# Positive Solutions for Third-Order Nonlinearp-Laplacian m-Point Boundary Value Problems on Time Scales

Authors: Fuyi Xu
Journal: Discrete Dynamics in Nature and Society
 (2008)

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## Abstract

We study the following third-orderp-Laplacian m-point boundary value problems
on time scales: (ϕp(uΔ∇))∇+a(t)f(t,u(t))=0, t∈[0,T]T, βu(0)−γuΔ(0)=0, u(T)=∑i=1m−2aiu(ξi), ϕp(uΔ∇(0))=∑i=1m−2biϕp(uΔ∇(ξi)), where ϕp(s) is p-Laplacian operator, that is, ϕp(s)=|s|p−2s, p>1,ϕp−1=ϕq, 1/p+1/q=1,0<ξ1<⋯<ξm−2<ρ(T). We obtain the existence of positive solutions by using fixed-point theorem in
cones. The conclusions in this paper essentially extend and improve the known results.


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## Body

## 1. Introduction

The theory
of
time scales was initiated by Hilger [1] as a means of unifying and extending
theories from differential and difference equations. The study of time scales
has lead to several important applications in the study of insect population
models, neural networks, heat transfer, and epidemic models, see, for example
[2–6]. Recently, the boundary value problems with p-Laplacian operator have also been discussed
extensively in the literature, for example, see [7–13].A time scaleT is a nonempty closed subset of ℝ. We make the
blanket assumption that 0, T are points in T.
By an interval (0, T), we always mean the intersection of the real interval (0,
T) with the given time scale; that is (0, T)∩T.In [14], Anderson
considered  the the following third-order nonlinear boundary value problem (BVP):
(1.1)x′′′(t)=f(t,x(t)),t1≤t≤t3,x(t1)=x′(t2)=0,γx(t3)+δx′′(t3)=0.Author studied the existence of
solutions for the nonlinear boundary value problem by using the Krasnoselskii's
fixed point theorem and Leggett and Williams fixed point theorem, respectively.In [8, 9], He considered the existence of positive
solutions of the p-Laplacian dynamic equations on time scales(1.2)(ϕp(uΔ))∇+a(t)f(u(t))=0,t∈[0,T]T,satisfying the boundary
conditions(1.3)u(0)−B0(uΔ(η))=0,uΔ(T)=0,or(1.4)uΔ(0)=0,u(T)−B1(uΔ(η))=0,where η∈(0,ρ(T)).
He obtained the existence of at least double and triple positive solutions of
the boundary value problems by using a new double fixed point theorem and
triple fixed point theorem, respectively.In [13], Zhou and Ma firstly studied the existence and
iteration of positive solutions for the following third-order generalized
right-focal boundary value problem with p-Laplacian operator:(1.5)(ϕp(u′′))′(t)=q(t)f(t,u(t)),0≤t≤1,u(0)=∑i=1mαiu(ξi),u′(η)=0,u′′(1)=∑i=1nβiu′′(θi).They established a corresponding
iterative scheme for the problem by using the monotone iterative technique.However, to the best of our knowledge, little work has
been done on the existence of positive solutions for third-orderp-Laplacian m-point boundary value problems on time scales.
This paper attempts to fill this gap in the literature.In this paper, by using different method, we are
concerned with the existence of positive solutions for the following
third-orderp-Laplacian m-point boundary value problems on time scales:(1.6)(ϕp(uΔ∇))∇+a(t)f(t,u(t))=0,t∈[0,T]T,βu(0)−γuΔ(0)=0,u(T)=∑i=1m−2aiu(ξi),ϕp(uΔ∇(0))=∑i=1m−2biϕp(uΔ∇(ξi)),where ϕp(s) is p-Laplacian operator, that is, ϕp(s)=|s|p−2s,p>1,ϕp−1=ϕq,(1/p)+(1/q)=1, and ai,bi,a,f satisfy(H1)
β,γ≥0,β+γ>0, ai∈[0,+∞),i=1,2,…,m−3,am−2>0,0<ξ1<⋯<ξm−2<ρ(T), 0<∑i=1m−2bi<1,0<∑i=1m−2aiξi<T, d=β(T−∑i=1m−2aiξi)+γ(1−∑i=1m−2ai)>0;(H2)
f:[0,T]T×[0,+∞)→R+ is continuous, a∈Cld((0,T)T,ℝ+) and there exists  t0∈[ξm−2,T)T such that a(t0)>0,
where ℝ+=[0,+∞).


