# Higher-Order Dynamic Delay Differential Equations on Time Scales

Authors: Hua Su; Lishan Liu; Xinjun Wang
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We study the existence of positive solutions for the nonlinear four-point singular boundary value problem with higher-orderp-Laplacian dynamic delay differential equations on time scales, subject to some boundary conditions. By using the fixed-point index theory, the existence of positive solution and many positive solutions for nonlinear four-point singular boundary value problem with p-Laplacian operator are obtained.


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## Body

## 1. Introduction

The study of dynamic equations on time scales goes back to its founder Stefan Hilger [1] and is a new area of still fairly theoretical exploration in mathematics. Boundary value problems for delay differential equations arise in a variety of areas of applied mathematics, physics, and variational problems of control theory (see [2, 3]). In recent years, many authors have begun to pay attention to the study of boundary value problems or with p-Laplacian equations or with p-Laplacian dynamic equations on time scales (see [4–19] and the references therein).In [7], Sun and Li considered the existence of positive solution of the following dynamic equations on time scales:
(1.1)uΔ∇(t)+a(t)f(t,u(t))=0,t∈(0,T),βu(0)-γuΔ(0)=0,αu(η)=u(T),
where β,γ≥0, β+γ>0, η∈(0,ρ(T)), 0<α<T/η. They obtained the existence of single and multiple positive solutions of the problem (1.1) by using fixed point theorem and Leggett-Williams fixed point theorem, respectively.In [10],  Avery and Anderson discussed the following dynamic equation on time scales:
(1.2)uΔ∇(t)+a(t)f(u(t))=0,t∈(0,T),u(0)=0,αu(η)=u(T).
He obtained some results for the existence of one positive solution of the problem (1.2) based on the limits f0=limu→0+(f(u)/u) and f∞=limu→∞(f(u)/u).In [11], Wang et al. discussed the following dynamic equation by using Avery-Peterson fixed theorem (see [10]):
(1.3)(ϕp(u′))′+q(t)f(t,u(t),u(t-1),u′(t))=0,t∈(0,1),(1.4)u(t)=ξ(t),-1≤t≤0,u(1)=0,(1.5)u(t)=ξ(t),-1≤t≤0,u′(1)=0.
They obtained some results for the existence, three positive solutions of the problem (1.3), (1.4) and (1.3), (1.5), respectively.However, there are not many concerning thep-Laplacian problems on time scales. Especially, for the singular multi point boundary value problems for higher-order p-Laplacian dynamic delay differential equations on time scales, with the author’s acknowledg, no one has studied the existence of positive solutions in this case.Recently, in [16], we study the existence of positive solutions for the following nonlinear two-point singular boundary value problem with p-Laplacian operator
(1.6)(ϕp(u′))′+a(t)f(u(t))=0,0<t<1,αϕp(u(0))-βϕp(u′(0))=0,γϕp(u(1))+δϕp(u′(1))=0,
by using the fixed point theorem of cone expansion and compression of norm type, the existence of positive solution and infinitely many positive solutions for nonlinear singular boundary value problem (1.6) with p-Laplacian operator are obtained.Now, motivated by the results mentioned above, in this paper, we study the existence of positive solutions for the following nonlinear four-point singular boundary value problem with higher-orderp-Laplacian dynamic delay differential equations operator on time scales (SBVP):
(1.7)(ϕp(uΔn-1(t)))∇+g(t)f(u(t),u(t-τ),uΔ(t),…,uΔn-2(t))=0,0<t<T,(1.8)u(t)=ζ(t),-τ≤t≤0,uΔi(0)=0,1≤i≤n-3,αϕp(uΔn-2(0))-βϕp(uΔn-1(ξ))=0,γϕp(uΔn-2(T))+δϕp(uΔn-1(η))=0,n≥3,
where ϕp(s) is p-Laplacian operator, that is, ϕp(s)=|s|p-2s, p>1, ϕq=ϕp-1, 1/p+1/q=1; ξ,η∈(0,T), τ∈[0,T] is prescribed and ξ<η, g:(0,T)→[0,∞), α>0, β≥0, γ>0, δ≥0.In this paper, by constructing one integral equation which is equivalent to the problem (1.7), (1.8), we research the existence of positive solutions for nonlinear singular boundary value problem (1.7), (1.8) when g and f satisfy some suitable conditions.Our main tool of this paper is the following fixed point index theory.Theorem 1.1 (see [18]).
