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

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


Introduction
In this paper, we consider 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), where n ≥ 3 and m ≥ 1 are integers, λ ∈ [0, ∞) is a parameter, and a i , ξ i , f satisfying 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][3][4][5][6]8,[15][16][17][18][19][20] and references therein.However, most of the above mentioned references only consider the cases in which f does not contain higher * E-mail address: xiedapeng9@yahoo.com.cnEJQTDE, 2010 No. 33, p. 1 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 u (n) (t) + µg(t)f (u(t)) = 0, t ∈ (0, 1), under the following assumptions: 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 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][11][12][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][3][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 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 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: Then T has at least three fixed points has a unique solution where and Proof.To prove this, we let Since Therefore, the problem (2.1) has a unique solution ) and G 2 (t, s) have the following properties , and max Proof.It is obvious that (i) holds.Next we check (ii).For all (t, s) Therefore, it follows from (2.3) and (2.4) that and Lemma 2.4.We assume that conditions (H 1 ) and (H 2 ) hold, the unique solution u(t) of the BVP (1.1) satisfies: EJQTDE, 2010 No. 33, p. 5 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, which contradicts λ ∈ [0, ∞), and so v(1) ≥ 0.
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) EJQTDE, 2010 No. 33, p. 7 and where τ is as in Lemma 2.3.

Main results
where We define the operator T on P by 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 where τ as in Lemma 2.3.We observe here that, for all u ∈ P , We use the following notations.Let To present our main results, we assume there exist constants 0 < a < b < τ n−1 d and the following assumptions hold.
Theorem 3.1.Assume that (H 1 ) − (H 5 ) hold, in addition, suppose λ satisfy Then the BVP (1.1) has at least two positive solutions u 1 , u 2 and one nonnegative solution Proof.First we show T : P (γ, d) → P (γ, d) is a completely continuous operator.
It can be shown that T : P → P is completely continuous by the Arzela-Ascoli theorem.
Next, we show all the conditions of Lemma 2.1 are satisfied.
To check condition (i) of Lemma 2.1, we take u and so

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