## 2. Preliminaries and Lemmas

For convenience, we list the following definitions
which can be found in [1–5].Definition 2.1.
A time scaleT is a nonempty closed subset of real numbers ℝ.
For t<supT and r>infT,
define the forward jump operator σ and backward jump operator ρ,
respectively, by(2.1)σ(t)=inf{τ∈T∣τ>t}∈T,ρ(r)=sup{τ∈T∣τ<r}∈T, for all t,r∈T.
If σ(t)>t, t is said to be right scattered; if ρ(r)<r, r is said to be left scattered; if σ(t)=t, t is said to be right dense; if ρ(r)=r, r is said to be left dense. If T has a right scattered minimum m,
define Tk=T−{m},
otherwise set Tk=T.
If T has a left scattered maximum M,
define Tk=T−{M},
otherwise set Tk=T.Definition 2.2.
Forf:T→ℝ and t∈Tk,
the delta derivative of f at the point t is defined to be the number fΔ(t),
(provided it exists), with the property that for each ϵ>0,
there is a neighborhood U of t such that(2.2)|f(σ(t))−f(s)−fΔ(t)(σ(t)−s)|≤ϵ|σ(t)−s|,for all s∈U.
Forf:T→ℝ and t∈Tk,
the nabla derivative of f at t,
denoted by f∇(t) (provided it exists), with the property that
for each ϵ>0,
there is a neighborhood U of t such that(2.3)|f(ρ(t))−f(s)−f∇(t)(ρ(t)−s)|≤ϵ|ρ(t)−s|,for all s∈U.Definition 2.3.
A functionf is left-dense continuous (i.e., ld-continuous), if f is continuous at each left-dense point in T and its right-sided limit exists at each
right-dense point in T.Definition 2.4.
IfϕΔ(t)=f(t),
then one defines the delta integral by(2.4)∫abf(t)Δt=ϕ(b)−ϕ(a).If F∇(t)=f(t),
then one defines the nabla integral by(2.5)∫abf(t)∇t=F(b)−F(a).Lemma 2.5.
Ifd=β(T−∑i=1m−2aiξi)+γ(1−∑i=1m−2ai)>0,
then for h∈Cld[0,T]T,
the boundary value problem (BVP) (2.6)uΔ∇+h(t)=0,t∈[0,T]T,βu(0)−γuΔ(0)=0,u(T)=∑i=1m−2aiu(ξi)has the unique solution(2.7)u(t)=−∫0t(t−s)h(s)∇s+βt+γd∫0T(T−s)h(s)∇s−βt+γd∑i=1m−2ai∫0ξi(ξi−s)h(s)∇s.Proof.
By
direct computation, we can easily get (2.7). So, we omit it.Lemma 2.6.
If0<∑i=1m−2bi<1,0<∑i=1m−2aiξi<T,d=β(T−∑i=1m−2aiξi)+γ(1−∑i=1m−2ai)>0, then for h∈Cld[0,T]T,
the boundary value problem (BVP) (2.8)(ϕp(uΔ∇))∇+h(t)=0,t∈[0,T]T,βu(0)−γuΔ(0)=0,u(T)=∑i=1m−2aiu(ξi),ϕp(uΔ∇(0))=∑i=1m−2biϕp(uΔ∇(ξi))has the unique solution(2.9)u(t)=−∫0t(t−s)ϕq(∫0sh(r)∇r+B)∇s+βt+γd∫0T(T−s)ϕq(∫0sh(r)∇r+B)∇s−βt+γd∑i=1m−2ai∫0ξi(ξi−s)ϕq(∫0sh(r)∇r+B)∇s,where B=∑i=1m−2bi∫0ξih(r)∇r/(1−∑i=1m−2bi).Proof.
