Electronic Journal of Qualitative Theory of Differential Equations 2010, No. 33, 1-13; http://www.math.u-szeged.hu/ejqtde/

Existence of multiple positive solutions of higher order multi-point nonhomogeneous boundary value problem

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

a Department of Mathematics, Hefei Normal University, Hefei, Anhui 230601, P R China b School of Mathematical Sciences, Huaiyin Normal University, Huaian, Jiangsu 223300, P R China

Abstract

In this paper, by using the Avery and Peterson fixed point theorem, we establish the existence of multiple positive solutions for the following higher order multi-point nonhomogeneous boundary value problem

u^{(n)}(t) + f(t, u(t), u′(t), …, u^{(n−2)}(t)) = 0, t ∈ (0, 1),

u(0) = u′(0) = ⋯ = u^{(n−3)}(0) = u^{(n−2)}(0) = 0, u^{(n−2)}(1) − ∑_{i=1}^{m} a_{i}u^{(n−2)}(ξ_{i}) = λ,

where n ≥ 3 and m ≥ 1 are integers, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m} < 1 are constants, λ ∈ [0, ∞) is a parameter, a_{i} > 0 for 1 ≤ i ≤ m and ∑_{i=1}^{m} a_{i}ξ_{i} < 1, f(t, u, u′, ⋯, u^{(n−2)}) ∈ C([0, 1] × [0, ∞)^{n−1}, [0, ∞)). We give an example to illustrate our result.

Keywords: Multi-point boundary value problem; Cone; Nonhomogeneous; Positive solution; Fixed point theorem.

1. Introduction

In this paper, we consider the existence of multiple positive solutions for the following higher order multi-point nonhomogeneous boundary value problem

\{ \begin{matrix}u^{(n)}(t) + f(t, u(t), u′(t), …, u^{(n−2)}(t)) = 0, t ∈ (0, 1), \\ u(0) = u′(0) = ⋯ = u^{(n−3)}(0) = u^{(n−2)}(0) = 0, u^{(n−2)}(1) − ∑_{i=1}^{m} a_{i}u^{(n−2)}(ξ_{i}) = λ, \end{matrix} (1.1)

where n ≥ 3 and m ≥ 1 are integers, λ ∈ [0, ∞) is a parameter, and a_{i}, ξ_{i}, f satisfying

(H_{1}) a_{i} > 0 for 1 ≤ i ≤ m, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m} < 1 and ∑_{i=1}^{m} a_{i}ξ_{i} < 1; (H_{2}) f : [0, 1] × [0, ∞)^{n−1} → [0, ∞) is continuous.

For the past few years, the existence of solutions for higher order ordinary differential equations has received a wide attention. We refer the reader to [2-6,8,15-20] and references therein. However, most of the above mentioned references only consider the cases in which f does not contain higher order derivatives of u and the parameter λ = 0. This is because the presence of higher order derivatives in the nonlinear function f and the parameter λ ≠ 0 make the study more difficult. For example, in [6], Graef and Yang obtained existence and nonexistence results for positive solutions of the following nth ordinary differential equation

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

\{ \begin{matrix}u^{(n)}(t) + μg(t)f(u(t)) = 0, t ∈ (0, 1), \\ u(0) = u′(0) = ⋯ = u^{(n−3)}(0) = u^{(n−2)}(0) = u^{(n−2)}(1) − ∑_{i=1}^{m} a_{i}u^{(n−2)}(ξ_{i}) = 0, \end{matrix} (1.2)

under the following assumptions:

(C_{1}) f : [0, 1] → [0, ∞) is a continuous function and μ > 0 is a parameter;

(C_{2}) g : [0, 1] → [0, ∞) is a continuous function with ∫_{0}^{1} g(t)dt > 0;

(C_{3}) n ≥ 3 and m ≥ 1 are integers;

(C_{4}) a_{i} > 0 for 1 ≤ i ≤ m and ∑_{i=1}^{m} a_{i} = 1;

(C_{5}) \frac{1}{2} ≤ ξ_{1} < ξ_{2} < ⋯ < ξ_{m} < 1.

Obviously, condition (H_{1}) in this paper is weaker than the conditions (C_{4}) and (C_{5}).

