Open Access

Existence of multiple positive solutions for fourth-order boundary value problems in Banach spaces

Boundary Value Problems20122012:107

DOI: 10.1186/1687-2770-2012-107

Received: 2 June 2012

Accepted: 24 September 2012

Published: 9 October 2012

Abstract

This paper deals with the positive solutions of a fourth-order boundary value problem in Banach spaces. By using the fixed-point theorem of strict-set-contractions, some sufficient conditions for the existence of at least one or two positive solutions to a fourth-order boundary value problem in Banach spaces are obtained. An example illustrating the main results is given.

MSC:34B15.

Keywords

u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq1_HTML.gif-positive operator boundary value problem positive solution fixed-point theorem measure of noncompactness

1 Introduction

In this paper, we consider the existence of multiple positive solutions for the fourth-order ordinary differential equation boundary value problem in a Banach space E
{ x ( 4 ) ( t ) = f ( t , x ( t ) , x ( t ) ) , t ( 0 , 1 ) , x ( 0 ) = x ( 1 ) = x ( 0 ) = x ( 1 ) = θ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ1_HTML.gif
(1.1)

where f : I × E × E E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq2_HTML.gif is continuous, I = [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq3_HTML.gif, θ is the zero element of E. This problem models deformations of an elastic beam in equilibrium state, whose two ends are simply supported. Owing to its importance in physics, the existence of this problem in a scalar space has been studied by many authors using Schauder’s fixed-point theorem and the Leray-Schauder degree theory (see [15] and references therein). On the other hand, the theory of ordinary differential equations (ODE) in abstract spaces has become an important branch of mathematics in last thirty years because of its application in partial differential equations and ODEs in appropriately infinite dimensional spaces (see, for example, [68]). For an abstract space, it is here worth mentioning that Guo and Lakshmikantham [9] discussed the multiple solutions of two-point boundary value problems of ordinary differential equations in a Banach space. Recently, Liu [10] obtained the sufficient condition for multiple positive solutions to fourth-order singular boundary value problems in an abstract space. In [11], by using the fixed-point index theory in a cone for a strict-set-contraction operator, the authors have studied the existence of multiple positive solutions for the singular boundary value problems with an integral boundary condition.

However, the above works in a Banach space were carried out under the assumption that the second-order derivative x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq4_HTML.gif is not involved explicitly in the nonlinear term f. This is because the presence of second-order derivatives in the nonlinear function f will make the study extremely difficult. As a result, the goal of this paper is to fill up the gap, that is, to investigate the existence of solutions for fourth-order boundary value problems of (1.1) in which the nonlinear function f contains second-order derivatives, i.e., f depends on x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq4_HTML.gif.

The main features of this paper are as follows. First, we discuss the existence results in an abstract space E, not E = R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq5_HTML.gif. Secondly, we will consider the nonlinear term which is more extensive than the nonlinear term of [10, 11]. Finally, the technique for dealing with fourth-order BVP is completely different from [10, 11]. Hence, we improve and generalize the results of [10, 11] to some degree, and so, it is interesting and important to study the existence of positive solutions of BVP (1.1). The arguments are based upon the u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq1_HTML.gif-positive operator and the fixed-point theorem in a cone for a strict-set-contraction operator.

The paper is organized as follows. In Section 2, we present some preliminaries and lemmas that will be used to prove our main results. In Section 3, various conditions on the existence of positive solutions to BVP (1.1) are discussed. In Section 4, we give an example to demonstrate our result.

2 Preliminaries

Let the real Banach space E with norm https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq6_HTML.gif be partially ordered by a cone P of E, i.e., x y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq7_HTML.gif if and only if y x P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq8_HTML.gif. P is said to be normal if there exists a positive constant N such that θ x y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq9_HTML.gif implies x N y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq10_HTML.gif. We consider problem (1.1) in C [ I , E ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq11_HTML.gif. Evidently, C [ I , E ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq11_HTML.gif is a Banach space with norm x C = max t I x ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq12_HTML.gif and Q = { x C [ I , E ] : x ( t ) θ ,  for  t I } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq13_HTML.gif is a cone of the Banach space C [ I , E ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq11_HTML.gif. In the following, x C 2 [ I , E ] C 4 [ ( 0 , 1 ) , E ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq14_HTML.gif is called a solution of problem (1.1) if it satisfies (1.1). x is a positive solution of (1.1) if, in addition, x is nonnegative and nontrivial, i.e., x C [ I , P ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq15_HTML.gif and x ( t ) θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq16_HTML.gif for t I https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq17_HTML.gif.

For a bounded set V in a Banach space, we denote by α ( V ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq18_HTML.gif the Kuratowski measure of noncompactness (see [68] for further understanding). In this paper, we denote by α ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq19_HTML.gif the Kuratowski measure of noncompactness of a bounded set in E and in C [ I , E ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq11_HTML.gif.

