# New Results for Multipoint Singular Boundary Value Problems on a Measure Chain

Authors: Yan Sun; Yongping Sun; Patricia J. Y. Wong
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We study the existence and uniqueness of positive solutions for a class of singularm-point boundary value problems of second order differential equations on a measure chain. A sharper sufficient condition for the existence and uniqueness of Crd⁡1[0,T] positive solutions as well as Crd⁡1[0,T] positive solutions is obtained by the technique of lower and upper solutions and the maximal principle theorem.


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## Body

## 1. Introduction

In this paper, we present the existence of positive solutions for the following second order singularm-point boundary value problem on a measure chain:
(1)xΔΔ(t)+a(t)f(t,x(t))=0,0<t<T,(2)x(T)=0,x(0)=∑i=1m-2‍αix(ηi),
where 0<αi<T, i=1,…,m-2, 0<η1<η2<⋯<ηm-2<T are constants and 0<∑i=1m-2αi<T, m≥3. We assume that 0<T<+∞ belong to T. We use (0,T) to denote (0,T)∩T, and similar notations are used for other intervals. The function a:(0,T)→[0,+∞) is rd-continuous, f:(0,T)×(0,+∞)→[0,+∞) is continuous, and f(t,x) may be singular at t=0 and t=T, x=0. Observe that when T=R or T=Z, the problem (1)-(2) reduces to boundary value problems of ordinary differential equations or difference equations.The existence of positive solutions for boundary value problem on a measure chain has been paid more attentions by many researchers. Related problems on measure chains can be found in [1–9].Recently, by making use of the Krasnosel’skii fixed point theorem, Goodrich [1] studied the existence of at least one positive solution for the following boundary value problem:
(3)uΔΔ(t)+λf(t,uσ(t))=0,t∈(a,σ2(b)),u(a)=ϕ(u),αu(σ2(b))=0,
where the nonlocal boundary condition ϕ:Crd⁡([a,σ2(b)]T,R)→[0,+∞) is continuous.Sun and Li [5] gave the existence results of the following three-point boundary value problem on time scale T:
(4)xΔ∇(t)+a(t)f(t,x(t))=0,t∈(0,T)⊂T,βx(0)-γxΔ(0)=0,αx(η)=x(T),
where β,γ≥0, β+γ>0, η∈(0,ρ(t)), 0<α<T/η, and d=β(T-αη)+γ(1-α)>0.By applying functional-type cone expansion-compression fixed point theorem, P. Wang and Y. Wang [6] established the existence of positive solutions of the following nonlinear boundary value problem given by the dynamic equation on time scales:
(5)xΔ∇(t)+a(t)f(t,x(t))=0,t∈(0,T)⊂T,βx(0)-γxΔ(0)=0,αx(T)-∑i=1m-2‍aix(ξi)=b,m≥3.By employing the Krasnosel’skii fixed point theorem, Hao et al. [2] discussed the existence of positive solutions of the following boundary value problem on a time scale:
(6)uΔΔ(t)+m(t)f(t,uσ(t))=0,t∈(a,b),αu(a)-βuΔ(a)=0,γu(σ(b))+δuΔ(σ(b))=0.Inspired and motivated greatly by the work of [1, 2, 5–9], we establish the existence and uniqueness of positive solution for singular m-point boundary value problem (1)-(2). By constructing lower and upper solutions and using the maximal theorem, we not only obtain a sharper sufficient condition for the existence and uniqueness of Crd⁡[0,T] positive solution, but also prove a sufficient condition for the existence of Crd⁡1[0,T] positive solution. Our technique is different from those of [1–9] and our results naturally complement/improve their work.We state some basic notions connected to time scales, which can be found in [9].Definition 1.
LetT be a time scale. For t∈T, theforward jump operatorσ:T→T is defined by
(7)σ(t)∶=inf⁡{s∈T:s>t},
and one defines thebackward jump operatorρ:T→T by
(8)ρ(t)∶=sup⁡{s∈T:s<t}.Definition 2.
One says thatt isright-scattered if σ(t)>t, and one says that t isleft-scattered if ρ(t)<t. Points are said to beisolated if they are both right-scattered and left-scattered.Definition 3.
