# Existence of2m-1 Positive Solutions for Sturm-Liouville Boundary Value Problems with Linear Functional Boundary Conditions on the Half-Line

Authors: Yanmei Sun; Zengqin Zhao
Journal: Journal of Applied Mathematics
 (2012)

---

## Abstract

By using the Leggett-Williams fixed theorem, we establish the existence of multiple positive solutions for second-order nonhomogeneous Sturm-Liouville boundary value problems with linear functional boundary conditions. One explicit example with singularity is presented to demonstrate the application of our main results.


---

## Body

## 1. Introduction

In this paper, we consider the following Sturm-Liouville boundary value problems on the half-line(1.1)(p(t)u′(t))′+Φ(t)f(t,u(t),u′(t))=0,0<t<+∞,α1u(0)-β1lim⁡t→0+p(t)u′(t)=T(u),α2lim⁡t→+∞u(t)+β2lim⁡t→+∞p(t)u′(t)=K(u),
where f:R+×R+×R→R+ is a continuous function, f≢0 on any subinterval of R+, here R+=[0,+∞);Φ:R+→R+ is a Lebesgue integrable function and may be singular at t=0;p∈C(R+,R+)⋂‍C1(R+),∫0+∞‍ds/p(s)<+∞;αi,βi≥0(i=1,2) with ρ=α1β2+α2β1+α1α2∫0+∞‍ds/p(s);T,K are linear positive functionals on C(R+)(T,K which are called positive if T(u),K(u)≥0 for u∈C(R+)).The theory of nonlocal boundary value problems for ordinary differential equations arises in different areas of applied mathematics and physics. There are many studies for nonlocal, including three-point, m-point, and integral boundary value problems on finite interval by applying different methods [1–3]. It is well known that boundary value problems on infinite interval arise in the study of radial solutions of nonlinear elliptic equations and models of gas pressure in a semi-infinite porous medium [4–6]. But the theory of Sturm-Liouville nonhomogeneous boundary value problems on infinite interval is yet rare.The linear functional boundary conditions cover some nonlocal three-point, m-point, and integral boundary conditions. In [7], Zhao and Li investigated some nonlinear singular differential equations with linear functional boundary conditions. However, the differential equations were defined only in a finite interval. Recently, Liu et al. [6] studied multiple positive solutions for Sturm-Liouville boundary value problems on the half-line
(1.2)(p(t)u′(t))′+m(t)f(t,u(t))=0,0≤t<+∞,α1u(0)-β1lim⁡t→0+p(t)u′(t)=0,α2lim⁡t→+∞u(t)+β2lim⁡t→+∞p(t)u′(t)=0.
However, the authors did not consider the case when Sturm-Liouville boundary value problems are nonhomogeneous. Therefore BVP(1.1) is the direct extension of [7]. So it is worthwhile to investigate BVP(1.1).We denote(1.3)a(t)=β1+α1∫0t‍dsp(s),b(t)=β2+α2∫t+∞‍dsp(s),(1.4)u(t)=ρ-1[1+a(t)b(t)]y1(t),u′(t)=y2(t),f(t,u,u′)=Ψ(t,y1,y2),a(0)=lim⁡t→0+a(t)=β1,a(∞)=lim⁡t→+∞a(t)=β1+α1∫0+∞‍dsp(s),b(0)=lim⁡t→0+b(t)=β2+α2∫0+∞‍dsp(s),b(∞)=lim⁡t→+∞b(t)=β2.In this paper, we always assume that the following conditions hold.(H1)
Ψ(t,y1,y2)≤q(t)Q(y1,y2),q(t)∈C(R+,R+),Q(y1,y2)∈C(R+×R,R+) and ∫0+∞‍Φ(s)q(s)ds<+∞.(H2)
For any constantτ∈[0,+∞),0<T(a(τ))<ρ,0<K(b(τ))<ρ and
(1.5)Δ=|ρ-T(b(τ))T(a(τ))K(b(τ))ρ-K(a(τ))|>0.Motivated and inspired by [5–9], we are concerned with the existence of multiple positive solutions for BVP(1.1) by applying Leggett-Williams fixed theorem. The main new features presented in this paper are as follows. Firstly, Sturm-Liouville nonhomogeneous boundary value problems with linear functional boundary conditions are seldom researched, it brings about many difficulties when we imply the integral equations of BVP(1.1). To solve the problem, we use a new method of undetermined coefficient to obtain the integral equations of boundary value problems with nonhomogeneous boundary conditions. Secondly, we discuss the existence of triple positive solutions and 2m-1 positive solutions of BVP(1.1). Finally, the methods used in this paper are different from [1, 6, 7] and the results obtained in this paper generalize and involve some results in [5].The rest of paper is organized as follows. In Section2, we present some preliminaries and lemmas. We state and prove the main results in Section 3. Finally, in Section 4, one example with a singular nonlinearity is presented to demonstrate the application of Theorem 3.1.