Often when authors deal with higher order boundary value problems in which the nonlinear function f contains higher order derivatives, they transform the higher order equation into a second order equation, see [7,9,14,22] and references therein. For instance, in [22], by using the fixed-point principle in a cone and the fixed-point index theory for a strict-set-contraction operator, Zhang, Feng and Ge established the existence and nonexistence of positive solutions for nth-order threepoint boundary value problems in Banach spaces

\{ \begin{matrix}u^{(n)}(t) + f(t, u(t), u′(t), …, u^{(n−2)}(t)) = θ, t ∈ J, \\ u(0) = u′(0) = ⋯ = u^{(n−3)}(0) = u^{(n−2)}(0) = θ, u^{(n−2)}(1) = ρu^{(n−2)}(η), \end{matrix} (1.3)

where J = [0, 1], f ∈ C(J × P^{n−1}, P), P is a cone of real Banach space, ρ ∈ (0, 1), and θ is the zero element of the real Banach space.

Nonhomogeneous boundary value problems have received special attention from many authors in recent years (see [10-13,17]). Recently, in the case of n = 3 and m = 1, by employing the Guo-Krasnosel’skii fixed point theorem and Schauder’s fixed point theorem, Sun [17] established existence and nonexistence of positive solutions to the problem (1.1) when f(t, u(t), u′(t)) = a(t)f(u(t)), λ ∈ (0, ∞) and the nonlinearity f is either superlinear or sublinear. However, our problem is more general than the problem of [2-4,6,8,15,17,22] and the aim of our paper is to investigate the existence of two or three positive solutions for the problem (1.1). The key tool in our approach is the Avery and Peterson fixed point theorem. We give an example to illustrate our result. To the best of our knowledge, no previous results are available for triple positive solutions for the nth-order multi-point boundary value problem with the higher order derivatives and the parameter λ by using the Avery and Peterson fixed point theorem. The goal of this paper is to fill this gap.

2. Preliminary Lemmas

Definition 2.1. The map α is said to be a nonnegative continuous concave functional on a cone K of a real Banach space E provided that α : K → [0, ∞) is continuous and

α(tx + (1 − t)y) ≥ tα(x) + (1 − t)α(y), ∀x, y ∈ K, 0 ≤ t ≤ 1.

Similarly, we say the map β is a nonnegative continuous convex functional on a cone K of a real Banach space E provided that β : K → [0, ∞) is continuous and

β(tx + (1 − t)y) ≤ tβ(x) + (1 − t)β(y), ∀x, y ∈ K, 0 ≤ t ≤ 1.

Let γ and θ be nonnegative continuous convex functionals on K, α 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:

P (γ, d) = {x ∈ K | γ(x) < d},
P (γ, α, b, d) = {x ∈ K | b ≤ α(x), γ(x) ≤ d},
P (γ, θ, α, b, c, d) = {x ∈ K | b ≤ α(x), θ(x) ≤ c, γ(x) ≤ d},
Q(γ, ψ, a, d) = {x ∈ K | a ≤ ψ(x), γ(x) ≤ d}.

Lemma 2.1.([1]) Let K be a cone in a real Banach space E. Let γ and θ be nonnegative continuous convex functionals on K, α be a nonnegative continuous functional on K, and ψ be a nonnegative continuous functional on K satisfying ψ(μx) ≤ μψ(x) for 0 ≤ μ ≤ 1, such that for some positive numbers M and d,

α(x) ≤ ψ(x) and ∥x∥ ≤ Mγ(x).

for all x ∈ \overline{P (γ, d)}. Suppose T : \overline{P (γ, d)} → \overline{P (γ, d)} is completely continuous and there exist positive numbers a, b and c with a < b such that

(i) {x ∈ P (γ, θ, α, b, c, d) | α(x) > b} ≠ ∅ and α(T x) > b for x ∈ P (γ, θ, α, b, c, d);

(ii) α(T x) > b for x ∈ P (γ, α, b, d) with θ(T x) > c;

(iii) 0 ∉ Q(γ, ψ, a, d) and ψ(T x) < a for x ∈ Q(γ, ψ, a, d) with ψ(x) = a.

Then T has at least three fixed points x_{1}, x_{2}, x_{3} ∈ \overline{P (γ, d)}, such that

γ(x_{i}) ≤ d for i = 1, 2, 3, b < α(x_{1}), a < ψ(x_{2}) with α(x_{2}) < b, ψ(x_{3}) < a.

