# Bifurcation from Interval and Positive Solutions of a Nonlinear Second-Order Dynamic Boundary Value Problem on Time Scales

Authors: Hua Luo
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

Let𝕋 be a time scale with 0,T∈𝕋. We give a global description of the branches of positive solutions to the nonlinear boundary value problem of second-order dynamic equation on a time scale 𝕋, uΔΔ(t)+f(t,uσ(t))=0,t∈[0,T]𝕋,u(0)=u(σ2(T))=0, which is not necessarily linearizable. Our approaches are based on topological degree theory and global bifurcation techniques.


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## Body

## 1. Introduction

Let𝕋 be a time scale with 0,T∈𝕋, we consider the existence of positive solutions, in this paper, for a nonlinear boundary value problem of second-order dynamic equation on a time scale 𝕋 as follows:
(1.1)uΔΔ(t)+f(t,uσ(t))=0,t∈[0,T]𝕋,u(0)=u(σ2(T))=0.
Research for the existence of solutions to the dynamic boundary value problem is rapidly growing in recent years. A great many existence results of positive solutions have been established for problem (1.1), see [1–5] and the references therein. The main tool used by them is the fixed point theorem in cones, and the key conditions in these papers do not depend on the first eigenvalue, λ1(a), of the following linear problem:
(1.2)uΔΔ(t)+λa(t)uσ(t)=0,t∈[0,T]𝕋,u(0)=u(σ2(T))=0,
and the corresponding existence conditions are not optimal.In 2006, for0,1∈𝕋, σ(0)=0,  ρ(1)=1,  f^∈C(ℝ) with sf^(s)>0 for s≠0, Luo and Ma [6] obtained the existence of at least one positive solution to problem:
(1.3)uΔΔ(t)+f^(uσ(t))=0,t∈[0,1]𝕋,u(0)=u(1)=0,
under the condition(H) if eitherf^0<μ1<f^∞ or f^∞<μ1<f^0, where
(1.4)f^0=lim|s|→0f^(s)s,f^∞=lim|s|→∞f^(s)s,and μ1 is the first eigenvalue of the linear problem:
(1.5)uΔΔ(t)+μuσ(t)=0,u(0)=u(1)=0.
The approaches adopted by Luo and Ma [6] are based on global bifurcation techniques. They obtained the existence of at least one positive solution by considering the branches of solutions, which bifurcate from one point. The key conditions in [6] depend on the first eigenvalue of the corresponding linear problem and the condition (H) is optimal!In this paper, we will use the following assumptions.(A1)f:[0,σ(T)]𝕋×[0,∞)→[0,∞) is continuous and there exist functions a0(·),a0(·),b∞(·),b∞(·)∈C([0,σ(T)]𝕋,(0,∞)), such that
(1.6)a0(t)u-ξ1(t,u)≤f(t,u)≤a0(t)u+ξ2(t,u)for some functionsξi∈C([0,σ(T)]𝕋×[0,∞),ℝ) with limu→0+(ξi(t,u)/u)=0  (i=1,2) uniformly for t∈[0,σ(T)]𝕋, and
(1.7)b∞(t)u-ζ1(t,u)≤f(t,u)≤b∞(t)u+ζ2(t,u)for some functionsζi∈C([0,σ(T)]𝕋×[0,∞),ℝ) with limu→+∞(ζi(t,u)/u)=0  (i=1,2) uniformly for t∈[0,σ(T)]𝕋.(A2)f(t,u)>0 for (t,u)∈[0,σ(T)]𝕋×(0,∞).(A3) There exists a functionc∈C([0,σ(T)]𝕋,(0,∞)) such that
(1.8)f(t,u)≥c(t)u,(t,u)∈[0,σ(T)]𝕋×[0,∞).Obviously, (A1) means thatf is not necessarily linearizable at 0 and +∞. We consider the existence of positive solutions of problem (1.1) in this paper by using bifurcation techniques. The difference from [6] is that the branches of positive solutions under consideration now bifurcate from not one point, but an interval. Our main idea is from [7], in which they considered positive solutions of fourth-order boundary value problems for differential equations. The main tool we will use is the following global bifurcation theorem for problems which is not necessarily linearizable.Theorem A (Rabinowitz, [8]).