SupposeE is a real Banach space, K⊂E is a cone, let Ωr={u∈K:∥u∥≤r}. Let operator T:Ωr→K be completely continuous and satisfy Tx≠x, for all x∈∂Ωr. Then(i)
if∥Tx∥≤∥x∥, for all x∈∂Ωr, then i(T,Ωr,K)=1;(ii)
if∥Tx∥≥∥x∥, for all x∈∂Ωr, then i(T,Ωr,K)=0.This paper is organized as follows. In Section2, we present some preliminaries and lemmas that will be used to prove our main results. In Section 3, we study the existence of at least two solutions of the systems (1.7), (1.8). In Section 4, we give an examples as the application.


## 2. Preliminaries and Lemmas

A time scaleT is an arbitrary nonempty closed subset of real numbers R+. In [1, 14, 20], we can find some basic definitions about time scale. The operators σ and ρ from T to T:
(2.1)σ(t)=inf⁡{τ∈T∣τ>t}∈T,ρ(t)=sup⁡{τ∈T∣τ<t}∈T
are called the forward jump operator and the backward jump operator, respectively.IfT=R, then xΔ(t)=x∇(t)=x′(t). If T=Z, then xΔ(t)=x(t+1)-x(t) is the forward difference operator, while x∇(t)=x(t)-x(t-1) is the backward difference operator.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. It is well known that f is ld-continuous.IfF∇(t)=f(t), then we define the nabla integral by
(2.2)∫abf(t)∇t=F(b)-F(a).
If FΔ(t)=f(t), then we define the delta integral by(2.3)∫abf(t)Δt=F(b)-F(a).In the rest of this paper,T is closed subset of R with 0∈Tk, T∈Tk. And let(2.4)B={u∈C[-τ,0]∩Cldn-2[0,T]:uΔi(0)=0,0≤i≤n-3}.Here,(2.5)Cldn-2[0,T]={u:[0,T]→R∣u(t)isleft-densen-2ordercontinuouslydifferentiable}.
Then B is a Banach space with the norm ∥u∥=maxt∈[0,T]|uΔn-2(t)|. And let(2.6)K={u∈B:uΔn-2(t)≥0,uΔn-2(t)isconcavefunction,t∈[0,T]}.Obviously,K is a cone in B. Set Kr={u∈K:∥u∥≤r}.Definition 2.1.
u(t) is called a solution of SBVP (1.7) and (1.8) if it satisfies the following:(1)
u∈C[-τ,0]∩Cldn-1(0,T);(2)
u(t)>0 for all t∈(0,T) and satisfy conditions (1.8);(3)
(ϕp(uΔn-1(t)))∇=-g(t)f(u(t),u(t-τ),uΔ(t),…,uΔn-2(t)) hold for t∈(0,T).
In the rest of the paper, we also make the following assumptions:
(H1)
f∈Cld([0,+∞)n,[0,+∞));(H2)
g(t)∈Cld((0,T),[0,+∞)) and there exists t0∈(0,T), such that(2.7)g(t0)>0,0<∫0Tg(s)∇s<+∞;(H3)
ζ(t)∈C([-τ,0], ζ(t)>0 on [-τ,0) and ζ(0)=0.It is easy to check that condition (H2) implies that
(2.8)0<∫0Tϕq(∫0sg(s1)∇s1)Δs<+∞.
We can easily get the following lemmas.Lemma 2.2.
Suppose condition(H2) holds. Then there exists a constant θ∈(0,1/2) satisfing
(2.9)0<∫θT-θg(t)∇t<∞.
Furthermore, the function
(2.10)A(t)=∫θtϕq(∫stg(s1)∇s1)Δs+∫tT-θϕq(∫tsg(s1)∇s1)∇s,t∈[θ,T-θ],
is positive continuous functions on [θ,T-θ]; therefore, A(t) has minimum on [θ,T-θ]. Hence we suppose, that there exists L>0 such that A(t)≥L, t∈[θ,T-θ].Proof.
