Existence of solutions for a class of fourth-order m-point boundary value problems ∗

Some existence criteria are established for a class of fourth-order m-point boundary value problem by using the upper and lower solution method and the Leray-Schauder continuation principle.

Although there are many works on fourth-order two-point, three-point or four-point BVPs, a little work has been done for more general fourth-order m-point BVPs [8].Motivated greatly by the above-mentioned excellent works, in this paper, we will investigate the following fourth-order m-point Throughout this paper, we always assume that 0 Some existence criteria are established for the BVP (1.1) by using the upper and lower solution method and the Leray-Schauder continuation principle.

Preliminaries
Then K is a cone in E and (E, K) is an ordered Banach space.For Banach space Then for any h ∈ E, the second-order m-point BVP has a unique solution where EJQTDE, 2010 No. 14, p. 2 Proof.If u is a solution of the BVP (2.1), then we may suppose that By the boundary conditions in (2.1), we know that Therefore, the unique solution of the BVP (2.1) In the remainder of this paper, we always assume that where Now, we define operators A and B : E → E as follows: has a solution, then does the BVP (1.1).
Proof.Suppose that v is a solution of the BVP (2.4).Then it is easy to prove that u = Av is a solution of the BVP (1.1).
then β is called an upper solution of the BVP (2.4).
Remark 2.2 If the inequality in Definition (2.1) then α is called a strict lower solution of the BVP (2.4).Similarly, we can also give the definition of a strict upper solution for the BVP (2.4).
We say that f satisfies Nagumo condition with respect to α and β provided that there exists a function h ∈ C ([0, +∞) , (0, +∞)) such that ) Lemma 2.3 Assume that α and β are, respectively, the lower and the upper solution of the BVP (2.4) with α (t) ≤ β (t) for t ∈ [0, 1] , and f satisfies the Nagumo condition with respect to α and β.
Then there exists N > 0 (depending only on α and β) such that any solution ω of the BVP (2.4) lying in Proof.It follows from the definition of λ and the mean-value theorem that there exists t 0 ∈ (0, 1) such that (2.10) In view of (2.8) and (2.10), we know that there exist t 2 , t 3 ∈ (0, 1) with t 2 < t 3 such that one of the following cases holds: Since the others is similar, we only consider Case 1.By the Nagumo condition, we have and so, which contradicts with (2.9) and the proof is complete.

Main result
Theorem 3.1 Assume that α and β are, respectively, the strict lower and the strict upper solution of the BVP (2.4) with α (t) ≤ β (t) for t ∈ [0, 1] , and f satisfies the Nagumo condition with respect to α and β.Then the BVP (2.4) has a solution v 0 and Proof.It follows from Lemma 2.3 that there exists N > 0 such that any solution ω of the BVP (2.4) |β ′ (t)| and define the auxiliary functions f 1 , f 2 , f 3 and F : [0, 1] × R 4 → R as follows: EJQTDE, 2010 No. 14, p. 5 and Consider the following auxiliary BVP If we define an operator T : X → X by then it is obvious that fixed points of T are solutions of the BVP (3.1).Now, we will apply the Leray-Schauder continuation principle to prove that the operator T has a fixed point.Since it is easy to verify that T : X → X is completely continuous by using the Arzela-Ascoli theorem, we only need to prove that the set of all possible solutions of the homotopy group problem v = λT v is a priori bounded in X by a constant independent of λ ∈ (0, 1) .Denote EJQTDE, 2010 No. 14, p. 6 and It is now immediate from the Leray-Schauder continuation principle that the operator T has a fixed point v 0 , which solves the BVP (3.1).Now, let us prove that v 0 is a solution of the BVP (2.4).Therefor, we only need to verify that α . First, we will verify that v 0 (t) ≤ β (t) for t ∈ [0, 1] .Suppose on the contrary that there exists We consider the following three cases: Case . Since β is a strict upper solution of the BVP (2.4), one has = −f (t 0 , (Aβ) (t 0 ) , (Bβ) (t 0 ) , β (t 0 ) , β ′ (t 0 )) > β ′′ (t 0 ) , which is a contradiction.{v 0 (t) − β (t)} > 0, which shows that v ′ 0 (1) ≥ β ′ (1) .On the other hand, v ′ 0 (1) = 0 ≤ β ′ (1) .Consequently, v ′ 0 (1) = β ′ (1) , and so, v ′′ 0 (1) ≤ β ′′ (1) .With the similar arguments as in Case 1, we can obtain a contradiction also. Thus