# Exact Multiplicity of Solutions for a Class of Singular Generalized One-Dimensionalp-Laplacian Problem

Authors: Youwei Zhang
Journal: Journal of Mathematics
 (2013)

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## Abstract

We describe the existence of positive solutions for a class of singular generalized one-dimensionalp-Laplacian problem. By applying the related fixed point theory in cone, some new and general results on the existence of positive solutions to the singular generalized p-Laplacian problem are obtained. Note that the nonlinear term f involves the first-order derivative explicitly.


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## Body

## 1. Introduction

Recently, increasing attention is paid to question of positive solution for singular boundary value problems [1–11]. One notices that the singular boundary value problems for ordinary differential equation describe many phenomena in applied mathematics and physical science, which can be found in the theory of nonlinear diffusion generated by nonlinear sources and in the thermal ignition of gases [12–17]. Moreover, there are excellent results of nonsingular problem, see [18–21] and the references therein.Moreover, the nonlocalp-Laplacian problems for ordinary differential equation have been studied extensively. There are many papers dealing with the existence of positive solutions for the nonlocal p-Laplacian boundary value problem, in which the nonlinear term f is independent of the first-order derivative with different boundary conditions. For example, Ma et al. [22] established some existence results of positive solutions for the problem
(1)(φp(u′))′+q(t)f(t,u)=0,t∈(0,1),u′(0)=∑i=1m-2‍αiu′(ξi),u(1)=∑i=1m-2‍βiu(ξi).
The main tool is the monotone iterative technique. By applications of fixed point index theory, Cheung and Ren [23] investigated two classes of quasilinear nonlocal problems:
(2)(φp(u′))′+h(t)f(t,u)=0,0<t<1,u′(0)=0,u(1)=∑i=1m-2‍αiu(ξi),u(0)=∑i=1m-2‍αiu(ξi),u′(1)=0.
Sufficient conditions for the existence of twin positive solutions are established. In [16], Feng and Ge studied nonlocal problem for the one-dimensional p-Laplacian
(3)(φp(u′))′+q(t)f(t,u,u′)=0,t∈(0,1),u(0)=0,u(1)=∑i=1m-2‍aiu(ξi).
They obtained sufficient conditions for the existence of at least three solutions to the above problem by using the Avery and Peterson fixed point theorem.However, there are not many concerning the boundary value problems, in which a generalizedp-Laplacian equation with a nonlocal boundary conditions. The motivation for the present work stems from many recent investigations in [16, 23–25] and references therein. Our purpose of this paper is to establish some sufficient conditions for the existence of triple positive solutions to a class of a singular generalized one-dimensional p-Laplacian problem
(𝒫)(r(t)φp(u′))′+h(t)f(t,u(t),u′(t))=0,t∈(a,b),α1u(a)-β1u′(a)=∑i=1m-2‍γiu(ξi),u′(b)=0,
where φp(u) is p-Laplacian operator; that is, φp(u)=|u|p-2u, p>1, (φp)-1(u)=φq(u), 1/p+1/q=1, and α1, β1, γi, ξi, r, h, f satisfy(A1)
α1,β1∈[0,∞), γi∈[0,∞) satisfy α1>∑i=1m-2‍γi, a<ξ1<⋯<ξm-2<b(m>3),(A2)
r(t)∈C1(a,b), r(t) is positive and nondecreasing on [a,b],(A3)
h∈L1((a,b),[0,∞)) may be singular at t=0 and/or t=1, and 0<∫Γ‍h(t)dt<∞, where Γ⊆[a,b],(A4)
f:[a,b]×[0,∞)×(-∞,∞)→[0,∞) is an L1-Carathédory function; that is, for each (t,x,y)∈[a,b]×[0,∞)×(-∞,∞), the mapping t→f(t,x,y) is Lebesgue measurable on [a,b]; for a.e. t∈[a,b], the mapping (t,x,y)→f(t,x,y) is continuous on [0,∞)×(-∞,∞).Under the above assumptions, some new and general results on the existence of multiple positive solutions to singular one-dimensionalp-Laplacian are obtained. Our results develop some results of Cheung and Ren [23] and include and improve the main results of Feng and Ge [16] and Zhang [25].The rest of the paper is organized as follows. In Section2, we provide some background material from the theory of cones in the Banach spaces, which are useful later. Some lemmas and criteria for the existence of three positive solutions for one-dimensional p-Laplacian problems are established in Section 3. Finally, we give an example to illustrate our main results.


