Electronic Journal of Qualitative Theory of Differential Equations

By applying the monotone iterative technique, we obtain the existence and unique- ness of C 1 (0,1) positive solutions in some set for singular boundary value problems of second


Introduction and the main result
In this paper, we consider the existence of positive solutions for the following nonlinear singular boundary value problem: where A is right continuous on [0, 1), left continuous at t = 1, and nondecreasing on [0, 1), with A(0) = 0.
1 0 u(t)dA(t) denotes the Riemann-Stieltjes integral of u with respect to A. k is a constant.Problems involving Riemann-Stieltjes integral boundary condition have been studied in [3,[7][8][9]13].These boundary conditions includes multipoint and integral boundary conditions, and sums of these, in a single framework.By changing variables t → 1 − t, studying (1.1) also covers the case For a comprehensive study of the case when there is a Riemann-Stieltjes integral boundary condition at both ends, see [7].
In recent years, there are many papers investigating nonlocal boundary value problems of the second order ordinary differential equation u ′′ + f (t, u) = 0.For example, we refer the reader to [1,[3][4][5][7][8][9]11,12] for some work on problems with integral type boundary conditions.However, there are fewer papers investigating boundary value problems of the equation −u ′′ +k 2 u = f (t, u).
In [6], Du and Zhao investigated the following multi-point boundary value problem They assumed f is decreasing in u and get existence of C[0, 1] positive solutions ω with the property that ω(t) ≥ m(1− t) for some m > 0. In a recent paper [5], Webb and Zima studied the problem (1.1) (and others) when dA is allowed to be a signed measure, and obtained existence of multiple positive solutions under suitable conditions on f (t, u).Here we only study the positive measure case.We impose stronger restrictions on f .We suppose f is increasing in u, satisfies a strong sublinear property and may be singular at t = 0, 1.By applying the monotone iterative technique, we obtain the existence and uniqueness of C 1 [0, 1] positive solutions in some set D. Also, we use iterative methods, we establish uniqueness, obtain error estimates and the convergence rate of C 1 [0, 1] positive solutions with the property that there exists M > m > 0 such that mt ≤ u(t) ≤ M t.
In this paper, we first introduce some preliminaries and lemmas in Section 2, and then we state our main results in Section 3.

Preliminaries and lemmas
We make the following assumptions: Our discussion is in the space E = C[0, 1] of continuous functions endowed with the usual supremum norm.Let P = {u ∈ C[0, 1] : u ≥ 0} be the standard cone of nonnegative continuous functions.
The Green's function for (1.1) is given in the following Lemma which was proved in [5] for the general case when dA is a signed measure.
Lemma 2.1 [5] Suppose that g ∈ C(0, 1) and (H 1 ) holds.Then the following linear boundary has a unique positive solution u and u can be expressed in the form where (2.3) Remark 2.2.We call F (t, s) the Green's function of problem (1.1).Suppose that (H 1 ), (H 2 ) hold.Then solutions of (1.1) are equivalent to continuous solutions of the integral equation where where e(s) = s(1 − s).

Main results
Now we state the main results as follows.converges uniformly to the unique solution u * (t) on [0, 1] as n → ∞.Furthermore, we have the error estimation where t 0 , v 0 are defined below, and F (t, s) is mentioned in (2.2).
Proof.From u(t) ∈ D we know there exists L u > 1 > l u > 0 such that This, together with (H 2 ), (1.2) and (1.3), implies that Let us define an operator T by From (3.1) and (3.3) and Lemma 2.2 we can have So the integral operator T makes sense.By (2.2), (2.3), (2.5), (2.6) and (2.7), we have that EJQTDE, 2010 No. 16, p. 5 Thus Thus, from (3.1), (3.7) and (3.8), we obtain It is known from Remark 2.2 that a fixed point of the operator T is a solution of BVP (1.1).
From condition (1.2) we obtain Obviously T is an increasing operator and from (1.3) we have Let x 0 ∈ D be given.Choose t 0 ∈ (0, 1) such that Let us define u 0 = t 0 x 0 , v 0 = 1 t 0 x 0 , t 0 ∈ (0, 1).Then u 0 ≤ v 0 and from (3.9) and (3.10) we have Now we define It is easy to verify from (3.11) that Clearly, u 0 = t 2 0 v 0 .By induction, we see that So {u n } is a cauchy sequence, therefore u n converges to some u * ∈ D. From this inequality it also follows that v n → u * .
We see that u * is a fixed point of T .Thus, We now prove t 1 ≥ 1.In fact, if 0 < t 1 < 1, then which contradicts the definition of t 1 since (t 1 ) b > t 1 .Thus t 1 ≥ 1 and x ≥ u * .In the same way, we can prove x ≤ u * and hence x = u * .The uniqueness of fixed points of A in D is proved.For Remark Suppose that β i (t)(i = 0, 1, 2, . . .m) are nonnegative continuous functions on (0, 1), which may be unbounded at the end points of (0, 1).Ω is the set of functions f (t, u) which satisfy the condition (H 2 ).Then we have the following conclusions: (1) β i (t) ∈ Ω, u b ∈ Ω, where 0 < b < 1; (2) If 0 < b i < +∞(i = 1, 2, . . .m) and b > max