POSITIVE SOLUTION FOR NONLINEAR THIRD-ORDER MULTI-POINT BOUNDARY VALUE PROBLEM AT RESONANCE

condition the associated linear operator is uninvertible. Third order differential equations arise in a variety of different areas of applied mathematics and physics, as the deflection of a curved beam having a constant or varying cross section, three layer beam and so on [20]. Recently much attention has been paid to the existence of solutions, especially for the positive solutions, of third-order multi-point boundary value problems at non-resonance (for details see [1, 21 30, 10, 7, 17, 22, 12, 27]). Anderson [1] established the existence of at least three positive solutions to


Introduction
This paper is motivated by the existence of positive solution for the third-order m-point boundary value problem where 0 < ξ 1 < ξ 2 < · · · < ξ m−2 < 1, 0 ≤ β i ≤ 1, i = 1, 2, · · · , m − 2, with the resonant condition m−2 ∑ i=1 β i = 1. It is well known that under this resonant condition the associated linear operator is uninvertible. Third order differential equations arise in a variety of different areas of applied mathematics and physics, as the deflection of a curved beam having a constant or varying cross section, three layer beam and so on [20]. Recently much attention has been paid to the existence of solutions, especially for the positive solutions, of third-order multi-point boundary value problems at non-resonance (for details see [1, 21 30, 10, 7, 17, 22, 12, 27]).
Anderson [1] established the existence of at least three positive solutions to problem    −x ′′′ (t) + f (x(t)) = 0, t ∈ (0, 1), where f : R → [0, +∞) is continuous and 1/2 ≤ t 2 < 1. By using the well-known Guo-Krasnoselskiȋ fixed point theorem [8], Palamides and Smyrlis [21] proved that there exist at least one positive solution for third-order three-point problem Yang [30] studied the existence of positive solutions for the third-order m-point boundary value problem By using the Avery-Peterson fixed point theorem, the author established the existence of at least three positive solutions of this problem. For resonant problem of second-order or higher-order differential equations, many existence results of solutions have been established, see [6,23,24,11,13,14,15,3,5,16,18,19,4]. In [4], the authors considered the third-order problem By using Mawhin continuation theorem, the existence results of solutions are established under the resonant condition β = 1, respectively. It is well known that the problem of existence for positive solution to nonlinear boundary value problem is very difficult when the resonant case is considered. Only few work gave the approach in this area for first and second-order differential equations [2,25,26,9,28,29]. To our best knowledge, few paper deal with the existence result of positive solution for resonant third-order boundary value problems. Motivated by the approach in [25,26,9], we study the positive solution for problem (1.1) under the resonant condition. By using the norm-type Leggett-Williams fixed point theorem, we establish the existence results of positive solutions.

Preliminaries
Operator L : domL ⊂ X → Y is called a Fredholm operator with index zero, that is, ImL is closed and dim Ker L=codim ImL< ∞, which implies that there exist continuous projections P : X → X and Q : Y → Y such that ImP = KerL and KerQ = ImL. Moreover, since dim Im Q=codim Im L, there exists an isomorphism J : ImQ → KerL. Denote by L P the restriction of L to KerP ∩ domL to ImL and its inverse by K P , so K P : ImL → KerP ∩ domL and the coincidence equation Denote γ : X → C be a retraction, that is, a continuous mapping such that γx = x for all x ∈ C and Ψ := P + JQN + K P (I − Q)N, Lemma 2.1 (Leggett-Williams norm-type theorem, [25]). Let C be a cone in

Main results
We define the spaces X = Y = C[0, 1] endowed with the maximum norm. It is well known that X and Y are Banach spaces. Define the linear operator L :

It is obvious that
Denote the function G(s), s ∈ [0, 1] as follow: Define the function k(t, s) as follow: Define the functions U (t, s) and positive number κ as follow: is continuous and satisfies the following conditions: , Proof. Firstly we may claim that Indeed, for each y ∈ {y ∈ Y | ∫ 1 0 G(s)y(s)ds = 0}, we choose We can check that which means x(t) ∈ domL. Thus On the other hand, for each y(t) ∈ ImL, there exists Integrating both sides on [0, t], we have

G(s)y(s)ds = 0.
Thus, It is obvious that dim KerL=1 and ImL is closed.
In fact, for each y(t) ∈ Y , we have This induces that y − y 1 ∈ ImL. Since Y 1 ∩ ImL = {0}, we have Y = Y 1 ⊕ ImL. Thus L is a Fredholm operator with index zero.
Define two projections P : Clearly, ImP = KerL, KerQ = ImL and KerP = {x ∈ X : ∫ 1 0 x(s)ds = 0}. Note that for y ∈ ImL, the inverse K P of L P is given by

s)y(s)ds.
In fact, It is easy to check that Next we will check that every condition of Lemma 2.1 is fulfilled. Remark that f can be extended continuously on [0, 1] × (−∞, +∞), condition (C1) of Lemma 2.1 is fulfilled.
Define the set of nonnegative functions C and subsets of X Ω 1 , Ω 2 by Remark that Ω 1 and Ω 2 are open and bounded sets. Furthermore Let the isomorphism J = I and (γx)(t) = |x(t)| for x ∈ X. Then γ is a retraction and maps subsets of Ω 2 into bounded subsets of C, which ensures that condition (C3) of Lemma 2.1 is fulfilled.
Then we prove that (C2) of Lemma 2.1 is fulfilled. For this purpose, suppose that there exists The proof is divided into two cases: (1) We show that t 0 ̸ = 0. Suppose, on the contrary, that x 0 (t) achieves maximum value R only at t 0 = 0. Then the boundary condition which is a contradiction.
Remark. The sign of third order derivative of a function y(t) at point t 0 can not be confirmed when t 0 is a maximal value of y(t). Thus the method in [29] are not applicable directly to problem (1.1-1.2). In our opinion, it is the key that the conditions (S2) in this paper are stronger than that in [29].

Example
We investigate the resonant third-order three-point boundary value problem     