Integrating both sides of (1.6) on [0,t],
we have(2.10)ϕp(uΔ∇(t))=ϕp(uΔ∇(0))−∫0th(r)∇r.So(2.11)ϕp(uΔ∇(ξi))=ϕp(uΔ∇(0))−∫0ξih(r)∇r.By boundary value condition ϕp(uΔ∇(0))=∑i=1m−2biϕp(uΔ∇(ξi)),
we have(2.12)ϕp(uΔ∇(0))=−∑i=1m−2bi∫0ξih(r)∇r1−∑i=1m−2bi.By (2.10) and (2.12), we know(2.13)uΔ∇(t)=−ϕq(∑i=1m−2bi∫0ξih(r)∇r1−∑i=1m−2bi+∫0th(r)∇r).This together with Lemma 2.5
implies that(2.14)u(t)=−∫0t(t−s)ϕq(∫0sh(r)∇r+B)∇s+βt+γd∫0T(T−s)ϕq(∫0sh(r)∇r+B)∇s−βt+γd∑i=1m−2ai∫0ξi(ξi−s)ϕq(∫0sh(r)∇r+B)∇s,where B=∑i=1m−2bi∫0ξih(r)∇r/(1−∑i=1m−2bi).
The proof is complete.Lemma 2.7.
Let0<∑i=1m−2aiξi<1,d>0.
If h∈Cld[0,T]T and h(t)≥0,
then the unique solution u of (2.8) satisfies (2.15)u(t)≥0.Proof.
ByuΔ∇(t)=−ϕq(∑i=1m−2bi∫0ξih(r)∇r/(1−∑i=1m−2bi))+∫0th(r)∇r)≤0,
we can know that the graph of u(t) is concave down on (0,T)T.
So we only prove u(0)≥0,u(T)≥0.
Firstly, we shall proveu(0)≥0 by the following two perspectives.
(i)
If0<∑i=1m−2ai≤1,
we have(2.16)u(0)=γd[∫0T(T−s)ϕq(∫0sh(r)∇r+B)∇s−∑i=1m−2ai∫0ξi(ξi−s)ϕq(∫0sh(r)∇r+B)∇s]≥γd[∫0T(T−s)ϕq(∫0sh(r)∇r+B)∇s−∑i=1m−2ai∫0T(T−s)ϕq(∫0sh(r)∇r+B)∇s]=γd(1−∑i=1m−2ai)∫0T(T−s)ϕq(∫0sh(r)∇r+B)∇s≥0.(ii)
If∑i=1m−2ai>1,
by (2.8), we have(2.17)u(0)=γd[∫0T(T−s)ϕq(∫0sh(r)∇r+B)∇s−∑i=1m−2ai∫0ξi(ξi−s)ϕq(∫0sh(r)∇r+B)∇s]≥γd[∫0T(T−s)ϕq(∫0sh(r)∇r+B)∇s−∑i=1m−2ai∫0T(ξi−s)ϕq(∫0sh(r)∇r+B)∇s]=γd∫0T[(T−∑i=1m−2aiξi)+(∑i=1m−2ai−1)s]ϕq(∫0sh(r)∇r+B)∇s≥0.On the other hand, we have(2.18)u(T)=−∫0T(T−s)ϕq(∫0sh(r)∇r+B)∇s+β+γd∫0T(T−s)ϕq(∫0sh(r)∇r+B)∇s−β+γd∑i=1m−2ai∫0ξi(ξi−s)ϕq(∫0sh(r)∇r+B)∇s≥βd[∑i=1m−2ai∫0ξi(ξi(T−s)−T(ξi−s))ϕq(∫0sh(r)∇r+B)∇s+∑i=1m−2aiξi∫ξiT(T−s)ϕq(∫0sh(r)∇r+B)∇s]+γd∑i=1m−2ai[∫0T(T−s)ϕq(∫0sh(r)∇r+B)∇s−∫0T(ξi−s)ϕq(∫0sh(r)∇r+B)∇s]=βd∑i=1m−2ai[∫0ξi(T−ξi)sϕq(∫0sh(r)∇r+B)∇s+ξi∫ξiT(T−s)ϕq(∫0sh(r)∇r+B)∇s]+γd∑i=1m−2ai[∫0T(T−ξi)ϕq(∫0sh(r)∇r+B)∇s]≥0.The proof is completed.Lemma 2.8.
Letai≥0,i=1,…,m−2,0<∑i=1m−2aiξi<T,d>0.
If h∈Cld[0,T]T and h(t)≥0,
then the unique positive solution u(t) of (BVP) (2.8) satisfies (2.19)inft∈[ξm−2,T]Tu(t)≥σ∥u∥,where σ=min{am−2(T−ξm−2)/(T−am−2ξm−2),am−2ξm−2/T,ξm−2/T},∥u∥=supt∈[0,T]T|u(t)|.Proof.
