Open Access

Positive solutions for a fourth-order p-Laplacian boundary value problem with impulsive effects

Boundary Value Problems20132013:120

DOI: 10.1186/1687-2770-2013-120

Received: 31 January 2013

Accepted: 15 April 2013

Published: 10 May 2013

Abstract

This paper is devoted to study the existence and multiplicity of positive solutions for the fourth-order p-Laplacian boundary value problem involving impulsive effects

{ ( | y | p 1 y ) = f ( t , y ) , t J , t t k , Δ y | t = t k = I k ( y ( t k ) ) , k = 1 , 2 , , m , y ( 0 ) = y ( 1 ) = y ( 0 ) = y ( 1 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equa_HTML.gif

where J = [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq1_HTML.gif, f C ( [ 0 , 1 ] × R + , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq2_HTML.gif, I k C ( R + , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq3_HTML.gif ( R + : = [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq4_HTML.gif). Based on a priori estimates achieved by utilizing the properties of concave functions and Jensen’s inequality, we adopt fixed point index theory to establish our main results.

MSC:34B18, 47H07, 47H11, 45M20, 26D15.

Keywords

p-Laplacian boundary value problem with impulsive effects positive solution fixed point index concave function Jensen inequality

1 Introduction

In this paper, we mainly investigate the existence and multiplicity of positive solutions for the fourth-order p-Laplacian boundary value problem with impulsive effects
{ ( | y | p 1 y ) = f ( t , y ) , t J , t t k , Δ y | t = t k = I k ( y ( t k ) ) , k = 1 , 2 , , m , y ( 0 ) = y ( 1 ) = y ( 0 ) = y ( 1 ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ1_HTML.gif
(1.1)

Here J = [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq1_HTML.gif, f C ( [ 0 , 1 ] × R + , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq2_HTML.gif, I k C ( R + , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq3_HTML.gif. Let 0 < t 1 < < t m < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq5_HTML.gif be fixed, Δ y | t = t k = y ( t k + ) y ( t k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq6_HTML.gif, where y ( t k + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq7_HTML.gif and y ( t k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq8_HTML.gif denote the right and left limit of y ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq9_HTML.gif at t = t k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq10_HTML.gif, respectively.

Fourth-order boundary value problems, including those with the p-Laplacian operator, have their origin in beam theory [1, 2], ice formation [3, 4], fluids on lungs [5], brain warping [6, 7], designing special curves on surfaces [6, 8], etc. In beam theory, more specifically, a beam with a small deformation, a beam of a material which satisfies a nonlinear power-like stress and strain law, and a beam with two-sided links which satisfies a nonlinear power-like elasticity law can be described by fourth-order differential equations along with their boundary value conditions. For the case of I k = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq11_HTML.gif, k = 1 , 2 , , m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq12_HTML.gif, and p = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq13_HTML.gif, problem (1.1) reduces to the differential equation y ( 4 ) ( t ) = f ( t , y ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq14_HTML.gif subject to boundary value conditions y ( 0 ) = y ( 1 ) = y ( 0 ) = y ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq15_HTML.gif, which can be used to model the deflection of elastic beams simply supported at the endpoints [911]. This explains the reason that the last two decades have witnessed an overgrowing interest in the research of such problems, with many papers in this direction published. We refer the interested reader to [1226] and references therein devoted to the existence of solutions for the equations with p-Laplacian operator.

In [17], Zhang et al. studied the existence and nonexistence of symmetric positive solutions of the following fourth-order boundary value problem with integral boundary conditions:
{ ( ϕ p ( u ( t ) ) ) = w ( t ) f ( t , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 g ( s ) u ( s ) d s , ϕ p ( u ( 0 ) ) = ϕ p ( u ( 1 ) ) = 0 1 h ( s ) ϕ p ( u ( s ) ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ2_HTML.gif
(1.2)

where w L 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq16_HTML.gif is nonnegative, symmetric on the interval [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq17_HTML.gif (i.e., w ( 1 t ) = w ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq18_HTML.gif for t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq19_HTML.gif), f C ( [ 0 , 1 ] × R + , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq20_HTML.gif, f ( 1 t , u ) = f ( t , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq21_HTML.gif for all ( t , x ) [ 0 , 1 ] × R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq22_HTML.gif, and g , h L 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq23_HTML.gif are nonnegative, symmetric on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq24_HTML.gif. The arguments are based upon a specially constructed cone and the fixed point theory for cones. Moreover, they also studied the nonexistence of a positive solution.

In [16], Luo and Luo considered the existence, multiplicity, and nonexistence of symmetric positive solutions for (1.2) with a ϕ-Laplacian operator and the term f involving the first derivative.