## 2. Preliminary

In order to discuss the main results, we need the following lemmas.Lemma 2.1.
Under the condition∫0+∞‍ds/p(s)<+∞ and ρ>0, the boundary value problem
(2.1)(p(t)u′(t))′+y(t)=0,0<t<+∞,α1u(0)-β1lim⁡t→0+p(t)u′(t)=T(u),α2lim⁡t→+∞u(t)+β2lim⁡t→+∞p(t)u′(t)=K(u),
has a unique solution for any y∈L[0,+∞). Moreover, this unique solution can be expressed in the form
(2.2)u(t)=∫0+∞‍G(t,s)y(s)ds+A(y)a(t)+B(y)b(t),
where G(t,s),A(y), and B(y) are defined by
(2.3)G(t,s)=ρ-1{a(t)b(s),0≤t≤s<∞,a(s)b(t),0≤s<t<∞,(a)A(y)=1Δ|T(∫0+∞‍G(τ,s)y(s)ds)ρ-T(b(τ))-K(∫0+∞‍G(τ,s)y(s)ds)K(b(τ))|,(b)B(y)=1Δ|K(∫0+∞‍G(τ,s)y(s)ds)ρ-K(b(τ))-T(∫0+∞‍G(τ,s)y(s)ds)T(b(τ))|.Proof.
a
(
t
) and b(t) in (1.3) are two linear independent solutions of the equation (p(t)u′(t))′=0, so the general solutions for the equation (p(t)u′(t))′+y(t)=0 can be expressed in the form
(2.4)u(t)=∫0+∞‍G(t,s)y(s)ds+Ca(t)+Db(t),
where C, D are undetermined constants. Through verifying directly, when C and D satisfy (a) and (b) separately, u(t) in (2.4) is a solution of BVP(2.1).
Now we need to prove that whenu(t) in (2.4) is a solution of BVP(2.1), C and D satisfy (a) and (b) separately.
Letu(t)=∫0+∞‍G(t,s)y(s)ds+Ca(t)+Db(t) be a solution of BVP(2.1), then
(2.5)u(t)=∫0t‍1ρa(s)b(t)y(s)ds+∫t+∞‍1ρa(t)b(s)y(s)ds+Ca(t)+Db(t),u′(t)=b′(t)ρ∫0t‍a(s)y(s)ds+a′(t)ρ∫t+∞‍b(s)y(s)ds+Cα1p(t)-Dα2p(t)=1p(t)(-α2ρ∫0t‍a(s)y(s)ds+α1ρ∫t+∞‍b(s)y(s)ds+Cα1-Dα2),(p(t)u′(t))′=-α2ρa(t)y(t)-α1ρb(t)y(t)=-α2a(t)+α1b(t)ρy(t)=-y(t).
That is, (p(t)u′(t))′+y(t)=0.
By (2.4), we have
(2.6)u(0)=β1ρ∫0+∞‍b(s)y(s)ds+Cβ1+Dβ2+Dα2∫0+∞‍dsp(s),u′(0)=1p(0)(α1ρ∫0+∞‍b(s)y(s)ds+Cα1-Dα2),u(∞)=β2ρ∫0+∞‍a(s)y(s)ds+Cβ1+Dβ2+Cα1∫0+∞‍dsp(s),u′(∞)=1p(∞)(-α2ρ∫0+∞‍a(s)y(s)ds+Cα1-Dα2),
then
(2.7)Dρ=T(∫0+∞‍G(τ,s)y(s)ds)+CT(a(τ))+DT(b(τ)),Cρ=K(∫0+∞‍G(τ,s)y(s)ds)+CK(a(τ))+DK(b(τ)).
From (2.7), we obtain that C and D satisfy (a) and (b) separately. The proof is completed.Remark 2.2.
Assume that(H2) holds. Then 0≤A(y)<+∞,0≤B(y)<+∞ for any y≥0 and any solution u(t) of BVP(2.1) is nonnegative.Lemma 2.3.
From (1.3) and (2.3), it is easy to get the following properties.(1)
G(t,s)/ρ-1[1+a(t)b(t)]≤1,a(t)/1+a(t)b(t)<1/b(t)≤1/β2,b(t)/1+a(t)b(t)<1/a(t)≤1/β1.(2)
G-(s)=limt→+∞G(t,s)=(β2/ρ)a(s)<+∞.(3)
G(t,s)≤G(s,s)≤a(s)b(s)/ρ<+∞.Lemma 2.4.
