# Uniqueness and Existence of Positive Solutions for Singular Differential Systems with Coupled Integral Boundary Value Problems

Authors: Yujun Cui; Lishan Liu; Xingqiu Zhang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper provides sufficient conditions for the existence and uniqueness of positive solutions to a singular differential system with integral boundary value. The emphasis here is that the boundary conditions are coupled and this is where the main novelty of this work lies. By mixed monotone method, the existence and uniqueness results of the problem are established. An example is given to demonstrate the main results.


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## Body

## 1. Introduction

In recent years, differential system has been studied extensively in the literature (see, for instance, [1–17] and their references). Most of the results told us that the equations had at least single and multiple positive solutions. In papers [6], the authors obtained some of the newest results for differential system with four-point coupled boundary conditions. But there is no result on the uniqueness of solution in them.In this paper, we discuss the existence and uniqueness of the positive solutions for a new class of boundary value problems of singular differential system. Precisely, we consider the following problem:(1)-x′′(t)=f(t,x(t),y(t)),t∈(0,1),-y′′(t)=g(t,x(t),y(t)),t∈(0,1),x(0)=∫01y(t)dα(t),y(0)=∫01x(t)dβ(t),x(1)=y(1)=0,
where α and β are right continuous on [0,1), left continuous at t=1, and nondecreasing on [0,1], with α(0)=β(0)=0; ∫01‍u(s)dα(s) and ∫01‍u(s)dβ(s) denote the Riemann-Stieltjes integrals of u with respect to α and β, respectively; f∈C((0,1)×[0,+∞)×(0,+∞),[0,+∞)), g∈C((0,1)×(0,+∞)×[0,+∞),[0,+∞)); that is, f(t,x,y) may be singular at t=0, t=1, and y=0 and g(t,x,y) may be singular at t=0, t=1, and x=0. By a positive solution of the system (1), we mean that (x,y)∈(C[0,1]∩C2(0,1))×(C[0,1]∩C2(0,1)), (x,y) satisfies (1), and x>0 and y>0 on [0,1).


## 2. Preliminaries

For eachu∈E:=C[0,1], we write ∥u∥=max{|u(t)|:t∈[0,1]}. Clearly, (E,∥·∥) is a Banach space. Similarly, for each (x,y)∈E×E, we write ∥(x,y)∥1=max{∥x∥,∥y∥}. Clearly, (E×E,∥·∥1) is a Banach space.Throughout this paper, we shall use the following notation:(2)k(t,s)={t(1-s),0≤t≤s≤1,s(1-t),0≤s≤t≤1.
It is well known that k(t,s) is the Green function of the following second order boundary value problem:
(3)-x′′(t)=0,0<t<1,x(0)=x(1)=0,
and k(t,s) is nonnegative continuous function. It is easy to verify that for t,s∈[0,1]×[0,1],
(4)k(t,t)k(s,s)=t(1-t)s(1-s)≤k(t,s)≤t(1-t)(ors(1-s)).We first list the following assumptions for convenience.(H1)
f∈C((0,1)×[0,+∞)×(0,+∞),[0,+∞)), f(t,x,y) is nondecreasing in x and nonincreasing in y, and there exist λ1,  μ1∈[0,1) such that
(5)cλ1f(t,x,y)≤f(t,cx,y),∀x,y>0,c∈(0,1),(6)f(t,x,cy)≤c-μ1f(t,x,y),∀x,y>0,c∈(0,1);g∈C((0,1)×(0,+∞)×[0,+∞),[0,+∞)), g(t,x,y) is nonincreasing in x and nondecreasing in y, and there exist λ2, μ2∈[0,1) such that
(7)cλ2g(t,x,y)≤g(t,x,cy),∀x,y>0,c∈(0,1),(8)g(t,cx,y)≤c-μ2g(t,x,y),∀x,y>0,c∈(0,1).(H2)
0<∫01‍f(t,1,1-t)dt<+∞,  0<∫01‍g(t,1-t,1)dt<+∞.(H3)
κ1>0,  κ2>0,  κ>0, where
(9)κ1=∫01(1-t)dα(t),κ2=∫01(1-t)dβ(t),κ=1-κ1κ2.Remark 1.
