# Positive Solutions for Second-Order Singular Semipositone Differential Equations Involving Stieltjes Integral Conditions

Authors: Jiqiang Jiang; Lishan Liu; Yonghong Wu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

By means of the fixed point theory in cones, we investigate the existence of positive solutions for the following second-order singular differential equations with a negatively perturbed term:−u′′(t)=λ[f(t,u(t))−q(t)], 0<t<1, αu(0)−βu′(0)=∫01u(s)dξ(s), γu(1)+δu′(1)=∫01u(s)dη(s),
where λ>0 is a parameter; f:(0,1)×(0,∞)→[0,∞)
is continuous; f(t,x)
may be singular at t=0, t=1, and x=0, and the perturbed term q:(0,1)→[0,+∞) is Lebesgue integrable and may have finitely many singularities in (0,1), which implies that the nonlinear term may change sign.


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## Body

## 1. Introduction

In this paper, we are concerned with positive solutions of the following second-order singular semipositone boundary value problem (BVP):(1.1)-u′′(t)=λ[f(t,u(t))-q(t)],0<t<1,αu(0)-βu′(0)=∫01u(s)dξ(s),γu(1)+δu′(1)=∫01u(s)dη(s),
where λ>0 is a parameter, α,γ≥0, β,δ>0 are constants such that ρ=αγ+αδ+βγ>0, and the integrals in (1.1) are given by Stieltjes integral with a signed measure, that is, ξ, η are suitable functions of bounded variation, q:(0,1)→[0,+∞) is a Lebesgue integral and may have finitely many singularities in [0,1], f:(0,1)×(0,∞)→[0,∞) is continuous, f(t,x) may be singular at t=0, t=1, and x=0.Semipositone BVPs occur in models for steady-state diffusion with reactions [1], and interest in obtaining conditions for the existence of positive solutions of such problems has been ongoing for many years. For a small sample of such work, we refer the reader to the papers of Agarwal et al. [2, 3], Kosmatov [4], Lan [5–7], Liu [8], Ma et al. [9, 10], and Yao [11]. In [12], the second-order m-point BVP,
(1.2)-u′′(t)=λf(t,u(t)),t∈(0,1),u′(0)=∑i=1m-2aiu′(ξi),u(1)=∑i=1m-2biu(ξi),
is studied, where ai,bi>0 (i=1,2,…,m-2), 0<ξ1<ξ2<⋯<ξm-2<1, λ is a positive parameter. By using the Krasnosel’skii fixed point theorem in cones, the authors established the conditions for the existence of at least one positive solution to (1.2), assuming that 0<∑i=1m-2ai<1, 0<∑i=1m-2bi<1, f:[0,1]×[0,+∞)→(-∞,+∞) is continuous, and there exists A>0 such that f(t,u)≥-A for (t,u)∈[0,1]×[0,+∞). If the constant A is replaced by any continuous function A(t) on [0,1], f also has a lower bound and the existence results are still true.Recently, Webb and Infante [13] studied arbitrary-order semipositone boundary value problems. The existence of multiple positive solutions is established via a Hammerstein integral equation of the form:
(1.3)u(t)=∫01k(t,s)g(s)f(s,u(s))ds,
where k is the corresponding Green function, g∈L1[0,1] is nonnegative and may have pointwise singularities, f:[0,1]×[0,+∞)→(-∞,+∞) satisfies the Carathéodory conditions and f(t,u)≥-A for some A>0. Although A is a constant, because of the term g, [13] includes nonlinearities that are bounded below by an integral function. It is worth mentioning that the boundary conditions cover both local and nonlocal types. Nonlocal boundary conditions are quite general, involving positive linear functionals on the space C[0,1], given by Stieltjes integrals.For the cases where the nonlinear term takes only nonnegative values, the existence of positive solutions of nonlinear boundary value problems with nonlocal boundary conditions, including multipoint and integral boundary conditions, has been extensively studied by many researchers in recent years [14–25]. Kong [17] studied the second-order singular BVP:(1.4)u′′(t)+λf(t,u(t))=0,t∈(0,1),u(0)=∫01u(s)dξ(s),u(1)=∫01u(s)dη(s),