At first, it is easily seen thatA(t) is continuous on [θ,T-θ]. Next, let
(2.11)A1(t)=∫θtϕq(∫stg(s1)∇s1)Δs,A2(t)=∫tT-θϕq(∫tsg(s1)∇s1)Δs.
Then, from condition (H2), we have that the function A1(t) is strictly monotone nondecreasing on [θ,T-θ] and A1(θ)=0, the function A2(t) is strictly monotone nonincreasing on [θ,T-θ] and A2(T-θ)=0, which implies L=mint∈[θ,T-θ]A(t)>0. The proof is complete.Lemma 2.3.
Letu∈K and θ of Lemma 2.2, then
(2.12)u(t)≥θ‖u‖,t∈[θ,T-θ].
The proof of the above lemma is similar to the proof in [17, Lemma 2.2], so we omit it.Lemma 2.4.
Suppose that conditions(H1),(H2),(H3) hold, u(t)∈B∩Cldn-1(0,1) is a solution of the following boundary value problems:
(2.13)(ϕp(uΔn-1(t)))∇+g(t)f(u(t),u(t-τ)+h(t-τ),uΔ(t),…,uΔn-2(t))=0,0<t<T,(2.14)u(t)=0,-τ≤t≤0,uΔi(0)=0,1≤i≤n-3,αϕp(uΔn-2(0))-βϕp(uΔn-1(ξ))=0,γϕp(uΔn-2(T))+δϕp(uΔn-1(η))=0,n≥3,
where
(2.15)h(t)={ζ(t),-τ≤t≤0,0,0≤t≤T.
Then, u¯(t)=u(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.7) and (1.8).Proof.
It is easy to check thatu¯(t) satisfies (1.7) and (1.8).So in the rest of the sections of this paper, we focus on SBVP (2.13) and (2.14).Lemma 2.5.
Suppose that conditions(H1),(H2),(H3) hold, u(t)∈B∩Cldn-1(0,1) is a solution of boundary value problems (2.13), (2.14) if and only if u(t)∈B is a solution of the following integral equation:
(2.16)u(t)={ζ(t),-τ≤t≤0,∫0t∫0s1⋯∫0sn-3w(sn-2)Δsn-2Δsn-3⋯Δs1,0≤t≤T,
where
(2.17)w(t)={ϕq(βα∫ξσg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)+∫0tϕq(∫sσg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs,0≤t≤σ,ϕq(δγ∫σηg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)+∫tTϕq(∫σsg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs,σ≤t≤T.Proof.
Necessity. Obviously, for t∈(-τ,0), we have u(t)=ζ(t). If t∈(0,1), by the equation of the boundary condition, we have uΔn-1(ξ)≥0, uΔn-1(η)≤0, then there exists a constant σ∈[ξ,η]⊂(0,T) such that uΔn-1(σ)=0.
Firstly, by integrating the equation of the problems (2.13) on (σ,T), we have(2.18)ϕp(uΔn-1(t))=ϕp(uΔn-1(σ))-∫σtg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s,
then
(2.19)uΔn-1(t)=-ϕq(∫σtg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s),
thus
(2.20)uΔn-2(t)=uΔn-2(σ)-∫σtϕq(∫σsg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs.
ByuΔn-1(σ)=0 and condition (2.18), let t=η on (2.18), we have(2.21)ϕp(uΔn-1(η))=-∫σηg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s.
By the equation of the boundary condition (2.14), we have
(2.22)ϕp(uΔn-2(T))=-δγϕp(uΔn-1(η)),
then
(2.23)uΔn-2(T)=ϕq(δγ∫σηg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s).
Then, by (2.20) and leting t=T on (2.20), we have
(2.24)uΔn-2(σ)=ϕq(δγ∫σηg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)+∫σTϕq(∫σsg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs.
Then
(2.25)uΔn-2(t)=ϕq(δγ∫σηg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)+∫tTϕq(∫σsg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs.
Then, by integrating (2.25) for n-2 times on (0,T), we have
(2.26)u(t)=∫0t∫0s1⋯∫0sn-3ϕq(σγ∫δηg(s)f(uΔn-2u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇sσγ∫δη)Δssn-2⋯Δs2Δs1+∫0t∫0s1⋯∫0sn-3(∫sn-2Tϕq(∫σsg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))×∇r∫σs)Δs∫sn-2T)Δssn-2⋯Δs2Δs1.
