1ds,(δ2-2)M2Γ(β-q)tβ-q,(δ1-2)M1Γ(α-ν1)tα-ν1,(δ1-2)M1Γ(α-ν2)tα-ν2,…,(δ1-2)M1Γ(α-νm)tα-νm)
for all t∈[0,1] and n≥1 and so
(40)0≤un′′(t)=∫01‍∂2∂t2Gα(t,s)fn1(D0+μ2vns,un(s),vn(s),(ϕ1un)(s),(ψ1vn)(s),D0+pun(s),D0+μ1vn(s),D0+μ2vn(s),…,D0+μmvn(s))ds≤1Γ(α-2)[∫01P1((δ2-2)M2Γ(β-μm)M1sα-1,M2sβ-1,M1∫0sγ1(s,τ)τα-1dτ,M2∫0sλ1(s,τ)τβ-1dτ,(δ1-2)M1Γ(α-p)sα-p,(δ2-2)M2Γ(β-μ1)sβ-μ1,(δ2-2)M2Γ(β-μ2)sβ-μ2,…,(δ2-2)M2Γ(β-μm)sβ-μm)ds+h1(∥vn∥*Γ(3-μm)1+∥un∥*,1+∥vn∥*,1+γ01∥un∥*,1+λ01∥vn∥*,1+∥un∥*Γ(3-p),1+∥vn∥*Γ(3-μ1),1+∥vn∥*Γ(3-μ2),…,1+∥vn∥*Γ(3-μm))×∫01k1(s)ds]=1Γ(α-2)×(Λ1+h1(∥vn∥*Γ(3-μm)1+∥un∥*,1+∥vn∥*,1+γ01∥un∥*,1+λ01∥vn∥*,1+∥un∥*Γ(3-p),1+∥vn∥*Γ(3-μ1),1+∥vn∥*Γ(3-μ2),…,1+∥vn∥*Γ(3-μm))∥k1∥1),0≤un′(t)=∫0t‍un′′(s)ds≤1Γ(α-2)×(Λ1+h1(∥vn∥*Γ(3-μm)1+∥un∥*,1+∥vn∥*,1+γ01∥un∥*,1+λ01∥vn∥*,1+∥un∥*Γ(3-p),1+∥vn∥*Γ(3-μ1),1+∥vn∥*Γ(3-μ2),…,1+∥vn∥*Γ(3-μm))∥k1∥1),0≤un(t)=∫0tun′(s)ds≤1Γ(α-2)×(Λ1+h1(∥vn∥*Γ(3-μm)1+∥un∥*,1+∥vn∥*,1+γ01∥un∥*,1+λ01∥vn∥*,1+∥un∥*Γ(3-p),1+∥vn∥*Γ(3-μ1),1+∥vn∥*Γ(3-μ2),…,1+∥vn∥*Γ(3-μm))∥k1∥1)
for all t∈[0,1] and n≥1 and also Λ1<∞, where
(41)Λ1=∫01P1((δ2-2)M2Γ(β-μm)M1sα-1,M2sβ-1,M1∫0sγ1(s,τ)τα-1dτ,M2∫0sλ1(s,τ)τβ-1dτ,(δ1-2)M1Γ(α-p)sα-p,(δ2-2)M2Γ(β-μ1)sβ-μ1,(δ2-2)M2Γ(β-μ2)sβ-μ2,…,(δ2-2)M2Γ(β-μm)sβ-μm)ds.
