Open Access

Positive solutions to second-order differential equations with dependence on the first-order derivative and nonlocal boundary conditions

Boundary Value Problems20132013:8

DOI: 10.1186/1687-2770-2013-8

Received: 20 September 2012

Accepted: 15 January 2013

Published: 17 January 2013

Abstract

In this paper, we consider the existence of positive solutions for second-order differential equations with deviating arguments and nonlocal boundary conditions. By the fixed point theorem due to Avery and Peterson, we provide sufficient conditions under which such boundary value problems have at least three positive solutions. We discuss our problem both for delayed and advanced arguments α and also in the case when α ( t ) = t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq1_HTML.gif, t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq2_HTML.gif. In all cases, the argument β can change the character on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq3_HTML.gif, see problem (1). It means that β can be delayed in some set J ¯ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq4_HTML.gif and advanced in [ 0 , 1 ] J ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq5_HTML.gif. An example is added to illustrate the results.

MSC:34B10.

Keywords

boundary value problems with delayed and advanced arguments nonlocal boundary conditions cone existence of positive solutions a fixed point theorem

1 Introduction

Put J = [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq6_HTML.gif, R + = [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq7_HTML.gif. Let us consider the following boundary value problem:
{ x ( t ) + h ( t ) f ( t , x ( α ( t ) ) , x ( β ( t ) ) ) = 0 , t ( 0 , 1 ) , x ( 0 ) = γ x ( η ) + λ 1 [ x ] , x ( 1 ) = ξ x ( η ) + λ 2 [ x ] , η ( 0 , 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equ1_HTML.gif
(1)
where λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq8_HTML.gif, λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq9_HTML.gif denote linear functionals on C ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq10_HTML.gif given by
λ 1 [ x ] = 0 1 x ( t ) d A ( t ) , λ 2 [ x ] = 0 1 x ( t ) d B ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equa_HTML.gif

involving Stieltjes integrals with suitable functions A and B of bounded variation on J. It is not assumed that λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq8_HTML.gif, λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq9_HTML.gif are positive to all positive x. As we see later, the measures dA, dB can be signed measures.

We introduce the following assumptions:

H1: f C ( J × R + × R , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq11_HTML.gif, α , β C ( J , J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq12_HTML.gif, A and B are functions of bounded variation;

H2: h C ( J , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq13_HTML.gif and h does not vanish identically on any subinterval;

H3: 1 γ λ 1 [ p ] > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq14_HTML.gif or 1 ξ λ 2 [ p ] > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq15_HTML.gif for p ( t ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq16_HTML.gif, t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq17_HTML.gif, γ , ξ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq18_HTML.gif.

Recently, the existence of multiple positive solutions for differential equations has been studied extensively; for details, see, for example, [131]. However, many works about positive solutions have been done under the assumption that the first-order derivative is not involved explicitly in nonlinear terms; see, for example, [3, 6, 814, 17, 20, 2527, 30]. From this list, only papers [912, 14, 20, 30] concern positive solutions to problems with deviating arguments. On the other hand, there are some papers considering the multiplicity of positive solutions with dependence on the first-order derivative; see, for example, [2, 4, 5, 7, 15, 16, 18, 19, 2124, 28, 29, 31]. Note that boundary conditions (BCs) in differential problems have important influence on the existence of the results obtained. In this paper, we consider problem (1) which is a problem with dependence on the first-order derivative with BCs involving Stieltjes integrals with signed measures of dA, dB appearing in functionals λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq8_HTML.gif, λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq9_HTML.gif; moreover, problem (1) depends on deviating arguments.

For example, in papers [2, 4, 15, 18, 22, 24], the existence of positive solutions to second-order differential equations with dependence on the first-order derivative (but without deviating arguments) has been studied with various BCs including the following:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equb_HTML.gif
by fixed point theorems in a cone (such as Avery-Peterson, an extension of Krasnoselskii’s fixed point theorem or monotone iterative method) with corresponding assumptions:
a i , b i ( 0 , 1 ) , i = 1 , 2 , , n , i = 1 n a i , i = 1 n b i ( 0 , 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equc_HTML.gif

or 1 α η > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq19_HTML.gif, respectively.

For example, in papers [811, 20, 22, 30], the existence of positive solutions to second-order differential equations including impulsive problems, but without dependence on the first-order derivative, has been studied with various BCs including the following:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equd_HTML.gif
under corresponding assumptions by fixed point theorems in a cone (such as Avery-Peterson, Leggett-Williams, Krasnoselskii or fixed point index theorem). See also paper [13], where positive solutions have been discussed for second-order impulsive problems with boundary conditions
x ( 0 ) = 0 , x ( 1 ) = 0 1 x ( s ) d A ( s ) ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Eque_HTML.gif

here λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq8_HTML.gif has the same form as in problem (1) with signed measure dA appearing in functional λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq8_HTML.gif.

Positive solutions to second-order differential equations with boundary conditions that involve Stieltjes integrals have been studied in the case of signed measures in papers [25, 26] with BCs including, for example, the following:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equf_HTML.gif

The main results of papers [25, 26] have been obtained by the fixed point index theory for problems without deviating arguments. The study of positive solutions to boundary value problems with Stieltjes integrals in the case of signed measures has also been done in papers [3, 7, 13, 14, 27] for second-order differential equations (also impulsive) or third-order differential equations by using the fixed point index theory, the Avery-Peterson fixed point theorem or fixed point index theory involving eigenvalues.

Note that BCs in problem (1) with functionals λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq8_HTML.gif, λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq9_HTML.gif cover some nonlocal BCs, for example,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equg_HTML.gif

for some constants a i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq20_HTML.gif, b i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq21_HTML.gif and some functions g 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq22_HTML.gif, g 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq23_HTML.gif. In our paper, the assumption that the measures dA, dB in the definitions of λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq8_HTML.gif, λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq9_HTML.gif are positive is not needed. More precisely, one needs to choose the above functions g 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq22_HTML.gif, g 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq23_HTML.gif in such a way that the assumption H4 holds. It means that g 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq22_HTML.gif, g 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq23_HTML.gif can change sign on J.