Lemma 2.2. Suppose that Δ =: ∑_{i=1}^{m} a_{i}ξ_{i} ≠ 0, then for y(t) ∈ C[0, 1], the boundary value problem

\{ \begin{matrix}u^{(n)}(t) + y(t) = 0, t ∈ (0, 1), \\ u(0) = u′(0) = ⋯ = u^{(n−3)}(0) = u^{(n−2)}(0) = 0, u^{(n−2)}(1) − ∑_{i=1}^{m} a_{i}u^{(n−2)}(ξ_{i}) = λ, \end{matrix} (2.1)

has a unique solution

u(t) = ∫_{0}^{1} G_{1}(t, s)y(s)ds + \frac{∑_{i=1}^{m} a_{i}t^{n−1}}{(n − 1)!(1 − Δ)} ∫_{0}^{1} G_{2}(ξ_{i}, s)y(s)ds + \frac{λt^{n−1}}{(n − 1)!(1 − Δ)}, (2.2)

where

G_{1}(t, s) = \frac{1}{(n − 1)!}\{ \begin{matrix}t^{n−1}(1 − s) − (t − s)^{n−1}, & 0 ≤ s ≤ t ≤ 1, \\ (1 − s)t^{n−1}, & 0 ≤ t ≤ s ≤ 1, \end{matrix}

and

G_{2}(t, s) = \frac{∂^{n−2}}{∂t^{n−2}} G_{1}(t, s) = \{ \begin{matrix}s(1 − t), & 0 ≤ s ≤ t ≤ 1, \\ t(1 − s), & 0 ≤ t ≤ s ≤ 1. \end{matrix}

Proof. To prove this, we let

u(t) = − ∫_{0}^{t} \frac{(t − s)^{n−1}}{(n − 1)!} y(s)ds + At^{n−1} + ∑_{i=1}^{n−2} A_{i}t^{i} + B.

Since u^{(i)}(0) = 0 for i = 0, 1, 2, ⋯, n − 2, we get B = 0 and A_{i} = 0 for i = 1, 2, ⋯, n − 2. Now we solve for A by u^{(n−2)}(1) − ∑_{i=1}^{m} a_{i}u^{(n−2)}(ξ_{i}) = λ, we see that

− ∫_{0}^{1}(1 − s)y(s)ds + (n − 1)!A + ∑_{i=1}^{m} a_{i} ∫_{0}^{ξ_{i}}(ξ_{i} − s)y(s)ds − (n − 1)!A · Δ = λ,

which implies

A = \frac{1}{(n − 1)!(1 − Δ)} (∫_{0}^{1}(1 − s)y(s)ds − ∑_{i=1}^{m} a_{i} ∫_{0}^{ξ_{i}}(ξ_{i} − s)y(s)ds + λ) .

Therefore, the problem (2.1) has a unique solution

u(t) = − ∫_{0}^{t} \frac{(t − s)^{n−1}}{(n − 1)!} y(s)ds + \frac{1}{(n − 1)!(1 − Δ)} (∫_{0}^{1}(1 − s)t^{n−1}y(s)ds

− ∑_{i=1}^{m} a_{i} ∫_{0}^{ξ_{i}}(ξ_{i} − s)t^{n−1}y(s)ds + λt^{n−1})

= − ∫_{0}^{t} \frac{(t − s)^{n−1}}{(n − 1)!} y(s)ds + \frac{1}{(n − 1)!} ∫_{0}^{1}(1 − s)t^{n−1}y(s)ds + \frac{Δ}{(n − 1)!(1 − Δ)}

× ∫_{0}^{1}(1 − s)t^{n−1}y(s)ds − \frac{∑_{i=1}^{m} a_{i}}{(n − 1)!(1 − Δ)} ∫_{0}^{ξ_{i}}(ξ_{i} − s)t^{n−1}y(s)ds + \frac{λt^{n−1}}{(n − 1)!(1 − Δ)}

= \frac{1}{(n − 1)!} ∫_{0}^{t}[(1−s)t^{n−1}−(t−s)^{n−1}]y(s)ds+\frac{1}{(n − 1)!} ∫_{t}^{1}(1−s)t^{n−1}y(s)ds+\frac{∑_{i=1}^{m} a_{i}t^{n−1}}{(n − 1)!(1 − Δ)}

× ∫_{0}^{1} ξ_{i}(1 − s)y(s)ds − \frac{∑_{i=1}^{m} a_{i}t^{n−1}}{(n − 1)!(1 − Δ)} ∫_{0}^{ξ_{i}}(ξ_{i} − s)y(s)ds + \frac{λt^{n−1}}{(n − 1)!(1 − Δ)}

= \frac{1}{(n − 1)!} ∫_{0}^{t}[(1 − s)t^{n−1} − (t − s)^{n−1}]y(s)ds + \frac{1}{(n − 1)!} ∫_{t}^{1}(1 − s)t^{n−1}y(s)ds

+\frac{∑_{i=1}^{m} a_{i}t^{n−1}}{(n − 1)!(1 − Δ)} (∫_{0}^{ξ_{i}} s(1 − ξ_{i})y(s)ds + ∫_{ξ_{i}}^{1} ξ_{i}(1 − s)y(s)ds) + \frac{λt^{n−1}}{(n − 1)!(1 − Δ)}

= ∫_{0}^{1} G_{1}(t, s)y(s)ds + \frac{∑_{i=1}^{m} a_{i}t^{n−1}}{(n − 1)!(1 − Δ)} ∫_{0}^{1} G_{2}(ξ_{i}, s)y(s)ds + \frac{λt^{n−1}}{(n − 1)!(1 − Δ)}.