Lemma 2.1 [6]

Let D E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq20_HTML.gif and D be a bounded set, f be uniformly continuous and bounded from I × D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq21_HTML.gif into E, then
α ( f ( I × V ) ) = max t I α ( f ( t , V ) ) , V D . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equa_HTML.gif

The key tool in our approach is the following fixed-point theorem of strict-set-contractions:

Theorem 2.1 [8]

Let P be a cone of a real Banach space E and P r , R = { u P | r u R } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq22_HTML.gif with 0 < r < R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq23_HTML.gif. Suppose that A : P r , R P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq24_HTML.gif is a strict set contraction such that one of the following two conditions is satisfied:
  1. (i)

    https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq25_HTML.gif for u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq26_HTML.gif, u = r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq27_HTML.gif and A u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq28_HTML.gif for u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq26_HTML.gif, u = R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq29_HTML.gif;

     
  2. (ii)

    A u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq30_HTML.gif for u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq26_HTML.gif, u = r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq27_HTML.gif and https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq25_HTML.gif for u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq26_HTML.gif, u = R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq29_HTML.gif;

     

then the operator A has at least one fixed point u P r , R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq31_HTML.gif such that r < u < R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq32_HTML.gif.

The following concept is due to Krasnosel’skii [12, 13], with a slightly more general definition in [12].

Definition 2.1 We say that a bounded linear operator T : C [ I , R ] C [ I , R ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq33_HTML.gif is u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq1_HTML.gif-positive on a cone K = { v C [ I , R ] : v ( t ) 0 , t [ 0 , 1 ] } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq34_HTML.gif if there exists u 0 K { θ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq35_HTML.gif such that for every u K { θ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq36_HTML.gif, there are positive constants k 1 ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq37_HTML.gif, k 2 ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq38_HTML.gif such that
k 1 ( u ) u 0 T u k 2 ( u ) u 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equb_HTML.gif
Lemma 2.2 Let T be u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq1_HTML.gif-positive on a cone K. If T is completely continuous, then r ( L ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq39_HTML.gif, the spectral radius of T, is the unique positive eigenvalue of T with its eigenfunction in K. Moreover, if λ 1 T u 0 = u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq40_HTML.gif holds with λ 1 = ( r ( T ) ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq41_HTML.gif, then for an arbitrary non-zero u K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq42_HTML.gif ( u k u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq43_HTML.gif) the elements u and λ 1 T u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq44_HTML.gif are incomparable
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equc_HTML.gif

In the following, the closed balls in spaces E and C [ I , E ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq11_HTML.gif are denoted, respectively, by Ω l = { x E : x l } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq45_HTML.gif ( l > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq46_HTML.gif) and B l = { x C [ I , E ] : x C l } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq47_HTML.gif ( l > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq46_HTML.gif).

For convenience, let us list the following assumptions:

( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq48_HTML.gif) f C [ I × P × P , P ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq49_HTML.gif, f is bounded and uniformly continuous in t on I × ( P Ω r ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq50_HTML.gif for any r > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq51_HTML.gif, and there exist two nonnegative constants l 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq52_HTML.gif, l 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq53_HTML.gif with 2 l 1 + 2 l 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq54_HTML.gif such that
α ( f ( t × D 1 × D 2 ) ) l 1 α ( D 1 ) + l 2 α ( D 2 ) , t I , D 1 , D 2 P Ω r . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equd_HTML.gif
( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq55_HTML.gif) There are three positive constants a 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq56_HTML.gif, b 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq57_HTML.gif, c 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq58_HTML.gif such that
f ( t , x , y ) a 1 x + b 1 y + c 1 N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Eque_HTML.gif

for all ( t , x , y ) I × P 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq59_HTML.gif and a 1 π 4 + b 1 π 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq60_HTML.gif.

( H 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq61_HTML.gif) There is a φ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq62_HTML.gif ( P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq63_HTML.gif denotes the dual cone of P) with φ ( x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq64_HTML.gif for any x > θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq65_HTML.gif, two nonnegative constants a 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq66_HTML.gif, b 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq67_HTML.gif and a real number r 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq68_HTML.gif such that
φ ( f ( t , x , y ) ) a 2 φ ( x ) + b 2 φ ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equf_HTML.gif

for all ( t , x , y ) I × ( P Ω r 1 ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq69_HTML.gif and a 2 π 4 + b 2 π 2 > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq70_HTML.gif.

( H 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq71_HTML.gif) There is a φ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq62_HTML.gif with φ ( x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq64_HTML.gif for any x > θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq65_HTML.gif and two nonnegative constants a 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq72_HTML.gif, b 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq73_HTML.gif and a real number R 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq74_HTML.gif such that
φ ( f ( t , x , y ) ) a 3 φ ( x ) + b 3 φ ( y ) , t I , x , y P , x , y R 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equg_HTML.gif

and a 3 π 4 + b 3 π 2 > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq75_HTML.gif.

( H 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq76_HTML.gif) There are three positive constants a 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq77_HTML.gif, b 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq78_HTML.gif, r 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq79_HTML.gif such that
f ( t , x , y ) a 4 x + b 4 y N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equh_HTML.gif

for all ( t , x , y ) I × ( P Ω r 2 ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq80_HTML.gif and a 4 π 4 + b 4 π 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq81_HTML.gif.

