# Solutions forp-Laplacian  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

LetT be a time scale. We study the existence of positive solutions for the nonlinear four-point singular boundary value problem with p-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 is obtained.


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## Body

## 1. Introduction

The study of dynamic equations on time scales goes back to its founder 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 [4–14] 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),(1.2)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) and (1.2) by using fixed-point theorem and Leggett-Williams fixed-point theorem (see [15]), respectively.In [4], Anderson discussed the following dynamic equation on time scales:(1.3)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 (1.3) based on the limits f0=limu→0+f(u)/u and f∞=limu→∞f(u)/u.In [5], Kaufmann studied the problem (1.3) and obtained the existence results of at least two positives solutions.In [14], Wang et al. discussed the following dynamic equation on time scales by using Avery-Peterson fixed theorem (see [14]):(1.4)(ϕp(u′))′+q(t)f(t,u(t),u(t-1),u′(t))=0,t∈(0,1),(1.5)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 (1.4), (1.5) and (1.4), and (1.5′), respectively.In [15], Lee and Sim discussed the following equation:(1.6)(ϕp(u′))′+λh(t)f(u(t))=0,a.e.t∈(0,1),u(0)=u(1)=0.
By applying the global bifurcation theorem and figuring the shape of unbounded subcontinua of solutions, they obtain many different types of global existence results of positive solutions.However, there are not many concerning thep-Laplacian problems on time scales. Especially, for the singular multipoint boundary value problems for p-Laplacian dynamic delay differential equations on time scales, with the author’s acknowledge, no one has studied the existence of positive solutions in this case.Recently, in [16], we have studied the existence of positive solutions for the following nonlinear two-point singular boundary value problem with p-Laplacian operator:(1.7)(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 (1.7) with p-Laplacian operator is 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):(1.8)(ϕp(uΔ(t)))∇+g(t)f(u(t-τ),u(t))=0,0<t<T,τ>0,(1.9)u(t)=ζ(t),-τ≤t≤0,αϕp(u(0))-βϕp(uΔ(ξ))=0,γϕp(u(T))+δϕp(uΔ(η))=0,
or(1.10)u(t)=ζ(t),-τ≤t≤0,u(0)-B0(uΔ(ξ))=0,u(T)+B1(uΔ(η))=0,
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 and B0,B1 are both nondecreasing continuous odd functions defined on (-∞,+∞).In this paper, by constructing one integral equation which is equivalent to the problem (1.8), (1.9) and (1.8), and (1.10), we research the existence of positive solutions for nonlinear singular boundary value problem (1.8), (1.9) and (1.8), and (1.10) 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 [17, 18]).
Suppose thatE 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,∀x∈∂Ωr. Then(i)
if∥Tx∥≤∥x∥,∀x∈∂Ωr, then i(T,Ωr,K)=1;(ii)
if∥Tx∥≥∥x∥,∀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 discuss the existence of single solution of the systems (1.8) and (1.9). In Section 4, we study the existence of at least two solutions of the systems (1.8) and (1.9). In Section 5, we discuss the existence of single and many solutions of the systems (1.8) and (1.10). In Section 6, we give two examples as the application.


