Open Access

Positive solutions for singular boundary value problems involving integral conditions

Boundary Value Problems20122012:72

DOI: 10.1186/1687-2770-2012-72

Received: 20 December 2011

Accepted: 1 June 2012

Published: 5 July 2012

Abstract

We are interested in the following singular boundary value problem:

{ u ( t ) + μ w ( t ) f ( u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equa_HTML.gif

where μ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq1_HTML.gif is a parameter and 0 1 u ( s ) d A ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq2_HTML.gif is the Stieltjes integral. The function w C ( ( 0 , 1 ) , ( 0 , + ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq3_HTML.gif and w may be singular at t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq4_HTML.gif and/or t = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq5_HTML.gif, f C ( [ 0 , + ) , ( 0 , + ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq6_HTML.gif and f = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq7_HTML.gif. Some a priori estimates and the existence, multiplicity and nonexistence of positive solutions are obtained. Our proofs are based on the method of global continuous theorem, the lower-upper solutions methods and fixed point index theory. Furthermore, we also discuss the interval of parameter μ such that the problem has a positive solution.

Keywords

singularity global continuous theorem solution of boundedness fixed point index positive solution

1 Introduction

We are concerned with the second order nonlocal boundary value problem:
{ u ( t ) + μ w ( t ) f ( u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ1_HTML.gif
(1.1)

where μ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq1_HTML.gif is a parameter and 0 1 u ( s ) d A ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq2_HTML.gif is a Stieltjes integral. The function w C ( ( 0 , 1 ) , ( 0 , + ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq3_HTML.gif and w may be singular at t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq4_HTML.gif and/or t = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq5_HTML.gif, f C ( [ 0 , + ) , ( 0 , + ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq6_HTML.gif and f = lim u f ( u ) u = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq8_HTML.gif.

Integral boundary conditions and multi-point boundary conditions for differential equations come from many areas of applied mathematics and physics [17]. Recently, singular boundary value problems have been extensively considered in a lot of literature [1, 2, 5, 8], since they model many physical phenomena including gas diffusion through porous media, nonlinear diffusion generated by nonlinear sources, chemically reacting systems as well as concentration in chemical or biological problems. In all these problems, positive solutions are very meaningful.

In [1, 2], Webb and Infante considered the existence of positive solutions of nonlinear boundary value problem:
u ( t ) + μ h ( t ) g ( t , u ( t ) ) = 0 , 0 < t < 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equb_HTML.gif
where h g are nonnegative functions, subjected to the nonlocal boundary conditions
u ( 0 ) = 0 , u ( 1 ) = α [ u ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equc_HTML.gif
Here α [ u ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq9_HTML.gif is a linear functional on C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq10_HTML.gif given by
α [ u ] = 0 1 u ( s ) d A ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equd_HTML.gif

involving a Stieltjes integral with a signed measure, that is, A has bounded variation. They dealt with many boundary conditions given in the literature in a unified way by utilizing the fixed point index theory in cones.

Recently, many researchers were interested in the global structure of positive solutions for the nonlinear boundary value problem (see, e.g., [3, 6, 7]). In 2009, Ma and An [3] considered the problem (1.1). Assume that

(A0) A : [ 0 , 1 ] R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq11_HTML.gif is nondecreasing and A ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq12_HTML.gif is not a constant on ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq13_HTML.gif, 0 κ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq14_HTML.gif with κ : = 0 1 t d A ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq15_HTML.gif, and 0 1 G ( t , s ) d A ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq16_HTML.gif for s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq17_HTML.gif (for the definition of G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq18_HTML.gif, see (2.1) below).

(A1) w : ( 0 , 1 ) [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq19_HTML.gif is continuous and w ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq20_HTML.gif on any subinterval of [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq21_HTML.gif, and w Φ L 1 [ 0 , 1 ] C ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq22_HTML.gif, where Φ ( s ) = s ( 1 s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq23_HTML.gif, s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq24_HTML.gif.

(A2) f C ( [ 0 , + ) , [ 0 , + ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq25_HTML.gif and f ( s ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq26_HTML.gif for s > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq27_HTML.gif.

(A3) f 0 = f = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq28_HTML.gif, where f 0 = lim u 0 + f ( u ) u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq29_HTML.gif and f = lim u f ( u ) u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq30_HTML.gif.

