# Twin Positive Solutions of a Nonlinearm-Point Boundary Value Problem for Third-Order p-Laplacian Dynamic Equations on Time Scales

Authors: Wei Han; Guang Zhang
Journal: Discrete Dynamics in Nature and Society
 (2008)

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## Abstract

Several existence theorems of twin positive solutions are established for a
nonlinearm-point boundary value problem of third-order p-Laplacian dynamic equations on time
scales by using a fixed point theorem. We present two theorems and four corollaries which generalize
the results of related literature. As an application, an example to demonstrate our results is given.
The obtained conditions are different from some known results.


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## Body

## 1. Introduction

A time scaleT is a nonempty closed subset of ℝ.
We make the blanket assumption that 0 and 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 this paper, we will be concerned with the existence of
positive solutions of thep-Laplacian dynamic equations on time scales:(1.1)(ϕp(uΔ∇))∇+a(t)f(t,u(t))=0,t∈(0,T),(1.2)ϕp(uΔ∇(0))=∑i=1m−2aiϕp(uΔ∇(ξi)),uΔ(0)=0,u(T)=∑i=1m−2biu(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),
and(H1)
ai,bi∈[0,+∞),i=1,2,…,
satisfy 0<∑i=1m−2ai<1 and ∑i=1m−2bi<1;(H2)
a(t)∈Cld([0,T],[0,+∞)) and there exists t0∈(ξm−2,T) such that a(t0)>0;(H3)
f∈C([0,T]×[0,+∞),[0,+∞)).We point out that the Δ-derivative and the ∇-derivative in (1.2) and the Cld space in (H2) are defined in Section 2.Recently, there has been much attention paid to the
existence of positive solutions for third-order nonlinear boundary value
problems of differential equations. For example, see [1–10] and the
listed references. Anderson [2] considered the following third-order nonlinear
problem:(1.3)x′′′(t)=f(t,x(t)),t1≤t≤t3,x(t1)=x′(t2)=0,γx(t3)+δx′′(t3)=0.He used the Krasnoselskii and
the Leggett and Williams fixed-point theorems to prove the existence of
solutions to the nonlinear problem (1.3). Li [6] considered the existence of
single and multiple positive solutions to the nonlinear singular third-order
two-point boundary value problem:(1.4)u′′′(t)+λa(t)f(u(t))=0,0<t<1,u(0)=u′(0)=u′′(1)=0.Under various assumptions on a and f,
they established intervals of the parameter λ which yield the existence of at least two and
infinitely many positive solutions of the boundary value problem by using
Krasnoselski's fixed-point theorem of cone expansion-compression type. Liu et
al. [7] discussed the existence of at least one or two nondecreasing positive solutions
for the following singular nonlinear
third-order differential equations:(1.5)x′′′(t)+λα(t)f(t,x(t))=0,a<t<b,x(a)=x′′(a)=x′(b)=0.Green's function and the
fixed-point theorem of cone expansion-compression type are utilized in their
paper. In [8], Sun considered the following nonlinear singular third-order
three-point boundary value problem:(1.6)u′′′(t)−λa(t)F(t,u(t))=0,0<t<1,u(0)=u′(η)=u′′(1)=0.He obtained various results on
the existence of single and multiple positive solutions to the boundary value
problem (1.6) by using a fixed-point theorem of cone expansion-compression type
due to Krasnosel'skii. In [10], Zhou and Ma studied the existence and iteration
of positive solutions for the following third-order generalized right-focal
boundary value problem with p-Laplacian operator:(1.7)(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 (1.7) by using the monotone iterative technique.On the other hand, the existence of positive solutions
for third-order nonlinear boundary value problems of difference equations is
also extensively studied by a number of authors (see [1, 3, 5, 9] and the
listed references). The present work is motivated by a recent paper [4]. In
[4], Henderson and Yin considered the existence of solutions for a third-order
boundary value problem on a time-scale equation of the form(1.8)uΔ3=f(t,u,uΔ,uΔΔ),t∈T,which is uniform for the
third-order difference equation and the third-order differential equation.