Letu(t¯)=maxt∈[0,T]Tu(t)=∥u∥,
we shall discuss it from the following two perspectives.
Case 1.
If0<∑i=1m−2ai<1.
Firstly, assumet¯<ξm−2<T,
then mint∈[ξm−2,T]Tu(t)=u(T). By u(T)=∑i=1m−2aiu(ξi)≥am−2u(ξm−2), we have(2.20)u(t¯)≤u(T)+u(T)−u(ξm−2)T−ξm−2(0−T)=u(T)−TT−ξm−2u(T)+TT−ξm−2u(ξm−2)≤u(T)(1−TT−ξm−2+Tam−2(T−ξm−2))=u(T)T−am−2ξm−2am−2(T−ξm−2).So(2.21)mint∈[ξm−2,T]Tu(t)≥am−2(T−ξm−2)T−am−2ξm−2∥u∥.Secondly, assume ξm−2<t¯<T,
then mint∈[ξm−2,T]Tu(t)=u(T).
Otherwise, we have mint∈[ξm−2,T]Tu(t)=u(ξm−2),
then t¯∈[ξm−2,T]T,u(ξm−2)≥u(ξm−1)≥⋯≥u(ξ2)≥u(ξ1).
By 0<∑i=1m−2ai<1, we have(2.22)u(T)=∑i=1m−2aiu(ξi)≤∑i=1m−2aiu(ξm−2)<u(ξm−2)≤u(T),a contradiction.
By concave ofu(t),  we get u(ξm−2)/ξm−2≥u(t¯)/t¯≥u(t¯)/T.
In fact, since u(T)≥am−2u(ξm−2),
then u(T)/am−2ξm−2≥u(t¯)/T,
which implies(2.23)mint∈[ξm−2,T]Tu(t)≥am−2ξm−2T∥u∥.
Case 2.
If∑i=1m−2ai>1.
Firstly, assumeu(ξm−2)≤u(T),
then mint∈[ξm−2,T]Tu(t)=u(ξm−2).
By concave of u(t), we have t¯∈[ξm−2,t]T,
which implies u(ξm−2)/ξm−2≥u(t¯)/t¯≥u(t¯)/T,
then(2.24)mint∈[ξm−2,T]Tu(t)≥ξm−2T∥u∥.
Secondly, assumeu(ξm−2)>u(t),
then mint∈[ξm−2,T]Tu(t)=u(T),
and t¯∈[ξ1,T]T.
If not, t¯∈[0,ξ1),
then u(ξ1)≥⋯≥u(ξm−2)>u(T).
So, we have(2.25)u(T)=∑i=1m−2aiu(ξi)>u(T)∑i=1m−2ai≥u(T),a contradiction. By ∑i=1m−2ai>1, there exists ξ¯∈{ξ1,ξ2,…,ξm−2} such that u(ξ¯)≤u(T),
then u(ξ1)≤u(ξ2)≤⋯≤u(ξm−2)≤u(1).
By concave of u(t), we have u(1)/ξ1≥u(ξ1)/ξ1≥u(t¯)/t¯≥u(t¯)/T,
then(2.26)mint∈[ξm−2,T]Tu(t)≥ξ1∥u∥.Therefore, by (2.21)–(2.26), we
have(2.27)inft∈[ξm−2,T]Tu(t)≥σ∥u∥,where σ=min{am−2(T−ξm−2)/(T−am−2ξm−2),am−2ξm−2/T,ξm−2/T}.
The proof is complete.LetE=Cld[0,T]T be endowed with the ordering x≤y if x(t)≤y(t), for all t∈[0,T]T, and ∥u∥=maxt∈[0,T]T|u(t)| is defined as usual by maximum norm. Clearly,
it follows that (E,∥u∥) is a Banach space.We define a cone by(2.28)K={u:u∈E,u(t)isconcave,nonnegativeon[0,T]T,inft∈[ξm−2,T]Tu(t)≥σ∥u∥}.Define an operator S:K→E by setting(2.29)Su(t)=−∫0t(t−s)ϕq(∫0sa(r)f(r,u(r))∇r+A)∇s+βt+γd∫0T(T−s)ϕq(∫0sa(r)f(r,u(r))∇r+A)∇s−βt+γd∑i=1m−2ai∫0ξi(ξi−s)ϕq(∫0sa(r)f(r,u(r))∇r+A)∇s,where A=∑i=1m−2bi∫0ξia(r)f(r,u(r))∇r/(1−∑i=1m−2bi).
