Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 35, 1-14; http://www.math.u-szeged.hu/ejqtde/

Triple positive solutions for second-order four-point boundary value problem with sign changing nonlinearities

Dapeng Xie^{a,}^{∗}, Yang Liu^{a}, Chuanzhi Bai^{b}

a Department of Mathematics, Hefei Teachers College, Hefei, Anhui 230061, P R China b Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsu 223300, P R China

Abstract

In this paper, we study the existence of triple positive solutions for second-order four-point boundary value problem with sign changing nonlinearities. We first study the associated Green’s function and obtain some useful properties. Our main tool is the fixed point theorem due to Avery and Peterson. The results of this paper are new and extent previously known results.

2000 Mathematics Subject Classication : 34B10, 34B18, 34B27.

Keywords: Four-point boundary value problem; Triple positive solutions; Cone; Sign changing nonlinearities.

1. Introduction

Boundary value problems (BVP) with sign changing nonlinearities have received special attention from many authors in recent years. Recently, existence results for positive solutions of second-order three-point boundary value problems with sign changing nonlinearities have been studied by some authors (see [3-6]). In [4], by applying the Krasnoseiskii fixed-point theorem in a cone, Liu have proved the existence of at least one positive solution for the following second-order three-point BVP

\{ \begin{matrix}u′′(t) + λa(t)f(u(t)) = 0, t ∈ (0, 1), \\ u(0) = 0, u(1) = βu(η). \end{matrix} (1.1)

where λ is a positive parameter, 0 < β < 1, 0 < η < 1, f : [0, +∞) → (0, +∞) is continuous and nondecreasing, a : [0, 1] → (−∞, +∞) is continuous and such that a(t) ≥ 0, t ∈ [0, η]; a(t) ≤ 0, t ∈ [η, 1]. Moreover, a(t) does not vanish identically on any subinterval of [0, 1].

Very recently, in [2], by applying the Krasnoseiskii fixed-point theorem in a cone, we improved the results obtained in [4] and established the existence of at least one or two positive solutions for second-order four-point BVP as follows

∗E-mail address: xiedapeng9@yahoo.com.cn

\{ \begin{matrix}u′′(t) + λa(t)f(u(t)) = 0, t ∈ (0, 1), \\ u(0) = αu(ξ), u(1) = βu(η). \end{matrix} (1.2)

where λ is a positive parameter, 0 ≤ α, β < 1, 0 < ξ < η < 1, f : [0, +∞) → (0, +∞) is continuous and nondecreasing, a : [0, 1] → (−∞, +∞) is continuous and such that a(t) ≤ 0, t ∈ [0, ξ]; a(t) ≥ 0, t ∈ [ξ, η]; a(t) ≤ 0, t ∈ [η, 1]. Moreover, a(t) does not vanish identically on any subinterval of [0, 1].

Inspired by [2], in this paper, by using the fixed point theorem due to Avery and Peterson, we consider the following second-order four-point boundary value problem with sign changing nonlinearities

\{ \begin{matrix}u′′(t) + a(t)f(u(t)) = 0, t ∈ (0, 1), \\ u(0) = αu′(1), u(ξ) = βu(η). \end{matrix} (1.3)

We make the following assumptions:

(H_{0}) 0 < max\{η, 1−η\} < ξ ≤ 1, − min\{η, 1−η\} ≤ α ≤ −\frac{(1−ξ)(ξ−βη)}{β(1−η)−(1−ξ)}, \frac{(1−ξ)(ξ+min\{η,1−η\})}{min\{η,1−η\}(1−η)−η(1−ξ)} < β < 1;

(H_{1}) f : [0, +∞) → (0, +∞) is continuous and nondecreasing;

(H_{2}) a : [0, 1] → (−∞, +∞) is continuous and such that a(t) ≥ 0, t ∈ [0, η]; a(t) ≤ 0, t ∈ [η, ξ]; a(t) ≥ 0, t ∈ [ξ, 1]. Moreover, a(t) does not vanish identically on any subinterval of [0, 1];

(H_{3}) There exists constant τ_{1} ∈ (βη, η), such that

g(t) = δa^{+}(η − δt) − \frac{1}{Λ}a^{−}(η + t) ≥ 0, t ∈ [0, ξ − η],

where a^{+}(t) = max{a(t), 0}, a^{−}(t) = − min{a(t), 0} and

δ = \frac{η − τ_{1}}{ξ − η} , Λ = \frac{min\{−α(1 − β), β(1 − η) − (1 − ξ)\}τ_{1}}{(ξ − η)[ξ + η − β(α + η)]} .