( H 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq82_HTML.gif) There exists η > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq83_HTML.gif such that
sup t I , x , y P Ω η f ( t , x , y ) < 4 η N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equi_HTML.gif
( H 7 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq84_HTML.gif) There is a φ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq62_HTML.gif with φ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq85_HTML.gif and φ ( x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq86_HTML.gif for any x > θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq65_HTML.gif and a real number η > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq83_HTML.gif such that
φ ( f ( t , x , y ) ) α η , t [ τ , 1 τ ] , x , y P , τ ( 1 τ ) η 30 N x , y η , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equj_HTML.gif

where τ ( 0 , 1 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq87_HTML.gif, α = ( τ ( 1 τ ) τ 1 τ s ( 1 s ) d s ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq88_HTML.gif.

Now, let G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq89_HTML.gif be the Green’s function of the linear problem v = 0 , t ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq90_HTML.gif together with v ( 0 ) = v ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq91_HTML.gif, which can be explicitly given by
G ( t , s ) = { t ( 1 s ) , 0 t s 1 , s ( 1 t ) , 0 s t 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equk_HTML.gif
Obviously, G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq89_HTML.gif have the following properties:
t ( 1 t ) s ( 1 s ) G ( t , s ) t ( 1 t ) , t , s [ 0 , 1 ] ; G ( t , s ) t ( 1 t ) G ( τ , s ) , τ , t , s [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ2_HTML.gif
(2.1)
Set
( S u ) ( t ) = 0 1 G ( t , s ) u ( s ) d s , t I , u C [ I , E ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equl_HTML.gif

Obviously, S : C [ I , E ] C [ I , E ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq92_HTML.gif is continuous.

Let u = x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq93_HTML.gif. Since x ( 0 ) = x ( 1 ) = θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq94_HTML.gif, we have
x ( t ) = ( S u ) ( t ) = 0 1 G ( t , s ) u ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ3_HTML.gif
(2.2)
Using the above transformation and (2.2), BVP (1.1) becomes
u ( t ) = f ( t , ( S u ) ( t ) , u ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ4_HTML.gif
(2.3)
with
u ( 0 ) = u ( 1 ) = θ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ5_HTML.gif
(2.4)
From (2.3) and (2.4), we have
u ( t ) = 0 1 G ( t , s ) f ( s , ( S u ) ( s ) , u ( s ) ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equm_HTML.gif
Now, define an operator A on Q by
( A u ) ( t ) = 0 1 G ( t , s ) f ( s , 0 1 G ( s , τ ) u ( τ ) d τ , u ( s ) ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ6_HTML.gif
(2.5)

The following Lemma 2.3 can be easily obtained.

Lemma 2.3 Assume that ( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq48_HTML.gif) holds. Then A : Q C 2 [ I , P ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq95_HTML.gif and
  1. (i)

    A : Q Q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq96_HTML.gif is continuous and bounded;

     
  2. (ii)

    BVP (1.1) has a solution in C 2 [ I , P ] C 4 [ ( 0 , 1 ) , P ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq97_HTML.gif if and only if A has a fixed point in Q.

     
Let
( T ψ ) ( t ) = 0 1 G ( t , s ) ψ ( s ) d s , ψ C [ I , R ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equn_HTML.gif

It is easy to see that T : C [ I , R ] C [ I , R ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq98_HTML.gif is a u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq1_HTML.gif-positive operator with u 0 ( t ) = sin π t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq99_HTML.gif and r ( T ) = 1 π 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq100_HTML.gif.

Lemma 2.4 Suppose that ( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq48_HTML.gif) holds. Then for any l > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq46_HTML.gif, A : Q B l Q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq101_HTML.gif is a strict set contraction.

Proof For any u Q B l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq102_HTML.gif and t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq103_HTML.gif, by the expression of S, we have
( S u ) ( t ) 0 1 G ( t , s ) u ( s ) d s l , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equo_HTML.gif
and thus S ( Q B l ) Q B l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq104_HTML.gif is continuous and bounded. By the uniformly continuous f and ( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq48_HTML.gif), and Lemma 2.1, we have
α ( f ( I × S ( D ) × D ) ) = max t I α ( f ( t × S ( D ) × D ) ) l 1 α ( S ( D ) ) + l 2 α ( D ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equp_HTML.gif
Since f is uniformly continuous and bounded on I × ( Q Ω l ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq105_HTML.gif, we see from (2.5) that A is continuous and bounded on Q B l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq106_HTML.gif. Let D Q B l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq107_HTML.gif, according to (2.5), it is easy to show that the functions { A u : u D } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq108_HTML.gif are uniformly bounded and equicontinuous, and so in [9] we have
α ( A ( D ) ) = max t I α ( A ( D ( t ) ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equq_HTML.gif
where
A ( D ( t ) ) = { A u ( t ) : u D , t  is fixed } P Ω l . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equr_HTML.gif
Using the obvious formula
0 1 u ( t ) d t co ¯ { u ( t ) : t I } , u C [ I , E ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equs_HTML.gif
and observing 0 G ( t , s ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq109_HTML.gif, we find
α ( A ( D ( t ) ) ) α ( co ¯ { G ( t , s ) f ( s , S u ( s ) , u ( s ) ) : u D , s I } ) α ( co ¯ { f ( s , S u ( s ) , u ( s ) ) θ : u D , s I } ) = α ( { f ( s , S u ( s ) , u ( s ) ) θ : u D , s I } ) α ( { f ( s , S u ( s ) , u ( s ) ) : u D , s I } ) α ( f ( I × S ( B ) × B ) ) l 1 α ( S ( B ) ) + l 2 α ( B ) l 1 α ( B ) + l 2 α ( B ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ7_HTML.gif
(2.6)

