Open Access

Eigenvalue criteria for existence of positive solutions of impulsive differential equations with non-separated boundary conditions

Boundary Value Problems20132013:3

DOI: 10.1186/1687-2770-2013-3

Received: 29 August 2012

Accepted: 28 December 2012

Published: 14 January 2013

Abstract

In this paper, we discuss the existence of positive solutions for second-order differential equations subject to nonlinear impulsive conditions and non-separated periodic boundary value conditions. Our criteria for the existence of positive solutions will be expressed in terms of the first eigenvalue of the corresponding nonimpulsive problem. The main tool of study is a fixed point theorem in a cone.

MSC:34B37, 34B18.

Keywords

impulsive differential equation positive solution fixed point theorem non-separated periodic boundary value condition

1 Introduction

Let ω be a fixed positive number. In this paper, we are concerned with the existence of positive solutions for the following boundary value problem (BVP) with impulses:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ1_HTML.gif
(1.1a)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ2_HTML.gif
(1.1b)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ3_HTML.gif
(1.1c)

Here, u [ 1 ] ( t ) = p ( t ) u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq1_HTML.gif denotes the quasi-derivative of u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq2_HTML.gif. The condition (1.1c) is called a non-separated periodic boundary value condition for (1.1a).

We assume throughout, and with further mention, that the following conditions hold.

(H1) Let J = [ 0 , ω ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq3_HTML.gif, and 0 < t 1 < t 2 < < t m < ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq4_HTML.gif, f C ( J × R + , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq5_HTML.gif, I i C ( R + , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq6_HTML.gif, R + = [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq7_HTML.gif. Δ ( u [ 1 ] ( t i ) ) = u [ 1 ] ( t i + ) u [ 1 ] ( t i ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq8_HTML.gif, where u [ 1 ] ( t i + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq9_HTML.gif (respectively u [ 1 ] ( t i ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq10_HTML.gif) denotes the right limit (respectively, the left limit) of u [ 1 ] ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq11_HTML.gif at t = t i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq12_HTML.gif.
  1. (H)
    0 ω 1 p ( t ) d t < , 0 ω q ( t ) d t < , p > 0 , q 0  and  q 0  a.e. on  [ 0 , ω ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equa_HTML.gif
     

A function u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq2_HTML.gif defined on J = J { t 1 , t 2 , , t m } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq13_HTML.gif is called a solution of BVP (1.1) ((1.1a)-(1.1c)) if its first derivative u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq14_HTML.gif exists for each t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq15_HTML.gif, p ( t ) u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq16_HTML.gif is absolutely continuous on each close subinterval of J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq17_HTML.gif, there exist finite values u [ 1 ] ( t i ± ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq18_HTML.gif, the impulse conditions (1.1b) and the boundary conditions (1.1c) are satisfied, and the equation (1.1a) is satisfied almost everywhere on J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq17_HTML.gif.

For the case of I i = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq19_HTML.gif ( i = 1 , 2 , , m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq20_HTML.gif), the problem (1.1) is related to a non-separated periodic boundary value problem of ODE. Atici and Guseinov [1] have proved the existence of a positive and twin positive solutions to BVP (1.1) by applying a fixed point theorem for the completely continuous operators in cones. In [2], Graef and Kong studied the following periodic boundary value problem:
{ ( p ( t ) u ) + q ( t ) u = h ( t ) f ( t , u ) , t ( 0 , ω ) , u ( 0 ) = u ( ω ) , u [ 1 ] ( 0 ) = u [ 1 ] ( ω ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ4_HTML.gif
(1.2)

where h ( t ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq21_HTML.gif. Based upon the properties of Green’s function obtained in [1], the authors extended and improved the work of [1] by using topological degree theory. They derived new criteria for the existence of non-trivial solutions, positive solutions and negative solutions of the problem (1.2) when f is a sign-changing function and not necessarily bounded from below even over [ 0 , ω ] × R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq22_HTML.gif. Very recently, He et al. [3] studied BVP (1.1) without impulses and generalized the results of [1, 4] via the fixed point index theory. The problem (1.2) in the case of p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq23_HTML.gif, the usual periodic boundary value problem, has been extensively investigated; see [47] for some results.