The aim of this paper is to improve the previous existence results in [4]. By using the fixed point theorem due to Avery and Peterson, we will study the existence of triple positive solutions for the BVP (1.3) under some conditions concerning the function a that is sign-changing on [0, 1]. To the best of our knowledge, to date no paper has appeared in the literature which discusses the existence of positive solutions for the BVP (1.3). This paper attempts to fill this gap in the literature. Remark 1.1. We point out that condition (H_{3}) is reasonable. For example, we take η = \frac{2}{5}, ξ = β = \frac{4}{5}, α = −\frac{3}{10}, τ_{1} = \frac{9}{25}, and

a(t) = \{ \begin{matrix}40(\frac{2}{5} − t), & t ∈ [0, \frac{2}{5}], \\ \frac{560}{27}(t − \frac{2}{5})(t − \frac{4}{5}), & t ∈ [\frac{2}{5}, \frac{4}{5}], \\ (t − \frac{4}{5}), & t ∈ [\frac{4}{5}, 1], \end{matrix}

Then, it is easy to check that (H_{0}) holds, and \frac{8}{25} = βη < τ_{1} < η = \frac{2}{5}, δ = \frac{1}{10}, Λ = \frac{27}{560}. Moreover, for t ∈ [0, \frac{2}{5}], we have

g(t) = δa^{+}(η − δt) − \frac{1}{Λ}a^{−}(η + t)

= \frac{1}{10}a^{+}(\frac{2}{5} − \frac{1}{10}t) − \frac{560}{27}a^{−}(\frac{2}{5} + t)

= t^{2} ≥ 0.

Thus, condition (H_{3}) holds, and condition (H_{2}) holds too.

2. Preliminary lemmas Let ϑ and θ be nonnegative continuous convex functionals on K, κ be a nonnegative continuous

concave functional on K, and ψ be a nonnegative continuous functional on K. Then for positive real numbers a, b, c and d, we define the following convex sets:

K(ϑ, d) = {u ∈ K : ϑ(u) < d};

K(ϑ, κ, b, d) = {u ∈ K : b ≤ κ(u), ϑ(u) ≤ d};
K(ϑ, θ, κ, b, c, d) = {u ∈ K : b ≤ κ(u), θ(u) ≤ c, ϑ(u) ≤ d};
K(ϑ, ψ, a, d) = {u ∈ K : a ≤ ψ(u), ϑ(u) ≤ d}.

Lemma 2.1 [1]. Let K be a cone in a Banach space E. Let ϑ and θ be nonnegative continuous convex functionals on K, κ be a nonnegative continuous concave functional on K, and ψ be a nonnegative continuous functional on K satisfying ψ(λu) ≤ λψ(u) for 0 ≤ λ ≤ 1, such that for some positive numbers M and d,

κ(u) ≤ ψ(u) and ∥u∥ ≤ Mϑ(u) (2.1) for all u ∈ \overline{K(ϑ, d)}. Suppose T : \overline{K(ϑ, d)} → \overline{K(ϑ, d)} is completely continuous and there exists positive numbers a, b and c with a < b such that (C_{1}) {u ∈ K(ϑ, θ, κ, b, c, d) : κ(u) > d} ≠ ∅ and κ(T u) > b for u ∈ K(ϑ, θ, κ, b, c, d); (C_{2}) κ(T u) > b, for u ∈ K(ϑ, κ, b, d) with θ(T u) > c; (C_{3}) 0 ∉ K(ϑ, ψ, a, d) and ψ(T u) < a for u ∈ K(ϑ, ψ, a, d), with ψ(u) = a. Then T has at least three fixed points u_{1}, u_{2} and u_{3} ∈ \overline{K(ϑ, d)} such that

ϑ(u_{i}) ≤ d for i = 1, 2, 3, b < κ(u_{1}), a < ψ(u_{2}) with κ(u_{2}) < b and ψ(u_{3}) < a. Lemma 2.2. If Δ := α + ξ − β(α + η) ≠ 0, then for y ∈ C[0, 1], the boundary value problem

u′′(t) + y(t) = 0, t ∈ (0, 1), (2.2)

u(0) = αu′(1), u(ξ) = βu(η), (2.3) has a unique solution

u(t) = \frac{α}{Δ}[(1 − β)t − (ξ − βη)] ∫_{0}^{1} y(s)ds + \frac{α + t}{Δ} ∫_{0}^{ξ}(ξ − s)y(s)ds

−\frac{β(α + t)}{Δ} ∫_{0}^{η}(η − s)y(s)ds − ∫_{0}^{t}(t − s)y(s)ds.

Proof. The proof follows by direct calculations.

Let G(t, s) be the Green’s function for the BVP (2.2)-(2.3). By direct calculation, we have

G(t, s) = \{ \begin{matrix}G_{1}(t, s), & s ∈ [0, η], \\ G_{2}(t, s), & s ∈ [η, ξ], \\ G_{3}(t, s), & s ∈ [ξ, 1], \end{matrix}

where

G_{1}(t, s) = \{ \begin{matrix}G_{11}(t, s) = \frac{s}{Δ}[ξ − βη − (1 − β)t], & s ≤ t, \\ G_{12}(t, s) = \frac{1}{Δ}\{α(1 − β)(t − s) + [ξ − βη − (1 − β)s]t\}, & t ≤ s, \end{matrix}

G_{2}(t, s) = \{ \begin{matrix}G_{21}(t, s) = \frac{1}{Δ}[αβ(η − s) + t(βη − s) + s(ξ − βη)], & s ≤ t, \\ G_{22}(t, s) = \frac{1}{Δ}[α(1 − β)t + α(βη − s) + t(ξ − s)], & t ≤ s, \end{matrix}