One says thatt isright-dense if t<sup⁡T and σ(t)=t, and one says that t isleft-dense if t>inf⁡T and ρ(t)=t. Points are said to bedense if they are both right-dense and left-dense. One defines thegraininess functionμ:T→[0,+∞) by
(9)μ(t)∶=σ(t)-t.For convenience, one lists the following conditions which will be referred to later:(H1)
0<αi<T, i=1,…,m-2, 0<η1<η2<⋯<ηm-2<σ(T) are constants satisfying 0<∑i=1m-2αi<T, m>3, and 0<∑i=1m-2αiηi<T;(H2)
a:(0,T)→[0,+∞) is rd-continuous, and there exists t0∈[ηm-2,T) such that a(t0)>0;(H3)
f:(0,T)×(0,+∞)→[0,+∞) is continuous, and f(t,x) is nonincreasing with respect to x, for all t∈(0,T), a(t)f(t,·)≢0, and 0<∫0T‍a(t)f(t,λt(T-t))dt<+∞, for all λ>0.


## 2. Preliminaries and Lemmas

Definition 4.
A functionx(t)∈Crd⁡[0,T]∩Crd⁡2(0,T) is said to be a Crd⁡[0,T]positive solution of the problem (1)-(2) if x(t) satisfies the problem (1)-(2) and x(t)>0, t∈(0,T); a Crd⁡[0,T] positive solution x(t) of the problem (1)-(2) is said to be a Crd⁡1[0,T]positive solution if xΔ(0+) and xΔ(T-) exist, and x(t)>0, t∈[0,T].Definition 5.
One says that a functionφ(t)∈Crd⁡2(0,T) is alower solution of the problem (1)-(2) on [0,T], if φ(t)∈Crd⁡[0,T]∩Crd⁡2[0,T] and satisfies
(10)φΔΔ(t)+a(t)f(t,φ(t))≥0,t∈(0,T),φ(T)≤0,φ(0)-∑i=1m-2‍αiφ(ηi)≤0.
Similarly, ψ(t)∈Crd⁡2[0,T] is said to be anupper solution of the problem (1)-(2) on [0,T], if ψ(t)∈Crd⁡[0,T]∩Crd⁡2(0,T) and satisfies
(11)ψΔΔ(t)+a(t)f(t,ψ(t))≤0,t∈(0,T),ψ(T)≥0,ψ(0)-∑i=1m-2‍αiψ(ηi)≥0.
One says (φ(t),ψ(t)) is acouple of lower and upper solutions of the problem (1)-(2), if there exist a lower solution φ(t) and an upper solution ψ(t) of the problem (1)-(2) such that
(12)φ(t)≤ψ(t),t∈[0,T].Lemma 6 (maximal principle).
Suppose that (H1) is satisfied. In addition, assume that 0<η1<η2<⋯<ηm-2<δn<T, n=1,2,…. Let
(13)Qn={x(t)∈Crd⁡[0,δn]∩Crd⁡2(0,δn):∑i=1m-2‍x(δn)≥0,x(0)-∑i=1m-2‍αix(ηi)≥0}.
For x∈Qn such that -xΔΔ(t)≥0 and for t∈(0,T), then x(t)≥0 for t∈[0,δn].Proof.
For allx∈Qn, let(14)-xΔΔ(t)=g(t),t∈(0,δn);(15)x(0)-∑i=1m-2‍αix(ηi)=r1,x(δn)=r2,
then r1≥0, r2≥0, and g(t)≥0, t∈(0,δn).
Integrating (14) from 0 to t, we obtain
(16)-xΔ(t)+xΔ(0)=∫0t‍g(s)Δs.
Again integrating (16) from 0 to t and exchanging integral sequence, we get
(17)-x(t)+x(0)+xΔ(0)t=∫0t‍(t-s)g(s)Δs,t∈(0,δn)⊂(0,T)⊂T.
