Bifurcation from interval and positive solutions for a class of fourth-order two-point boundary value problem

Boundary Value Problems20132013:170

DOI: 10.1186/1687-2770-2013-170

Received: 14 November 2012

Accepted: 8 July 2013

Published: 22 July 2013

Abstract

We consider the fourth-order two-point boundary value problem x ′′′′ + k x + l x = f ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq1_HTML.gif, 0 < t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq2_HTML.gif, x ( 0 ) = x ( 1 ) = x ( 0 ) = x ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq3_HTML.gif, which is not necessarily linearizable. We give conditions on the parameters k, l and f ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq4_HTML.gif that guarantee the existence of positive solutions. The proof of our main result is based upon topological degree theory and global bifurcation techniques.

MSC:34B15.

Keywords

topological degree fourth-order ordinary differential equation bifurcation positive solution eigenvalue

1 Introduction

The deformations of an elastic beam in an equilibrium state with fixed both endpoints can be described by the fourth-order boundary value problem
x ′′′′ + l x = λ h ( t ) f ( x ) , 0 < t < 1 , x ( 0 ) = x ( 1 ) = x ( 0 ) = x ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ1_HTML.gif
(1.1)

where f : R R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq5_HTML.gif is continuous, λ R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq6_HTML.gif is a parameter and l is a given constant. Since problem (1.1) cannot transform into a system of second-order equations, the treatment method of the second-order system does not apply to it. Thus, the existing literature on problem (1.1) is limited. When l = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq7_HTML.gif, the existence of positive solutions of problem (1.1) has been studied by several authors, see [15]. Especially, when l 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq8_HTML.gif, Xu and Han [6] studied the existence of nodal solutions of problem (1.1) by applying disconjugate operator theory and bifurcation techniques.

Recently, motivated by [6], when k, l satisfy (A1), Shen [7] studied the existence of nodal solutions of a general fourth-order boundary value problem by applying disconjugate operator theory [8, 9] and Rabinowitz’s global bifurcation theorem
x ′′′′ + k x + l x = f ( t , x ) , 0 < t < 1 , x ( 0 ) = x ( 1 ) = x ( 0 ) = x ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ2_HTML.gif
(1.2)

where

(A1) one of following conditions holds:
  1. (i)
    k, l satisfying ( k , l ) { ( k , l ) | k ( , 0 ] , l ( 0 , ) } { ( 0 , π 4 64 ) } { ( k , l ) | k ( , π 2 ) , l ( , 0 ] } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq9_HTML.gif are given constants with
    π 2 ( k π 2 ) < l 1 4 ( k π 2 4 ) 2 ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ3_HTML.gif
    (1.3)
     
  2. (ii)
    k, l satisfying ( k , l ) { ( k , l ) | k ( 0 , π 2 2 ) , l ( 0 , ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq10_HTML.gif are given constants with
    1 4 ( π 2 k π 4 4 ) < l 1 4 k 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ4_HTML.gif
    (1.4)
     

In this paper, we consider bifurcation from interval and positive solutions for problem (1.2). In order to prove our main result, condition (A1) and the following weaker conditions are satisfied throughout this paper:

(H1) f : [ 0 , 1 ] × [ 0 , ) [ 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq11_HTML.gif is continuous and there exist functions a 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq12_HTML.gif, a 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq13_HTML.gif, b ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq14_HTML.gif, and b ( t ) C ( [ 0 , 1 ] , [ 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq15_HTML.gif such that
a 0 ( t ) x ξ 1 ( t , x ) f ( t , x ) a 0 ( t ) x + ξ 2 ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ5_HTML.gif
(1.5)
for some functions ξ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq16_HTML.gif, ξ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq17_HTML.gif defined on [ 0 , 1 ] × [ 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq18_HTML.gif with
ξ 1 ( t , x ) = o ( x ) , ξ 2 ( t , x ) = o ( x ) as  x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ6_HTML.gif
(1.6)
uniformly for t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq19_HTML.gif, and
b ( t ) x ζ 1 ( t , x ) f ( t , x ) b ( t ) x + ζ 2 ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ7_HTML.gif
(1.7)
for some functions ζ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq20_HTML.gif, ζ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq21_HTML.gif defined on [ 0 , 1 ] × [ 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq18_HTML.gif with
ζ 1 ( t , x ) = o ( x ) , ζ 2 ( t , x ) = o ( x ) as  x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ8_HTML.gif
(1.8)

uniformly for t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq19_HTML.gif.