LetV be a real reflexive Banach space. Let F:ℝ×V→V be completely continuous such that F(λ,0)=0, for all λ∈ℝ. Let a,b∈ℝ(a<b) be such that u=0 is an isolated solution of the equation
(1.9)u-F(λ,u)=0,u∈V,
for λ=a, and λ=b, where (a,0), (b,0) are not bifurcation points of (1.9). Furthermore, assuming that
(1.10)deg(I-F(a,·),Br(0),0)≠deg(I-F(b,·),Br(0),0),
where Br(0) is an isolating neighborhood of the nontrivial solution, and deg(I-F,Br(0),0) denote the degree of I-F on Br(0) with respect to 0. Let
(1.11)𝒮={(λ,u):(λ,u)isasolutionof(1.9)withu≠0}¯∪([a,b]×{0}).
Then there exists a connected component 𝒞 of 𝒮 containing [a,b]×{0}, and either(i)
𝒞 is unbounded, or(ii)
𝒞∩[(ℝ∖[a,b])×{0}]≠∅.The rest of the paper is organized as follows. In Section2, we firstly introduce the time scales concepts and notations that we will use in this paper. Next, Section 3 states some notations and proves some necessary preliminary results, and Section 4 studies the bifurcation from the trivial solution for a nonlinear problem which is not necessarily linearizable and then establishes our main result.


## 2. Introduction for Time Scales

A time scale𝕋 is a nonempty closed subset of ℝ, assuming that 𝕋 has the topology that it inherits from the standard topology on ℝ. Define the forward and backward jump operators σ,ρ:𝕋→𝕋 by
(2.1)σ(t)=inf{τ>t∣τ∈𝕋},ρ(t)=sup{τ<t∣τ∈𝕋}.
Here, we put inf∅=sup𝕋, sup∅=inf𝕋. Let 𝕋k which is derived from the time scale 𝕋 be
(2.2)𝕋k∶={t∈𝕋:tisnonmaximalorρ(t)=t},
and 𝕋k2:=𝕋kk. Define interval I on 𝕋 by I𝕋=I∩𝕋.Definition 2.1.
Ifu:𝕋→ℝ is a function and t∈𝕋k, then the Δ-derivative of u at the point t is defined to be the number uΔ(t) (provided that it exists) with the property that for each ε>0, there is a neighborhood U of t such that
(2.3)|u(σ(t))-u(s)-uΔ(t)(σ(t)-s)|⩽ε|σ(t)-s|
for all s∈U. The function u is called Δ-differentiable on 𝕋 if uΔ(t) exists for all t∈𝕋k.The second
Δ
-derivative of
u
at
t
∈
𝕋
k
2, if it exists, is defined to be uΔ2(t)=uΔΔ(t):=(uΔ)Δ(t). We also define the function uσ:=u∘σ and uρ:=u∘ρ.Definition 2.2.
IfUΔ=u holds on 𝕋k, we define the Cauchy Δ-integral by
(2.4)∫st‍u(τ)Δτ=U(t)-U(s),s,t∈𝕋k.Lemma 2.3 (see [2, Theorems 2.7 and 2.8]).
Assumea,b∈𝕋, then
(2.5)∫ab‍fΔ(t)g(t)Δt=f(t)g(t)|ab-∫ab‍fσ(t)gΔ(t)Δt.
Furthermore, if a≤b, f(t) is a continuous function on [a,b], then
(2.6)∫ab‍f(t)Δt=∫aρ(b)‍f(t)Δt+[b-ρ(b)]f(ρ(b)).Define the Banach spaceC(𝕋,ℝ)  (denoted by C(𝕋)) to be the set of continuous functions u:𝕋→ℝ with the norm
(2.7)∥u∥0=sup{|u(t)|∣t∈𝕋}.