## 2. Preliminaries and Lemmas

For
convenience, we list the following definitions which can be found in [4, 11–15].Definition 2.1.
LetT be a time scale. For t<sup⁡T and r>inf⁡T,
define the forward jump operator σ and the backward jump operator ρ,
respectively, by(2.1)σ(t)=inf⁡{τ∈T∣τ>t}∈T,ρ(r)=sup⁡{τ∈T∣τ<r}∈Tfor all t,r∈T.
If σ(t)>t, t is said to be right-scattered, and if ρ(r)<r,r  is said to be left-scattered; if σ(t)=t, t is said to be right-dense, and 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 that 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
is denoted by f∇(t) (provided that 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).To prove the main results in this paper, we will
employ several lemmas. These lemmas are based on the linear BVP(2.6)(ϕp(uΔ∇))∇+h(t)=0,t∈(0,T),(2.7)ϕp(uΔ∇(0))=∑i=1m−2aiϕp(uΔ∇(ξi)),uΔ(0)=0,u(T)=∑i=1m−2biu(ξi).Lemma 2.5.
If∑i=1m−2ai⁢≠1 and ∑i=1m−2bi⁢≠1,
then for h∈Cld[0,T] the BVP (2.6)-(2.7) has the unique solution(2.8)u(t)=−∫0t(t−s)ϕq(∫0sh(τ)∇τ−A)∇s+C,where(2.9)A=−∑i=1m−2ai∫0ξih(τ)∇τ1−∑i=1m−2ai,C=∫0T(T−s)ϕq(∫0sh(τ)∇τ−A)∇s−∑i=1m−2bi∫0ξi(ξi−s)ϕq(∫0sh(τ)∇τ−A)∇s1−∑i=1m−2bi.Proof.
(i) Letu be a solution, then we will show that (2.8)
holds. By taking the nabla integral of problem (2.6) on (0,t),
we have(2.10)ϕp(uΔ∇(t))=−∫0th(τ)∇τ+Athen(2.11)uΔ∇(t)=ϕq(−∫0th(τ)∇τ+A)=−ϕq(∫0th(τ)∇τ−A).By taking the nabla integral of
(2.11) on (0,t),
we can get(2.12)uΔ(t)=−∫0tϕq(∫0sh(τ)∇τ−A)∇s+B.By taking the delta integral of
(2.12) on (0,t),
we can get(2.13)u(t)=−∫0t(t−s)ϕq(∫0sh(τ)∇τ−A)∇s+Bt+C.Similarly, let t=0 on (2.10), then we have ϕp(uΔ∇(0))=A;
let t=ξi on (2.10), then we have(2.14)ϕp(uΔ∇(ξi))=−∫0ξih(τ)∇τ+A.
Let t=0 on (2.12), then we have(2.15)uΔ(0)=B.Let t=T on (2.13), then we have(2.16)u(T)=−∫0T(T−s)ϕq(∫0sh(τ)∇τ−A)∇s+BT+C.Similarly, let t=ξi on (2.13), then we have(2.17)u(ξi)=−∫0ξi(ξi−s)ϕq(∫0sh(τ)∇τ−A)∇s+Bξi+C.By the boundary condition (2.7),
we can get(2.18)B=0,(2.19)A=∑i=1m−2ai(−∫0ξih(τ)∇τ+A).Solving
(2.19), we get(2.20)A=−∑i=1m−2ai∫0ξih(τ)∇τ1−∑i=1m−2ai.By the boundary condition (2.7),
we can obtain(2.21)−∫0T(T−s)ϕq(∫0sh(τ)∇τ−A)∇s+C=∑i=1m−2bi[−∫0ξi(ξi−s)ϕq(∫0sh(τ)∇τ−A)∇s+C].Substituting (2.20) in the above
expression, one has(2.22)C=∫0T(T−s)ϕq(∫0sh(τ)∇τ−A)∇s−∑i=1m−2bi∫0ξi(ξi−s)ϕq(∫0sh(τ)∇τ−A)∇s1−∑i=1m−2bi.
(ii) We show that the
functionu given in (2.8) is a solution.
Letu be as in (2.8). By [12, Theorem 2.10(iii)] and
taking the delta derivative of (2.8), we have(2.23)uΔ(t)=−∫0tϕq(∫0sh(τ)∇τ−A)∇s;moreover, we get
(2.24)uΔ∇(t)=−ϕq(∫0th(τ)∇τ−A),ϕp(uΔ∇)=−(∫0th(τ)∇τ−A).Taking the nabla derivative of
this expression yields (ϕp(uΔ∇))∇=−h(t).
Also, routine calculation verifies that u satisfies the boundary value conditions in
(2.7) so that u given in (2.8) is a solution of (2.6) and
(2.7). The proof is complete.Lemma 2.6.
Assume
(H1)
holds. Forh∈Cld[0,T] and h≥0,
the unique solution u of (2.6) and (2.7) satisfies(2.25)u(t)≥0fort∈[0,T].Proof.