## 2. Preliminaries and Lemmas

For convenience, we can found some basic definitions in [1, 19, 20].In the rest of this paper,T is closed subset of R with 0∈Tk,T∈Tk. And let B={u∈Cld[-τ,T]}, then B is a Banach space with the norm ∥u∥=maxt∈[-τ,T]|u(t)|. And let(2.1)K={u∈B:u(t)≥0,u(t)is concave function,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 (1.8) and (1.9) if it satisfies the following:(1)
u∈C[-τ,0]∩Cld(0,T),(2)
u(t)>0 for all t∈(0,T) and satisfies conditions (1.9),(3)
(ϕp(uΔ(t)))∇=-g(t)f(u(t-τ),u(t)) holds for t∈(0,T). In the rest of the paper, we also make the following assumptions:(H1)
f∈Cld([0,+∞)2,[0,+∞)),(H2)
g(t)∈Cld((0,T),[0,+∞)) and there exists t0∈(0,T), such that
(2.2)g(t0)>0,0<∫0Tg(s)∇s<+∞,(H3)
ζ(t)∈C([-τ,0], ζ(t)>0 on [-τ,0) and ζ(0)=0,(H4)
B0,B1 are both increasing, continuous, odd functions defined on (-∞,+∞), and at least one of them satisfies the condition that there exists one b>0 such that
(2.3)0<Bi(v)≤bv,∀v≥0,i=0or1.It is easy to check that condition (H2) implies that
(2.4)0<∫0Tϕq(∫0sg(s1)∇s1)Δs<+∞.
We can easily get the following Lemmas.Lemma 2.2.
Suppose that condition(H2) holds. Then there exists a constant θ∈(0,1/2) that satisfies
(2.5)0<∫θT-θg(t)∇t<∞.
Furthermore, the function
(2.6)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]. Nest, let
(2.7)A1(t)=∫θtϕq(∫stg(s1)∇s1)Δs,A2(t)=∫tT-θϕq(∫tsg(s1)∇s1)Δs.
Then, from condition (H2), we have 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 (see [16]).
Letu∈K and θ of Lemma 2.2, then
(2.8)u(t)≥θ‖u‖,t∈[θ,T-θ].Lemma 2.4.
Suppose that conditions(H1),(H2),(H3), and (H4) hold, u(t)∈B∩Cld(0,1) is a solution of the following boundary value problems:
(2.9)(ϕp(uΔ(t)))∇+g(t)f(u(t-τ)+h(t-τ),u(t))=0,0<t<T,(2.10)u(t)=0,-τ≤t≤0,αϕp(u(0))-βϕp(uΔ(ξ))=0,γϕp(u(T))+δϕp(uΔ(η))=0,
or
(2.10′)u(t)=0,-τ≤t≤0,u(0)-B0(uΔ(ξ))=0,u(T)+B1(uΔ(η))=0,
where
(2.11)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 (1.8) and (1.9) or (1.8) and (1.10).Proof.
It is easy to check thatu¯(t) satisfies (1.8) and (1.9) or (1.8) and (1.10).So in the rest section of this paper, we focus on SBVP (2.9), (2.10), and (2.9), (2.10′).Lemma 2.5.
Suppose that conditions(H1),(H2),(H3),or (H1),(H2),(H3),(H4), hold, u(t)∈B∩Cld(0,1) is a solution of boundary value problems (2.9), (2.10) or (2.9), (2.10′), respectively, if and only if u(t)∈B is a solution of the following integral equation, respectively:
(2.12)u(t)={ζ(t),-τ≤t≤0,∫0tw(s)Δs,0≤t≤T,u(t)={ζ(t),-τ≤t≤0,∫0tw¯(s)Δs,0≤t≤T,
where
(2.13)w(t)={ϕq(βα∫ξσg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫0tϕq(∫sσg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs,0≤t≤σ,ϕq(δγ∫σηg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫tTϕq(∫σsg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs,σ≤t≤T,(2.13′)w¯(t)={B0∘ϕq(∫ξϱg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫0tϕq(∫sϱg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs,0≤t≤ϱ,B1∘ϕq(∫ϱηg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫tTϕq(∫ϱsg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs,ϱ≤t≤T.
Here σ,ϱ is unique solution of the equation, respectively,
(2.14)g1(t)=g2(t),g¯1(t)=g¯2(t),
where
(2.15)g1(t)=ϕq(βα∫ξσg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫0tϕq(∫sσg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs,g2(t)=ϕq(δγ∫σηg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫tTϕq(∫σsg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs,g¯1(t)=B0∘ϕq(∫ξϱg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫0tϕq(∫sϱg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs,g¯2(t)=B1∘ϕq(∫ϱηg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫tTϕq(∫ϱsg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs.