For any constant0<a*<b*<∞, there exists 0<c*<1, such that, for τ,s∈[0,∞),G(t,s)/ρ-1[1+a(t)b(t)]≥c*G(τ,s)/ρ-1[1+a(τ)b(τ)],a(t)/ρ-1[1+a(t)b(t)]≥c*a(τ)/ρ-1[1+a(τ)b(τ)],b(t)/ρ-1[1+a(t)b(t)]≥c*b(τ)/ρ-1[1+a(τ)b(τ)],t∈[a*,b*].Proof.
By (1.3), it is obvious that a(t) is increasing, and b(t) is decreasing on t∈[0,+∞); therefore, by (2.3), we have
(2.8)G(t,s)ρ-1[1+a(t)b(t)]={a(t)b(s)1+a(t)b(t)≥a(a*)β21+a(b*)b(a*),t≤s,a(s)b(t)1+a(t)b(t)≥b(b*)β11+a(b*)b(a*),s≤t.
We take c*=min{a(a*)β2/(1+a(b*)b(a*)),b(b*)β1/(1+a(b*)b(a*))}, then 0<c*<1; this is because that
(2.9)a(a*)β21+a(b*)b(a*)≤a(b*)b(a*)1+a(b*)b(a*)<1,b(b*)β11+a(b*)b(a*)≤a(b*)b(a*)1+a(b*)b(a*)<1.
By Lemma 2.3(1), we have G(τ,s)/ρ-1[1+a(τ)b(τ)]≤1, then
(2.10)G(t,s)ρ-1[1+a(t)b(t)]≥c*≥c*G(τ,s)ρ-1[1+a(τ)b(τ)],a(t)1+a(t)b(t)≥a(a*)1+a(b*)b(a*)=a(a*)β21+a(b*)b(a*)1β2≥c*1b(t)>c*a(τ)1+a(τ)b(τ).
Similarly, we can obtain that b(t)/(1+a(t)b(t))≥c*(b(τ)/(1+a(τ)b(τ))). The proof is completed.In this paper, we use the space(2.11)E={u∈C1(R+):sup⁡t∈[0,+∞)|u(t)|ρ-1[1+a(t)b(t)]<+∞,sup⁡t∈[0,+∞)|u′(t)|<+∞}
with the norm ∥u∥=max{∥u∥1,∥u′∥∞}, where ∥u∥1=supt∈[0,+∞)|u(t)|/ρ-1[1+a(t)b(t)] and ∥u∥∞=supt∈[0,+∞)|u(t)|, then (E,∥u∥) is a Banach space.Let(2.12)P={u∈E:u(t)≥0,min⁡t∈[a*,b*]u(t)ρ-1[1+a(t)b(t)]≥c*u(τ)ρ-1[1+a(τ)b(τ)],τ∈R+}.
Clearly P is a cone of E.Lemma 2.5 (see [10]).
LetM⊆Cl(R+,R)={x∈C(R+,R)∣limt→+∞x(t) exists}, then M is precompact if the following conditions hold:(a)
M is bounded in Cl;(b)
the functions belonging toM are locally equicontinuous on any interval of R+;(c)
the functions fromM are equiconvergent; that is, given ɛ>0, there corresponds T(ɛ)>0 such that |x(t)-x(∞)|<ɛ for any t≥T(ɛ) and x∈M.We shall consider nonnegative continuous and concave functionalα on P; that is, α:P→[0,∞) is continuous and satisfies
(2.13)α(tx+(1-t)y)≥tα(x)+(1-t)α(y),∀x,y∈P,0≤t≤1.
We denote the set {x∈P∣a≤α(x),∥x∥≤b}(b>a>0) by P(α,a,b) and
(2.14)Pr={x∈P∣‖x‖<r}.
The key tool in our approach is the following Leggett-Williams fixed point theorem.Theorem 2.6 (see [11]).
LetT:Pc-→Pc be completely continuous and α a nonnegative continuous concave functional on P with α(x)≤∥x∥ for any x∈Pc-. Suppose that there exist 0<a<b<d≤c such that(c1)
{x∈P(α,b,d)∣α(x)>b}≠ϕ, and α(Tx)>b, for x∈P(α,b,d);(c2)
∥Tx∥<a, for x∈Pa-;(c3)
α(Tx)>b for x∈P(α,b,c) with ∥Tx∥>d.Then T has at least three fixed points x1,x2,x3, with
(2.15)‖x1‖<a,b<α(x2),‖x3‖>a,α(x3)<b.