## 2. Preliminaries

Letγ and θ be nonnegative continuous convex functionals on a cone K, α a nonnegative continuous concave functional on K, β a nonnegative continuous functional on K, and m1,  m2,  m3,  and  m4 positive numbers. We define the following convex sets and a closed set
(4)P(γ,m4)={u∈K:γ(u)<m4},P(γ,α,m2,m4)={u∈K:m2≤α(u),γ(u)≤m4},P(γ,θ,α,m2,m3,m4)={u∈K:m2≤α(u),θ(u)≤m3,γ(u)≤m4},Q(γ,β,m1,m4)={u∈K:m1≤β(u),γ(u)≤m4}.Now we state the fixed point theorem due to Avery and Peterson [26].Lemma 1.
LetK be a cone in a real Banach space E. Let γ and θ be nonnegative continuous convex functionals on K, α a nonnegative continuous concave functional on K, and β a nonnegative continuous functional on K satisfying β(ρu)≤ρβ(u) for 0≤ρ≤1, such that, for some positive numbers ϵ and m4,
(5)α(u)≤β(u),∥u∥≤ϵγ(u)∀u∈P(γ,m4)¯.
Suppose that T:P(γ,m4)¯→P(γ,m4)¯ is completely continuous and there are positive numbers m1, m2, and m3 with m1<m2 such that(B1)
{u∈P(γ,θ,α,m2,m3,m4):α(u)>m2}≠∅, α(Tu)>m2 for u∈P(γ,θ,α,m2,m3,m4),(B2)
α(Tu)>m2 for u∈P(γ,α,m2,m4) with θ(Tu)>m3,(B3)
0∉Q(γ,β,m1,m4) and β(Tu)<m1 for u∈Q(γ,β,m1,m4) with β(u)=m1. Then T has at least three fixed points u1,u2,u3∈P(γ,m4)¯ such that
(6)γ(ui)≤m4(i=1,2,3),m2<α(u1),m1<β(u2)withα(u2)<m2,β(u3)<m1.Denote thatL+[a,b]={v∈L[a,b]:v(t)≥0,t∈[a,b]}.Lemma 2.
Assume that(A1) and (A2) hold. If g∈L+[a,b], then the boundary value problem
(7)(r(t)φp(u′))′+g(t)=0,t∈(a,b),(8)α1u(a)-β1u′(a)=∑i=1m-2‍γiu(ξi),u′(b)=0,
has a unique solution
(9)u(t)=∫at‍φq(1r(s)∫sb‍g(τ)dτ)ds+1α1-∑i=1m-2γi×(β1φq(1r(a)∫ab‍g(s)ds)+∑i=1m-2‍γi∫aξi‍φq(1r(s)∫sb‍g(τ)dτ)ds).Proof.
For a.e.t∈[a,b], integrating (7) from t to b, in view of (8), we have
(10)u′(t)=φq(1r(t)∫tb‍g(s)ds).
Integrating this from a to t yields
(11)u(t)=u(a)+∫at‍φq(1r(s)∫sb‍g(τ)dτ)ds.
Again, it follows from (8) that
(12)u(a)=1α1-∑i=1m-2γi×(∑i=1m-2‍γi∫aξi‍φq(1r(s)∫sb‍g(τ)dτ)dsβ1φq(1r(a)∫ab‍g(s)ds)+∑i=1m-2‍γi∫aξi‍φq(1r(s)∫sb‍g(τ)dτ)ds).
The proof is complete.Lemma 3.
Assume that (A1) and (A2) hold. If g∈L+[a,b], then the unique solution of the problems (7) and (8) satisfies
(13)u(t)≥0,mint∈[σ,b-σ]u(t)≥τ1maxt∈[a,b]u(t),
where τ1=β1/(β1+(b-a)(α1-∑i=1m-2‍γi)),σ∈(a,(a+b)/2).Proof.