They obtained the following main result:

Theorem 1.1 ([3], Theorem 4.1])

Assume that (A 0)-(A 3) hold. Then there exists a component T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq31_HTML.gifinwhich joins ( , θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq32_HTML.gifto ( , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq33_HTML.gif, and
Proj R T = [ ρ , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Eque_HTML.gif
for some ρ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq34_HTML.gif. Moreover, there exists μ ρ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq35_HTML.gifsuch that (1.1) has at least two positive solutions for μ ( μ , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq36_HTML.gif. Here T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq31_HTML.gifjoins ( , θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq37_HTML.gifto ( , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq38_HTML.gifsuch that
lim ( μ , u ) T , u 1 μ u = 0 , lim ( μ , u ) T , u > 1 μ u = . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equf_HTML.gif

Here, ∑ is the closure of the set of positive solutions of (1.1) on [ 0 , ) × X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq39_HTML.gif, X = { u C 1 [ 0 , 1 ] : u ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d s } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq40_HTML.gif, and the component of a set M is a maximal connected subset of M.

A natural problem arises: How can we consider the global structure of positive solutions for the case f 0 = f = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq41_HTML.gif?

In this paper, we first obtain the global structure of positive solutions by the use of global continuous theorem, and some a priori estimates. Applying the analysis technique, we construct the lower and upper solutions. Those combined with the fixed point index theory, the existence, multiplicity and nonexistence of positive solutions to (1.1) in the case f 0 = f = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq41_HTML.gif are investigated. Finally, we discuss the interval of parameter μ such that the problem (1.1) has positive solutions. The proof of the method which is based on the construction of some bounds of the solution together with global continuous theorem and fixed point index is of independent interest, and is different from the other papers.

This paper is arranged as follows. We will give some hypotheses and lemmas in Section 2. In Section 3, new criteria of the existence, multiplicity and nonexistence of a positive solution are obtained. Moreover, an example is given to illustrate our result.

2 Preliminaries and lemmas

Let X denote the Banach space C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq10_HTML.gif with the maximum norm
u = max t [ 0 , 1 ] | u ( t ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equg_HTML.gif
Define K : = { u C [ 0 , 1 ] : u is concave in [ 0 , 1 ] and u ( t ) 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq42_HTML.gif, then K is a cone. Let
G ( t , s ) = { ( 1 t ) s , 0 s t 1 , ( 1 s ) t , 0 t s 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ2_HTML.gif
(2.1)
Denote
κ = 0 1 t d A ( t ) , G ( s ) : = 0 1 G ( t , s ) d A ( t ) , for s [ 0 , 1 ] , G 1 ( t , s ) = G ( t , s ) + t 1 κ G ( s ) , κ 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ3_HTML.gif
(2.2)

Throughout this paper, we suppose that the following conditions hold:

(H0) A : [ 0 , 1 ] R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq11_HTML.gif is nondecreasing, d A ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq43_HTML.gif on ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq13_HTML.gif and κ : = 0 1 t d A ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq15_HTML.gif with 0 κ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq14_HTML.gif.

(H1) w C ( ( 0 , 1 ) , ( 0 , + ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq3_HTML.gif and w may be singular at t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq4_HTML.gif and/or 1, satisfying
0 < 0 1 G 1 ( t , s ) w ( s ) d s < , t ( 0 , 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ4_HTML.gif
(2.3)

(H2) f C ( [ 0 , ) , ( 0 , ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq44_HTML.gif ( f ( 0 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq45_HTML.gif obviously holds).

(H3) f = lim u f ( u ) u = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq46_HTML.gif.

Remark 2.1 It is easy to see from (H0) that G ( s ) [ A ( 1 ) A ( 0 ) ] G ( s , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq47_HTML.gif. Therefore, if we assume that 0 < 0 1 G ( s , s ) w ( s ) d s < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq48_HTML.gif, then (2.3) obviously holds.

Define an operator T : K X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq49_HTML.gif as follows:
T u ( t ) = T ( μ , u ) ( t ) : = μ 0 1 G 1 ( t , s ) w ( s ) f ( u ( s ) ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equh_HTML.gif

Assume that the conditions (H0)-(H2) hold, then it is easy to verify that T : K K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq50_HTML.gif is well defined and completely continuous.