Let(2.26)φ0(s)=ϕq(∫0sh(τ)∇τ−A).Since(2.27)∫0sh(τ)∇τ−A=∫0sh(τ)∇τ+∑i=1m−2ai∫0ξih(τ)∇τ1−∑i=1m−2ai≥0,then φ0(s)≥0.
According to Lemma2.5, we get(2.28)u(0)=C=∫0T(T−s)φ0(s)∇s−∑i=1m−2bi∫0ξi(ξi−s)φ0(s)∇s1−∑i=1m−2bi≥∫0T(T−s)φ0(s)∇s−∑i=1m−2bi∫0ξi(T−s)φ0(s)∇s1−∑i=1m−2bi=∫0T(T−s)φ0(s)∇s−∑i=1m−2bi(∫0T(T−s)φ0(s)∇s−∫ξiT(T−s)φ0(s)∇s)1−∑i=1m−2bi=∫0T(T−s)φ0(s)∇s+∑i=1m−2bi∫ξiT(T−s)φ0(s)∇s1−∑i=1m−2bi≥0,u(T)=−∫0T(T−s)φ0(s)∇s+C=−∫0T(T−s)φ0(s)∇s+∫0T(T−s)φ0(s)∇s−∑i=1m−2bi∫0ξi(ξi−s)φ0(s)∇s1−∑i=1m−2bi≥−∫0T(T−s)φ0(s)∇s+∫0T(T−s)φ0(s)∇s−∑i=1m−2bi∫0ξi(T−s)φ0(s)∇s1−∑i=1m−2bi=∑i=1m−2bi∫ξiT(T−s)φ0(s)∇s1−∑i=1m−2bi≥0.If t∈(0,T),
we have(2.29)u(t)=−∫0t(t−s)φ0(s)∇s+11−∑i=1m−2bi[∫0T(T−s)φ0(s)∇s−∑i=1m−2bi∫0ξi(ξi−s)φ0(s)∇s]≥−∫0T(T−s)φ0(s)∇s+11−∑i=1m−2bi[∫0T(T−s)φ0(s)∇s−∑i=1m−2bi∫0ξi(T−s)φ0(s)∇s]=11−∑i=1m−2bi[−(1−∑i=1m−2bi)∫0T(T−s)φ0(s)∇s+∫0T(T−s)φ0(s)∇s−∑i=1m−2bi∫0ξi(T−s)φ0(s)∇s]=11−∑i=1m−2bi∑i=1m−2bi∫ξiT(T−s)φ0(s)∇s≥0.
So u(t)≥0,⁢t∈[0,T].Lemma 2.7.
Assume
(H1)
holds. Ifh∈Cld[0,T] and h≥0,
then the unique solution u of (2.6) and (2.7) satisfies(2.30)inf⁡t∈[0,T]u(t)≥γ∥u∥,where(2.31)γ=∑i=1m−2bi(T−ξi)T−∑i=1m−2biξi,∥u∥=max⁡t∈[0,T]|u(t)|.Proof.
It is easy to
check thatuΔ(t)=−∫0tφ0(s)∇s≤0;
this implies that(2.32)∥u∥=u(0),min⁡t∈[0,T]u(t)=u(T).It is easy to see that uΔ(t2)≤uΔ(t1) for any t1,t2∈[0,T] with t1≤t2.
Hence, uΔ(t) is a decreasing function on [0,T].
This means that the graph of u(t) is concave down on (0,T).
For eachi∈{1,2,…,m−2},
we have(2.33)u(T)−u(0)T−0≥u(T)−u(ξi)T−ξi,that is,(2.34)Tu(ξi)−ξiu(T)≥(T−ξi)u(0),so that(2.35)T∑i=1m−2biu(ξi)−∑i=1m−2biξiu(T)≥∑i=1m−2bi(T−ξi)u(0).With the boundary condition u(T)=∑i=1m−2biu(ξi),
we have(2.36)u(T)≥∑i=1m−2bi(T−ξi)T−∑i=1m−2biξiu(0).This completes the proof.Let the norm onCld[0,T] be the maximum norm. Then, the Cld[0,T] is a Banach space. It is easy to see that BVP
(1.1)-(1.2) has a solution u=u(t) if and only if u is a fixed point of the operator(2.37)(Au)(t)=−∫0t(t−s)ϕq(∫0sa(τ)f(τ,u(τ))∇τ−A)∇s+C,where(2.38)A=−∑i=1m−2ai∫0ξia(τ)f(τ,u(τ))∇τ1−∑i=1m−2ai,C=∫0T(T−s)ϕq(∫0sa(τ)f(τ,u(τ))∇τ−A)∇s−∑i=1m−2bi∫0ξi(ξi−s)ϕq(∫0sa(τ)f(τ,u(τ))∇τ−A)∇s1−∑i=1m−2bi.Denote(2.39)K={u∣u∈Cld[0,T],u(t)≥0,inf⁡t∈[0,T]u(t)≥γ∥u∥},where γ is the same as in Lemma 2.7. It is obvious
that K is a cone in Cld[0,T].
By Lemma 2.7, A(K)⊂K.
So by applying Arzela-Ascoli theorem on time scales [16], we can obtain that A(K) is relatively compact. In view of Lebesgue's
dominated convergence theorem on time scales [13], it is easy to prove that A is continuous. Hence, A:K→K is completely continuous.Lemma 2.8.
A:K→K is completely continuous.Proof.
First, we
show thatA maps bounded set into bounded
set.
Assumec>0 is a constant and u∈Kc¯={u∈K:∥u∥≤c}.
Note that the continuity of f guarantees that there is c′>0 such that f(t,u(t))≤ϕp(c′) for t∈[0,T].
So(2.40)∥Au∥=max⁡t∈[0,T]|Au(t)|≤C≤∫0T(T−s)ϕq(∫0sa(τ)f(τ,u(τ))∇τ−A)∇s1−∑i=1m−2bi≤c′∫0T(T−s)ϕq(∫0sa(τ)∇τ+∑i=1m−2ai∫0ξia(τ)∇τ/(1−∑i=1m−2ai))∇s1−∑i=1m−2bi.That is, AKc¯ is uniformly bounded.