where λ is a positive parameter, f:(0,1)×(0,+∞)→[0,+∞) is continuous, ξ(s) and η(s) are nondecreasing, and the integrals in (1.4) are Riemann-Stieltjes integrals. Sufficient conditions are obtained for the existence and uniqueness of a positive solution by using the mixed monotone operator theory.Inspired by the above work, the purpose of this paper is to establish the existence of positive solutions to BVP (1.1). By using the fixed point theorem on a cone, some new existence results are obtained for the case where the nonlinearity is allowed to be sign changing. We will address here that the problem tackled has several new features. Firstly, as q∈L1[0,1], the perturbed effect of q on f may be so large that the nonlinearity may tend to negative infinity at some singular points. Secondly, the BVP (1.1) possesses singularity, that is, the perturbed term q may has finitely many singularities in [0,1], and f(t,x) is allowed to be singular at t=0, t=1, and x=0. Obviously, the problem in question is different from those in [2–13]. Thirdly, ∫01u(s)dξ(s) and ∫01u(s)dη(s) denote the Stieltjes integrals where ξ,η are of bounded variation, that is, dξ and dη can change sign. This includes the multipoint problems and integral problems as special cases.The rest of this paper is organized as follows. In Section2, we present some lemmas and preliminaries, and we transform the singularly perturbed problem (1.1) to an equivalent approximate problem by constructing a modified function. Section 3 gives the main results and their proofs. In Section 4, two examples are given to demonstrate the validity of our main results.LetK be a cone in a Banach space E. For 0<r<R<+∞, let Kr={x∈K:∥x∥<r}, ∂Kr={x∈K:∥x∥=r}, and K¯r,R={x∈K:r≤∥x∥≤R}. The proof of the main theorem of this paper is based on the fixed point theory in cone. We list here one lemma [26, 27] which is needed in our following argument.Lemma 1.1.
LetK be a positive cone in real Banach space  E,  T:K¯r,R→K is a completely continuous operator. If the following conditions hold: (i)
∥Tx∥≤∥x∥ for  x∈∂KR,(ii)
there existse∈∂K1 such that  x≠Tx+me for any  x∈∂Kr and  m>0,then, T has a fixed point in K¯r,R.Remark 1.2.
If (i) and (ii) are satisfied forx∈∂Kr and x∈∂KR, respectively, then Lemma 1.1 is still true.


## 2. Preliminaries and Lemmas

Denote(2.1)ϕ1(t)=1ρ(δ+γ(1-t)),ϕ2(t)=1ρ(β+αt),e(t)=G(t,t),t∈[0,1],k1=1-∫01ϕ1(t)dξ(t),k2=∫01ϕ2(t)dξ(t),k3=∫01ϕ1(t)dη(t),k4=1-∫01ϕ2(t)dη(t),k=k1k4-k2k3,σ=ρ(α+β)(γ+δ),
where
(2.2)G(t,s)=1ρ{(β+αs)(δ+γ(1-t)),0≤s≤t≤1,(β+αt)(δ+γ(1-s)),0≤t≤s≤1.
Obviously,
(2.3)e(t)=ρϕ1(t)ϕ2(t)=1ρ(β+αt)(δ+γ(1-t)),t∈[0,1],(2.4)σe(t)e(s)≤G(t,s)≤e(s)(ore(t))≤σ-1,t,s∈[0,1].Throughout this paper, we adopt the following assumptions.(H1)
k1,k4∈(0,1], k2≥0, k3≥0, k>0, and
(2.5)Gξ(s)=∫01G(t,s)dξ(t)≥0,Gη(s)=∫01G(t,s)dη(t)≥0,s∈[0,1].(H2)
q:(0,1)→[0,+∞) is a Lebesgue integral and ∫01q(t)dt>0.(H3)
For any(t,x)∈(0,1)×(0,+∞),
(2.6)0≤f(t,x)≤p(t)(g(x)+h(x)),
where p∈C(0,1) with p>0 on (0,1) and ∫01p(t)dt<+∞, g>0 is continuous and nonincreasing on (0,+∞), h≥0 is continuous on [0,+∞), and for any constant r>0,
(2.7)0<∫01p(s)g(re(s))ds<+∞.Remark 2.1.
Ifdξ and dη are two positive measures, then the assumption (H1) can be replaced by a weaker assumption:(H1′)
k1>0, k4>0, k>0.Remark 2.2.
It follows from (2.4) and (H3) that
(2.8)∫01e(s)p(s)g(re(s))ds≤σ-1∫01p(s)g(re(s))ds<+∞.
For convenience, in the rest of this paper, we define several constants as follows:
(2.9)L1=1+k4(γ+δ)+k3(α+β)ρk∫01dξ(τ)+k2(γ+δ)+k1(α+β)ρk∫01dη(τ),L2=σ[1+k4(γ+δ)+k3(α+β)ρk∫01e(τ)dξ(τ)+k2(γ+δ)+k1(α+β)ρk∫01e(τ)dη(τ)],L3=1+k4δ+k3βkβδ∫01e(τ)dξ(τ)+k2δ+k1βkβδ∫01e(τ)dη(τ).Remark 2.3.