(H2) f ( t , x ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq22_HTML.gif for t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq19_HTML.gif and x ( 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq23_HTML.gif.

(H3) There exists a function c ( t ) C ( [ 0 , 1 ] , [ 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq24_HTML.gif with c ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq25_HTML.gif in any subinterval of [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq26_HTML.gif such that
f ( t , x ) c ( t ) x , ( t , x ) [ 0 , 1 ] × [ 0 , ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ9_HTML.gif
(1.9)

It is the purpose of this paper to study the existence of positive solutions of (1.2) under conditions (A1), (H1), (H2) and (H3). The main tool we use is the following global bifurcation theorem for the problem which is not necessarily linearizable.

Theorem A (Rabinowitz [10])

Let V be a real reflexive Banach space. Let F : R × V V http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq27_HTML.gif be completely continuous such that F ( λ , 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq28_HTML.gif, λ R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq29_HTML.gif. Let a , b R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq30_HTML.gif ( a < b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq31_HTML.gif) be such that u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq32_HTML.gif is an isolated solution of the following equation:
u F ( λ , u ) = 0 , u V http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ10_HTML.gif
(1.10)
for λ = a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq33_HTML.gif and λ = b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq34_HTML.gif, where ( a , 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq35_HTML.gif, ( b , 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq36_HTML.gif are not bifurcation points of (1.10). Furthermore, assume that
d ( I F ( a , ) , B r ( 0 ) , 0 ) d ( I F ( b , ) , B r ( 0 ) , 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ11_HTML.gif
(1.11)
where B r ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq37_HTML.gif is an isolating neighborhood of the trivial solution. Let
S = { ( λ , u ) : ( λ , u ) is a solution of  (1.10)  with u 0 } ¯ ( [ a , b ] × { 0 } ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equa_HTML.gif
Then there exists a continuum (i.e., a closed connected set) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq38_HTML.gif of http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq39_HTML.gif containing [ a , b ] × { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq40_HTML.gif, and either
  1. (i)

    http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq38_HTML.gif is unbounded in V × R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq41_HTML.gif, or

     
  2. (ii)

    C [ ( R [ a , b ] ) × { 0 } ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq42_HTML.gif.

     

Remark 1.1 For other results on the existence and multiplicity of positive solutions and nodal solutions for boundary value problems of fourth-order ordinary differential equations based on bifurcation techniques, see [1120].

2 Hypotheses and lemmas

Let
L [ x ] : = x ′′′′ + k x + l x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ12_HTML.gif
(2.1)

Theorem 2.1 (see [[7], Theorem 2.4])

Let (A1) hold. Then
  1. (i)
    L [ x ] = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq43_HTML.gif is disconjugate on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq26_HTML.gif, and L [ x ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq44_HTML.gif has a factorization
    L [ x ] : = ρ 4 ( ρ 3 ( ρ 2 ( ρ 1 ( ρ 0 x ) ) ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ13_HTML.gif
    (2.2)
     
where ρ k C 4 k [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq45_HTML.gif with ρ k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq46_HTML.gif ( k = 0 , 1 , 2 , 3 , 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq47_HTML.gif);
  1. (ii)
    x ( 0 ) = x ( 1 ) = x ( 0 ) = x ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq3_HTML.gif if and only if
    ( L 0 x ) ( 0 ) = ( L 0 x ) ( 1 ) = ( L 1 x ) ( 0 ) = ( L 1 x ) ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ14_HTML.gif
    (2.3)
     
where
L 0 x = ρ 0 x , L i x = ρ i ( L i 1 x ) , i = 1 , 2 , 3 , 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ15_HTML.gif
(2.4)