Lemma 2.1 ([9] Global continuation theorem)

Let X be a Banach space and let K be an order cone in X. Consider the equation
u = T ( μ , u ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ5_HTML.gif
(2.4)

where μ R + = [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq51_HTML.gifand u X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq52_HTML.gif. Suppose that T : R + × K K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq53_HTML.gifis completely continuous and T ( 0 , u ) = θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq54_HTML.giffor all u K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq55_HTML.gif, then L + ( K ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq56_HTML.gif, the component of the solution set of (2.4) containing ( 0 , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq57_HTML.gifis unbounded.

Remark 2.2
  1. (1)

    We note that u is a positive solution of the problem (1.1) if and only if u = T ( μ , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq58_HTML.gif on K.

     
  2. (2)

    If T ( μ , θ ) θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq59_HTML.gif for μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq60_HTML.gif and T ( 0 , u ) = θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq61_HTML.gif for all u K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq55_HTML.gif, then we get from Lemma 2.1 that there exists an unbounded continuum L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq62_HTML.gif emanating from ( 0 , θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq63_HTML.gif in the closure of the set of positive solutions (1.1) in R + × K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq64_HTML.gif.

     

Lemma 2.2 ([10])

Let X be a Banach space, K an order cone in X and S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq65_HTML.gifan open bounded set in X with θ S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq66_HTML.gif. Suppose that T : S ¯ K K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq67_HTML.gifis a completely continuous operator. If T y λ y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq68_HTML.gif, for all y S K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq69_HTML.gifand all λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq70_HTML.gif, then i ( T , S K , K ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq71_HTML.gif.

3 Main results

Lemma 3.1 Let (H 0)-(H 3) hold and let J = [ μ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq72_HTML.gifwith μ 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq73_HTML.gif. Then there exists a constant Q J > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq74_HTML.gifsuch that for all μ J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq75_HTML.gifand all possible positive solutions u μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq76_HTML.gifof (1.1), the inequality u μ < Q J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq77_HTML.gifholds.

Proof Suppose on the contrary that there exist a sequence { μ n } J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq78_HTML.gif and a sequence { u μ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq79_HTML.gif of the positive solutions of (1.1) corresponding to μ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq80_HTML.gif such that
u μ n , as n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equi_HTML.gif
Denote u n : = u μ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq81_HTML.gif. From the concavity of u n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq82_HTML.gif, it follows that
u n ( t ) u n 4 > 0 , for t [ 1 4 , 3 4 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ6_HTML.gif
(3.1)
Choose η : = 10 π 2 μ 0 w https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq83_HTML.gif, where w = min t [ 1 4 , 3 4 ] w ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq84_HTML.gif. Then we find from (H3) that there exists a constant R > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq85_HTML.gif such that
f ( u ) > η u , for all u > R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ7_HTML.gif
(3.2)
Since lim n u n = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq86_HTML.gif, we get
u N > 4 R , for enough large N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equj_HTML.gif
This, together with (3.1) and (3.2), implies
f ( u N ( t ) ) > η u N ( t ) , for t ( 1 4 , 3 4 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ8_HTML.gif
(3.3)
Put ψ ( t ) : = cos 2 π t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq87_HTML.gif, for t [ 1 4 , 3 4 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq88_HTML.gif. Hence, we know from (3.3) that
μ N 1 4 3 4 w ( t ) f ( u N ( t ) ) ψ ( t ) d t η μ N 1 4 3 4 w ( t ) u N ( t ) ψ ( t ) d t η w μ N 1 4 3 4 u N ( t ) ψ ( t ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equk_HTML.gif
On the other hand, multiplying (1.1) by ψ and integrating by parts, we obtain that
μ N 1 4 3 4 w ( t ) f ( u N ( t ) ) ψ ( t ) d t = 1 4 3 4 u N ( t ) ψ ( t ) d t = u N ( t ) ψ ( t ) | 1 4 3 4 + 1 4 3 4 u N ( t ) ψ ( t ) d t = 1 4 3 4 u N ( t ) ψ ( t ) d t + u N ( t ) ψ ( t ) | 1 4 3 4 4 π 2 1 4 3 4 u N ( t ) ψ ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equl_HTML.gif
leads to
η w μ N 1 4 3 4 u N ( t ) ψ ( t ) d t 4 π 2 1 4 3 4 u N ( t ) ψ ( t ) ] d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equm_HTML.gif
i.e.,
10 π 2 μ 0 w = η 4 π 2 μ N w < 10 π 2 μ 0 w . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equn_HTML.gif

This is a contradiction. □

Lemma 3.2 Assume that the hypotheses (H 0)-(H 3) hold. Then there exists ρ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq89_HTML.gifsuch that if the problem (1.1) has a positive solution for parameter μ, then μ ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq90_HTML.gif.

