# Multiplicity of Positive Solutions for a Singular Second-Order Three-Point Boundary Value Problem with a Parameter

Authors: Jian Liu; Hanying Feng; Xingfang Feng
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

This paper is concerned with the following second-order three-point boundary value problemu″t+β2ut+λqtft,ut=0, t∈0,1, u0=0, u(1)=δu(η), where β∈(0,π/2), δ>0, η∈(0,1), and λ is a positive parameter. First, Green’s function for the associated linear boundary value problem is constructed, and then some useful properties of Green’s function are obtained. Finally, existence, multiplicity, and nonexistence results for positive solutions are derived in terms of different values of λ by means of the fixed point index theory.


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## Body

## 1. Introduction

For given positive numbersη∈(0,1) and β∈(0,π/2), the existence, multiplicity, and nonexistence of positive solutions for the following boundary value problem (BVP for short)
(1)u′′(t)+β2u(t)+λq(t)f(t,u(t))=0,t∈(0,1)u(0)=0,u(1)=δu(η)
are considered, where λ is a positive parameter, f∈C([0,1]×[0+∞),[0+∞)), and q:(0,1)→[0,+∞) may be singular at t=0 and 1.A functionu(t)∈C2(0,1) is said to be a solution of BVP (1) if u satisfies BVP (1). Moreover, if u(t)>0 for any t∈(0,1), then u is said to be a positive solution of BVP (1).Due to a wide range of applications in physics and engineering, second-order boundary value problems have been extensively investigated by numerous researchers in recent years. The study of multipoint boundary value problems was initiated by Il’in and Moiseev [1]. Gupta studied three-point boundary value problems for nonlinear ordinary differential equations in [2]. Since then, nonlinear three-point boundary value problems have been studied by many authors using the fixed point index theorem, Leray-Schauder continuation theorem, nonlinear alternative of Leray-Schauder, coincidence degree theory, and fixed point theorem in cones. For details, the readers are referred to [3–7] and the references therein.In [8], positive solutions for the following three-point boundary value problem at resonance
(2)x′′(t)=f(t,x(t)),t∈(0,1),x′(0)=0,x(η)=x(1)
were studied. Han’s approach is to rewrite the original BVP as an equivalent one so that the Krasnosel’skii-Guo fixed point theorem can be applied and then the existence and multiplicity of positive solutions are investigated.Then in [9], Han considered the following three-point boundary value problem:
(3)x′′(t)+β2x(t)=h(t)f(t,x(t)),t∈(0,1),x′(0)=0,x(η)=x(1),
under some conditions concerning the first eigenvalue of the relevant linear operator, where η∈(0,1) is a constant and h(t) is allowed to be singular at t=0 and t=1. The existence of positive solutions is studied by means of fixed point index theory.Motivated by the above work, here we study the second-order three-point BVP (1). Under certain suitable conditions, the results of existence, multiplicity, and nonexistence of positive solutions for BVP (1) were established via the fixed point index theory.We make the following assumptions:(H1)
0<β<(π/2), sinβ-δsinβη>0, and δcos⁡βη-cos⁡β≥0;(H2)
sinβ(1-η)-δsinβη>0 and sinβα-δsinβη>0, where 0<α<(1/2);(H3)
q(t)≥0,q(t)≢0 for t∈(0,1) and ∫01‍q(s)ds<∞;(H4)
f(t,x) is nondecreasing in x and f(t,x)>0 for any (t,x)∈[0,1]×(0,+∞).The main results of the present paper are summarized as follows.Theorem 1.
Let(H1)–(H4) be fulfilled and suppose that
(4)f0:=lim⁡x→0+min⁡t∈(0,1)f(t,x)x=∞,f∞:=lim⁡x→+∞min⁡t∈(0,1)f(t,x)x=∞.
Then, there exists λ*>0 such that BVP (1) has at least two positive solutions for λ∈(0,λ*), at least one positive solution for λ=λ*, and no positive solution for λ>λ*.The remainder of this paper is arranged as follows. Green’s function of BVP (1) and its properties are given in Section 2, and some preliminaries are also presented. The proof of Theorem 1 is given in Section 3.