G_{3}(t, s) = \{ \begin{matrix}G_{31}(t, s) = \frac{1}{Δ}\{[α(1 − β) + ξ − βη]s − α(ξ − βη) − (ξ − βη)t\}, & s ≤ t, \\ G_{32}(t, s) = \frac{α}{Δ}[(1 − β)t − (ξ − βη)], & t ≤ s. \end{matrix}

Lemma 2.3. Suppose (H_{0}) holds. Then G(t, s) has the following properties

(i) G(t, s) ≥ 0, for all t, s ∈ [0, 1]; (ii) G(t, s_{1}) ≥ ΛG(t, s_{2}), t ∈ [0, 1], s_{1} ∈ [τ_{1}, η], s_{2} ∈ [η, ξ], where τ_{1} and Λ are given as in (H_{3}). Proof. (i) By calculating, we obtain

β > \frac{(1 − ξ)(ξ + min\{η, 1 − η\})}{min\{η, 1 − η\}(1 − η) + η(1 − ξ)} > \frac{1 − ξ}{1 − η},

and

α ≤ −\frac{(1 − ξ)(ξ − βη)}{β(1 − η) − (1 − ξ)} ≤ −\frac{(1 − ξ)(ξ − βη)}{β(ξ − η)}.

From this and (H_{0}), we have

G_{11}(t, s) = \frac{s}{Δ}[ξ − βη − (1 − β)t] ≥ \frac{s}{Δ}[β(1 − η) − (1 − ξ)] ≥ 0,

G_{12}(t, s) = \frac{1}{Δ}\{α(1 − β)(t − s) + [ξ − βη − (1 − β)s]t\}

EJQTDE, 2009 No. 35, p. 4

≥ \frac{1}{Δ}\{−α(1 − β)(s − t) + [ξ − βη − (1 − β)η]t\}

= \frac{1}{Δ}[−α(1 − β)(s − t) + (ξ − η)t] ≥ 0,

G_{21}(t, s) = \frac{1}{Δ}[αβ(η − s) + t(βη − s) + s(ξ − βη)]

≥ \frac{1}{Δ}[αβ(η − s) + (βη − s) + s(ξ − βη)]

= \frac{1}{Δ}[βη(1 + α) − s[β(α + η) + (1 − ξ)]\}

≥ \frac{1}{Δ}\{βη(1 + α) − ξ[β(α + η) + (1 − ξ)]\}

= \frac{1}{Δ}[−αβ(ξ − η) − (1 − ξ)(ξ − βη)] ≥ 0,

G_{22}(t, s) = \frac{1}{Δ}[α(1 − β)t + α(βη − s) + t(ξ − s)]

≥ \frac{1}{Δ}[α(1 − β)s + α(βη − s) + t(ξ − s)]

= \frac{1}{Δ}[−αβ(s − η) + t(ξ − s)] ≥ 0,

G_{31}(t, s) = \frac{1}{Δ}\{[α(1 − β) + ξ − βη]s − α(ξ − βη) − (ξ − βη)t\}

≥ \frac{1}{Δ}[α(1 − β) + (ξ − βη)ξ − α(ξ − βη) − (ξ − βη)]

= \frac{1}{Δ}\{α[(1 − β) − (ξ − βη)] − (ξ − βη)(1 − ξ)\}

= \frac{1}{Δ}\{α[(1 − ξ) − β(1 − η)] − (ξ − βη)(1 − ξ)\} ≥ 0,

G_{32}(t, s) = \frac{α}{Δ}[(1 − β)t − (ξ − βη)] ≥ \frac{α}{Δ}[(1 − β) − (ξ − βη)] = \frac{−α}{Δ}[β(1 − η) − (1 − ξ)] ≥ 0.

Thus, (i) holds.

(ii) By β > \frac{1−ξ}{1−η}, we get

α(1 − β) + ξ − η ≥ α(1 − \frac{1 − ξ}{1 − η}) + ξ − η = \frac{α + 1 − η}{1 − η} ≥ 0. (2.4)

If t ≤ s_{1}, in view of (2.4) and (H_{0}), we have

G_{12}(t, s_{1}) = \frac{1}{Δ}\{α(1 − β)(t − s_{1}) + [ξ − βη − (1 − β)s_{1}]t\}

= \frac{1}{Δ}\{−α(1 − β)s_{1} + [α(1 − β) + ξ − βη − (1 − β)s_{1}]t\}

≥ \frac{1}{Δ}\{−α(1 − β)s_{1} + [α(1 − β) + ξ − βη − (1 − β)η]t\}

> \frac{1}{Δ}\{−α(1 − β)τ_{1} + [α(1 − β) + ξ − η]t\}

≥ \frac{−α(1 − β)τ_{1}}{Δ} ,

G_{22}(t, s_{2}) = \frac{1}{Δ}[α(1 − β)t + α(βη − s_{2}) + t(ξ − s_{2})]

≤ \frac{1}{Δ}\{−α(s_{2} − βη) + t[α(1 − β) + ξ − η]\}

≤ \frac{1}{Δ}\{−α(ξ − βη) + η[α(1 − β) + ξ − η]\}

= \frac{1}{Δ}(ξ − η)(η − α)

≤ \frac{1}{Δ}(ξ − η)[(ξ + η) − β(α + η)].