Lemma 2.3. Let 0 < τ < \frac{1}{2}. G_{1}(t, s) and G_{2}(t, s) have the following properties

(i) G_{2}(t, t)G_{2}(s, s) ≤ G_{2}(t, s) ≤ G_{2}(s, s), for all (t, s) ∈ [0, 1] × [0, 1], and max_{t∈[0,1]} ∫_{0}^{1} G_{2}(t, s)ds = \frac{1}{8};

(ii) G_{1}(t, s) ≥ 0, for all (t, s) ∈ [0, 1] × [0, 1], G_{1}(t, s) ≥ τ^{n−1}G_{1}(1, s), for all (t, s) ∈ [τ, 1 − τ] × [0, 1]. Proof. It is obvious that (i) holds. Next we check (ii). For all (t, s) ∈ [0, 1] × [0, 1], if s ≤ t, we have

G_{1}(t, s) = \frac{1}{(n − 1)!}[(1 − s)t^{n−1} − (t − s)^{n−1}]

≥ \frac{1}{(n − 1)!}[(1 − s)t^{n−1} − (t − ts)^{n−1}]

= \frac{t^{n−1}}{(n − 1)!}[(1 − s) − (1 − s)^{n−1}]. (2.3)

If t ≤ s, we get

G_{1}(t, s) = \frac{t^{n−1}}{(n − 1)!}(1 − s) ≥ \frac{t^{n−1}}{(n − 1)!}[(1 − s) − (1 − s)^{n−1}]. (2.4)

Therefore, it follows from (2.3) and (2.4) that

G_{1}(t, s) ≥ 0, for all (t, s) ∈ [0, 1] × [0, 1], and

G_{1}(t, s) ≥ \frac{τ^{n−1}}{(n − 1)!}[(1 − s) − (1 − s)^{n−1}] = τ^{n−1}G_{1}(1, s), for all (t, s) ∈ [τ, 1 − τ] × [0, 1].

Lemma 2.4. We assume that conditions (H_{1}) and (H_{2}) hold, the unique solution u(t) of the BVP (1.1) satisfies:

u^{(i)}(t) ≥ 0 (i = 0, 1, ⋯, n − 2), ∀t ∈ [0, 1]. (2.5) Proof. Let v(t) = u^{(n−2)}(t), for 0 ≤ t ≤ 1. Thus,

v′′(t) = u^{(n)}(t) = −f(t, u(t), u′(t), …, u^{(n−2)}(t)) ≤ 0, for 0 ≤ t ≤ 1,

v(0) = 0, v(1) − ∑_{i=1}^{m} a_{i}v(ξ_{i}) = λ.

In the following we will show that v(1) ≥ 0. If otherwise, v(1) < 0. From v(0) = 0 and v(t) is concave downward, we obtain

v(t) ≥ tv(1), for 0 ≤ t ≤ 1. Hence,

λ = v(1) − ∑_{i=1}^{m} a_{i}v(ξ_{i}) ≤ v(1) − ∑_{i=1}^{m} a_{i}ξ_{i}v(1) = (1 − ∑_{i=1}^{m} a_{i}ξ_{i})v(1) < 0,

which contradicts λ ∈ [0, ∞), and so v(1) ≥ 0.

Since v(0) = 0, v(1) ≥ 0 and v(t) is concave downward, we have

v(t) = u^{(n−2)}(t) ≥ 0, for 0 ≤ t ≤ 1. (2.6) Combining (2.6) with u^{(i)}(0) = 0 (i = 0, 1, ⋯, n − 3), we obtain

u^{(i)}(t) ≥ 0 (i = 0, 1, ⋯, n − 3), for all t ∈ [0, 1]. (2.7) Therefore, we have by (2.6) and (2.7) that (2.5) holds. Lemma 2.5. We assume that conditions (H_{1}) and (H_{2}) hold, the unique solution u(t) of the BVP (1.1) satisfies:

(i) u^{(n−2)}(t) ≥ τ(1 − τ)|u^{(n−2)}|_{0}, ∀t ∈ [τ, 1 − τ];

(ii) u(t) ≥ τ^{n−1}|u|_{0}, ∀t ∈ [τ, 1 − τ],

where |u^{(i)}|_{0} = max_{t∈[0,1]} |u^{(i)}(t)| (i = 0, 1, ⋯, n − 2), τ as in Lemma 2.3.

