Positive solutions of second-order semipositone singular three-point boundary value problems

In this paper we prove the existence of positive solutions for a class of second order semipositone singular three-point boundary value problems. The results are obtained by the use of a GuoKrasnoselskii’s fixed point theorem in cones.

The second-order boundary value problem arises in the study of draining and coating flows.Choi [1] obtained the following results in 1991.
Wong [2] later obtained the similar results in 1993 when f (t, u) = p(t)q(u), α = 0, h(t) ≡ 1 where p(t) > 0 is singular at 0 and at most O( 1 t α ) as t → 0 + for some α ∈ [0, 2); h is locally Lipschitz continuous, increasing.Ha and Lee [3] obtained in 1997 the similar results.Recently Agarwal et al. [4] improved the above results and obtained the results when 0 is a positive constant for each given η > 0 and satisfying in the existing literature, few people considered the BVP (1.1).Only a handful of papers [8][9][10] have appeared when the nonlinearity term is allowed to change sign; Moreover most of them treated with semipositone problems of the form α = 0, h(t) ≡ 1, f (t, u) + M ≥ 0 for some M > 0. It is value to point out that g may not to be nonnegative in this paper.we obtain an interval of λ which ensures the existence of at least one positive solution of BVP (1.1).Our results are new and different from those of [1][2][3][4][5][6].Particularly, we do not use the method of lower and upper solutions which was essential for the technique used in [1][2][3][4][5] .
This paper is organized as follows.In Section 2, we present some lemmas that will be used to prove our main results.In section 3, by using Krasnoselskii's fixed point theorem in cones, we discuss the existence of positive solutions of the BVP(1.1).In each theorem, an interval of eigenvalues is determined to ensure the existence of positive solutions of the BVP(1.1)
has a unique solution where Proof.From u = −h(t) we have For t ∈ [0, 1], integrating from 0 to t we get Combining this with u(0) = u(1) = αu(η) we conclude that Therefore, the three-point BVP has a unique solution This completes the proof.
Proof.For any x ∈ K, let y(t) = T x(t).By definition of the operator T , we have x(0) = x(1) and x ≤ 0, so there exists a t 0 ∈ (0, 1], such that y = y(t 0 ).By Remark 2.3 (ii), we have Define the function h n for n ≥ 2, by EJQTDE, 2009 No. 5, p. 6 Then h n : [0, 1] → [0, +∞) is continuous and h n ≤ h(t), t ∈ (0, 1).Following, for n ≥ 2, let By the same method as in the beginning, we get T n (K) ⊂ K. Obviously, T n is also completely continuous on K for any n ≥ 2 by an application of Ascoli Arzela theorem (see [11]).Define Then, for any t ∈ [0, 1], for each fixed r > 0 and x ∈ D r , . By Lemma 2.2, T n converges uniformly to T as n → ∞, and therefore T is completely continuous.This completes the proof.
Lemma 2.4. [7,13]Let X be a Banach space, and let K ⊂ X be a cone in X. Assume that Ω 1 , Ω 2 are open bounded subsets of K with 0 ∈ Ω 1 ⊂ Ω 1 ⊂ Ω 2 .If T : K → K be a completely continuous operator such that either then T has a fixed point in Ω 2 \ Ω 1 .

Main results
In this section, we present and prove our main results.Then for any x ∈ K ∩ ∂Ω 1 , we have On the other hand, choose [m, n] ⊂ (0, 1) and a constant L > 0 such that By Remark 2.2, for any t ∈ [m, n], there exists a constant D > 0 such that and let Ω 2 = {x ∈ C[0, 1] : x < R}, then for any x ∈ K ∩ ∂Ω 2 , we have  Let u(t) = x(t) − φ(t), then u(t) is a C[0, 1] ∩ C 2 [0, 1] positive solution of the BVP(1.1).We complete the proof.