Ifx∈C[0,1]∩C2(0,1) satisfies (1.1), and x(t)>0 for any t∈(0,1), then we say that x is a C[0,1]∩C2(0,1) positive solution of BVP (1.1).Lemma 2.4.
Assume that(H1)  holds. Then, for any  y∈L1[0,1], the problem,
(2.10)-u′′(t)=y(t),t∈(0,1),αu(0)-βu′(0)=∫01u(s)dξ(s),γu(1)+δu′(1)=∫01u(s)dη(s),
has a unique solution
(2.11)u(t)=∫01H(t,s)y(s)ds,
where
(2.12)H(t,s)=G(t,s)+k4ϕ1(t)+k3ϕ2(t)k∫01G(τ,s)dξ(τ)+k2ϕ1(t)+k1ϕ2(t)k∫01G(τ,s)dη(τ).Proof.
The proof is similar to Lemma  2.2 of [28], so we omit it.Lemma 2.5.
Suppose that(H1) holds, then Green’s function  H(t,s) defined by  (2.12) possesses the following properties: (i)
H(t,s)≤L1e(s), t,s∈[0,1];(ii)
L2e(t)e(s)≤H(t,s)≤L3e(t), t,s∈[0,1],where L1, L2, and L3 are defined by (2.9).Proof.
(i) It follows from (2.4) that
(2.13)H(t,s)≤e(s)+k4(γ+δ)+k3(α+β)ρk∫01e(s)dξ(τ)+k2(γ+δ)+k1(α+β)ρk∫01e(s)dη(τ)=L1e(s),t,s∈[0,1].
(ii) By the monotonicity ofϕ1, ϕ2 and the definition of G(t,s), we have
(2.14)e(t)α+β≤ϕ1(t)=e(t)ρϕ2(t)=e(t)αt+β≤e(t)β,t∈[0,1],e(t)γ+δ≤ϕ2(t)=e(t)ρϕ1(t)=e(t)γ(1-t)+δ≤e(t)δ,t∈[0,1].
By (2.4) and the left-hand side of inequalities (2.14), we have
(2.15)H(t,s)≥σe(t)e(s)+σe(t)e(s)k4/(α+β)+k3/(γ+δ)k∫01e(τ)dξ(τ)+σe(t)e(s)k2/(α+β)+k1/(γ+δ)k∫01e(τ)dη(τ)=σe(t)e(s)[1+k4(γ+δ)+k3(α+β)ρk∫01e(τ)dξ(τ)+k2(γ+δ)+k1(α+β)ρk∫01e(τ)dη(τ)]=L2e(t)e(s),t,s∈[0,1].
Similarly, by (2.4) and the right-hand side of inequalities (2.14), we have
(2.16)H(t,s)=G(t,s)+k4ϕ1(t)+k3ϕ2(t)k∫01G(τ,s)dξ(τ)+k2ϕ1(t)+k1ϕ2(t)k∫01G(τ,s)dη(τ)≤e(t)+e(t)[k4/β+k3/δk∫01G(τ,s)dξ(τ)+k2/β+k1/δk∫01G(τ,s)dη(τ)]≤e(t)+e(t)[k4δ+k3βkβδ∫01e(τ)dξ(τ)+k2δ+k1βkβδ∫01e(τ)dη(τ)]=e(t)[1+k4δ+k3βkβδ∫01e(τ)dξ(τ)+k2δ+k1βkβδ∫01e(τ)dη(τ)]=L3e(t),t,s∈[0,1].
The proof of Lemma 2.5 is completed.Lemma 2.6.
Suppose that(H1) and  (H2) hold. Then, the boundary value problem,
(2.17)-w′′(t)=2λp(t),t∈(0,1),αw(0)-βw′(0)=∫01w(s)dξ(s),γw(1)+δw′(1)=∫01w(s)dη(s),
has unique solution
(2.18)w(t)=2λ∫01H(t,s)p(s)ds,
which satisfies
(2.19)w(t)≤2λL3e(t)∫01p(s)ds,t∈[0,1].Proof.