Proof Let u be a positive solution of (1.1) corresponding to μ. From the hypotheses (H2) and (H3), it follows that there exists ϱ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq91_HTML.gif such that f ( u ) ϱ u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq92_HTML.gif for all u > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq93_HTML.gif. Consequently, we have
μ ϱ w ( t ) u ( t ) μ w ( t ) f ( u ( t ) ) = u ( t ) , t ( 0 , 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ9_HTML.gif
(3.4)
Let λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq94_HTML.gif be the first eigenvalue of
{ φ ( t ) + λ w ( t ) φ ( t ) = 0 , t ( 0 , 1 ) , φ ( 0 ) = φ ( 1 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equo_HTML.gif
and let φ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq95_HTML.gif be the positive eigenfunction corresponding to λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq94_HTML.gif (see [8]). It is easy to see that φ 1 ( 1 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq96_HTML.gif. Multiplying (3.4) by φ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq95_HTML.gif and integrating by parts, we get that
μ ϱ 0 1 w ( t ) u ( t ) φ 1 ( t ) d t 0 1 u ( t ) φ 1 ( t ) d t = φ 1 ( t ) u ( t ) | 0 1 + 0 1 u ( t ) φ 1 ( t ) d t = u ( 1 ) φ 1 ( 1 ) u ( 0 ) φ 1 ( 0 ) 0 1 u ( t ) φ 1 ( t ) d t 0 1 u ( t ) φ 1 ( t ) d t = λ 1 0 1 w ( t ) u ( t ) φ 1 ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equp_HTML.gif
implies
μ λ 1 ϱ 1 ρ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equq_HTML.gif

This completes the proof. □

Lemma 3.3 Assume that (H 0)-(H 3) hold. Then we have
lim ( μ , u μ ) L u μ 1 , μ 0 u μ = . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equr_HTML.gif
Proof Suppose this fails, that is, there exists { ( μ , u μ ) } L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq97_HTML.gif such that
μ 0 and 1 u μ M 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equs_HTML.gif
where M 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq98_HTML.gif is a positive constant. Since ( μ , u μ ) L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq99_HTML.gif, we get that for all t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq100_HTML.gif
u μ ( t ) = μ 0 1 G 1 ( t , s ) w ( s ) f ( u μ ( s ) ) d s = ( μ 0 1 G 1 ( t , s ) w ( s ) f ( u μ ( s ) ) u μ d s ) u μ μ ( M 1 max t [ 0 , 1 ] 0 1 G 1 ( t , s ) w ( s ) d s ) u μ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equt_HTML.gif
where
M 1 = max { f ( u μ ( t ) ) u μ : 1 u μ M 0 , t [ 0 , 1 ] } ( 0 , ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equu_HTML.gif
Hence, we find that
μ { M 1 max t [ 0 , 1 ] 0 1 G 1 ( t , s ) w ( s ) d s } 1 > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equv_HTML.gif

Thus, it implies that μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq101_HTML.gif. This is contradiction. □

Theorem 3.1 Assume that the conditions (H 0)-(H 3) hold and 0 1 d A ( t ) < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq102_HTML.gif. Then there exists a constant μ 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq103_HTML.gifsuch that the problem (1.1) has at least two positive solutions for 0 < μ < μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq104_HTML.gif, and at least one positive solution for μ = μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq105_HTML.gif, and no positive solution for μ > μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq106_HTML.gif.

Proof Define
μ 1 : = sup { μ > 0 : for all μ ( 0 , μ ) , there exist at least two positive solutions of (1.1) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ10_HTML.gif
(3.5)

From Lemma 2.1 and Remark 2.2, we can find that there exists an unbounded continuum L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq62_HTML.gif emanating from ( 0 , θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq63_HTML.gif in the closure of the set of positive solutions in R + × K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq64_HTML.gif and T ( 0 , u ) = θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq54_HTML.gif for all u K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq55_HTML.gif. Meanwhile, Lemma 3.1 and Lemma 3.3 respectively imply that u μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq76_HTML.gif is bounded ( μ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq1_HTML.gif, μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq101_HTML.gif) and unbounded ( μ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq1_HTML.gif, μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq107_HTML.gif and u μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq108_HTML.gif). Therefore, we conclude that the set of (3.5) is nonempty. Those combined with Lemma 3.2 follows that μ 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq103_HTML.gif is well defined and μ 1 ( 0 , ρ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq109_HTML.gif. From the definition of μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq110_HTML.gif, it is easy to see that the problem (1.1) has at least two positive solutions for μ ( 0 , μ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq111_HTML.gif. Again, since the continuum is a compact connected set and T is a completely continuous operator, the problem (1.1) has at least one positive solution at μ = μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq105_HTML.gif.