## 3. Existence Results

Define the operatorT:P→P by
(3.1)(Tu)(t)=∫0+∞‍G(t,s)Φ(s)f(s,u(s),u′(s))ds+A(Φf)a(t)+B(Φf)b(t),0<t<∞.
Then u(t) is a fixed point of operator T if and only if u(t) is a solution of BVP(1.1).For convenience, we denoteδ,α(x) by
(3.2)0<δ≤a(a*)b(b*)1+a(b*)b(a*)∫a*b*‍Φ(s)ds,α(u)=min⁡t∈[a*,b*]u(t)ρ-1[1+a(t)b(t)],∀u∈P.Theorem 3.1.
Suppose that(H1),(H2) hold, and assume there exist 0<r1<b1<l1<r2 with l1=max{b1/c*,supt∈[0,+∞)(b1/c*p(t))}, such that(H3)
Q(y1,y2)≤min{r2/∫0+∞‍Φ(s)q(s)ds+A(Φq)/ρ-1β2+B(Φq)/ρ-1β1, r2/supt∈[0,+∞)(1/p(t))(∫0+∞‍Φ(s)q(s)ds+A(Φq)α1+B(Φq)α2)}, 0≤y1≤r2, |y2|≤r2,(H4)
Ψ(t,y1,y2)>b1/δ,t∈[a*,b*],b1≤y1≤r2,|y2|≤r2,(H5)
Q(y1,y2)<min{r1/∫0+∞‍Φ(s)q(s)ds+(A(Φq)/ρ-1β2)+(B(Φq)/ρ-1β1), r1/supt∈[0,+∞)(1/p(t))(∫0+∞‍Φ(s)q(s)ds+A(Φq)α1+B(Φq)α2)}, 0≤y1≤r1, |y2|≤r1.Then BVP(1.1) has at least three positive solutions u1, u2, and u3 with
(3.3)‖u1‖<r1,b1<α(u2),‖u3‖>r1,α(u3)<b1.Proof.
Firstly we prove thatT:P→P is continuous.
We will show thatT:P→P is well defined and T(P)⊂P. For all u(t)∈P, by (H2), Φ(t) and f are nonnegative functions, and we have Tu(t)≥0. From (H1), (H2), we obtain
(A)A(Φf)=1Δ|T(∫0+∞‍G(τ,s)Φ(s)f(s,u(s),u′(s))ds)ρ-T(b(τ))-K(∫0+∞‍G(τ,s)Φ(s)f(s,u(s),u′(s))ds)K(b(τ))|≤max⁡y1∈[0,‖u‖],|y2|≤‖u‖⁡Q(y1,y2)Δ|T(∫0+∞‍G(τ,s)Φ(s)q(s)ds)ρ-T(b(τ))-K(∫0+∞‍G(τ,s)Φ(s)q(s)ds)K(b(τ))|=A(Φq)max⁡y1∈[0,‖u‖],|y2|≤‖u‖Q(y1,y2).
In the same way, we have
(B)B(Φf)≤B(Φq)max⁡y1∈[0,‖u‖],|y2|≤‖u‖Q(y1,y2).
By Lemma 2.3(1), (A), (B), and (H1), for all u(t)∈P, we have
(3.4)(Tu)(t)ρ-1[1+a(t)b(t)]=∫0+∞‍G(t,s)ρ-1[1+a(t)b(t)]Φ(s)f(s,u(s),u′(s))ds+A(Φf)a(t)ρ-1[1+a(t)b(t)]+B(Φf)b(t)ρ-1[1+a(t)b(t)]≤∫0+∞‍Φ(s)f(s,u(s),u′(s))ds+A(Φf)ρ-1β2+B(Φf)ρ-1β1≤max⁡y1∈[0,‖u‖],|y2|≤‖u‖Q(y1,y2)(∫0+∞‍Φ(s)q(s)ds+A(Φq)ρ-1β2+B(Φq)ρ-1β1)<+∞,(3.5)|(Tu)′(t)|=1p(t)|∫0t‍-α2a(s)ρΦ(s)f(s,u(s),u′(s))ds1p(t)+∫t+∞‍α1b(s)ρΦ(s)f(s,u(s),u′(s))ds+A(Φf)α1-B(Φf)α2|≤sup⁡t∈[0,+∞)1p(t)max⁡y1∈[0,‖u‖],|y2|≤‖u‖Q(y1,y2)×(∫0+∞‍Φ(s)q(s)ds+A(Φq)α1+B(Φq)α2)<+∞.
Hence, T:P→P is well defined. By (3.1), (H1), the Lebesgue dominated convergence theorem and the continuity of p(t), for any u∈P,t1,t2∈R+, we have
(3.6)|(Tu)′(t1)-(Tu)′(t2)|≤α2a(∞)ρ|1p(t1)-1p(t2)|∫0t1‍Φ(s)f(s,u(s),u′(s))ds+α2a(∞)ρp(t2)∫t1t2‍Φ(s)f(s,x(s),x′(s))ds+α1b(0)ρ|1p(t1)-1p(t2)|∫0+∞‍Φ(s)f(s,u(s),u′(s))ds+α1b(0)ρp(t2)∫t1t2‍Φ(s)f(s,x(s),x′(s))ds+(A(Φf)α1+B(Φf)α2)|1p(t1)-1p(t2)|⟶0,ast1⟶t2.