By (H1) and (H2), we can get
(10)0<∫01f(t,1-t,1)dt<+∞,0<∫01g(t,1,1-t)dt<+∞.Remark 2.
(i) (5) and (6) imply that
(11)f(t,cx,y)≤cλ1f(t,x,y),∀x,y>0,c>1,(12)f(t,x,y)≤cμ1f(t,x,cy),∀x,y>0,c>1.
Conversely, (11) implies (5) and (12) implies (6).
(ii) (7) and (8) implies that
(13)g(t,cx,y)≤cλ2g(t,x,y),∀x,y>0,c>1,(14)g(t,x,y)≤cμ2g(t,x,cy),∀x,y>0,c>1.
Conversely, (13) implies (7) and (14) implies (8).Lemma 3.
Assume that (H3) holds. Let u,v∈L[0,1]∩C(0,1); then the system of BVPs
(15)-x′′(t)=u(t),-y′′(t)=v(t),t∈(0,1),x(0)=∫01y(t)dα(t),y(0)=∫01x(t)dβ(t),x(1)=y(1)=0has integral representation
(16)x(t)=∫01G1(t,s)u(s)ds+∫01H1(t,s)v(s)ds,y(t)=∫01G2(t,s)v(s)ds+∫01H2(t,s)u(s)ds,
where
(17)G1(t,s)=k1(1-t)κ∫01k(s,τ)dβ(τ)+k(t,s),H1(t,s)=1-tκ∫01k(s,τ)dα(τ),G2(t,s)=k2(1-t)κ∫01k(s,τ)dα(τ)+k(t,s),H2(t,s)=1-tκ∫01k(s,τ)dβ(τ).Proof.
It is easy to see that (15) is equivalent to the system of integral equations
(18)x(t)=x(0)(1-t)+∫01k(t,s)u(s)ds,t∈[0,1],(19)y(t)=y(0)(1-t)+∫01k(t,s)v(s)ds,t∈[0,1].
Integrating (18) and (19) with respect to dβ(t) and dα(t), respectively, on [0,1] gives
(20)∫01‍x(t)dβ(t)=x(0)∫01(1-t)dβ(t)+∬01k(t,s)u(s)dsdβ(t),∫01‍y(t)dα(t)=y(0)∫01(1-t)dα(t)+∬01k(t,s)v(s)dsdα(t).
Therefore,
(21)(-κ211-κ1)(x(0)y(0))=(∬01k(t,s)u(s)dsdβ(t)∬01k(t,s)v(s)dsdα(t)),
and so
(22)(x(0)y(0))=1κ(-κ111-κ2)(∬01k(t,s)u(s)dsdβ(t)∬01k(t,s)v(s)dsdα(t)).
Substituting (22) into (18) and (19), we have
(23)x(t)=κ1(1-t)κ∬01k(t,s)u(s)dsdβ(t)+1-tκ∬01k(t,s)v(s)dsdα(t)+∫01k(t,s)u(s)ds,y(t)=1-tκ∬01k(t,s)u(s)dsdβ(t)+κ2(1-t)κ∬01k(t,s)v(s)dsdα(t)+∫01k(t,s)v(s)ds,
which is equivalent to the system (16).Remark 4.
