Open Access

On positive solutions for nonhomogeneous m-point boundary value problems with two parameters

Boundary Value Problems20122012:87

DOI: 10.1186/1687-2770-2012-87

Received: 14 May 2012

Accepted: 27 July 2012

Published: 6 August 2012

Abstract

This paper is concerned with the existence, multiplicity, and nonexistence of positive solutions for nonhomogeneous m-point boundary value problems with two parameters. The proof is based on the fixed-point theorem, the upper-lower solutions method, and the fixed-point index.

MSC:34B10, 34B18.

Keywords

nonhomogeneous BVP positive solutions upper-lower solutions fixed-point theorem fixed point index

1 Introduction

Many authors have studied the existence, nonexistence, and multiplicity of positive solutions for multipoint boundary value problems by using the fixed-point theorem, the fixed point index theory, and the lower and upper solutions method. We refer the readers to the references [14]. Recently, Hao, Liu and Wu [5] studied the existence, nonexistence, and multiplicity of positive solutions for the following nonhomogeneous boundary value problems:
{ u ( t ) = a ( t ) f ( t , u ( t ) ) , u ( 0 ) = 0 , u ( 1 ) i = 1 m 2 k i u ( ξ i ) = b , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equa_HTML.gif

where b > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq1_HTML.gif, k i > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq2_HTML.gif ( i = 1 , 2 , , m 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq3_HTML.gif), 0 < ξ 1 < ξ 2 < < ξ m 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq4_HTML.gif, i = 1 m 2 k i ξ i < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq5_HTML.gif, a ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq6_HTML.gif may be singular at t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq7_HTML.gif and/or t = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq8_HTML.gif. They showed that there exists a positive number b > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq9_HTML.gif such that the problem has at least two positive solutions for 0 < b < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq10_HTML.gif, at least one positive solution for b = b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq11_HTML.gif and no solution for b > b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq12_HTML.gif by using the Krasnosel’skii-Guo fixed-point theorem, the upper-lower solutions method, and the topological degree theory.

Inspired by the above references, the purpose of this paper is to study the following more general nonhomogeneous boundary value problems:
{ u ( t ) = λ h ( t ) f ( u ( t ) ) , u ( 0 ) = 0 , u ( 1 ) i = 1 m 2 k i u ( ξ i ) = μ 0 1 g ( u ( s ) ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equ1_HTML.gif
(1)

where λ, μ are positive parameters, k i > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq2_HTML.gif, 0 < ξ 1 < ξ 2 < < ξ m 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq4_HTML.gif. The main result of the present paper is summarized as follows.

Theorem 1.1 Assume the following conditions hold:

(H1) ( λ , μ ) R + 2 { ( 0 , 0 ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq13_HTML.gif are nonnegative parameters;

(H2) h : [ 0 , 1 ] [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq14_HTML.gif is continuous, h ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq15_HTML.gif does not vanish identically on any subinterval of [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq16_HTML.gif and 0 1 G ( s , s ) h ( s ) d s < + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq17_HTML.gif, where G ( s , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq18_HTML.gif is given in Sect. 2;

(H3) f , g C ( R + , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq19_HTML.gif is nondecreasing with respect to u, respectively, that is,
f ( u 1 ) f ( u 2 ) if  u 1 u 2 , g ( u 1 ) g ( u 2 ) if  u 1 u 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equb_HTML.gif

And either f ( 0 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq20_HTML.gif or g ( 0 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq21_HTML.gif;

(H4) There exist constants m 1 , m 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq22_HTML.gif such that f ( u ) m 1 u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq23_HTML.gif and g ( u ) m 2 u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq24_HTML.gif, respectively, for all u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq25_HTML.gif;

(H5) lim | u | + f ( u ) u = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq26_HTML.gif, lim | u | + g ( u ) u = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq27_HTML.gif.