A standard approach (see, for example, [2527]) to studying positive solutions of boundary value problems such as (1) is to translate problem (1) to a Hammerstein integral equation
x ( t ) = Γ 1 ( t ) λ 1 [ x ] + Γ 2 ( t ) λ 2 [ x ] + Γ 3 ( t ) 0 1 G ( η , s ) h ( s ) f ( s , x ( α ( s ) ) , x ( β ( s ) ) ) d s + 0 1 G ( t , s ) h ( s ) f ( s , x ( α ( s ) ) , x ( β ( s ) ) ) d s W x ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equ2_HTML.gif
(2)
to find a solution as a fixed point of the operator W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq24_HTML.gif by using a fixed point theorem in a cone. Γ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq25_HTML.gif, Γ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq26_HTML.gif, Γ 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq27_HTML.gif are corresponding continuous functions while λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq8_HTML.gif and λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq9_HTML.gif have the same form as in problem (1). G denotes a Green function connected with our problem, so in our case it is given by
G ( t , s ) = { s ( 1 t ) , 0 s t , t ( 1 s ) , t s 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equh_HTML.gif

In our paper, we eliminate λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq8_HTML.gif and λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq9_HTML.gif from problem (2) to obtain the equation x = W ¯ x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq28_HTML.gif with a corresponding operator W ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq29_HTML.gif, and then we seek solutions as fixed points of this operator  W ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq29_HTML.gif.

Note that if we put γ = ξ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq30_HTML.gif in the BCs of problem (1), then this new problem is more general than the previous one because in this case someone, for example, can take λ 1 [ x ] = γ x ( η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq31_HTML.gif, λ 2 [ x ] = ξ x ( η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq32_HTML.gif. In this paper, we try to explain why for some cases we have to discuss problem (1) with constants γ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq33_HTML.gif or ξ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq34_HTML.gif.

To apply such a fixed point theorem in a cone to problem (1), we have to construct a suitable cone K. Usually, we need to find a nonnegative function κ and a constant ρ ¯ ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq35_HTML.gif such that G ( t , s ) κ ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq36_HTML.gif for t , s J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq37_HTML.gif and G ( t , s ) ρ ¯ κ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq38_HTML.gif for t [ η , η ¯ ] [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq39_HTML.gif and s J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq40_HTML.gif (see, for example, [2527]) to work with the inequality
min [ η , η ¯ ] | x ( t ) | ρ ¯ max t J | x ( t ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equi_HTML.gif
Indeed, for problems without deviating arguments, someone can use any interval [ η , η ¯ ] [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq41_HTML.gif. It means that when α ( t ) = t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq1_HTML.gif on J, then we can take γ = ξ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq30_HTML.gif in the boundary conditions of problem (1) to work with the inequality
min [ ζ , ϱ ] | x ( t ) | κ max t J | x ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equj_HTML.gif

for ζ, ϱ such that ζ + ϱ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq42_HTML.gif, 0 < ζ < ϱ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq43_HTML.gif with κ = min ( ζ , 1 ϱ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq44_HTML.gif; see Section 5.

Note that for problems with delayed or advanced arguments, we have to use interval [ 0 , η ] [ 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq45_HTML.gif or [ η , 1 ] ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq46_HTML.gif, respectively. We see that if γ = ξ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq30_HTML.gif, then ρ ¯ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq47_HTML.gif for problem (1) with deviated arguments. It shows that the approach from papers [2527] needs a little modification to problems with delayed or advanced arguments. Consider the situation α ( t ) t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq48_HTML.gif on J. In this case, we can put ξ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq49_HTML.gif in the boundary conditions of problem (1) to find a constant ρ ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq50_HTML.gif to work with the inequality
min [ 0 , η ] | x ( t ) | ρ max t J | x ( t ) | ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equk_HTML.gif

see Section 3. For the case α ( t ) t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq51_HTML.gif on J, we can put γ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq52_HTML.gif to work similarly as in Section 3; see Section 4. Note that in the above three cases for the argument β, we need only the assumption β C ( J , J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq53_HTML.gif, which means that β can change the character in J.

Note that in cited papers, positive solutions to differential equations with dependence on the first-order derivative have been investigated only for problems without deviating arguments, see [2, 4, 5, 7, 15, 16, 18, 19, 2124, 28, 29, 31]. Moreover, BCs in problem (1) cover some nonlocal BCs discussed earlier.

Motivated by [2527], in this paper, we apply the fixed point theorem due to Avery-Peterson to obtain sufficient conditions for the existence of multiple positive solutions to problems of type (1). In problem (1), an unknown x depends on deviating arguments which can be both of advanced or delayed type. To the author’s knowledge, it is the first paper when positive solutions have been investigated for such general boundary value problems with functionals λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq8_HTML.gif, λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq9_HTML.gif and with deviating arguments α, β in differential equations in which f depends also on the first-order derivative. It is important to indicate that problems of type (1) have been discussed with signed measures of dA, dB appearing in Stieltjes integrals of functionals λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq8_HTML.gif, λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq9_HTML.gif.

The organization of this paper is as follows. In Section 2, we present some necessary lemmas connected with our main results. In Section 3, we first present some definitions and a theorem of Avery and Peterson which is useful in our research. Also in Section 3, we discuss the existence of multiple positive solutions to problems with delayed argument α, by using the above mentioned Avery-Peterson theorem. At the end of this section, an example is added to verify theoretical results. In Section 4, we formulate sufficient conditions under which problems with advanced argument α have positive solutions. In the last section, we discuss problems of type (1) when α ( t ) = t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq1_HTML.gif on J.

