Existence of solutions for second-order integral boundary value problems ∗

Abstract. In this paper, using a new comparison result and monotone iterative method, we consider the existence of solution of integral boundary value problem for second-order differential equation. To obtain corresponding results, we also discuss second-order differential inequalities. The interesting point is that the one-sided Lipschitz constant is related to the first eigenvalues corresponding to the relevant operators.


Introduction
We will devote to considering the existence of solution of the following integral boundary value problem for second-order differential equation, using the method of upper and lower solutions and its associated monotone iterative technique −x (t) = f t, x(t) , t ∈ (0, 1), x(s) dA(s), x where f ∈ C([0, 1]×R, R); A and B are right continuous on [0, 1), left continuous at t = 1; and nondecreasing on [0, 1], with A(0) = B(0) = 0, 1 0 u(s) dA(s) and 1 0 u(s) dB(s) denote the Riemann-Stieltjes integrals of u with respect to A and B, respectively.
The theory of integral boundary value problems for differential equations is an important and significant branch of nonlinear analysis [1,6,[10][11][12][18][19][20]22,23,[25][26][27][28][30][31][32].It is worth mentioning that integral boundary value problems for differential equations appear often in investigations connected with applied mathematics and physics such as heat conduction, chemical engineering, underground water flow, thermo-elasticity, and plasma physics [8,9,21].One of the basic problems considered in the theory of integral boundary value problems for differential equations is to establish convenient conditions guaranteeing the existence of solutions of those equations.
To obtain existence results for differential equations, someone used the monotone iterative method [2,5,14].There is a vast literature devoted to the applications of this method to differential equations with different boundary conditions, for details, see [4,7,15,16,24,29].In [3], Alberto Cabada and Susana Lois successfully investigated different maximum and anti-maximum principles for the operator L[M ]u = −u + M u with separated boundary conditions.Motivated by [3], in this paper, we first present a new comparison theorem for the operator −u − λu with integral boundary value condition, and then, by using the monotone iterative technique, we investigate the extremal solutions of (1).We should note that the constant λ is related to the first eigenvalues corresponding to the relevant operators.
Throughout this paper, we always suppose that where t dA(t),

Preliminaries and lemmas
Let X be the Banach space C[0, 1] with x = sup t∈[0,1] |x(t)|.Define a set P ⊂ X by It can be easily verified that P is indeed a cone in X.
For σ ∈ X and µ 1 , µ 2 , λ ∈ R. Consider the following linear integral boundary value problems − x (t) = λx(t) + σ(t), t ∈ (0, 1), Nonlinear Anal.Model.Control, 21 (6):828-838 To study (2), consider the operator T : X → X defined by where k(t, s) is given by Then if (H1) holds, by [27,28], For the function k(t, s), it is easy to know that Take For sake of simplicity, we only prove the following Lemma 2 in the case that A(t) ≡ 0 and B(t) ≡ 0 hold.Similar arguments applies when the other condition hold with cones http://www.mii.lt/NANow, the operator T can be simplified as Define a set K ⊂ X by where γ = It can be easily verified that K is indeed a cone in X and K ⊂ P .Lemma 1. T (P ) ⊂ K and the map T : K → K is completely continuous.
Proof.Inequality (4) and the definition of T imply that T (P ) ⊂ K.The completely continuity of the integral operator T is well known.This completes the proof.Definition 1. (See [13].)Let e be a fixed nonzero element in the positive cone P of the Banach space X.The linear operator T is said to be increasing if T (P ) ⊂ P .The linear operator T is said to be e-bounded if, for every nonzero x ∈ P , a natural number n and two positive numbers α, β can be found such that αe T n x βe.
Lemma 2. The operator T defined by (5) is a e-bounded operator, in which e is given by e(t) = t.
By Lemma 2 and Krein-Rutman theorem [13], we know that the operator T defined by ( 5), the spectral radius r(T ) = 0 and T has a positive eigenfunction corresponding to its first eigenvalue λ 1 = (r(T )) −1 .
We present a new comparison result, which is crucial for our discussion.
Proof.To obtain the required results, we only need to prove that the operator equation has an unique fixed point in X.From Lemma 3 operator equation (I − λT )x = θ has only a zero solution.Then by Lemma 1 and the Fredholm alternative theorem for linear compact operator, the operator equation ( 8) has an unique solution in X for any given σ ∈ X and ρ ∈ X.This completes the proof.

Main results
In this section, on the basis of Lemma 3 and Lemma 4, using the monotone iterative technique, we shall show an existence theorem of extremal solutions of (1).