If 0 < i = 1 m 2 k i < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq28_HTML.gif, then there exists a bounded and continuous curve Γ separating R + 2 { ( 0 , 0 ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq29_HTML.gif into two disjoint subsets Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq30_HTML.gif and Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq31_HTML.gif such that (1) has at least two positive solutions for ( λ , μ ) Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq32_HTML.gif, one positive solution for ( λ , μ ) Γ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq33_HTML.gif, and no solution for ( λ , μ ) Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq34_HTML.gif. Moreover, let Γ + Γ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq35_HTML.gif be the parametric representation of Γ, where
Γ + : μ = μ ( λ ) > 0 , Γ 0 : μ = μ ( λ ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equc_HTML.gif

Then on Γ + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq36_HTML.gif, the function μ = μ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq37_HTML.gif is continuous and nonincreasing, that is, if λ λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq38_HTML.gif, we have μ ( λ ) μ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq39_HTML.gif.

For the proof of Theorem 1.1, we also need the following lemmas.

Lemma 1.2 [6]

Let E be a Banach space, K a cone in E and Ω bounded open in E. Let 0 Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq40_HTML.gif and T : K Ω ¯ K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq41_HTML.gif be condensing. Suppose that T x λ x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq42_HTML.gif for all x K Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq43_HTML.gif and all λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq44_HTML.gif. Then
i ( T , K Ω , K ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equd_HTML.gif

Lemma 1.3 [6]

Let E be a Banach space and K a cone in E. For r > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq45_HTML.gif, define K r = { x K : x < r } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq46_HTML.gif. Assume that T : K r ¯ K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq47_HTML.gif is a compact map such that T x x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq48_HTML.gif for x K r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq49_HTML.gif. If x T x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq50_HTML.gif for all x K r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq49_HTML.gif, then
i ( T , K r , K ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Eque_HTML.gif

2 Preliminaries

Lemma 2.1 [5]

Assume that 0 < i = 1 m 2 k i ξ i < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq51_HTML.gif. If y ( t ) C ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq52_HTML.gif with 0 1 G ( s , s ) y ( s ) d s < + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq53_HTML.gif, then the Green function for the homogeneous BVP
{ u ( t ) = y ( t ) , u ( 0 ) = 0 , u ( 1 ) i = 1 m 2 k i u ( ξ i ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equf_HTML.gif
is given by
G ( t , s ) = 1 1 i = 1 m 2 k i ξ i { s ( 1 t ) i = 1 m 2 k i ( ξ i t ) s , s t , s ξ 1 , t [ ( 1 s ) i = 1 m 2 k i ( ξ i s ) ] , t s ξ 1 , s ( 1 t ) + i = 1 j k i ξ i ( t s ) i = j + 1 m 2 k i ( ξ i t ) s , ξ j s ξ j + 1 , s t , j = 1 , 2 , , m 3 , t [ ( 1 s ) i = j + 1 m 2 k i ( ξ i s ) ] , ξ j s ξ j + 1 , t s , j = 1 , 2 , , m 3 , s ( 1 t ) + i = 1 m 2 k i ξ i ( t s ) , ξ m 2 s t , t ( 1 s ) , ξ m 2 s , t s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equg_HTML.gif
Moreover, the Green function satisfies the following properties:
  1. (i)

    G ( t , s ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq54_HTML.gif for t , s ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq55_HTML.gif, and G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq56_HTML.gif is continuous on [ 0 , 1 ] × [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq57_HTML.gif;

     
  2. (ii)

    G ( t , s ) G ( s , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq58_HTML.gif for all t , s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq59_HTML.gif.