From (17) and boundary condition (15), we obtain
(18)x(t)=1A0(((T-∑i=1m-2‍αi)t+∑i=1m-2‍αiηi)r2+(δn-t)r1)+δn-tA0∑i=1m-2‍(αi∫0δn‍Gn(ηi,s)g(s)Δs)+∫0δn‍Gn(t,s)g(s)Δs,t∈(0,δn),
where A0=δn(T-∑i=1m-2αi)+∑i=1m-2αiηi>0 and
(19)Gn(t,s)=1δn{s(δn-t),0≤s<t<δn<T,t(δn-s),0≤t<s<δn<T.
Consequently, from (18) and the definition of Gn(t,s), we see that x(t)≥0, t∈[0,δn].Let(20)E={x:[0,T]⟶R∣x(t)isanonnegativecontinuous∑hhhhhhhhhhhhhhhfunction,xΔ(t)iscontinuoushhhhhhhhhhhhhhhon⁡[0,T],andxΔΔ(t)ishhhhhhhhhhhhhhh∑right-densecontinuouson⁡[0,T)}.
We will use the Banach space (E,∥·∥) equipped with the norm∥x∥=max⁡t∈[0,T]⁡|x(t)|. Let
(21)P={x∈E:thereexistsarealnumbermx,suchthatx(t)≥mxk(t),t∈[0,T]},
where k(t)=G(t,t). Then P is a positive cone of E. Define the nonlinear operator B as follows:
(22)Bx(t)=∫0T‍G(t,s)a(s)f(s,x(s))Δs+T-tT(T-∑i=1m-2αi)+∑i=1m-2αiηi×∑i=1m-2‍αi∫0T‍G(ηi,s)a(s)f(s,x(s))Δs,
which for notational simplicity will be written as
(23)Bx(t)=∫0T‍G(t,s)a(s)f(s,x(s))Δs+A(T-t)∫0T‍G(ηi,s)a(s)f(s,x(s))Δs,
where ηi∈(0,T) with 0<η1<⋯<ηm-2<T and
(24)A=∑i=1m-2αiT(T-∑i=1m-2αi)+∑i=1m-2αiηi,G(t,s)=1T{t(T-s),0≤t≤s≤T,s(T-t),0≤s≤t≤T,G(ηi,s)=1T{ηi(T-s),0≤ηi≤s≤T,s(T-ηi),0≤s≤ηi≤T.It is easy to see that(25)k(t)=G(t,t)=1Tt(T-t)≤T4,t∈[0,T],1Tk(t)k(s)≤G(t,s)≤k(t).It is clear that the existence of a positive solution of (1)-(2) is equivalent to the existence of a nontrivial fixed point of B in P.Lemma 7.
Suppose that (H1)–(H3) hold. Then BP⊂P, and B is a decreasing operator.Proof.
It is obvious thatk(t)∈P, so P is not empty: P°≠ϕ. For all x(t)∈P, by the definition of P, there exists a real number mx, such that x(t)≥mxk(t), t∈[0,T]. From (H3), we know that
(26)∫0T‍k(s)a(s)f(s,x(s))Δs≤∫0T‍k(s)a(s)f(s,mxk(s))Δs<+∞.
From the definition ofG(t,s) and (23), we find
(27)Bx(t)=∫0T‍G(t,s)a(s)f(s,x(s))Δs+A(T-t)∫0T‍G(ηi,s)a(s)f(s,x(s))Δs≤∫0T‍k(s)a(s)f(s,x(s))Δs+A(T-t)∫0T‍k(s)a(s)f(s,x(s))Δs=T2-t∑i=1m-2αi+∑i=1m-2αiηiT(T-∑i=1m-2αi)+∑i=1m-2αiηi×∫0T‍k(s)a(s)f(s,x(s))Δs≤A(T2+∑i=1m-2αiηi)∑i=1m-2αi×∫0T‍k(s)a(s)f(s,x(s))Δs<+∞.
Let w=max⁡t∈[0,T]⁡x(t). From (H2), we see ∫0Tk(s)a(s)f(s,w)Δs>0. Since f(t,x) is continuous on (0,T)×(0,+∞), thus
(28)∫0T‍k(s)a(s)f(s,x(s))Δs≥∫0T‍k(s)a(s)f(s,w)Δs>0.