In addition, notice that for anyt1,t2∈[0,T],
we have(2.41)|Au(t1)−Au(t2)|=|∫0t1(t2−t1)ϕq(∫0sa(τ)f(τ,u(τ))∇τ−A)∇s+∫t1t2(t2−s)ϕq(∫0sa(τ)f(τ,u(τ))∇τ−A)∇s|≤c′|t1−t2|[∫0Tϕq(∫0sa(τ)∇τ+∑i=1m−2ai∫0ξia(τ)∇τ1−∑i=1m−2ai)∇s+Tmax⁡s∈[0,T]ϕq(∫0sa(τ)∇τ+∑i=1m−2ai∫0ξia(τ)∇τ1−∑i=1m−2ai)].So, by applying Arzela-Ascoli
theorem on time scales [16], we obtain that AKc¯ is relatively compact.
Finally, we prove thatA:Kc¯→K is continuous. Suppose that {un}n=1∞⊂Kc¯ and un(t) converges to u*(t) uniformly on [0,T]. Hence, {Aun(t)}n=1∞ is uniformly bounded and equicontinuous on [0,T]. The Arzela-Ascoli theorem on time scales [16]
tells us that there exists uniformly convergent subsequence in {Aun(t)}n=1∞. Let {Aun(m)(t)}m=1∞ be a subsequence which converges to v(t) uniformly on [0,T]. In addition,(2.42)0≤Aun(t)≤c′∫0T(T−s)ϕq(∫0sa(τ)∇τ+∑i=1m−2ai∫0ξia(τ)∇τ/(1−∑i=1m−2ai))∇s1−∑i=1m−2bi.Observe the expression of {Aun(m)(t)}, and then letting m→∞, we obtain(2.43)v(t)=−∫0t(t−s)ϕq(∫0sa(τ)f(τ,u*(τ))∇τ−A*)∇s+C*,where(2.44)A*=−∑i=1m−2ai∫0ξia(τ)f(τ,u*(τ))∇τ1−∑i=1m−2ai,C*=11−∑i=1m−2bi[∫0T(T−s)ϕq(∫0sa(τ)f(τ,u*(τ))∇τ−A*)∇s−∑i=1m−2bi∫0ξi(ξi−s)ϕq(∫0sa(τ)f(τ,u*(τ))∇τ−A*)∇s]. Here, we have used the Lebesgue
dominated convergence theorem on time scales [13]. From the
definition of A,
we know that v(t)=Au*(t) on [0,T]. This shows that each subsequence of {Aun(t)}n=1∞ uniformly converges to Au*(t). Therefore, the sequence {Aun(t)}n=1∞ uniformly converges to Au*(t). This means that A is continuous at u*∈Kc¯. So, A is continuous on Kc¯ since u* is arbitrary. Thus, A is completely continuous. This proof is
complete.Lemma 2.9.
Let(2.45)φ(s)=ϕq(∫0sa(τ)f(τ,u(τ))∇τ−A).For ξi⁢(i=1,…,m−2),(2.46)∫0ξi(ξi−s)φ(s)∇s≤ξiT∫0T(T−s)φ(s)∇s.Proof.
Since(2.47)∫0sa(τ)f(τ,u(τ))∇τ−A=∫0sa(τ)f(τ,u(τ))∇τ+∑i=1m−2ai∫0ξia(τ)f(τ,u(τ))∇τ1−∑i=1m−2ai≥0,then φ(s)≥0.
For all t∈(0,T],
we have(2.48)(∫0t(t−s)φ(s)∇st)∇=t∫0tφ(s)∇s−∫0t(t−s)φ(s)∇stρ(t)≥0.In fact, let ψ(t)=t∫0tφ(s)∇s−∫0t(t−s)φ(s)∇s;
taking the nabla derivative of this expression, we have(2.49)ψ∇(t)=∫0tφ(s)∇s+tφ(t)−∫0tφ(s)∇s=tφ(t)≥0.Hence, ψ(t) is a nondecreasing function on [0,T].
That is,(2.50)ψ(t)≥0.For all t∈(0,T],(2.51)∫0t(t−s)φ(s)∇st≤∫0T(T−s)φ(s)∇sT.By (2.51), for ξi⁢(i=1,…,m−2),
we have(2.52)∫0ξi(ξi−s)φ(s)∇s≤ξiT∫0T(T−s)φ(s)∇s.Lemma 2.10 (See [17]).
LetE be a Banach space, and let K⊂E be a cone. Assume Ω1,Ω2 are open bounded
subsets of E with 0∈Ω1,Ω1¯⊂Ω2, and let
F:K∩(Ω2¯∖Ω1)→K be a completely continuous operator such that
(i)
∥Fu∥≤∥u∥,⁢u∈K∩∂Ω1, and ∥Fu∥≥∥u∥,⁢u∈K∩∂Ω2,
or(ii)
∥Fu∥≥∥u∥,⁢u∈K∩∂Ω1,
and ∥Fu∥≤∥u∥,⁢u∈K∩∂Ω2.Then, F has a fixed point in K∩(Ω2¯∖Ω1).Now, we introduce the following notations. Let(2.53)A0={11−∑i=1m−2bi∫0T(T−s)ϕq[∫0sa(τ)∇τ+∑i=1m−2ai∫0ξia(τ)∇τ1−∑i=1m−2ai]∇s}−1,B0={T∑i=1m−2bi−∑i=1m−2biξiT(1−∑i=1m−2bi)∫0T(T−s)ϕq[∫0sa(τ)∇τ+∑i=1m−2ai∫0ξia(τ)∇τ1−∑i=1m−2ai]∇s}−1.For l>0,Ωl={u∈K:∥u∥<l},
and ∂Ωl={u∈K:∥u∥=l},(2.54)α(l)=sup⁡{∥Au∥:u∈∂Ωl},β(l)=inf⁡{∥Au∥:u∈∂Ωl},by Lemma 2.6, where
α and β are well defined.