Equation g1(t)=g2(t), g¯1(t)=g¯2(t) has unique solution in (0,T). Because g1(t), g¯1(t) is strictly monotone increasing on [0,T), and g1(0)=0, g¯1(0)=0, g2(t),g¯2(t) is strictly monotone decreasing on (0,T], and g2(T)=0, g¯2(T)=0.Proof.
We only proof the first section of the results.
Necessity.
Obviously, fort∈(-τ,0), we have u(t)=ζ(t).
Ift∈(0,T), by the equation of the boundary condition and we have uΔ(ξ)≥0,uΔ(η)≤0, then there exist is a constant σ∈[ξ,η]⊂(0,T) such that uΔ(σ)=0.
Firstly, by integrating the equation of the problems (2.9) on (σ,t), we have
(2.16)ϕp(uΔ(t))=ϕp(uΔ(σ))-∫σtg(s)f(u(s-τ)+h(s-τ),u(s))∇s,
then
(2.17)uΔ(t)=-ϕq(∫σtg(s)f(u(s-τ)+h(s-τ),u(s))∇s),
thus
(2.18)u(t)=u(σ)-∫σtϕq(∫σsg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs.
ByuΔ(σ)=0 and condition (2.16), t=η on (2.16), we have
(2.19)ϕp(uΔ(η))=-∫σηg(s)f(u(s-τ)+h(s-τ),u(s))∇s.
By the equation of the boundary condition (2.10), we have
(2.20)ϕp(u(T))=-δγϕp(uΔ(η)),
then
(2.21)u(T)=ϕq(δγ∫σηg(s)f(u(s-τ)+h(s-τ),u(s))∇s).
Then, by (2.18) and let t=T on (2.18), we have
(2.22)u(σ)=ϕq(δγ∫σηg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫σTϕq(∫σsg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs.
Then
(2.23)u(t)=ϕq(δγ∫σηg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫tTϕq(∫σsg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs.
Similarly, for t∈(0,σ), by integrating the equation of problems (2.9) on (0,σ), we have
(2.24)u(t)=ϕq(βα∫ξσg(s)f(u(s-τ)+h(s-τ),u(s))∇s)Δs+∫0tϕq(∫sσg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs.
Therefore, for any t∈[0,T], u(t) can be expressed as equation
(2.25)u(t)={ζ(t),-τ≤t≤0,∫0tw(s)Δs,0≤t≤T,
where w(t) is expressed as (2.13).Sufficiency.
Suppose thatu(t)=∫0tw(s)Δsn-2Δs,0≤t≤T. Then by (2.13), we have
(2.26)uΔ(t)={ϕq(∫tσg(s)f(u(s-τ)+h(s-τ),u(s))∇s)≥0,0≤t≤σ,-ϕq(∫σtg(s)f(u(s-τ)+h(s-τ),u(s))∇s)≤0,σ≤t≤T,
So, (ϕp(uΔ)∇+g(t)f(u(t-τ)+h(t-τ),u(t))=0,0<t<T. These imply that (2.9) holds. Furthermore, by letting t=0 and t=T on (2.13) and (2.26), we can obtain the boundary value equations of (2.10). The proof is complete.Now, we define an operetor equationT given by(2.27)(Tu)(t)={ζ(t),-τ≤t≤0,∫0tw(s)Δs,0≤t≤T,(T¯u)(t)={ζ(t),-τ≤t≤0,∫0tw¯(s)Δs,0≤t≤T,
where w(t),w¯(t) is given by (2.13) and (2.13′).From the definition ofT,T¯ and above discussion, we deduce that for each u∈K,Tu,T¯u∈K. Moreover, we have the following Lemma.Lemma 2.6.
T,T¯:K→K is completely continuous.Proof.
We only proof the completely continuous ofT.
Because(2.28)(Tu)Δ(t)=wΔ(t)={ϕq(∫tσg(s)f(u(s-τ)+h(s-τ),u(s))∇s)≥0,0≤t≤σ-ϕq(∫σtg(s)f(u(s-τ)+h(s-τ),u(s))∇s)≤0,σ≤t≤T,
is continuous, decreasing on [0,T], and satisfies that (Tu)Δ(σ)=0, then, Tu∈K for each u∈K and (Tu)(σ)=maxt∈[0,T](Tu)(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), and (H3) hold, the solution u(t)∈K of problem (2.9) and (2.10) satisfy
(2.29)max⁡0≤t≤T⁡|u(t-τ)+h(t-τ)|≤max⁡-τ≤t≤0⁡|ζ(t)|.Proof.
Firstly, we can have(2.30)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)|.
The proof is complete.For convenience, we set(2.31)H=max⁡-τ≤t≤0|ζ(t)|,θ*=2L,θ*=1(T+ϕq(β/α))ϕq(∫0Tg(r)∇r),θ**=1(b+1)ϕq(∫0Tg(r)∇r),
where L is the constant from Lemma 2.2. By Lemma 2.5, we can also set(2.32)f0=lim⁡u2→0max⁡0≤u1≤Hf(u1,u2)u2p-1,f∞=lim⁡u2→∞max⁡0≤u1≤Hf(u1,u2)u2p-1,f0=lim⁡u2→0min⁡0≤u1≤Hf(u1,u2)u2p-1f∞=lim⁡u2→∞min⁡0≤u1≤Hf(u1,u2)u2p-1.