That is, (Tu)(t)∈C1(R0+); therefore, (Tu)(t)∈E.
By Lemma2.4, we have
(3.7)min⁡t∈[a*,b*](Tu)(t)ρ-1[1+a(t)b(t)]=min⁡t∈[a*,b*](∫0+∞‍G(t,s)ρ-1[1+a(t)b(t)]Φ(s)f(s,u(s),u′(s))ds=min⁡t∈[a*,b*]+a(t)A(Φf)ρ-1[1+a(t)b(t)]+b(t)B(Φf)ρ-1[1+a(t)b(t)])≥c*(∫0+∞‍G(τ,s)ρ-1[1+a(τ)b(τ)]Φ(s)f(s,u(s),u′(s))dsc*=+a(τ)A(Φf)ρ-1[1+a(t)b(t)]+b(τ)B(Φf)ρ-1[1+a(t)b(t)])=c*(Tu)(τ)ρ-1[1+a(τ)b(τ)],
therefore T:P→P.
We show thatT:P→P is continuous. In fact suppose {um}⊆P,u0∈P and um→u0(m→+∞), then there exists M>0, such that ∥um∥≤M. By (H1), we have
(3.8)∫0+∞‍Φ(s)|f(s,um(s),um′(s))-f(s,u0(s),u0′(s))|ds≤2∫0+∞‍Φ(s)f(s,u(s),u′(s))ds≤2max⁡y1∈[0,M],|y2|≤MQ(y1,y2)×∫0+∞‍Φ(s)q(s)ds<+∞.
Therefore, by Lemma 2.3(1), the continuity of f and the Lebesgue dominated convergence theorem imply that
(3.9)|(Tum)(t)-(Tu0)(t)ρ-1[1+a(t)b(t)]|=|∫0+∞‍G(t,s)ρ-1[1+a(t)b(t)]=×Φ(s)[f(s,um(s),um′(s))-f(s,u0(s),u0′(s))]ds∫0+∞‍|≤∫0+∞‍Φ(s)|f(s,um(s),um′(s))-f(s,u0(s),u0′(s))|ds⟶0,m⟶+∞,|(Tum)′(t)-(Tu0)′(t)|≤sup⁡t∈[0,+∞)1p(t)∫0+∞‍Φ(s)|f(s,um(s),um′(s))-f(s,u0(s),u0′(s))|ds⟶0,m⟶+∞.
Thus, ∥Tum-Tu0∥→0(m→+∞). Therefore T:P→P is continuous.
Secondly we show thatT:P→P is compact operator.
For any bounded setB⊂P, there exists a constant L>0 such that ∥u∥≤L, for all u∈B. By Lemma 2.3(1), (A), (B), and (H1), we have
(3.10)(Tu)(t)=ρ-1[1+a(t)b(t)](Tu)(t)ρ-1[1+a(t)b(t)]≤ρ-1[1+a(∞)b(0)](∫0+∞‍G(t,s)ρ-1[1+a(t)b(t)]Φ(s)f(s,u(s),u′(s))ds+A(Φf)a(t)ρ-1[1+a(t)b(t)]+B(Φf)b(t)ρ-1[1+a(t)b(t)])≤ρ-1[1+a(∞)b(0)](∫0+∞‍Φ(s)f(s,u(s),u′(s))ds+A(Φf)ρ-1β2+B(Φf)ρ-1β1)≤ρ-1[1+a(∞)b(0)]max⁡y1∈[0,L],|y2|≤LQ(y1,y2)(∫0+∞‍Φ(s)q(s)ds+A(Φq)ρ-1β2+B(Φq)ρ-1β1)<+∞,(Tu)(∞)=∫0+∞‍G-(s)Φ(s)f(s,u(s),u′(s))ds+A(Φf)a(∞)+B(Φf)b(∞)=β2ρ∫0+∞‍a(s)Φ(s)f(s,u(s),u′(s))ds+A(Φf)a(∞)+B(Φf)b(∞)≤max⁡y1∈[0,L],|y2|≤LQ(y1,y2)(β2a(∞)ρ∫0+∞‍Φ(s)q(s)ds+A(Φq)a(∞)+B(Φq)b(∞))<+∞.
Therefore, (Tu)(t)⊆Cl(R+,R).
By (3.4) and (3.5), we have
(3.11)‖Tu‖1=sup⁡t∈[0,+∞)(Tu)(t)ρ-1[1+a(t)b(t)]≤max⁡y1∈[0,L],|y2|≤LQ(y1,y2)(∫0+∞‍Φ(s)q(s)ds+A(Φq)ρ-1β2+B(Φq)ρ-1β1)<+∞,‖(Tu)′‖∞=max⁡t∈[0,+∞)|(Tu)′(t)|≤sup⁡t∈[0,+∞)1p(t)max⁡y1∈[0,L],|y2|≤LQ(y1,y2)(∫0+∞‍Φ(s)q(s)ds+A(Φq)α1+B(Φq)α2)<+∞,
so TB is bounded.
GivenT>0,t1,t2∈[0,T], by (H1) and Lemma 2.3(1), we have
(3.12)|G(t1,s)ρ-1[1+a(t1)b(t1)]-G(t2,s)ρ-1[1+a(t2)b(t2)]|Φ(s)f(s,u(s),u′(s))≤2max⁡y1∈[0,L],|y2|≤LQ(y1,y2)×Φ(s)q(s).