On the other hand, impulsive differential equations are a basic tool to study processes that are subjected to abrupt changes in their state. There has been a significant development in the last two decades. Boundary problems of second-order differential equations with impulse have received considerable attention and much literature has been published; see, for instance, [817] and their references. However, there are fewer results about positive solutions for second-order impulsive differential equations. To our best knowledge, there is no result about nonlinear impulsive differential equations with non-separated periodic boundary conditions.

Motivated by the work above, in this paper we study the existence of positive solutions for the boundary value problem (1.1). By using fixed point theorems in a cone, criteria are established under some conditions on f ( t , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq24_HTML.gif concerning the first eigenvalue corresponding to the relevant linear operator. More important, the impulsive terms are different from those of papers [8, 9].

2 Preliminaries

In this section, we collect some preliminary results that will be used in the subsequent section. We denote by φ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq25_HTML.gif and ψ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq26_HTML.gif the unique solutions of the corresponding homogeneous equation
( p ( t ) u ( t ) ) + q ( t ) u ( t ) = 0 , t J , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ5_HTML.gif
(2.1)
under the initial boundary conditions
φ ( 0 ) = 1 , φ [ 1 ] ( 0 ) = 0 ; ψ ( 0 ) = 0 , ψ [ 1 ] ( 0 ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ6_HTML.gif
(2.2)

Put D = φ ( ω ) + ψ [ 1 ] ( ω ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq27_HTML.gif, then by [[1], Lemma 2.3], D > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq28_HTML.gif.

Definition 2.1 For two differential functions y and z, we define their Wronskian by
W t ( y , z ) = y ( t ) z [ 1 ] ( t ) y [ 1 ] ( t ) z ( t ) = p ( t ) [ y ( t ) z ( t ) y ( t ) z ( t ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equb_HTML.gif

Theorem 2.1 The Wronskian of any two solutions for equations (2.1) is constant. Especially, W t ( φ , ψ ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq29_HTML.gif.

Proof Suppose that y and z are two solutions of (2.1), then
{ W t ( y , z ) } = { p ( t ) [ y ( t ) z ( t ) y ( t ) z ( t ) ] } = y ( t ) [ p ( t ) z ( t ) ] [ p ( t ) y ( t ) ] z ( t ) = 0 ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equc_HTML.gif

therefore, the Wronskian is constant. Further, from the initial conditions (2.2), we have W t ( φ , ψ ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq30_HTML.gif. The proof is complete. □

Consider the following equation:
{ ( p ( t ) u ( t ) ) + q ( t ) u ( t ) = 0 , t J , u ( 0 ) = u ( ω ) , u [ 1 ] ( 0 ) = u [ 1 ] ( ω ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ7_HTML.gif
(2.3)

From Theorem 2.5 in [1], equation (2.3) has a Green function G ( t , s ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq31_HTML.gif for all s , t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq32_HTML.gif, which has the following properties:

( G 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq33_HTML.gif) G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq34_HTML.gif is continuous in t and s for all t , s J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq35_HTML.gif.

( G 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq36_HTML.gif) If A = min 0 t , s ω G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq37_HTML.gif and B = max 0 t , s ω G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq38_HTML.gif, then B > A > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq39_HTML.gif.

( G 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq40_HTML.gif)
G ( t , s ) = ψ ( ω ) D φ ( t ) φ ( s ) φ [ 1 ] ( ω ) D ψ ( t ) ψ ( s ) + { ψ [ 1 ] ( ω ) 1 D φ ( t ) ψ ( s ) φ ( ω ) 1 D φ ( s ) ψ ( t ) , 0 s t ω , ψ [ 1 ] ( ω ) 1 D φ ( s ) ψ ( t ) φ ( ω ) 1 D φ ( t ) ψ ( s ) , 0 t s ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equd_HTML.gif