Theorem 2.2 (see [[7], Theorem 2.7])

Let (A1) hold and h C ( [ 0 , 1 ] , [ 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq48_HTML.gif with h ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq49_HTML.gif on any subinterval of [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq26_HTML.gif. Then
  1. (i)
    the problem
    { x ′′′′ ( t ) + k x ( t ) + l x ( t ) = λ h ( t ) x , 0 < t < 1 , x ( 0 ) = x ( 1 ) = x ( 0 ) = x ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ16_HTML.gif
    (2.5)
     
has an infinite sequence of positive eigenvalues
0 < λ 1 ( h ) < λ 2 ( h ) < < λ k ( h ) < λ k + 1 ( h ) < ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ17_HTML.gif
(2.6)
  1. (ii)

    λ k ( h ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq50_HTML.gif as k + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq51_HTML.gif;

     
  2. (iii)

    to each eigenvalue λ k ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq52_HTML.gif, there corresponds an essential unique eigenfunction ψ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq53_HTML.gif which has exactly k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq54_HTML.gif simple zeros in ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq55_HTML.gif and is positive near 0;

     
  3. (iv)

    given an arbitrary subinterval of [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq26_HTML.gif, an eigenfunction that belongs to a sufficiently large eigenvalue changes its sign in that subinterval;

     
  4. (v)

    for each k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq56_HTML.gif, the algebraic multiplicity of λ k ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq52_HTML.gif is 1.

     

Theorem 2.3 (see [[7], Theorem 2.8]) (Maximum principle)

Let (A1) hold. Let e C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq57_HTML.gif with e 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq58_HTML.gif on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq26_HTML.gif and e 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq59_HTML.gif in [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq26_HTML.gif. If x C 4 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq60_HTML.gif satisfies
{ x ′′′′ ( t ) + k x ( t ) + l x = e ( t ) , 0 < t < 1 , x ( 0 ) = x ( 1 ) = x ( 0 ) = x ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ18_HTML.gif
(2.7)

then x > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq61_HTML.gif on ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq55_HTML.gif.

Let Y = C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq62_HTML.gif with the norm x = max t [ 0 , 1 ] | x | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq63_HTML.gif. Let E = C 2 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq64_HTML.gif with its usual norm x = max { x , x , x } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq65_HTML.gif. By a positive solution of (1.2), we mean x is a solution of (1.2) with x > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq61_HTML.gif (i.e., x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq66_HTML.gif in ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq55_HTML.gif and x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq67_HTML.gif).

Let H : = L 2 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq68_HTML.gif with the inner product , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq69_HTML.gif and the norm L 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq70_HTML.gif. Further, define the linear operator L ˆ : D ( L ˆ ) E Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq71_HTML.gif
L ˆ x = x ′′′′ + k x + l x , x D ( L ˆ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ19_HTML.gif
(2.8)
with
D ( L ˆ ) = { x C 4 [ 0 , 1 ] | x ( 0 ) = x ( 1 ) = x ( 0 ) = x ( 1 ) = 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ20_HTML.gif
(2.9)

Then L ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq72_HTML.gif is a closed operator and L ˆ 1 : Y E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq73_HTML.gif is completely continuous.