     
Lemma 2.2 Assume that (H 1)-(H 5) hold. If 0 < i = 1 m 2 k i ξ i < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq51_HTML.gif, then u C 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq60_HTML.gif is a solution of (1) if and only if u C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq61_HTML.gif satisfies the following nonlinear integral equation:
u ( t ) = λ 0 1 G ( t , s ) h ( s ) f ( u ( s ) ) d s + μ 0 1 g ( u ( s ) ) d s 1 i = 1 m 2 k i ξ i t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equh_HTML.gif
Proof Integrating both sides of (1) from 0 to t twice and applying the boundary conditions, then we can obtain
u ( t ) = μ 0 1 g ( u ( s ) ) d s 1 i = 1 m 2 k i ξ i t λ 0 t ( t s ) h ( s ) f ( u ( s ) ) d s + λ t 1 i = 1 m 2 k i ξ i [ 0 1 ( 1 s ) h ( s ) f ( u ( s ) ) d s i = 1 m 2 k i 0 ξ i ( ξ i s ) h ( s ) f ( u ( s ) ) d s ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equi_HTML.gif
Furthermore, by Lemma 2.1, we can obtain
u ( t ) = λ 0 1 G ( t , s ) h ( s ) f ( u ( s ) ) d s + μ 0 1 g ( u ( s ) ) d s 1 i = 1 m 2 k i ξ i t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equj_HTML.gif

 □

Let E denote the Banach space C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq62_HTML.gif with the norm u = max t [ 0 , 1 ] | u ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq63_HTML.gif. A function u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq64_HTML.gif is said to be a solution of (1) if u C [ 0 , 1 ] C 2 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq65_HTML.gif satisfies (1). Moreover, from Lemma 2.2, it is clear to see that u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq64_HTML.gif is a solution of (1) is equivalent to the fixed point of the operator T defined as
T u ( t ) = λ 0 1 G ( t , s ) h ( s ) f ( u ( s ) ) d s + μ 0 1 g ( u ( s ) ) d s 1 i = 1 m 2 k i ξ i t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equk_HTML.gif
In addition, define a cone K E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq66_HTML.gif as
K = { u E : u ( t ) 0 , t [ 0 , 1 ] , inf t [ ξ 1 , ξ m 2 ] u ( t ) θ u } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equl_HTML.gif

where θ = k 1 ξ 1 min [ 1 ξ 1 , ξ 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq67_HTML.gif. Then we have

Lemma 2.3 If (H 1)-(H 3) hold, then T : K K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq68_HTML.gif is completely continuous.

The proof procedure of Lemma 2.3 is standard, so we omit it.

Now, we will establish the classical lower and upper solutions method for our problem. As usual, we say that x ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq69_HTML.gif is a lower solution for (1) if
{ x ( t ) + λ h ( t ) f ( x ( t ) ) 0 , x ( 0 ) 0 , x ( 1 ) i = 1 m 2 k i x ( ξ i ) 0 1 g ( x ( s ) ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equm_HTML.gif
Similarly, we define the upper solution y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq70_HTML.gif of the problem (1):
{ y ( t ) + λ h ( t ) f ( y ( t ) ) 0 , y ( 0 ) 0 , y ( 1 ) i = 1 m 2 k i y ( ξ i ) 0 1 g ( y ( s ) ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equn_HTML.gif

Lemma 2.4 Let x ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq69_HTML.gif, y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq70_HTML.gif be lower and upper solutions, respectively, of (1) such that 0 x ( t ) y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq71_HTML.gif. Then (1) has a nonnegative solution u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq64_HTML.gif satisfying x ( t ) u ( t ) y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq72_HTML.gif for t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq73_HTML.gif.

Proof Define
D x y = { u R : x ( t ) u ( t ) y ( t ) , t [ 0 , 1 ] } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equo_HTML.gif

It is clear to see that D x y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq74_HTML.gif is a bounded, convex and closed subset in Banach space E. Now we can prove that T : D x y D x y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq75_HTML.gif.

For any u ( t ) D x y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq76_HTML.gif, from (H3), we have
T u ( t ) = λ 0 1 G ( t , s ) h ( s ) f ( u ( s ) ) d s + μ 0 1 g ( u ( s ) ) d s 1 i = 1 m 2 k i ξ i t λ 0 1 G ( t , s ) h ( s ) f ( y ( s ) ) d s + μ 0 1 g ( y ( s ) ) d s 1 i = 1 m 2 k i ξ i t = y ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equp_HTML.gif
On the other hand, we also have
T u ( t ) = λ 0 1 G ( t , s ) h ( s ) f ( u ( s ) ) d s + μ 0 1 g ( u ( s ) ) d s 1 i = 1 m 2 k i ξ i t λ 0 1 G ( t , s ) h ( s ) f ( x ( s ) ) d s + μ 0 1 g ( x ( s ) ) d s 1 i = 1 m 2 k i ξ i t = x ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equq_HTML.gif

From above inequalities, we obtain that T : D x y D x y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq75_HTML.gif.