## 3. The Existence of Single Positive Solution to (1.8) and (1.9)

In this section, we present our main results.Theorem 3.1.
Suppose that condition(H1),(H2), and (H3) hold. Assume that f also satisfies(A1):
f(u1,u2)≥(mr)p-1, for θr≤u2≤r, 0≤u1≤H,(A2):
f(u1,u2)≤(MR)p-1, for 0≤u2≤R,0≤u1≤H,where m∈(θ*,∞), M∈(0,θ*).
Then, the SBVP (2.9), (2.10) has a solution u such that ∥u∥ lies between r and R. Furthermore by Lemma 2.4, u¯(t)=u(t)+h(t),-τ≤t≤T is a positive solution to the SBVP (1.8) and (1.9).Proof of Theorem3.1.
Without loss of generality, we suppose thatr<R. For any u∈K, by Lemma 2.3, we have
(3.1)u(t)≥θ‖u‖,t∈[θ,T-θ].
We define two open subset Ω1 and Ω2 of E:
(3.2)Ω1={u∈K:‖u‖<r},Ω2={u∈K:‖u‖<R}.
For any u∈∂Ω1, by (3.1), we have
(3.3)r=‖u‖≥u(t)≥θ‖u‖=θr,t∈[θ,T-θ].
For t∈[θ,T-θ] and u∈∂Ω1, we shall discuss it from three perspectives.(i)
Ifσ∈[θ,T-θ], thus for u∈∂Ω1, by (A1) and Lemma 2.4, we have(3.4)2‖Tu‖=2(Tu)(σ)≥∫0σϕq(∫sσg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs+∫σTϕq(∫σsg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs≥∫θσϕq(∫sσg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs+∫σT-θϕq(∫σsg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs≥mrA(σ)≥mrL≥2r=2‖u‖.(ii)
Ifσ∈(T-θ,T], thus for u∈∂Ω1, by (A1) and Lemma 2.4, we have(3.5)‖Tu‖=(Tu)(σ)≥ϕq(βα∫ξσg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫0σϕq(∫sσg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs≥∫θT-θϕq(∫sT-θg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs≥mrA(T-θ)≥mrL≥2r>r=‖u‖.(iii)
Ifσ∈(0,θ), thus for u∈∂Ω1, by (A1) and Lemma 2.4, we have(3.6)‖Tu‖=(Tu)(σ)≥ϕq(δγ∫σηg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫σTϕq(∫σsg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs≥∫θT-θϕq(∫θsg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs≥mrA(θ)≥mrL≥2r>r=‖u‖.
Therefore, no matter under which condition, we all have(3.7)‖Tu‖>‖u‖,∀u∈∂Ω1.
Then by Theorem 1.1, we have
(3.8)i(T,Ω1,K)=0.
On the other hand, foru∈∂Ω2, we have u(t)≤∥u∥=R, and by (A2), we know that
(3.9)‖Tu‖=(Tu)(σ)≤ϕq(βα∫ξσg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫0Tϕq(∫sσg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs≤(T+ϕq(βα))MRϕq(∫0Tg(r)∇r)≤R=‖u‖,
thus
(3.10)‖Tu‖≤‖u‖,∀u∈∂Ω2.
Then, by Theorem 1.1, we have
(3.11)i(T,Ω2,K)=1.
Therefore, by (3.8), and (3.11), r<R, we have
(3.12)i(T,Ω2∖Ω¯1,K)=1.
Then operator T has a fixed point u∈(Ω1∖Ω¯2), and r≤∥u∥≤R. This completes the proof of Theorem 3.1.Theorem 3.2.