On the other hand, for all x∈P, by making use of (25), we obtain
(29)Bx(t)=∫0T‍G(t,s)a(s)f(s,x(s))Δs+A(T-t)∫0T‍G(ηi,s)a(s)f(s,x(s))Δs≥1Tk(t)∫0T‍k(s)a(s)f(s,x(s))Δs+1TA(T-t)k(ηi)∫0T‍k(s)a(s)f(s,x(s))Δs≥1T(1+∑i=1m-2αiηi(T-ηi)T(T-∑i=1m-2αi)+∑i=1m-2αiηi)×∫0T‍k(s)a(s)f(s,x(s))Δs·k(t)=mBxk(t),0000000000000000000000000000000∀t∈[0,T],
where
(30)mBx=1T(1+∑i=1m-2αiηi(T-ηi)T(T-∑i=1m-2αi)+∑i=1m-2αiηi)×∫0T‍k(s)a(s)f(s,x(s))Δs.
Therefore Bx is well defined on P, and Bx∈P for all x∈P. So BP⊂P.
For allx1,x2∈P with x1<x2, from (H2), we see that
(31)(Bx1)(t)=∫0T‍G(t,s)a(s)f(s,x1(s))Δs+A(T-t)∫0T‍G(ηi,s)a(s)f(s,x1(s))Δs≥∫0T‍G(t,s)a(s)f(s,x2(s))Δs+A(T-t)∫0T‍G(ηi,s)a(s)f(s,x2(s))Δs=(Bx2)(t).
Therefore, (Bx1)(t)≥(Bx2)(t) and hence B is a decreasing operator.Lemma 8.
Suppose that (H1)–(H3) hold. Then for any x∈P, the problem (1)-(2) has an upper solution x¯ and a lower solution x_, and (x_(t),x¯(t)) is a couple of lower and upper solution of the problem (1)-(2).Proof.
For allx∈P, we know that
(32)(Bx)(t)=∫0T‍G(t,s)a(s)f(s,x(s))Δs+A(T-t)∫0T‍G(ηi,s)a(s)f(s,x(s))Δs=1T(T-t)∫0t‍sa(s)f(s,x(s))Δs+tT∫tT‍(T-s)a(s)f(s,x(s))Δs+A(T-t)∫0T‍G(ηi,s)a(s)f(s,x(s))Δs,(Bx)Δ(t)=-1T∫0t‍sa(s)f(s,x(s))Δs+1T∫tT‍(T-s)a(s)f(s,x(s))Δs-A∫0T‍G(ηi,s)a(s)f(s,x(s))Δs,(Bx)ΔΔ(t)=-a(t)f(t,x(t)),t∈(0,T).
By simple computation, we obtain
(33)(Bx)(0)-∑i=1m-2‍αi(Bx)(ηi)=0,(Bx)(T)=0,(34)(Bx)ΔΔ(t)+a(t)f(t,x(t))=0.
Let
(35)x_(t)=min⁡{k(t),(Bk)(t)},x¯(t)=max⁡{k(t),(Bk)(t)}.
Obviously x_(t), x¯(t) are well defined, and
(36)x_(t)≤x¯(t).
Since BP⊂P, then for any k(t)∈P, we see that (Bk)(t)∈P. Thus there exists positive real number mBk, such that (Bk)(t)≥mBkk(t). From (35) and (36), we know that
(37)x¯(t)≥x_(t)=min⁡{k(t),(Bk)(t)}≥min⁡{1,mBk}k(t)=m1k(t),
where m1=min⁡{1,mBk}. Therefore x_,x¯∈P. Thus (Bx_)(t) and (Bx¯)(t) are well defined, and
(38)(Bx¯)(t)≤(Bx_)(t)≤(Bm1k)(t).
Since B is a nonincreasing operator, from (35), we know that
(39)(Bx¯)(t)≤(Bk)(t)≤x¯(t),(Bx_)(t)≥(Bk)(t)≥x_(t).
From (34) and the above discussion, we know that
(40)(Bx¯)ΔΔ(t)+a(t)f(t,Bx¯(t))≥(Bx¯)ΔΔ(t)+a(t)f(t,x¯(t))=0,(Bx_)ΔΔ(t)+a(t)f(t,Bx_(t))≤(Bx_)ΔΔ(t)+a(t)f(t,x_(t))=0.