Hence,
(42)∥un∥*≤1Γ(α-2)(Λ1+h1(∥vn∥*Γ(3-μm)1+∥un∥*,1+∥vn∥*,1+γ01∥un∥*,1+λ01∥vn∥*,1+∥un∥*Γ(3-p),1+∥vn∥*Γ(3-μ1),1+∥vn∥*Γ(3-μ2),…,1+∥vn∥*Γ(3-μm))∥k1∥1)≤1Γ(α-2)(Λ1+h1(∥(un,vn)∥**Γ(3-μm)1+∥(un,vn)∥**,1+∥(un,vn)∥**,1+γ01∥(un,vn)∥**,1+λ01∥(un,vn)∥**,1+∥(un,vn)∥**Γ(3-p),1+∥(un,vn)∥**Γ(3-μ1),1+∥(un,vn)∥**Γ(3-μ2),…,1+∥(un,vn)∥**Γ(3-μm))∥k1∥1)=Gn1
for all n≥1. Similarly, we have
(43)∥vn∥*≤1Γ(β-2)×(1+∥(un,vn)∥**Γ(3-νm))∥k2∥1Λ2+h2(1+∥(un,vn)∥**Γ(3-νm))∥k2∥11+∥(un,vn)∥**,1+∥(un,vn)∥**,1+γ02∥(un,vn)∥**,1+λ02∥(un,vn)∥**,1+∥(un,vn)∥**Γ(3-q),1+∥(un,vn)∥**Γ(3-ν1),1+∥(un,vn)∥**Γ(3-ν2),…,1+∥(un,vn)∥**Γ(3-νm))∥k2∥1)=Gn2
for all n≥1, where
(44)Λ2=∫01P2((δ1-2)M1Γ(α-νm)sα-νmM1sα-1,M2sβ-1,M1∫0sγ2(s,τ)τα-1dτ,M2∫0s‍λ2(s,τ)τβ-1dτ,(δ2-2)M2Γ(β-q)sβ-q,(δ1-2)M1Γ(α-ν1)sα-ν1,(δ1-2)M1Γ(α-ν2)sα-ν2,…,(δ1-2)M1Γ(α-νm)sα-νm)ds.
Thus, ∥(un,vn)∥**≤max{Gn1,Gn2} for all n≥1. Since limx→∞hj(x,x,…,x)/x=0 for j=1,2, there exists L>0 such that
(45)1Γ(α-2)(Λ1+h1(1+υΓ(3-μm))∥k1∥11+υ,1+υ,1+γ01υ,1+λ01υ,1+υΓ(3-p),1+υΓ(3-μ1),1+υΓ(3-μ2),…,1+υΓ(3-μm))∥k1∥1)<υ,1Γ(β-2)(Λ2+h2(υΓ(3-νm)1+υ,1+υ,1+γ02υ,1+λ02υ,1+υΓ(3-q),1+υΓ(3-ν1),1+υΓ(3-ν2),…,1+υΓ(3-νm))∥k2∥1)<υ
for all υ≥L. Consequently, ∥(un,vn)∥**<L for all n≥1 and so {(un,vn)}n≥1 is a bounded sequence in X×X. It remains to prove that {(un,vn)′′} is equicontinuous on [0,1]. Put
(46)V1=h1(LΓ(3-μm)1+L,1+L,1+γ01L,1+λ01L,1+LΓ(3-p),1+LΓ(3-μ1),1+LΓ(3-μ2),…,1+LΓ(3-μm)),V2=h2(LΓ(3-νm)1+L,1+L,1+γ02L,1+λ02L,1+LΓ(3-q),1+LΓ(3-ν1),1+LΓ(3-ν2),…,1+LΓ(3-νm)),Φ1(t)=P1((δ2-2)M2Γ(β-μm)M1tα-1,M2tβ-1,M1∫0tγ1(t,s)sα-1ds,M2∫0tλ1(t,s)sβ-1ds(δ1-2)M1Γ(α-p)tα-p,(δ2-2)M2Γ(β-μ1)tβ-μ1,(δ2-2)M2Γ(β-μ2)tβ-μ2,…,(δ2-2)M2Γ(β-μm)tβ-μm),Φ2(t)=P2((δ1-2)M1Γ(α-νm)tα-νmM1tα-1,M2tβ-1,M1∫0tγ2(t,s)sα-1ds,M2∫0tλ2(t,s)sβ-1ds,(δ2-2)M2Γ(β-q)tβ-q,(δ1-2)M1Γ(α-ν1)tα-ν1,(δ1-2)M1Γ(α-ν2)tα-ν2,…,(δ1-2)M1Γ(α-νm)tα-νm)
for all t∈(0,1]. Note that, Λj=∫01‍Φj(t)dt for j=1,2 and
(47)fn1(D0+μmD0+νmt,un(t),vn(t),(ϕ1un)(t),(ψ1vn)(t),D0+pun(t),D0+μ1vn(t),D0+μ2vn(t),…,D0+μmvn(t))≤Φ1(t)+V1k1(t),fn2(D0+νmt,un(t),vn(t),(ϕ2un)(t),(ψ2vn)(t),D0+qvn(t),D0+ν1un(t),D0+ν2un(t),…,D0+νmun(t))≤Φ2(t)+V2k2(t)
for almost all t∈[0,1] and n≥1. If 0≤t1<t2≤1, then
(48)|un′′(t2)-un′′(t1)|=|∫01(∂2∂t2Gα(t2,s)-∂2∂t2Gα(t1,s))fn1(s,un(s),vn(s),(ϕ1un)(s),(ψ1vn)(s),D0+pun(s),D0+μ1vn(s),D0+μ2vn(s),…,D0+μmvn(s))ds∫01(∂2∂t2Gα(t2,s)-∂2∂t2Gα(t1,s))|≤1Γ(α-2)[(t2α-3-t1α-3)∫01(Φ1(s)+V1k1(s))ds+∫t1t2(t2-s)α-3(Φ1(s)+V1k1(s))ds+∫0t1((t2-s)α-3-(t1-s)α-3)×(Φ1(s)+V1k1(s))ds∫01(Φ1(s)+V1k1(s))ds]≤1Γ(α-2)[(Φ1(s)+V1k1(s))(t2α-3-t1α-3)(Λ1+V1∥k1∥1)+(t2-t1)α-3(Λ1+V1∥k1∥1)+∫0t1((t2-s)α-3-(t1-s)α-3)×(Φ1(s)+V1k1(s))ds].