Proof. (i) From (2.2) and Lemma 2.3 (i), we have u^{(n−2)}(t) = ∫_{0}^{1} G_{2}(t, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}t}{1 − Δ} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds + \frac{λt}{1 − Δ}

≤ ∫_{0}^{1} G_{2}(s, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}}{1 − Δ} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds + \frac{λ}{1 − Δ},

which implies |u^{(n−2)}|_{0} ≤ ∫_{0}^{1} G_{2}(s, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}}{1 − Δ} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds + \frac{λ}{1 − Δ}. (2.8)

On the other hand, for each t ∈ [τ, 1 − τ], we obtain by (2.2), (2.8) and Lemma 2.3 (i) that

u^{(n−2)}(t) = ∫_{0}^{1} G_{2}(t, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}t}{1 − Δ} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds + \frac{λt}{1 − Δ}

≥ τ(1 − τ) ∫_{0}^{1} G_{2}(s, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}τ}{1 − Δ} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds + \frac{λτ}{1 − Δ}

≥ τ(1 − τ) (∫_{0}^{1} G_{2}(s, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}}{1 − Δ} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds + \frac{λ}{1 − Δ})

≥ τ(1 − τ)|u^{(n−2)}|_{0}.

(ii) It follows from (2.2), (2.5) and Lemma 2.3 (ii) that

|u|_{0} = max_{t∈[0,1]} |u(t)| = u(1) = ∫_{0}^{1} G_{1}(1, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}}{(n − 1)!(1 − Δ)} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds + \frac{λ}{(n − 1)!(1 − Δ)}. (2.9)

Thus, for any t ∈ [τ, 1 − τ], in view of Lemma 2.3 (ii), (2.2) and (2.9), we get

u(t) ≥ τ^{n−1} (∫_{0}^{1} G_{1}(1, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}}{(n − 1)!(1 − Δ)} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), ⋯, u^{(n−2)}(s))ds + \frac{λ}{(n − 1)!(1 − Δ)})

≥ τ^{n−1}|u|_{0}.

From the above discussion, the proof is complete.

From Lemma 2.4, Lemma 2.5 (i) and the proof of lemma 2.4 in [21], we can easily check that the following Lemma holds. Lemma 2.6. Suppose that (H_{1}) and (H_{2}) hold, then the unique solution u(t) of the BVP (1.1) satisfies

u(t) ≤ u′(t) ≤ ⋯ ≤ u^{(n−3)}(t) ≤ |u^{(n−2)}|_{0}, ∀t ∈ [0, 1], EJQTDE, 2010 No. 33, p. 7 and

u^{(n−3)}(t) ≤ \frac{1}{τ(1−τ)}u^{(n−2)}(t), ∀t ∈ [τ, 1 − τ],

where τ is as in Lemma 2.3.

3. Main results Let C^{n−2}[0, 1] be endowed with the norm ∥u∥ = max{|u|_{0}, |u′|_{0}, ⋯, |u^{(n−2)}|_{0}},

where |u^{(i)}|_{0} = max_{t∈[0,1]} |u^{(i)}(t)| (i = 0, 1, ⋯, n − 2). Denote

E = {u ∈ C^{n−2}[0, 1], u^{(i)}(t) ≥ 0 (i = 0, 1, ⋯, n − 2) and u^{(i)}(t) ≤ u^{(i+1)}(t) ≤ |u^{(n−2)}|_{0} (i = 0, 1, ⋯, n − 4), ∀t ∈ [0, 1]},

P = \{u ∈ E : u(t) ≥ τ^{n−1}|u|_{0} and u^{(n−3)}(t) ≤ \frac{1}{τ(1−τ)}u^{(n−2)}(t), ∀t ∈ [τ, 1 − τ]\}. It is obvious that P is a cone in C^{(n−2)}[0, 1].

We define the operator T on P by T u(t) = ∫_{0}^{1} G_{1}(t, s)f(s, u(s), u′(s), …, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}t^{n−1}}{(n − 1)!(1 − Δ)} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), u′(s), …, u^{(n−2)}(s))ds + \frac{λt^{n−1}}{(n − 1)!(1 − Δ)},

where G_{1}(t, s) and G_{2}(t, s) are given in Lemma 2.2. It is easy to see that the BVP (1.1) has a solution u(t) if and only if u(t) is a fixed point of the operator T.