From (4) and (H3), for t∈[0,1], we have
(24)Gi(t,s)≤ρs(1-s)(orρ(1-t)),Hi(t,s)≤ρs(1-s)(orρ(1-t)),i=1,2,Gi(t,s)≥ν(1-t)s(1-s),Hi(t,s)≥ν(1-t)s(1-s),i=1,2,
where
(25)ρ=max{κ1κ∫01dβ(τ)+1,κ2κ∫01dα(τ)+1,1κ∫01dβ(τ),1κ∫01dα(τ)},ν=min{κ1κ∫01τ(1-τ)dβ(τ),κ2κ∫01τ(1-τ)dα(τ),1κ∫01τ(1-τ)dβ(τ),1κ∫01τ(1-τ)dα(τ)}.Denote(26)P={(x,y)∈E×E:x(t)≥γ(1-t)∥(x,y)∥1,y(t)≥γ(1-t)∥(x,y)∥1,t∈[0,1]},
where γ=ν/ρ∈(0,1). It can be easily seen that P is a cone in E×E. For any real constant r>0, define Pr={(x,y)∈P:∥(x,y)∥1<r}.Define an operatorT:P∖{θ}→P by
(27)T(x,y)=(T1(x,y),T2(x,y)),
where operators T1,T2:P∖{θ}→Q={u∈E∣u(t)≥0,t∈[0,1]} are defined by
(28)T1(x,y)(t)=∫01G1(t,s)f(s,x(s),y(s))ds+∫01H1(t,s)g(s,x(s),y(s))ds,t∈[0,1],T2(x,y)(t)=∫01G2(t,s)g(s,x(s),y(s))ds+∫01H2(t,s)f(s,x(s),y(s))ds,t∈[0,1].
Now we claim that T(x,y) is well defined for (x,y)∈P∖{θ}. In fact, since (x,y)∈P∖{θ}, we can see that
(29)γ(1-t)∥(x,y)∥1≤x(t),y(t)≤∥(x,y)∥1,t∈[0,1].Letc be a positive number such that ∥(x,y)∥1/c<1 and c>1. From (H1) and Remark 2, we have
(30)f(t,x(t),y(t))≤f(t,c,γ∥(x,y)∥1(1-t))≤cλ1f(t,1,γ∥(x,y)∥1c(1-t))≤cλ1+μ1(γ∥(x,y)∥1)-μ1f(t,1,1-t),g(t,x(t),y(t))≤cλ2+μ2(γ∥(x,y)∥1)-μ2f(t,1,1-t).
Hence, for any t∈[0,1], by Remark 4 and equation (30), we get
(31)Ti(x,y)(t)≤ρ∫01f(s,x(s),y(s))ds+ρ∫01g(s,x(s),y(s))ds≤ρcλ1+μ1(γ∥(x,y)∥1)-μ1∫01f(s,1,1-s)ds+ρcλ2+μ2(γ∥(x,y)∥1)-μ2∫01g(s,1-s,1)ds<+∞,i=1,2.
Thus, T is well defined on P∖{θ}.Lemma 5.
Assume that (H1), (H2), and (H3) hold. Then, for any 0<a<b<+∞, T:(Pb¯∖Pa)→P is a completely continuous operator.Proof.
Firstly, we show thatT(Pb¯∖Pa)⊂P. By Remark 4, for τ,t,s∈[0,1], we obtain
(32)Gi(t,s)≥γ(1-t)Gi(τ,s),Hi(t,s)≥γ(1-t)Hi(τ,s),i=1,2,H1(t,s)≥γ(1-t)G2(τ,s),G1(t,s)≥γ(1-t)H2(τ,s),H2(t,s)≥γ(1-t)G1(τ,s),G2(t,s)≥γ(1-t)H1(τ,s).
Hence, for (x,y)∈Pb¯∖Pa, t∈[0,1], we have
(33)T1(x,y)(t)=∫01G1(t,s)f(s,x(s),y(s))ds+∫01H1(t,s)g(s,x(s),y(s))ds≥γ(1-t)∫01G1(τ,s)f(s,x(s),y(s))ds+γ(1-t)∫01H1(τ,s)g(s,x(s),y(s))ds=γ(1-t)T1(x,y)(τ),∀τ∈[0,1],T1(x,y)(t)=∫01G1(t,s)f(s,x(s),y(s))ds+∫01H1(t,s)g(s,x(s),y(s))ds≥γ(1-t)∫01H2(τ,s)f(s,x(s),y(s))ds+γ(1-t)∫01G2(τ,s)g(s,x(s),y(s))ds=γ(1-t)T2(x,y)(τ),∀τ∈[0,1].