The proof is similar to Lemma  3.6 in [25], we omit it.LetE be the Banach space C1[a,b] with the norm ∥u∥=max{∥u∥0,∥u′∥0}, where ∥u∥0=max0≤t≤1|u(t)|. Define a cone K⊂E by
(14)K={∑i=1m-2‍u∈E:u(t)≥0fort∈[a,b],u′′(t)≤0,u′(t)≥0fort∈(a,b),α1u(a)-β1u′(a)=∑i=1m-2‍γiu(ξi),u′(b)=0,mint∈[σ,b-σ]u(t)≥τ1maxt∈[a,b]u(t)∑i=1m-2‍}.
Let the assumptions (A1) and (A2) hold. Denote that
(15)λ=∫ab‍(b-sr(s))1/(p-1)ds+1α1-∑i=1m-2γi×(β1(b-ar(a))1/(p-1)+∑i=1m-2‍γi∫aξi‍(b-sr(s))1/(p-1)ds).
Evidently, λ≥0. We take
(16)u0(t)=∫at‍(m-b-sr(s))1/(p-1)ds+1α1-∑i=1m-2γi×(∑i=1m-2‍β1(m-b-ar(a))1/(p-1)+∑i=1m-2‍γi∫aξi‍(m-b-sr(s))1/(p-1)ds),
where m- satisfies (m-)1/(p-1)∶=m2/τ12λ, m2>0 is constant. It is not difficult to check that u0(t)≥0 for a.e. t∈[a,b], u′′(t)≤0, u′(t)≥0 for a.e. t∈(a,b), and u0(t) satisfies α1u0(a)-β1u0′(a)=∑i=1m-2‍γiu0(ξi), u0′(b)=0. That is to say u0∈K. Hence, K∖{0}≠∅.To obtain the existence of solutions for the problem(𝒫), the following priori estimate is needful.Lemma 4.
Assume that (A1) holds. If u∈K, then
(17)maxt∈[a,b]u(t)≤τ2maxt∈[a,b]|u′(t)|,
where τ2=(β1+α1(b-a))/(α1-∑i=1m-2‍γi).Proof.
The proof is similar to Lemma  3.1 in [21], we omit it.Let the nonnegative continuous convex functionalsγ and θ, the nonnegative continuous concave functional α, and the nonnegative continuous functional β be defined on cone K by
(18)α(u)=mint∈[σ,b-σ]u(t),β(u)=θ(u)=maxt∈[a,b]u(t),γ(u)=maxt∈[a,b]|u′(t)|.In the view of assumption (A3), that is, 0<∫ab‍h(τ)dτ<∞ and (A3). Thus, we can define an operator T by
(19)(Tu)(t)∶=∫at‍φq(1r(s)∫sb‍h(τ)f(τ,u(τ),u′(τ))dτ)ds+1α1-∑i=1m-2γi×(β1φq(1r(a)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ)+∑i=1m-2‍γi∫aξi‍φq(1r(s)∫sb‍h(τ)×f(τ,u(τ),u′(τ))dτ∫sb‍)ds)fora.e.t∈[a,b].Lemma 5.
Assume that(A1)–(A4) hold. Then T:K→K is completely continuous.Proof.
For anyu∈K, by the definition of Tu, we see that Tu∈C1[a,b], (φp((Tu)′))′∈L1[a,b]. From (A1)–(A4) and Lemma 3, we have (Tu)(t)≥0, (φp((Tu)′))′≤0 implies that (Tu)(t) is concave on [a,b], -(φp((Tu)′))′=h(t)f(t,(Tu)(t),(Tu)′(t))≥0, α1Tu(a)-β1Tu′(a)=∑i=1m-2‍γiTu(ξi), and Tu′(b)=0. So we can conclude that Tu∈K. In what follows, we prove that T is compact. Let R>0, there exists a set Ω such that Ω⊂{u∈K:∥u′∥0≤R}. For any u∈Ω, the assumptions imply that
(20)0≤∫ab‍h(t)f(t,u(t),u′(t))dt≤max(t,u,u′)∈[a,b]×[0,τ2R]×[-R,R]f(t,u,u′)∫ab‍h(t)dt<∞.