Except that, many researchers considered and studied the existence of positive solutions for a lot of impulsive boundary value problems; see, for example, [2129] and the references therein.

In [21], Feng considered the problem (1.2) with impulsive effects and he obtained the existence and multiplicity of positive solutions. The fundamental tool in this paper is Guo-Krasnosel’skii fixed point theorem on a cone. Moreover, the nonlinearity f can be allowed to grow both sublinear and superlinear. Therefore, he improved and generalized the results of [17] to some degree. However, we can easily find that these papers do only simple promotion based on their original papers, and no substantial changes.

Motivated by the works mentioned above, in this paper, we study the existence and multiplicity of positive solutions for (1.1). Nevertheless, our methodology and results in this paper are different from those in the papers cited above. The main features of this paper are as follows. Firstly, we convert the boundary value problem (1.1) into an equivalent integral equation. Next, we consider impulsive effect as a perturbation to the corresponding problem without the impulsive terms, so that we can construct an integral operator for an appropriate linear Dirichlet boundary value problem and obtain its first eigenvalue and eigenfunction. Our main results are formulated in terms of spectral radii of the linear integral operator, and our a priori estimates for positive solutions are derived by developing some properties of positive concave functions and using Jensen’s inequality. It is of interest to note that our nonlinearity f may grow superlinearly and sublinearly. The main tool used in the proofs is fixed point index theory, combined with the a priori estimates of positive solutions. Although our problem (1.1) merely involves Dirichlet boundary conditions, both our methodology and the results in this work improve and extend the corresponding ones from [2129].

2 Preliminaries

Let E : = C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq25_HTML.gif, u : = sup t [ 0 , 1 ] | u ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq26_HTML.gif. Then ( E , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq27_HTML.gif is a real Banach space. Let J : = J { t 1 , t 2 , , t m } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq28_HTML.gif and introduce the following space:
P C [ 0 , 1 ] : = { y C [ 0 , 1 ] , y | ( t k , t k + 1 ) C ( t k , t k + 1 ) , y ( t k ) = y ( t k ) , y ( t k + ) , k = 1 , 2 , , m } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equb_HTML.gif

with the norm y P C = max { y , y } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq29_HTML.gif. Then ( P C [ 0 , 1 ] , P C ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq30_HTML.gif is also a Banach space.

A function y P C [ 0 , 1 ] C 4 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq31_HTML.gif is called a solution of (1.1) if it satisfies the differential equation
( | y | p 1 y ) = f ( t , y ) , t J , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equc_HTML.gif

and the function y satisfies the conditions Δ y | t = t k = y ( t k + ) y ( t k ) = I k ( y ( t k ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq32_HTML.gif, and the Dirichlet boundary conditions y ( 0 ) = y ( 1 ) = y ( 0 ) = y ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq33_HTML.gif.

Lemma 2.1 (see [21])

If y is a solution of the integral equation
y ( t ) = 0 1 G ( t , s ) ( 0 1 G ( s , τ ) f ( τ , y ( τ ) ) d τ ) 1 p d s + k = 1 m G ( t , t k ) I k ( y ( t k ) ) : = ( A y ) ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ3_HTML.gif
(2.1)

then y is a solution of (1.1), where G ( t , s ) = min { t , s } min { 1 s , 1 t } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq34_HTML.gif, t , s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq35_HTML.gif. Note that if f C ( [ 0 , 1 ] × R + , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq2_HTML.gif, I k C ( R + , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq36_HTML.gif, then A : C [ 0 , 1 ] C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq37_HTML.gif is a completely continuous operator, and the existence of positive solutions for (1.1) is equivalent to that of positive fixed points of A.

Remark 2.1 By (2.1), we easily find y is concave on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq24_HTML.gif. Indeed,
y ( t ) = ( 0 1 G ( t , s ) f ( s , y ( s ) ) d s ) 1 p 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equd_HTML.gif

implies y is concave on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq24_HTML.gif. Furthermore, y ( t k ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq38_HTML.gif ( k = 1 , 2 , , m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq12_HTML.gif) leads to y ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq39_HTML.gif, t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq40_HTML.gif.

Let P be a cone in C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq41_HTML.gif which is defined as
P : = { y C [ 0 , 1 ] : y ( t ) t ( 1 t ) y , t J } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Eque_HTML.gif

In what follows, we prove that A ( P ) P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq42_HTML.gif.

Lemma 2.2 A ( P ) P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq42_HTML.gif.