Lemma 2.4 Let ψ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq74_HTML.gif be the first eigenfunction of (2.5). Then, for all x D ( L ˆ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq75_HTML.gif, we get
L ˆ x , ψ 1 = x , L ˆ ψ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ21_HTML.gif
(2.10)
Proof Obviously, x D ( L ˆ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq76_HTML.gif, we have
ψ 1 ( 0 ) = ψ 1 ( 1 ) = ψ 1 ( 0 ) = ψ 1 ( 1 ) = 0 , x ( 0 ) = x ( 1 ) = x ( 0 ) = x ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equb_HTML.gif
Integrating by parts, we obtain
L ˆ x , ψ 1 = 0 1 [ x ′′′′ ( t ) + k x ( t ) + l x ( t ) ] ψ 1 ( t ) d t = 0 1 x ( t ) [ ψ 1 ( t ) + k ψ 1 ( t ) + l ψ 1 ( t ) ] d t = x , L ˆ ψ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equc_HTML.gif

 □

Let Σ R + × E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq77_HTML.gif be the closure of the set of positive solutions of the problem
L ˆ x = λ f ( t , x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ22_HTML.gif
(2.11)
We extend the function f to a continuous function f ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq78_HTML.gif defined on [ 0 , 1 ] × R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq79_HTML.gif by
f ¯ ( t , x ) = { f ( t , x ) , ( t , x ) [ 0 , 1 ] × [ 0 , ] , f ( t , 0 ) , ( t , x ) [ 0 , 1 ] × ( , 0 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ23_HTML.gif
(2.12)
Then f ¯ ( t , x ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq80_HTML.gif for ( t , x ) [ 0 , 1 ] × R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq81_HTML.gif. For λ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq82_HTML.gif, let x be an arbitrary solution of the problem
L ˆ x = λ f ¯ ( t , x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ24_HTML.gif
(2.13)

Since λ f ¯ ( t , x ( t ) ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq83_HTML.gif for t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq19_HTML.gif, we have x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq66_HTML.gif for t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq19_HTML.gif. Thus x is a nonnegative solution of (2.11), and the closure of the set of nontrivial solutions ( λ , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq84_HTML.gif of (2.13) in R + × E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq85_HTML.gif is exactly Σ.

Let N : E Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq86_HTML.gif be the Nemytskii operator associated with the function f ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq78_HTML.gif
N ( x ) ( t ) = f ¯ ( t , x ) , x E . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ25_HTML.gif
(2.14)
Then (2.13), with λ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq82_HTML.gif, is equivalent to the operator equation
x = λ L ˆ 1 N ( x ) , x E . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ26_HTML.gif
(2.15)
In the following, we shall apply the Leray-Schauder degree theory, mainly to the mapping Φ λ : E E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq87_HTML.gif,
Φ λ ( x ) = x λ L ˆ 1 N ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ27_HTML.gif
(2.16)

For R > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq88_HTML.gif, let B R = { x E : x R } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq89_HTML.gif, and let deg ( Φ λ , B R , 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq90_HTML.gif denote the degree of Φ λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq91_HTML.gif on B R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq92_HTML.gif with respect to 0.

Lemma 2.5 Let Λ R + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq93_HTML.gif be a compact interval with [ λ 1 ( a 0 ) , λ 1 ( a 0 ) ] Λ = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq94_HTML.gif. Then there exists a number δ 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq95_HTML.gif with the property
Φ λ ( x ) 0 , x E : 0 < x δ 1 , λ Λ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ28_HTML.gif
(2.17)

Proof Suppose to the contrary that there exist sequences { μ n } Λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq96_HTML.gif and { x n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq97_HTML.gif in E : μ n μ Λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq98_HTML.gif, x n 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq99_HTML.gif in E, such that Φ μ n ( x n ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq100_HTML.gif for all n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq101_HTML.gif, then x n 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq102_HTML.gif in [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq26_HTML.gif.