2 Some lemmas

Let us introduce the following notations:
x = max ( x 1 , x 1 ) with  z 1 = max t J | z ( t ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equl_HTML.gif
Lemma 1 Let x C 1 ( J , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq54_HTML.gif, p ( t ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq16_HTML.gif, t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq17_HTML.gif. Assume that A and B are functions of bounded variation and, moreover,
x ( 0 ) = γ x ( η ) + λ 1 [ x ] , x ( 1 ) = ξ x ( η ) + λ 2 [ x ] , γ , ξ 0 , η ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equm_HTML.gif
with
  1. (i)

    1 γ λ 1 [ p ] 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq55_HTML.gif or

     
  2. (ii)

    1 ξ λ 2 [ p ] 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq56_HTML.gif.

     
Then
x 1 M x 1 , M = 1 + { Var A + γ | 1 γ λ 1 [ p ] | , in case  (i) , Var B + ξ | 1 ξ λ 2 [ p ] | , in case  (ii) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equn_HTML.gif

Here, VarA denotes the variation of a function A on J.

Proof Note that in case (i), we have
x ( 0 ) = γ x ( η ) + λ 1 [ x ] = γ [ x ( η ) x ( 0 ) ] + γ x ( 0 ) + 0 1 ( x ( t ) x ( 0 ) ) d A ( t ) + λ 1 [ p ] x ( 0 ) = γ 0 η x ( s ) d s + 0 1 ( 0 t x ( s ) d s ) d A ( t ) + γ x ( 0 ) + λ 1 [ p ] x ( 0 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equo_HTML.gif
so
x ( 0 ) = 1 1 γ λ 1 [ p ] [ γ 0 η x ( s ) d s + 0 1 ( 0 t x ( s ) d s ) d A ( t ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equp_HTML.gif
Hence,
| x ( 0 ) | 1 | 1 γ λ 1 [ p ] | ( γ + Var A ) x 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equq_HTML.gif
Combining this with the relation
x ( t ) = x ( 0 ) + 0 t x ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equr_HTML.gif
we obtain
x 1 | x ( 0 ) | + x 1 M x 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equs_HTML.gif

This proves case (i).

In case (ii), similarly,
x ( 1 ) = ξ x ( η ) + λ 2 [ x ] = ξ [ x ( η ) x ( 1 ) ] + ξ x ( 1 ) 0 1 ( x ( 1 ) x ( t ) ) d B ( t ) + λ 2 [ p ] x ( 1 ) = ξ η 1 x ( s ) d s 0 1 ( t 1 x ( s ) d s ) d B ( t ) + ξ x ( 1 ) + λ 2 [ p ] x ( 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equt_HTML.gif
so
x ( 1 ) = 1 1 ξ λ 2 [ p ] [ ξ η 1 x ( s ) d s + 0 1 ( 0 1 x ( s ) d s ) d B ( t ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equu_HTML.gif
Hence,
| x ( 1 ) | 1 | 1 ξ λ 2 [ p ] | ( ξ + Var B ) x 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equv_HTML.gif
Adding to this the relation
x ( t ) = x ( 1 ) t 1 x ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equw_HTML.gif

we get the result in case (ii). This ends the proof. □

Remark 1 If we assume that A and B are increasing functions, then there exists σ J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq57_HTML.gif such that
x ( 0 ) = 1 1 γ λ 1 [ p ] [ γ 0 η x ( s ) d s + 0 1 ( 0 t x ( s ) d s ) d A ( t ) ] = 1 1 γ λ 1 [ p ] [ γ 0 η x ( s ) d s + 0 σ x ( s ) d s 0 1 d A ( t ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equx_HTML.gif
Hence,
| x ( 0 ) | 1 | 1 γ λ 1 [ p ] | ( γ + | 0 1 d A ( t ) | ) x 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equy_HTML.gif
Similarly, we can show that
| x ( 1 ) | 1 | 1 ξ λ 2 [ p ] | ( ξ + | 0 1 d B ( t ) | ) x 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equz_HTML.gif
Now, the constant M from Lemma 1 has the form
M = 1 + { 1 | 1 γ λ 1 [ p ] | ( γ + | 0 1 d A ( t ) | ) , in case (i) , 1 | 1 ξ λ 2 [ p ] | ( ξ + | 0 1 d B ( t ) | ) , in case (ii) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equaa_HTML.gif
Consider the following problem:
{ u ( t ) + y ( t ) = 0 , t ( 0 , 1 ) , u ( 0 ) = γ u ( η ) + λ 1 [ u ] , u ( 1 ) = ξ u ( η ) + λ 2 [ u ] , η ( 0 , 1 ) , γ , ξ 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equ3_HTML.gif
(3)

Let us introduce the assumption.

H0: A and B are functions of bounded variation and
δ 1 γ + η ( γ ξ ) 0 , Δ A 1 ( B 2 1 + ξ η ) + A 2 ( 1 ξ B 1 ) + δ η γ B 1 ( 1 γ ) B 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equab_HTML.gif
for
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equac_HTML.gif

We require the following result.