Suppose that condition(H1),(H2), and (H3) hold. Assume that f also satisfies(A3):
f0=φ∈[0,(θ*/4)p-1),(A4):
f∞=λ∈((2θ*/θ)p-1,∞).Then, the SBVP (2.9), (2.10) has a solution u which is bounded in ∥·∥. Furthermore, by Lemma 2.4, u¯(t)=u(t)+h(t),-τ≤t≤T is a positive solution to the SBVP (1.8), (1.9).Proof of Theorem3.2.
First, byf0=φ∈[0,(θ*/4)p-1), for ϵ=(θ*/4)p-1-φ, there exists an adequately small positive number ρ, as 0≤u2≤ρ,u2≠0,u1≤H, we have
(3.13)f(u1,u2)≤(φ+ϵ)(u2)p-1≤(θ*4)p-1ρp-1=(θ*4ρ)p-1.
Then let R=ρ,M=θ*/4∈(0,θ*), thus by (3.13),
(3.14)f(u1,u2)≤(MR)p-1,0≤u2≤R.
So condition (A2) holds.
Next, by condition (A4), f∞=λ∈((2θ*/θ)p-1,∞), then for ϵ=λ-(2θ*/θ)p-1, there exists an appropriately big positive number r≠R, as u2≥θr,u1≤H, we have
(3.15)f(u1,u2)≥(λ-ϵ)(u2)p-1≥(2θ*θ)p-1(θr)p-1=(2θ*r)p-1.
Let m=2θ*>θ*, thus by (3.15), condition (A1) holds. Therefore, by Theorem 3.1, we know that the results of Theorem 3.2 hold. The proof of Theorem 3.2 is complete.Theorem 3.3.
Suppose that conditions(H1),(H2),(H3) hold. Assume that f also satisfies(A5):
f∞=λ∈[0,(θ*/4)p-1),(A6):
f0=φ∈((2θ*/θ)p-1,∞).Then, the SBVP (2.9), (2.10) has a solution u which is bounded in ∥·∥. Furthermore by Lemma 2.4, u¯(t)=u(t)+h(t),-τ≤t≤T is a positive solution to the SBVP (1.8), (1.9).Proof of Theorem3.3.
First, by condition (A6), f0=φ∈((2θ*/θ)p-1,∞), then for ϵ=φ-(2θ*/θ)p-1, there exists an adequately small positive number r, as 0≤u2≤r,u2≠0,u1≤H, we have
(3.16)f(u1,u2)≥(φ-ϵ)(u2)p-1=(2θ*θ)p-1(u2)p-1,
thus when θr≤u2≤r,u1≤H, we have
(3.17)f(u1,u2)≥(2θ*θ)p-1(θr)p-1=(2θ*r)p-1.
Let m=2θ*>θ*, so by (3.17), condition (A1) holds.
Next, by condition (A5): f∞=λ∈[0,(θ*/4)p-1), then for ϵ=(θ*/4)p-1-λ, there exists an suitably big positive number ρ≠r, as u2≥ρ,u1≤H, we have
(3.18)f(u1,u2)≤(λ+ϵ)(u2)p-1≤(θ*4)p-1(u2)p-1.
If f is unbounded, by the continuity of f on [0,∞)2, then exists constant R(≠r)≥ρ, and a point (u01,u02)∈[0,∞)2 such that
(3.19)ρ≤u02≤R,f(u1,u2)≤f(u01,u02),0≤u2≤R,u1≤H.
Thus, by ρ≤u02≤R,u1≤H, we know
(3.20)f(u1,u2)≤f(u01,u02)≤(θ*4)p-1(u02)p-1≤(θ*4R)p-1.
Choose M=θ*/4∈(0,θ*). Then, we have
(3.21)f(u1,u2)≤(MR)p-1,0≤u2≤R,u1≤H.
If f is bounded, we suppose f(u1,u2)≤M¯p-1,u2∈[0,∞),M¯∈R+, there exists an appropriately big positive number R>(4/θ*)M¯, then choose M=θ*/4∈(0,θ*), we have
(3.22)f(u1,u2)≤M¯p-1≤(θ*4R)p-1=(MR)p-1,0≤u2≤R,u1≤H.
Therefore, condition (A2) holds. Therefore, by Theorem 3.1, we know that the results of Theorem 3.3 holds. The proof of Theorem 3.3 is complete.