It follows from (2.11), Lemma 2.5, (H1) and (H2) that (2.18) and (2.19) hold.LetX=C[0,1] be a real Banach space with the norm ∥x∥=maxt∈[0,1]|x(t)| for x∈X. We let
(2.20)K={x∈X:xisconcaveon⁡[0,1],x(t)≥Λe(t)‖x‖fort∈[0,1]},
where Λ=L2/L1. Clearly, K is a cone of X.For anyu∈X, let us define a function [·]+:
(2.21)[u(t)]+={u(t),u(t)≥0,0,u(t)<0.Next, we consider the following approximate problem of (1.1):
(2.22)-x′′(t)=λ[f(t,[x(t)-w(t)]+)+q(t)],t∈(0,1),αx(0)-βx′(0)=∫01x(s)dξ(s),γx(1)+δx′(1)=∫01x(s)dη(s).Lemma 2.7.
Ifx∈C[0,1]∩C2(0,1) is a positive solution of problem  (2.22) with  x(t)≥w(t) for any  t∈[0,1], then  x-w is a positive solution of the singular semipositone differential equation  (1.1).Proof.
Ifx is a positive solution of (2.22) such that x(t)≥w(t) for any t∈[0,1], then from (2.22) and the definition of [u(t)]+, we have
(2.23)-x′′(t)=λ[f(t,x(t)-w(t))+q(t)],t∈(0,1),αx(0)-βx′(0)=∫01x(s)dξ(s),γx(1)+δx′(1)=∫01x(s)dη(s).
Let u=x-w, then u′′=x′′-w′′, which implies that
(2.24)-x′′=-u′′-w′′=-u′′+2λq(t).
Thus, (2.23) becomes
(2.25)-u′′(t)=λ[f(t,u(t))-q(t)],t∈(0,1),αu(0)-βu′(0)=∫01u(s)dξ(s),γu(1)+δu′(1)=∫01u(s)dη(s),
that is, x-w is a positive solution of (1.1). The proof is complete.To overcome singularity, we consider the following approximate problem of (2.22):
(2.26)-x′′(t)=λ[f(t,[x(t)-w(t)]++n-1)+q(t)],t∈(0,1),αx(0)-βx′(0)=∫01x(s)dξ(s),γx(1)+δx′(1)=∫01x(s)dη(s),
where n is a positive integer. For any n∈ℕ, let us define a nonlinear integral operator Tnλ:K→X as follows:
(2.27)Tnλx(t)=λ∫01H(t,s)[f(s,[x(s)-w(s)]++n-1)+q(s)]ds.
It is obvious that solving (2.26) in C[0,1]∩C2(0,1) is equivalent to solving the fixed point equation Tnλx=x in the Banach space C[0,1].Lemma 2.8.
Assume that(H1)–(H3) hold, then for each  n∈ℕ,  λ>0,  R>r≥4λL3Λ-1∫01q(s)ds,  Tnλ:K¯r,R→K is a completely continuous operator.Proof.
Letn∈ℕ be fixed. For any x∈K, by (2.27) we have
(2.28)(Tnλx)′′(t)=-λ[f(s,[x(s)-w(s)]++n-1)+q(s)]≤0,Tnλx(0)=λ∫01H(0,s)[f(s,[x(s)-w(s)]++n-1)+q(s)]ds≥0,Tnλx(1)=λ∫01H(1,s)[f(s,[x(s)-w(s)]++n-1)+q(s)]ds≥0,
which implies that Tnλ is nonnegative and concave on [0,1]. For any x∈K and t∈[0,1], it follows from Lemma 2.5 that
(2.29)Tnλx(t)=λ∫01H(t,s)[f(s,[x(s)-w(s)]++n-1)+q(s)]ds≤λL1∫01e(s)[f(s,[x(s)-w(s)]++n-1)+q(s)]ds.
Thus,
(2.30)‖Tnλx‖≤λL1∫01e(s)[f(s,[x(s)-w(s)]++n-1)+q(s)]ds.
On the other hand, from Lemma 2.5, we also obtain
(2.31)Tnλx(t)=λ∫01H(t,s)[f(s,[x(s)-w(s)]++n-1)+q(s)]ds≥λL2e(t)∫01e(s)[f(s,[x(s)-w(s)]++n-1)+q(s)]ds.
So,
(2.32)Tnλx(t)≥Λe(t)‖Tnλx‖,t∈[0,1].
This yields that Tnλ(K)⊂K.
Next, we prove thatTnλ:K¯r,R→K is completely continuous. Suppose xm∈K¯r,R and x0∈K¯r,R with ∥xm-x0∥→0(m→∞). Notice that
(2.33)|[xm(s)-w(s)]+-[x0(s)-w(s)]+|=||xm(s)-w(s)|+xm(s)-w(s)2-|x0(s)-w(s)|+x0(s)-w(s)2|=||xm(s)-w(s)|-|x0(s)-w(s)|2+xm(s)-x0(s)2|≤|xm(s)-x0(s)|.