Thus,

\frac{G(t, s_{1})}{G(t, s_{2})} = \frac{G_{12}(t, s_{1})}{G_{22}(t, s_{2})} ≥ Λ.

If s_{1} ≤ t ≤ s_{2}, we have by (2.4) and (H_{0}) that

G_{11}(t, s_{1}) = \frac{1}{Δ}\{s_{1}[ξ − βη − (1 − β)t]\} ≥ \frac{1}{Δ}\{τ_{1}[ξ − βη − (1 − β)ξ]\}

= \frac{1}{Δ}[βτ_{1}(ξ − η)] ≥ \frac{τ_{1}}{Δ}[β(1 − η) − (1 − ξ)],

G_{22}(t, s_{2}) = \frac{1}{Δ}[α(t − s_{2}) + αβ(η − t) + t(ξ − s_{2})]

≤ \frac{1}{Δ}\{−α(s_{2} − βη) + t[α(1 − β) + ξ − η]\}

≤ \frac{1}{Δ}\{−α(ξ − βη) + ξ[α(1 − β) + ξ − η]\}

= \frac{1}{Δ}(ξ − η)(ξ − αβ)

≤ \frac{1}{Δ}(ξ − η)[(ξ + η) − β(α + η)]\}.

So,

\frac{G(t, s_{1})}{G(t, s_{2})} = \frac{G_{11}(t, s_{1})}{G_{22}(t, s_{2})} ≥ Λ.

If t ≥ s_{2}, it follows from (H_{0}) that EJQTDE, 2009 No. 35, p. 6

G_{11}(t, s_{1}) ≥ \frac{τ_{1}}{Δ}[β(1 − η) − (1 − ξ)],

G_{21}(t, s_{2}) = \frac{1}{Δ}[αβ(η − s_{2}) + βη(t − s_{2}) + s_{2}(ξ − t)]

= \frac{1}{Δ}[−αβ(s_{2} − η) + (βη − s_{2})t + s_{2}(ξ − βη)]

≤ \frac{1}{Δ}[−αβ(ξ − η) + (βη − η)η + ξ(ξ − βη)]

= \frac{1}{Δ}(ξ − η)[(ξ + η) − β(α + η)].

Then,

\frac{G(t, s_{1})}{G(t, s_{2})} = \frac{G_{11}(t, s_{1})}{G_{21}(t, s_{2})} ≥ Λ.

This completes the proof.

Let C[0, 1] be the Banach space with norm ∥u∥ = max_{0≤t≤1} |u(t)|. Denote

X={ u ∈ C[0, 1] : min_{0≤t≤1} u(t) ≥ 0 and u(0) = αu′(1), u(ξ) = βu(η) },

and define the cone K ⊂ X and the operator T : C[0, 1] → C[0, 1] by

K={ u ∈ X: u(t) is concave on [0, η] ∪ [ξ, 1], u(t) is convex on [η, ξ] }, and

(T u)(t) = ∫_{0}^{1} G(t, s)a(s)f(u(s))ds.

Lemma 2.4. Assume that u ∈ K, then

u(t) ≥ ϕ(t)u(η), t ∈ [0, η], u(t) ≤ ϕ(t)u(η), t ∈ [η, ξ], (2.5) where

ϕ(t) = \{ \begin{matrix}\frac{t}{η} , & t ∈ [0, η], \\ \frac{ξ − βη + (β − 1)t}{ξ − η}, & t ∈ [η, ξ]. \end{matrix}

Proof. Since u ∈ K, then u(t) is concave on [0, η], u(t) is convex on [η, ξ]. Thus, we have

u(t) ≥ u(0) + \frac{u(η) − u(0)}{η}t ≥ \frac{t}{η} u(η), for all t ∈ [0, η],

and

u(t) ≤ u(ξ) + \frac{u(ξ) − u(η)}{ξ − η}(t − ξ) = \frac{ξ − βη + (β − 1)t}{ξ − η}u(η), for each t ∈ [η, ξ].

So, (2.5) holds. Lemma 2.5. Assume that u ∈ K.

(i) If max_{0≤t≤η} u(t) ≥ max_{ξ<t≤1} u(t), then u(t) ≥ γ∥u∥, for t ∈ [βη, τ_{1}], (2.6) (ii) If max_{ξ<t≤1} u(t) ≥ max_{0≤t≤η} u(t), then u(t) ≥ γ∥u∥, for t ∈ [ξ + β(1 − ξ), τ_{2}], (2.7)

where τ_{1} ∈ (βη, η), τ_{2} ∈ (ξ + β(1 − ξ), 1), γ = min \{β, 1 − \frac{τ_{1}}{η} , 1 − \frac{τ_{2} − ξ}{1 − ξ} \}.

