The Upper and Lower Solution Method for Nonlinear Third-order Three-point Boundary Value Problem *

This paper is concerned with the following nonlinear third-order three-point boundary value problem u ′′′ (t) + f (t, u (t) , u ′ (t)) = 0, t ∈ [0, 1] , u (0) = u ′ (0) = 0, u ′ (1) = αu ′ (η) , where 0 < η < 1 and 0 ≤ α < 1. A new maximum principle is established and some existence criteria are obtained for the above problem by using the upper and lower solution method.


Introduction
Third-order differential equations arise in a variety of different areas of applied mathematics and physics, e.g., in the deflection of a curved beam having a constant or varying cross section, a three-layer beam, electromagnetic waves or gravity driven flows and so on [6].
Their main tool was the well-known Guo-Krasnoselskii fixed point theorem.
Motivated greatly by [5,7], in this paper, we will investigate the following nonlinear third-order three-point BVP where 0 < η < 1 and 0 ≤ α < 1.A new maximum principle is established and some existence criteria are obtained for the BVP (1.1) by using the upper and lower solution method.In order to obtain our main results, we need the following fixed point theorem [1].

Preliminaries
In this section, we will present some fundamental definitions and several important lemmas.
then y is called an upper solution of the BVP (1.1).
Let G (t, s) be the Green ′ s function of the second-order three-point BVP Then For G (t, s), we have the following two lemmas.
it is easy to obtain that which implies that M < 1 2 .
Lemma 2.3 Assume that λ 1 and λ 2 are two nonnegative constants with Proof.We consider two cases: , which implies that the graph of m (t) is concave up.Since m (0) ≤ 0, we only need to prove m (1) ≤ 0. Suppose on the contrary that m (1 . Suppose on the contrary that there exists Case 2. λ 1 = 0. Suppose on the contrary that there exists Noting that m 1 < 0, we obtain EJQTDE, 2010 No. 26, p. 3 And so, which implies that λ 1 + λ 2 > 2. This contradicts the fact that λ 1 + λ 2 ≤ 2.

Main results
In the remainder of this paper, we always assume that the following condition is satisfied: R is continuous and there exist two nonnegative constants λ 1 and λ 2 with Theorem 3.1 If the BVP (1.1) has a lower solution x and an upper solution y with x ′ (t) ≤ y ′ (t) for t ∈ [0, 1] , then the BVP (1.1) has a solution u ∈ C 3 [0, 1], which satisfies (3.1) |v (t)| and Then K is a cone in E and (E, K) is an ordered Banach space.Now, if we define operators L : D ⊂ E → E and N : E → E as follows: and where Now, we shall show that the operator equation (3.2) is solvable.The proof will be given in several steps.

EJQTDE, 2010 No. 26, p. 4
Suppose h ∈ E. We will find unique v ∈ D such that Lv = h.Since Lv = h is equivalent to the integral equation we define a mapping A : Noting that M < 1 2 and 0 ≤ λ 1 + λ 2 ≤ 2, it is easy to verify that A : E → E is a contraction mapping.And so, there exists unique v ∈ D such that Av = v, which implies that Lv = h.This shows that L is invertible.
Step 2. and So, Since f and L −1 are continuous, we only need to prove that L −1 : E → E is compact.Let X be a bounded subset in E. Then there exists a constant C > 0 such that h ≤ C for any h ∈ X.For any On the one hand, for any v ∈ L −1 (X), we have . This shows that L −1 (X) is uniformly bounded.
On the other hand, in view of the uniform continuity of G (t, s) , we know that for any ǫ > 0, there exists a δ > 0 such that for any t 1 , t 2 ∈ [0, 1] and By the Arzela-Ascoli theorem, we know that L −1 (X) is relatively compact, which implies that L −1 : E → E is a compact mapping.
Proof.Since the proof of ( 2) and ( 3) is similar, we only prove (1).Let x (t) ≡ 0 and y Then it is easy to verify that x and y are lower and upper solutions of the BVP (1.1), respectively.By Theorem 3.1, we know that the BVP (1.1) has a solution u satisfying x ′ (t) ≤ u ′ (t) ≤ y ′ (t) for t ∈ [0, 1] , which together with x(0) = u(0) = y(0) = 0 implies that u is nonnegative and u ≤ c.

Theorem 1 . 1
Let (E, K) be an ordered Banach space and [a, b] be a nonempty interval in E. If T : [a, b] → E is an increasing compact mapping and a ≤ T a, T b ≤ b, then T has a fixed point in [a, b].