Set y n = x n / x n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq103_HTML.gif. Then L y n = μ n x n 1 N ( x n ) = μ n x n 1 f ( t , x n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq104_HTML.gif and y n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq105_HTML.gif. Now, from condition (H1), we have the following:
a 0 ( t ) x n ξ 1 ( t , x n ) f ( t , x n ) a 0 ( t ) x n + ξ 2 ( t , x n ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ29_HTML.gif
(2.18)
and, accordingly,
μ n ( a 0 ( t ) y n ξ 1 ( t , x n ) x n ) μ n f ( t , x n ) x n μ n ( a 0 ( t ) y n + ξ 2 ( t , x n ) x n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ30_HTML.gif
(2.19)
Let φ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq106_HTML.gif and φ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq107_HTML.gif denote the nonnegative eigenfunctions corresponding to λ 1 ( a 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq108_HTML.gif and λ 1 ( a 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq109_HTML.gif, respectively. Then we have, from the first inequality in (2.19),
μ n ( a 0 ( t ) y n ξ 1 ( t , x n ) x n ) , φ 0 μ n f ( t , x n ) x n , φ 0 = L ˆ y n , φ 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ31_HTML.gif
(2.20)
From Lemma 2.4, we have
L ˆ y n , φ 0 = y n , L ˆ φ 0 = λ 1 ( a 0 ) y n , a 0 ( t ) φ 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ32_HTML.gif
(2.21)
Since x n 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq99_HTML.gif in E, from (1.6) we have
ξ 1 ( t , x n ) x n 0 as  x n 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ33_HTML.gif
(2.22)
By the fact that y n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq105_HTML.gif, we conclude that y n y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq110_HTML.gif in E. Thus,
y n , a 0 ( t ) φ 0 y , a 0 ( t ) φ 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ34_HTML.gif
(2.23)
Combining this and (2.21) and letting n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq111_HTML.gif in (2.20), we get
μ a 0 ( t ) y , φ 0 λ 1 ( a 0 ) a 0 ( t ) φ 0 , y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ35_HTML.gif
(2.24)
and consequently
μ λ 1 ( a 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ36_HTML.gif
(2.25)
Similarly, we deduce from the second inequality in (2.19) that
λ 1 ( a 0 ) μ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ37_HTML.gif
(2.26)

Thus, λ 1 ( a 0 ) μ λ 1 ( a 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq112_HTML.gif. This contradicts μ Λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq113_HTML.gif. □

Corollary 2.6 For λ ( 0 , λ 1 ( a 0 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq114_HTML.gif and δ ( 0 , δ 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq115_HTML.gif, deg ( Φ λ , B δ , 0 ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq116_HTML.gif.

Proof Lemma 2.5, applied to the interval Λ = [ 0 , λ ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq117_HTML.gif, guarantees the existence of δ 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq95_HTML.gif such that for δ ( 0 , δ 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq115_HTML.gif,
x τ λ L ˆ 1 N ( x ) 0 , x E : 0 < x δ , τ [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ38_HTML.gif
(2.27)
Hence, for any δ ( 0 , δ 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq115_HTML.gif,
deg ( Φ λ , B δ , 0 ) = deg ( I , B δ , 0 ) = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ39_HTML.gif
(2.28)

which ends the proof. □

Lemma 2.7 Suppose λ > λ 1 ( a 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq118_HTML.gif. Then there exists δ 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq119_HTML.gif such that x E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq120_HTML.gif with 0 < x δ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq121_HTML.gif, τ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq122_HTML.gif,
Φ λ ( x ) τ φ 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ40_HTML.gif
(2.29)

where φ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq107_HTML.gif is the nonnegative eigenfunction corresponding to λ 1 ( a 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq109_HTML.gif.

Proof We assume to the contrary that there exist τ n 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq123_HTML.gif and a sequence { x n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq97_HTML.gif, with x n > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq124_HTML.gif and x n 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq125_HTML.gif in E, such that Φ λ ( x n ) = τ n φ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq126_HTML.gif for all n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq101_HTML.gif. As
L ˆ x n = λ N ( x n ) + τ n λ 1 ( a 0 ) a 0 ( t ) φ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ41_HTML.gif
(2.30)
and τ n λ 1 ( a 0 ) a 0 ( t ) φ 0 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq127_HTML.gif in ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq55_HTML.gif, it follows that
x n 0 , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ42_HTML.gif
(2.31)
Notice that x n D ( L ˆ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq128_HTML.gif has a unique decomposition
x n = ω n + s n φ 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ43_HTML.gif
(2.32)

where s n R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq129_HTML.gif and ω n , a 0 ( t ) φ 0 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq130_HTML.gif. Since x n 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq102_HTML.gif on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq26_HTML.gif and x n 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq131_HTML.gif, we have from (2.32) that s n > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq132_HTML.gif.