Lemma 2 Let the assumption H0 hold and let y L 1 ( J , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq58_HTML.gif. Then problem (3) has a unique solution given by
u ( t ) = 1 Δ [ 1 ξ η B 2 ( 1 ξ B 1 ) t ] λ 1 [ F ¯ y ] + 1 Δ [ η γ + A 2 + ( 1 γ A 1 ) t ] λ 2 [ F ¯ y ] + F ¯ y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equad_HTML.gif
with
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equae_HTML.gif
Proof Integrating the differential equation in (3) two times, we have
u ( t ) = u ( 0 ) + t u ( 0 ) 0 t ( t s ) y ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equ4_HTML.gif
(4)
Put t = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq59_HTML.gif and use the boundary conditions from problem (3) to obtain
ξ u ( η ) + λ 2 [ u ] = γ u ( η ) + λ 1 [ u ] + u ( 0 ) 0 1 ( 1 s ) y ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equaf_HTML.gif
Now, finding from this u ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq60_HTML.gif and then substituting it to formula (4), we have
u ( t ) = [ γ + t ( ξ γ ) ] u ( η ) + ( 1 t ) λ 1 [ u ] + t λ 2 [ u ] + 0 1 G ( t , s ) y ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equ5_HTML.gif
(5)
Next, putting t = η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq61_HTML.gif, we can find u ( η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq62_HTML.gif, and then substitute it to formula (5) to obtain
u ( t ) = 1 δ ( [ 1 ξ η ( 1 ξ ) t ] λ 1 [ u ] + [ η γ + ( 1 γ ) t ] λ 2 [ u ] ) + F ¯ y ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equ6_HTML.gif
(6)
Now, we have to eliminate λ 1 [ u ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq63_HTML.gif and λ 2 [ u ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq64_HTML.gif from (6). If u is a solution of (6), then
{ λ 1 [ u ] = 1 δ [ ( 1 ξ η ) A 1 ( 1 ξ ) A 2 ] λ 1 [ u ] + 1 δ [ η γ A 1 + ( 1 γ ) A 2 ] λ 2 [ u ] + λ 1 [ F ¯ y ] , λ 2 [ u ] = 1 δ [ ( 1 ξ η ) B 1 ( 1 ξ ) B 2 ] λ 1 [ u ] + 1 δ [ η γ B 1 + ( 1 γ ) B 2 ] λ 2 [ u ] + λ 2 [ F ¯ y ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equag_HTML.gif

Solving this system with respect to λ 1 [ u ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq63_HTML.gif, λ 2 [ u ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq64_HTML.gif and then substituting to (6), we have the assertion of this lemma. This ends the proof. □

Define the operator T by
T u ( t ) = 1 Δ [ 1 ξ η B 2 ( 1 ξ B 1 ) t ] λ 1 [ F u ] + 1 Δ [ η γ + A 2 + ( 1 γ A 1 ) t ] λ 2 [ F u ] + F u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equah_HTML.gif
with
F u ( t ) = γ + t ( ξ γ ) δ 0 1 G ( η , s ) h ( s ) f ( s , u ( α ( s ) ) , u ( β ( s ) ) ) d s + 0 1 G ( t , s ) h ( s ) f ( s , u ( α ( s ) ) , u ( β ( s ) ) ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equai_HTML.gif
We consider the Banach space E = ( C 1 ( J , R ) , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq65_HTML.gif with the maximum norm x = max ( x 1 , x 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq66_HTML.gif. Define the cone K E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq67_HTML.gif by
K = { x E : x ( t ) 0 , t J , λ 1 [ x ] 0 , λ 2 [ x ] 0 , min [ 0 , η ] x ( t ) ρ x 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equaj_HTML.gif
with
ρ = min ( γ ( 1 η ) , 1 η , η γ 1 + γ ( η 1 ) ) , γ > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equak_HTML.gif

Let us introduce the following assumption.

H4: A and B are functions of bounded variation and
  1. (i)

    δ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq68_HTML.gif, Δ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq69_HTML.gif, A j 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq70_HTML.gif, B j 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq71_HTML.gif, G j ( s ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq72_HTML.gif for j = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq73_HTML.gif where A j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq74_HTML.gif, B j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq75_HTML.gif, G j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq76_HTML.gif, δ, Δ are defined as in the assumption H0,

     
  2. (ii)

    γ ( A 1 A 2 ) + ξ A 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq77_HTML.gif, γ ( B 1 B 2 ) + ξ B 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq78_HTML.gif, η γ B 1 + ( 1 γ ) B 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq79_HTML.gif, ( 1 ξ η ) A 1 ( 1 ξ ) A 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq80_HTML.gif, B 1 B 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq81_HTML.gif, δ η γ B 1 ( 1 γ ) B 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq82_HTML.gif, η γ A 1 + ( 1 γ ) A 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq83_HTML.gif, 1 ξ η B 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq84_HTML.gif, δ ( 1 ξ η ) A 1 + ( 1 ξ ) A 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq85_HTML.gif, ( 1 ξ η ) B 1 ( 1 ξ ) B 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq86_HTML.gif.

     

Lemma 3 Let the assumptions H1-H4 hold. Then T : K K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq87_HTML.gif.

Proof Clearly, u K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq88_HTML.gif is a positive solution of problem (1) if and only if u K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq88_HTML.gif solves the operator equation u = T u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq89_HTML.gif. Then
{ λ 1 [ F u ] = 1 δ [ γ ( A 1 A 2 ) + ξ A 2 ] 0 1 G ( η , s ) h ( s ) f ( s , u ( α ( s ) ) , u ( β ( s ) ) ) d s λ 1 [ F u ] = + 0 1 G 1 ( s ) h ( s ) f ( s , u ( α ( s ) ) , u ( β ( s ) ) ) d s , λ 2 [ F u ] = 1 δ [ γ ( B 1 B 2 ) + ξ B 2 ] 0 1 G ( η , s ) h ( s ) f ( s , u ( α ( s ) ) , u ( β ( s ) ) ) d s λ 2 [ F u ] = + 0 1 G 2 ( s ) h ( s ) f ( s , u ( α ( s ) ) , u ( β ( s ) ) ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equ7_HTML.gif
(7)

Note that λ 1 [ F u ] 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq90_HTML.gif, λ 2 [ F u ] 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq91_HTML.gif in view of the assumptions H1, H2, H4 and the positivity of Green’s function G.

Note that ( T u ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq92_HTML.gif. Moreover,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equal_HTML.gif

Hence, Tu is concave and T u ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq93_HTML.gif on J.