Therefore, by Schauder’s fixed theorem, the operator T has a fixed point u ( t ) D x y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq76_HTML.gif, which is the solution of (1). □

3 Proof of Theorem 1.1

Lemma 3.1 Assume (H 1)-(H 5) hold and Σ be a compact subset of R + 2 { ( 0 , 0 ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq29_HTML.gif. Then there exists a constant C Σ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq77_HTML.gif such that for all ( λ , μ ) Σ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq78_HTML.gif and all possible positive solutions u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq64_HTML.gif of (1) at ( λ , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq79_HTML.gif, one has u C Σ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq80_HTML.gif.

Proof Suppose on the contrary that there exists a sequence { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq81_HTML.gif of positive solutions of Eq. (1) at ( λ n , μ n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq82_HTML.gif such that ( λ n , μ n ) Σ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq83_HTML.gif for all n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq84_HTML.gif and
u n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equr_HTML.gif
Then u n ( t ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq85_HTML.gif, and thus
inf t [ ξ 1 , ξ m 2 ] u n ( t ) θ u n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equ2_HTML.gif
(2)
Since Σ is compact, the sequence { ( λ n , μ n ) } n = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq86_HTML.gif has a convergent subsequence which we denote without loss of generality still by { ( λ n , μ n ) } n = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq86_HTML.gif such that
lim n λ n = λ , lim n μ n = μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equs_HTML.gif

and at least λ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq87_HTML.gif or μ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq88_HTML.gif.

Case (I). If λ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq87_HTML.gif, we have λ n λ / 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq89_HTML.gif for n sufficient large. Then by (H5), there exists a R > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq90_HTML.gif such that
f ( u ) L u , u R , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equt_HTML.gif
where L satisfies
λ 2 L θ min t [ 0 , 1 ] ξ 1 ξ m 2 G ( t , s ) h ( s ) d s > 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equu_HTML.gif
Since u n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq91_HTML.gif, for n sufficient large, we
u n ( t ) = λ n 0 1 G ( t , s ) h ( s ) f ( u n ( s ) ) d s + μ n 0 1 g ( u n ( s ) ) d s 1 i = 1 m 2 k i ξ i t λ n 0 1 G ( t , s ) h ( s ) f ( u n ( s ) ) d s λ n 0 1 G ( t , s ) h ( s ) L u n ( s ) d s λ 2 ξ 1 ξ m 2 G ( t , s ) h ( s ) L u n ( s ) d s λ 2 L θ u n min t [ 0 , 1 ] ξ 1 ξ m 2 G ( t , s ) h ( s ) d s > u n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equv_HTML.gif

This is a contradiction.

Case (II). If μ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq88_HTML.gif, then we have μ n μ / 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq92_HTML.gif for n sufficient large. Since lim | u | + g ( u ) u = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq27_HTML.gif, there exists a R > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq90_HTML.gif such that
g ( u ) M u , u R , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equw_HTML.gif
where M satisfies
μ 2 M θ ξ 1 ( ξ m 2 ξ 1 ) 1 i = 1 m 2 k i ξ i > 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equx_HTML.gif
Since u n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq91_HTML.gif, then for n sufficient large, we have
u n ( ξ 1 ) = λ n 0 1 G ( ξ 1 , s ) h ( s ) f ( u n ( s ) ) d s + μ n 0 1 g ( u n ( s ) ) d s 1 i = 1 m 2 k i ξ i ξ 1 μ n 0 1 g ( u n ( s ) ) d s 1 i = 1 m 2 k i ξ i ξ 1 μ n 0 1 M u n ( s ) d s 1 i = 1 m 2 k i ξ i ξ 1 μ 2 M θ u ξ 1 ( ξ m 2 ξ 1 ) 1 i = 1 m 2 k i ξ i > u n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equy_HTML.gif