For i=1,2, we define the Banach space Ci(𝕋) to be the set of the ith Δ-differential functions u:𝕋→ℝ for which uΔi∈C(𝕋ki) with the norm
(2.8)∥u∥i=sup{∥u∥∞,∥uΔ∥0,…,∥uΔi∥0},
where
(2.9)∥uΔj∥0=sup{|uΔj(t)|∣t∈𝕋kj},j=0,1,…,i.


## 3. Preliminaries and Necessary Lemmas

Assuming thata∈C([0,σ2(T)]𝕋,(0,∞)), then from [9, Theorem 2.9], linear problem
(3.1)uΔΔ(t)+λa(t)uσ(t)=0,t∈[0,T]𝕋,u(0)=u(σ2(T))=0
has a unique principal eigenvalue λ1(a), with a corresponding positive eigenfunction.LetE=C2[0,T]𝕋,  X=C[0,T]𝕋, and
(3.2)Y={y∈C1[0,σ2(T)]𝕋∣y(0)=y(σ2(T))=0}.
We will work essentially in the Banach space Y with the norm
(3.3)∥y∥=max{∥y∥0,∥yΔ∥0},
where
(3.4)∥y∥0=sup{|y(t)|∣t∈[0,σ2(T)]𝕋},∥yΔ∥0=sup{|yΔ(t)|∣t∈[0,σ(T)]𝕋}.
By a positive solution of problem (1.1), we mean u is a solution of (1.1) with u≥0 in (0,σ2(T))𝕋 and u≢0.Lemma 3.1.
Fory∈Y, we have
(3.5)∥y∥0≤σ(T)∥yΔ∥0,max{1,1σ(T)}∥y∥0≤∥y∥≤max{1,σ(T)}∥yΔ∥0.Proof.
Byy(0)=0, we have that
(3.6)y(t)=∫0t‍yΔ(s)Δs,t∈[0,σ(T)]𝕋
and so
(3.7)∥y∥0≤σ(T)∥yΔ∥0.
Therefore,
(3.8)∥y∥=max{∥y∥0,∥yΔ∥0}≤max{1,σ(T)}∥yΔ∥0,∥y∥≥max{∥y∥0,1σ(T)∥y∥0}=max{1,1σ(T)}∥y∥0.Define the linear operatorL:D(L)⊂E→X,
(3.9)Lu=-uΔΔ,u∈D(L)
with
(3.10)D(L)={u∈E∣u(0)=u(σ2(T))=0}.
Then L is a closed operator, and L-1:X→Y is completely continuous, see [10, Lemma 3.7].LetΣ⊂ℝ+×Y be the closure of the set of positive solutions to the problem
(3.11)Lu(t)=λf(t,uσ(t)),t∈[0,σ(T)]𝕋.
We extend the function f to a continuous function f- defined on [0,σ(T)]𝕋×ℝ by
(3.12)f-(t,u)={f(t,u),(t,u)∈[0,σ(T)]𝕋×[0,∞),f(t,0),(t,u)∈[0,σ(T)]𝕋×(-∞,0).
Then f-(t,u)≥0 on [0,σ(T)]𝕋×ℝ. For λ≥0, the arbitrary solution u to the eigenvalue problem
(3.13)uΔΔ(t)+λf-(t,uσ(t))=0,t∈[0,T]𝕋,u(0)=u(σ2(T))=0
satisfies that uΔΔ(t)≤0 on [0,T]𝕋, and consequently, the graph of u is concave down on [0,σ2(T)]𝕋. This together with the boundary conditions u(0)=u(σ2(T))=0 imply that
(3.14)u(t)≥0,t∈[0,σ2(T)]𝕋.
Thus, (3.14) implies that u is a nonnegative solution of problem (3.13), and the closure of the set of nontrivial solutions (λ,u) of (3.13) in ℝ+×Y is exactly Σ.Letg∈C([0,σ(T)]𝕋×ℝ,ℝ), and let N^:Y→X be the Nemytskii operator associated with the function g:
(3.15)N^(u)(t)=g(t,uσ(t)),u∈Y.Lemma 3.2.