Similarly, for t∈(0,σ), by integrating the equation of problems (2.13) on (0,σ), we have
(2.27)u(t)=∫0t∫0s1⋯∫0sn-3ϕq(βα∫ξσg(s)f(uΔn-2u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇sβα∫ξσ)Δssn-2⋯Δs2Δs1+∫0t∫0s1⋯∫0sn-3(∫0sn-2ϕq(∫sσg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))×∇r∫sσ)Δs∫0sn-2)Δssn-2⋯Δs2Δs1.
Therefore, for any t∈[0,T], u(t) can be expressed as equation
(2.28)u(t)={ζ(t),-τ≤t≤0,∫0t∫0s1⋯∫0sn-3w(sn-2)Δsn-2Δsn-3⋯Δs1,0≤t≤T,
where w(t) is expressed as (2.17). Then the results of Lemma 2.3 hold.
Sufficiency. Suppose that u(t)=∫0t∫0s1⋯∫0sn-3w(sn-2)Δsn-2Δsn-3⋯Δs1, 0≤t≤T. Then by (2.17), we have(2.29)uΔn-1(t)={ϕq(∫tσg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)≥0,0≤t≤σ,-ϕq(∫σtg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)≤0,σ≤t≤T.
So, (ϕp(uΔn-1))∇+g(t)f(u(t),u(t-τ)+h(t-τ),uΔ(t),…,uΔn-2(t))=0, 0<t<T. These imply that (2.13) holds. Furthermore, by letting t=0 and t=T on (2.17) and (2.29), we can obtain the boundary value equations of (2.14). The proof is complete.Now, we define an operator equationT given by
(2.30)(Tu)(t)={ζ(t),-τ≤t≤0,∫0t∫0s1⋯∫0sn-3w(sn-2)Δsn-2Δsn-3⋯Δs1,0≤t≤T,
where w(t) is given by (2.17).From the definition ofT and the previous discussion, we deduce that, for each u∈K, Tu∈K. Moreover, we have the following lemmas.Lemma 2.6.
T:K→K is completely continuous.Proof.
Because(2.31)(Tu)Δn-1(t)=wΔ(t)={ϕq(∫tσg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)≥0,0≤t≤σ,-ϕq(∫σtg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)≤0,σ≤t≤T,
is continuous, decreasing on [0,T] and satisfies (Tu)Δn-1(σ)=0, then, Tu∈K for each u∈K and (Tu)Δn-2(σ)=maxt∈[0,T](Tu)Δn-2(t). This shows that TK⊂K. Furthermore, it is easy to check by Arzela-ascoli Theorem that T:K→K is completely continuous.Lemma 2.7.
Suppose that conditions(H1),(H2),(H3) hold, the solution u(t) of problem (2.13), (2.14) satisfies
(2.32)max⁡0≤t≤T|u(t-τ)+h(t-τ)|≤max⁡-τ≤t≤0|ζ(t)|,u(t)≤TuΔ(t)≤⋯≤Tn-3uΔn-3(t),t∈[0,T],
and for θ∈(0,T/2) in Lemma 2.2, one has
(2.33)uΔn-3(t)≤TθuΔn-2(t),t∈[θ,T-θ].Proof.
Firstly, we can have(2.34)max⁡0≤t≤T|u(t-τ)+h(t-τ)|≤max⁡0≤t≤T|u(t-τ)|+max⁡0≤t≤T|h(t-τ)|=max⁡-τ≤t≤T-τ|u(t)|+max⁡-τ≤t≤T-τ|h(t)|=max⁡-τ≤t≤0|ζ(t)|.
Next, ifu(t) is the solution of problem (2.13), (2.14), then uΔn-2(t) is concave function, and uΔi(t)≥0(i=0,1,…,n-2), t∈[0,T]. Thus, we have(2.35)uΔi(t)=∫0tuΔi+1(s)Δs≤tuΔi+1(t)≤TuΔi+1(t),i=0,1,…,n-4,
that is, u(t)≤TuΔ(t)≤⋯≤Tn-3uΔn-3(t), t∈[0,T].
Finally, by Lemma2.3, for t∈[θ,T-θ], we have uΔn-2(t)≥θ∥uΔn-2∥. By uΔn-3(t)=∫0tuΔn-2(s)Δs≤T∥uΔn-2∥, we have
(2.36)uΔn-3(t)≤TθuΔn-2(t),t∈[θ,T-θ].