where B = { u ( s ) | s I , u D } P l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq110_HTML.gif, S ( B ) = { 0 1 G ( t , s ) u ( s ) d s | t I , u D } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq111_HTML.gif.

From the fact of [9], we know
α ( B ) 2 α ( D ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ8_HTML.gif
(2.7)
It follows from (2.6) and (2.7) that
α ( A ( D ( t ) ) ) 2 ( l 1 + l 2 ) α ( D ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equt_HTML.gif

and consequently, A is a strict set contraction on Q B l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq106_HTML.gif because 2 ( l 1 + l 2 ) < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq112_HTML.gif. □

3 Main results

Theorem 3.1 Let a cone P be normal and condition ( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq48_HTML.gif) be satisfied. If ( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq55_HTML.gif) and ( H 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq61_HTML.gif) or ( H 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq71_HTML.gif) and ( H 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq76_HTML.gif) are satisfied, then BVP (1.1) has at least one positive solution.

Proof Set
Q 1 = { u Q : u ( t ) t ( 1 t ) u ( s ) , t , s [ 0 , 1 ] } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equu_HTML.gif
It is clear that Q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq113_HTML.gif is a cone of the Banach space C [ I , E ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq11_HTML.gif and Q 1 Q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq114_HTML.gif. For any u Q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq115_HTML.gif, by (2.1), we can obtain A ( Q ) Q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq116_HTML.gif, then
A ( Q 1 ) Q 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equv_HTML.gif
We first assume that ( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq55_HTML.gif) and ( H 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq61_HTML.gif) are satisfied. Let
W = { u Q 1 | A u u } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equw_HTML.gif

In the following, we prove that W is bounded.

For any u W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq117_HTML.gif, we have θ u A u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq118_HTML.gif, that is, θ u ( t ) A u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq119_HTML.gif, t I https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq17_HTML.gif. And so u ( t ) N A u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq120_HTML.gif, set v ( t ) = u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq121_HTML.gif, by ( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq55_HTML.gif)
v ( t ) N A u ( t ) N 0 1 G ( t , s ) f ( s , 0 1 G ( s , τ ) u ( τ ) d τ , u ( s ) ) d s 0 1 G ( t , s ) ( a 1 0 1 G ( s , τ ) u ( τ ) d τ + b 1 u ( s ) + c 1 ) d s 0 1 G ( t , s ) ( a 1 0 1 G ( s , τ ) u ( τ ) d τ + b 1 u ( s ) ) d s + c 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ9_HTML.gif
(3.1)
For ψ C [ I , R ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq122_HTML.gif, let L 1 ψ = a 1 T 2 ψ + b 1 T ψ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq123_HTML.gif, then L 1 : C [ I , R ] C [ I , R ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq124_HTML.gif is a bounded linear operator. From (3.1), one deduces that
( ( I L 1 ) v ) ( t ) c 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equx_HTML.gif
Since π 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq125_HTML.gif is the first eigenvalue of T, by ( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq55_HTML.gif), the first eigenvalue of L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq126_HTML.gif, r ( L 1 ) = a 1 π 4 + b 1 π 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq127_HTML.gif. Therefore, by [14], the inverse operator ( I L 1 ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq128_HTML.gif exists and
( I L 1 ) 1 = I + L 1 + L 1 2 + + L 1 n + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equy_HTML.gif

It follows from L 1 ( K ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq129_HTML.gif that ( I L 1 ) 1 ( K ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq130_HTML.gif. So, we know that v ( t ) ( I L 1 ) 1 c 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq131_HTML.gif, t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq103_HTML.gif and W is bounded.