We next show that λ 1 [ T u ] 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq94_HTML.gif, λ 2 [ T u ] 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq95_HTML.gif. Indeed,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equam_HTML.gif
Finally, we show that
min [ 0 , η ] T u ( t ) ρ T u 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equan_HTML.gif

To do it, we consider two steps. Let T u 1 = T u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq96_HTML.gif.

Step 1. Let T u ( 0 ) < T u ( η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq97_HTML.gif. Then t ( 0 , η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq98_HTML.gif or t ( η , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq99_HTML.gif and min [ 0 , η ] T u ( t ) = T u ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq100_HTML.gif.

Let t ( 0 , η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq98_HTML.gif. Then
T u 1 T u ( 1 ) T u ( η ) T u ( 1 ) 1 t 1 η , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equao_HTML.gif
so
T u 1 T u ( 1 ) + 1 1 η [ T u ( η ) T u ( 1 ) ] < 1 1 η T u ( η ) = 1 γ ( 1 η ) ( T u ( 0 ) λ 1 [ u ] ) 1 γ ( 1 η ) T u ( 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equap_HTML.gif
It yields
min [ 0 , η ] T u ( t ) γ ( 1 η ) T u 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equaq_HTML.gif
Let t ( η , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq101_HTML.gif. Then
T u 1 T u ( 0 ) T u ( η ) T u ( 0 ) t 0 η 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equar_HTML.gif
so
T u 1 1 η [ T u ( η ) + ( η 1 ) T u ( 0 ) ] = 1 η [ 1 γ ( T u ( 0 ) λ 1 [ u ] ) + ( η 1 ) T u ( 0 ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equas_HTML.gif
It yields
min [ 0 , η ] T u ( t ) γ η 1 + γ ( η 1 ) T u 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equat_HTML.gif
Step 2. Let T u ( 0 ) T u ( η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq102_HTML.gif. Then t ( 0 , η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq98_HTML.gif and min [ 0 , η ] T u ( t ) = T u ( η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq103_HTML.gif. Then
T u 1 T u ( 1 ) T u ( η ) T u ( 1 ) 1 t 1 η , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equau_HTML.gif
so
T u 1 T u ( 1 ) + 1 1 η [ T u ( η ) T u ( 1 ) ] < 1 1 η T u ( η ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equav_HTML.gif
Hence,
min [ 0 , η ] T u ( t ) ( 1 η ) T u 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equaw_HTML.gif

It shows T : K K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq87_HTML.gif. This ends the proof. □

Remark 2 Take d B ( t ) = ( b t 1 ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq104_HTML.gif, b > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq105_HTML.gif. Note that the measure changes the sign and is increasing. It is easy to show that
B 1 = 1 2 ( b 2 ) , B 2 = 1 6 ( 2 b 3 ) , G 2 ( s ) = s ( 1 s ) 6 ( b s + b 3 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equax_HTML.gif

If we assume that b 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq106_HTML.gif, then B 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq107_HTML.gif, B 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq108_HTML.gif, G 2 ( s ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq109_HTML.gif, s J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq40_HTML.gif.

Remark 3 Take d A ( t ) = ( a t 2 1 ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq110_HTML.gif, a > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq111_HTML.gif. Note that the measure changes the sign and is increasing. It is easy to show that
A 1 = 1 3 ( a 3 ) , A 2 = 1 4 ( a 2 ) , G 1 ( s ) = s ( 1 s ) 12 ( a s 2 + a s + a 6 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equay_HTML.gif

If we assume that a 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq112_HTML.gif, then A 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq113_HTML.gif, A 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq114_HTML.gif, G 1 ( s ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq115_HTML.gif, s J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq40_HTML.gif.

Remark 4 Let d A ( t ) = ( 3 t 1 ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq116_HTML.gif, d B ( t ) = ( 7 2 t 1 ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq117_HTML.gif, t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq17_HTML.gif. Then the assumptions H3, H4 hold if one of the following conditions is satisfied:
  1. (i)

    ξ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq49_HTML.gif, 0 < γ < 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq118_HTML.gif,

     
  2. (ii)

    γ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq52_HTML.gif, 0 < ξ < 1 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq119_HTML.gif,

     
  3. (iii)

    γ = ξ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq30_HTML.gif.

     
We consider only case (i). First of all, we see that dA, dB change the sign and are increasing. Indeed, for p = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq120_HTML.gif, t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq17_HTML.gif, we have
A 1 = A 2 = λ 1 [ p ] = 1 2 , B 1 = λ 2 [ p ] = 3 4 , B 2 = 2 3 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equaz_HTML.gif
It means that the assumption H3 holds. Moreover,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equba_HTML.gif

It proves that the assumption H4 holds.

By a similar way, we prove the assertion in case (ii) or (iii).

3 Positive solutions to problem (1) with delayed arguments

Now, we present the necessary definitions from the theory of cones in Banach spaces.

Definition 1 Let E be a real Banach space. A nonempty convex closed set P E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq121_HTML.gif is said to be a cone provided that
  1. (i)

    k u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq122_HTML.gif for all u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq123_HTML.gif and all k 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq124_HTML.gif, and

     
  2. (ii)

    u , u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq125_HTML.gif implies u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq126_HTML.gif.

     

Note that every cone P E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq121_HTML.gif induces an ordering in E given by x y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq127_HTML.gif if y x P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq128_HTML.gif.

Definition 2 A map Φ is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E if Φ : P R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq129_HTML.gif is continuous and
Φ ( t x + ( 1 t ) y ) t Φ ( x ) + ( 1 t ) Φ ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbb_HTML.gif

for all x , y P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq130_HTML.gif and t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq2_HTML.gif.

Similarly, we say the map φ is a nonnegative continuous convex functional on a cone P of a real Banach space E if φ : P R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq131_HTML.gif is continuous and
φ ( t x + ( 1 t ) y ) t φ ( x ) + ( 1 t ) φ ( y ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbc_HTML.gif

for all x , y P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq130_HTML.gif and t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq2_HTML.gif.