Choose σ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq133_HTML.gif such that
σ < λ λ 1 ( a 0 ) λ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ44_HTML.gif
(2.33)
By (H1), there exists r 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq134_HTML.gif such that
| ξ 1 ( t , x ) | σ a 0 ( t ) x , t [ 0 , 1 ] , x [ 0 , r 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ45_HTML.gif
(2.34)
Therefore, for t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq19_HTML.gif, x [ 0 , r 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq135_HTML.gif,
f ( t , x ) a 0 ( t ) x ξ 1 ( t , x ) ( 1 σ ) a 0 ( t ) x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ46_HTML.gif
(2.35)
Since x n 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq125_HTML.gif, there exists N > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq136_HTML.gif such that
0 x n r 1 , n N , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ47_HTML.gif
(2.36)
and consequently
f ( t , x n ) ( 1 σ ) a 0 ( t ) x n , n N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ48_HTML.gif
(2.37)
Applying Lemma 2.4 and (2.37), it follows that
s n λ 1 ( a 0 ) φ 0 , a 0 ( t ) φ 0 = x n , L ˆ φ 0 = L ˆ x n , φ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ49_HTML.gif
(2.38)
= λ N ( x n ) , φ 0 + τ n λ 1 ( a 0 ) a 0 ( t ) φ 0 , φ 0 λ N ( x n ) , φ 0 λ ( 1 σ ) a 0 ( t ) x n , φ 0 = λ ( 1 σ ) a 0 ( t ) φ 0 , x n = λ ( 1 σ ) s n a 0 ( t ) φ 0 , φ 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ50_HTML.gif
(2.39)
Thus,
λ 1 ( a 0 ) λ ( 1 σ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ51_HTML.gif
(2.40)

This contradicts (2.33). □

Corollary 2.8 For λ > λ 1 ( a 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq118_HTML.gif and δ ( 0 , δ 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq137_HTML.gif, deg ( Φ λ , B δ , 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq138_HTML.gif.

Proof Let 0 < δ δ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq139_HTML.gif, where δ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq140_HTML.gif is the number asserted in Lemma 2.7. As Φ λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq91_HTML.gif is bounded in B ¯ δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq141_HTML.gif, there exists c > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq142_HTML.gif such that Φ λ ( x ) c φ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq143_HTML.gif for all x B ¯ δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq144_HTML.gif. By Lemma 2.7, one has
Φ λ ( x ) τ c φ 0 , x B δ , τ [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ52_HTML.gif
(2.41)
Hence
deg ( Φ λ , B δ , 0 ) = deg ( Φ λ c φ 0 , B δ , 0 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ53_HTML.gif
(2.42)

 □

Now, using Theorem A, we may prove the following.

Proposition 2.9 [ λ 1 ( a 0 ) , λ 1 ( a 0 ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq145_HTML.gif is a bifurcation interval from the trivial solution for (2.15). There exists an unbounded component C of a positive solution of (2.15), which meets [ λ 1 ( a 0 ) , λ 1 ( a 0 ) ] × { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq146_HTML.gif. Moreover,
C [ ( R [ λ 1 ( a 0 ) , λ 1 ( a 0 ) ] ) × { 0 } ] = . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ54_HTML.gif
(2.43)
Proof For fixed n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq101_HTML.gif with λ 1 ( a 0 ) 1 n > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq147_HTML.gif, let us take that a n = λ 1 ( a 0 ) 1 n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq148_HTML.gif, b n = λ 1 ( a 0 ) + 1 n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq149_HTML.gif and δ ¯ = min { δ 1 , δ 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq150_HTML.gif. It is easy to check that for 0 < δ < δ ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq151_HTML.gif, all of the conditions of Theorem A are satisfied. So, there exists a connected component C n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq152_HTML.gif of solutions of (2.15) containing [ a n , b n ] × { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq153_HTML.gif, and either
  1. (i)