Taking R 2 > max { r 1 , sup W } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq132_HTML.gif, we have
A u u , u B R 2 Q 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ10_HTML.gif
(3.2)
Next, we are going to verify that for any r 3 ( 0 , r 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq133_HTML.gif,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ11_HTML.gif
(3.3)
If this is false, then there exists u 1 B r 3 Q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq134_HTML.gif such that A u 1 u 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq135_HTML.gif. This together with ( H 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq61_HTML.gif) yields
φ ( u 1 ( t ) ) φ ( ( A u 1 ) ( t ) ) = φ ( 0 1 G ( t , s ) f ( s , ( S u 1 ) ( s ) , u 1 ( s ) ) d s ) = 0 1 G ( t , s ) φ ( f ( s , ( S u 1 ) ( s ) , u 1 ( s ) ) ) d s 0 1 G ( t , s ) ( a 2 φ ( ( S u 1 ) ( s ) ) + b 2 φ ( u 1 ( s ) ) ) d s = 0 1 G ( t , s ) ( a 2 0 1 G ( s , τ ) φ ( u 1 ( τ ) ) d τ + b 2 φ ( u 1 ( s ) ) ) d s = ( a 2 T 2 + b 2 T ) φ ( u 1 ( t ) ) , t I . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equz_HTML.gif
For ψ C [ I , R ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq122_HTML.gif, let L 2 ψ = a 2 T 2 ψ + b 2 T ψ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq136_HTML.gif, then the above inequality can be written in the form
φ ( u 1 ( t ) ) L 2 ( φ ( u 1 ( t ) ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ12_HTML.gif
(3.4)
It is easy to see that
φ ( u 1 ( t ) ) 0 , t I . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equaa_HTML.gif
In fact, φ ( u 1 ( t ) ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq137_HTML.gif implies u 1 ( t ) = θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq138_HTML.gif for t I https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq17_HTML.gif, and consequently, u 1 C = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq139_HTML.gif in contradiction to u 1 C = r 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq140_HTML.gif. Now, notice that L 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq141_HTML.gif is a u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq1_HTML.gif-positive operator with u 0 ( t ) = sin π t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq142_HTML.gif, then by Lemma 2.2, we have φ ( u 1 ( t ) ) = μ 0 sin π t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq143_HTML.gif for some μ 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq144_HTML.gif. This together with ( a 2 π 4 + b 2 π 2 ) sin π t = L 2 ( sin π t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq145_HTML.gif and (3.4) implies that
μ 0 sin π t = φ ( u 1 ( t ) ) L 2 ( φ ( u 1 ( t ) ) ) = L 2 ( μ 0 sin π t ) = μ 0 ( a 2 π 4 + b 2 π 2 ) sin π t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equab_HTML.gif

which is a contradiction to a 2 π 4 + b 2 π 2 > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq146_HTML.gif. So, (3.3) holds.

By Lemma 2.4, A is a strict set contraction on Q r 3 , R 2 = { u Q 1 : r 3 u C R 2 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq147_HTML.gif. Observing (3.2) and (3.3) and using Theorem 2.1, we see that A has a fixed point on Q r 3 , R 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq148_HTML.gif.

Next, in the case that ( H 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq71_HTML.gif) and ( H 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq76_HTML.gif) are satisfied, by the method as in establishing (3.3), we can assert from ( H 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq76_HTML.gif) that for any r 4 ( 0 , r 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq149_HTML.gif,
A u u , u B r 4 Q 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ13_HTML.gif
(3.5)
Let
( L 3 v ) ( t ) = 0 1 G ( t , s ) ( a 3 0 1 G ( s , τ ) v ( τ ) d τ + b 3 v ( s ) ) d s , v C [ I , R ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equac_HTML.gif

It is clear that L 3 : C [ I , R ] C [ I , R ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq150_HTML.gif is a completely continuous linear u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq1_HTML.gif-operator with u 0 ( t ) = sin π t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq99_HTML.gif and L 3 : K 1 K 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq151_HTML.gif in which K 1 = { v K C [ I , R ] : v ( t ) t ( 1 t ) v ( s ) , t , s I } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq152_HTML.gif. In addition, the spectral radius r ( L 3 ) = a 3 π 4 + b 3 π 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq153_HTML.gif and sin π t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq154_HTML.gif is the positive eigenfunction of L 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq155_HTML.gif corresponding to its first eigenvalue λ 1 = ( r ( L 3 ) ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq156_HTML.gif.

Let
( L δ v ) ( t ) = δ 1 δ G ( t , s ) ( a 3 0 1 G ( s , τ ) v ( τ ) d τ + b 3 v ( s ) ) d s , v C [ I , R ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equad_HTML.gif

where δ ( 0 , 1 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq157_HTML.gif. It is clear that L δ : C [ I , R ] C [ I , R ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq158_HTML.gif is a completely continuous linear u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq1_HTML.gif-operator with u 0 ( t ) = t ( 1 t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq159_HTML.gif and L δ ( K 1 ) K 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq160_HTML.gif. Thus, the spectral radius r ( L δ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq161_HTML.gif and L δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq162_HTML.gif has a positive eigenfunction corresponding to its first eigenvalue λ δ = ( r ( L δ ) ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq163_HTML.gif.

Take δ n ( 0 , 1 / 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq164_HTML.gif ( n = 1 , 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq165_HTML.gif) satisfying δ 1 δ 2 δ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq166_HTML.gif and δ n 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq167_HTML.gif ( n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq168_HTML.gif). For m > n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq169_HTML.gif, v K 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq170_HTML.gif, we have
( L δ n v ) ( t ) ( L δ m v ) ( t ) ( L 3 v ) ( t ) , t I . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equae_HTML.gif

By [12], we have r ( L δ n ) r ( L δ m ) r ( L 3 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq171_HTML.gif. Let λ δ n = ( r ( T δ n ) ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq172_HTML.gif, by Gelfand’s formula, we have λ δ n λ δ m λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq173_HTML.gif. Let λ δ n λ ˜ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq174_HTML.gif as n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq168_HTML.gif.