## 2. Preliminaries

In this section we collect some preliminary results that will be used in subsequent sections.Consider the linear boundary value problem(5)-u′′(t)-β2u(t)=h(t),t∈(0,1),u(0)=0,u(1)=δu(η).Lemma 2.
Assume that(H1) holds. Then, for each h∈C[0,1], BVP (5) has a unique solution
(6)u(t)=∫01G(t,s)h(s)ds,
where
(7)G(t,s)=1β(sinβ-δsinβη)×{[sin⁡β(1-s)+δsin⁡β(s-η)]sin⁡βt,ghhhhhhhhhhhhhhhhh0≤t≤s≤η,sin⁡βssinβ(1-t)+δsin⁡βssinβ(t-η),ghhhhhhhhhhhhhhhhhhhhhs≤t,s≤η,sin⁡β(1-s)sin⁡βt,hggght≤s,η≤s,sin⁡βssinβ(1-t)+δsin⁡βηsinβ(t-s),ghhhhhhhhhhhhhhhhhhhhη≤s≤t≤1.Proof.
Suppose that(8)G(t,s)=-{a1cos⁡βt+a2sinβt,0≤t≤s≤η,a3cos⁡βt+a4sinβt,s≤t,s≤η,a5cos⁡βt+a6sinβt,t≤s,η≤s,a7cos⁡βt+a8sinβt,η≤s≤t≤1.
According to the definition and properties of Green’s function, for anys∈[0,η], we have
(9)a1cos⁡⁡βs+a2sin⁡βs=a3cos⁡⁡βs+a4sin⁡βs,(-βa1sin⁡βs+a2βcos⁡⁡βs)-(-a3βsin⁡βs+a4βcos⁡⁡βs)=-1,
and thus
(10)a1-a3=1βsin⁡βs,a2-a4=-1βcos⁡⁡βs.
Then by using the boundary conditions, we have
(11)a1=0,a3cos⁡⁡β+a4sin⁡β=δ(a3cos⁡⁡βη+a4sin⁡βη).
Therefore
(12)a1=0,a2=a4-1βcos⁡βs=-sin⁡β(s-1)+δsin⁡β(η-s)β(δsin⁡βη-sin⁡β),a3=-1βsin⁡βs,a4=-sin⁡βs(cos⁡⁡β-δcos⁡⁡βη)β(δsin⁡βη-sin⁡β).
For any s∈[η,1], we have
(13)a5cos⁡⁡βs+a6sin⁡βs=a7cos⁡⁡βs+a8sin⁡βs,(-βa5sin⁡βs+a6βcos⁡⁡βs)-(-a7βsin⁡βs+a8βcos⁡⁡βs)=-1,
and hence
(14)a5-a7=1βsin⁡βs,a6-a8=-1βcos⁡⁡βs.
By using the boundary conditions, we have
(15)a5=0,a7cos⁡⁡β+a8sin⁡β=δ(a5cos⁡⁡βη+a6sin⁡βη).
Then
(16)a5=0,a6=sin⁡β(1-s)β(δsin⁡βη-sin⁡β),a7=-1βsin⁡βs,a8=δsin⁡βηcos⁡⁡βs-sin⁡βscos⁡⁡ββ(δsin⁡βη-sin⁡β).
Consequently, we can get Green’s function G(t,s), and the lemma is proved.Lemma 3.
There exist a continuous functiong:[0,1]→[0,∞) and a constant γ∈(0,1] such that(i)
if(H1) holds, 0≤G(t,s)≤g(s), t,s∈[0,1];(ii)
if(H1) and (H2) hold, G(t,s)≥γg(s), (t,s)∈[α,1-α]×[0,1].Proof.
Firstly, it is obvious thatG(t,s)≥0 for any (t,s)∈[0,1]×[0,1].
Next, we will give the continuous functiong(s) and the constant γ.
Let(17)ϕ(s)=s(1-s),H(t,s)=μϕ(s)-G(t,s).
In the first step, we try finding the upper bounds.