We take the arguments to show that the operator T is completely continuous. In view of Lemma 4 and the continuity of f in u and u′, assume that un,u0∈K satisfy ∥un-u0∥→0(n→∞). Moreover, φq(υ) is an increasing and continuously function, and we obtain that
(21)∥Tun-Tu0∥0=maxa≤t≤b|Tun(t)-Tu0(t)|≤∫ab‍(φq(1r(s)∫sb‍h(τ)f(τ,un(τ),un′(τ))dτ)-φq(1r(s)∫sb‍h(τ)f(τ,u0(τ),u0′(τ))dτ))ds+β1α1-∑i=1m-2γi×(φq(1r(a)∫ab‍h(τ)f(τ,un(τ),un′(τ))dτ)-φq(1r(a)∫ab‍h(τ)f(τ,u0(τ),u0′(τ))dτ))+1α1-∑i=1m-2γi∑i=1m-2‍γi×∫aξi‍(φq(1r(s)∫sb‍h(τ)f(τ,un(τ),un′(τ))dτ)ds-φq(1r(s)∫sb‍h(τ)×f(τ,u0(τ),u0′(τ))dτ1r(s)∫sb‍)ds)⟶0(n⟶∞),∥(Tun)′-(Tu0)′∥0=maxa≤t≤b|(Tun)′(t)-(Tu0)′(t)|≤|φq(1r(a)∫ab‍h(τ)f(τ,un(τ),un′(τ))dτ)-φq(1r(a)∫ab‍h(τ)f(τ,u0(τ),u0′(τ))dτ)|⟶0(n⟶∞).
Therefore,
(22)∥Tun-Tu0∥⟶0(n⟶∞).
This means that the operator T:K→K is continuous.
Sinceh(t) may be singular at t=a and/or t=b, choose two sequences {μn}n=1∞, {νn}n=1∞⊂(a,b) satisfying μn≤νn for any n≥1, such that μn→a and αn→b as n→∞, respectively. Define
(23)hn(t)={infa≤t≤μnh(t),a≤t≤μn,h(t),μn≤t≤νn,infνn≤t≤bh(t),νn≤t≤b,
and an operator sequence Tn by
(24)(Tnu)(t)=∫at‍φq(1r(s)∫sb‍hn(τ)f(τ,u(τ),u′(τ))dτ)ds+1α1-∑i=1m-2γi×(β1φq(1r(a)∫ab‍hn(τ)f(τ,u(τ),u′(τ))dτ)+∑i=1m-2‍γi×∫aξi‍φq(1r(s)∫sb‍hn(τ)×f(τ,u(τ),u′(τ))dτ1r(s)∫sb‍)ds)fora.e.t∈[a,b].
Clearly, hn(t):[a,b]→[0,∞) is a piecewise continuous function, and the operator Tn:K→K is well defined. Further, we can see that Tn:K→K is completely continuous.
Letr>0, 𝔹r:={u∈K:∥u′∥0≤r} and Mr=max{f(t,u,u′):u∈K,t∈[a,b], u∈[0,τ2r],  u′∈[-r,r]}. We will prove that Tn approaches T uniformly on 𝔹r. From the absolute continuity of integral, we obtain
(25)limn→∞∫Γ(n)‍φq(1r(s)∫sb‍h(τ)f(τ,u(τ),u′(τ))dτ)ds=0,
where Γ(n)=[a,μn]∪[νn,b]. For any u∈𝔹r and a.e. t∈[a,μn], one has that
(26)∥Tnu-Tu∥0=maxa≤t≤μn|Tnu(t)-Tu(t)|≤∫aμn‍φq(1r(s)∫sb‍hn(τ)f(τ,u(τ),u′(τ))dτ)ds-φq(1r(s)∫sb‍h(τ)f(τ,u(τ),u′(τ))dτ)ds+β1α1-∑i=1m-2γi×(φq(1r(a)∫ab‍hn(τ)f(τ,u(τ),u′(τ))dτ)-φq(1r(a)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ))+1α1-∑i=1m-2γi∑i=1m-2‍γi×∫aξi‍(φq(1r(s)∫sb‍hn(τ)f(τ,u(τ),u′(τ))dτ)ds-φq(1r(s)∫sb‍h(τ)f(τ,u(τ),u′(τ))dτ)ds)≤∫Γ(n)‍φq(1r(s)∫sb‍hn(τ)f(τ,u(τ),u′(τ))dτ)ds-φq(1r(s)∫sb‍h(τ)f(τ,u(τ),u′(τ))dτ)ds+β1α1-∑i=1m-2γi×(φq(1r(a)∫ab‍hn(τ)f(τ,u(τ),u′(τ))dτ)-φq(1r(a)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ))+1α1-∑i=1m-2γi∑i=1m-2‍γi×∫aξi‍(φq(1r(s)∫sb‍hn(τ)f(τ,u(τ),u′(τ))dτ)ds-φq(1r(s)∫sb‍h(τ)f(τ,u(τ),u′(τ))dτ)ds).