Combining with Theorem 2.1, we can also prove that

( G 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq41_HTML.gif)
G ( 0 , s ) = G ( ω , s ) , p ( t ) G t | ( 0 , s ) = p ( t ) G t | ( ω , s ) , 0 ω q ( t ) G ( t , s ) d s = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Eque_HTML.gif
Remark 1 From paper [1], we can get G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq34_HTML.gif when q ( t ) = c 2 p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq42_HTML.gif ( c > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq43_HTML.gif) and p ( t ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq44_HTML.gif,
G ( t , s ) = 1 2 c ( e c 0 ω ( d x / p ( x ) ) 1 ) { e c s t ( d x / p ( x ) ) + e c [ 0 ω ( d x / p ( x ) ) + t s ( d x / p ( x ) ) ] , 0 s t ω , e c t s ( d x / p ( x ) ) + e c [ 0 ω ( d x / p ( x ) ) + s t ( d x / p ( x ) ) ] , 0 t s ω , A = e c / 2 0 ω ( d x / p ( x ) ) c [ e c 0 ω ( d x / p ( x ) ) 1 ] , B = 1 + e c 0 ω ( d x / p ( x ) ) 2 c [ e c 0 ω ( d x / p ( x ) ) 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equf_HTML.gif
Especially, in the case of p ( t ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq45_HTML.gif, q ( t ) c 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq46_HTML.gif ( c > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq43_HTML.gif), Green’s function G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq34_HTML.gif has the form
G ( t , s ) = 1 2 c ( e c ω 1 ) { e c ( t s ) + e c ( ω + s t ) , 0 s t ω , e c ( s t ) + e c ( ω + t s ) , 0 t s ω , A = e ( c ω / 2 ) c ( e c ω 1 ) , B = 1 + e c ω 2 c ( e c ω 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equg_HTML.gif
Define an operator
( T u ) ( t ) = 0 ω G ( t , s ) u ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equh_HTML.gif

then it is easy to check that T : C ( J ) C ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq47_HTML.gif is a completely continuous operator. By virtue of the Krein-Rutman theorem, the authors in [3] got the following result.

Lemma 2.1 The spectral radius r ( T ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq48_HTML.gif and T has a positive eigenfunction corresponding to its first eigenvalue λ 1 = ( r ( T ) ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq49_HTML.gif.

In what follows, we denote the positive eigenfunction corresponding to λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq50_HTML.gif by ϕ and max t J ϕ ( t ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq51_HTML.gif. Define a mapping Φ and a cone K in a Banach space C ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq52_HTML.gif by
( Φ u ) ( t ) = λ 0 ω G ( t , s ) f ( s , u ( s ) ) d s + i = 1 m G ( t , t i ) I i ( u ( t i ) ) , t J , K = { u C ( J ) , u ( t ) δ u } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equi_HTML.gif

where δ = A B https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq53_HTML.gif, u = max t J | u ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq54_HTML.gif.

Lemma 2.2 The fixed point of the mapping Φ is a solution of (1.1).

Proof Clearly, Φu is continuous in t. For t t k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq55_HTML.gif,
( Φ u ) ( t ) = λ 0 ω G t f ( s , u ( s ) ) d s + i = 1 m G t ( t , t i ) I i ( u ( t i ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equj_HTML.gif
Using ( G 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq40_HTML.gif) and ( G 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq41_HTML.gif), we have ( Φ u ) ( 0 ) = ( Φ u ) ( ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq56_HTML.gif, ( Φ u ) [ 1 ] ( 0 ) = ( Φ u ) [ 1 ] ( ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq57_HTML.gif and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equk_HTML.gif

which implies that the fixed point of Φ is the solution of (1.1). The proof is complete. □

The proofs of the main theorems of this paper are based on fixed point theory. The following two well-known lemmas in [18] are needed in our argument.