We only need to show that there existsμ=μ*>0 such that
(18)H(t,s)s≥t≥0,H(t,s)s≤t≥0,(t,s)∈[0,1]×[0,1].Case1. s∈[0,η].Ifs=0, then G(t,s)=0 and ϕ(s)=0; the conclusion is true.Ifs∈(0,η], then
(19)H(t,s)s≥t=μs(1-s)-[sin⁡β(1-s)+δsin⁡β(s-η)]sin⁡βtβ(sin⁡β-δsin⁡βη)≥μt(1-s)-sin⁡β(1-s)sin⁡βtβ(sin⁡β-δsin⁡βη)≥[μ-sin⁡β(1-s)sin⁡βt(1-s)(sin⁡β-δsin⁡βη)βt]t(1-s)≥[μ-sin⁡β(1-η)(1-η)(sin⁡β-δsin⁡βη)]t(1-s),so, for
(20)μ≥μ1:=sin⁡β(1-η)(1-η)(sin⁡β-δsin⁡βη),
we have H(t,s)s≥t≥0. Consider
(21)H(t,s)s≤t=μs(1-s)-[sinβ(1-t)+δsinβ(t-η)]sinβsβ(sinβ-δsinβη)≥μs(1-s)-[sin⁡β(1-s)+δsin⁡β(1-s)]sin⁡βsβ(sin⁡β-δsin⁡βη)≥μs(1-s)-(1+δ)sinβssinβ(1-s)β(sin⁡β-δsin⁡βη)≥[μ-(1+δ)sinβssinβ(1-s)βs(sin⁡β-δsin⁡βη)(1-s)]s(1-s)≥[μ-(1+δ)sin⁡β(1-η)(sinβ-δsin⁡βη)(1-η)]s(1-s),
so, for
(22)μ≥μ2:=(1+δ)sin⁡β(1-η)(sin⁡β-δsin⁡βη)(1-η),
we have H(t,s)s≤t≥0.
Case
2. s∈[η,1].Ifs=1, then G(t,s)=0 and ϕ(s)=0; the conclusion is true.Ifs∈[η,1), then
(23)H(t,s)s≥t=μs(1-s)-sin⁡β(1-s)sin⁡βtβ(sin⁡β-δsin⁡βη)≥μt(1-s)-sin⁡β(1-s)sin⁡βtβ(sin⁡β-δsin⁡βη)≥[μ-sinβ(1-s)sinβtβt(1-s)(sinβ-δsinβη)]t(1-s)≥[μ-sinβtt(sinβ-δsinβη)]t(1-s)≥[μ-βsinβ-δsinβη]t(1-s),so, for
(24)μ≥μ3:=βsinβ-δsinβη,
we have H(t,s)s≥t≥0. Consider
(25)H(t,s)s≤t=μs(1-s)-sinβssinβ(1-t)+δsin⁡βηsin⁡β(t-s)β(sin⁡β-δsin⁡βη)≥μs(1-s)-sinβssinβ(1-s)+δsinβηsinβ(1-s)β(sin⁡β-δsin⁡βη)≥[μ-(sin⁡βs+δsin⁡βη)sin⁡β(1-s)(sin⁡β-δsin⁡βη)βs(1-s)]s(1-s)≥[μ-(1+δ)sin⁡βs(sin⁡β-δsin⁡βη)s]s(1-s)≥[μ-(1+δ)sin⁡βη(sin⁡β-δsin⁡βη)η]s(1-s),
so, for
(26)μ≥μ4:=(1+δ)sin⁡βη(sin⁡β-δsin⁡βη)η,
we have H(t,s)s≤t≥0.
Thus, we takeμ=μ*≥max⁡⁡{μ1,μ2,μ3,μ4} and then H(t,s)≥0 for (t,s)∈[0,1]×[0,1], and accordingly μ*ϕ(s)≥G(t,s).
Setg(s):=μ*ϕ(s) and then G(t,s)≤g(s), t,s∈[0,1].
In the next step, we try finding the lower bounds.