Definition 3 An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.

Let φ and Θ be nonnegative continuous convex functionals on P, let Φ be a nonnegative continuous concave functional on P, and let Ψ be a nonnegative continuous functional on P. Then, for positive numbers a, b, c, d, we define the following sets:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbd_HTML.gif

We will use the following fixed point theorem of Avery and Peterson to establish multiple positive solutions to problem (1).

Theorem 1 (see [1])

Let P be a cone in a real Banach space E. Let φ and Θ be nonnegative continuous convex functionals on P, let Φ be a nonnegative continuous concave functional on P, and let Ψ be a nonnegative continuous functional on P satisfying Ψ ( k x ) k Ψ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq132_HTML.gif for 0 k 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq133_HTML.gif such that for some positive numbers M ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq134_HTML.gif and d,
Φ ( x ) Ψ ( x ) and x M ¯ φ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Eqube_HTML.gif
for all x P ( φ , d ) ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq135_HTML.gif. Suppose
T : P ( φ , d ) ¯ P ( φ , d ) ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbf_HTML.gif

is completely continuous and there exist positive numbers a, b, c with a < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq136_HTML.gif such that

(S1): { x P ( φ , Θ , Φ , b , c , d ) : Φ ( x ) > b } 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq137_HTML.gif and Φ ( T x ) > b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq138_HTML.gif for x P ( φ , Θ , Φ , b , c , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq139_HTML.gif,

(S2): Φ ( T x ) > b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq138_HTML.gif for x P ( φ , Φ , b , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq140_HTML.gif with Θ ( T x ) > c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq141_HTML.gif,

(S3): 0 R ( φ , Ψ , a , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq142_HTML.gif and Ψ ( T x ) < a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq143_HTML.gif for x R ( φ , Ψ , a , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq144_HTML.gif with Ψ ( x ) = a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq145_HTML.gif.

Then T has at least three fixed points x 1 , x 2 , x 3 P ( φ , d ) ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq146_HTML.gif such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbg_HTML.gif
and
Ψ ( x 3 ) < a . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbh_HTML.gif
We apply Theorem 1 with the cone K instead of P and let P ¯ r = { x K : x r } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq147_HTML.gif. Now, we define the nonnegative continuous concave functional Φ on K by
Φ ( x ) = min [ 0 , η ] | x ( t ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbi_HTML.gif

Note that Φ ( x ) x 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq148_HTML.gif. Put Ψ ( x ) = Θ ( x ) = x 1 , φ ( x ) = x 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq149_HTML.gif.

Now, we can formulate the main result of this section.

Theorem 2 Let the assumptions H1-H4 hold with ξ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq49_HTML.gif, γ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq33_HTML.gif. Let α ( t ) t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq48_HTML.gif, t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq17_HTML.gif. In addition, we assume that there exist positive constants a, b, c, d, M, a < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq136_HTML.gif and such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbj_HTML.gif
with
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbk_HTML.gif

and

(A1): f ( t , u , v ) d μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq150_HTML.gif for ( t , u , v ) J × [ 0 , M d ] × [ d , d ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq151_HTML.gif,

(A2): f ( t , u , v ) b L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq152_HTML.gif for ( t , u , v ) [ 0 , η ] × [ b , b ρ ] × [ d , d ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq153_HTML.gif,

(A3): f ( t , u , v ) a μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq154_HTML.gif for ( t , u , v ) J × [ 0 , a ] × [ d , d ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq155_HTML.gif.

Then problem (1) has at least three nonnegative solutions x 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq156_HTML.gif, x 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq157_HTML.gif, x 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq158_HTML.gif satisfying x i 1 d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq159_HTML.gif, i = 1 , 2 , 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq160_HTML.gif,
b Φ ( x 1 ) , a < x 2 1 with Φ ( x 2 ) < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbl_HTML.gif

and x 3 1 < a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq161_HTML.gif.

Proof Basing on the definitions of T, we see that T P ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq162_HTML.gif is equicontinuous on J, so T is completely continuous.

Let x P ( φ , d ) ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq135_HTML.gif, so φ ( x ) = x 1 d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq163_HTML.gif. By Lemma 1, x 1 M d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq164_HTML.gif, so 0 x ( t ) M d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq165_HTML.gif, t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq17_HTML.gif. Assumption (A1) implies f ( t , x ( α ( t ) ) , x ( β ( t ) ) ) d μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq166_HTML.gif.

Moreover, in view of (7),
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbm_HTML.gif
Combining it, we have
φ ( T x ) = max [ 0 , 1 ] | ( T x ) ( t ) | 1 Δ ( 1 B 1 ) λ 1 [ F x ] + 1 Δ ( 1 γ A 1 ) λ 2 [ F x ] + max [ 0 , 1 ] | ( F x ) ( t ) | d μ ( 1 Δ ( 1 B 1 ) D 1 + 1 Δ ( 1 γ A 1 ) D 2 + D 3 ) < d . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbn_HTML.gif

This proves that T : P ( φ , d ) ¯ P ( φ , d ) ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq167_HTML.gif.