Letg(t,u)≥0 on [0,σ(T)]𝕋×ℝ. Let u∈D(L) be such that Lu≥λN^(u) in [0,T]𝕋, λ≥0. Then
(3.16)u(t)≥0,t∈[0,σ2(T)]𝕋.
Moreover,
(3.17)u(t)>0,t∈(0,σ2(T))𝕋,
whenever u≢0.LetN:Y→X be the Nemytskii operator associated with the function f-(3.18)N(u)(t)=f-(t,uσ(t)),u∈Y.
Then (3.13), with λ≥0, is equivalent to the operator equation
(3.19)u=λL-1N(u),u∈Y.
In the following we will apply the Brouwer degree theory, mainly to the mapping Φλ:Y→Y,
(3.20)Φλ(u)=u-λL-1N(u).
For R>0, let BR={u∈Y:∥u∥<R}.Lemma 3.3.
LetΛ⊂ℝ+ be a compact interval with [λ1(a0),λ1(a0)]∩Λ=∅. Then there exists a number δ1>0 with the property
(3.21)Φλ(u)≠0,∀u∈Y:0<∥u∥≤δ1,∀λ∈Λ.Proof.
Suppose to the contrary that there exist sequences{μn} in Λ and {un} in Y:μn→μ*∈Λ,∥un∥>0 and un→0  (n→∞) in Y, such that Φμn(un)=0 for all n∈ℕ. By Lemma 3.2, un(t)≥0 for t∈[0,σ2(T)]𝕋.
Setvn=un/∥un∥. Then from Lun=μnN(un), we have Lvn=μn∥un∥-1N(un). Since ∥un∥-1N(un) is bounded in X, we infer that {vn} is relatively compact in Y, hence (for a subsequence) vn→v-(n→∞) with v-≥0 in Y, ∥v-∥=1. Let Iσ:Crd[0,σ2(T)]𝕋→Crd[0,σ(T)]𝕋 be defined as Iσ(y)=yσ. Then Iσ is linear. For u∈Y, ∥Iσu∥0=supt∈[0,σ(T)]𝕋|uσ(t)|=∥u∥0.
Now, from condition (A1), we have that(3.22)a0(t)unσ(t)-ξ1(t,unσ(t))≤f(t,unσ(t))≤a0(t)unσ(t)+ξ2(t,unσ(t)).
According to
(3.23)unσ(t)∥un∥=Iσ(un)∥un∥=Iσ(un∥un∥)=Iσ(vn)=vnσ,
we get
(3.24)μn(a0(t)vnσ(t)-ξ1(t,unσ(t))∥un∥)≤μnf(t,unσ(t))∥un∥≤μn(a0(t)vnσ(t)+ξ2(t,unσ(t))∥un∥).
Letφ0 and φ0 denote the eigenfunctions corresponding to λ1(a0) and λ1(a0), respectively. Denote ξ-1(y):=max(t,s)∈[0,σ(T)]𝕋×[0,y]|ξ1(t,s)|. Then ξ-1(y) is nondecreasing. From limu→0+(ξ1(t,u)/u)=0 uniformly for t∈[0,σ(T)]𝕋, we have
(3.25)limy→0+ξ-1(y)y=0.
According to Lun=μnf-(t,unσ(t)) and un(t)≥0, we have from the first inequality in (3.24) that
(3.26)∫0σ(T)‍[μn(a0(t)vnσ(t)-ξ1(t,unσ(t))∥un∥)]φ0σ(t)Δt≤∫0σ(T)‍Lun(t)∥un∥φ0σ(t)Δt=∫0σ(T)‍Lvn(t)φ0σ(t)Δt.