Therefore for any u∈B, by (3.1), the Lebesgue dominated convergence theorem and the continuity of G(t,s), a(t), and b(t), we have
(3.13)|(Tu)(t1)ρ-1[1+a(t1)b(t1)]-(Tx)(t2)ρ-1[1+a(t2)b(t2)]|≤∫0+∞‍|G(t1,s)ρ-1[1+a(t1)b(t1)]-G(t2,s)ρ-1[1+a(t2)b(t2)]|×Φ(s)f(s,u(s),u′(s))ds+A(Φf)|a(t1)ρ-1[1+a(t1)b(t1)]-a(t2)ρ-1[1+a(t2)b(t2)]|+B(Φf)|b(t1)ρ-1[1+a(t1)b(t1)]-b(t2)ρ-1[1+a(t2)b(t2)]|⟶0,ast1⟶t2.
By a similar proof as (3.6), we obtain |(Tu)′(t1)-(Tu)′(t2)|→0, as t1→t2. Thus, TB is equicontinuous on [0,T]. Since T>0 is arbitrary, TB is locally equicontinuous on [0,+∞).
By Lemma2.3(2), (H2) and the Lebesgue dominated convergence theorem, we obtain
(3.14)lim⁡t→+∞|(Tu)(t)ρ-1[1+a(t)b(t)]|=1ρ-1[1+a(∞)b(∞)]|∫0+∞‍β2a(s)Φ(s)f(s,x(s),x′(s))ds+A(Φf)a(∞)+B(Φf)b(∞)|≤max⁡y1∈[0,L],|y2|≤L⁡Q(y1,y2)ρ-1(1+β1β2)(β2a(∞)∫0+∞‍Φ(s)q(s)ds+A(Φq)a(∞)+B(Φq)b(∞))<+∞,|(Tu)(t)ρ-1[1+a(t)b(t)]-(Tu)(∞)ρ-1[1+a(∞)b(∞)]|≤∫0t‍a(s)b(t)|11+a(t)b(t)-11+a(∞)b(∞)|Φ(s)f(s,x(s),x′(s))ds+∫0t‍a(s)1+a(∞)b(∞)|b(t)-β2|Φ(s)f(s,x(s),x′(s))ds+∫t+∞‍b(s)|a(t)-a(s)|1+a(t)b(t)Φ(s)f(s,x(s),x′(s))ds+∫t+∞‍a(s)b(s)|11+a(t)b(t)-11+a(∞)b(∞)|Φ(s)f(s,x(s),x′(s))ds+∫t+∞‍a(s)1+a(∞)b(∞)|b(s)-β2|Φ(s)f(s,x(s),x′(s))ds+A(Φf)|a(t)-a(∞)|ρ-1[1+a(t)b(t)]+B(Φf)|b(t)-b(∞)|ρ-1[1+a(t)b(t)]+[A(Φf)a(∞)+B(Φf)b(∞)]×|1ρ-1[1+a(t)b(t)]-1ρ-1[1+a(∞)b(∞)]|≤max⁡y1∈[0,M],|y2|≤MQ(y1,y2)×{b(0)a(∞)∫0t‍|11+a(t)b(t)-11+a(∞)b(∞)|Φ(s)q(s)ds+a(∞)1+a(∞)b(∞)∫0t‍|b(t)-β2|Φ(s)q(s)ds+b(0)1+β1β2∫t∞‍|a(t)-a(s)|Φ(s)q(s)ds+a(∞)b(0)∫t∞‍|11+a(t)b(t)-11+a(∞)b(∞)|Φ(s)q(s)ds+a(∞)1+a(∞)b(∞)∫t+∞‍|b(s)-β2|Φ(s)q(s)ds+A(Φq)|a(t)-a(∞)|ρ-1[1+a(t)b(t)]+B(Φq)|b(t)-b(∞)|ρ-1[1+a(t)b(t)]+(A(Φq)a(∞)+B(Φq)b(∞))×|1ρ-1[1+a(t)b(t)]-1ρ-1[1+a(∞)b(∞)]|}⟶0,ast⟶+∞.
By (3.5), we know that limt→+∞|(Tu)′(t)|<+∞, then
(3.15)|(Tu)′(t)-(Tu)′(∞)|=|1p(t)∫0t‍-α2a(s)ρΦ(s)f(s,u(s),u′(s))ds+1p(t)∫t+∞‍α1b(s)ρΦ(s)f(s,u(s),u′(s))ds+1p(t)A(Φf)α1-1p(t)B(Φf)α2+1p(∞)∫0t‍α2a(s)ρΦ(s)f(s,u(s),u′(s))ds+1p(∞)∫t+∞‍α2a(s)ρΦ(s)f(s,u(s),u′(s))ds-1p(∞)A(Φf)α1+1p(∞)B(Φf)α2|≤max⁡y1∈[0,L],|y2|≤LQ(y1,y2)|1p(t)-1p(∞)|×{∫0t‍α2a(s)ρΦ(s)q(s)ds+1p(t)∫t+∞‍α1b(s)ρΦ(s)q(s)ds+1p(∞)××∫t+∞‍α2a(s)ρΦ(s)q(s)ds+(A(Φq)α1+B(Φq)α2)|1p(t)-1p(∞)|}⟶0,ast⟶+∞.
Therefore, TB is equiconvergent at ∞. By Lemma 2.5, TB is completely continuous.
Finally we will show that all conditions of Theorem2.6 hold.