Lemma 2.3 [18]

Let X be a Banach space and K be a cone in X. Suppose Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq58_HTML.gif and Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq59_HTML.gif are open subsets of X such that 0 Ω 1 Ω ¯ 1 Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq60_HTML.gif, and suppose that
Φ : K ( Ω ¯ 2 Ω 1 ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equl_HTML.gif

is a completely continuous operator such that

  • inf u K Ω 1 Φ u > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq61_HTML.gif, u μ Φ u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq62_HTML.giffor u K Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq63_HTML.gifand μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq64_HTML.gif, and u μ Φ u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq62_HTML.giffor u K Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq65_HTML.gifand 0 < μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq66_HTML.gif, or

  • inf u K Ω 2 Φ u > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq67_HTML.gif, u μ Φ u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq62_HTML.giffor u K Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq65_HTML.gifand μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq64_HTML.gif, and u μ Φ u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq62_HTML.giffor u K Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq63_HTML.gifand 0 < μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq66_HTML.gif.

Then Φ has a fixed point in K ( Ω ¯ 2 Ω 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq68_HTML.gif.

Lemma 2.4 [18]

Let X be a Banach space and K be a cone in X. Suppose Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq58_HTML.gif and Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq59_HTML.gif are open subsets of X such that 0 Ω 1 Ω ¯ 1 Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq60_HTML.gif, and suppose that
Φ : K ( Ω ¯ 2 Ω 1 ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equm_HTML.gif

is a completely continuous operator such that

  • There exists u 0 K { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq69_HTML.gifsuch that u Φ u + μ u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq70_HTML.giffor u K Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq63_HTML.gifand μ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq71_HTML.gif, Φ u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq72_HTML.giffor u K Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq65_HTML.gif, or

  • There exists u 0 K { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq69_HTML.gifsuch that u Φ u + μ u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq70_HTML.giffor u K Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq65_HTML.gifand μ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq71_HTML.gif, Φ u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq72_HTML.giffor u K Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq63_HTML.gif.

Then Φ has a fixed point in Ω ¯ 2 Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq73_HTML.gif.

3 Main results

Recalling that δ was defined after Lemma 2.1, for convenience, we introduce the following notations. Assume that the constant r > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq74_HTML.gif and γ is some positive function on J,
f ¯ γ r = sup { f ( t , u ) γ ( t ) u , t J , u [ δ r , r ] } , f γ r = inf { f ( t , u ) γ ( t ) u , t J , u [ δ r , r ] } , f ¯ γ 0 = lim r 0 + f ¯ γ r , f γ 0 = lim r 0 + f γ r , f ¯ γ = lim r + f ¯ γ r , f γ = lim r + f γ r , I ¯ i r = sup { I i ( u ) u , u [ δ r , r ] } , I i r = inf { I i ( u ) u , u [ δ r , r ] } , I ¯ i 0 = lim r 0 + I ¯ i r , I i 0 = lim r 0 + I i r , I ¯ i = lim r + I ¯ i r , I i = lim r + I i r . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equn_HTML.gif
Theorem 3.1 Assume that there exist positive constants α, β such that f q α 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq75_HTML.gif, f ¯ q β 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq76_HTML.gif, I i α 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq77_HTML.gif, I i β 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq78_HTML.gif and
0 < λ ( 1 A i = 1 m I i α f q α , 1 B i = 1 m I ¯ i β f ¯ q β ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ8_HTML.gif
(3.1)

Then (1.1) has at least one positive solution u such that min { α , β } u max { α , β } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq79_HTML.gif.