This is a contradiction. □

Lemma 3.2 Assume (H 1)-(H 4) hold. If (1) has a positive solution at ( λ ¯ , μ ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq93_HTML.gif, then Eq. (1) has a positive solution at ( λ , μ ) R + 2 { ( 0 , 0 ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq13_HTML.gif for all ( λ , μ ) ( λ ¯ , μ ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq94_HTML.gif.

Proof Let u ¯ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq95_HTML.gif be the solution of Eq. (1) at ( λ ¯ , μ ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq93_HTML.gif, then u ¯ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq95_HTML.gif be the upper solution of (1) at ( λ , μ ) R + 2 { ( 0 , 0 ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq13_HTML.gif with ( λ , μ ) ( λ ¯ , μ ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq94_HTML.gif. Since f ( 0 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq20_HTML.gif or g ( 0 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq21_HTML.gif, u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq96_HTML.gif is not a solution of (1), but it is the lower solution of (1) at ( λ , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq79_HTML.gif. Therefore, by Lemma 2.4, we obtain the result. □

Lemma 3.3 Assume (H 1)-(H 5) hold. Then there exists ( λ , μ ) > ( 0 , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq97_HTML.gif such that Eq. (1) has a positive solution for all ( λ , μ ) ( λ , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq98_HTML.gif.

Proof Let β ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq99_HTML.gif be the unique solution of
{ u ( t ) = h ( t ) , u ( 0 ) = 0 , u ( 1 ) i = 1 m 2 k i u ( ξ i ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equ3_HTML.gif
(3)
It is clear to see that β ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq99_HTML.gif is a positive solution of (3). Let M f = max t [ 0 , 1 ] f ( β ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq100_HTML.gif, M g = max t [ 0 , 1 ] g ( β ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq101_HTML.gif, then by (H4), we know that M f > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq102_HTML.gif and M g > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq103_HTML.gif. Set ( λ , μ ) = ( 1 / M f , 1 / M g ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq104_HTML.gif, we have
{ β ( t ) + λ h ( t ) f ( β ( t ) ) = h ( t ) + λ h ( t ) f ( β ( t ) ) = h ( t ) ( λ f ( β ( t ) ) 1 ) 0 , β ( 0 ) = 0 , β ( 1 ) i = 1 m 2 k i β ( ξ i ) μ 0 1 g ( β ( s ) ) d s = 0 1 1 μ g ( β ( s ) ) d s 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equz_HTML.gif

which implies that β ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq99_HTML.gif is an upper solution of (3) at ( λ , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq105_HTML.gif. On the other hand, 0 is a lower solution of (1) and 0 β ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq106_HTML.gif. By (H3), 0 is not a solution of (1). Hence, (1) has a positive solution at ( λ , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq105_HTML.gif, Lemma 3.2 now implies the conclusion of Lemma 3.3. □

Define a set S by
S = { ( λ , μ ) R + 2 { ( 0 , 0 ) } : ( 1 )  has a positive solution at  ( λ , μ ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equaa_HTML.gif

Then it follows from Lemma 3.3 that S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq107_HTML.gif and ( S , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq108_HTML.gif is a partially ordered set.

Lemma 3.4 Assume (H 1)-(H 5) hold. Then ( S , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq108_HTML.gif is bounded above.

