three-point

In this paper, we investigate the existence of positive solutions for a class of singular nth-order three-point boundary value problem. The associated Green’s function for the boundary value problem is given at first, and some useful properties of the Green’s function are obtained. The main tool is fixed-point index theory. The results obtained in this paper essentially improve and generalize some well-known results.

In this paper, we study the existence of positive solutions for a singular nth-order three-point boundary value problem as follows where a < η < b, 0 [0, +∞)) may be singular at t = a and t = b.
Up to now, no paper has appeared in the literature which discusses the existence of positive solutions for the problem (1.2).This paper attempts to fill this gap in the literature.In order to obtain our result, we give at first the associated Green's function for the problem (1.2), which is the base for further discussion.Our results extend and improve the results of Eloe and Ahmad [3] (α = a = 0, b = 1 and f (t, u) = f (u)).Our results are obtained under certain suitable weaker conditions than that in [3].It is also noted that our method here is different from that of Eloe and Ahmad [3].

Expression and properties of Green
has a unique solution where (2.3) Proof.In fact, if w(t) is a solution of the problem (2.1), then we may suppose that Since Therefore, the problem (2.1) has a unique solution where Proof.It is obvious that H(t, s) is nonnegative.Moreover, Thus, (i) holds.
If s = a or s = b, we easily see that (ii) holds.If s ∈ (a, b) and t ∈ [a, b], we have and then Thus, (ii) holds.The proof is completed.
the problem has a solution where here H(t, s) is given by (2.3).
By Theorem 2.3, we obtain the following corollary.where G(t, s) is given as in (2.5).
Theorem 2.5.G(t, s) has the following properties where k(s) as in Lemma 2.2, and here φ(t) and k(s) as in Lemma 2.2.
Proof.It is clear that (i) holds.Next we will divide the proof of (ii) into two cases.
Case (1) If a ≤ t ≤ η, then, by Lemma 2.2 (i), we have By the inequality above, we know that (ii) holds.

Main results
Throughout this paper, we assume the following conditions hold.
where τ ∈ (a, a+b 2 ), γ := M 2 M 1 (here M 1 and M 2 are defined as in Theorem 2.5).Obviously, K and P are cones of C[a, b].Define the operators A 1 , A 2 and T by and Proof.It is obvious that (i) and (ii) hold.By Theorem 2.3, we know that T (P ) ⊂ P .Next, we will prove that the operator T is completely continuous.
For m ≥ 2, define h m by and define the operator T m as follows It is easy to show that the operator T m is compact on P for all m ≥ 2 by using Arzela-Ascoli theorem.
In addition, the continuity of G(t, s)h m (s) on Hence, the completely continuous operator T m converges uniformly to T as m → ∞ on any bounded subset of P , and T : P → P is completely continuous.
By Virtue of Krein-Rutmann theorems, It is easy to see that the following lemma holds.From this and (H 1 ), we know that there exists t 0 ∈ (a, b), such that G(t 0 , t 0 )h(t 0 ) > 0, then there is

Proof.
The three-point boundary value problem (2.4) can be obtained from replacing u(a) = 0 by u(a) = αu(η) and u(b) = 0 by u(b) = βu(η) in (2.1).Thus, we suppose the solution of the three-point boundary value problem (2.4) can be expressed by

Lemma 3 . 1 .
It is clear that the problem (1.2) has a positive solution u = u(t) if and only if u is a fixed point of T .Suppose (H 0 ) − (H 2 ) hold.Then (i) The operator A 1 : C[a, b] → C[a, b] is completely continuous and satisfies A 1 (K) ⊂ K. (ii) The operator A 2 : C[a, b] → C[a, b] is completely continuous and satisfies A 2 (K) ⊂ K. (iii) The operator T : P → C[a, b] is completely continuous and satisfies T (P ) ⊂ P .