We only need to show that there existsμ=μ*>0 such that
(27)H(t,s)s≥t≤0,H(t,s)s≤t≤0,(t,s)∈[α,1-α]×[0,1].Case1. s∈[0,η].Ifs=0, then G(t,s)=0 and ϕ(s)=0; the conclusion is true.Ifs∈(0,η], then
(28)H(t,s)s≥t=μs(1-s)-[sin⁡β(1-s)+δsin⁡β(s-η)]sin⁡βtβ(sin⁡β-δsin⁡βη)≤μ4-sin⁡β(1-η)-δsin⁡βηβ(sin⁡β-δsin⁡βη)sin⁡βα,so, for
(29)μ≤μ5:=4sin⁡βαsin⁡β(1-η)-δsin⁡βηβ(sin⁡β-δsin⁡βη),
we have H(t,s)s≥t≤0. Consider
(30)H(t,s)s≤t=μs(1-s)-[sin⁡β(1-t)+δsinβ(t-η)]sin⁡βsβ(sin⁡β-δsin⁡βη)≤[μ-(sinβα-δsinβη)sinβs(sinβ-δsinβη)βs]s≤[μ-(sinβα-δsinβη)sinβη(sinβ-δsinβη)βη]s,
so, for
(31)μ≤μ6:=(sin⁡βα-δsin⁡βη)sin⁡βη(sin⁡β-δsin⁡βη)βη,
we have H(t,s)s≤t≤0.
Case
2. s∈[η,1].
Ifs=1, then G(t,s)=0 and ϕ(s)=0; the conclusion is true.
Ifs∈[η,1), then
(32)H(t,s)s≥t=μs(1-s)-sin⁡β(1-s)sin⁡βtβ(sin⁡β-δsin⁡βη)≤[μ-sinβ(1-s)sinβαβ(1-s)(sinβ-δsinβη)](1-s)≤[μ-sin⁡β(1-η)sin⁡βαβ(1-η)(sin⁡β-δsin⁡βη)](1-s),
so, for
(33)μ≤μ7:=sinβ(1-η)sinβαβ(1-η)(sinβ-δsinβη),
we have H(t,s)s≥t≤0. Consider
(34)H(t,s)s≤t=μs(1-s)-sinβssinβ(1-t)+δsin⁡βηsin⁡β(t-s)β(sin⁡β-δsin⁡βη)≤[μ-sinβssinβαβs(sin⁡β-δsin⁡βη)]s≤[μ-sin⁡βsin⁡βαβ(sin⁡β-δsin⁡βη)]s,
so, for
(35)μ≤μ8:=sin⁡βsin⁡βαβ(sin⁡β-δsin⁡βη),
we have H(t,s)s≤t≤0.
Letμ=μ*, where  0<μ*≤min⁡⁡{μ5,μ6,μ7,μ8},   we have H(t,s)≤0  for  (t,s)∈[α,1-α]×[0,1]. Thus, μ*ϕ(s)≤G(t,s), Namely
(36)γg(s)≤G(t,s),(t,s)∈[α,1-α]×[0,1],
where γ:=μ*/μ*∈(0,1].
This completes the proof of the lemma.LetE=C[0,1] be equipped with norm ∥u∥=max⁡t∈[0,1]⁡|u(t)|; then (E,∥·∥) is a real Banach space.Define the coneP by
(37)P={u∈E:u(t)≥0,min⁡t∈[α,1-α]u(t)≥γ∥u∥};
then P is a nonempty closed subset of E.Foru,v∈E, we write u≤v if u(t)≤v(t) for any t∈[0,1]. For any r>0, let Kr={u∈E:∥u∥<r} and ∂Kr={u∈E:∥u∥=r}.Define the operatorT:P→E by
(38)(Tu)(t)=∫01G(t,s)q(s)f(s,u(s))ds.Lemma 4.
Assume that(H1)–(H4) hold; then the operator T:P→P is completely continuous.Proof.
For for allu∈P, it follows from the definition of T and Lemma 3 that
(39)0≤Tu(t)=∫01G(t,s)q(s)f(s,u(s))ds≤∫01g(s)q(s)f(s,u(s))ds,t∈[0,1].
So,
(40)∥Tu∥≤∫01g(s)q(s)f(s,u(s))ds.
