SINGULAR HIGHER ORDER BOUNDARY VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL EQUATIONS

We study singular boundary value problems for differential equations with left focal boundary conditions of the form,


Introduction
This paper is somewhat of an extension of the recent work done by Kunkel [6].Kunkel looked at an extension of Rach unková and Rach unek's work where they studied a second order singular boundary value problem for the discrete p-Laplacian, φ p (x) = |x| p−2 x [7].Kunkel's results extend theirs to the second order differential case, but only for p = 2, i.e. φ 2 (x) = x.In this paper, we extend Kunkel's work to a higher order boundary value problem that is a generalization of the lower order case.
For our purposes, we will define what it means to be a lower solution and an upper solution and, along with the Brouwer fixed point theorem, create a lower and upper C. Kunkel solutions method that can be used in our higher order differential case.We can then apply this method to our higher order boundary value problem to achieve the desired result.
The use of an upper and lower solutions method for proving the existence of a solution to a boundary value problem has been studied by others, including [2], [3], [4], and [5].Singular boundary value problems for differential and difference equations have been investigated by several authors in recent years and an exhaustive survey of many of these results is found in [1].

Preliminaries
In this section we will state some of the definitions that are used throughout the remainder of the paper.
We consider the singular higher order differential equation, satisfying left focal boundary conditions, Our goal is to prove the existence of a positive solution of problem (1), (2).
We also define what is considered a regular problem and what is a singular problem.2) is called regular.If D = R n−1 and f has singularities on the boundary of D, then problem (1), ( 2) is singular.
We will assume throughout this paper that the following hold: C: f (t, x 0 , . . ., x n−2 ) has a singularity at x 0 = 0, i.e. lim sup Let us first consider the regular differential equation, where h is continuous on [0, 1] × R n−1 .In this section, we establish a lower and upper solutions method for the regular problem (3) satisfying boundary conditions (2).We first define our lower and upper solutions. ( Theorem 3.1 (Lower and Upper Solutions Method) Let α and β be given lower and upper solutions of ( 3), (2), respectively, and 2) has a solution u(t) satisfying, Proof.
We proceed through a sequence of steps involving modifications of the function h.Step 1.For t ∈ (0, 1) and (x 0 , . . .,

C. Kunkel
Thus, h is continuous on (0, 1) × R n−1 and there exists M > 0 so that, We now study the auxiliary equation, satisfying boundary conditions (2).Our immediate goal is to prove the existence of a solution to problem (9), (2).
Step 2. We lay the foundation to use the Brouwer fixed point theorem.To this end, define and also define for u ∈ E, E is a Banach space.Further, we define an operator T : T is a continuous operator.Moreover, from the bounds placed on h in Step 1 and from (10), if r > M, then T (B(r)) ⊂ B(r), where B(r) := {u ∈ E : u < r}.Therefore, by the Brouwer fixed point theorem [8], there exists u ∈ B(r) such that u = T u.
Step 3. We now show that u is a fixed point of T if and only if u is a solution of (9), First assume u = T u.Then, from (10), we have that Thus, Also, we have that Continuing in this manner, we have that and that Thus making, Also, we have that u (n−1) (t) = (−1) n−1 t 0 h(r 0 , u(r 0 ), . . ., u (n−2) (r 0 ))dr 0 , making Therefore, (2) is satisfied, and differentiating one more time yields, making (9) satisfied.Now assume u(t) solves ( 9), (2).Then clearly, Thus, by integrating back and applying appropriate boundary conditions, we get that Also, Continuing in this process of integrating and applying appropriate boundary conditions, we get that i.e. u = T u.
Step 4. We now show that solutions u(t) of ( 9), (2) satisfy, Consider the case of obtaining u(t) ≤ β(t).Let v(t) = u(t) − β(t).For the sake of establishing a contradiction, assume that Conditions ( 2) and (7) On the other hand, A similar argument shows that α(t) ≤ u(t).Thus, the conclusion of the theorem holds and our proof is complete.

Main Result
In this section, we make use of Theorem 3.1 to obtain positive solutions of the singular problem (1), (2).
We proceed through a sequence of steps.
For the sake of establishing a contradiction, assume for k ε ≥ k 0 , we have that We therefore, need only consider two cases: 0 < t < 1 − δ and 1 − δ < t < 1.