    C n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq152_HTML.gif is unbounded, or

     
  2. (ii)

    C n [ ( R [ a n , b n ] ) × { 0 } ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq154_HTML.gif.

     

By Lemma 2.5, the case (ii) cannot occur. Thus C n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq152_HTML.gif is unbounded bifurcated from [ a n , b n ] × { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq153_HTML.gif in R × E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq155_HTML.gif. Furthermore, we have from Lemma 2.5 that for any closed interval I [ a n , b n ] [ λ 1 ( a 0 ) , λ 1 ( a 0 ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq156_HTML.gif, if x { x E | ( λ , x ) Σ , λ I } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq157_HTML.gif, then x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq158_HTML.gif in E is impossible. So, C n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq152_HTML.gif must be bifurcated from [ λ 1 ( a 0 ) , λ 1 ( a 0 ) ] × { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq146_HTML.gif in R × E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq159_HTML.gif. □

3 Main results

Theorem 3.1 Let (A1), (H1), (H2), (H3) hold. Assume that either
λ 1 ( b ) < 1 < λ 1 ( a 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ55_HTML.gif
(3.1)
or
λ 1 ( a 0 ) < 1 < λ 1 ( b ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ56_HTML.gif
(3.2)

then problem (1.2) has at least one positive solution.

Proof of Theorem 3.1 It is clear that any solution of (2.15) of the form ( 1 , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq160_HTML.gif yields a solution x of (1.2). We will show that C crosses the hyperplane { 1 } × E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq161_HTML.gif in R × E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq159_HTML.gif. To do this, it is enough to show that C joins [ λ 1 ( a 0 ) , λ 1 ( a 0 ) ] × { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq146_HTML.gif to [ λ 1 ( b ) , λ 1 ( b ) ] × { } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq162_HTML.gif. Let ( μ n , x n ) C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq163_HTML.gif satisfy
μ n + x n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ57_HTML.gif
(3.3)

We note that μ n > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq164_HTML.gif for all n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq101_HTML.gif since ( 0 , 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq165_HTML.gif is the only solution of (2.15) for λ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq166_HTML.gif and C ( { 0 } × E ) = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq167_HTML.gif.

Case 1. λ 1 ( b ) < 1 < λ 1 ( a 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq168_HTML.gif.

In this case, we show that
( λ 1 ( b ) , λ 1 ( a 0 ) ) { λ R | ( λ , x ) C } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equd_HTML.gif

We divide the proof into two steps.

Step 1. We show that { μ n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq169_HTML.gif is bounded.

Since ( μ n , x n ) C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq163_HTML.gif, L x n = μ n f ( t , x n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq170_HTML.gif. From (H3), we have
L x n μ n c ( t ) x n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ58_HTML.gif
(3.4)

Let φ ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq171_HTML.gif denote the nonnegative eigenfunction corresponding to λ 1 ( c ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq172_HTML.gif.

From (3.4), we have
L x n , φ ¯ μ n c ( t ) x n , φ ¯ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ59_HTML.gif
(3.5)
By Lemma 2.4, we have
λ 1 ( c ) x n , c ( t ) φ ¯ = x n , L φ ¯ μ n c ( t ) φ ¯ , x n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ60_HTML.gif
(3.6)
Thus
μ n λ 1 ( c ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ61_HTML.gif
(3.7)

Step 2. We show that C joins [ λ 1 ( a 0 ) , λ 1 ( a 0 ) ] × { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq146_HTML.gif to [ λ 1 ( b ) , λ 1 ( b ) ] × { } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq173_HTML.gif.