In view of Lemma3 and (40), we have
(41)Tu(t)=∫01G(t,s)q(s)f(s,u(s))ds≥γ∫01g(s)q(s)f(s,u(s))ds,t∈[α,1-α].
And so
(42)min⁡α≤t≤1-αTu(t)≥γ∥Tu∥,
which shows that T(P)⊂P. By the Ascoli-Arzela theorem, it is easy to show that T:P→P is completely continuous.
In view of Lemmas2 and 3, it is easy to see that u∈E is a solution of BVP (5) if and only if u∈E is a fixed point of the operator λT.
The proofs of our main results are based on the fixed point index theory. The following three well-known lemmas in [10, 11] are needed in our argument.Lemma 5.
LetE be a Banach space and P⊂E a cone in E. Assume that Ω is a bounded open subset of E. Suppose that T:P∩Ω¯→P is a completely continuous operator. If there exists x0∈P∖{θ} such that x-Tx≠μx0, for all x∈P∩∂Ω and μ≥0, then the fixed point index i(T,P∩Ω,P)=0.Lemma 6.
LetE be a Banach space and P⊂E a cone in E. Assume that Ω is a bounded open subset of E. Suppose that T:P∩Ω¯→P is a completely continuous operator. If inf⁡x∈P∩∂Ω∥Tx∥>0 and μTx≠x, for x∈P∩∂Ω and μ≥1, then the fixed point index i(T,P∩Ω,P)=0.Lemma 7.
LetE be a Banach space and P⊂E a cone in E. Assume that Ω is a bounded open subset of E with θ∈Ω. Suppose that T:P∩Ω¯→P is a completely continuous operator. If Tx≠μx for all x∈P∩∂Ω and μ≥1, then the fixed point index i(T,P∩Ω,P)=1.


## 3. Proofs of the Main Results

For convenience, we firstly introduce the following notations.Φ={(λ,u):λ>0 and u∈PisapositivesolutionofBVP(1)};Λ={λ>0:thereexistsu∈Psuchthat(λ,u)∈Φ};λ*=sup⁡Λ;λ*=inf⁡Λ;A=∫α1-αg(s)q(s)ds.Lemma 8.
Suppose that(H1)–(H3) hold and f0=∞. Then Φ≠∅.Proof.
LetR>0 be fixed; then we can choose λ0>0 small enough such that λ0sup⁡u∈P∩K¯R∥Tu∥<R. It is easy to see that
(43)λ0Tu≠μu,∀u∈P∩∂KR,μ≥1.
By Lemma 7, it follows that
(44)i(λ0T,P∩KR,P)=1.
From f0=∞, it follows that there exists r∈(0,R) such that
(45)f(t,x)≥1λ0γ2Ax,∀x∈[0,r],t∈[α,1-α].
We may suppose that λ0T has no fixed point on P∩∂Kr. Otherwise, the proof is finished. Let e(t)≡1 for t∈[0,1]. Then e∈∂K1. We claim that
(46)u≠λ0Tu+μe,∀u∈P∩∂Kr,μ≥0.
In fact, if not, there exist u1∈P∩∂Kr and μ1≥0 such that u1=λ0Tu1+μ1e; then μ1>0. For u1∈P∩∂Kr and μ1>0, by Lemma 3 and (45), for t∈[α,1-α], we have
(47)u1(t)=(λ0Tu1)(t)+μ1e(t)=λ0∫01G(t,s)q(s)f(s,u1(s))ds+μ1≥γλ0∫α1-αg(s)q(s)f(s,u1(s))ds+μ1≥γλ01λ0γ2A∫α1-αg(s)q(s)u1(s)ds+μ1≥1A∥u1∥∫01g(s)q(s)ds+μ1=∥u1∥+μ1=r+μ1;
we get r≥r+μ1, which is a contradiction. Thus, (46) holds. It follows from Lemma 5 that
(48)i(λ0T,P∩Kr,P)=0.
By virtue of the additivity of the fixed point index, from (44) and (48), we have
(49)i(λ0T,P∩(KR∖K¯r),P)=i(λ0T,P∩KR,P)-i(λ0T,P∩Kr,P)=1,
which implies that the nonlinear operator λ0T has one fixed point u0∈P∩(KR∖K¯r). Therefore, (λ0,u0)∈Φ. The proof is complete.Lemma 9.