For u∈𝔹r and a.e. t∈[νn,b], we have the same result. It is easy to see that, for any u∈𝔹r and a.e. t∈[μn,νn], there is ∥Tnu-Tu∥0=0. Similarly, we can obtain that, for any u∈𝔹r and a.e. t∈[a,μn], t∈[μn,νn], t∈[νn,b], respectively,
(27)∥(Tnu)′-(Tu)′∥0=maxt|(Tnu)′(t)-(Tu)′(t)|≤|φq(1r(a)∫ab‍hn(τ)f(τ,u(τ),u′(τ))dτ)-φq(1r(a)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ)|.
In view of the continuity of function φq, from the above arguments, we have
(28)∥Tnu-Tu∥⟶0(n⟶∞).
That is to say, the sequence Tn is uniformly an approximate T on any bounded subset of K. Applying the Arzela-Ascoli lemma to the operator T, we can conclude that TΩ is relatively compact; that is, T is completely continuous.


## 3. Main Results

We are ready to apply the Avery and Peterson fixed point theorem to the operatorT to give sufficient conditions for the existence of at least three positive solutions to the problem (𝒫). For convenience, we introduce the following notation:
(29)L=1r(a)∫ab‍h(τ)dτ,M=α1-∑i=1m-2γiβ1,N=b-a+β1+∑i=1m-2γi(ξi-a)α1-∑i=1m-2γi.Theorem 6.
Suppose(A1)–(A4) hold and f(t,0,0)≢0 for a.e. t∈[a,b]. If there exist positive numbers m1, m2, and m4 with m1<m2≤τ12λ((b-a)/r(a))-1/(p-1)m4 such that the following conditions are satisfied:(C1)
f(t,μ,ν)≤(1/L)φp(m4),  (t,μ,ν)∈[a,b]×[0,τ2m4]×[-m4,m4],(C2)
f(t,μ,ν)>(1/L)φp(m2/τ1M),  (t,μ,ν)∈[σ,b-σ]×[m2,m2τ1-2]×[-m4,m4],(C3)
f(t,μ,ν)≤(1/L)φp(m1/N),  (t,μ,ν)∈[a,b]×[0,m1]×[-m4,m4], then the problem (𝒫) has at least three positive solutions u1, u2, and u3 satisfying
(30)maxt∈[a,b]|ui′(t)|≤m4fori=1,2,3,m2<mint∈[σ,b-σ]u1(t),m1<maxt∈[a,b]u2(t)withmint∈[σ,b-σ]u2(t)<m2,maxt∈[a,b]u3(t)<m1.Proof.
It is known that boundary value problem(𝒫) has a solution u if and only if u solves the operator equation u=Tu. By the definition of operator T, from Lemma 4 and the concavity of u, the functionals defined above satisfy σ(b-σ)θ(u)≤α(u)≤θ(u)=β(u), ∥u∥=max{θ(u),γ(u)}≤max{1,τ2}γ(u) for any u∈P(γ,m4)¯⊂K. We also note that β(ρu)≤ρβ(u) for 0≤ρ≤1, which suffices to show that the conditions of Lemma 1 hold with respect to T. First of all, we show that if the assumption (C1) is satisfied, then
(31)T:P(γ,m4)¯⟶P(γ,m4)¯.
