POSITIVE SOLUTIONS OF A THIRD ORDER NONLOCAL BOUNDARY VALUE PROBLEM

. We consider a nonlocal boundary value problem for a third order diﬀerential equation. Suﬃcient conditions for the existence and nonexistence of positive solutions for the problem are obtained. The results are illustrated with some examples.

Boundary value problems for ordinary differential equations are important from a theoretical perspective as well as for their wide variety of applications in engineering and the physical and biological sciences. The works of Dulácska [7] on the effects of soil settlement, the classic work of Love [15] and the more recent monograph by Timoshenko [18] on elasticity, and that of Mansfield [16] on the deformation of structures are but a few examples of the rich sources of applications of these kinds of problems. Surveys of known theoretical results can be found in the monographs by Agarwal [2] and Agarwal, O'Regan, and Wong [3]. In addition, recent contributions to the literature include the papers of Anderson and Davis [1], Chu and Zhou [4], Davis et al. [5,6], Graef and Yang [8,9], Henderson and Wang [11], Karakostas and Tsamatos [12,13], Nowakowski and Orpel [17], Wang [19], Webb [20], and Yang [21].
In order to prove some of our results, we will use the following fixed point theorem which is known as the Guo-Krasnosel'skii fixed point theorem [10,14]. Theorem 1.1. Let X be a Banach space over the reals, and let P ⊂ X be a cone in X. Assume that Ω 1 and Ω 2 are bounded open subsets of X with 0 ∈ Ω 1 ⊂ Ω 1 ⊂ Ω 2 , and let L : P ∩ (Ω 2 − Ω 1 ) → P be a completely continuous operator such that either Then L has a fixed point in P ∩ (Ω 2 − Ω 1 ).
Clearly, X is a Banach space over the reals. We also define the constants 2. Green's function and estimates to positive solutions. We need the characteristic function χ to write the expression for the Green function for the problem (1)- (2). Recall that if I ⊂ R is an interval, then the characteristic function χ on I is given by We also define The problem (1)-(2) is equivalent to the integral equation in the sense that if u is a solution of the boundary value problem (1)-(2), then it is a solution of the integral equation (3), and conversely. We will take different cases to prove the positivity of the Green function. If s ≥ p, then If s ≤ p and s ≥ t, then If s ≤ p and s ≤ t, then It is easy to see from the above expressions that The following two lemmas provide information about functions that satisfy the boundary conditions.
In the remainder of the paper, we will make use of the function a : It is easy to see that Our next lemma provides a lower estimate on u(t). (2) and (4), then Proof. Suppose that u(t) ∈ C 3 [0, 1] satisfies (2) and (4). Without loss of generality, we may assume that u = u(p) = 1. If we define and y ′′′ (t) = u ′′′ (t) ≥ 0. Clearly, we have y(0) = 0, y(p) = 0, and y ′ (p) = 0. By the Mean Value Theorem, there exists r 1 ∈ (0, p) such that y ′ (r 1 ) = 0. Since y ′ (t) is concave upward, we have We see that P is a positive cone in the Banach space X. Define the operator T : P → X by It is well known that T : P → X is a completely continuous operator, and the same type of arguments as those used in Section 2 shows that T (P ) ⊂ P . Now, finding a positive solution of the integral equation (3) is equivalent to solving That is, in order to solve the problem (1)-(2), we only need to find a fixed point of T in P . Our first existence result is the following. Proof. Choose ε > 0 such that (F 0 + ε)B ≤ 1. There exists H 1 > 0 such that For each u ∈ P with u = H 1 , we have which means T u ≤ u . Hence, if we let Now choose c ∈ (0, 1/4) and δ > 0 such that There exists H 3 > 0 such that Let Thus, for u ∈ P with u = H 2 , we have which means that T u ≥ u . If we let then Ω 1 ⊂ Ω 2 and T u ≥ u for u ∈ P ∩ ∂Ω 2 . Therefore, condition (K1) of Theorem 1.1 is satisfied, and so there exists a fixed point of T in P . This in turn implies that there is a positive solution to the boundary value problem (1)-(2) and completes the proof of the theorem.
The following theorem is a companion result to Theorem 3.1. Proof. Choose ε > 0 such that (f 0 − ε)A ≥ 1. There exists H 1 > 0 such that For u ∈ P with u = H 1 , we have

then
T u ≥ u for u ∈ P ∩ ∂Ω 1 . To construct Ω 2 , we choose δ ∈ (0, 1) such that There exists an H 3 > 0 such that which means T u ≤ u . Thus, if we let then Ω 1 ⊂ Ω 2 and T u ≤ u for u ∈ P ∩ ∂Ω 2 . We can then conclude from part (K2) of Theorem 1.1 that the problem (1)-(2) has at least one positive solution.
4. Nonexistence results. In this section, we give some results that ensure that the problem (1)-(2) has no positive solutions. Proof. Assume to the contrary that x(t) is a positive solution of the problem (1)-(2). Then, x ∈ P , x(p) > 0, and which is a contradiction.
The proof of the following theorem is quite similar to that of Theorem 4.1, and in the interest of space, the details will be omitted.

5.
Examples. We conclude this paper with some examples that illustrate the above theorems.
In conclusion, we wish to point out that we have not required that f 0 = F 0 = 0 and f ∞ = F ∞ = +∞ (the superlinear case), or that f 0 = F 0 = +∞ and f ∞ = F ∞ = 0 (the sublinear case), or that f (u)/u even has a limit at 0 or ∞. Notice that if f is superlinear, then Theorem 3.1 applies, while if f is sublinear, then Theorem 3.2 should be used. Also, we do not require that g(t) not vanish identically on any subinterval of [0, 1] as is often done.