Proof Let ( λ , μ ) S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq109_HTML.gif and u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq64_HTML.gif be a positive solution of (1) at ( λ , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq79_HTML.gif, then we have
u u ( ξ 1 ) = λ 0 1 G ( ξ 1 , s ) h ( s ) f ( u ( s ) ) d s + μ 0 1 g ( u ( s ) ) d s 1 i = 1 m 2 k i ξ i ξ 1 λ 0 1 G ( ξ 1 , s ) h ( s ) m 1 u ( s ) d s + μ 0 1 m 2 u ( s ) d s 1 i = 1 m 2 k i ξ i ξ 1 λ ξ 1 ξ m 2 G ( ξ 1 , s ) h ( s ) m 1 u ( s ) d s + μ m 2 θ u ξ 1 ( ξ m 2 ξ 1 ) 1 i = 1 m 2 k i ξ i λ m 1 θ u ξ 1 ξ m 2 G ( ξ 1 , s ) h ( s ) d s + μ m 2 θ u ξ 1 ( ξ m 2 ξ 1 ) 1 i = 1 m 2 k i ξ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equab_HTML.gif
by (H4). Furthermore, we can obtain that
λ m 1 θ ξ 1 ξ m 2 G ( ξ 1 , s ) h ( s ) d s + μ m 2 θ ξ 1 ( ξ m 2 ξ 1 ) 1 i = 1 m 2 k i ξ i 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equac_HTML.gif

 □

Lemma 3.5 Assume (H 1)-(H 5) hold. Then every chain in S has a unique supremum in S.

Lemma 3.6 Assume (H 1)-(H 5) hold. Then there exists a λ ˜ [ λ , λ ¯ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq110_HTML.gif such (1) has a positive solution at ( λ , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq111_HTML.gif for all 0 < λ λ ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq112_HTML.gif, no solution at ( λ , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq111_HTML.gif for all λ > λ ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq113_HTML.gif. Similarly, there exists a μ ˜ [ μ , μ ¯ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq114_HTML.gif such that (1) has a positive solution at ( 0 , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq115_HTML.gif for all 0 < μ μ ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq116_HTML.gif, and no solution at ( 0 , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq115_HTML.gif for all μ > μ ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq117_HTML.gif.

Lemma 3.7 Assume (H 1)-(H 5) hold. Then there exists a continuous curve Γ separating R + 2 { ( 0 , 0 ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq29_HTML.gif into two disjoint subsets Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq30_HTML.gif and Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq31_HTML.gif such that Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq30_HTML.gif is bounded and Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq31_HTML.gif is unbounded, Eq. (1) has at least one solution for ( λ , μ ) Ω 1 Γ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq118_HTML.gif, and no solution for ( λ , μ ) Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq34_HTML.gif. The function μ = μ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq37_HTML.gif is nonincreasing, that is, if
λ λ λ ˜ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equad_HTML.gif
then
μ ( λ ) μ ( λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equae_HTML.gif
Lemma 3.8 Let ( λ , μ ) Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq32_HTML.gif. Then there exists ε 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq119_HTML.gif such that ( u + ε , v + ε ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq120_HTML.gif is an upper solution of (1) at ( λ , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq79_HTML.gif for all 0 < ε ε 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq121_HTML.gif, where ( u , v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq122_HTML.gif is the positive solution of Eq. (1) corresponding to some ( λ , μ ) Γ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq123_HTML.gif satisfying
( λ , μ ) ( λ , μ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equaf_HTML.gif
Proof From (H4), there exists constant M > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq124_HTML.gif such that
f ( u ( t ) ) M > 0 , g ( u ( t ) ) M > 0 , for all  t [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equag_HTML.gif
Then by the uniform continuity of f and g on a compact set, there exist ϵ 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq125_HTML.gif such that
| f ( u ( t ) + ϵ ) f ( u ( t ) ) | < M ( λ λ ) λ , | g ( u ( t ) + ϵ ) g ( u ( t ) ) | < M ( μ μ ) μ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equah_HTML.gif

for all t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq73_HTML.gif and 0 < ϵ ϵ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq126_HTML.gif.