In fact, for u∈P(γ,m4)¯, there is γ(u)=maxt∈[a,b]|u′(t)|≤m4. With Lemma 4, there is maxt∈[a,b]u(t)≤τ2maxt∈[a,b]|u′(t)|≤τ2m4, and the assumption (C1) implies that f(t,u(t),u′(t))≤(1/L)φp(m4) for a.e. t∈[a,b]. On the other hand, for u∈K, we obtain that Tu∈K, and so (Tu)(t) is concave on [a,b], and there is maxt∈[a,b]|(Tu)′(t)|=|(Tu)′(a)|. By using of r(t) is positive and monotone increasing on a.e. [a,b], and so
(32)γ(Tu)=maxt∈[a,b]|(Tu)′(t)|=φq(1r(a)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ)≤φq(1Lφp(m4)1r(a)∫ab‍h(τ)dτ)=m4.
Therefore, (31) is satisfied.
To check if the condition(B1) in Lemma 1 is fulfilled, we choose u0(t), where u0(t) is defined by (16). From the hypothesis conditions of the theorem, we have
(33)α(u0)=mint∈[σ,b-σ]u0(t)=u0(σ)≥τ1(m-)1/(p-1)λ≥m2τ1-1>m2,θ(u0)=maxt∈[a,b]u0(t)=u0(b)=(m-)1/(p-1)λ>m2τ1-2>m2,(34)γ(u0)=maxt∈[a,b]|(u0)′(t)|=|(u0)′(a)|=(m-b-ar(a))1/(p-1)=m2τ1-2λ(b-ar(a))1/(p-1)≤m4.
Equation (33) implies that α(u0)≥τ1θ(u0). That is, mint∈[σ,b-σ]u0(t)≥τ1maxt∈[a,b]u0(t). So for u0∈K, u0∈P(γ,θ,α,m2,m2τ1-2,m4) and {u0∈P(γ,θ,α,m2,m2τ1-2,m4):α(u0)>m2}≠∅ hold. Hence, for u∈P(γ,θ,α,m2,m2τ1-2,m4), we see that m2≤u(t)≤m2τ1-2 and |u′(t)|≤m4 for a.e. t∈[σ,b-σ]. Hence by the assumption (C2), one has that f(t,u(t),u′(t))>(1/L)φp(m2τ1-1/M) for a.e. t∈[σ,b-σ]. By the definition of the functional α, we obtain
(35)α(Tu)=mint∈[σ,b-σ](Tu)(t)≥τ1maxt∈[a,b](Tu)(t)=τ1(Tu)(b)=τ1(∫ab‍φq(1r(s)∫sb‍h(τ)f(τ,u(τ),u′(τ))dτ)ds+β1α1-∑i=1m-2γiφq×(1r(a)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ)+1α1-∑i=1m-2γi∑i=1m-2‍γi×∫aξi‍φq(1r(s)∫sb‍h(τ)×f(τ,u(τ),u′(τ))dτ∫sb‍)ds)≥τ1β1α1-∑i=1m-2γiφq(1Lφp(m2τ1-1M)1r(a)×∫ab‍h(τ)dτ)=m2.
Therefore, we get α(Tu)>m2 for u∈P(γ,θ,α,m2,m2τ1-2,m4).
Next, we show that the condition(B2) in Lemma 1 holds. In fact, if u∈P(γ,α,m2,m4) with θ(Tu)>m2τ1-2, then
(36)α(Tu)=mint∈[σ,b-σ](Tu)(t)≥τ1maxt∈[a,b](Tu)(t)=τ1θ(Tu)>m2τ1-1>m2.