This, together with the continuity of f, implies
(2.34)|f(s,[xm(s)-w(s)]++n-1)+q(s)-[f(s,[x0(s)-w(s)]++n-1)+q(s)]|=|f(s,[xm(s)-w(s)]++n-1)-f(s,[x0(s)-w(s)]++n-1)|⟶0,m⟶∞.
Using the Lebesgue dominated convergence theorem, we have
(2.35)‖Tnλxm-Tnλx0‖≤λL1∫01e(s)|f(s,[xm(s)-w(s)]++n-1)-f(s,[x0(s)-w(s)]++n-1)|ds⟶0,m⟶∞.
So, Tnλ:K¯r,R→K is continuous.
LetB⊂K¯r,R be any bounded set, then for any x∈B, we have x∈K, r≤∥x∥≤R. Therefore, we have
(2.36)[x(t)-w(t)]+≤x(t)≤‖x‖≤R,t∈[0,1],x(t)-w(t)≥x(t)-2λL3e(t)∫01q(s)ds≥x(t)-2λL3x(t)Λr∫01q(s)ds≥12x(t)≥12rΛe(t)>0,t∈[0,1].
By (H3), we have
(2.37)Lr:=∫01p(s)g(12Λre(s))ds<+∞.
It is easy to show that Tnλ(B) is uniformly bounded. In order to show that Tnλ is a compact operator, we only need to show that Tnλ(B) is equicontinuous. By the continuity of H(t,s) on [0,1]×[0,1], for any ε>0, there exists δ1>0 such that for any t1,t2,s∈[0,1] and |t1-t2|<δ1, we have
(2.38)|H(t1,s)-H(t2,s)|<ε.
By (2.36)–(2.37), and (2.27), we have
(2.39)|Tnλx(t1)-Tnλx(t2)|≤λ∫01|H(t1,s)-H(t2,s)|[f(s,[x(s)-w(s)]++n-1)+q(s)]ds<ελ∫01[f(s,[x(s)-w(s)]++n-1)+q(s)]ds≤ελ∫01[p(s)(g([x(s)-w(s)]++n-1)+h([x(s)-w(s)]++n-1))+q(s)(g([x(s)-w(s)]++n-1)+h([x(s)-w(s)]++n-1))]ds≤ελ[Lr+(1+h*(R))∫01[p(s)+q(s)]ds],
where
(2.40)h*(r)=max⁡y∈[0,1+r]h(y).
This means that Tnλ(B) is equicontinuous. By the Arzela-Ascoli theorem, Tnλ(B) is a relatively compact set. Now since λ and n are given arbitrarily, the conclusion of this lemma is valid.


## 3. Main Results

Theorem 3.1.
Assume that conditions(H1)–(H3) are satisfied. Further assume that the following condition holds.(H4)
There exists an interval[a,b]⊂(0,1) such that(3.1)liminf⁡u→+∞min⁡t∈[a,b]f(t,u)u=+∞.
Then, there exists λ*>0 such that the BVP (1.1) has at least one positive solution u(t)∈C[0,1]∩C2(0,1) provided λ∈(0,λ*). Furthermore, the solution also satisfies u(t)≥l̃e(t) for some positive constant l̃.Proof.
Taker>4L3Λ-1∫01q(s)ds. Let
(3.2)λ*=min⁡{1,r2L1∫01e(s)p(s)g((1/2)rΛe(s))ds,r2L1(h*(r)+1)∫01e(s)[p(s)+q(s)]ds},
where h* is defined by (2.40). For any λ∈(0,λ*), x∈∂Kr, noticing that λ*≤1, we have
(3.3)[x(t)-w(t)]+≤x(t)≤‖x‖≤r,t∈[0,1],x(t)-w(t)≥x(t)-2λL3e(t)∫01q(s)ds≥x(t)-2L3e(t)∫01q(s)ds≥x(t)-2L3x(t)Λr∫01q(s)ds≥12x(t)≥12rΛe(t)>0,t∈[0,1].
For any λ∈(0,λ*), by (3.3), we have
(3.4)|Tnλx(t)|=λ∫01H(t,s)[f(s,[x(s)-w(s)]++n-1)+q(s)]ds≤λL1∫01e(s)[p(s)(g([x(s)-w(s)]++n-1)+h([x(s)-w(s)]++n-1))+q(s)]ds≤λL1∫01e(s)[p(s)(g(12rΛe(s))+h*(r))+q(s)]ds≤λL1∫01e(s)p(s)g(12rΛe(s))ds+λL1(h*(r)+1)∫01e(s)[p(s)+q(s)]ds≤r2+r2=r,
which means that
(3.5)‖Tnλx‖≤‖x‖,x∈∂Kr.