Let u ϵ = u + ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq127_HTML.gif, then we have
u ϵ ( t ) + λ h ( t ) f ( u ϵ ( t ) ) = λ h ( t ) f ( u ϵ ( t ) ) + λ h ( t ) f ( u ϵ ( t ) ) h ( t ) ( λ λ ) ( M f ( u ) ) 0 , u ϵ ( 0 ) = ϵ > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equai_HTML.gif
and
u ϵ ( 1 ) i = 1 m 2 k i u ϵ ( ξ i ) μ 0 1 g ( u ϵ ) d s = u ( 1 ) + ϵ i = 1 m 2 k i ( u ( ξ i ) + ϵ ) μ 0 1 g ( u + ϵ ) d s = u ( 1 ) i = 1 m 2 k i ( u ( ξ i ) ) + ϵ i = 1 m 2 k i ϵ μ 0 1 g ( u + ϵ ) d s = μ 0 1 g ( u ) d s μ 0 1 g ( u + ϵ ) d s + ( 1 i = 1 m 2 k i ) ϵ = ( μ μ ) 0 1 g ( u ) d s + μ ( 0 1 g ( u ) g ( u + ϵ ) ) d s + ( 1 i = 1 m 2 k i ) ϵ ( 1 i = 1 m 2 k i ) ϵ > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equaj_HTML.gif

From above inequalities, it is clear to see that u ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq128_HTML.gif, is an upper solution of (1) at ( λ , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq79_HTML.gif for all 0 < ϵ ϵ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq126_HTML.gif. □

Proof of Theorem 1.1 From above lemmas, we need only to show the existence of the second positive solution of (1) for ( λ , μ ) Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq32_HTML.gif. Let ( λ , μ ) Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq32_HTML.gif, then there exists ( λ , μ ) Γ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq123_HTML.gif such that
( λ , μ ) ( λ , μ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equak_HTML.gif
Let ( u , v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq122_HTML.gif be the positive solution of (1) at ( λ , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq105_HTML.gif. Then for ε 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq119_HTML.gif given by Lemma 3.8 and for all ε : 0 < ε ε 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq129_HTML.gif, denote
u ˜ = u + ε , v ˜ = v + ε . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equal_HTML.gif
Define the set
D = { u E : ε < u < u ˜ } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equam_HTML.gif
Then D is bounded open set in E and 0 D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq130_HTML.gif. The map T satisfies K D ¯ K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq131_HTML.gif and is condensing, since it is completely continuous. Now let ( u , v ) K D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq132_HTML.gif, then there exists ξ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq133_HTML.gif such that either u ( ξ ) = u ˜ ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq134_HTML.gif. Then by (H) and Lemma 3.8, we obtain
T u ( ξ ) = λ 0 1 G ( ξ , s ) h ( s ) f ( u ( s ) ) d s + μ 0 1 g ( u ( s ) ) d s 1 i = 1 m 2 k i ξ i ξ λ 0 1 G ( ξ , s ) h ( s ) f ( u ˜ ( s ) ) d s + μ 0 1 g ( u ˜ ( s ) ) d s 1 i = 1 m 2 k i ξ i ξ < u ˜ ( ξ ) = u ( ξ ) ϑ u ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equan_HTML.gif
for all ϑ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq135_HTML.gif. Thus, T ( u ) ϑ u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq136_HTML.gif for all u K D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq137_HTML.gif and ϑ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq135_HTML.gif, Lemma 1.2 now implies that
i ( T , K D , K ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equao_HTML.gif
Now for some fixed λ and μ, it follows from assumption (H4) that there exists a R > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq90_HTML.gif such that
f ( u ) L u , and g ( u ) L u , u R , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equ4_HTML.gif
(4)
where L satisfies
L θ ( λ ξ 1 ξ m 2 G ( ξ 1 , s ) h ( s ) d s + μ ξ 1 ( ξ m 2 ξ 1 ) 1 i = 1 m 2 k i ξ i ) > 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equap_HTML.gif
Let R = max { C Σ , θ 1 R , ( u ˜ , v ˜ ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq138_HTML.gif where C Σ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq139_HTML.gif is given by Lemma 3.1 with Σ a compact set in R + 2 { ( 0 , 0 ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq29_HTML.gif containing ( λ , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq79_HTML.gif. Let
K R = { u K : u < R } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equaq_HTML.gif
Then it follows from Lemma 3.1,
T ( u ) u u K R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equar_HTML.gif
Moreover, for u K R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq140_HTML.gif, we have
inf t [ ξ 1 , ξ m 2 ] u ( t ) θ u R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equas_HTML.gif
Furthermore, we have
T u ( ξ 1 ) = λ 0 1 G ( ξ 1 , s ) h ( s ) f ( u ( s ) ) d s + μ 0 1 g ( u ( s ) ) d s 1 i = 1 m 2 k i ξ i ξ 1 0 1 G ( ξ 1 , s ) h ( s ) L u ( s ) d s + μ 0 1 L u ( s ) d s 1 i = 1 m 2 k i ξ i ξ 1 λ L θ u ξ 1 ξ m 2 G ( ξ 1 , s ) h ( s ) d s + μ L θ u ξ 1 ( ξ m 2 ξ 1 ) 1 i = 1 m 2 k i ξ i = L θ ( λ ξ 1 ξ m 2 G ( ξ 1 , s ) h ( s ) d s + μ ξ 1 ( ξ m 2 ξ 1 ) 1 i = 1 m 2 k i ξ i ) u > u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equat_HTML.gif
Thus, T u > u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq141_HTML.gif and it follows from Lemma 1.3 that
i ( T , K R , K ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equau_HTML.gif
By the additivity of the fixed-point index,
0 = i ( T , K R , K ) = i ( T , K D , K ) + i ( T , K R K D ¯ , K ) = 1 + i ( T , K R K D ¯ , K ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equav_HTML.gif
which yields
i ( T , K R K D ¯ , K ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equaw_HTML.gif