In the following, we prove that λ ˜ 1 = λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq175_HTML.gif.

Let v δ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq176_HTML.gif be the positive eigenfunction of T δ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq177_HTML.gif corresponding to λ δ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq178_HTML.gif, i.e.,
v δ n ( t ) = λ δ n ( L δ v δ n ) ( t ) = λ δ n δ n 1 δ n G ( t , s ) ( a 3 0 1 G ( s , τ ) v δ n ( τ ) d τ + b 3 v δ n ( s ) ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ14_HTML.gif
(3.6)
satisfying v δ n = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq179_HTML.gif. Without loss of generality, by standard argument, we may suppose by the Arzela-Ascoli theorem and λ δ n λ ˜ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq174_HTML.gif that v δ n v ˜ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq180_HTML.gif as n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq181_HTML.gif. Thus, v ˜ 0 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq182_HTML.gif and by (3.6), we have
v ˜ 0 ( t ) = λ ˜ 1 0 1 G ( t , s ) ( a 3 0 1 G ( s , τ ) v ˜ 0 ( τ ) d τ + b 3 v ˜ 0 ( s ) ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equaf_HTML.gif

that is, v ˜ 0 ( t ) = λ ˜ 1 ( L 3 v ˜ 0 ) ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq183_HTML.gif. This together with Lemma 2.2 guarantees that λ ˜ 1 = λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq184_HTML.gif.

By the above argument, it is easy to see that there exists a δ ( 0 , 1 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq185_HTML.gif such that
1 < r ( L δ ) < a 3 π 4 + b 3 π 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equag_HTML.gif
Choose
R 3 > max { R 1 , r 2 , 30 N R 1 δ ( 1 δ ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ15_HTML.gif
(3.7)
Now, we assert that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ16_HTML.gif
(3.8)
If this is not true, then there exists u 2 Q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq186_HTML.gif with u C = R 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq187_HTML.gif such that A u 2 u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq188_HTML.gif, then A u 2 ( t ) u 2 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq189_HTML.gif. Moreover, by the definition of Q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq113_HTML.gif, we know
u 2 ( t ) t ( 1 t ) u 2 ( s ) , t , s I , ( S u 2 ) ( t ) = 0 1 G ( t , s ) u 2 ( s ) d s t ( 1 t ) 0 1 s 2 ( 1 s ) 2 d s u 2 ( τ ) , t , τ I . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equah_HTML.gif
Thus, N u 2 ( t ) t ( 1 t ) u 2 C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq190_HTML.gif, N ( S u 2 ) ( t ) t ( 1 t ) 30 u 2 C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq191_HTML.gif, which implies by (3.7) we have
min t [ δ , 1 δ ] u ( t ) δ ( 1 δ ) N u C R 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equai_HTML.gif
and
min t [ δ , 1 δ ] ( S u ) ( t ) δ ( 1 δ ) 30 N u C R 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equaj_HTML.gif
So, by ( H 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq71_HTML.gif), we get
φ ( u 2 ( t ) ) φ ( ( A u 2 ) ( t ) ) = φ ( 0 1 G ( t , s ) f ( s , ( S u 2 ) ( s ) , u 2 ( s ) ) d s ) δ 1 δ G ( t , s ) φ ( f ( s , ( S u 2 ) ( s ) , u 2 ( s ) ) ) d s δ 1 δ G ( t , s ) ( a 3 φ ( ( S u 2 ) ( s ) ) + b 3 φ ( u 2 ( s ) ) ) d s δ 1 δ G ( t , s ) ( a 3 0 1 G ( s , τ ) φ ( u 2 ( τ ) ) d τ + b 3 φ ( u 2 ( s ) ) ) d s = L δ ( φ ( u 2 ( t ) ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equak_HTML.gif
It is easy to see that
φ ( u 2 ( t ) ) 0 , t I . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equal_HTML.gif
In fact, φ ( u 2 ( t ) ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq192_HTML.gif implies u 2 ( t ) = θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq193_HTML.gif for t I https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq17_HTML.gif, and consequently, u 2 C = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq194_HTML.gif in contradiction to u 2 C = R 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq195_HTML.gif. Now, notice that L δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq162_HTML.gif is a u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq1_HTML.gif-positive operator with u 0 ( t ) = t ( 1 t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq159_HTML.gif. Then by Lemma 2.2, we have φ ( u 2 ( t ) ) = μ 0 v δ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq196_HTML.gif for some μ 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq144_HTML.gif, where v δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq197_HTML.gif is the positive eigenfunction of L δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq162_HTML.gif corresponding to λ δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq198_HTML.gif. This together with r ( L δ ) v δ ( t ) = L δ ( v δ ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq199_HTML.gif implies that
μ 0 v δ ( t ) = φ ( u 2 ( t ) ) L δ ( φ ( u 2 ( t ) ) ) = L 3 ( μ 0 v δ ( t ) ) = μ 0 r ( L δ ) v δ ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equam_HTML.gif

which is a contradiction to r ( L δ ) > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq200_HTML.gif. So, (3.8) holds.

