Iterative Technique for a Third-order Differential Equation with Three-point Nonlinear Boundary Value Conditions

In this paper, we study the existence of extremal solutions for a nonlinear third-order differential equation with three-point nonlinear boundary value conditions. By means of the method of upper and lower solutions and different monotone iterative techniques, the sufficient conditions which guarantee the existence of extremal solutions are given. An example illustrates the main results.


Introduction
Nonlinear boundary value conditions in differential equations can describe many phenomena in applied mathematics, engineering, physical or biological processes.In this paper, we consider the following third-order differential equation with three-point nonlinear boundary value conditions where ξ ∈ (0, 1), f : [0, 1] × R → R and p : R × R → R are continuous.
The main contributions are as follows: (a) we present problems with linear boundary value conditions, and on this basis we obtain the existence of the extremal solutions for problem (1.1) by applying the method of upper and lower solutions and monotone iterative technique; (b) the iterative technique is not unique and an example illustrates the result.

Notations and preliminaries
In this section, we present some definitions and lemmas that will be used throughout the paper.
Proof.Integrating the equation over [0, t] for three times, we have And then Putting t = ξ, we have Substituting it into (2.2),we get Remark 2.3.It is easy to see that G(t, s) > 0 for all (t, s) ∈ (0, 1) × (0, 1) and where where

Main result
In this section, we shall apply the method of upper and lower solutions and monotone iterative technique to consider the existence of extremal solutions for problem (1.1).
Theorem 3.1.Assume that the following conditions hold.
(A 1 ) f (t, u) is increasing with respect to u.
(A 3 ) There exist constants ς, τ such that 0 < τξ < ς and for v 0 (1) Then problem (1.1) has extremal solutions in the sector [v 0 , w 0 ], where Proof.For n = 0, 1, . . ., define Then due to Lemma 2.2, it is easy to show that v n+1 (t), w n+1 (t) are solutions of the following boundary value problems, respectively: and Moreover, from (2.3) we have where G (t, s) is given as in (2.4).Claim 1.The sequences v n (t), w n (t)(n ≥ 1) are lower and upper solutions of problem (1.1), respectively and the following relation holds First, we prove that Let x(t) = w 0 (t) − w 1 (t).From (3.2) and (A 1 ), we have In view of Lemma 2.5, we have That is, By Lemma 2.5, we have w 1 (t) ≥ v 1 (t), t ∈ [0, 1].And then, by induction, ( In what follows, we show that v 1 (t), w 1 (t) are lower and upper solutions of problem (1.1), respectively.From (3.1), (3.2) and (A 1 ), (A 2 ), it follows that which prove that v 1 (t) is a lower solution of problem (1.1).Similarly, it can be obtained that w 1 (t) is an upper solution of problem (1.1).
Analogously to the above arguments, using the induction method, we can show that the sequences v n (t), w n (t) (n ≥ 1) are lower and upper solutions of problem (1.1), respectively and the following relation holds Claim 2. The sequences {v n (t)}, {w n (t)} uniformly converge to their limit functions v(t), w(t), respectively.
Proof.Using Lemmas 2.6 and 2.7, we can complete the proof by the same way as in Theorem 3. It is not difficult to show that v 0 = 0, w 0 (t) = t