The proof is complete.For convenience, we set(2.37)H=max⁡-τ≤t≤0|ζ(t)|,θ*=2L,θ*=1(T+ϕq(β/α))ϕq(∫0Tg(r)∇r),m∈(θ*,∞),M∈(0,θ*),
where L is the constant from Lemma 2.2. By Lemma 2.5, we can also set(2.38)f0=lim⁡un→0max⁡(u1,u2,…,un)∈ℵf(u1,u2,…,un)unp-1,f∞=lim⁡un→∞min⁡(u1,u2,…,un)∈ℵf(u1,u2,…,un)unp-1,
where ℵ={(u1,u2,…,un)|0≤u1≤Tu3⋯≤Tn-3un-1≤(Tn-2/θ)un,u2≤H}.


## 3. The Existence of Multiple Positive Solutions

In this section, we also make the following conditions:(A1)
f(u1,u2,…,un)≥(mr)p-1, for θr≤un≤r, (u1,u2,…,un)∈ℵ;(A2)
f(u1,u2,…,un)≤(MR)p-1, for 0≤un≤R, (u1,u2,…,un)∈ℵ.Next, we will discuss the existence of multiple positive solutions.Theorem 3.1.
Suppose that conditions (H1), (H2), (H3), and (A2) hold. Assume that f also satisfies(A3)
f0=+∞;(A4)
f∞=+∞.Then, the SBVP (2.13), (2.14) hase at last two solutions u1, u2 such that
(3.1)0<‖u1‖<R<‖u2‖.Proof.
For anyu∈K, by Lemma 2.3, we have
(3.2)uΔn-2(t)≥θ‖u‖,t∈[θ,T-θ].
First, by condition (A3), for any N>2/θL, there exists a constant ρ*∈(0,R) such that
(3.3)f(u1,u2,…,un)≥(Nun)p-1,0<un≤ρ*,un≠0.
Set Ωρ*={u∈K:∥u∥<ρ*}. For any u∈∂Ωρ*, by (3.2) we have
(3.4)ρ*=‖u‖≥uΔn-2(t)≥θ‖u‖=θρ*,t∈[θ,T-θ].
For any u∈∂Ωρ*, by (3.3) and Lemmas 2.3–2.6, we will discuss it from three perspectives.(i)
Ifσ∈[θ,T-θ], we have
(3.5)2‖Tu‖=2(Tu)Δn-2(σ)≥∫0σϕq(∫sσg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs+∫σTϕq(∫σsg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs≥∫θσϕq(∫sσg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs+∫σT-θϕq(∫σsg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs≥NθA(σ)‖u‖≥2‖u‖.(ii)
Ifσ∈(T-θ,T], we have(3.6)‖Tu‖=(Tu)Δn-2(σ)≥ϕq(βα∫ξσg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)+∫0σϕq(∫sσg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs≥∫θT-θϕq(∫sT-θg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs≥NθA(T-θ)‖u‖>‖u‖.(iii)
Ifσ∈(0,θ), we have
(3.7)‖Tu‖=(Tu)Δn-2(σ)≥ϕq(δγ∫σηg(s)f(u(s),u(s-τ)+h(r-τ),uΔ(s),…,uΔn-2(s))∇s)+∫σTϕq(∫σsg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs≥∫θT-θϕq(∫θsg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs≥NθA(θ)‖u‖>‖u‖.
Therefore, no matter under which condition, we all have
(3.8)‖Tu‖≥‖u‖,∀u∈∂Ωρ*.
Then, by Theorem 1.1, we have
(3.9)i(T,Ωρ*,K)=0.Next, by condition (A4), for any N¯>2/θL, there exists a constant ρ0>0 such that
(3.10)f(u1,u2,…,un)≥(N¯un)p-1,un>ρ0.
We choose a constant ρ*>max{R,ρ0/θ}, obviously ρ*<R<ρ*. Set Ωρ*={u∈K:∥u∥<ρ*}. For any u∈∂Ωρ*, by Lemma 2.3, we have
(3.11)u(t)≥θ‖u‖=θρ*>ρ0,t∈[θ,T-θ].