Hence, T has at least one fixed point in K D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq142_HTML.gif and another one in K R K D ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq143_HTML.gif; this shows that in Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq30_HTML.gif, (1) has at least two positive solution. □

Example Consider the following boundary value problem:
{ u ( t ) = λ ( u + 1 ) 2 , u ( 0 ) = 0 , u ( 1 ) i = 1 m 2 k i u ( ξ i ) = μ 0 1 ( u ( s ) + 2 ) 3 d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_Equ5_HTML.gif
(5)

where f ( u ) = ( u + 1 ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq144_HTML.gif, g ( u ) = ( u + 2 ) 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq145_HTML.gif, and h ( t ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-87/MediaObjects/13661_2012_Article_199_IEq146_HTML.gif.

Declarations

Acknowledgements

The authors would like to thank the referees for valuable comments and suggestions for improving this paper. The first author is supported financially by the Fundamental Research Funds for the Central Universities.

Authors’ Affiliations

(1)
College of Science, Hohai University
(2)
Department of Mathematics, Nanjing University of Aeronautics and Astronautics

References

  1. Kong L, Kong Q: Higher order boundary value problems with nonhomogeneous boundary conditions. Nonlinear Anal. 2010, 72: 240-261. 10.1016/j.na.2009.06.050MathSciNetView Article
  2. Kong L, Wong JSW: Positive solutions for higher order multi-point boundary value problems with nonhomogeneous boundary conditions. J. Math. Anal. Appl. 2010, 367: 588-611. 10.1016/j.jmaa.2010.01.063MathSciNetView Article
  3. Ma R: Positive solutions for nonhomogeneous m-point boundary value problems. Comput. Math. Appl. 2004, 47: 689-698. 10.1016/S0898-1221(04)90056-9MathSciNetView Article
  4. Yang X: Existence of positive solutions for 2m-order nonlinear differential systems. Nonlinear Anal. 2005, 61: 77-95. 10.1016/j.na.2004.11.013MathSciNetView Article
  5. Hao X, Liu L, Wu Y: On positive solutions of an m-point nonhomogeneous singular boundary value problem. Nonlinear Anal. 2010, 73: 2532-2540. 10.1016/j.na.2010.06.028MathSciNetView Article
  6. Guo D, Lakshmikantham V: Nonlinear Problems in Abstract Cones. Academic Press, New York; 1988.

Copyright

© Wang and An; licensee Springer 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.