## 3. Main Results

Theorem 3.1.
Assume
(H1), (H2), and (H3)
hold, and assume that the following conditions
hold:
(A1)
pi∈C([0,+∞),[0,+∞)),i=1,2, and (3.1)lim⁡l→0+¯⁢p1(l)lp−1<A0p−1,lim⁡l→∞¯⁢p2(l)lp−1<A0p−1;(A2)
ki∈L1([0,T],[0,+∞)),i=1,2;(A3)
there exist0<c1≤c2 and 0≤λ2<p−1<λ1 such that(3.2)f(t,l)≤p1(l)+k1(t)lλ1,(t,l)∈[0,T]×[0,c1],f(t,l)≤p2(l)+k2(t)lλ2,(t,l)∈[0,T]×[c2,+∞);(A4)
there existsb>0 such that(3.3)min⁡{f(t,l):(t,l)∈[0,T]×[γb,b]}≥(bB0)p−1.Then, problem (1.1)-(1.2) has at least two positive solutionsu1*,u2* satisfying 0<∥u1*∥<b<∥u2*∥.Theorem 3.2.
Assume
(H1), (H2),
and (H3)
hold, and assume that the following conditions
hold:
(B1)
pi∈C([0,+∞),[0,+∞)),i=3,4, and (3.4)lim⁡l→0+¯⁢p3(l)lp−1>(B0γ)p−1,lim⁡l→∞¯⁢p4(l)lp−1>(B0γ)p−1;(B2)
ki∈L1([0,T],[0,+∞)),i=3,4;(B3)
there exist0<c3≤c4 and 0≤λ4<p−1<λ3 such that(3.5)f(t,l)≥p3(l)−k3(t)lλ3,(t,l)∈[0,T]×[0,c3],f(t,l)≥p4(l)−k4(t)lλ4,(t,l)∈[0,T]×[c4,+∞);(B4)
there existsa>0 such that(3.6)max⁡{f(t,l):(t,l)∈[0,T]×[0,a]}≤(aA0)p−1.Then, problem (1.1)-(1.2) has at least two positive solutionsu3*,u4* satisfying 0<∥u3*∥<a<∥u4*∥.Proof  of Theorem3.1.
Let(3.7)ϵ=12min⁡[A0p−1−lim⁡l→0+¯⁢p1(l)lp−1,⁢A0p−1−lim⁡l→∞¯⁢p2(l)lp−1],then there exist 0<a¯1≤c1 and c2≤a¯2<+∞ such that(3.8)p1(l)≤(A0p−1−ϵ)lp−1,0≤l≤a¯1,p2(l)≤(A0p−1−ϵ)lp−1,a¯2≤l≤+∞.If 0≤l≤a¯1,⁢u∈∂Ωl,
then 0≤u(t)≤l,⁢0≤t≤T.
By condition (A3),
we have(3.9)f(t,u(t))≤p1(u(t))+k1(t)uλ1(t)≤(A0p−1−ϵ)up−1(t)+k1(t)uλ1(t)≤(A0p−1−ϵ)∥u∥p−1+k1(t)∥u∥λ1=(A0p−1−ϵ)lp−1+k1(t)lλ1so that(3.10)∫0sa(τ)f(τ,u(τ))∇τ−A=∫0sa(τ)f(τ,u(τ))∇τ+∑i=1m−2ai∫0ξia(τ)f(τ,u(τ))∇τ1−∑i=1m−2ai≤∫0sa(τ)[(A0p−1−ϵ)lp−1+k1(τ)lλ1]∇τ+∑i=1m−2ai∫0ξia(τ)[(A0p−1−ϵ)lp−1+k1(τ)lλ1]∇τ1−∑i=1m−2ai.Therefore,(3.11)∥Au∥≤C=11−∑i=1m−2bi(∫0T(T−s)φ(s)∇s−∑i=1m−2bi∫0ξi(ξi−s)φ(s)∇s)≤11−∑i=1m−2bi∫0T(T−s)φ(s)∇s≤11−∑i=1m−2bi×∫0T(T−s)ϕq{∫0sa(τ)[(A0p−1−ϵ)lp−1+k1(τ)lλ1]∇τ⁢+∑i=1m−2ai∫0ξia(τ)[(A0p−1−ϵ)lp−1+k1(τ)lλ1]∇τ1−∑i=1m−2ai}∇s.It follows that(3.12)α(l)l≤11−∑i=1m−2bi×∫0T(T−s)ϕq{∫0sa(τ)[A0p−1−ϵ+k1(τ)lλ1−p+1]∇τ⁢+∑i=1m−2ai∫0ξia(τ)[A0p−1−ϵ+k1(τ)lλ1−p+1]∇τ1−∑i=1m−2ai}∇s.Noticing λ1−p+1>0,