In order to obtain the results, we define the nonnegative continuous functional α, the nonnegative continuous convex functional θ, γ, and the nonnegative continuous functional ψ be defined on the cone P by

γ(u) = max_{t∈[0,1]} |u^{(n−2)}(t)|, ψ(u) = θ(u) = max_{t∈[0,1]} |u(t)|, α(u) = min_{t∈[τ,1−τ]} |u(t)|,

where τ as in Lemma 2.3. We observe here that, for all u ∈ P,

τ^{n−1}θ(u) ≤ α(u) ≤ θ(u) = ψ(u), ∥u∥ = max{|u|_{0}, |u′|_{0}, ⋯, |u^{(n−2)}|_{0}} = γ(u). (3.1) We use the following notations. Let

M = \frac{1}{4} + \frac{2 ∑_{i=1}^{i=m} a_{i}}{1 − Δ} ∫_{0}^{1} G_{2}(ξ_{i}, s)ds,

N = 2 ∫_{0}^{1} G_{1}(1, s)ds + \frac{2 ∑_{i=1}^{i=m} a_{i}}{(n − 1)!(1 − Δ)} ∫_{0}^{1} G_{2}(ξ_{i}, s)ds,

R = ∫_{0}^{1} G_{1}(τ, s)ds + \frac{∑_{i=1}^{i=m} a_{i}τ^{n−1}}{(n − 1)!(1 − Δ)} ∫_{0}^{1} G_{2}(ξ_{i}, s)ds.

To present our main results, we assume there exist constants 0 < a < b < τ^{n−1}d and the following assumptions hold.

(H_{3}) f(t, u, u′, ⋯, u^{(n−2)}) ≤ \frac{d}{M}, for (t, u, u′, ⋯, u^{(n−2)}) ∈ [0, 1] × [0, d]^{n−1};

(H_{4}) f(t, u, u′, ⋯, u^{(n−2)}) > \frac{b}{R}, for (t, u, u′, ⋯, u^{(n−2)}) ∈ [τ, 1−τ]×[b, \frac{b}{τ^{n−1}} ]×[b, d]^{n−3} ×[τ(1−τ)b, d];

(H_{5}) f(t, u, u′, ⋯, u^{(n−2)}) < \frac{a}{N}, for (t, u, u′, ⋯, u^{(n−2)}) ∈ [0, 1] × [0, a] × [0, d]^{n−2}.

Theorem 3.1. Assume that (H_{1}) − (H_{5}) hold, in addition, suppose λ satisfy

0 ≤ λ ≤ min\{d, (n − 1)!a\}\frac{1 − Δ}{2}. (3.2)

Then the BVP (1.1) has at least two positive solutions u_{1}, u_{2} and one nonnegative solution u_{3} such that

max_{t∈[0,1]} |u_{i}^{(n−2)}(t)| ≤ d, for i = 1, 2, 3; min_{t∈[τ,1−τ]} |u_{1}(t)| > b;

a < max_{t∈[0,1]} |u_{2}(t)|, with min_{t∈[τ,1−τ]} |u_{2}(t)| < b; max_{t∈[0,1]} |u_{3}(t)| < a.

Proof. First we show T : \overline{P (γ, d)} → \overline{P (γ, d)} is a completely continuous operator.

If u ∈ P, then from Lemma 2.4 and Lemma 2.6, (T u)^{(i)}(t) ≥ 0 (i = 0, 1, ⋯, n − 2), (T u)^{(i)}(t) ≤ (T u)^{(i+1)}(t) ≤ |(T u)^{(n−2)}|_{0} (i = 0, 1, ⋯, n − 4), ∀t ∈ [0, 1] and (T u)^{(n−3)}(t) ≤ \frac{1}{τ(1−τ)}(T u)^{(n−2)}(t), ∀t ∈ [τ, 1−τ], and by Lemma 2.5 (ii), T u(t) ≥ τ^{n−1}|T u|_{0}, ∀t ∈ [τ, 1−τ]. This shows that T : P → P. It can be shown that T : P → P is completely continuous by the Arzela-Ascoli theorem.