## 4. The Existence of Many Positive Solutions to (1.8) and (1.9)

Next, we will discuss the existence of many positive solutions.Theorem 4.1.
Suppose that conditions(H1),(H2),(H3), and (A2) in Theorem 3.1 hold. Assume that f also satisfies(A7):
f0=+∞,(A8):
f∞=+∞.Then, the SBVP (2.9), (2.10) has at last two solutions u1,u2 such that
(4.1)0<‖u1‖<R<‖u2‖.
Furthermore, by Lemma 2.4, u¯1(t)=u1(t)+h(t),u¯2(t)=u2(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.8), (1.9).Proof of Theorem4.1.
First, by condition (A7), for any N>2/θL, there exists a constant ρ*∈(0,R) such that
(4.2)f(u1,u2)≥(Nu2)p-1,0<u2≤ρ*,u1≤H.
Set Ωρ*={u∈K:∥u∥<ρ*}, for any u∈∂Ωρ*, by (4.2) and Lemma 2.3, similar to the previous proof of Theorem 3.1, we can have from three perspectives
(4.3)‖Tu‖≥‖u‖,∀u∈∂Ωρ*.
Then by Theorem 1.1, we have
(4.4)i(T,Ωρ*,K)=0.
Next, by condition (A8), for any N¯>2/θL, there exists a constant ρ0>0 such that
(4.5)f(u1,u2)≥(N¯u2)p-1,u2>ρ0,u1≤H.
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
(4.6)u(t)≥θ‖u‖=θρ*>ρ0,t∈[θ,T-θ].
Then by (4.5) and also similar to the previous proof of Theorem 3.1, we can also have from three perspectives
(4.7)‖Tu‖≥‖u‖,∀u∈∂Ωρ*.
Then by Theorem 1.1, we have
(4.8)i(T,Ωρ*,K)=0.
Finally, setΩR={u∈K:∥u∥<R}, For any u∈∂ΩR, by (A2), Lemma 2.3 and also similar to the latter proof of Theorem 3.1, we can also have
(4.9)‖Tu‖≤‖u‖,∀u∈∂ΩR.
Then by Theorem 1.1, we have
(4.10)i(T,ΩR,K)=1.
Therefore, by (4.4), (4.8), (4.10), ρ*<R<ρ*, we have
(4.11)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 (2.9), (2.10) and ρ*<∥u1∥<R<∥u2∥<ρ*. The proof of Theorem 4.1 is complete.Theorem 4.2.
Suppose that conditions(H1),(H2),(H3), and (A1) in Theorem 3.1 hold. Assume that f also satisfies(A9):
f0=0,(A10):
f∞=0.Then, the SBVP (2.9), (2.10) has at last two solutions u1,u2 such that 0<∥u1∥<r<∥u2∥. Furthermore, by Lemma 2.4, u¯1(t)=u1(t)+h(t),u¯2(t)=u2(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.8), (1.10).Proof of Theorem4.2.
First, byf0=0, for ϵ1∈(0,θ*), there exists a constant ρ*∈(0,r) such that
(4.12)f(u1,u2)≤(ϵ1u2)p-1,0<u2≤ρ*,u1≤H.
Set Ωρ*={u∈K:∥u∥<ρ*}, for any u∈∂Ωρ*, by (4.12), we have
(4.13)‖Tu‖=(Tu)(σ)≤ϕq(βα∫ξσg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫0Tϕq(∫sσg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs≤ϕq(βα∫ξσg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+Tϕq(∫0Tg(r)f(u(r-τ)+h(r-τ),u(r))∇r)≤(T+ϕq(βα))ϵ1ρ*ϕq(∫0Tg(r)∇r)≤ρ*=‖u‖,
that is
(4.14)‖Tu‖≤‖u‖,∀u∈∂Ωρ*.
Then by Theorem 1.1, we have
(4.15)i(T,Ωρ*,K)=1.
Next, letf*(x)=max0≤u2≤x,u1≤Hf(u1,u2), and note that f*(x) is monotone increasing with respect to x≥0. Then from f∞=0, it is easy to see that
(4.16)lim⁡x→∞⁡f*(x)xp-1=0.
Therefore, for any ϵ2∈(0,θ*), there exists a constant ρ*>r such that
(4.17)f*(x)≤(ϵ2x)p-1,x≥ρ*.
Set Ωρ*={u∈K:∥u∥<ρ*}, for any u∈∂Ωρ*, by (4.17), we have
(4.18)‖Tu‖=(Tu)(σ)≤ϕq(βα∫ξσg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+∫0Tϕq(∫sσg(r)f(u(r-τ)+h(r-τ),u(r))∇r)Δs≤ϕq(βα∫ξσg(s)f(u(s-τ)+h(s-τ),u(s))∇s)+Tϕq(∫0Tg(r)f(u(r-τ)+h(r-τ),u(r))∇r)≤(T+ϕq(βα))ϕq(∫0Tg(r)f*(ρ*)∇r)≤(T+ϕq(βα))ϵ2ρ*ϕq(∫0Tg(r)∇r)≤r*=‖u‖,
that is
(4.19)‖Tu‖≤‖u‖,∀u∈∂Ωρ*.
Then by Theorem 1.1, we have
(4.20)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
(4.21)‖Tu‖≥‖u‖,∀u∈∂Ωr.
Then by Theorem 1.1, we have
(4.22)i(T,Ωr,K)=0.
Therefore, by (4.15), (4.20), (4.22), ρ*<r<ρ*, we have
(4.23)i(T,Ωr∖Ω¯ρ*,K)=-1,i(T,Ωρ*∖Ω¯r,K)=1.
Then T have fixed point u1∈Ωr∖Ω¯ρ*, and fixed point u2∈Ωρ*∖Ω¯r. Obviously, u1,u2 are all positive solutions of problem (1.8), (1.9) and ρ*<∥u1∥<r<∥u2∥<ρ*. The proof of Theorem 4.2 is complete.Similar to Theorem3.1, we also obtain the following Theorems.Theorem 4.3.
Suppose that conditions(H1),(H2),(H3) and (A2) in Theorem 3.1, (A4) in Theorem 3.2 and (A6) in Theorem 3.3 hold. Then, the SBVP (2.9), (2.10) has at last two solutions u1,u2 such that 0<∥u1∥<R<∥u2∥. Furthermore by Lemma 2.4, u¯1(t)=u1(t)+h(t),u¯2(t)=u2(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.8), (1.10).Theorem 4.4.
Suppose that conditions(H1),(H2),(H3) and (A1) in Theorem 3.1, (A3) in Theorem 3.2 and (A5) in Theorem 3.3 hold. Then, the SBVP (2.9), (2.10) have at last two solutions u1,u2 such that 0<∥u1∥<r<∥u2∥. Furthermore by Lemma 2.4, u¯1(t)=u1(t)+h(t), u¯2(t)=u2(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.8), (1.10).