By Lemma 2.4, A is a strict set contraction on Q r 4 , R 3 = { u Q 1 : r 4 u C R 3 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq201_HTML.gif. Observing (3.5) and (3.8) and using Theorem 2.1, we see that A has a fixed point on Q r 4 , R 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq202_HTML.gif. This together with Lemma 2.3 implies that BVP (1.1) has at least one positive solution. □

Theorem 3.2 Let a cone P be normal. Suppose that conditions ( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq48_HTML.gif), ( H 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq61_HTML.gif), ( H 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq71_HTML.gif) and ( H 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq82_HTML.gif) are satisfied. Then BVP (1.1) has at least two positive solutions.

Proof We can take the same Q 1 C [ I , E ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq203_HTML.gif as in Theorem 3.1. As in the proof of Theorem 3.1, we can also obtain that A ( Q 1 ) Q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq204_HTML.gif. And we choose r 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq205_HTML.gif, R 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq206_HTML.gif with R 4 > η > r 5 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq207_HTML.gif such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ17_HTML.gif
(3.9)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ18_HTML.gif
(3.10)
On the other hand, it is easy to see that
A u u , u B η Q 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ19_HTML.gif
(3.11)
In fact, if there exists u 3 Q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq208_HTML.gif with u 3 C = η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq209_HTML.gif such that A u 3 u 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq210_HTML.gif, then observing max t , s I G ( t , s ) = 1 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq211_HTML.gif and S u 2 C η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq212_HTML.gif, we get
θ u 3 ( t ) 0 1 G ( t , s ) f ( s , ( S u 3 ) ( s ) , u 3 ( s ) ) d s 1 4 0 1 f ( s , ( S u 3 ) ( s ) , u 3 ( s ) ) d s , t I , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equan_HTML.gif
and so
u 3 ( t ) N 4 0 1 f ( s , ( S u 3 ) ( s ) , u 3 ( s ) ) d s 1 4 N M , t I , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ20_HTML.gif
(3.12)
where, by virtue of ( H 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq76_HTML.gif),
M = sup t I , x , y P Ω η f ( t , x , y ) < 4 η N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ21_HTML.gif
(3.13)
It follows from (3.12) and (3.13) that
η = u 3 C 4 η N M < η , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equao_HTML.gif

a contradiction. Thus (3.11) is true.

By Lemma 2.4, A is a strict set contraction on Q r 5 , η = { u Q 1 | r 5 u C η } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq213_HTML.gif, and also on Q η , R 4 = { u Q 1 | η u C R 4 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq214_HTML.gif. Observing (3.9), (3.10), (3.11) and applying, respectively, Theorem 2.1 to A, Q r 5 , η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq215_HTML.gif and Q η , R 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq216_HTML.gif, we assert that there exist u 1 Q r 5 , η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq217_HTML.gif and u 2 Q η , R 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq218_HTML.gif such that A u 1 = u 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq219_HTML.gif and A u 2 = u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq220_HTML.gif and, by Lemma 2.3 and (3.11), S u 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq221_HTML.gif, S u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq222_HTML.gif are positive solutions of BVP (1.1). □

Theorem 3.3 Let a cone P be normal. Suppose that conditions ( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq48_HTML.gif), ( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq55_HTML.gif) and ( H 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq76_HTML.gif) and ( H 7 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq84_HTML.gif) are satisfied. Then BVP (1.1) has at least two positive solutions.

Proof We can take the same Q 1 C [ I , E ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq203_HTML.gif as in Theorem 3.1. As in the proof of Theorem 3.1, we can also obtain that A ( Q 1 ) Q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq204_HTML.gif. And we choose r 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq223_HTML.gif, R 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq224_HTML.gif with R 5 > η > r 6 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq225_HTML.gif such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ22_HTML.gif
(3.14)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ23_HTML.gif
(3.15)
On the other hand, it is easy to see that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ24_HTML.gif
(3.16)
In fact, if there exists u 4 Q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq226_HTML.gif with u 4 C = η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq227_HTML.gif such that A u 4 u 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq228_HTML.gif, then
θ ( A u 4 ) ( t ) u 4 ( t ) , t I . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equap_HTML.gif
Observing u 4 ( t ) Q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq229_HTML.gif and t ( 1 t ) 30 u 4 ( s ) ( S u 4 ) ( t ) u 4 ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq230_HTML.gif, we get
η = φ u 4 C φ ( u 4 ( t ) ) φ ( ( A u 4 ) ( t ) ) = 0 1 G ( t , s ) φ ( f ( s , ( S u 4 ) ( s ) , u 4 ( s ) ) ) d s τ 1 τ G ( t , s ) φ ( f ( s , ( S u 4 ) ( s ) , u 4 ( s ) ) ) d s > α η τ ( 1 τ ) τ 1 τ s ( 1 s ) d s = η , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equaq_HTML.gif

which is a contradiction. Hence, (3.16) holds.