we have(3.13)lim⁡l→0+¯⁢α(l)l≤11−∑i=1m−2bi∫0T(T−s)ϕq[∫0sa(τ)(A0p−1−ϵ)∇τ+∑i=1m−2ai∫0ξia(τ)(A0p−1−ϵ)∇τ1−∑i=1m−2ai]∇s=(A0p−1−ϵ)1/(p−1)1−∑i=1m−2bi∫0T(T−s)ϕq[∫0sa(τ)∇τ+∑i=1m−2ai∫0ξia(τ)∇τ1−∑i=1m−2ai]∇s=(A0p−1−ϵ)1/(p−1)A0−1=(1−A0−(p−1)ϵ)1/(p−1)<1.Therefore, there exist 0<a1<a¯1 such that α(a1)<a1.
It implies that ∥Au∥<∥u∥,u∈∂Ωa1.
Ifa¯2≤l<+∞ and u∈∂Ωl,
then 0≤u(t)≤l.
Similar to the above argument, noticing that λ2−p+1<0,
we can get lim¯⁡l→∞(α(l)/l)<1.
Therefore, there exist 0<a¯2<a2 such that α(a2)<a2.
It implies that ∥Au∥<∥u∥,⁢u∈∂Ωa2.
On the other hand, sincef:[0,T]×[0,+∞)→[0,+∞) is continuous, by condition (A4),
there exist a1<b1<b<b2<a2 such that(3.14)min⁡{f(t,l):(t,l)∈[0,T]×[γbi,bi]}≥(biB0)p−1,i=1,2.If u∈∂Ωb1,
then γb1≤u(t)≤b1,⁢0≤t≤T.
Applying Lemma 2.9, it follows that(3.15)∥Au∥=max⁡0≤t≤T|(Au)(t)|≥−∫0T(T−s)φ(s)∇s+11−∑i=1m−2bi(∫0T(T−s)φ(s)∇s−∑i=1m−2bi∫0ξi(ξi−s)φ(s)∇s)=∑i=1m−2bi1−∑i=1m−2bi∫0T(T−s)φ(s)∇s−∑i=1m−2bi∫0ξi(ξi−s)φ(s)∇s1−∑i=1m−2bi≥∑i=1m−2bi1−∑i=1m−2bi∫0T(T−s)φ(s)∇s−∑i=1m−2biξiT(1−∑i=1m−2bi)∫0T(T−s)φ(s)∇s=T∑i=1m−2bi−∑i=1m−2biξiT(1−∑i=1m−2bi)∫0T(T−s)φ(s)∇s≥T∑i=1m−2bi−∑i=1m−2biξiT(1−∑i=1m−2bi)∫0T(T−s)ϕq[∫0sa(τ)(b1B0)p−1∇τ+∑i=1m−2ai∫0ξia(τ)(b1B0)p−1∇τ1−∑i=1m−2ai]∇s=T∑i=1m−2bi−∑i=1m−2biξiT(1−∑i=1m−2bi)b1B0∫0T(T−s)ϕq[∫0sa(τ)∇τ+∑i=1m−2ai∫0ξia(τ)∇τ1−∑i=1m−2ai]∇s=b1B0B0−1=b1=∥u∥.
In the same way, we
can prove
that if u∈∂Ωb2,
then ∥Au∥≥∥u∥.Now, we consider the operatorA on Ωb1¯∖Ωa1 and Ωa2¯∖Ωb2,
respectively. By Lemma 2.10, we assert that the operator A has two fixed points u1*,u2*∈K such that a1≤∥u1*∥≤b1 and b2≤∥u2*∥≤a2.
Therefore, ui*,i=1,2,
are positive solutions of problem (1.1)-(1.2).Proof  of Theorem3.2.