From (3.3) and (3.7), we have that x n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq174_HTML.gif. Notice that (2.15) is equivalent to the integral equation
x n ( t ) = μ n 0 1 G ( t , s ) f ( s , x n ( s ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ62_HTML.gif
(3.8)
which implies that
μ n 0 1 G ( t , s ) [ b ( s ) x n ( s ) ζ 1 ( s , x n ( s ) ) ] d s x n ( t ) μ n 0 1 G ( t , s ) [ b ( s ) x n ( s ) + ζ 2 ( s , x n ( s ) ) ] d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ63_HTML.gif
(3.9)
We divide both of (3.9) by x n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq175_HTML.gif and set y n = x n x n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq176_HTML.gif. Since y n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq177_HTML.gif is bounded in E, there exists a subsequence of { y n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq178_HTML.gif and y E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq179_HTML.gif, with y 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq180_HTML.gif and y 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq181_HTML.gif on ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq55_HTML.gif, such that
μ n μ , y n y in  E , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ64_HTML.gif
(3.10)
relabeling if necessary. Thus, (3.9) yields that
μ 0 1 G ( t , s ) b ( s ) y ( s ) d s y ( t ) μ 0 1 G ( t , s ) b ( s ) y ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ65_HTML.gif
(3.11)
which implies that
μ b ( t ) y L y μ b ( t ) y L y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ66_HTML.gif
(3.12)
Let φ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq182_HTML.gif and φ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq183_HTML.gif denote the nonnegative eigenfunction corresponding to λ 1 ( b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq184_HTML.gif and λ 1 ( b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq185_HTML.gif, respectively. Then we have, from the first inequality in (3.12),
μ b ( t ) y , φ L y , φ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Eque_HTML.gif
From Lemma 2.4, integrating by parts, we obtain that
μ b ( t ) y , φ L y , φ = L φ , y = λ 1 ( b ) b ( t ) φ , y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equf_HTML.gif
and consequently
μ λ 1 ( b ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ67_HTML.gif
(3.13)
Similarly, we deduce from the second inequality in (3.12) that
λ 1 ( b ) μ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ68_HTML.gif
(3.14)
Thus
λ 1 ( b ) μ λ 1 ( b ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ69_HTML.gif
(3.15)

So, C joins [ λ 1 ( a 0 ) , λ 1 ( a 0 ) ] × { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq146_HTML.gif to [ λ 1 ( b ) , λ 1 ( b ) ] × { } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq173_HTML.gif.

Case 2. λ 1 ( a 0 ) < 1 < λ 1 ( b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq186_HTML.gif.

In this case, if ( μ n , x n ) C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq163_HTML.gif is such that
lim n ( μ n + x n ) = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equg_HTML.gif
and
lim n μ n = , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equh_HTML.gif
then
( λ 1 ( a 0 ) , λ 1 ( b ) ) { λ ( 0 , ) | ( λ , x ) C } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ70_HTML.gif
(3.16)
and, moreover,
( { 1 } × E ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ71_HTML.gif
(3.17)
Assume that { μ n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq169_HTML.gif is bounded; applying a similar argument to that used in Step 2 of Case 1, after taking a subsequence and relabeling if necessary, we obtain
μ n μ [ λ 1 ( a 0 ) , λ 1 ( b ) ] , x n as  n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_Equ72_HTML.gif
(3.18)

Again C joins [ λ 1 ( a 0 ) , λ 1 ( a 0 ) ] × { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq146_HTML.gif to [ λ 1 ( b ) , λ 1 ( b ) ] × { } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-170/MediaObjects/13661_2012_Article_599_IEq173_HTML.gif and the result follows. □

Declarations

Acknowledgements

This work is supported by the NSF of Gansu Province (No. 1114-04).

Authors’ Affiliations

(1)
Department of Basic Courses, Lanzhou Institute of Technology

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