Proof. Since u ∈ K, then u(t) ≥ 0, for all t ∈ [0, 1], u(t) is concave on [0, η] ∪ [ξ, 1], u(t) is convex on [η, ξ], and u(ξ) = βu(η) < u(η). Thus, we have

∥u∥ = max_{0≤t≤1} |u(t)| = max{ max_{0≤t≤η} u(t), max_{ξ<t≤1} u(t)}.

There are two cases to consider

(i) max_{0≤t≤η} u(t) ≥ max_{ξ<t≤1} u(t), so max_{0≤t≤1} |u(t)| = max_{0≤t≤η} u(t). Let

ω_{1} = inf{ε ∈ [0, η] : max_{0≤t≤η} u(t) = u(ε)}.

If t ∈ [0, ω_{1}], then by the concavity of u(t), we have

u(t) ≥ u(0) + \frac{u(ω_{1}) − u(0)}{ω_{1}}t = \frac{ω_{1} − t}{ω_{1}}u(0) + \frac{t}{ω_{1}} u(ω_{1}) ≥ \frac{t}{ω_{1}} u(ω_{1}) ≥ \frac{t}{η} ∥u∥. (2.8)

If t ∈ [ω_{1}, η], similarly, we obtain

u(t) ≥ u(ω_{1}) + \frac{u(ω_{1}) − u(η)}{ω_{1} − η}(t − ω_{1}) = \frac{η − t}{η − ω_{1}} u(ω_{1}) + \frac{t − ω_{1}}{η − ω_{1}} u(η)

≥ \frac{η − t}{η − ω_{1}} u(ω_{1}) ≥ \frac{η − t}{η}∥u∥. (2.9)

In view of (2.8) and (2.9), we get

u(t) ≥ min \{\frac{t}{η} , 1 − \frac{t}{η} \} ∥u∥, t ∈ [0, η],

which yields

min_{t∈[βη,τ_{1}]} u(t) ≥ min \{β, 1 − \frac{τ_{1}}{η} \} ∥u∥ ≥ γ∥u∥.

(ii) max_{ξ<t≤1} u(t) ≥ max_{0≤t≤η} u(t), then max_{0≤t≤1} |u(t)| = max_{ξ≤t≤1} u(t). Let

ω_{2} = inf{ϵ ∈ (ξ, 1] : max_{ξ<t≤1} u(t) = u(ϵ)}.

If t ∈ (ξ, ω_{2}], by the concavity of u(t), we have

u(t) ≥ u(ξ) + \frac{u(ω_{2}) − u(ξ)}{ω_{2} − ξ}(t − ξ) = \frac{ω_{2} − t}{ω_{2} − ξ} u(ξ) + \frac{t − ξ}{ω_{2} − ξ} u(ω_{2})

≥ \frac{t − ξ}{ω_{2} − ξ} u(ω_{2}) ≥ \frac{t − ξ}{1 − ξ} ∥u∥. (2.10)

If t ∈ [ω_{2}, 1], similarly, we get

u(t) ≥ u(ω_{2}) + \frac{u(ω_{2}) − u(1)}{ω_{2} − 1}(t − ω_{2}) = \frac{1 − t}{1 − ω_{2}} u(ω_{2}) + \frac{t − ω_{2}}{1 − ω_{2}} u(1)

≥ \frac{1 − t}{1 − ω_{2}} u(ω_{2}) ≥ \frac{1 − t}{1 − ξ} ∥u∥. (2.11)

It follows from (2.10) and (2.11) that

u(t) ≥ min \{\frac{t − ξ}{1 − ξ} , 1 − \frac{t − ξ}{1 − ξ} \} ∥u∥, t ∈ [ξ, 1].

From this

min_{t∈[ξ+β(1−ξ),τ_{2}]} u(t) ≥ min \{β, 1 − \frac{τ_{2} − ξ}{1 − ξ} \} ∥u∥ ≥ γ∥u∥.

Lemma 2.6. Assume that (H_{0})-(H_{3}) hold. Then,

∫_{τ_{1}}^{η} G(t, s)a^{+}(s)f(u(s))ds ≥ ∫_{η}^{ξ} G(t, s)a^{−}(s)f(u(s))ds. (2.12)

Proof. If x ∈ [0, ξ − η], then η − δx ∈ [τ_{1}, η], η + x ∈ [η, ξ], where τ_{1} and δ as in (H_{3}). By the definition of ϕ ( see Lemma 2.4 ), we have

ϕ (η − δx) = 1 − \frac{η − τ_{1}}{η(ξ − η)}x and ϕ(η + x) = 1 − \frac{1 − β}{ξ − η} x, for x ∈ [0, ξ − η].