Let(3.16)ϵ=12min⁡[lim⁡l→0+¯⁢p3(l)lp−1−(B0γ)p−1,⁢lim⁡l→∞¯⁢p4(l)lp−1−(B0γ)p−1],then there exist 0<b¯3≤c3 and c4≤b¯4<+∞ such that(3.17)p3(l)≥[(B0γ)p−1+ϵ]lp−1,0≤l≤b¯3,p4(l)≥[(B0γ)p−1+ϵ]lp−1,b¯4≤l≤+∞. If 0≤l≤b¯3,⁢u∈∂Ωl,
then γl≤u(t)≤l,⁢0≤t≤T.
By Lemma 2.9 and condition (B3),
we have(3.18)∥Au∥=max⁡0≤t≤T|(Au)(t)|≥T∑i=1m−2bi−∑i=1m−2biξiT(1−∑i=1m−2bi)∫0T(T−s)φ(s)∇s≥T∑i=1m−2bi−∑i=1m−2biξiT(1−∑i=1m−2bi)×∫0T(T−s)ϕq{∫0sa(τ)[p3(u(τ))−k3(τ)uλ3]∇τ⁢+∑i=1m−2ai∫0ξia(τ)[p3(u(τ))−k3(τ)uλ3]∇τ1−∑i=1m−2ai}∇s≥T∑i=1m−2bi−∑i=1m−2biξiT(1−∑i=1m−2bi)×∫0T(T−s)ϕq{∫0sa(τ)[((B0γ)p−1+ϵ)(γl)p−1−k3(τ)lλ3]∇τ+∑i=1m−2ai∫0ξia(τ)[((B0/γ)p−1+ϵ)(γl)p−1−k3(τ)lλ3]∇τ1−∑i=1m−2ai}∇s.It follows that(3.19)β(l)l≥T∑i=1m−2bi−∑i=1m−2biξiT(1−∑i=1m−2bi)×∫0T(T−s)ϕq{∫0sa(τ)[B0p−1+γp−1ϵ−k3(τ)lλ3−p+1]∇τ+∑i=1m−2ai∫0ξia(τ)[B0p−1+γp−1ϵ−k3(τ)lλ3−p+1]∇τ1−∑i=1m−2ai}∇s.Noticing λ3−p+1>0,
we get(3.20)lim⁡l→0+¯⁢β(l)l≥T∑i=1m−2bi−∑i=1m−2biξiT(1−∑i=1m−2bi)×∫0T(T−s)ϕq[∫0sa(τ)(B0p−1+γp−1ϵ)∇τ+∑i=1m−2ai∫0ξia(τ)(B0p−1+γp−1ϵ)∇τ1−∑i=1m−2ai]∇s=(B0p−1+γp−1ϵ)1/(p−1)B0−1=(1+γp−1B0−(p−1)ϵ)1/(p−1)>1.Therefore, there
exists
b3 with 0<b3<a such that β(b3)>b3.
It implies that ∥Au∥>∥u∥ for u∈∂Ωb3.
Ifb¯4≤γl<+∞ and u∈∂Ωl,
then b¯4≤γl≤u(t)≤l,⁢0≤t≤T.
Similar to the above argument, noticing that λ4−p+1<0,
we can get lim̲l→+∞⁢(β(l)/l)>1.
Therefore, there existb4 with 0<b4<+∞ such that β(b4)>b4.
It implies that ∥Au∥>∥u∥ for u∈∂Ωb4.
By condition(B4),
we can see that there exist b3<a3<a<a4<b4 such that(3.21)max⁡{f(t,l):(t,l)∈[0,T]×[0,ai]}≤(aiA0)p−1,i=3,4.If u∈∂Ωa3,
then 0≤u(t)≤a3, 0≤t≤T,
and f(t,u(t))≤(a3A0)p−1.
It follows that(3.22)∥Au∥≤11−∑i=1m−2bi∫0T(T−s)φ(s)∇s≤11−∑i=1m−2bia3A0∫0T(T−s)ϕq[∫0sa(τ)∇τ+∑i=1m−2ai∫0ξia(τ)∇τ1−∑i=1m−2ai]∇s=a3=∥u∥.Similarly, if u∈∂Ωa4,
then ∥Au∥≤∥u∥.Now, we study the operatorA on Ωa3¯∖Ωb3 and Ωb4¯∖Ωa4,
respectively. By Lemma 2.10, we assert that the operator A has two fixed points u3*,u4*∈K such that b3≤∥u3*∥≤a3 and a4≤∥u4*∥≤b4.
Therefore, ui*,⁢i=3,4,
are positive solutions of problem (1.1)-(1.2).