On the other hand, choose a real numberM>0 such that λML2l12Λ∫abe(s)ds>2, where l1=min0≤t≤1e(t)>0, L2 is defined by (2.9). By (H4), there exists N>0 such that for any t∈[a,b], we have
(3.6)f(t,u)≥Mu,u≥N.
Take R>max{r,2N/l1Λ}. Next, we take φ1≡1∈∂K1={x∈K:∥x∥=1}, and for any x∈∂KR, m>0, n∈ℕ, we will show
(3.7)x≠Tnλx+mφ1.
Otherwise, there exist x0∈∂KR and m0>0 such that
(3.8)x0=Tnλx0+m0φ1.
From x0∈∂KR, we know that ∥x0∥=R. Then, for t∈[a,b], we have
(3.9)x0(t)-w(t)≥x0(t)-2λL3e(t)∫01q(s)ds≥x0(t)-2L3x0(t)ΛR∫01q(s)ds≥12x0(t)≥12RΛe(t)≥l1RΛ2≥N>0.
So, by (3.6), (3.9), we have
(3.10)x0(t)=λ∫01H(t,s)[f(s,[x0(s)-w(s)]++n-1)+q(s)]ds+m0≥λL2e(t)∫01e(s)f(s,[x0(s)-w(s)]++n-1)ds+m0≥λL2e(t)∫abe(s)f(s,[x0(s)-w(s)]++n-1)ds+m0≥λML2e(t)∫abe(s)([x0(s)-w(s)]++n-1)ds+m0≥12λML2l12ΛR∫abe(s)ds+m0≥R+m0>R.
This implies that R>R, which is a contradiction. This yields that (3.7) holds. By (3.5), (3.7), and Lemma 1.1, for any n∈ℕ and λ∈(0,λ*), we obtain that Tnλ has a fixed point xn in K¯r,R.
Let{xn}n=1∞ be the sequence of solutions of the boundary value problems (2.26). It is easy to see that they are uniformly bounded. Next, we show that {xn}n=1∞ are equicontinuous on [0,1]. From xn∈K¯r,R, we know that
(3.11)[xn(t)-w(t)]+≤xn(t)≤‖xn‖≤R,t∈[0,1],xn(t)-w(t)≥xn(t)-2λL3e(t)∫01q(s)ds≥xn(t)-2λL3xn(t)Λ‖xn‖∫01q(s)ds≥12xn(t)≥12Λe(t)‖xn‖≥12Λre(t)>0,t∈[0,1].
For any ε>0, by the continuity of H(t,s) in [0,1]×[0,1], there exists δ2>0 such that for any t1,t2,s∈[0,1] and |t1-t2|<δ2, we have
(3.12)|H(t1,s)-H(t2,s)|<ε.
This, combined with (2.11) and (2.37), implies that for any t1,t2∈[0,1] and |t1-t2|<δ2, we have
(3.13)|xn(t1)-xn(t2)|≤∫01|H(t1,s)-H(t2,s)|[f(s,[x(s)-w(s)]++n-1)+q(s)]ds<ε∫01[f(s,[x(s)-w(s)]++n-1)+q(s)]ds≤ε∫01[p(s)(g([x(s)-w(s)]++n-1)+h([x(s)-w(s)]++n-1))+q(s)]ds≤ε[Lr+(1+h*(R))∫01(p(s)+q(s))ds].
By the Ascoli-Arzela theorem, the sequence {xn}n=1∞ has a subsequence being uniformly convergent on [0,1]. Without loss of generality, we still assume that {xn}n=1∞ itself uniformly converges to x on [0,1]. Since {xn}n=1∞∈K¯r,R⊂K, we have xn≥0. By (2.26), we have
(3.14)xn(t)=xn(12)+(t-12)xn′(12)-λ∫1/2tds∫1/2s[f(τ,xn(τ)-w(τ)+n-1)+q(τ)]dτ,t∈(0,1).
From (3.14), we know that {xn′(1/2)}n=1∞ is bounded sets. Without loss of generality, we may assume xn′(1/2)→c1 as n→∞. Then, by (3.14) and the Lebesgue dominated convergence theorem, we have
(3.15)x(t)=x(12)+c1(t-12)-λ∫1/2tds∫1/2s[f(τ,x(τ)-w(τ))+q(τ)]dτ,t∈(0,1).
By (3.15), direct computation shows that
(3.16)-x′′(t)=λ[f(t,x(t)-w(t))+q(t)],0<t<1.