Then T1(x,y)(t)≥γ(1-t)∥T1(x,y)∥ and T1(x,y)(t)≥γ(1-t)∥T2(x,y)∥; that is, T1(x,y)(t)≥γ(1-t)∥(T1(x,y),T2(x,y))∥1. In the same way, we can prove that T2(x,y)(t)≥γ(1-t)∥(T1(x,y),T2(x,y))∥1. Therefore, T(Pb¯∖Pa)⊂P.
Next, we prove thatT is a compact operator. That is, for any bounded subset V⊂Pb¯∖Pa, we show that T(V) is relatively compact in E×E. Since V⊂Pb¯∖Pa is a bounded subset, there exists a constant c>1 such that ∥(x,y)∥1=max{∥x∥,∥y∥}≤c for all (x,y)∈V. Notice that, for any (x,y)∈V, we have
(34)∥T(x,y)∥1=max{∥T1(x,y)∥,∥T2(x,y)∥}
and from (H1), (H2), Remarks 2 and 4, (16), and (18), we obtain
(35)Ti(x,y)(t)≤ρ∫01f(s,x(s),y(s))ds+ρ∫01g(s,x(s),y(s))ds≤ρcλ1+μ1(γa)-μ1∫01f(s,1,1-s)ds+ρcλ2+μ2(γa)-μ2∫01g(s,1-s,1)ds<+∞,i=1,2.
Therefore, T(V) is uniformly bounded.
In the following, we shall show thatT(V) is equicontinuous on [0,1].
For(x,y)∈V, t∈[0,1], using Lemma 3, we have
(36)T1(x,y)(t)=∫01G1(t,s)f(s,x(s),y(s))ds+∫01H1(t,s)g(s,x(s),y(s))ds=κ1(1-t)κ∫01(∫01k(s,τ)dβ(τ))f(s,x(s),y(s))ds+∫0ts(1-t)f(s,x(s),y(s))ds+∫t1t(1-s)f(s,x(s),y(s))ds+1-tκ∫01(∫01k(s,τ)dα(τ))g(s,x(s),y(s))ds.
Differentiating with respect to t and combining (H1) and (H2), we obtain
(37)|T1(x,y)′(t)|=|-κ1κ∫01(∫01k(s,τ)dβ(τ))f(s,x(s),y(s))ds-∫0tsf(s,x(s),y(s))ds+∫t1(1-s)f(s,x(s),y(s))ds-1κ∫01(∫01k(s,τ)dα(τ))g(s,x(s),y(s))ds|≤κ1κ∫01(∫01k(s,τ)dβ(τ))f(s,x(s),y(s))ds+∫0tsf(s,x(s),y(s))ds+∫t1(1-s)f(s,x(s),y(s))ds+1κ∫01(∫01k(s,τ)dα(τ))g(s,x(s),y(s))ds≤cλ1+μ1(γa)-μ1(ρκ1κ∫01f(s,1,1-s)ds+∫0tsf(s,1,1-s)ds+∫t1(1-s)f(s,1,1-s)ds)+ρκcλ2+μ2(γa)-μ2∫01g(s,1-s,1)ds=:K(t).
Exchanging the integral order, we have
(38)∫01K(t)dt=cλ1+μ1(γa)-μ1(ρκ1κ∫01f(s,1,1-s)ds+2∫01s(1-s)f(s,1,1-s)ds)+ρκcλ2+μ2(γa)-μ2∫01g(s,1-s,s)ds<+∞.