Equation (33) implies (Bx¯)(t) and (Bx_)(t) satisfy conditions (2). From (38) and (40), we know that
(41)(x_(t),x¯(t))=((Bx¯)(t),(Bx_)(t))
is a couple of lower and upper solution of the problem (1)-(2), and x¯,x_∈P. Therefore
(42)x¯,x_∈Crd⁡[0,T]∩Crd⁡2(0,T).
Consequently
(43)x¯ΔΔ(t)+a(t)f(t,x¯(t))≤0,t∈(0,T).x¯(T)≥0,x¯(0)-∑i=1m-2‍αix¯(ηi)≥0.


## 3. Main Results

Theorem 9.
Suppose that (H1)–(H3) hold. Then the problem (1)-(2) has a unique positive solution x*∈Crd[0,T]∩Crd2(0,T) satisfying x*(t)≥νk(t), where ν is a positive constant.Proof of Theorem9.
We have the following.
(I) Existence of Positive Solution to the Problem (1)-(2). From Lemma 8, we know that the problem (1)-(2) has a couple of lower and upper solution. Let (ϕ_(t),ϕ¯(t)) be a couple of lower and upper solution of the problem (1)-(2). Then for any t∈(0,T), we have that k(t)>0, G(t,s), ϕ_(t), ϕ¯(t) are strictly positive continuous function.
Define auxiliary functionH(t,x) and operator Q as follows:
(44)H(t,x)={f(t,ϕ_(t)),ifx<ϕ_(t),f(t,x),ifϕ_(t)≤x≤ϕ¯(t),f(t,ϕ¯(t)),ifx>ϕ¯(t),(45)(Qx)(t)=∫0T‍G(t,s)a(s)H(s,x(s))Δs+A(T-t)∫0T‍G(ηi,s)a(s)H(s,x(s))Δs,0000000000000000000000000000000∀x∈P.
Obviously, we see that H(t,x):(0,T)×(0,+∞)→[0,+∞) is continuous.
Consider the following second order differential equationm-point singular boundary value problem:
(46)yΔΔ(t)+a(t)H(t,y)=0,t∈(0,T),y(T)=0,y(0)-∑i=1m-2‍αiy(ηi)=0.
It is well known that the existence of a positive solution of problem (46) is equivalent to the existence of a nontrivial fixed point of Q in P.
Now we prove thatQ is a completely continuous operator.
Fromϕ_(t)∈P, there exists a positive real number m1 such that ϕ_(t)≥m1k(t), for all t∈[0,T]. Combining (H2)~(H3) with (44) and (42), we know that
(47)∫01‍k(s)a(s)f(s,ϕ_(s))Δs
is definitely a finite real number. Denote
(48)Γ=∫0T‍a(s)f(s,ϕ_(s))Δs.
Let constant C>0, for all x∈P¯c={x∈P:∥x∥≤c}. Thus, we obtain
(49)(Qx)(t)=∫0T‍G(t,s)a(s)H(s,x(s))Δs+A(T-t)∫0T‍G(ηi,s)a(s)H(s,x(s))Δs≤∫0T‍k(s)a(s)H(s,ϕ_(s))Δs+A(T-t)∫0T‍G(ηi,s)a(s)H(s,ϕ_(s))Δs≤∫0T‍k(s)a(s)H(s,m1k(s))Δs+A(T-t)∫0T‍k(s)a(s)H(s,m1k(s))Δs<+∞.
Consequently, Q(P¯c) is uniformly bounded.