If u ∈ \overline{P (γ, d)}, then γ(u) = max_{t∈[0,1]} |u^{(n−2)}(t)| ≤ d, and so 0 ≤ u^{(i)}(t) ≤ d (i = 0, 1, ⋯, n − 2), for

all t ∈ [0, 1], then assumption (H_{3}) implies f(t, u, u′, ⋯, u^{(n−2)}) ≤ \frac{d}{M}. Then, it follows by Lemma 2.3 (i) and (3.2) that

γ(T u) = max_{t∈[0,1]} |(T u)^{(n−2)}(t)| = max_{t∈[0,1]}(T u)^{(n−2)}(t)

= max_{t∈[0,1]} {∫_{0}^{1} G_{2}(t, s)f(s, u(s), u′(s), …, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}t}{1 − Δ} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), u′(s), …, u^{(n−2)}(s))ds + \frac{λt}{1 − Δ}\}

≤ \frac{d}{M} · ( max_{t∈[0,1]} ∫_{0}^{1} G_{2}(t, s)ds + \frac{∑_{i=1}^{m} a_{i}}{1 − Δ} ∫_{0}^{1} G_{2}(ξ_{i}, s)ds) + \frac{λ}{1 − Δ}

≤ \frac{d}{M} · (\frac{1}{8} + \frac{∑_{i=1}^{m} a_{i}}{1 − Δ} ∫_{0}^{1} G_{2}(ξ_{i}, s)ds) + \frac{d}{2}

= \frac{d}{2} + \frac{d}{2} = d.

Therefore, T : \overline{P (γ, d)} → \overline{P (γ, d)}.

Next, we show all the conditions of Lemma 2.1 are satisfied.

To check condition (i) of Lemma 2.1, we take u(t) = \frac{b}{τ^{n−1}} , for t ∈ [0, 1]. It is easy to see that u(t) = \frac{b}{τ^{n−1}} ∈ P (γ, θ, α, b, \frac{b}{τ^{n−1}} , d) and α(u) = α(\frac{b}{τ^{n−1}} ) > b, and so \{u ∈ P (γ, θ, α, b, \frac{b}{τ^{n−1}} , d)|α(u) > b\} ≠ ∅. For u ∈ P (γ, θ, α, b, \frac{b}{τ^{n−1}} , d), we have

b ≤ u(t) ≤ \frac{b}{τ^{n−1}} , for t ∈ [τ, 1 − τ], and max_{t∈[0,1]} |u^{(n−2)}(t)| ≤ d,

then,

b ≤ u^{(i)}(t) ≤ d (i = 1, ⋯, n − 3), for t ∈ [τ, 1 − τ],

and so

τ(1 − τ)b ≤ τ(1 − τ)u^{(n−3)}(t) ≤ u^{(n−2)}(t) ≤ d, for t ∈ [τ, 1 − τ].

Thus, assumption (H_{4}) implies f(t, u, u′, ⋯, u^{(n−2)}) > \frac{b}{R}, and by the definitions of α and the cone P, we have

α(T u) = min_{t∈[τ,1−τ]} |T u(t)| = T u(τ) = ∫_{0}^{1} G_{1}(τ, s)f(s, u(s), u′(s), …, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}τ^{n−1}}{(n − 1)!(1 − Δ)} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), u′(s), …, u^{(n−2)}(s))ds + \frac{λτ^{n−1}}{(n − 1)!(1 − Δ)}

> \frac{b}{R} · (∫_{0}^{1} G_{1}(τ, s)ds + \frac{∑_{i=1}^{m} a_{i}τ^{n−1}}{(n − 1)!(1 − Δ)} ∫_{0}^{1} G_{2}(ξ_{i}, s)ds) = b.

So, condition (i) of Lemma 2.1 is satisfied.

Secondly, we show (ii) of Lemma 2.1 is satisfied. From (3.1) and b ≤ τ^{n−1}d, we get

α(T u) ≥ τ^{n−1}θ(T u) > τ^{n−1} \frac{b}{τ^{n−1}} = b, for all u ∈ P (γ, α, a, d) with θ(T u) > \frac{b}{τ^{n−1}} .

Finally, we show condition (iii) of Lemma 2.1 is also satisfied. Obviously, as ψ(0) = 0 < a, there holds 0 ∉ Q(α, ψ, a, d). Suppose u ∈ Q(α, ψ, a, d) with ψ(u) = a. Then we have by (3.2), the assumption (H_{5}), the definitions of ψ and the cone P that

ψ(T u) = max_{t∈[0,1]} |T u(t)| = T u(1)

= ∫_{0}^{1} G_{1}(1, s)f(s, u(s), u′(s), …, u^{(n−2)}(s))ds

+\frac{∑_{i=1}^{m} a_{i}}{(n − 1)!(1 − Δ)} ∫_{0}^{1} G_{2}(ξ_{i}, s)f(s, u(s), u′(s), …, u^{(n−2)}(s))ds + \frac{λ}{(n − 1)!(1 − Δ)}

< \frac{a}{N} · (∫_{0}^{1} G_{1}(1, s)ds + \frac{∑_{i=1}^{m} a_{i}}{(n − 1)!(1 − Δ)} ∫_{0}^{1} G_{2}(ξ_{i}, s)ds) + \frac{a}{2}

= \frac{a}{2} + \frac{a}{2} = a.