From the absolute continuity of the integral, we know that T1(V) is equicontinuous on [0,1]. Thus, according to the Ascoli-Arzela theorem, T1(V) is a relatively compact set. In the same way, we can prove that T2(V) is relatively compact. Therefore, T(V) is relatively compact.
Finally, it remains to show thatT is continuous. We need to prove only that T1,T2:Pb¯∖Pa→Q are continuous. Suppose that (xm,ym),(x0,y0)∈Pb¯∖Pa, and ∥(xm,ym)-(x0,y0)∥1→0(m→∞). Let L=sup{∥(xm,ym)∥1,m=0,1,2,…}. Then we may still choose positive constants c such that L/c<1 and c>1. From (H1) and Remark 2, we get
(39)f(t,xm(t),ym(t))≤cλ1+μ1(γa)-μ1f(t,1,1-t),m=0,1,2…,g(t,xm(t),ym(t))≤cλ2+μ2(γa)-μ2g(t,1-t,1),m=0,1,2…,|T1(xm,ym)(t)-T1(x0,y0)(t)|≤ρ∫01|f(s,xm(s),ym(s))-f(s,x0(s),y0(s))|ds+ρ∫01|g(s,xm(s),ym(s))-g(s,x0(s),y0(s))|ds.
For anyϵ>0, by (H2), there exists a positive number δ∈(0,1/2) such that
(40)∫[0,δ]∪[1-δ,1]ρcλ1+μ1(γa)-μ1f(s,1,1-s)ds<ɛ4,∫[0,δ]∪[1-δ,1]‍ρcλ2+μ2(γa)-μ2g(s,1-s,1)ds<ɛ4.
On the other hand, for (x,y)∈Pb¯∖Pa and t∈[δ,1-δ], we have
(41)aγδ≤x(t),y(t)≤b.
Since f(t,x,y) and g(t,x,y) are uniformly continuous in [δ,1-δ]×[aγδ,b]×[aγδ,b], we have
(42)limm→+∞|f(t,xm(t),ym(t))-f(t,x0(t),y0(t))|=limm→+∞|g(t,xm(t),ym(t))-g(t,x0(t),y0(t))|=0
holds uniformly on t∈[δ,1-δ]. Then the Lebesgue dominated convergence theorem yields that
(43)∫δ1-δ|f(s,xm(s),ym(s))-f(s,x0(s),y0(s))|ds→0∫δ1-δ|g(s,xm(s),ym(s))-g(s,x0(s),y0(s))|ds→0,m→+∞.
Thus, for above ɛ>0, there exists a natural number N, for m>N; we have
(44)∫δ1-δρ|f(s,xm(s),ym(s))-f(s,x0(s),y0(s))|ds<ɛ4,∫δ1-δρ|g(s,xm(s),ym(s))-g(s,x0(s),y0(s))|ds<ɛ4.
It follows from (39)–(44) that when m>N(45)∥T1(xm,ym)-T1(x0,y0)∥≤ρ∫01|f(s,xm(s),ym(s))-f(s,x0(s),y0(s))|ds+ρ∫01|g(s,xm(s),ym(s))-g(s,x0(s),y0(s))|ds≤∫[0,δ]∪[1-δ,1]ρcλ1+μ1(γa)-μ1f(s,1,1-s)ds+∫[0,δ]∪[1-δ,1]ρcλ2+μ2(γa)-μ2g(s,1-s,1)ds+∫δ1-δρ|f(s,xm(s),ym(s))-g(s,x0(s),y0(s))|ds+∫δ1-δρ|g(s,xm(s),ym(s))-g(s,x0(s),y0(s))|ds<ɛ.
This implies that T1:Pb¯∖Pa→Q is continuous. Similarly, we can prove that T2:Pb¯∖Pa→Q is continuous. So, T:Pb¯∖Pa→P is continuous. Summing up, T:Pb¯∖Pa→P is completely continuous.Our main tool of this paper is the following cone compression and expansion fixed point theorem.Lemma 6 (see [18]).