## 5. The Existence of Many Positive Solutions to (1.8) and (1.10)

In the following, we will deal with problem (1.8), (1.10), the method is similar to that in Sections 3 and 4, so we omit many proof in this section.Theorem 5.1.
Suppose that condition(H1),(H2),(H3),(H4) hold. Assume that f also satisfies(A1′):
f(u1,u2)≥(mr)p-1, for θr≤u2≤r,0≤u1≤H,(A2′):
f(u1,u2)≤(MR)p-1, for 0≤u2≤R,0≤u1≤H,where m∈(θ*,∞),M∈(0,θ**). Then, the SBVP (2.9), (2.13) has a solution u such that ∥u∥ lies between r and R. Furthermore by Lemma 2.4, u¯(t)=u(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.8), (1.10).Theorem 5.2.
Suppose that condition(H1),(H2),(H3),(H4) hold. Assume that f also satisfies(A3′):
f0=φ∈[0,(θ**/4)p-1),(A4′):
f∞=λ∈((2θ*/θ)p-1,∞).Then, the SBVP (2.9), (2.13) has a solution u which is bounded in ∥·∥. Furthermore by Lemma 2.4, u¯(t)=u(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.8), (1.10).Theorem 5.3.
Suppose that condition(H1),(H2),(H3),(H4) hold. Assume that f also satisfies(A5′):
f∞=λ∈[0,(θ**/4)p-1),(A6′):
f0=φ∈((2θ*/θ)p-1,∞).Then, the SBVP (2.9), (2.13) has a solution u which is bounded in ∥·∥. Furthermore by Lemma 2.4, u¯(t)=u(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.8), (1.10).Theorem 5.4.
Suppose that conditions(H1),(H2),(H3),(H4) and (A2′) in Theorem 5.1 hold. Assume that f also satisfies(A7′):
f0=+∞,(A8′):
f∞=+∞.Then, the SBVP (2.9), (2.13) has at least two solutions u1, u2 such that
(5.1)0<‖u1‖<R<‖u2‖.
Furthermore by Lemma 2.4, u¯1(t)=u1(t)+h(t),u¯2(t)=u2(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.8), (1.10).Theorem 5.5.
Suppose that conditions(H1),(H2),(H3),(H4) and (A1′) in Theorem 5.1 hold. Assume that f also satisfies(A9′):
f0=0,(A10′):
f∞=0.Then, the SBVP (2.9), (2.13) has at least two solutions u1, u2 such that 0<∥u1∥<r<∥u2∥. Furthermore by Lemma 2.4, u¯1(t)=u1(t)+h(t),u¯2(t)=u2(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.8), (1.10).Theorem 5.6.
Suppose that conditions(H1),(H2),(H3),(H4) and (A2) in Theorem 5.1, (A4) in Theorem 5.2 and (A6′) in Theorem 3.3 hold. Then, the SBVP (2.9), (2.13) has at least two solutions u1,u2 such that 0<∥u1∥<R<∥u2∥. Furthermore by Lemma 2.4, u¯1(t)=u1(t)+h(t),u¯2(t)=u2(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.8), (1.10).Theorem 5.7.
Suppose that conditions(H1),(H2),(H3),(H4) and (A1′) in Theorem 5.1, (A3′) in Theorem 5.2 and (A5′) in Theorem 3.3 hold. Then, the SBVP (2.9), (2.13) has at least two solutions u1,u2 such that 0<∥u1∥<r<∥u2∥. Furthermore by Lemma 2.4, u¯1(t)=u1(t)+h(t),u¯2(t)=u2(t)+h(t), -τ≤t≤T is a positive solution to the SBVP (1.8), (1.10).