This shows condition (iii) of Lemma 2.1 is also satisfied. Therefore, the hypotheses of Lemma 2.1 are satisfied and there exist two positive solutions u_{1}, u_{2} and one nonnegative solution u_{3} for the BVP (1.1) such that

max_{t∈[0,1]} |u_{i}^{(n−2)}(t)| ≤ d, for i = 1, 2, 3; min_{t∈[τ,1−τ]} |u_{1}(t)| > b;

a < max_{t∈[0,1]} |u_{2}(t)|, with min_{t∈[τ,1−τ]} |u_{2}(t)| < b; max_{t∈[0,1]} |u_{3}(t)| < a.

Remark 3.2. By Theorem 3.1, there are three non-negative solutions, two positive and a third u_{3} which may be zero.

4. Example

In this section, in order to illustrate our main result, we consider an example. Example 4.1. Consider the boundary value problem

\{ \begin{matrix}u′′′(t) + f(t, u(t), u′(t)) = 0, t ∈ (0, 1), \\ u(0) = u′(0) = 0, u′(1) − \frac{1}{4}u′(\frac{1}{3}) − \frac{3}{4}u′(\frac{2}{3}) = λ, \end{matrix} (4.1)

where

f(t, u, v) = \{ \begin{matrix}sin πt + u^{6} + \frac{\sqrt{v}}{30} , & 0 ≤ t ≤ 1, 0 ≤ u < 2, v ≥ 0, \\ sin πt + 64 + \frac{15}{2}\sqrt{u − 2} + \frac{\sqrt{v}}{30} , & 0 ≤ t ≤ 1, 2 ≤ u < 18, v ≥ 0, \\ sin πt + 94 + \sqrt{u − 18} + \frac{\sqrt{v}}{30} , & 0 ≤ t ≤ 1, u ≥ 18, v ≥ 0. \end{matrix}

To show the problem (4.1) has at least two positive solutions and one nonnegative solution, we apply Theorem 3.1 with n = 3, m = 2, a_{1} = \frac{1}{4}, a_{2} = \frac{3}{4}, ξ_{1} = \frac{1}{3} and ξ_{2} = \frac{2}{3}. Clearly (H_{1}) and (H_{2}) are satisfied. We take τ = \frac{1}{3}. After some simple calculations, we get M = \frac{47}{60}, N = \frac{13}{30} and R = \frac{59}{1620}. If we take a = 1, b = 2 and d = 81, we obtain

f(t, u, v) ≤ 103.237 < 103.404 = \frac{d}{M}, for 0 ≤ t ≤ 1, 0 ≤ u ≤ 81, 0 ≤ v ≤ 81,

f(t, u, v) ≥ 64.888 > 54.915 = \frac{b}{R}, for \frac{1}{3} ≤ t ≤ \frac{2}{3}, 2 ≤ u ≤ 18, \frac{4}{9} ≤ v ≤ 81,

f(t, u, v) ≤ 2.3 < 2.308 = \frac{a}{N}, for 0 ≤ t ≤ 1, 0 ≤ u ≤ 1, 0 ≤ v ≤ 81.

Thus, for 0 ≤ λ ≤ min\{d, (n − 1)!a\}\frac{1−Δ}{2} = \frac{1}{2}, by Theorem 3.1, the problem (4.1) has at least two positive solutions u_{1}, u_{2} and one nonnegative solution u_{3} such that

max_{t∈[0,1]} |u_{i}^{(n−2)}(t)| ≤ 81, for i = 1, 2, 3; min_{t∈[ \frac{1}{3} , \frac{2}{3} ]} |u_{1}(t)| > 2;

1 < max_{t∈[0,1]} |u_{2}(t)|, with min_{t∈[ \frac{1}{3} , \frac{2}{3} ]} |u_{2}(t)| < 2; max_{t∈[0,1]} |u_{3}(t)| < 1.

Since 0 is not a solution for any λ ∈ [0, 1/2], it follows that u_{3} is also a positive solution.

Acknowledgements

The authors are very grateful to the referees for careful reading of the original manuscript and

for valuable suggestions on improving this paper. 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 April 8, 2010)