Notice that
(3.27)ξ1(t,unσ(t))∥un∥≤ξ-1(∥Iσun∥0)max{1,1/σ(T)}∥un∥0≤ξ-1(∥Iσun∥0)∥Iσun∥0→0(n→∞)
by Lemma 3.1. Let n→∞, by integration by parts (2.5), we have
(3.28)∫0σ2(T)‍μ*a0(t)v-σ(t)φ0σ(t)Δt≤∫0σ2(T)‍Lv-(t)φ0σ(t)Δt=∫0σ2(T)‍Lφ0(t)v-σ(t)Δt=∫0σ2(T)‍λ1(a0)a0(t)φ0σ(t)v-σ(t)Δt,
and consequently
(3.29)μ*≤λ1(a0).
Similarly, we deduce from the second inequality in (3.24) that
(3.30)λ1(a0)≤μ*.
Thus, λ1(a0)≤μ*≤λ1(a0). This contradicts μ*∈Λ.Remark 3.4.
Ifσ(T)<σ2(T), then the value of σ2(T) do not contribute to the value of (3.28). Thus we can discuss the integral from 0 to σ2(T). For Lv- and Lφ0 are not defined at σ(T), we may define the values of (Lv-)φ0σ and (Lφ0)v-σ to be zero at σ(T), since the functions φ0σ and v-σ are zero at σ(T). The details of the discussion can be found in [9, Page 497].Corollary 3.5.
Forλ∈(0,λ1(a0)) and δ∈(0,δ1), deg(Φλ,Bδ,0)=1.Proof.
Lemma3.3, applied to the interval Λ=[0,λ], guarantees the existence of δ1>0 such that for δ∈(0,δ1)(3.31)u-τλL-1N(u)≠0,∀u∈Y:0<∥u∥≤δ,τ∈[0,1].
Hence, for any δ∈(0,δ1),
(3.32)deg(Φλ,Bδ,0)=deg(I,Bδ,0)=1,
which implies the assertion.On the other hand, we have the following.Lemma 3.6.
Supposeλ>λ1(a0). Then there exists δ2>0 such that for allu∈Y with 0<∥u∥≤δ2, for allτ≥0,
(3.33)Φλ(u)≠τφ0,
where φ0 is the positive eigenfunction corresponding to λ1(a0).Proof.
We assume again to the contrary that there existτn≥0 and a sequence {un} with ∥un∥>0 and un→0 in Y such that Φλ(un)=τnφ0 for all n∈ℕ. As
(3.34)Lun=λN(un)+τnλ1(a0)a0(t)φ0σ(t)
and τnλ1(a0)a0(t)φ0σ(t)≥0 in [0,T]𝕋, we can conclude from Lemma 3.2 that un(t)≥0 for t∈[0,σ2(T)]𝕋.
Notice thatun∈D(L) has a unique orthogonal decomposition
(3.35)un=wn+snφ0,
with sn∈ℝ. Since un≥0 on [0,σ2(T)]𝕋 and ∥un∥>0, we have from (3.35) that sn>0.
Chooseσ>0 such that
(3.36)σ<λ-λ1(a0)λ.
By (A1), there exists r4>0, such that
(3.37)f(t,u)≥(1-σ)a0(t)u,∀(t,u)∈[0,σ(T)]𝕋×[0,r4].
Since ∥un∥→0, there exists N*>0, such that
(3.38)0≤un≤r4,∀n≥N*,
and consequently
(3.39)f(t,unσ(t))≥(1-σ)a0(t)unσ(t),∀n≥N*.
Applying (3.35) and (3.39), it follows that
(3.40)snλ1(a0)∫0σ2(T)‍a0(t)[φ0σ(t)]2Δt=∫0σ2(T)‍[wnσ(t)+snϕ0σ(t)]Lφ0(t)Δt=∫0σ2(T)‍unσ(t)Lφ0(t)Δt=∫0σ2(T)‍φ0σ(t)Lun(t)Δt=λ∫0σ2(T)‍N(un)φ0σ(t)Δt+τnλ1(a0)∫0σ2(T)‍a0(t)[φ0σ(t)]2Δt≥λ∫0σ2(T)‍N(un)φ0σ(t)Δt≥λ(1-σ)∫0σ2(T)‍a0(t)unσ(t)φ0σ(t)Δt=λ(1-σ)sn∫0σ2(T)‍a0(t)[φ0σ(t)]2Δt.