Proof Clearly, α β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq80_HTML.gif, let α = min { α , β } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq81_HTML.gif, β = max { α , β } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq82_HTML.gif. Define the open sets
Ω α = { u C ( J ) : u < α } , Ω β = { u C ( J ) : u < β } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equo_HTML.gif
Then Φ : K ( Ω ¯ β Ω α ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq83_HTML.gif is completely continuous. By (3.1) and the definition of f q α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq84_HTML.gif, I i α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq85_HTML.gif, f ¯ q β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq86_HTML.gif, I ¯ i β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq87_HTML.gif, there exists ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq88_HTML.gif such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ9_HTML.gif
(3.2)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ10_HTML.gif
(3.3)
and
f ( t , u ) ( 1 + ε ) f ¯ q β q ( t ) u , I i ( u ) ( 1 + ε ) I ¯ i β u , i = 1 , 2 , , m , δ β u β . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ11_HTML.gif
(3.4)
Let u 0 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq89_HTML.gif. We show that
u Φ u + μ , u K Ω α  and  μ > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ12_HTML.gif
(3.5)
If not, there exist u 1 K Ω α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq90_HTML.gif and μ 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq91_HTML.gif such that u 1 = Φ u 1 + μ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq92_HTML.gif. Let u 1 ( ρ ) = min t J u 1 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq93_HTML.gif. Noting that δ α u 1 α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq94_HTML.gif for any t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq95_HTML.gif, we obtain that for t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq95_HTML.gif,
u 1 ( t ) = ( Φ u 1 ) ( t ) + μ 1 = λ 0 ω G ( t , s ) f ( s , u 1 ( s ) ) d s + i = 1 m G ( t , t i ) I i ( u 1 ( t i ) ) + μ 1 ( 1 ε ) λ f q α 0 ω G ( t , s ) q ( s ) u 1 ( s ) d s + A ( 1 ε ) i = 1 m I i α u 1 ( t i ) + μ 1 ( 1 ε ) λ f q α u 1 ( ρ ) 0 ω G ( t , s ) q ( s ) d s + A u 1 ( ρ ) ( 1 ε ) i = 1 m I i α + μ 1 ( 1 ε ) ( λ f q α + A i = 1 m I i α ) u 1 ( ρ ) + μ 1 u 1 ( ρ ) + μ 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equp_HTML.gif

which implies that u 1 ( ρ ) > u 1 ( ρ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq96_HTML.gif, a contradiction.

On the other hand, for u K Ω β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq97_HTML.gif, δ β u ( t ) β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq98_HTML.gif, we have
( Φ u ) ( t ) = λ 0 ω G ( t , s ) f ( s , u ( s ) ) d s + i = 0 m G ( t , t i ) I i ( u i ( t ) ) ( 1 + ε ) λ f ¯ q β 0 ω G ( t , s ) q ( s ) u ( s ) d s + B ( 1 + ε ) i = 1 m I ¯ i β u ( t i ) ( 1 + ε ) λ f ¯ q β u 0 ω G ( t , s ) q ( s ) d s + B ( 1 + ε ) u i = 1 m I ¯ i β ( 1 + ε ) ( λ f ¯ q β + B i = 1 m I ¯ i β ) u u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equq_HTML.gif

From Lemma 2.4 it follows that Φ has a fixed point u K ( Ω ¯ β Ω α ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq99_HTML.gif. Furthermore, α u β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq100_HTML.gif and u ( t ) δ α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq101_HTML.gif, which means that u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq2_HTML.gif is a positive solution of Eq. (1.1). The proof is complete. □

In the next theorem, we make use of the eigenvalue λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq50_HTML.gif and the corresponding eigenfunction ϕ introduced in Lemma 2.1.

Theorem 3.2 Assume that there exist positive constants α, β such that f γ α 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq102_HTML.gif, f γ β 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq103_HTML.gif, I i α 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq77_HTML.gif, I i β 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq78_HTML.gif and
0 < λ ( λ 1 0 ω ϕ ( s ) d s δ i = 1 m I i α ϕ ( t i ) δ f γ α 0 ω ϕ ( s ) d s , δ λ 1 0 ω ϕ ( s ) d s δ i = 1 m I ¯ i β ϕ ( t i ) δ f ¯ γ β 0 ω ϕ ( s ) d s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ13_HTML.gif
(3.6)

here γ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq104_HTML.gif on J. Then (1.1) has at least one positive solution u such that min { α , β } u max { α , β } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq105_HTML.gif.