Suppose(H1)–(H4) hold and f0=f∞=∞. Then 0<λ*<∞.Proof.
By Lemma8, it is easy to see that λ*>0. It follows from (H4) and f0=f∞=∞ that there exists C>0 such that f(t,x)≥Cx for all x≥0 and t∈[0,1]. Let (λ,u)∈Φ; by the definition of cone P and Lemma 2, for t∈[α,1-α], we obtain that
(50)u(t)=(λTu)(t)=λ∫01G(t,s)q(s)f(s,u(s))ds≥λγC∫01g(s)q(s)u(s)ds≥λγ2C∥u∥∫α1-αg(s)q(s)ds=λγ2AC∥u∥.
So, ∥u∥≥λγ2AC∥u∥. We get λ≤(γ2AC)-1. This completes the proof of lemma.Lemma 10.
Suppose(H1)–(H4) hold and f0=f∞=∞. Then (0,λ*)⊂Λ. Moreover, for any λ∈(0,λ*), BVP (1) has at least two positive solutions.Proof.
For any fixedλ∈(0,λ*), we prove that λ∈Λ. By the definition of λ*, there exists λ2∈Λ, such that λ<λ2≤λ* and (λ2,u2)∈Φ. Let R<min⁡t∈[0,1]u2(t) be fixed. From the proof of Lemma 8, we see that there exist λ1<λ, r<R, and u1(t)∈P∩(KR∖K¯r) such that (λ1,u1)∈Φ. It is easy to see that 0<u1(t)<u2(t) for all t∈[0,1]. Then, by (H2), we have
(51)-u1′′(t)-β2u1(t)=λ1q(t)f(t,u1(t)),t∈(0,1),-u2′′(t)-β2u2(t)=λ2q(t)f(t,u2(t)),t∈(0,1).
Consider now the modified BVP:(52)-u′′(t)-β2u(t)=λq(t)f1(t,u(t)),t∈(0,1)u(0)=0,u(1)=δu(η),
where
(53)f1(t,u(t))={f(t,u1(t)),u(t)≤u1(t),f(t,u(t)),u1(t)<u(t)<u2(t),f(t,u2(t)),u(t)≥u2(t).
Clearly, the function λf1 is bounded for t∈[0,1] and u∈P and is continuous in u. Define the operator T1:E→E by
(54)(T1u)(t)=∫01G(t,s)q(s)f1(s,u(s))ds,mmmmmmmmmmmmu∈E,t∈[0,1].
Then T1:P→P is completely continuous and all the fixed points of operator λT1 are the solutions for BVP (52). It is easy to see that there exists r0>∥u2∥ such that ∥λT1u∥<r0 for any u∈P. From Lemma 7, we have
(55)i(λT1,P∩Kr0,P)=1.
Let
(56)U={u∈P:u1(t)<u(t)<u2(t),∀t∈[0,1]}.
We claim that if u∈P is a fixed point of operator λT1, then u∈U. In fact, if u=λT1u, then
(57)u(t)=(λT1u)(t)=λ∫01G(t,s)q(s)f1(s,u(s))ds<λ2∫01G(t,s)q(s)f(s,u2(s))ds=(λ2Tu2)(t)=u2(t),u(t)=(λT1u)(t)=λ∫01G(t,s)q(s)f1(s,u(s))ds>λ1∫01G(t,s)q(s)f(s,u1(s))ds=(λ1Tu1)(t)=u1(t).
From the excision property of the fixed point index and (55), we obtain that
(58)i(λT1,U,P)=i(λT1,P∩Kr0,P)=1.
From the definition of T1, we know that T1=T on U¯. Then,
(59)i(λT,U,P)=1.
Hence, the nonlinear operator λT has at least fixed point v1∈U. Then v1 is one positive solution of BVP (1). This gives λ∈Λ,(λ,v1)∈Φ and (0,λ)⊂Λ.