Obviously, u is a solution of boundary value problem (1.6)
if and only if u is a fixed point of operator S.Lemma 2.9.
S:K→K  is completely continuous.Proof.
By(H2) and  Lemmas
2.7-2.8, we easily get SK⊂K.
By Arzela-Ascoli theorem and Lebesgue dominated convergence theorem, we can
easily prove S is completely continuous.Lemma 2.10 (see [15]).
LetK be a cone in a Banach space X.
Let D be an open bounded subset of X with DK=D∩K≠∅ and D¯K≠K.
Assume that A:D¯K→K is a compact map such that x≠Ax for x∈∂DK.
Then the following results hold.
(1)
If∥Ax∥≤∥x∥,  x∈∂DK,
then  iK(A,DK)=1.(2)
If there existsx0∈K∖{0} such that  x≠Ax+λx0, for all x∈∂DK and all λ>0,
then iK(A,DK)=0.(3)
LetU be open in X such that U¯⊂DK.
If iK(A,DK)=1 and iK(A,UK)=0,
then A has a fixed point in DK∖U¯K.
The same result holds if iK(A,DK)=0 and iK(A,UK)=1,
where iK(A,DK) denotes fixed point index.
We  define(2.30)Kρ={u(t)∈K:∥u∥<ρ},Ωρ={u(t)∈K:minξm−2≤t≤Tx(t)<σρ}.Lemma 2.11 (see [15]).
Ωρ  defined above has the following properties:
(a)
Kσρ⊂Ωρ⊂Kρ;(b)
Ωρ  is open relative to K;(c)
x∈∂Ωρ  if and only if minξm−2≤t≤Tx(t)=σρ;(d)
ifx∈∂Ωρ,
then σρ≤x(t)≤ρ,  for t∈[ξm−2,T]T.
For the convenience, we introduce the following
notations:(2.31)fσρρ=min{minξm−2≤t≤Tf(t,u)ϕp(ρ):u∈[σρ,ρ]},f0ρ=max{max0≤t≤Tf(t,u)ϕp(ρ):u∈[0,ρ]},fα=limu→αsupmax0≤t≤Tf(t,u)ϕp(u),fα=limu→αinfmaxξm−2≤t≤Tf(t,u)ϕp(u)(α:=∞or0+),1m=(βT+γ)d∫0T(T−s)∇sϕq(∫0Ta(r)∇r+∑i=1m−2bi∫0ξia(r)∇r1−∑i=1m−2bi),1M=1d∫ξm−2T(T−s)ϕq(∫ξm−2sa(r)∇r)∇smin{βξm−2+γ,βmax{∑i=1m−2aiξ1,am−2ξm−2}+γ∑i=1m−2ai}.Lemma 2.12.
Iff satisfies the following condition: (2.32)f0ρ≤ϕp(m),u≠Su,u∈∂Kρ,then(2.33)iK(S,Kρ)=1.Proof.
Foru∈∂Kρ,
then from (2.32), we have(2.34)∫0sa(r)f(r,u(r))∇r+A=∫0sa(r)f(r,u(r))∇r+∑i=1m−2bi∫0ξia(r)f(r,u(r))∇r1−∑i=1m−2bi≤∫0Ta(r)f(r,u(r))∇r+∑i=1m−2bi∫0ξia(r)f(r,u(r))∇r1−∑i=1m−2bi≤ϕp(mρ)(∫0Ta(r)∇r+∑i=1m−2bi∫0ξia(r)∇r1−∑i=1m−2bi).So that(2.35)ϕq(∫0sa(r)f(r,u(r))∇r+A)≤mρϕq(∫0Ta(r)∇r+∑i=1m−2bi∫0ξia(r)∇r1−∑i=1m−2bi).Therefore,(2.36)Su(t)≤βt+γd∫0T(T−s)ϕq(∫0sa(r)f(r,u(r))∇r+A)∇s≤(βT+γ)mρd∫0T(T−s)∇sϕq(∫0Ta(r)∇r+∑i=1m−2bi∫0ξia(r)∇r1−∑i=1m−2bi)=ρ.This implies that ∥Su∥≤∥u∥ for u∈∂Kρ.