Next, we only show that the problem (1.1) has no positive solution for any μ > μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq106_HTML.gif. Suppose on the contrary that there exists some μ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq112_HTML.gif ( > μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq113_HTML.gif) such that the problem (1.1) has a positive solution u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq114_HTML.gif corresponding to μ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq112_HTML.gif. Then we will prove that the problem (1.1) has at least two positive solutions for any μ [ μ 1 , μ 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq115_HTML.gif which contradicts the definition of (3.5).

For the sake of obtaining the contradiction, we divide the proof into four steps.

Step 1. Constructing a modified boundary value problem.

Choose arbitrarily a constant μ [ μ 1 , μ 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq116_HTML.gif. Since f is uniformly continuous on [ 0 , u 2 + 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq117_HTML.gif, there exists a constant σ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq118_HTML.gif such that
f ( u + σ ) < f ( u ) + ϵ , for u [ 0 , u 2 + 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ11_HTML.gif
(3.6)
where
ϵ : = ( μ 2 μ ) min u [ 0 , u 2 + 1 ] f ( u ) 2 μ > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equw_HTML.gif
Denote ζ ( t ) : = u 2 ( t ) + σ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq119_HTML.gif, for t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq120_HTML.gif. Then we claim that
ζ ( t ) + μ w ( t ) f ( ζ ( t ) ) < 0 , t ( 0 , 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ12_HTML.gif
(3.7)
and
ζ ( 0 ) = σ , ζ ( 1 ) = 0 1 u 2 ( s ) d A ( s ) + σ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ13_HTML.gif
(3.8)
Indeed, using (3.6), we have
ζ ( t ) + μ w ( t ) f ( ζ ( t ) ) = u 2 ( t ) + μ w ( t ) f ( u 2 ( t ) + σ ) = μ 2 w ( t ) f ( u 2 ( t ) ) + μ w ( t ) f ( u 2 ( t ) + σ ) < w ( t ) [ μ 2 f ( u 2 ( t ) ) + μ f ( u 2 ( t ) ) + μ ϵ ] μ 2 μ 2 × w ( t ) f ( u 2 ( t ) ) < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equx_HTML.gif
Define a set S : = { u C [ 0 , 1 ] | 0 < u ( t ) < ζ ( t ) , t ( 0 , 1 ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq121_HTML.gif. Then the set S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq65_HTML.gif is bounded and open in C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq10_HTML.gif. Now, we construct the modified second-order boundary value problem:
{ u ( t ) + μ w ( t ) f ( ξ ( u ( t ) ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ14_HTML.gif
(3.9)
where ξ : R R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq122_HTML.gif is defined by
ξ ( u ( t ) ) = { ζ ( t ) , if u ( t ) > ζ ( t ) , u ( t ) , if 0 u ( t ) ζ ( t ) , 0 , if u ( t ) < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equy_HTML.gif

Step 2. We will show that if u is a positive solution of (3.9), then u S K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq123_HTML.gif.

Let u be a positive solution of (3.9), then we claim that
u S K . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ15_HTML.gif
(3.10)
Suppose this fails, that is, u S K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq124_HTML.gif. Clearly, we only show that u ( t ) ζ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq125_HTML.gif, for t ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq126_HTML.gif. Comparing the boundary conditions (3.8) and (3.9), the only following three cases need to be considered. Case I. There exists t 1 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq127_HTML.gif such that u ( t 1 ) = ζ ( t 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq128_HTML.gif and 0 < u ( t ) < ζ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq129_HTML.gif, for t ( t 1 σ 1 , t 1 ) ( t 1 , t 1 + σ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq130_HTML.gif and some σ 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq131_HTML.gif; Case II. There exists t 2 ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq132_HTML.gif such that u ( t 2 ) = ζ ( t 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq133_HTML.gif and u ( t ) ζ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq134_HTML.gif for t ( t 2 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq135_HTML.gif. Case III. There exists [ t 3 , t 4 ] ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq136_HTML.gif such that u ( t ) ζ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq137_HTML.gif, t [ t 3 , t 4 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq138_HTML.gif, u ( t 3 ) = ζ ( t 3 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq139_HTML.gif and u ( t 4 ) = ζ ( t 4 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq140_HTML.gif. See the three Figures 1, 2 and 3.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_fig1.jpg
Figure 1

Case I.