We now find the second positive solution of BVP (1). By f∞=∞ and the continuity of f(t,x) with respect to x, there exists C>0 such that
(60)f(t,x)≥2xλγ2A-CγA,∀x≥0,t∈[0,1].
For e(t)≡1, let
(61)Ω={u∈P:thereexistsτ≥0suchthatu=λTu+τe}.
We claim that Ω is bounded in E. In fact, for any u∈Ω, it follows from Lemma 3 and (60) that
(62)u(t)=(λTu)(t)+τe(t)=(λTu)(t)+τ≥λ∫01G(t,s)q(s)f(s,u(s))ds≥λγ∫α1-αg(s)q(s)[2u(s)λγ2A-CγA]ds≥λγ∫α1-αg(s)q(s)[2γ∥u∥λγ2A-CγA]ds=2∥u∥-λC,t∈[α,1-α].
This implies ∥u∥≤λC. Thus Ω is bounded in E. Therefore there exists R1>∥u2∥ such that
(63)u≠λTu+τe,∀u∈P∩∂KR1,τ≥0.
By Lemma 5, we get that
(64)i(λT,P∩KR1,P)=0.
Using a similar argument as in deriving (48), we have that
(65)i(λT,P∩Kr1,P)=0,
where 0<r1<min⁡t∈[0,1]u1(t). According to the additivity of the fixed point index and by (59), (64), and (65), we have
(66)i(λT,P∩(KR1∖(U¯∪K¯r1),P))=i(λT,P∩KR1,P)-i(λT,U,P)-i(λT,P∩Kr1,P)=-1,
which implies that the nonlinear operator λT has at least one fixed point v2∈P∩(KR1∖(U¯∪K¯r1)). Thus, BVP (1) has another positive solution. The proof is complete.Lemma 11.
Suppose(H1)–(H4) hold and f0=f∞=∞. Then Λ=(0,λ*].Proof.
In view of Lemma10, it suffices to prove that λ*∈Λ. By the definition of λ*, we can choose {λn}⊂Λ with λn≥(λ*/2)(n=1,2,…) such that λn→λ* as n→∞. By the definition of Λ, there exists {un}⊂P∖{θ} such that (λn,un)∈Φ. We now show that {un} is bounded. Supposing the contrary, then there exists a subsequence of {un} (still denoted by {un}) such that ∥un∥→∞ as n→∞. It follows from {un}⊂P∖{θ} that un≥γ∥un∥ for all t∈[0,1]. Choose sufficiently large τ such that
(67)λ*γ2Aτ2>1.
By f∞=∞, there exists R>0 such that f(t,u)≥τu for all u>γR and t∈[0,1]. Since ∥un∥→∞ as n→∞, there exists sufficiently large n0 such that ∥un0∥≥R. Thus, for t∈[α,1-α], we have
(68)un0(t)=(λn0Tun0)(t)=λn0∫01G(t,s)q(s)f(s,un0(s))ds≥λ*2γτ∫α1-αg(s)q(s)un0(s)ds,≥λ*2γ2τ∥un0∥∫α1-αg(s)q(s)ds=λ*2γ2τA∥un0∥.
This gives
(69)λ*γ2Aτ2≤1,
which contradicts the choice ofτ. Hence, {un} is bounded. It follows from the completely continuity of T that {Tun} is equicontinuous; that is, for each ɛ>0, there is a δ>0 such that
(70)|un(t1)-un(t2)|=λn|(Tun)(t1)-(Tun)(t2)|<λnɛ≤λ*ɛ,
where n=1,2,…,t1,t2∈[0,1] and |t1-t2|<δ. Then {un} is equicontinuous. According to the Ascoli-Arzela theorem, {un} is relatively compact. Hence, there exists a subsequence of {un} (still denoted by {un}) and u*∈P such that un→u* as n→∞. By un=λnTun, letting n→∞, we obtain that u*=λ*Tu*. If u*=θ, using a similar argument as in deriving (69) and by f0=∞, we also get a contradiction. Then u*∈P∖{θ}, and so λ*∈Λ. This completes the proof.Proof of Theorem1.
Theorem1 readily follows from Lemmas 8, 9, 10, and 11.



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*Source: 603203-2014-08-12.xml*