Proof Obviously, α β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq80_HTML.gif, put α = min { α , β } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq106_HTML.gif, β = max { α , β } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq82_HTML.gif. Define the open sets
Ω α = { u C ( J ) : u < α } , Ω β = { u C ( J ) : u < β } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equr_HTML.gif
At first, we show that Φ : K ( Ω ¯ β Ω α ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq107_HTML.gif. For any u K ( Ω ¯ β Ω α ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq99_HTML.gif, from ( G 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq36_HTML.gif), we have
0 < ( Φ u ) ( t ) B ( λ 0 ω f ( s , u ( s ) ) d s + i = 1 m I i ( u ( t i ) ) ) < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equs_HTML.gif
On the other hand,
( Φ u ) ( t ) A ( λ 0 ω f ( s , u ( s ) ) d s + i = 1 m I i ( u ( t i ) ) ) A B Φ u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equt_HTML.gif

It is easy to check that Φ : K ( Ω ¯ β Ω α ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq108_HTML.gif is completely continuous.

Next, we show that
μ Φ u u , u K Ω β  and  0 < μ 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ14_HTML.gif
(3.7)
If not, there exist μ 0 ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq109_HTML.gif and u 0 K Ω β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq110_HTML.gif such that μ 0 Φ u 0 = u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq111_HTML.gif. Hence,
{ ( p ( t ) u 0 ( t ) ) + q ( t ) u 0 ( t ) = μ 0 λ f ( t , u 0 ( t ) ) , t J , Δ ( u 0 [ 1 ] ( t k ) ) = μ 0 I k ( u 0 ( t k ) ) , k = 1 , , m , u 0 ( 0 ) = u 0 ( ω ) , u 0 [ 1 ] ( 0 ) = u 0 [ 1 ] ( ω ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ15_HTML.gif
(3.8)
Multiplying the first equation of (3.8) by ϕ and integrating from 0 to ω, we obtain that
0 ω ( p ( t ) u 0 ( t ) ) ϕ ( t ) d t + 0 ω q ( t ) u 0 ( t ) ϕ ( t ) d t = μ 0 λ 0 ω f ( s , u 0 ( s ) ) ϕ ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ16_HTML.gif
(3.9)
One can find that
0 ω ( p ( t ) u 0 ( t ) ) ϕ ( t ) d t = μ 0 i = 1 m I i ( u 0 ( t i ) ) ϕ ( t i ) + 0 ω ( q ( t ) λ 1 ) ϕ ( t ) u 0 ( t ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ17_HTML.gif
(3.10)
Substituting (3.10) into (3.9), we get
μ 0 i = 1 m I i ( u 0 ( t i ) ) ϕ ( t i ) + λ 1 0 ω ϕ ( t ) u 0 ( t ) d t = μ 0 λ 0 ω f ( t , u 0 ( t ) ) ϕ ( t ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equu_HTML.gif
Noting that δ u 0 u 0 u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq112_HTML.gif, therefore,
u 0 i = 1 m I ¯ i β ϕ ( t i ) + λ 1 δ u 0 0 ω ϕ ( t ) d t λ f ¯ γ β 0 ω ϕ ( t ) d t u 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equv_HTML.gif
which implies that
λ δ λ 1 0 ω ϕ ( s ) d s i = 1 m I ¯ i β ϕ ( t i ) f ¯ γ β 0 ω ϕ ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equw_HTML.gif

a contradiction.