Now, we need to show that condition (S1) is satisfied. Take
x 0 ( t ) = 1 2 ( b + b ρ ) , t J . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbo_HTML.gif
Then x 0 ( t ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq168_HTML.gif, t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq17_HTML.gif, and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbp_HTML.gif
for p ( t ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq16_HTML.gif, t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq17_HTML.gif. Moreover,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbq_HTML.gif
This proves that
{ x 0 P ( φ , Θ , Φ , b , b ρ , d ) : b < Φ ( x 0 ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbr_HTML.gif
Let b x ( t ) b ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq169_HTML.gif for t [ 0 , η ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq170_HTML.gif. Then 0 α ( t ) t η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq171_HTML.gif for t [ 0 , η ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq170_HTML.gif, so b x ( α ( t ) ) b ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq172_HTML.gif, t [ 0 , η ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq170_HTML.gif. Assumption (A2) implies f ( t , x ( α ( t ) ) , x ( β ( t ) ) ) b L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq173_HTML.gif. Hence,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbs_HTML.gif
Moreover,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbt_HTML.gif
It yields
Φ ( T x ) = min [ 0 , η ] ( T x ) ( t ) = min ( ( T x ) ( 0 ) , ( T x ) ( η ) ) γ ( T x ) ( η ) = γ Δ [ 1 B 2 ( 1 B 1 ) η ] λ 1 [ F x ] + γ Δ [ η γ + A 2 + ( 1 γ A 1 ) η ] λ 2 [ F x ] + γ δ 0 1 G ( η , s ) h ( s ) f ( s , x ( α ( s ) ) , x ( β ( s ) ) ) d s b γ L ( 1 Δ ( [ 1 B 2 ( 1 B 1 ) η ] D 1 + [ η γ + A 2 + ( 1 γ A 1 ) η ] D 2 ) + D 4 δ ) > b . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbu_HTML.gif

This proves that condition (S1) holds.

Now, we need to prove that condition (S2) is satisfied. Take x P ( φ , Φ , b , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq140_HTML.gif and T x 1 > b ρ = c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq174_HTML.gif. Then
Φ ( T x ) = min [ 0 , η ] ( T x ) ( t ) ρ T x 1 > ρ b ρ = b , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbv_HTML.gif

so condition (S2) holds.

Indeed, φ ( 0 ) = 0 < a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq175_HTML.gif, so 0 R ( φ , Ψ , a , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq142_HTML.gif. Suppose that x R ( φ , Ψ , a , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq144_HTML.gif with Ψ ( x ) = x 1 = a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq176_HTML.gif. Note that G ( t , s ) G ( s , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq177_HTML.gif, t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq17_HTML.gif. Then
F x 1 γ δ 0 1 G ( η , s ) h ( s ) f ( s , x ( α ( s ) ) , x ( β ( s ) ) ) d s + 0 1 G ( s , s ) h ( s ) f ( s , x ( α ( s ) ) , x ( β ( s ) ) ) d s a μ [ γ δ 0 1 G ( η , s ) h ( s ) d s + 0 1 G ( s , s ) h ( s ) d s ] a μ D 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbw_HTML.gif
and finally,
Ψ ( T x ) = max t J ( T x ) ( t ) 1 Δ [ 1 B 2 ] λ 1 [ F x ] + 1 Δ [ η γ + A 2 + 1 γ A 1 ] λ 2 [ F x ] + F x 1 a μ ( 1 Δ ( [ 1 B 2 ] D 1 + [ η γ + A 2 + 1 γ A 1 ] D 2 ) + D 5 ) < a . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbx_HTML.gif

This shows that condition (S3) is satisfied.

Since all the conditions of Theorem 1 are satisfied, problem (1) has at least three nonnegative solutions x 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq156_HTML.gif, x 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq157_HTML.gif, x 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq158_HTML.gif such that x i d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq178_HTML.gif for i = 1 , 2 , 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq160_HTML.gif, and
b min [ 0 , η ] x 1 ( t ) , a < x 2 1 with  min [ 0 , η ] x 2 ( t ) < b , x 3 1 < a . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equby_HTML.gif

This ends the proof. □

Example Consider the following problem:
{ x ( t ) + h f ( t , x ( α ( t ) ) , x ( β ( t ) ) ) = 0 , t ( 0 , 1 ) , x ( 0 ) = 1 4 x ( 1 2 ) , x ( 1 ) = 1 2 0 1 x ( t ) ( 7 t 2 ) d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equ8_HTML.gif
(8)
where
f ( t , u , v ) = { 1 100 cos t + ( v 20 , 000 ) 2 , ( t , u , v ) [ 0 , 1 ] × [ 0 , 1 ] × [ d , d ] , 1 100 cos t + 2 ( u 1 ) + ( v 20 , 000 ) 2 , ( t , u , v ) [ 0 , 1 ] × [ 1 , 16 ] × [ d , d ] , 1 100 cos t + 30 + ( v 20 , 000 ) 2 , ( t , v ) [ 0 , 1 ] × [ d , d ] , u 16 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equbz_HTML.gif
with d = 2 , 000 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq179_HTML.gif. For example, we can take α ( t ) = ρ ¯ t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq180_HTML.gif, β ( t ) = t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq181_HTML.gif on J with fixed ρ ¯ ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq182_HTML.gif. Indeed, f C ( [ 0 , 1 ] × R + × [ d , d ] , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq183_HTML.gif, γ = 1 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq184_HTML.gif, η = 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq185_HTML.gif, h ( t ) = h > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq186_HTML.gif, ξ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq49_HTML.gif and
λ 1 [ x ] = 0 , λ 2 [ x ] = 1 2 0 1 x ( t ) ( 7 t 2 ) d t , ρ = 1 8 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equca_HTML.gif
Note that d B ( t ) = 1 2 ( 7 t 2 ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq187_HTML.gif, so the measure changes the sign on J. Moreover,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equcb_HTML.gif
so the assumption H4 holds; see Remark 4. Next,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equcc_HTML.gif
Put a = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq188_HTML.gif, b = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq189_HTML.gif, h = 30 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq190_HTML.gif, then c = 16 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq191_HTML.gif, μ > 37.18 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq192_HTML.gif, L < 1.94 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq193_HTML.gif. Let μ = 40 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq194_HTML.gif, L = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq195_HTML.gif. Then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equcd_HTML.gif
and
f ( t , u , v ) 1 100 + 30 + ( 2 , 000 20 , 000 ) 2 = 30.02 < 50 = d μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equce_HTML.gif

for ( t , u , v ) [ 0 , 1 ] × [ 0 , 2 d ] × [ d , d ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq196_HTML.gif.