By βη < τ_{1} < η, we get

ϕ(η − δx) ≥ ϕ(η + x), for all x ∈ [0, ξ − η]. (2.13)

Let s = η − δx, ∀x ∈ [0, ξ − η]. It follows from Lemmas 2.3 and 2.4, (2.13) and f is nondecreasing that

∫_{τ_{1}}^{η} G(t, s)a^{+}(s)f(u(s))ds = δ ∫_{0}^{ξ−η} G(t, η − δx)a^{+}(η − δx)f(u(η − δx))dx
≥ δΛ ∫_{0}^{ξ−η} G(t, η + x)a^{+}(η − δx)f(ϕ(η − δx)u(η))dx
≥ ∫_{0}^{ξ−η} G(t, η + x)a^{+}(η − δx)f(ϕ(η + x)u(η))dx. (2.14)

On the other hand, let s = η + x, x ∈ [0, ξ − η], by Lemma 2.4, one has

∫_{η}^{ξ} G(t, s)a^{−}(s)f(u(s))ds = ∫_{0}^{ξ−η} G(t, η + x)a^{−}(η + x)f(u(η + x))dx

≤ ∫_{0}^{ξ−η} G(t, η + x)a^{−}(η + x)f(ϕ(η + x)u(η))dx. (2.15)

Thus, in view of (2.14) and (2.15), we know that (2.12) holds. Lemma 2.7. Assume that (H_{0})-(H_{3}) hold, then T : K → K is completely continuous. Proof. Firstly, we claim T (K) ⊂ K. In fact, for all u ∈ K, we have by Lemma 2.6 that

(T u)(t) = ∫_{0}^{1} G(t, s)a(s)f(u(s))ds

= (∫_{0}^{τ_{1}} G(t, s)a^{+}(s)f(u(s))ds + ∫_{τ_{1}}^{η} G(t, s)a^{+}(s)f(u(s))ds
− ∫_{η}^{ξ} G(t, s)a^{−}(s)f(u(s))ds + ∫_{ξ}^{1} G(t, s)a^{+}(s)f(u(s))ds) ≥ 0.

Moreover, (T u)(0) = α(T u)′(1), (T u)(ξ) = β(T u)(η), Then, T : K → X. On the other hand, (T u)′′(t) = −a^{+}(t)f(u(t)) ≤ 0, t ∈ [0, η],

(T u)′′(t) = a^{−}(t)f(u(t)) ≥ 0, t ∈ [η, ξ],
(T u)′′(t) = −a^{+}(t)f(u(t)) ≤ 0, t ∈ [ξ, 1],

This show that T : K → K. It can be shown that T : K → K is completely continuous by Arzela-Ascoli theorem.

3. Main result Let the nonnegative continuous concave functional κ on K, the nonnegative continuous convex

functionals θ, ϑ and the nonnegative continuous functional ψ be defined on the cone K by

κ(u) = min{ min_{βη≤t≤τ_{1}} |u(t)|, min_{ξ+β(1−ξ)≤t≤τ_{2}} |u(t)|}, ϑ(u) = θ(u) = ψ(u) = max_{0≤t≤1} |u(t)|, for u ∈ K.

For the sake of brevity, we denote

N = max_{0≤t≤1} (∫_{0}^{η} G(t, s)a^{+}(s)ds + ∫_{ξ}^{1} G(t, s)a^{+}(s)ds),

M = min{M_{1}, M_{2}, M_{3}, M_{4}}, where

M_{1} = ∫_{βη}^{τ_{1}} G(βη, s)a^{+}(s)ds, M_{2} = ∫_{βη}^{τ_{1}} G(τ_{1}, s)a^{+}(s)ds,

M_{3} = ∫_{ξ+β(1−ξ)}^{τ_{2}} G(ξ + β(1 − ξ), s)a^{+}(s)ds, M_{4} = ∫_{ξ+β(1−ξ)}^{τ_{2}} G(τ_{2}, s)a^{+}(s)ds.

Therorem 3.1. Suppose that (H_{0})-(H_{3}) hold. Let 0 < a < b < γd, and the following conditions hold,

(H_{4}) f(u(t)) ≤ \frac{d}{N}, for all t ∈ [0, 1], u ∈ [0, d],

(H_{5}) f(u(t)) ≥ \frac{b}{M}, for all t ∈ [βη, τ_{1}] ∪ [ξ + β(1 − ξ), τ_{2}], u ∈ [b, \frac{b}{γ}],

(H_{6}) f(u(t)) ≤ \frac{a}{N}, for all t ∈ [0, 1], u ∈ [0, a].

Then the BVP (1.3) have at least three positive solutions u_{1}, u_{2} and u_{3} such that

max_{0≤t≤1} |u_{i}(t)| ≤ d for i = 1, 2, 3, b < min_{βη≤t≤τ_{1}} |u_{1}(t)| ( or b < min_{ξ+β(1−ξ)≤t≤τ_{2}} |u_{1}(t)| ), max_{0≤t≤1} |u_{1}| ≤ d,

a < max_{0≤t≤1} |u_{2}(t)| < \frac{b}{γ} with min_{βη≤t≤τ_{1}} |u_{2}(t)| < b (or min_{ξ+β(1−ξ)≤t≤τ_{2}} |u_{2}(t)| < b ), and max_{0≤t≤1} |u_{3}(t)| < a.

Proof. Firstly, we check T : \overline{K(ϑ, d)} → \overline{K(ϑ, d)} is completely continuous operator.