Finally, we show that
inf u K Ω α Φ u > 0 , μ Φ u u , u K Ω α  and  μ 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equx_HTML.gif
Since f ( t , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq24_HTML.gif and I i ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq113_HTML.gif are negative for u [ δ α , α ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq114_HTML.gif and t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq95_HTML.gif, the condition (3.6) implies that f γ α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq115_HTML.gif. Hence, 0 ω f ( s , u ( s ) ) d s > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq116_HTML.gif for u K Ω α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq117_HTML.gif and for any u K Ω α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq117_HTML.gif,
( Φ u ) ( t ) = λ 0 ω G ( t , s ) f ( s , u ( s ) ) d s + i = 1 m G ( t , t i ) I i ( u ( t i ) ) A λ 0 ω f ( s , u ( s ) ) d s > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equy_HTML.gif
Suppose that there exist μ 0 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq118_HTML.gif and u 0 K Ω α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq119_HTML.gif such that μ 0 Φ u 0 = u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq111_HTML.gif, that is,
{ ( p ( t ) u 0 ( t ) ) + q ( t ) u 0 ( t ) = μ 0 λ f ( t , u 0 ( t ) ) , t J , Δ ( u 0 [ 1 ] ( t k ) ) = μ 0 I k ( u 0 ( t k ) ) , k = 1 , , m , u 0 ( 0 ) = u 0 ( ω ) , u 0 [ 1 ] ( 0 ) = u 0 [ 1 ] ( ω ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ18_HTML.gif
(3.11)
Multiplying the first equation of (3.11) by ϕ and integrating from 0 to ω, we obtain that
0 ω ( p ( t ) u 0 ( t ) ) ϕ ( t ) d t + 0 ω q ( t ) u 0 ( t ) ϕ ( t ) d t = μ 0 λ 0 ω f ( s , u 0 ( s ) ) ϕ ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ19_HTML.gif
(3.12)
One can get that
0 ω ( p ( t ) u 0 ( t ) ) ϕ ( t ) d t = μ 0 i = 1 m I i ( u 0 ( t i ) ) ϕ ( t i ) + 0 ω ( p ( t ) ϕ ( t ) ) u 0 ( t ) d t = μ 0 i = 1 m I i ( u 0 ( t i ) ) ϕ ( t i ) + 0 ω ( q ( t ) λ 1 ) ϕ ( t ) u 0 ( t ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ20_HTML.gif
(3.13)
Substituting (3.13) into (3.12), we get
μ 0 i = 1 m I i ( u 0 ( t i ) ) ϕ ( t i ) + λ 1 0 ω ϕ ( t ) u 0 ( t ) d t = μ 0 λ 0 ω f ( t , u 0 ( t ) ) ϕ ( t ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equz_HTML.gif
Noting that δ u 0 u 0 u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq112_HTML.gif, therefore,
δ u 0 i = 1 m I i α ϕ ( t i ) + λ 1 u 0 0 ω ϕ ( t ) d t μ 0 λ 0 ω f ( s , u 0 ( s ) ) d s λ δ f γ α 0 ω u 0 ( s ) ϕ ( s ) d s λ δ f γ α 0 ω ϕ ( s ) d s u 0 > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equaa_HTML.gif
It is impossible for λ 1 0 ω ϕ ( s ) d s δ i = 1 m I i α ϕ ( t i ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq120_HTML.gif. When λ 1 0 ω ϕ ( s ) d s δ i = 1 m I i α ϕ ( t i ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq121_HTML.gif,
λ < λ 1 0 ω ϕ ( s ) d s δ i = 1 m I i α ϕ ( t i ) δ f γ α 0 ω ϕ ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equab_HTML.gif

a contradiction.

From Lemma 2.3 it follows that Φ has a fixed point u K ( Ω ¯ β Ω α ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq122_HTML.gif. Furthermore, α u β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq100_HTML.gif and u δ α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq123_HTML.gif, which means that u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq2_HTML.gif is a positive solution of Eq. (1.1). The proof is complete. □

Corollary 3.1 Assume that f γ 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq124_HTML.gif, f γ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq125_HTML.gif, I i 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq126_HTML.gif, I i > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq127_HTML.gif and
0 < λ ( λ 1 0 ω ϕ ( s ) d s δ i = 1 m I i 0 ϕ ( t i ) δ f γ 0 0 ω ϕ ( s ) d s , δ λ 1 0 ω ϕ ( s ) d s i = 1 m I ¯ i ϕ ( t i ) f ¯ γ 0 ω ϕ ( s ) d s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equac_HTML.gif
or
0 < λ ( λ 1 0 ω ϕ ( s ) d s δ i = 1 m I i ϕ ( t i ) δ f γ 0 ω ϕ ( s ) d s , δ λ 1 0 ω ϕ ( s ) d s i = 1 m I ¯ i 0 ϕ ( t i ) f ¯ γ 0 0 ω ϕ ( s ) d s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equad_HTML.gif

here γ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq104_HTML.gif on J. Then (1.1) has at least one positive solution.