Thus,
(3.41)λ1(a0)≥λ(1-σ).
This contradicts (3.36).Corollary 3.7.
Forλ>λ1(a0) and δ∈(0,δ2), deg(Φλ,Bδ,0)=0.Proof.
Let0<ϵ≤δ2, where δ2 is the number asserted in Lemma 3.6. As Φλ is bounded in B-ϵ, there exists c>0 such that Φλ(u)≠cφ0, for allu∈B-ϵ. By Lemma 3.6,
(3.42)Φλ(u)≠tcφ0,u∈∂Bϵ,t∈[0,1].
Hence,
(3.43)deg(Φλ,Bϵ,0)=deg(Φλ-cφ0,Bϵ,0)=0.Now, using Theorem A, we may prove the following.Lemma 3.8.
[
λ
1
(
a
0
)
,
λ
1
(
a
0
)
] is a bifurcation interval from the trivial solution for (3.19). There exists an unbounded component 𝒞 of positive solutions of (3.19) which meets [λ1(a0),λ1(a0)]×{0}. Moreover,
(3.44)𝒞∩[(ℝ∖[λ1(a0),λ1(a0)])×{0}]=∅.Proof.
For fixedn∈ℕ with λ1(a0)-1/n>0, set an=λ1(a0)-1/n<λ1(a0),  bn=λ1(a0)+1/n>λ1(a0) and δ^=min{δ1,δ2}. It is easy to check that for 0<δ<δ^, all of the conditions of Theorem A are satisfied. So there exists a connected component 𝒞n of solutions of (3.19) containing [an,bn]×{0}, and either(i)
𝒞n is unbounded, or(ii)
𝒞n∩[(ℝ∖[an,bn])×{0}]≠∅.
By Lemma3.3, the case (ii) cannot occur. Thus, 𝒞n is unbounded bifurcated from [an,bn]×{0} in ℝ×Y. Furthermore, we have also from Lemma 3.3 that for any closed interval I⊂[an,bn]∖[λ1(a0),λ1(a0)], if u∈{y∈Y∣(λ,y)∈𝒞n,λ∈I}, then the fact that ∥u∥→0 in Y is impossible. So 𝒞n must be bifurcated from [λ1(a0),λ1(a0)]×{0} in ℝ×Y.


## 4. The Main Result

We obtain the following main result in this paper.Theorem 4.1.
Let (A1), (A2), and (A3) hold. Assuming that either(4.1)λ1(b∞)<1<λ1(a0)
or
(4.2)λ1(a0)<1<λ1(b∞).
Then problem (1.1) has at least one positive solution.Proof.
It is clear that any solution to (3.19) of the form (1,u) yields a solution u of problem (1.1). We will show that 𝒞 crosses the hyperplane {1}×Y in ℝ×Y. To do this, it is enough to show that 𝒞 joins [λ1(a0),λ1(a0)]×{0} to [λ1(b∞),λ1(b∞)]×{∞}. Let (ηn,yn)∈𝒞 satisfy
(4.3)ηn+∥yn∥→∞.
We note that ηn>0 for all n∈ℕ since (0,0) is the only solution to (3.19) for λ=0, and 𝒞∩({0}×Y)=∅.
Case  1 (λ1(b∞)<1<λ1(a0)). In this case, we show that the interval(4.4)(λ1(b∞),λ1(a0))⊆{λ∈ℝ∣(λ,u)∈𝒞}.
We divide the proof into two steps.
Step  1. We show that {ηn} is bounded.
Since(ηn,yn)∈𝒞,  Lyn=ηnf(t,ynσ(t)). From (A3), we have
(4.5)Lyn≥ηnc(t)ynσ(t).