On the other hand, letting n→∞ in the following boundary conditions:
(3.17)αxn(0)-βxn′(0)=∫01xn(s)dξ(s),γxn(1)+δxn′(1)=∫01xn(s)dη(s),
we deduce that x is a positive solution of BVP (2.22).
Letu(t)=x(t)-w(t) and l̃=(1/2)Λr. By (3.11) and the convergence of sequence {xn}n=1∞, we have u(t)≥l̃e(t)>0, t∈[0,1]. It then follows from Lemma 2.7 that BVP (1.1) has at least one positive solution u satisfying u≥l̃e(t) for any t∈[0,1]. The proof is completed.Theorem 3.2.
Assume that conditions(H1)–(H3) are satisfied. In addition, assume that the following condition holds.(H5)
There exists an interval[c,d]⊂(0,1) such that
(3.18)liminf⁡u→+∞min⁡t∈[c,d]f(t,u)>4L3∫01q(s)dsΛL2l2∫cde(s)ds,
where l2=minc≤t≤de(t) and
(3.19)lim⁡u→+∞h(u)u=0.
Then there exists λ*>0 such that the BVP (1.1) has at least one positive solution u(t)∈C[0,1]∩C2(0,1) provided λ∈(λ*,+∞). Furthermore, the solution also satisfies u(t)≥l̃e(t) for some positive constant l̃.Proof.
By (3.18), there exists N0>0 such that, for any t∈[c,d], u≥N0, we have
(3.20)f(t,u)≥4L3∫01q(s)dsΛL2l2∫cde(s)ds.
Choose λ*=N0/2l2L3∫01q(s)ds. Let r=4λL3Λ-1∫01q(s)ds as λ>λ*. Next, we take φ1≡1∈∂K1={x∈K:∥x∥=1}, and for any x∈∂Kr, m>0, n∈ℕ, we will show that
(3.21)x≠Tnλx+mφ1.
Otherwise, there exist x0∈∂Kr and m0>0 such that
(3.22)x0=Tnλx0+m0φ1.
From x0∈∂Kr, we know that ∥x0∥=r, and
(3.23)x0(t)-w(t)≥Λre(t)-2λL3e(t)∫01q(s)ds=2λL3e(t)∫01q(s)ds≥N0l2e(t).
So, we have x0(t)-w(t)≥N0 on t∈[c,d], x0∈∂Kr. Then, by (3.20) we have
(3.24)x0(t)=λ∫01H(t,s)[f(s,[x0(s)-w(s)]++n-1)+q(s)]ds+m0≥λL2e(t)∫01e(s)f(s,[x0(s)-w(s)]++n-1)ds+m0≥λL2e(t)∫cde(s)f(s,[x0(s)-w(s)]++n-1)ds+m0≥λL2e(t)4L3∫01q(s)dsΛL2l2∫cde(s)ds∫cde(s)ds+m0≥4λΛ-1L3∫01q(s)ds+m0=r+m0>r.
This implies that r>r, which is a contradiction. This yields that (3.21) holds.
On the other hand, by (3.19) and the continuity of h(u) on [0,+∞), we have
(3.25)lim⁡u→+∞h*(u)u=0,
where h*(u) is defined by (2.40). For
(3.26)ϵ=[4λL1∫01e(s)[p(s)+q(s)]ds]-1,
there exists M0>0 such that when x>M0, for any 0≤y≤x, we have h(y)≤ϵx. Take
(3.27)R>max⁡⁡{2,r,M0,2λL1∫01e(s)p(s)g(Λe(s))ds}.
Then, for any x∈∂KR,t∈[0,1], we have
(3.28)[x(t)-w(t)]+≤x(t)≤‖x‖≤R,x(t)-w(t)≥x(t)-2λL3e(t)∫01q(s)ds≥x(t)-2λL3x(t)ΛR∫01q(s)ds≥12x(t)≥12RΛe(t)>Λe(t)>0,t∈[0,1].
It follows from (3.19) and (3.28) that
(3.29)|Tnλx(t)|=λ∫01H(t,s)[f(s,[x(s)-w(s)]++n-1)+q(s)]ds≤λL1∫01e(s)p(s)g([x(s)-w(s)]++n-1)ds+λL1∫01e(s)[p(s)h([x(s)-w(s)]++n-1)+q(s)]ds≤λL1∫01e(s)p(s)g(Λe(s))ds+λL1∫01e(s)[p(s)ϵ(R+1)+q(s)]ds≤λL1∫01e(s)p(s)g(Λe(s))ds+ϵλL1(R+2)∫01e(s)[p(s)+q(s)]ds≤R2+R2=R,
which means that
(3.30)‖Tnλx‖≤‖x‖,x∈∂KR.