If u ∈ \overline{K(ϑ, d)}, then ϑ(u) = max_{0≤t≤1} |u(t)| ≤ d. It follows from (H_{4}) that

ϑ(T u) = max_{0≤t≤1} |T u(t)| = max_{0≤t≤1} ∫_{0}^{1} G(t, s)a(s)f(u(s))ds

= max_{0≤t≤1} (∫_{0}^{η} G(t, s)a^{+}(s)f(u(s))ds − ∫_{η}^{ξ} G(t, s)a^{−}(s)f(u(s))ds

+ ∫_{ξ}^{1} G(t, s)a^{+}(s)f(u(s))ds)

≤ max_{0≤t≤1} (∫_{0}^{η} G(t, s)a^{+}(s)f(u(s))ds + ∫_{ξ}^{1} G(t, s)a^{+}(s)f(u(s))ds)

≤ \frac{d}{N} · max_{0≤t≤1} (∫_{0}^{η} G(t, s)a^{+}(s)ds + ∫_{ξ}^{1} G(t, s)a^{+}(s)ds) = d.

Therefore, T : \overline{K(ϑ, d)} → \overline{K(ϑ, d)}. From this and Lemma 2.7, we know that T : \overline{K(ϑ, d)} → \overline{K(ϑ, d)} is completely continuous operator.

To check condition (C_{1}) of Lemma 2.1, we choose u(t) = \frac{b}{γ}, 0 ≤ t ≤ 1. It is easy to check that u(t) = \frac{b}{γ} ∈ K(ϑ, θ, κ, b, \frac{b}{γ} , d) and κ(u) = κ(\frac{b}{γ}) > b, and so \{u ∈ K(ϑ, θ, κ, b, \frac{b}{γ} , d) : κ(u) > b\} ≠ ∅. Thus, for all u ∈ K(ϑ, θ, κ, b, \frac{b}{γ} , d), we have that b ≤ u(t) ≤ \frac{b}{γ}, for t ∈ [βη, τ_{1}] ∪ [ξ + β(1 − ξ), τ_{2}]. We consider the following two cases.

Case (i) Suppose that max_{0≤t≤η} u(t) ≥ max_{ξ<t≤1} u(t). For all u ∈ K(ϑ, θ, κ, b, \frac{b}{γ} , d), then we have T u ∈ K, and so T u is concave on [0, η], we have to distinguish two cases: (1) min_{βη≤t≤τ_{1}} T u(t) = T u(βη); (2) min_{βη≤t≤τ_{1}} T u(t) = T u(τ_{1}). For case (1), we have from Lemma 2.6 that

min_{βη≤t≤τ_{1}} T u(t) = T u(βη) = ∫_{0}^{1} G(βη, s)a(s)f(u(s))ds

= ∫_{0}^{τ_{1}} G(βη, s)a^{+}(s)f(u(s))ds + ∫_{τ_{1}}^{η} G(βη, s)a^{+}(s)f(u(s))ds

− ∫_{η}^{ξ} G(βη, s)a^{−}(s)f(u(s))ds + ∫_{ξ}^{1} G(βη, s)a^{+}(s)f(u(s))ds

≥ ∫_{0}^{τ_{1}} G(βη, s)a^{+}(s)f(u(s))ds + ∫_{ξ}^{1} G(βη, s)a^{+}(s)f(u(s))ds

≥ ∫_{βη}^{τ_{1}} G(βη, s)a^{+}(s)f(u(s))ds

≥ \frac{b}{M} · ∫_{βη}^{τ_{1}} G(βη, s)a^{+}(s)ds ≥ b.

For case (2), similarly, we get

min_{βη≤t≤τ_{1}} T u(t) = T u(τ_{1}) ≥ ∫_{βη}^{τ_{1}} G(τ_{1}, s)a^{+}(s)f(u(s))ds ≥ \frac{b}{M} · ∫_{βη}^{τ_{1}} G(τ_{1}, s)a^{+}(s)ds ≥ b.

Thus,

min_{βη≤t≤τ_{1}} T u(t) ≥ b, for all u ∈ K(ϑ, θ, κ, b, \frac{b}{γ} , d) with b ≤ u(t) ≤ \frac{b}{γ}, t ∈ [βη, τ_{1}]. (3.1)

Case (ii) Suppose that max_{ξ<t≤1} u(t) ≥ max_{0≤t≤η} u(t). For all u ∈ K(ϑ, θ, κ, b, \frac{b}{γ} , d), then we have T u ∈ K, and so T u is concave on [ξ, 1], we have to distinguish two cases: (1) min_{ξ+β(1−ξ)≤t≤τ_{2}} T u(t) = T u(ξ + β(1 − ξ)); (2) min_{ξ+β(1−ξ)≤t≤τ_{2}} T u(t) = T u(τ_{2}). For case (1), it follows from Lemma 2.6 that

min_{ξ+β(1−ξ)≤t≤τ_{2}} T u(t) = T u(ξ + β(1 − ξ)) ≥ ∫_{ξ+β(1−ξ)}^{τ_{2}} G(ξ + β(1 − ξ), s)a^{+}(s)f(u(s))ds

≥ \frac{b}{M} · ∫_{ξ+β(1−ξ)}^{τ_{2}} G(ξ + β(1 − ξ), s)a^{+}(s)ds ≥ b.