All the assumptions of Theorem 2 hold, so problem (8) has at least three positive solutions.

Remark 5 We can also construct an example in which, for example, λ 1 [ x ] = 0 1 x ( t ) ( 3 t 1 ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq197_HTML.gif to use the results of Remark 4. Note that also this measure changes the sign.

4 Positive solutions to problem (1) with advanced arguments

In this section, we consider the case when α ( t ) t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq51_HTML.gif on J, so the interval [ 0 , η ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq198_HTML.gif is now replaced by [ η , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq199_HTML.gif. It means that we can put γ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq52_HTML.gif with ξ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq34_HTML.gif in the boundary conditions of problem (1) because someone can take λ 1 [ x ] = γ ¯ x ( η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq200_HTML.gif as an example. Let us introduce the cone K 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq201_HTML.gif by
K 2 = { x E : x ( t ) 0 , t J , λ 1 [ x ] 0 , λ 2 [ x ] 0 , min [ η , 1 ] x ( t ) Γ x 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equcf_HTML.gif
with
Γ = min ( ξ ( 1 η ) 1 ξ η , ξ η , η ) , ξ > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equcg_HTML.gif

Now Φ ( x ) = min [ η , 1 ] | x ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq202_HTML.gif. Functionals Ψ, Θ, φ are defined as in Section 3. We formulate only the main result using the cone K 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq201_HTML.gif instead of K (see Theorem 2); the proof is similar to the previous one.

Theorem 3 Let the assumptions H1-H4 hold with γ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq52_HTML.gif, ξ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq34_HTML.gif. Let α ( t ) t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq51_HTML.gif, t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq17_HTML.gif. In addition, we assume that there exist positive constants a, b, c, d, M, a < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq136_HTML.gif and such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equch_HTML.gif
with
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equci_HTML.gif

and

(B1): f ( t , u , v ) d μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq150_HTML.gif for ( t , u , v ) J × [ 0 , M d ] × [ d , d ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq151_HTML.gif,

(B2): f ( t , u , v ) b L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq152_HTML.gif for ( t , u , v ) [ η , 1 ] × [ b , b Γ ] × [ d , d ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq203_HTML.gif,

(B3): f ( t , u , v ) a μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq154_HTML.gif for ( t , u , v ) J × [ 0 , a ] × [ d , d ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq155_HTML.gif.

Then problem (1) has at least three nonnegative solutions x 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq156_HTML.gif, x 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq157_HTML.gif, x 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq158_HTML.gif satisfying x i 1 d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq159_HTML.gif, i = 1 , 2 , 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq160_HTML.gif,
b Φ ( x 1 ) , a < x 2 1 with Φ ( x 2 ) < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equcj_HTML.gif

and x 3 1 < a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq161_HTML.gif.

5 Positive solutions to problem (1) for the case when α ( t ) = t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq1_HTML.gifon J

In this section, we consider problem (1) when α ( t ) = t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq1_HTML.gif on J and γ = ξ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq30_HTML.gif. It means that now Φ ( x ) = min [ ζ , ϱ ] | x ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq204_HTML.gif for some fixed constants ζ, ϱ such that 0 < ζ < ϱ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq43_HTML.gif. For 0 < ζ + ϱ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq205_HTML.gif we can show that G ( t , s ) κ G ( s , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq206_HTML.gif, t [ ζ , ϱ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq207_HTML.gif, s J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq40_HTML.gif. Now, for κ = min ( ζ , 1 ϱ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq44_HTML.gif, we introduce the cone K 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq208_HTML.gif by
K 3 = { x E : x ( t ) 0 , t J , λ 1 [ x ] 0 , λ 2 [ x ] 0 , min [ ζ , ϱ ] x ( t ) κ x 1 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equck_HTML.gif

Functionals Ψ, Θ, φ are defined as in Section 3; the cone K is now replaced by K 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq208_HTML.gif.

Theorem 4 Let the assumptions H1-H4 hold with γ = ξ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq30_HTML.gif. Let 0 < ζ + ϱ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq205_HTML.gif, α ( t ) = t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq1_HTML.gif, t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq17_HTML.gif. In addition, we assume that there exist positive constants a, b, c, d, M, a < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq136_HTML.gif and such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equcl_HTML.gif
with
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equcm_HTML.gif

and

(C1): f ( t , u , v ) d μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq150_HTML.gif for ( t , u , v ) J × [ 0 , M d ] × [ d , d ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq151_HTML.gif,

(C2): f ( t , u , v ) b L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq152_HTML.gif for ( t , u , v ) [ ζ , ϱ ] × [ b , b κ ] × [ d , d ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq209_HTML.gif,

(C3): f ( t , u , v ) a μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq154_HTML.gif for ( t , u , v ) J × [ 0 , a ] × [ d , d ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq155_HTML.gif.

Then problem (1) has at least three nonnegative solutions x 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq156_HTML.gif, x 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq157_HTML.gif, x 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq158_HTML.gif satisfying x i 1 d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq159_HTML.gif, i = 1 , 2 , 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq160_HTML.gif,
b Φ ( x 1 ) , a < x 2 1 with Φ ( x 2 ) < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_Equcn_HTML.gif

and x 3 1 < a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-8/MediaObjects/13661_2012_Article_275_IEq161_HTML.gif.

6 Conclusions

In this paper, we have discussed boundary value problems for second-order differential equations with deviating arguments and with dependence on the first-order derivative. In our research, the deviating arguments can be both delayed and advanced. By using the fixed point theorem of Avery and Peterson, new sufficient conditions for the existence of positive solutions to such boundary problems have been derived. An example is provided for illustration.

Declarations

Authors’ Affiliations

(1)
Department of Differential Equations and Applied Mathematics, Gdańsk University of Technology

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© Jankowski; licensee Springer. 2013

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