By (3.21), (3.30), and Lemma 1.1, for any n∈ℕ and λ>λ*, we obtain that Tnλ has a fixed point xn in K¯r,R satisfying r≤∥xn∥≤R. The rest of proof is similar to Theorem 3.1. The proof is complete.Remark 3.3.
From the proof of Theorem3.2, we can see that if (H5) is replaced by the following condition.(H5′)
There exists an interval[c,d]⊂(0,1) such that
(3.31)liminf⁡u→+∞min⁡t∈[c,d]f(t,u)=+∞,lim⁡u→+∞h(u)u=0,
then, the conclusion of Theorem 3.2 is still true.


## 4. Applications

In this section, we construct two examples to demonstrate the application of our main results.Example 4.1.
Consider the following 4-point boundary value problem:(4.1)-u′′(t)=λ[1t(1-t)(1u2+u2+1)-q(t)],0<t<1,u(0)-u′(0)=14u(13)+19u(23),u(1)+u′(1)=38u(13)+u(23),
where λ>0 is a parameter and
(4.2)q(t)=14+343[1t+11-t+1(t-1/2)23].
The BVP (4.1) can be regarded as a boundary value problem of the form of (1.1). In this situation, α=β=γ=δ=1 and
(4.3)ξ(s)={0,s∈[0,13),14,s∈[13,23),1336,s∈[23,1],η(s)={0,s∈[0,13),38,s∈[13,23),118,s∈[23,1].
Let
(4.4)f(t,x)=1t(1-t)(1x2+x2+1),for(t,x)∈(0,1)×(0,+∞)
and let p(t)=1/t(1-t), g(x)=1/x2, h(x)=x2+1. By direct calculation, we have ∫01p(t)dt=π, ∫01q(t)dt=1, and
(4.5)ρ=3,σ=34,ϕ1(t)=2-t3,ϕ2(t)=1+t3,e(t)=13(2-t)(1+t),k1=263324,k2=1481,k3=4772,k4=518,k=73648,L1=11,L2=22736,L3=1099.
Clearly, the conditions (H1)–(H3) hold. Taking t∈[1/4,3/4]⊂[0,1], we have
(4.6)liminf⁡x→+∞min⁡t∈[1/4,3/4]f(t,x)x=liminf⁡x→+∞min⁡t∈[1/4,3/4](1/t(1-t))(1/x2+x2+1)x=+∞.
Thus (H4) also holds. Consequently, by Theorem 3.1, we infer that the singular BVP (4.1) has at least one positive solution provided λ is small enough.Example 4.2.
Consider the following problem:(4.7)-u′′(t)=λ[1t(1-t)(1u2+u+1)-2t3(1-t)4],0<t<1,u(0)-u′(0)=-∫01u(t)cos⁡2πtdt,u(1)+u′(1)=15u(14)+u(34),
where λ>0 is a parameter. Let
(4.8)f(t,x)=1t(1-t)(1x2+x+1),(t,x)∈(0,1)×(0,+∞),q(t)=2t3(1-t)4,p(t)=1t(1-t),g(x)=1x2,h(x)=x+1.
Then, ∫01p(t)dt=π, ∫01q(t)dt=2π. Here, dξ(t)=-cos2πtdt, so the measure dξ changes sign on [0,1]. By direct calculation, we have
(4.9)k1=1,k2=0,k3=815,k4=13,k=13,Gξ(s)=14π2(1-cos⁡2πs)≥0,Gη(s)≥0,
where
(4.10)Gη(s)={15G2(14,s)+G2(34,s),0≤s<14,15G1(14,s)+G2(34,s),14≤s≤34,15G1(14,s)+G1(34,s),34<s≤1,G(t,s)={G1(t,s)=(2-s)(1+t)3,0≤t≤s≤1,G2(t,s)=(2-t)(1+s)3,0≤s≤t≤1.
Taking t∈[1/4,3/4]⊂[0,1], we have
(4.11)liminf⁡x→+∞min⁡t∈[1/4,3/4]f(t,x)=liminf⁡x→+∞min⁡t∈[1/4,3/4]1t(1-t)(1x2+x+1)=+∞,lim⁡x→+∞h(x)x=lim⁡x→+∞x+1x=0.
So all assumptions of Theorem 3.2 are satisfied. By Theorem 3.2, we know that BVP (4.7) has at least one positive solution provided λ is large enough.



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*Source: 696283-2012-06-28.xml*