Finally, we assert that the condition(B3) in Lemma 1 also holds. Since β(0)=0<m1, there holds that 0∉Q(γ,β,m1,m4). Assume that u∈Q(γ,β,m1,m4) with β(u)=m1. Then, by the assumption (C3) we get
(37)β(Tu)=maxt∈[a,b](Tu)(t)=(Tu)(b)=∫ab‍φq(1r(s)∫sb‍h(τ)f(τ,u(τ),u′(τ))dτ)ds+β1α1-∑i=1m-2γiφq×(1r(a)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ)+1α1-∑i=1m-2γi∑i=1m-2‍γi×∫aξi‍φq(1r(s)∫sb‍h(τ)f(τ,u(τ),u′(τ))dτ)ds≤∫ab‍φq(1r(a)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ)ds+β1α1-∑i=1m-2γi×φq(1r(a)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ)+1α1-∑i=1m-2γi∑i=1m-2‍γi×∫aξi‍φq(1r(a)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ)ds≤∫ab‍φq(1Lφp(m1N)1r(a)∫ab‍h(τ)dτ)ds+β1α1-∑i=1m-2γiφq(1Lφp(m1N)1r(a)∫ab‍h(τ)dτ)+1α1-∑i=1m-2γi∑i=1m-2‍γi×∫aξi‍φq(1Lφp(m1N)1r(a)∫ab‍h(τ)dτ)ds≤m1N(b-a+β1+∑i=1m-2γi(ξi-a)α1-∑i=1m-2γi)=m1.
Thus, the conditions (B1)–(B3) in Lemma 1 hold, and so the problem (𝒫) has at least three positive solutions u1, u2, and u3 such that (30) holds.Similar to the above arguments, we will discuss the problem(𝒫*)(r(t)φp(u′))′+h(t)f(t,u(t),u′(t))=0,t∈(a,b),u′(a)=0,α2u(b)+β2u′(b)=∑i=1m-2‍δiu(ξi).
Now, we replace the assumption (A1) with the following:(A1*) α2,β2∈[0,∞), δi∈[0,∞) satisfy α2>∑i=1m-2‍δi, a<ξ1<⋯<ξm-2<b(m>3).We only give the main results of the problem(𝒫*), and the proofs are omitted.Lemma 7.
Assume that(A1*), (A2), (A3), and (A4) hold. Then the problem (𝒫*) has a unique solution
(38)u(t)=∫tb‍φq(1r(s)∫as‍h(τ)f(τ,u(τ),u′(τ))dτ)ds+1α2-∑i=1m-2δi×(β2φq(1r(b)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ)+∑i=1m-2‍δi∫ξib‍φq×(1r(s)∫as‍h(τ)f(τ,u(τ),u′(τ))dτ)ds).The cone is defined by(39)K*={maxt∈[a,b]u∈E:u(t)≥0fort∈[a,b],u′′(t)≤0,u′(t)≥0fort∈(a,b),u′(a)=0,α2u(b)+β2u′(b)=∑i=1m-2‍δiu(ξi)andmint∈[σ,b-σ]u(t)≥τ1maxt∈[a,b]u(t)}.Lemma 8.
Assume that(A1) holds. If u∈K*, then
(40)mint∈[σ,b-σ]u(t)≥τ3∥u∥0,∥u∥0≤τ4∥u′∥0,
where τ3=β2/(β2+(b-a)(α2-∑i=1m-2‍δi)) and τ4=(β2+α2(b-a))/(α2-∑i=1m-2‍δi).In what follows, we define integral operatorsT and L*, M*, and N*:
(41)(Tu)(t)∶=∫tb‍φq(1r(s)∫as‍h(τ)f(τ,u(τ),u′(τ))dτ)ds+1α2-∑i=1m-2δi×(β2φq(1r(b)∫ab‍h(τ)f(τ,u(τ),u′(τ))dτ)+∑i=1m-2‍δi×∫ξib‍φq(1r(s)∫as‍h(τ)f(τ,u(τ),u′(τ))dτ)ds)fora.e.t∈[a,b],L*=1r(b)∫ab‍h(τ)dτ,M*=α2-∑i=1m-2δiβ2,N*=b-a+β2+∑i=1m-2δi(b-ξi)α2-∑i=1m-2δi.Theorem 9.