## 4. Further Discussion

If the
conditions of Theorems3.1 and
3.2 are weakened, we will get the existence of
single positive solution of problem (1.1)-(1.2).Corollary 4.1.
Assume
(H1), (H2),
and (H3)
hold, and assume that the following conditions
hold:
(C1)
p1∈C([0,+∞),[0,+∞)),
and lim¯l→0+(p1(l)/lp−1)<A0p−1;(C2)
k1∈L1([0,T],[0,+∞));(C3)
there existc1>0 and λ1>p−1 such that(4.1)f(t,l)≤p1(l)+k1(t)lλ1,(t,l)∈[0,T]×[0,c1];(C4)
there existsb>0 such that(4.2)min⁡{f(t,l):(t,l)∈[0,T]×[γb,b]}≥(bB0)p−1.Then, problem (1.1)-(1.2) has at least one positive solution.Corollary 4.2.
Assume
(H1), (H2),
and (H3)
hold, and assume that the following conditions
hold:
(D1)
p2∈C([0,+∞),[0,+∞)), and lim¯l→∞(p2(l)/lp−1)<A0p−1;(D2)
k2∈L1([0,T],[0,+∞));(C3)
there existc2>0 and 0≤λ2<p−1 such that(4.3)f(t,l)≤p2(l)+k2(t)lλ2,(t,l)∈[0,T]×[c2,+∞);(D4)
there existsb>0 such that(4.4)min⁡{f(t,l):(t,l)∈[0,T]×[γb,b]}≥(bB0)p−1.Then, problem (1.1)-(1.2) has at least one positive solution.Corollary 4.3.
Assume
(H1), (H2),
and (H3)
hold, and assume that the following conditions
hold:
(E1)
p3∈C([0,+∞),[0,+∞)), and lim̲l→0+(p3(l)/lp−1)>(B0/γ)p−1;(E2)
k3∈L1([0,T],[0,+∞));(E3)
there existc3>0 and λ3>p−1 such that(4.5)f(t,l)≥p3(l)−k3(t)lλ3,(t,l)∈[0,T]×[0,c3];(E4)
there existsa>0 such that(4.6)max⁡{f(t,l):(t,l)∈[0,T]×[0,a]}≤(aA0)p−1.Then, problem (1.1)-(1.2) has at least one positive solution.Corollary 4.4.
Assume
(H1), (H2),
and (H3)
hold, and assume that the following conditions
hold:
(F1)
p4∈C([0,+∞),[0,+∞)), and lim̲l→∞(p4(l)/lp−1)>(B0/γ)p−1;(F2)
k4∈L1([0,T],[0,+∞));(F3)
there existc4>0 and 0≤λ4<p−1 such that(4.7)f(t,l)≥p4(l)−k4(t)lλ4,(t,l)∈[0,T]×[c4,+∞);(F4)
there existsa>0 such that(4.8)max⁡{f(t,l):(t,l)∈[0,T]×[0,a]}≤(aA0)p−1.Then, problem (1.1)-(1.2) has at least one positive solution.The proof of the above results is similar to those of
Theorems3.1 and
3.2; thus we omit it.


## 5. Some Examples

In this
section, we present a simple example to explain our results. We only study the
caseT=R,(0,T)=(0,1).Letf(t,0)≡0.
Consider the following BVP:(5.1)(ϕ3(u′′))′+f(t,u(t))=0,t∈(0,1),ϕ3(u′′(0))=12ϕ3(u′′(12)),u′(0)=0,u(1)=12u(12), where(5.2)f(t,u)={223u3+min⁡{1t(1−t),2u}u5,(t,u)∈[0,1]×[0,1],225,(t,u)∈[0,1]×[1,3],74u+36min⁡{1t(1−t),2u3}u3,(t,u)∈[0,1]×[3,+∞).It is easy to check that f:[0,1]×[0,+∞)→[0,+∞) is continuous. In this case, p=3,a(t)≡1,⁢m=3,a1=b1=1/2,
and ξ1=1/2,
and it follows from a direct calculation that(5.3)A0=[11−1/2∫01(1−s)ϕq(s+(1/2)⋅(1/2)1−1/2)ds]−1=1.1062,γ=b1(1−ξ1)1−b1ξ1=(1/2)(1−1/2)1−(1/2)⋅(1/2)=13.We have(5.4)B0=[b1−b1ξ11−b1∫0T(1−s)ϕq(s+a1ξ11−a1)ds]−1=[1/2−(1/2)⋅(1/2)1−1/2∫01(1−s)(s+(1/2)⋅(1/2)1−1/2)1/2ds]−1=4.4248<5.Choosing c1=1,c2=3,b=3,λ1=5/2,⁢λ2=3/2,⁢p1(u)=223u3,p2(u)=74u,
and k1(t)=k2(t)=1/t(1−t),
it is easy to check that(5.5)f(t,u)≤p1(u)+k1(t)u5/2,(t,u)∈[0,1]×[0,1],f(t,u)≤p2(u)+k2(t)u3/2,(t,u)∈[0,1]×[3,+∞),lim⁡u→0¯⁢p1(u)u2=lim⁡l→0¯⁢223u3u2=0<A02=(1.1062)2,lim⁡u→∞¯⁢p2(u)u2=lim⁡l→∞¯⁢74uu2=0<A02=(1.1062)2,min⁡{f(t,u):(t,u)∈[0,1]×[1,3]}=225>(13.274)2=(bB0)2.It follows that f satisfies the conditions (A1)–(A4) of Theorem 3.1; then problem (5.1) has
at least two positive solutions.



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*Source: 257680-2008-07-28.xml*