By Lemma 2.4, A is a strict set contraction on Q r 6 , η = { u Q 1 | r 6 u C η } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq231_HTML.gif and also on Q η , R 5 = { u Q 1 | η u C R 5 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq232_HTML.gif. Observing (3.14), (3.15), (3.16) and applying, respectively, Theorem 2.1 to A, Q r 6 , η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq233_HTML.gif and Q η , R 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq234_HTML.gif, we assert that there exist u 1 Q r 6 , η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq235_HTML.gif and u 2 Q η , R 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq236_HTML.gif such that A u 1 = u 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq219_HTML.gif and A u 2 = u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq220_HTML.gif and, by Lemma 2.3 and (3.16), S u 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq221_HTML.gif, S u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq222_HTML.gif are positive solutions of BVP (1.1). □

4 One example

Now, we consider an example to illustrate our results.

Example 4.1 Consider the following boundary value problem of the finite system of scalar differential equations:
{ x n ( 4 ) ( t ) = f n ( t , x ( t ) , x ( t ) ) , t ( 0 , 1 ) , x ( 0 ) = x ( 1 ) = x ( 0 ) = x ( 1 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ25_HTML.gif
(4.1)
where
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ26_HTML.gif
(4.2)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equ27_HTML.gif
(4.3)
Claim (4.1) has at least two positive solutions x ( t ) = ( x 1 ( t ) , x 2 ( t ) , , x 10 ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq237_HTML.gif and x ( t ) = ( x 1 ( t ) , x 2 ( t ) , , x 10 ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq238_HTML.gif such that
0 < max 1 i 10 , t [ 0 , 1 ] | x i ( t ) | < 0.51 < max 1 i 10 , t [ 0 , 1 ] | x i ( t ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equar_HTML.gif
Proof Let E = R 10 = { x = ( x 1 , x 2 , , x 10 ) : x n R , n = 1 , 2 , , 10 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq239_HTML.gif with the norm x = max 1 n 10 | x n | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq240_HTML.gif, and P = { x = ( x 1 , x 2 , , x 10 ) : x n 0 , n = 1 , 2 , , 10 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq241_HTML.gif. Then P is a normal cone in E, and the normal constant is N = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq242_HTML.gif. System (4.1) can be regarded as a boundary value problem of (1.1) in E, where x = ( x 1 , x 2 , , x 10 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq243_HTML.gif, y = ( y 1 , y 2 , , y 10 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq244_HTML.gif,
f ( t , x , y ) = ( f 1 ( t , x , y ) , f 2 ( t , x , y ) , , f 10 ( t , x , y ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equas_HTML.gif
Evidently, f : I × P 2 P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq245_HTML.gif is continuous. In this case, condition ( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq48_HTML.gif) is automatically satisfied. Since α ( f ( t , D 1 , D 2 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq246_HTML.gif is identical zero for any t I https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq17_HTML.gif and D 1 , D 2 P Ω l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq247_HTML.gif. Obviously, P = P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq248_HTML.gif, so we may choose φ = ( 1 , 1 , , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq249_HTML.gif, then for any x , y P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq250_HTML.gif, we have
φ ( f ( t , x , y ) ) = n = 1 10 f n ( t , x , y ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equat_HTML.gif
Noticing x φ ( x ) 10 x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq251_HTML.gif, we have
φ ( f ( t , x , y ) ) = n = 1 10 f n ( t , x , y ) 100 y 10 φ ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equau_HTML.gif
for all ( t , x , y ) I × P 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq59_HTML.gif with x 0.02 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq252_HTML.gif and y 0.02 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq253_HTML.gif, and
φ ( f ( t , x , y ) ) = n = 1 10 f n ( t , x , y ) 100 y 10 φ ( y ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equav_HTML.gif

for all ( t , x , y ) I × P 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq59_HTML.gif with x 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq254_HTML.gif and y 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq255_HTML.gif. So, the conditions ( H 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq61_HTML.gif) and ( H 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq71_HTML.gif) are satisfied with a 2 = a 3 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq256_HTML.gif and b 2 = b 3 = 10 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq257_HTML.gif.

Choosing η = 0.51 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq258_HTML.gif for t I https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq17_HTML.gif and x , y P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq259_HTML.gif with x , y 0.51 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq260_HTML.gif, we have
f ( t , x , y ) 2.0102 < 2.04 = 4 η . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_Equaw_HTML.gif

So, condition ( H 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-107/MediaObjects/13661_2012_Article_206_IEq82_HTML.gif) is satisfied. Thus, our conclusion follows from Theorem 3.2. □

Declarations

Acknowledgements

The authors would like to thank the referees for carefully reading this article and making valuable comments and suggestions. This work is supported by the Foundation items: NSFC (10971179), NSF (BS2010SF023, BS2012SF022) of Shandong Province.

Authors’ Affiliations

(1)
Department of Mathematics, Shandong University of Science and Technology
(2)
Department of Mathematics, Xuzhou Normal University

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