For case (2), similarly, we obtain

min_{ξ+β(1−ξ)≤t≤τ_{2}} T u(t) = T u(τ_{2}) ≥ ∫_{ξ+β(1−ξ)}^{τ_{2}} G(ξ + β(1 − ξ), s)a^{+}(s)f(u(s))ds

≥ \frac{b}{M} · ∫_{ξ+β(1−ξ)}^{τ_{2}} G(ξ + β(1 − ξ), s)a^{+}(s)ds ≥ b.

Hence,

min_{ξ+β(1−ξ)≤t≤τ_{2}} T u(t) ≥ b, for all u ∈ K(ϑ, θ, κ, b, \frac{b}{γ} , d) with b ≤ u(t) ≤ \frac{b}{γ}, t ∈ [ξ + β(1 − ξ), τ_{2}].(3.2)

In view of (3.1) and (3.2), we have

κ(T u) = min\{ min_{βη≤t≤τ_{1}} T u(t), min_{ξ+β(1−ξ)≤t≤τ_{2}} T u(t)\} ≥ b, for all u ∈ K(ϑ, θ, κ, b, \frac{b}{γ} , d).

This shows that condition (C_{1}) of Lemma 2.1 is sastisfied.

Secondly, we show (C_{2}) of Lemma 2.1 is satisfied. If max_{0≤t≤η} u(t) ≥ max_{ξ<t≤1} u(t). By Lemma 2.5 (i) and Lemma 2.7, we have

min_{βη≤t≤τ_{1}} T u(t) ≥ γ∥T u∥ = γθ(T u) > b, for all u ∈ K(ϑ, κ, b, d) with θ(T u) > \frac{b}{γ}. (3.3)

If max_{ξ<t≤1} u(t) ≥ max_{0≤t≤η} u(t). It follows from Lemma 2.5 (ii) and Lemma 2.7 that

min_{ξ+β(1−ξ)≤t≤τ_{2}} T u(t) ≥ γ∥T u∥ = γθ(T u) > b, for all u ∈ K(ϑ, κ, b, d) with θ(T u) > \frac{b}{γ}. (3.4)

Therefore, we obtain by (3.3) and (3.4) that

κ(T u) = min\{ min_{βη≤t≤τ_{1}} T u(t), min_{ξ+β(1−ξ)≤t≤τ_{2}} T u(t)\} ≥ b, for all u ∈ K(ϑ, κ, b, d) with θ(T u) > \frac{b}{γ}.

Finally, we show condition (C_{3}) of Lemma 2.1 is also satisfied. Obviously, as ψ(0) = 0 < a, there holds 0 ∉ R(ϑ, ψ, a, d). Suppose u ∈ R(ϑ, ψ, a, d) with ψ(u) = a. Then we have by condition (H_{6}) that

ψ(T u) = max_{0≤t≤1} |T u(t)| = max_{0≤t≤1} ∫_{0}^{1} G(t, s)a(s)f(u(s))ds

= max_{0≤t≤1} (∫_{0}^{η} G(t, s)a^{+}(s)f(u(s))ds − ∫_{η}^{ξ} G(t, s)a^{−}(s)f(u(s))ds

+ ∫_{ξ}^{1} G(t, s)a^{+}(s)f(u(s))ds)

≤ max_{0≤t≤1} (∫_{0}^{η} G(t, s)a^{+}(s)f(u(s))ds + ∫_{ξ}^{1} G(t, s)a^{+}(s)f(u(s))ds)

≤ \frac{a}{N} · max_{0≤t≤1} (∫_{0}^{η} G(t, s)a^{+}(s)f(u(s))ds + ∫_{ξ}^{1} G(t, s)a^{+}(s)f(u(s))ds)

= a.

So, condition (C_{3}) of Lemma 2.1 is satisfied. Therefore, according to Lemma 2.1, there exist three positive solutions u_{1}, u_{2} and u_{3} for the BVP (1.3) such that

max_{0≤t≤1} |u_{i}(t)| ≤ d for i = 1, 2, 3, b < min_{βη≤t≤τ_{1}} |u_{1}(t)| ( or b < min_{ξ+β(1−ξ)≤t≤τ_{2}} |u_{1}(t)| ), max_{0≤t≤1} |u_{1}| ≤ d,

a < max_{0≤t≤1} |u_{2}(t)| < \frac{b}{γ} with min_{βη≤t≤τ_{1}} |u_{2}(t)| < b ( or min_{ξ+β(1−ξ)≤t≤τ_{2}} |u_{2}(t)| < b ), and max_{0≤t≤1} |u_{3}(t)| < a.

Acknowledgements

Project supported by the National Natural Science Foundation of China (10771212) and the Natural Science Foundation of Jiangsu Education Office (06KJB110010).

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(Received February 25, 2009)