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_fig2.jpg
Figure 2

Case II.

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_fig3.jpg
Figure 3

Case III.

Case I. From (3.7), it implies that there exists a constant ϵ 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq141_HTML.gif such that
max t [ t 1 σ 1 , t 1 + σ 1 ] { ζ ( t ) + μ w ( t ) f ( ζ ( t ) ) } = ϵ 1 < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ16_HTML.gif
(3.11)
Since f is uniformly continuous on [ 0 , u 2 + 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq117_HTML.gif, there exists a σ 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq142_HTML.gif such that if u , v [ 0 , ζ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq143_HTML.gif and | u v | < σ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq144_HTML.gif, then we get
| f ( u ) f ( v ) | < ϵ 1 W 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ17_HTML.gif
(3.12)
where W : = μ max t [ t 1 σ 1 , t 1 + σ 1 ] w ( t ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq145_HTML.gif. From the assumption of Case I, it follows that there exists a subinterval [ r , s ] ( t 1 σ 1 , t 1 + σ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq146_HTML.gif with t 1 ( r , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq147_HTML.gif such that
σ 1 < u ( t ) ζ ( t ) 0 , for t [ r , s ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equz_HTML.gif
and
( u ζ ) ( r ) > 0 and ( u ζ ) ( s ) < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equaa_HTML.gif
This together with (3.11) and (3.12) leads to
0 > ( u ζ ) ( s ) ( u ζ ) ( r ) = r s [ u ( t ) ζ ( t ) ] d t = r s [ μ w ( t ) f ( u ( t ) ) + ζ ( t ) ] d t = r s [ μ w ( t ) ( f ( u ( t ) ) f ( ζ ( t ) ) ) + ζ ( t ) + μ w ( t ) f ( ζ ( t ) ) ] d t r s [ μ w ( t ) ϵ 1 W 1 ϵ 1 ] d t = ε 1 r s [ 1 μ w ( t ) W ] d t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equab_HTML.gif

This is a contradiction.

Case II. Let x ( t ) : = u ( t ) ζ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq148_HTML.gif, for t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq120_HTML.gif. Then we have that for t ( t 2 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq135_HTML.gif
x ( t 2 ) = 0 , x ( t ) 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ18_HTML.gif
(3.13)
x ( t ) = ( u ( t ) ζ ( t ) ) = μ w ( t ) f ( ζ ( t ) ) ζ ( t ) > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ19_HTML.gif
(3.14)
Obviously, we have x ( t 2 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq149_HTML.gif. From (3.13), it implies that x ( t ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq150_HTML.gif, for any t ( t 2 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq151_HTML.gif. Hence, the function x ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq152_HTML.gif is strictly increasing in [ t 2 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq153_HTML.gif. From (3.8) and the boundary condition (3.9), we find
x ( 1 ) = u ( 1 ) ζ ( 1 ) = 0 1 u ( s ) d A ( s ) 0 1 u 2 ( s ) d A ( s ) σ = 0 1 ( u ( s ) ζ ( s ) ) d A ( s ) + σ ( 0 1 d A ( s ) 1 ) = 0 1 x ( s ) d A ( s ) σ ( 1 0 1 d A ( s ) ) < x ( 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equac_HTML.gif
We obtain a contradiction. In particular, if t 2 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq154_HTML.gif, then we obtain
x ( 1 ) = u ( 1 ) ζ ( 1 ) = 0 1 x ( s ) d A ( s ) + σ ( 0 1 d A ( s ) 1 ) < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equad_HTML.gif

This contradicts (3.13).

Case III. Since ξ ( u ( t ) ) = ζ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq155_HTML.gif, for t ( t 3 , t 4 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq156_HTML.gif, we have that ( u ( t ) ζ ( t ) ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq157_HTML.gif, for t ( t 3 , t 4 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq156_HTML.gif. Again since ( u ζ ) ( t 3 ) = ( u ζ ) ( t 4 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq158_HTML.gif, we know from the maximum principle that u ( t ) < ζ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq159_HTML.gif, for t ( t 3 , t 4 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq160_HTML.gif. This contradicts the assumption of Case III.