Now we prove thatQ is compact. Let D⊂P be a bounded set. Then there exists G1>0 such that ∥x∥≤G1, for all x∈D. It is easy to prove that Q(D) is a bounded set in P. Since G(t,s) is continuous on [0,T]×[0,T], thus it is uniformly continuous. Therefore, choose 0<δ<2ɛ/AΓT, for all ɛ>0, and t′,t′′,s∈[0,T] such that |t′-t′′|<δ; we have |G(t1,s)-G(t2,s)|<(ɛ/2)Γ-1. Thus, for any x∈D,t′,t′′∈[0,T] such that |t′-t′′|<δ, we obtain
(50)|(Qx)(t′)-(Qx)(t′′)|≤∫0T‍|G(t′,s)-G(t′′,s)|a(s)H(s,x(s))Δs+A|t′-t′′|∫0T‍k(s)a(s)H(s,x(s))Δs<∫0T‍ɛ2Γ-1a(s)H(s,x(s))Δs+2ɛAΓT·A·T4·∫0T‍a(s)H(s,x(s))Δs≤ɛ2Γ-1·∫0T‍a(s)f(s,ϕ_(s))Δs+ɛ2Γ·∫0T‍a(s)f(s,ϕ_(s))Δs<ɛ2Γ-1·Γ+ɛ2Γ·Γ=ɛ.
Therefore Q(D) is equicontinuous on P. Then by making use of Arzela-Ascoli theorem [8] on time scale, we know that Q(D) is relatively compact. Consequently Q is compact.
On the other hand, letB={x∈P:ϕ_(t)≤x(t)≤ϕ¯(t),forallt∈[0,T]}. Denote M1=max⁡t∈[0,T]⁡k(t). Then for any ϵ>0, from (49), we know that there exist μ∈(0,T/2) such that
(51)∫0μ‍k(s)a(s)f(s,ϕ_(s))Δs+∫T-μT‍k(s)a(s)f(s,ϕ_(s))Δs<ϵ4(1+TA).
For R>0, since H(t,x) is continuous on [μ,T-μ]×[0,R], consequently uniformly continuous. Thus there exists μ′>0 satisfying μ>μ′>0, for any t∈[μ,T-μ], x1,x2∈[0,R] such that |x1-x2|<μ′; we have
(52)|a(t)(H(t,x1(t))-H(t,x2(t)))|<ϵ2M1(1+TA).
Thus, from (51) and (52), for all x1,x2∈B such that |x1-x2|<μ′, we obtain
(53)|(Qx2)(t)-(Qx1)(t)|≤∫0T‍k(s)a(s)|H(s,x1(s))-H(s,x2(s))|Δs+A(T-t)∫0T‍k(s)a(s)|H(s,x1(s))-H(s,x2(s))|Δs≤(1+TA)M1∫μT-μ‍a(s)|H(s,x1(s))-H(s,x2(s))|Δs+2(1+TA)(∫0μ‍k(s)a(s)f(s,ϕ_(s))Δshihhhhhhhhhhhhh+∫T-μT‍k(s)a(s)f(s,ϕ_(s))Δs)<M1(1+TA)·ϵ2M1(1+TA)+2(1+TA)·ϵ4(1+TA)=ϵ
which implies Q is continuous on P. Consequently Q is a completely continuous operator.
From Schauder fixed point theorem [7] on time scale, we know that Q has at least one positive fixed point φ∈Crd⁡[0,T]∩Crd⁡1(0,T) in P, and φ(t) satisfies φ=Qφ. Thus φ(t) satisfies the following differential equation m-point boundary value problem:
(54)φΔΔ(t)+a(t)H(t,φ(t))=0,φ(0)-∑i=1m-2‍αiφ(ηi)=0,φ(T)=0.
Now we will prove thatϕ_(t)≤φ(t)≤ϕ¯(t), for all t∈[0,T]. First, we will prove that φ(t)≤ϕ¯(t), for all t∈[0,T]. In fact, if not, then there exists t*∈(0,T) such that
(55)ϕ¯(t*)<φ(t*).
Let ψ=ϕ¯(t)-φ(t), for all t∈[0,T]. Denote
(56)α=inf⁡{t1:0≤t1<t*,ψ(t)<0,t∈(t1,t*)},β=sup⁡{t2:t*≤t2<T,ψ(t)<0,t∈(t*,t2)}.
Then ϕ¯(t)<φ(t), for all t∈(α,β). Thus from (44), we see that H(t,φ(t))=f(t,ϕ¯(t)). Combining (43) with (44), we get
(57)ϕ¯ΔΔ(t)+a(t)H(t,ϕ¯(t))=ϕ¯ΔΔ(t)+a(t)f(t,ϕ¯(t))≤0,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh∀t∈(α,β).