Corollary 3.2 Assume that there exists a constant α such that f γ ρ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq128_HTML.gif, I i ρ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq129_HTML.gif ( ρ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq130_HTML.gif, α and ∞) and
f ¯ γ α 0 ω ϕ ( s ) d s + i = 1 m I ¯ i α ϕ ( t i ) δ 0 ω ϕ ( s ) d s < λ 1 < min { δ f γ 0 + δ i = 1 m I i 0 ϕ ( t i ) 0 ω ϕ ( s ) d s , δ f γ + δ i = 1 m I i ϕ ( t i ) 0 ω ϕ ( s ) d s } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equae_HTML.gif

here γ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq131_HTML.gif on J. Then there exists one open interval Θ : 1 Θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq132_HTML.gif such that (1.1) has at least two positive solutions for λ Θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq133_HTML.gif.

Example 1 Consider the equation
{ u ( t ) + 3 u ( t ) = λ f ( t , u ) , t J , t t i , Δ u ( t ) = I i ( u ( t i ) ) , i = 1 , 2 , , m , u ( 0 ) = u ( 1 ) , u ( 0 ) = u ( 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ21_HTML.gif
(3.14)
where p ( t ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq134_HTML.gif, q ( t ) = 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq135_HTML.gif and
f ( t , u ) = { u ρ , u 1 , 1 , u > 1 , I i ( u ) = { 3 , u 1 , 3 u , u > 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equaf_HTML.gif

here ρ > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq136_HTML.gif and i = 1 , 2 , , m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq137_HTML.gif. Since I i 0 = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq138_HTML.gif, I ¯ i = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq139_HTML.gif and f ¯ q = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq140_HTML.gif, by Theorem 3.1, (3.14) has at least one positive solution for any λ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq141_HTML.gif.

Example 2 Consider the equation
{ u ( t ) + 20 u ( t ) = λ f ( t , u ) , t J , t t i , Δ u ( t i ) = I i ( u ( t i ) ) , i = 1 , 2 , , m , u ( 0 ) = u ( 1 ) , u ( 0 ) = u ( 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ22_HTML.gif
(3.15)

where f ( t , u ) = e u 10 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq142_HTML.gif, I i ( u ) = u 2 100 ( m + i ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq143_HTML.gif.

It is well known that, for the problem consisting of the equation u = λ u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq144_HTML.gif, t ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq145_HTML.gif, and the boundary condition
u ( 0 ) = u ( 1 ) , u ( 0 ) = u ( 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equ23_HTML.gif
(3.16)
the first eigenvalue is 0 (see, for example, [[19], p.428]). It follows that the first eigenvalue is λ 1 = 20 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq146_HTML.gif for the problem consisting of the equation
u + 20 u = λ u , t ( 0 , 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_Equag_HTML.gif

and the boundary condition (3.16). Meanwhile, we can obtain the positive eigenfunction ϕ ( t ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq147_HTML.gif corresponding to λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq50_HTML.gif. It is also easy to check that δ = 2 e 1 + e https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq148_HTML.gif, f γ 0 = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq149_HTML.gif, f γ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq150_HTML.gif and I i = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq151_HTML.gif (here γ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq152_HTML.gif). So, the right-hand side of the inequality in Corollary 3.2 is obviously satisfied. Considering the monotonicity of f ( t , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq24_HTML.gif and I i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq153_HTML.gif, we can choose a sufficiently small positive constant α such that the left-hand side of the inequality is true. Therefore, by a direct application of Corollary 3.2, there exists one open interval Θ : 1 Θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq132_HTML.gif such that (3.15) has at least two positive solutions for λ Θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-3/MediaObjects/13661_2012_Article_256_IEq133_HTML.gif.

Declarations

Acknowledgements

The authors would like to thank anonymous referees very much for helpful comments and suggestions which led to the improvement of presentation and quality of work. This research was partially supported by the NNSF of China (No. 11001274, 11171085) and the Postdoctoral Science Foundation of Central South University and China (No. 2011M501280).

Authors’ Affiliations

(1)
Department of Mathematics and Statistics, Central South University
(2)
Department of Mathematics, Hangzhou Normal University

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© Liang and Shen; licensee Springer. 2013

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