From the definition ofα, we can get α(u)≤∥u∥ for all u∈P. For all u∈Pr2-, we have ∥u∥≤r2; therefore 0≤y1≤r2, |y2|≤r2. By (3.4), (3.5), and (H3), we have(3.16)|(Tu)(t)|ρ-1[1+a(t)b(t)]≤max⁡y1∈[0,r2],|y2|≤r2Q(y1,y2)(∫0+∞‍Φ(s)q(s)ds+A(Φq)ρ-1β2+B(Φq)ρ-1β1)≤r2,|(Tu)′(t)|≤sup⁡t∈[0,+∞)1p(t)max⁡y1∈[0,r2],|y2|≤r2Q(y1,y2)×(∫0+∞‍Φ(s)q(s)ds+A(Φq)α1+B(Φq)α2)≤r2,
that is, ∥Tu∥≤r2 for u∈Pr2-. Thus T:Pr2-→Pr2-.
Similarly for anyu∈Pr1-, we have ∥Tu∥<r1, which means that condition (c2) of Theorem 2.6 holds.
In order to apply condition(c1) of Theorem 2.6, we choose u(t)=b1ρ-1[1+a(t)b(t)]/c*,t∈R0+, then ∥u∥≤l1; this is because
(3.17)‖u‖1=b1c*≤l1,‖u′‖∞=sup⁡t∈[0,+∞)|u′(t)|=sup⁡t∈[0,+∞)|b1ρ-1[a′(t)b(t)+a(t)b′(t)]c*|≤sup⁡t∈[0,+∞)1p(t)b1c*≤l1,
and α(u)=mint∈[a*,b*](u(t)/ρ-1[1+a(t)b(t)])=b1/c*>b1, which means that {u∈P(α,b1,l1)|α(u)>b1}≠ϕ. For all u∈P(α,b1,l1), we have α(u)≥b1 and ∥u∥≤l1, thus b1≤u(t)/ρ-1[1+a(t)b(t)]≤l1,|u′(t)|≤l1, that is, b1≤y1≤l1,|y2|≤l1. By (H4), we can get
(3.18)α(Tu(t))=min⁡t∈[a*,b*](Tu)(t)ρ-1[1+a(t)b(t)]≥min⁡t∈[a*,b*]1ρ-1[1+a(t)b(t)]×(∫0a*‍a(s)b(t)ρΦ(s)f(s,u(s),u′(s))ds=××+∫a*t‍a(s)b(t)ρΦ(s)f(s,u(s),u′(s))ds+∫tb*‍a(t)b(s)ρΦ(s)f(s,u(s),u′(s))ds=××+∫b*+∞‍a(t)b(s)ρΦ(s)f(s,u(s),u′(s))ds)>a(a*)b(b*)1+a(b*)b(a*)∫a*b*‍Φ(s)f(s,u(s),u′(s))ds>a(a*)b(b*)1+a(b*)b(a*)∫a*b*‍Φ(s)dsb1δ≥b1.
Consequently condition (c1) of Theorem 2.6 holds.
We will prove that condition(c3) of Theorem 2.6 holds. If u∈P(α,b1,r2), and ∥Tu(t)∥>l1, by (H4), we have
(3.19)α(Tu(t))=min⁡t∈[a*,b*](Tu)(t)ρ-1[1+a(t)b(t)]>a(a*)b(b*)1+a(b*)b(a*)∫a*b*‍Φ(s)dsb1δ≥b1.
Therefore, condition (c3) of Theorem 2.6 is satisfied. Then we can complete the proof of this theorem by Leggett-Williams fixed point theorem.Theorem 3.2.
Suppose that(H1), (H2) hold, and assume there exist 0<r1<b1<l1<r2<b2<l2<r3<···<rm with li=max{bi/c*,supt∈[0,+∞)bi/c*p(t)}, such that(H6)
Q(y1,y2)<min{ri/∫0+∞‍Φ(s)q(s)ds+A(Φq)/ρ-1β2+B(Φq)/ρ-1β1, ri/supt∈[0,+∞)(1/p(t))(∫0+∞‍Φ(s)q(s)ds+A(Φq)α1+B(Φq)α2)}, 0≤y1≤ri, |y2|≤ri, 1≤i≤m,(H7)
Ψ(t,y1,y2)>bi/δ, t∈[a*,b*], bi≤y1≤ri+1, |y2|≤ri+1, 1≤i≤m-1.Then BVP(1.1) has at least 2m-1 positive solutions.Proof.
Whenm=1, it follows from (H6) that T has at least one positive solution by the Schauder fixed point theorem. When m=2, it is clear that Theorem 3.1 holds. Then we can obtain three positive solutions. In this way, we can finish the proof by the method of induction.