Then, by (3.10), Lemmas 2.3–2.6 and also similar to the previous proof, we can also have from three perspectives that
(3.12)‖Tu‖≥‖u‖,∀u∈∂Ωρ*.
Then, by Theorem 1.1, we have
(3.13)i(T,Ωρ*,K)=0.
Finally, set ΩR={u∈K:∥u∥<R}. For any u∈∂ΩR, we have u(t)≤∥u∥=R, by (A2) we know
(3.14)‖Tu‖=(Tu)Δn-2(σ)≤ϕq(βα∫ξσg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)+∫0Tϕq(∫sσg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs≤(T+ϕq(βα))MRϕq(∫0Tg(r)∇r)≤R=‖u‖.
Thus,
(3.15)‖Tu‖≤‖u‖,∀u∈∂ΩR.
Then, by Theorem 1.1, we have
(3.16)i(T,ΩR,K)=1.
Therefore, by (3.9), (3.13), (3.16), ρ*<R<ρ* we have
(3.17)i(T,ΩR∖Ω¯ρ*,K)=1,i(T,Ωρ*∖Ω¯R,K)=-1.
Then T has fixed point u1∈ΩR∖Ω¯ρ* and fixed point u2∈Ωρ*∖Ω¯R. Obviously, u1,u2 are all positive solutions of problem (2.13), (2.14) and ρ*<∥u1∥<R<∥u2∥<ρ*. Proof of Theorem 3.1 is complete.Theorem 3.2.
Suppose that conditions (H1), (H2), (H3), (A1) hold. Assume that f also satisfies(A5)
f0=0;(A6)
f∞=0.Then, the SBVP (2.13), (2.14) has at last two solutions u1,u2 such that 0<∥u1∥<r<∥u2∥.Proof.
First, byf0=0, for ϵ1∈(0,θ*), there exists a constant ρ*∈(0,r) such that
(3.18)f(u1,u2,…,un)≤(ϵ1un)p-1,0<un≤ρ*.
Set Ωρ*={u∈K:∥u∥<ρ*}, for any u∈∂Ωρ*, by (3.18), we have
(3.19)‖Tu‖=(Tu)Δn-2(σ)≤ϕq(βα∫ξσg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)+∫0Tϕq(∫sσg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs≤ϕq(βα∫ξσg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)+Tϕq(∫0Tg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)≤(T+ϕq(βα))ϵ1ρ*ϕq(∫0Tg(r)∇r)≤ρ*=‖u‖,
that is,
(3.20)‖Tu‖≤‖u‖,∀u∈∂Ωρ*
Then, by Theorem 1.1, we have
(3.21)i(T,Ωρ*,K)=1.
Next, letf*(x)=max0≤un-1≤xf(u1,u2,…,un-1); note that f*(x) is monotone increasing with respect to x≥0. Then, from f∞=0, it is easy to see that(3.22)lim⁡x→∞f*(x)xp-1=0.
Therefore, for any ϵ2∈(0,θ*), there exists a constant ρ*>r such that
(3.23)f*(x)≤(ϵ2x)p-1,x≥ρ*.
Set Ωρ*={u∈K:∥u∥<ρ*}, for any u∈∂Ωρ*, by (3.23), we have
(3.24)‖Tu‖=(Tu)Δn-2(σ)≤ϕq(βα∫ξσg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)+∫0Tϕq(∫sσg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)Δs≤ϕq(βα∫ξσg(s)f(u(s),u(s-τ)+h(s-τ),uΔ(s),…,uΔn-2(s))∇s)+Tϕq(∫0Tg(r)f(u(r),u(r-τ)+h(r-τ),uΔ(r),…,uΔn-2(r))∇r)≤(T+ϕq(βα))ϕq(∫0Tg(r)f*(ρ*)∇r)≤(T+ϕq(βα))ϵ2ρ*ϕq(∫0Tg(r)∇r)≤r*=‖u‖,
that is,
(3.25)‖Tu‖≤‖u‖,∀u∈∂Ωρ*.
Then, by Theorem 1.1, we have
(3.26)i(T,Ωρ*,K)=1.
Finally, setΩr={u∈K:∥u∥<r}. For any u∈∂Ωr, by (A1), Lemma 2.3 and also similar to the previous proof of Theorem 3.1, we can also have(3.27)‖Tu‖≥‖u‖,∀u∈∂Ωr.