Similarly, we have
(49)|vn′′(t2)-vn′′(t1)|≤1Γ(β-2)[(t2β-3-t1β-3)(Λ2+V2∥k2∥1)+(t2-t1)β-3(Λ2+V2∥k2∥1)+∫0t1((t2-s)β-3-(t1-s)β-3)×(Φ2(s)+V2k2(s))ds].
Let ϵ>0 be given. Choose δ0>0 such that 0<t2α-3-t1α-3<ϵ, 0<t2β-3-t1β-3<ϵ,
(50)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<δ0 and 0≤s≤t1. Thus, for each 0≤t1<t2≤1 and t2-t1<min{δ0,ϵα-3,ϵβ-3}, we have |un′′(t2)-un′′(t1)|≤(3ϵ/Γ(α-2))(Λ1+V1∥k1∥1) and |vn′′(t2)-vn′′(t1)|≤(3ϵ/Γ(β-2))(Λ2+V2∥k2∥1). Therefore, {(un,vn)′′}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)≥M1tα-1, v(t)≥M2tβ-1, D0+pu(t)≥((δ1-2)M1/Γ(α-p))tα-p, D0+qv(t)≥((δ2-2)M2/Γ(β-q))tβ-q, D0+νiu(t)≥((δ1-2)M1/Γ(α-νi))tα-νi, and D0+μiv(t)≥((δ2-2)M2/Γ(β-μi))tβ-μi for i=1,2,…,m and t∈[0,1] and (ϕju)(t)≥M1∫0t‍γj(t,s)sα-1ds and
(51)(ψjv)(t)≥M2∫0t‍λj(t,s)sβ-1ds
for j=1,2 and t∈[0,1].Proof.
By using Theorem4, for each natural number n, the system (15), (2), and (3) has a solution (un,vn) in P. Also by using Lemma 5, the set {(un,vn)}n≥1 is a relatively compact subset of X×X. By using the Arzela-Ascoli theorem, without loss of generality we can assume that {(un,vn)}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 limn→∞D0+pun=D0+pu, limn→∞D0+qvn=D0+qv, limn→∞D0+νiun=D0+νiu, and limn→∞D0+μivn=D0+μiv for i=1,2,…,m and limn→∞ϕjun=ϕju and limn→∞ψjvn=ψjv for j=1,2. Thus, it is easy to see that (u,v) satisfy the desired conditions. Also,
(52)limn→∞fn1(D0+νmt,un(t),vn(t),(ϕ1un)(t),(ψ1vn)(t),D0+pun(t),D0+μ1vn(t),D0+μ2vn(t),…,D0+μmvn(t))=f1(D0+νmt,u(t),v(t),(ϕ1u)(t),(ψ1v)(t),D0+pu(t),D0+μ1v(t),D0+μ2v(t),…,D0+μmv(t)),limn→∞fn2(D0+νmt,un(t),vn(t),(ϕ2un)(t),(ψ2vn)(t),D0+qvn(t),D0+ν1un(t),D0+ν2un(t),…,D0+νmun(t))=f2(D0+νmt,u(t),v(t),(ϕ2u)(t),(ψ2v)(t),D0+qv(t),D0+ν1u(t),D0+ν2u(t),…,D0+νmu(t))
for almost all t∈[0,1]. If ξ=sup{∥(un,vn)∥**:n≥1}, then ∥D0+pun∥≤ξ/Γ(3-p), ∥D0+qvn∥≤ξ/Γ(3-q), ∥D0+νiun∥≤ξ/Γ(3-νi), and ∥D0+μivn∥≤ξ/Γ(3-μi) for i=1,2,…,m and n≥1∥ϕjun∥≤ξγ0j and ∥ψjvn∥≤ξλ0j for j=1,2 and n≥1. Thus,