In view of the above discussion and (54), we have ψΔΔ(t)=ϕ¯ΔΔ(t)-φΔΔ(t)≤0. For above α,β,ψ(t), there are two cases: (1) ψ(α)=ψ(β)=0; (2) ψ(α)>0, ψ(β)=0.
If (1) holds, thenψ(t)>0, for all t∈[α,β] which contradicts (55).
If (2) holds, fromψ(α)>0, we know that α=0, ψΔ(t*)=0. With the aid of ψΔ(t) increasing on [α,β] it follows that ψΔ(t)≥0, t∈[t*,β], that is, ψ(t) increasing on [t*,β]. From ψ(β)=0 it follows that ψ(t*)≤0, which contradicts (55). Thus we have ϕ¯(t)≥φ(t).
Similarly, we can verify thatϕ_(t)≤φ(t). Consequently φ(t) is a positive solution of the problem (1)-(2).
(II) Unique Positive Solution of the Problem (1)-(2).Let x1(t), x2(t) be two Crd⁡[0,T] positive solutions of the problem (1)-(2), and x1(t)≠x2(t). Without loss of generality, we assume that t*∈(0,T) such that x2(t*)>x1(t*).
Let(58)α=inf⁡{t1:0≤t1<t*,x2(t)>x1(t),t∈(t1,t*)},β=sup⁡{t2:t*≤t2≤T,x2(t)>x1(t),t∈(t*,t2)},
and φ(t)=x2(t)-x1(t) for all t∈[0,T]. It is easy to see that there are two cases for above α, β, φ: (i) φ(α)=φ(β)=0; (ii) φ(α)>0, φ(β)=0;
Fromt*∈(α,β), we have x2(t)>x1(t), f(t,x2(t))≤f(t,x1(t)). Thus, φΔΔ(t)=x2ΔΔ(t)-x1ΔΔ(t)≥0 for all t∈(α,β).
Case (i). From φΔΔ(t)>0 and φ(α)=φ(β)=0, we obtain φ(t)≤0, t∈[α,β], which contradicts x2(t*)>x1(t*).
Case (ii). From φ(α)>0, we know that α=0,φΔ(t*)=0. With the aid of φΔ(t) increasing on [α,β], we get that φΔ(t)≥0, t∈[t*,β]; that is, φ(t) is increasing on [t*,β]. From φ(β)=0, we see that φ(t*)≤0, which contradicts x2(t*)>x1(t*). Therefore the problem (1)-(2) has a unique positive solution.Theorem 10.
Suppose that (H1)–(H3) hold. In addition, one assumes that the following condition is satisfied:
(59)(H3′)a(t)f(t,r)≢0,0<∫01‍a(s)f(s,rk(s))Δs<+∞,000000000000000000000000000∀r>0.
Then, the problem (1)-(2) has a unique positive solution φ∈Crd1[0,T]∩Crd2(0,T), and there exist positive real numbers L>l>0 such that
(60)l(T-t)≤φ(t)≤L(T-t).Proof.
By making use of Lemma8, we know that the problem (1)-(2) has a couple of lower and upper solution. Applying Theorem 9 we see that the problem (1)-(2) has a unique positive solution φ∈Crd⁡1[0,T]∩Crd⁡2(0,T). From (H3), we know that f(t,φ(t)) is integrable on (0,T). Thus φΔΔ(t) is integrable on (0,T). It follows from the fact that φΔ(0+) and φΔ(T-) exist that we see that φ(t) is a Crd⁡[0,T] positive solution of the problem (1)-(2). Clearly, φ(t) may be expressed by
(61)φ(t)=∫0T‍G(t,s)a(s)f(s,φ(s))Δs+A(T-t)∫0T‍G(ηi,s)a(s)f(s,φ(s))Δs.
Thus, for any t∈[0,T], we get
(62)φ(t)≤A(T-t)∫0T‍k(ηi)a(s)f(s,φ(s))Δs+k(t)∫0T‍a(s)f(s,φ(s))Δs≤(Ak(ηi)+1)∫0T‍a(s)f(s,φ(s))Δs(T-t)=L(T-t),
where L=(1+Ak(ηi))∫0Ta(s)f(s,φ(s))Δs.