LetE be a Banach space and P a cone in E. Suppose that Ω1 and Ω2 are two bounded open subsets of E with θ∈Ω1, Ω1¯⊂Ω2. If  T:P∩(Ω2¯∖Ω1)→P is a completely continuous operator satisfying
(46)∥Tx∥≥∥x∥,forx∈P∩∂Ω1,∥Tx∥≤∥x∥,forx∈P∩∂Ω2,
then T has a fixed point in P∩(Ω2¯∖Ω1).


## 3. Main Results

In this section, we present our main results.Theorem 7.
Suppose that conditions(H1), (H2), and (H3) hold. Then, if λ1+μ1<1 and λ2+μ2<1, the differential system (1) has a unique positive solution (x*,y*).Proof.
We divide the rather long proof into three steps.(i) The differential system (1) has at least one positive solution (x*,y*).
Chooser, R such that
(47)0<r≤min{×f(s,1-s,1)dsν4γλ1∫01‍)1/1-λ1(ν4γλ1∫01s(1-s)×f(s,1-s,1)dsν4γλ1∫01‍)1/(1-λ1),12},R≥max{+ρ∫01‍g(s,1-s,1)ds)1/1-max{λ1,λ2}1γ(ρ∫01f(s,1,1-s)ds+ρ∫01‍g(s,1-s,1)ds)1/(1-max{λ1,λ2}),1γ,2}.
Clearly0<r<1<R. By Lemma 5, T:PR¯∖Pr→P is completely continuous.
ExtendT (denote T yet) to T:PR¯→P which is completely continuous.
Then, for(x,y)∈∂Pr, we have
(48)rγ(1-t)≤x(t),y(t)≤r,t∈[0,1].
By Remarks 1 and 2, (H1), and (H2), we get
(49)Ti(x,y)(t)≥ν4∫01s(1-s)f(s,γr(1-s),r)ds≥ν4∫01s(1-s)f(s,γr(1-s),1)ds≥ν4γλ1rλ1∫01s(1-s)f(s,1-s,1)ds≥r=∥(x,y)∥1,i=1,2,t∈[0,34].
This guarantees that
(50)∥T(x,y)∥1≥∥(x,y)∥1,∀(x,y)∈∂Pr.
On the other hand, for (x,y)∈∂PR, we have
(51)Rγ(1-t)≤x(t),y(t)≤R,t∈[0,1].
Therefore,
(52)Ti(x,y)(t)≤ρ∫01f(s,R,γR(1-s))ds+ρ∫01g(s,γR(1-s),R)ds≤ρ∫01f(s,R,1-s)ds+ρ∫01g(s,1-s,R)ds≤ρRλ1∫01f(s,1,1-s)ds+ρRλ2∫01g(s,1-s,1)ds≤ρRmax{λ1,λ2}(∫01f(s,1,1-s)ds+∫01g(s,1-s,1)ds)≤R=∥(x,y)∥1,i=1,2,t∈[0,1].This guarantees that
(53)∥T(x,y)∥1≤∥(x,y)∥1,∀(x,y)∈∂PR.
By the complete continuity of T, (50) and (53), and Lemma 6, we obtain that T has a fixed point (x*,y*) in PR¯∖Pr. Consequently, (1) has a positive solution (x*,y*) in PR¯∖Pr.(ii) Suppose that(x,y) is a positive solution of the differential system (1).
Then there exist real numbers0<m<1 such that
(54)m(1-t)≤x(t)≤1m(1-t),m(1-t)≤y(t)≤1m(1-t),∀t∈[0,1].
From Lemma 5, we know that (x,y)∈P∖{θ}. So, we have
(55)γ∥(x,y)∥1(1-t)≤x(t),y(t)≤∥(x,y)∥1.