Then, by Theorem 1.1, we have
(3.28)i(T,Ωr,K)=0.
Therefore, by (3.21), (3.28), (3.26), ρ*<r<ρ*, we have
(3.29)i(T,Ωr∖Ω¯ρ*,K)=-1,i(T,Ωρ*∖Ω¯r,K)=1.
Then T has fixed point u1∈Ωr∖Ω¯ρ* and fixed point u2∈Ωρ*∖Ω¯r. Obviously, u1,u2 are all positive solutions of problem (2.13), (2.14) and ρ*<∥u1∥<r<∥u2∥<ρ*. The proof of Theorem 3.2 is complete.Similar to Theorems3.1 and 3.2, we also obtain the following theorems.Theorem 3.3.
Suppose that conditions (H1), (H2), (H3), and (A2) hold and(A7)
f∞=λ∈((2θ*/θ)p-1,∞),(A8)
f0=φ∈((2θ*/θ)p-1,∞).Then, the SBVP (2.13), (2.14) has at last two solutions u1,u2 such that 0<∥u1∥<R<∥u2∥.Theorem 3.4.
Suppose that conditions (H1), (H2), (H3), and (A1) hold and(A9)
f0=φ∈[0,(θ*/4)p-1);(A10)
f∞=λ∈[0,(θ*/4)p-1).Then, the SBVP (2.13), (2.14) has at last two solutions u1,u2 such that 0<∥u1∥<r<∥u2∥.


## 4. An Example

Example 4.1.
Consider the following 3-order singular boundary value problem (SBVP) withp-Laplacian:
(4.1)(ϕp(uΔΔ))∇(t)+164π4t-1/2(1-t)[u(t)+u(t-1)+(uΔ)2(t)+(uΔ)4(t)]=0,0<t<1,u(t)=-tet,-1≤t≤0,2ϕp(uΔ(0))-ϕp(uΔΔ(14))=0,ϕp(uΔ(1))+δϕp(uΔΔ(12))=0,
where
(4.2)β=γ=1,α=2,p=4,δ≥0,p=4,ξ=14,η=13,θ=14,τ=T=1.
So, by Lemma 2.4, we discuss the following SBVP:
(4.3)(ϕp(uΔΔ))∇(t)+164π4t-1/2(1-t)[(uΔ)4u(t)+[u(t-1)+h(t-1)]+(uΔ)2(t)+(uΔ)4(t)]=0,0<t<1,u(t)=0,-1≤t≤0,2ϕp(uΔ(0))-ϕp(uΔΔ(14))=0,ϕp(uΔ(1))+δϕp(uΔΔ(12))=0,
where
(4.4)h(t)={ζ(t),-1≤t≤0,0,0≤t≤1,ζ(t)=-tet,g(t)=164π4t-1/2(1-t),f(u1,u2,u3)=u1+u2+u32+u34.
Then, obviously,
(4.5)q=43,∫01g(t)∇t=164π3,H=max⁡-1≤t≤0|ζ(t)|=e,f∞=+∞,f0=+∞,
so conditions (H1), (H2), (H3), (A2), and (A3) hold.
Next,(4.6)ϕq(∫01a(t)∇t)=14π,θ*=4π1+43,
we choose R=3, M=2 and for θ=1/4, because of the monotone increasing of f(u1,u2,u3) on [0,∞)3, then
(4.7)f(u1,u2,u3)≤f(34,e,3)=34+e+90,0≤u3≤3,0≤u1≤14u3,0≤u2≤e.
Therefore, by
(4.8)M∈(0,θ*),(MR)p-1=(6)3=216,
we know
(4.9)f(u1,u2,u3)≤(MR)p-1,0≤u3≤3,0≤u1≤14u3,0≤u2≤e,
so condition (A2) holds. Then, by Theorem 3.1, SBVP (4.3) has at least two positive solutions v1,v2 and 0<∥v1∥<3<∥v2∥. Then, by Lemma 2.4, v¯1(t)=v1(t)+h(t), v¯2(t)=v2(t)+h(t), t∈(-1,1) are the positive solutions of the SBVP (4.1).



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*Source: 939162-2012-04-24.xml*