Letφ- denote the nonnegative eigenfunction corresponding to λ1(c). From (4.5), by integration by parts formula (2.5), we have
(4.6)λ1(c)∫0σ2(T)‍ynσ(t)c(t)φ-σ(t)Δt=∫0σ2(T)‍(Lφ-)(t)ynσ(t)Δt=∫0σ2(T)‍(Lyn)(t)φ-σ(t)Δt≥ηn∫0σ2(T)‍c(t)ynσ(t)φ-σ(t)Δt.
Thus,
(4.7)ηn≤λ1(c).
Step  2. We show that 𝒞 joins [λ1(a0),λ1(a0)]×{0} to [λ1(b∞),λ1(b∞)]×{∞}.
From (4.3) and (4.7), we have that ∥yn∥→∞. Notice that (3.19) is equivalent to the equation
(4.8)yn(t)=ηn∫0σ2(T)‍H(t,s)f(s,ynσ(s))Δs,
where H(t,s) is the Green’s function for problem -uΔΔ=0,  u(0)=u(σ2(T))=0. So we have from (A1),
(4.9)ηn∫0σ2(T)‍H(t,s)[b∞(s)ynσ(s)+ζ2(s,ynσ(s))]Δs≥yn(t)≥ηn∫0σ2(T)‍H(t,s)[b∞(s)ynσ(s)+ζ1(s,ynσ(s))]Δs.
We divide the both sides of (4.9) by ∥yn∥ and set vn=yn/∥yn∥. Since vn is bounded in Y, choosing a subsequence and relabelling if necessary, we see that vn→v~  (n→∞) for some v~∈Y with v~≥0 in Y, ∥v~∥=1 and ηn→η*  (n→∞). Similar to the proof of Lemma 3.3, we have that
(4.10)limn→∞ζi(s,ynσ(s))∥yn∥=0,i=1,2,
and n→∞, it is easy to verify that
(4.11)η*∫0σ2(T)‍H(t,s)b∞(s)v~σ(s)Δs≥v~(t)≥η*∫0σ2(T)‍H(t,s)b∞(s)v~σ(s)Δs,
which implies that
(4.12)η*b∞(t)v~σ(t)≤Lv~≤η*b∞(t)v~σ(t).
Letφ∞ and φ∞ denote the nonnegative eigenfunctions corresponding to λ1(b∞) and λ1(b∞), respectively. Then we have from the first inequality in (4.12) that
(4.13)∫0σ2(T)‍η*b∞(t)v~σ(t)φ∞σ(t)Δt≤∫0σ2(T)‍Lv~(t)φ∞σ(t)Δt.
By integration by parts formula (2.5), we obtain that
(4.14)η*∫0σ2(T)‍b∞(t)φ∞σ(t)v~σ(t)Δt≤∫0σ2(T)‍Lφ∞(t)v~σ(t)Δt=λ1(b∞)∫0σ2(T)‍b∞(t)φ∞σ(t)v~σ(t)Δt,
and consequently
(4.15)η*≤λ1(b∞).
Similarly, we deduce from the second inequality in (4.12) that
(4.16)λ1(b∞)≤η*.
Thus,
(4.17)λ1(b∞)≤η*≤λ1(b∞).
So 𝒞 joins [λ1(a0),λ1(a0)]×{0} to [λ1(b∞),λ1(b∞)]×{∞}.
Case  2 (λ1(a0)<1<λ1(b∞)). In this case, if (ηn,yn)∈𝒞 is such that
(4.18)limn→∞(ηn+yn)=∞,limn→∞ηn=∞,
then
(4.19)(λ1(a0),λ1(b∞))⊆{λ∈(0,∞)∣(λ,u)∈𝒞}
and, moreover,
(4.20)({1}×Y)∩𝒞≠∅.
Assuming that{ηn} is bounded, applying a similar argument to that used in Step  2 of Case  1, after taking a subsequence and relabelling if necessary, it follows that
(4.21)ηn→η*∈[λ1(a0),λ1(b∞)],yn→∞,asn→∞.
Again 𝒞 joins [λ1(a0),λ1(a0)]×{0} to [λ1(b∞),λ1(b∞)]×{∞} and the result follows.



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*Source: 316080-2012-12-04.xml*