Let c be a constant such that ∥(x,y)∥1/c<1 and c>1/γ. By Lemma 3, we get
(56)x(t)≤ρ(1-t)∫01f(s,c,γ∥(x,y)∥1c(1-s))ds+ρ(1-t)∫01g(s,γ∥(x,y)∥1c(1-s),c)ds≤cλ1+μ1(γ∥(x,y)∥1)-μ1ρ(1-t)∫01f(s,1,1-s)ds+cλ2+μ2(γ∥(x,y)∥1)-μ2ρ(1-t)∫01g(s,1-s,1)ds=:C(1-t),t∈[0,1].
In the same way, we can prove that y(t)≤C(1-t), t∈[0,1]. Then we may pick out m such that m=min{γ∥(x,y)∥1,1/C,1/2}, which implies that (54) holds.(iii) The differential system (1) has a unique positive solution (x*,y*).
Assuming the contrary, we find that the differential system (1) has a positive solution (x*,y*) different from (x*,y*). By (54), there exist δ1,δ2>0, such that
(57)δ1(1-t)≤x*(t),y*(t)≤1δ1(1-t),∀t∈[0,1],δ2(1-t)≤x*(t),y*(t)≤1δ2(1-t),∀t∈[0,1].
Hence, we have
(58)δ1δ2x*(t)≤x*(t)≤1δ1δ2x*(t),δ1δ2y*(t)≤y*(t)≤1δ1δ2y*(t),∀t∈[0,1].
Clearly, δ1δ2≠1. Put
(59)δ*=sup{δ∣δx*(t)≤x*(t)≤1δx*(t),δy*(t)≤y*(t)≤1δy*(t),∀t∈[0,1]}.
It is easy to see that 1>δ*≥δ1δ2>0, and
(60)δ*x*(t)≤x*(t)≤1δ*x*(t),δ*y*(t)≤y*(t)≤1δ*y*(t),∀t∈[0,1].
So, by (H1), we have
(61)f(t,x*(t),y*(t))≥f(t,δ*x*(t),1δ*y*(t))≥δ*λ1+μ1f(t,x*(t),y*(t))≥δ*σf(t,x*(t),y*(t)),g(t,x*(t),y*(t))≥δ*λ2+μ2g(t,x*(t),y*(t))≥δ*σg(t,x*(t),y*(t)),
where σ=max{λ1+μ1,λ2+μ2} such that σ<1. Therefore, we have
(62)x*(t)=T1(x*,y*)(t)=∫01G1(t,s)f(s,x*(s),y*(s))ds+∫01H1(t,s)g(s,x*(s),y*(s))ds≥δ*σ∫01G1(t,s)f(s,x*(s),y*(s))ds+∫01H1(t,s)g(s,x*(s),y*(s))ds=δ*σT1(x*,y*)(t)=δ*σx*(t).
Similarly, we can get
(63)y*(t)≥δ*σy*(t),x*(t)≥δ*σx*(t),y*(t)≥δ*σy*(t).
Noticing that 0<δ*, σ<1, we get to a contradiction with the maximality of δ*. Thus, the differential system (1) has a unique positive solution (x*,y*). This completes the proof of Theorem 7.


## 4. An Example

In this section, we give an example to illustrate the usefulness of our main results. Let us consider the singular differential system with couple boundary value problem(64)-x′′=xyt(1-t)3,-y′′=y3x,x(1)=y(1)=0,x(0)=y(13)+y(12),y(0)=∫01x(t)dt2.Let(65)f(t,x,y)=xyt(1-t)3,g(t,x,y)=y3x,α(t)={0,t∈[0,13),1,t∈[13,12),2,t∈[12,1],β(t)=t2,λ1=μ2=12,λ2=μ1=13;
then
(66)κ1=76,κ2=13,κ=1-κ1κ2=1118,∫01‍f(s,1,1-s)ds=B(23,16),∫01‍g(s,1-s,1)ds=B(1,12).
So all conditions of Theorem 7 are satisfied for (64), and our conclusion follows from Theorem 7.



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*Source: 340487-2013-11-20.xml*