On the other hand,(63)φ(t)≥A(T-t)∫0T‍G(ηi,s)a(s)f(s,φ(s))Δs≥A(T-t)min⁡s∈[0,T]G(ηi,s)∫0T‍a(s)f(s,φ(s))Δs=l(T-t),
where l=Amin⁡s∈[0,T]⁡G(ηi,s)∫0Ta(s)f(s,φ(s))Δs. Therefore, from (62) and (63), we see that (60) holds.Corollary 11.
Suppose that (H1)–(H3) hold. In addition, assume that the following condition is satisfied: (H4)
f(t,r)≢0, 0<∫0Tk(s)f(s,r)Δs<+∞, for all r>0.
Then the following problem(64)xΔΔ(t)+f(t,x)=0,t∈(0,T),x(T)=0,x(0)-∑i=1m-2‍αix(ηi)=0
has a unique positive solution φ∈Crd[0,T]∩Crd2(0,T), and there exists a constant l>0 such that φ(t)≥l(T-t).Corollary 12.
Suppose that (H1)–(H3) hold. In addition, assume that the following condition is satisfied: (H5)
f(t,r)≢0,∫0Tf(s,r)Δs<+∞, for all r>0.Then the problem (64) has a unique positive solution ψ∈Crd[0,T]∩Crd2(0,T), and there exist positive constants L>l>0 such that l(T-t)≤ψ(t)≤L(T-t).Ifa(t) is nonsingular at t=0 or t=T, x=0, and for all x≥0, one has f(t,x)≤f(t,0), t∈(0,T), then the following conclusion holds.Theorem 13.
Suppose thata∈Crd[0,T], f(t,x):[0,T]×[0,+∞)→[0,+∞) is continuous, a(t)f(t,r)≢0, for all r≥0, and f(t,x) is nonincreasing with respect to x, for all t∈[0,T]. Then the problem (1)-(2) has a unique positive solution.Remark 14.
Under some weaker condition, we not only establish the existence of positive solution of the problem (1)-(2), but also obtain the uniqueness of the positive solution.Remark 15.
Without the cavity off(t,x) and other stronger conditions imposed on f(t,x), only f(t,x) nonincreasing with respect to x, we obtain new results. The main results hold even if the problem is nonsingular.


## 4. Examples

In this section, we will present two examples to illustrate the main result in this paper.Example 1.
LetT={0,1/4,1/3}∪[1/2,1]. Consider the following boundary value problem:
(65)xΔΔ(t)+1t(1-t)x3=0,0<t<1,x(1)=0,x(0)=13x(14)+16x(23).
Then, the four-point boundary value problem (65) has at least one positive solution.
Note thata(t)=1, for t∈(0,1), f(t,x)=1/t(1-t)x3 is nonincreasing in x, and f(t,x) is singular at t=0 and t=1.
Obviously,(H1) and (H2) are satisfied. Moreover, for any fixed λ>0, (H3) follows immediately from
(66)0<∫01‍a(t)f(t,λt(1-t))dt=∫01‍dtt(1-t)λt(1-t)3<+∞.
Thus, the existence of a positive solution follows from Theorem 9.Example 2.
LetT={0,1/9,1/7,1/6,1/5,1/3}∪[1/2,1]. Consider the following boundary value problem:
(67)xΔΔ(t)+tt(1-t)x5=0,0<t<1,x(1)=0,x(0)=13x(16)+15x(13)+17x(34)+19x(89).
Then, the four-point boundary value problem (65) has at least one positive solution.
Note thata(t)=t, for t∈(0,1), f(t,x)=1/t(1-t)x5 is nonincreasing in x, and f(t,x) is singular at t=0 and t=1.
Obviously,(H1) and (H2) are satisfied. Moreover, for any fixed λ>0, (H3) follows immediately from
(68)0<∫01‍a(t)f(t,λt(1-t))dt=∫01‍tdtt(1-t)λt(1-t)5<+∞.
Thus, the existence of a positive solution follows from Theorem 9.



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