Suppose that(A1*), (A2), (A3), and (A4) hold and f(t,0,0)≢0 for a.e. t∈[a,b]. If there exist positive numbers λ′, m1, m2, and m4 with m1<m2≤τ32λ′((b-a)/r(b))-1/(p-1)m4 such that the following conditions are satisfied:(D1)
f(t,μ,ν)≤(1/L*)φp(m4/L*),(t,μ,ν)∈[a,b]×[0,τ4m4]×[-m4,m4],(D2)
f(t,μ,ν)>(1/L*)φp(m2/M*),(t,μ,ν)∈[a,b]×[m2,m2τ3-1]×[-m4,m4],(D3)
f(t,μ,ν)≤(1/L*)φp(m1/N*),(t,μ,ν)∈[a,b]×[0,m1]×[-m4,m4],then the problem (𝒫*) has at least three positive solutions u1, u2, and u3 satisfying
(42)maxt∈[a,b]|ui′(t)|≤m4fori=1,2,3,m2<mint∈[a,b]u1(t),m1<maxt∈[a,b]u2(t)withmint∈[a,b]u2(t)<m2,maxt∈[a,b]u3(t)<m1.From Theorem6 or Theorem 9, we can see that, when assumptions like (C1), (C2), and (C3) are imposed appropriately on the nonlinear term f, we also establish the existence results of an arbitrary odd number of positive solutions of the problem (𝒫) or (𝒫*). Here, we only present the form of Theorem 6.Theorem 10.
Suppose that(A1)–(A4) hold and f(t,0,0)≢0 for a.e. t∈[a,b]. If there exist positive numbers λ, m1i, m2i, and m4ii=1,2,…, with
(43)m11<m21≤τ12λ((b-a)/r(a))1/(p-1)m41<m12<m22≤τ12λ((b-a)/r(a))1/(p-1)m42<⋯<m1n<m2n≤τ12λ((b-a)/r(a))1/(p-1)m4n,
such that the following conditions are satisfied:(C1)
f(t,μ,ν)≤(1/L)φp(m4i),  (t,μ,ν)∈[a,b]×[0,τ2m4i]×[-m4i,m4i],(C2)
f(t,μ,ν)>(1/L)φp(m2i/τ1M),  (t,μ,ν)∈[σ,b-σ]×[m2i,m2iτ1-2]×[-m4i,m4i],(C3)
f(t,μ,ν)≤(1/L)φp(m1/N),  (t,μ,ν)∈[a,b]×[0,m1i]×[-m4i,m4i], then the problem (𝒫) has at least 2n-1 positive solutions.


## 4. Example

Letp=3, m=4, and σ=1/4. Consider the boundary value problem
(44)(r(t)φp(u′))′+h(t)f(t,u(t),u′(t))=0,0<t<1,2u(0)-u′(0)=12u(14)+12u(12),u′(1)=0,
where
(45)r(t)=t+1,h(t)=1t(1-t),f(t,μ,ν)={10-4t+2μ8+(ν3×104)3×10-5,μ≤4,10-4t+131072+(ν3×104)3×10-5,μ>4.It is easy to check that(A1)–(A4) hold. By some calculations, we have τ1=1/2, τ2=3, λ≈1.615, L≈3.142, M=1, and N=2.375. If we choose m1=1/2, m2=1, and m4=3×104, then f(t,μ,ν) satisfies
(46)f(t,μ,ν)≤1Lφ3(m4)≈2.866×108,(t,μ,ν)∈[0,1]×[0,9×104]×[-3×104,3×104],f(t,μ,ν)>1Lφ3(m2τ1M)≈1.274,(t,μ,ν)∈[14,34]×[1,4]×[-3×104,3×104],f(t,μ,ν)≤1Lφ3(m1N)≈0.014,(t,μ,ν)∈[0,1]×[0,12]×[-3×104,3×104].
Thus all assumptions of Theorem 6 hold, and so the problem (44) has at least three positive solutions u1, u2, and u3 such that maxt∈[0,1]|ui′(t)|≤3×104 for i=1,2,3, 1<min(1/4)≤t≤(3/4)u1(t), 1/2<maxt∈[0,1]u2(t)  with min(1/4)≤t≤(3/4)u2(t)<1, and maxt∈[0,1]u3(t)<1/2.



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*Source: 593285-2013-07-17.xml*