Therefore, we conclude that the claim (3.10) holds.

Step 3. i ( T ˜ μ , S K , K ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq161_HTML.gif (the definition of T ˜ μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq162_HTML.gif see below).

Using (2.2), we define an operator T ˜ μ : K X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq163_HTML.gif by
T ˜ μ ( u ) ( t ) = μ 0 1 G 1 ( t , s ) w ( s ) f ( ξ ( u ( s ) ) ) d s , t [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equae_HTML.gif
Then T ˜ μ : K K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq164_HTML.gif is completely continuous and u is a positive solution of (3.9) if and only if u = T ˜ μ ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq165_HTML.gif on K. From the definition of ξ ( u ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq166_HTML.gif, it implies that there exists R 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq167_HTML.gif such that T ˜ μ u < R 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq168_HTML.gif, for all u K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq55_HTML.gif. Consequently, we get from Lemma 2.2 that
i ( T ˜ μ , B R 1 K , K ) = 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equaf_HTML.gif
where B R 1 = { u X : u < R 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq169_HTML.gif. Applying the conclusion of Step 2 and the excision property of fixed point index, we find that
i ( T ˜ μ , S K , K ) = i ( T ˜ μ , B R 1 K , K ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ20_HTML.gif
(3.15)

Step 4. We conclude that the problem (1.1) has at least two positive solutions corresponding to μ.

Since the problem (1.1) is equivalent to the problem (3.9) on S K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq170_HTML.gif, we get that the problem (1.1) has a positive solution in S K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq170_HTML.gif. Without loss of generality, we may suppose that T has no fixed point on S K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq171_HTML.gif (otherwise the proof is completed). Then i ( T , S K , K ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq172_HTML.gif is well defined and (3.15) implies
i ( T , S K , K ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ21_HTML.gif
(3.16)
On the other hand, from Lemma 3.2, we choose μ > ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq173_HTML.gif such that the problem (1.1) has no positive solution in K. By a priori estimate in J = [ μ , μ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq174_HTML.gif, there exists R 2 ( > R 1 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq175_HTML.gif such that for all possible positive solutions u λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq176_HTML.gif of (1.1) with λ [ μ , μ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq177_HTML.gif, we know that u λ < R 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq178_HTML.gif. Define G : [ 0 , 1 ] × ( B ¯ R 2 K ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq179_HTML.gif by
G ( ν , u ) = T ( ν μ + ( 1 ν ) μ , u ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equag_HTML.gif
Then it is easy to verify that G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq180_HTML.gif is completely continuous on [ 0 , 1 ] × K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq181_HTML.gif and G ( ν , u ) u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq182_HTML.gif for all ( ν , u ) [ 0 , 1 ] × ( B R 2 K ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq183_HTML.gif. From the property of homotopy invariance, it follows that
i ( T , B R 2 K , K ) = i ( T ( μ , ) , B R 2 K , K ) = i ( T ( μ , ) , B R 2 K , K ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equah_HTML.gif
Hence, by the additivity property and (3.16), we have
i ( T , ( B R 2 S ¯ ) K , K ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equai_HTML.gif

Then we conclude that the problem (1.1) has at least two positive solutions corresponding to μ. □

Remark 3.1
  1. (i)
    From the hypotheses (H2) and (H3), it implies that there exists L > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq184_HTML.gif such that
    D = f ( L ) L = min u > 0 f ( u ) u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ22_HTML.gif
    (3.17)
     
Let f attain its maximum at the point L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq185_HTML.gif of [ 0 , L ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq186_HTML.gif. If 0 1 G 1 ( t , s ) w ( s ) d s < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq187_HTML.gif, then adopting the similar method as in [11], Theorem 1], we get that for 0 < μ < ( f ( L ) L 0 1 G 1 ( t , s ) w ( s ) d s ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq188_HTML.gif, the problem (1.1) has at least two positive solutions u 1 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq189_HTML.gif and u 2 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq190_HTML.gif such that 0 < u 1 < L < u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq191_HTML.gif by the use of compression of conical shells in [12], Corollary 20.1]. Consequently, we know that μ 1 ( f ( L ) L 0 1 G 1 ( t , s ) w ( s ) d s ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq192_HTML.gif.
  1. (ii)
    If u is a positive solution of the equation (1.1) corresponding to μ, then we have
    u = max t [ 0 , 1 ] | u ( t ) | = μ max t [ 0 , 1 ] 0 1 G 1 ( t , s ) w ( s ) f ( u ( s ) ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equaj_HTML.gif
     