(53)0≤Gα(t,s)fn1(D0+μ2s,un(s),vn(s),(ϕ1un)(s),(ψ1vn)(s),D0+pun(s),D0+μ1vn(s),D0+μ2vn(s),…,D0+μmvn(s))≤1Γ(α)(1+ξΓ(3-μm))k1(s)Φ1(s)+h1(1+ξΓ(3-μm))1+ξ,1+ξ,1+γ01ξ,1+λ01ξ,1+ξΓ(3-p),1+ξΓ(3-μ1),1+ξΓ(3-μ2),…,1+ξΓ(3-μm))k1(s)),0≤Gβ(t,s)fn2(D0+νms,un(s),vn(s),(ϕ2un)(s),(ψ2vn)(s),D0+qvn(s),D0+ν1un(s),D0+ν2un(s),…,D0+νmun(s))≤1Γ(β)(ξΓ(3-νm)Φ2(s)+h2(ξΓ(3-νm)1+ξ,1+ξ,1+γ02ξ,1+λ02ξ,1+ξΓ(3-q),1+ξΓ(3-ν1),1+ξΓ(3-ν2),…,1+ξΓ(3-νm))k2(s))
for almost all (t,s)∈[0,1]×[0,1] and n≥1. Hence, by using the Lebesgue dominated convergence theorem we get
(54)u(t)=∫01Gα(t,s)f1(D0+μms,u(s),v(s),(ϕ1u)(s),(ψ1v)(s),D0+pu(s),D0+μ1v(s),D0+μ2v(s),…,D0+μmv(s))ds,v(t)=∫01‍Gβ(t,s)f2(D0+νmu(s)s,u(s),v(s),(ϕ2u)(s),(ψ2v)(s),D0+qv(s),D0+ν1u(s),D0+ν2u(s),…,D0+νmu(s))ds
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.
Letm1>0, m2>0, n≥4, n-1<α,β<n, 0<p,q<1, 1<μi,νi<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)), ai′∈(0,1/(β-μi)), and ci′∈(0,1/(α-νi)) for i=1,2,…,m and also b1,b2,b3,b4,b5,b1′,b2′,…,bm′,d1,d2,d3,d4,d5,d1′,d2′,…, dm′∈(0,1). Assume that a3, a4, c3 and c4 are positive real numbers such that ∫01‍(∫0s‍γ1(s,τ)τα-1dτ)-a3ds<∞, ∫01‍(∫0s‍γ2(s,τ)τα-1dτ)-c3ds<∞, ∫01‍(∫0s‍λ1(s,τ)τβ-1dτ)-a4ds<∞, and ∫01‍(∫0s‍λ2(s,τ)τβ-1dτ)-c4ds<∞. Define the functions f1 and f2 on [0,1]×𝔇 by
(55)f1(t,x,y,z,w,v,u1,u2,…,um)=1xa1+1ya2+1za3+1wa4+1va5+1u1a1′+1u2a2′+⋯+1umam′+|ρ1(t)|(xb1+yb2+zb3+wb4+vb5+u1b1′+u2b2′+⋯+umbm′)+σ1(t),f2(t,x,y,z,w,v,u1,u2,…,um)=1xc1+1yc2+1zc3+1wc4+1vc5+1u1c1′+1u2c2′+⋯+1umcm′+|ρ2(t)|(umdm′xd1+yd2+zd3+wd4+vd5+u1d1′+u2d2′+⋯+umdm′)+σ2(t).
Note that the functions f1 and f2 satisfy the conditions (H1) and (H2), where
(56)P1(x,y,z,w,v,u1,u2,…,um)=1xa1+1ya2+1za3+1wa4+1va5+1u1a1′+1u2a2′+⋯+1umam′,P2(x,y,z,w,v,u1,u2,…,um)=1xc1+1yc2+1zc3+1wc4+1vc5+1u1c1′+1u2c2′+⋯+1umcm′,h1(x,y,z,w,v,u1,u2,…,um)=1+xb1+yb2+zb3+wb4+vb5+u1b1′+u2b2′+⋯+umbm′,h2(x,y,z,w,v,u1,u2,…,um)=1+xd1+yd2+zd3+wd4+vd5+u1d1′+u2d2′+⋯+umdm′,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.



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