Hence, by Lemma 2.10(1) it follows that iK(S,Kρ)=1.Lemma 2.13.
Iff satisfies the following
condition: (2.37)fσρρ≥ϕp(Mσ),u≠Su,u∈∂Ωρ,then(2.38)iK(S,Ωρ)=0.Proof.
Lete(t)≡1 for t∈[0,T]T.
Then e∈∂K1.
We claim that(2.39)u≠Su+λe,u∈∂Ωρ,λ>0.In fact, if not, there exist u0∈∂Ωρ and λ0>0 such that u0=Su0+λ0e.
By fσρρ≥ϕp(Mσ),
we have(2.40)∫0sa(r)f(r,u0(r))∇r+A=∫0sa(r)f(r,u0(r))∇r+∑i=1m−2bi∫0ξia(r)f(r,u0(r))∇r1−∑i=1m−2bi≥∫ξm−2sa(r)f+(r,u(r))∇r≥ϕp(Mσρ)(∫ξm−2sa(r)∇r).So that(2.41)ϕq(∫0sa(r)f(r,u(r))∇r+A)≥Mσρϕq(∫ξm−2sa(r)∇r).By [16, Theorem 2.2(iv)], for t>0, we have(2.42)(∫0t(t−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇st)Δ=∫0tsϕq(∫0sa(r)f(r,u0(r))∇r+A)∇stσ(t)≥0.So, for i=1,2,…,m−2, we have(2.43)∫0ξm−2(ξm−2−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇sξm−2≥∫0ξi(ξi−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇sξi.Therefore,(2.44)Su0(ξm−2)≥βd[ξm−2∫0T(t−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇s−T∫0ξm−2(ξm−2−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇s]+γd[∫0T(t−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇s−∫0ξm−2(ξm−2−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇s]≥βξm−2+γd∫ξm−2T(T−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇s≥(βξm−2+γ)Mσρd∫ξm−2T(T−s)ϕq(∫ξm−2sa(r)∇r)∇s,(2.45)Su0(T)≥βd∑i=1m−2ai[∫0ξi(t−ξi)sϕq(∫0sa(r)f(r,u0(r))∇r+A)∇s+ξi∫ξiT(T−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇s]+γd∑i=1m−2[∫0T(t−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇s−∫0ξi(T−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇s]≥βd∑i=1m−2aiξi∫ξiT(t−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇s+γd∑i=1m−2∫ξiT(t−s)ϕq(∫0sa(r)f(r,u0(r))∇r+A)∇s≥Mσρd(βmax{∑i=1m−2aiξ1,am−2ξm−2}+γ∑i=1m−2)∫ξm−2T(T−s)ϕq(∫ξm−2sa(r)∇r)∇s.Obviously, we can know(2.46)mint∈[ξm−2,T]TSu0(t)=min{Su0(ξm−2),Su0(T)}≥Mσρd∫ξm−2T(T−s)ϕq(∫ξm−2sa(r)∇r)∇s×min{βξm−2+γ,βmax{∑i=1m−2aiξ1,am−2ξm−2}+γ∑i=1m−2}≥σρ.For t∈[ξm−2,T]T,
then(2.47)u0(t)=Su0(t)+λ0e(t)≥mint∈[ξm−2,T]Su0(t)+λ0=min{Su0(ξm−2),Su0(T)}+λ0≥σρ+λ0.This together with Lemma 2.11(c)
implies that(2.48)σρ≥σρ+λ0,a contradiction. Hence, by Lemma
2.10(2), it follows that iK(S,Ωρ)=0.