## 4.  Example

Consider the following singular Sturm-Liouville singular boundary value problems for second-order differential equation on the half-line(4.1)((1+t)2u′(t))′+e-t(1+t)tf(t,u(t),u′(t))=0,0<t<+∞,u(0)-lim⁡t→0+p(t)u′(t)=∑i=1m-2‍(13)iu(ξi),0<ξi<+∞,lim⁡t→+∞u(t)+lim⁡t→+∞p(t)u′(t)=∫0+∞‍13e-s(1+s)u(s)ds,
where
(4.2)f(t,u(t),u′(t))=Ψ(t,y1,y2)={y14+1550|y2|,y1≤1,1+1550|y2|,y1≥1,p(t)=(1+t)2,α1=α2=β1=β2=1,a(t)=2-1/(1+t),b(t)=1+1/(1+t),Φ(t)=e-t(1+t)/t which is singular at t=0,ρ=3,T(u)=∑i=1m-2‍(1/3)iu(ξi),K(u)=∫0+∞‍(1/3)e-s(1+s)u(s)ds.Setq(t)=1 and
(4.3)Q(y1,y2)={y14+1550|y2|,y1≤1,1+1550|y2|,y1≥1,
then ∫0+∞‍Φ(s)q(s)ds<3,a(0)=1,a(∞)=2,b(0)=2,b(∞)=1,1/2<T(a(τ))<1,1/2<T(b(τ))<1,K(a(τ))=K(b(τ))=1,Δ>3,A(Φq)<26/9,B(Φq)<20/9.Chooser1=1/3,b1=7/5,r2=19. When a*=1,b*=2, by the definition of δ, we may choose δ=8/5. By direct calculations, we imply that
(4.4)min⁡{r1∫0+∞‍Φ(s)q(s)ds+A(Φq)/ρ-1β2+B(Φq)/ρ-1β1,minminr1sup⁡t∈[0,+∞)(1/p(t))(∫0+∞‍Φ(s)q(s)ds+A(Φq)α1+B(Φq)α2)}>3r155,min⁡{r2∫0+∞‍Φ(s)q(s)ds+A(Φq)/ρ-1β2+B(Φq)/ρ-1β1,minminr2sup⁡t∈[0,+∞)(1/p(t))(∫0+∞‍Φ(s)q(s)ds+A(Φq)α1+B(Φq)α2)}>3r255,Q(y1,y2)≤(13)4+1550×13<155=3r155,for0≤y1≤13,|y2|≤13,Q(y1,y2)≤1+19550<3×1955=3r255,for0≤y1≤19,|y2|≤19,Ψ(t,y1,y2)≥1>7/58/5=b1δ,fort∈[1,2],7/5≤y1≤19,|y2|≤19.Therefore, the conditions(H1)–(H5) hold. Applying Theorem 3.1 we conclude that BVP(4.1) has at least three positive solutions.



---

*Source: 653675-2012-07-11.xml*