## 6. Application

In the section, we present two simple examples to explain our result.Example 6.1.
LetT={1-(1/2)N}∪{1}, where N denotes the set of all nonnegative integers. Consider the following 3-order singular boundary value problem (SBVP) with p-Laplacian
(6.1)(ϕp(uΔ))∇(t)+120t-1/2u1/2(t)⋅[15+(94/5)e2u(t)120u(t-1)+7eu(t)+e2u(t)]=0,0<t<1,u(t)=t2-1,-1≤t≤0,ϕp(u(0))-ϕp(uΔ(14))=0,ϕp(u(1))+δϕp(uΔ(12))=0,
where
(6.2)α=γ=1,β=1,p=32,δ≥0,ξ=14,η=12,θ=14,τ=T=1.
So, by Lemma 2.4, we discuss the following SBVP:
(6.3)(ϕp(uΔ))∇(t)+120t-1/2u1/2(t)⋅[15+(94/5)e2u(t)120[u(t-1)+h(t-1)]+7eu(t)+e2u(t)]=0,0<t<1,u(t)=0,-1≤t≤0,ϕp(u(0))-ϕp(uΔ(14))=0,ϕp(u(1))+δϕp(uΔ(12))=0,
where
(6.4)h(t)={t2-1,-1≤t≤0,0,0≤t≤1,g(t)=120t-1/2,ζ(t)=t2-1,f(u1,u2)=(u2)1/2[15+(94/5)e2u2120u1+7eu2+e2u2].
Then obviously,
(6.5)q=3,H=max⁡-1≤t≤0|ζ(t)|=1,f0=φ=lim⁡u2→0+max⁡0≤u1≤1f(u1,u2)u2p-1=5120,f∞=λ=lim⁡u2→∞min⁡0≤u1≤1f(u1,u2)u2p-1=955,∫0Tg(t)∇t=110,
so conditions (H1),(H2),(H3) hold.
Next,(6.6)θ*=1(T+ϕq(β/α))ϕq(∫0Tg(r)∇r)=50,
then (θ*/4)p-1=52/2>51/20, that is, φ∈[0,(θ*/4)p-1), so condition (A3) holds.
Forθ=1/4, it is easy to see by calculating that
(6.7)L=min⁡t∈[θ,T-θ]A(t)=116(736+33).
Because of
(6.8)(2θ*θ)p-1=96×(17+123)1/2<955,
then
(6.9)λ∈((2θ*θ)p-1,∞),
so condition (A4) holds. Then by Theorem 3.2, SBVP (6.3) has at least a positive solution u(t). So, u¯(t)=u(t)+h(t),-1<t<1 is the positive solution of SBVP (6.1).Example 6.2.
Consider the following 3-order singular boundary value problem (SBVP) withp-Laplacian:
(6.10)(ϕp(uΔ))∇(t)+164π4t-1/2(1-t)[u(t-1)+u2(t)+u4(t)]=0,0<t<1,u(t)=-tet,-1≤t≤0,2ϕp(u(0))-ϕp(uΔ(14))=0,ϕp(u(1))+δϕp(uΔ(12))=0,
where
(6.11)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:
(6.12)(ϕp(uΔ))∇(t)+164π4t-1/2(1-t)[[u(t-1)+h(t-1)]+u2(t)+u4(t)]=0,0<t<1,u(t)=0,-1≤t≤0,2ϕp(u(0))-ϕp(uΔ(14))=0,ϕp(u(1))+δϕp(uΔ(12))=0,
where
(6.13)h(t)={ζ(t),-1≤t≤0,0,0≤t≤1,ζ(t)=-tet,g(t)=164π4t-1/2(1-t),f(u1,u2)=u1+u22+u24.
Then obviously,
(6.14)q=43,∫01g(t)∇t=164π3,H=max⁡-1≤t≤0|ζ(t)|=e,f∞=+∞,f0=+∞,
so conditions (H1),(H2),(H3),(A7),(A8) hold.
Next,(6.15)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
(6.16)f(u1,u2)≤f(e,3)=e+90,0≤u2≤3,0≤u1≤e.
Therefore, by
(6.17)M∈(0,θ*),(MR)p-1=(6)3=216,
we know that
(6.18)f(u1,u2,u3)≤(MR)p-1,0≤u2≤3,0≤u1≤e,
so condition (A2) holds. Then by Theorem 4.1, SBVP (6.12) 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 (6.10).



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*Source: 652465-2012-05-10.xml*