## 3. Main Results

We now give our results on the
existence of positive solutions of BVP (1.6).Theorem 3.1.
Suppose conditions(H1) and (H2) hold, and assume that one of the following
conditions hold.
(H3)
There existρ1,ρ2∈(0,+∞)  with ρ1<σρ2  such  that f0ρ1≤ϕp(m),fσρ2ρ2≥ϕp(Mσ).(H4)
There existρ1,ρ2∈(0,+∞)  with ρ1<ρ2  such  that f0ρ2≤ϕp(m),fσρ1ρ1≥ϕp(Mσ). Then, the
boundary value problem (1.6) has at least one positive solution.Proof.
Assume
that(H3) holds, we show that S has a fixed point u1 in Ωρ2∖K¯ρ1.
By f0ρ1≤ϕp(m) and Lemma 2.12, we have that(3.1)iK(S,Kρ1)=1.By fσρ2ρ2≥ϕp(Mσ) and Lemma 2.13, we have that(3.2)iK(S,Kρ2)=0.By Lemma 2.11(a) and ρ1<σρ2,
we have K¯ρ1⊂Kσρ2⊂Ωρ2.
It follows from Lemma 2.10(3) that S has a fixed point u1 in Ωρ2∖K¯ρ1.
When condition (H4) holds, the proof is similar to the above, so
we omit it here.
As a special case of Theorem3.1, we obtain the
following result.Corollary 3.2.
Suppose conditions(H1), (H2) hold, and assume that one of the following
conditions holds.
(H5)
0≤f0<ϕp(m)  and ϕp(M)<f∞≤∞.(H6)
0≤f∞<ϕp(m) and ϕp(M)<f0≤∞. Then, the
boundary value problem (1.6) has at least one positive solution.Theorem 3.3.
Assume conditions(H1) and (H2) hold, and suppose that one of the following
conditions holds.
(H7)
There existρ1,ρ2,
and ρ3∈(0,+∞) with ρ1<σρ2 and ρ2<ρ3 such that (3.3)f0ρ1≤ϕp(m),fσρ2ρ2≥ϕp(Mσ),u≠Su,∀u∈∂Ωρ2,f0ρ3≤ϕp(m).(H8)
There existρ1,ρ2,
and ρ3∈(0,+∞) with ρ1<ρ2<σρ3 such that (3.4)f0ρ2≤ϕp(m),fσρ1ρ1≥ϕp(Mσ),u≠Su,∀u∈∂Kρ2,fσρ3ρ3≥ϕp(Mσ). Then, the boundary value problem (1.6) has at least two positive
solutions. Moreover, if in (H7)f0ρ1≤ϕp(m) is replaced by f0ρ1<ϕp(m),
then the BVP (1.6) has a third positive solution u3∈Kρ1.Proof.
Assume
that condition(H7) holds, we show that either S has a fixed point u1 in ∂Kρ1 or Ωρ2∖K¯ρ1.
If u≠Su for u∈∂Kρ1∪∂Kρ3.
By Lemma 2.12 and Lemma 2.13, we have that(3.5)iK(S,Kρ1)=1,iK(S,Kρ3)=1,iK(S,Kρ2)=0. By Lemma 2.11(a) and ρ1<σρ2,
we have K¯ρ1⊂Kσρ2⊂Ωρ2.
It follows from Lemma 2.10(3) that S has a fixed point u1 in Ωρ2∖K¯ρ1.
Similarly, S has a fixed point u2 in Kρ3∖Ω¯ρ2.
When condition (H8) holds, the proof is similar to the above, so
we omit it here.
As a special case of Theorem3.3, we obtain the
following result.Corollary 3.4.
Assume conditions(H1) and (H2) hold, if there exists ρ>0 such that one of the following conditions
holds.
(H9)
0≤f0<ϕp(m),fσρρ≥ϕp(Mσ),u≠Su,∀u∈∂Ωρ  and 0≤f∞<ϕp(m).(H10)
ϕp(M)<f0≤∞,f0ρ≤ϕp(m),u≠Su,∀u∈∂Kρ and ϕp(M)<f∞≤∞. Then, the
boundary value problem (1.6) has at least two positive solutions.


## 4. Some Examples

In this section, we present
some simple examples to explain our results. We only study the caseT=ℝ,(0,T)T=(0,1).Example 4.1.
Consider the following three-point boundary value problem withp-Laplacian:(4.1)(ϕp(u′′))′+a(t)f(t,u)=0,0<t<1,u′(0)=0,u(1)=12u(13),(ϕp(u′′)(0))=14(ϕp(u′′)(13)),where β=0,γ=1,a1=1/2,b1=1/4,ξ1=1/3,a(t)=1,p=q=2. By computing, we can know σ=1/6,M=819/16,m=9/10. Let ρ1=1,ρ2=208,
then σρ1<ρ1<σρ2<ρ2.
We define a nonlinearity f as follows:(4.2)f(t,u)={9t310(16−u)3,0<t<1,u∈[0,16],9t310sin(65π2u−15π2),0<t<1,u∈[16,1],9t310(208202−6202u)+81996(6202u−6202),0<t<1,u∈[1,2086],81996+t3(u−2086)2,0<t<1,u∈[2086,208],81996+t3(208−2086)2[1+(u−208)],0<t<1,u∈[208,+∞].Then, by the definition of f,  we have
(i)
f0ρ1≤ϕp(m)=9/10;(ii)
fσρ2ρ2≥ϕp(Mσ)=819/19968.
So condition(H3) holds, by Theorem 3.1, boundary value problem
(4.1) has at least one positive solution.



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