i.e., from (3.17),
1 = μ max t [ 0 , 1 ] 0 1 G 1 ( t , s ) w ( s ) f ( u ( s ) ) u d s μ D max t [ 0 , 1 ] 0 1 G 1 ( t , s ) w ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equak_HTML.gif

Therefore, we get that μ ( D max t [ 0 , 1 ] 0 1 G 1 ( t , s ) w ( s ) d s ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq193_HTML.gif.

Corollary 3.1 Assume that (H 1)-(H 3) hold. Consider the following m-point boundary value problem
{ u ( t ) + μ w ( t ) f ( u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = i = 1 m 2 α i u ( η i ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ23_HTML.gif
(3.18)

where μ is a positive parameter, α i [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq194_HTML.gif, i = 1 , 2 , , m 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq195_HTML.gif, 0 < η 1 < < η m 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq196_HTML.gifand 0 i = 1 m 2 α i η i < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq197_HTML.gif. Then there exists a constant μ ¯ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq198_HTML.gifsuch that the problem (3.18) has at least two positive solutions for 0 < μ < μ ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq199_HTML.gif, and at least one positive solution for μ = μ ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq200_HTML.gif, and no positive solution for μ > μ ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq201_HTML.gif.

Proof In the boundary condition of (1.1), if we let
A ( s ) = i = 1 m 2 α i χ ( s η i ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equal_HTML.gif
where χ ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq202_HTML.gif is the characteristic function on [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq203_HTML.gif, i.e.,
χ ( s ) = { 1 , if s 0 , 0 , if s < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equam_HTML.gif

Then the boundary condition of (1.1) reduces to the m-point boundary condition of (3.18). Applying the method of Theorem 3.1, we get the conclusion. □

Example 3.1 We consider the following singular boundary value problem:
{ u ( t ) + μ 1 t 3 ( 2 + sin u ( t ) + e u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equ24_HTML.gif
(3.19)

where A ( s ) = s 2 ( 2 s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq204_HTML.gif.

Computing yields
κ = 0 1 t d A ( t ) = 0 1 t d ( 2 t 2 t 3 ) = 7 12 , G ( s ) = 0 1 G ( t , s ) d A ( t ) G ( s ) = 0 s ( 1 s ) t ( 4 t 3 t 2 ) d t + s 1 ( 1 t ) ( 4 t 3 t 2 ) s d t G ( s ) = 5 12 s 2 3 s 3 + 1 4 s 4 , G 1 ( t , s ) = G ( t , s ) + t 7 ( 5 s 8 s 3 + 3 s 4 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equan_HTML.gif
We find that for any t ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq205_HTML.gif,
0 < 0 1 G 1 ( t , s ) w ( s ) d s = 0 1 G ( t , s ) s 3 d s + t 7 0 1 s 2 3 ( 5 8 s 2 + 3 s 3 ) d s = t ( 2 , 988 2 , 695 9 10 t 2 ) < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_Equao_HTML.gif

Thus, it implies that (2.3) holds. It is easy to verify that the conditions (H2) and (H3) hold. Therefore, by Theorem 3.1, we obtain that there exists a constant μ 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq103_HTML.gif such that the problem (3.19) has at least two positive solutions for 0 < μ < μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq104_HTML.gif, and at least one positive solution for μ = μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq105_HTML.gif, and no positive solution for μ > μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-72/MediaObjects/13661_2011_Article_187_IEq106_HTML.gif.

Author’s contributions

The author typed, read and approved the final manuscript.

Declarations

Acknowledgement

The author would like to thank the anonymous referees very much for helpful comments and suggestions which led to the improvement of presentation and quality of the work. The work was supported partly by NSCF of Tianyuan Youth Foundation (No. 11126125), K. C. Wong Magna Fund of Ningbo University, Subject Foundation of Ningbo University (No. xkl11044) and Hulan’s Excellent Doctor Foundation of Ningbo University.

Authors’ Affiliations

(1)
Department of Mathematics, Ningbo University

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