Proof We easily see that t ( 1 t ) G ( s , s ) G ( t , s ) G ( s , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq43_HTML.gif, t , s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq35_HTML.gif. Consequently, on the one hand, we find
( A y ) ( t ) 0 1 G ( s , s ) ( 0 1 G ( s , τ ) f ( τ , y ( τ ) ) d τ ) 1 p d s + k = 1 m G ( t k , t k ) I k ( y ( t k ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equf_HTML.gif
On the other hand,
( A y ) ( t ) t ( 1 t ) [ 0 1 G ( s , s ) ( 0 1 G ( s , τ ) f ( τ , y ( τ ) ) d τ ) 1 p d s + k = 1 m G ( t k , t k ) I k ( y ( t k ) ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equg_HTML.gif

Therefore, ( A y ) ( t ) t ( 1 t ) A y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq44_HTML.gif, for any t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq19_HTML.gif, as required. This completes the proof. □

We denote B ρ : = { u E : u < ρ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq45_HTML.gif for ρ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq46_HTML.gif in the sequel.

Lemma 2.3 (see [30])

Suppose A : P P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq47_HTML.gif is a completely continuous operator and has no fixed points on B ρ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq48_HTML.gif.
  1. 1.

    If A y y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq49_HTML.gif for all y B ρ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq50_HTML.gif, then i ( A , B ρ P , P ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq51_HTML.gif, where i is fixed point index on P.

     
  2. 2.

    If A y y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq52_HTML.gif for all y B ρ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq50_HTML.gif, then i ( A , B ρ P , P ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq53_HTML.gif.

     

Lemma 2.4 (see [30])

If A : B ¯ ρ P P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq54_HTML.gif is a completely continuous operator. If there exists y 0 P { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq55_HTML.gif such that y A y λ y 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq56_HTML.gif, λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq57_HTML.gif, y B ρ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq58_HTML.gif, then i ( A , B ρ P , P ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq59_HTML.gif.

Lemma 2.5 (see [30])

If 0 B ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq60_HTML.gif and A : B ¯ ρ P P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq54_HTML.gif is a completely continuous operator. If y λ A y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq61_HTML.gif, y B ρ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq62_HTML.gif, 0 λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq63_HTML.gif, then i ( A , B ρ P , P ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq64_HTML.gif.

Lemma 2.6 Let ψ ( t ) : = sin ( π t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq65_HTML.gif. Then
0 1 G ( t , s ) ψ ( t ) d t = 1 π 2 ψ ( s ) , 0 1 G ( t , s ) ψ ( s ) d s = 1 π 2 ψ ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ4_HTML.gif
(2.2)

Lemma 2.7 (Jensen’s inequalities)

Let θ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq66_HTML.gif, n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq67_HTML.gif, a i 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq68_HTML.gif ( i = 1 , 2 , , n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq69_HTML.gif), and φ C ( [ 0 , 1 ] , R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq70_HTML.gif. Then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equh_HTML.gif

3 Main results

Let p : = max { 1 , p } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq71_HTML.gif, p : = min { 1 , p } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq72_HTML.gif, κ 1 : = 2 p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq73_HTML.gif, κ 2 : = 2 m ( p 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq74_HTML.gif, κ 3 : = 2 p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq75_HTML.gif, κ 4 : = 2 m ( p 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq76_HTML.gif, κ 5 : = 2 p p + p 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq77_HTML.gif, κ 6 : = 2 ( m + 1 ) ( p 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq78_HTML.gif. We now list our hypotheses.

(H1) There is a ρ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq79_HTML.gif such that 0 y < ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq80_HTML.gif and 0 t 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq81_HTML.gif imply
f ( t , y ) η p ρ p , I k ( y ) η k ρ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equi_HTML.gif
where η , η k 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq82_HTML.gif satisfy
η + k = 1 m η k > 0 , η 0 1 G ( s , s ) ( 0 1 G ( s , τ ) d τ ) 1 p d s + k = 1 m G ( t k , t k ) η k < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equj_HTML.gif
(H2) There exist 0 < r 0 < ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq83_HTML.gif and a 1 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq84_HTML.gif, a 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq85_HTML.gif satisfying
a 1 p p κ 1 + π 3 2 σ p a 2 p κ 2 k = 1 m sin ( π t k ) > π 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equk_HTML.gif
such that
f ( t , y ) a 1 y p , I k ( y ) a 2 y , t [ 0 , 1 ] , 0 < y < r 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ5_HTML.gif
(3.1)

where σ : = min t [ t 1 , t m ] t ( 1 t ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq86_HTML.gif.

(H3) There exist c > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq87_HTML.gif and a 3 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq88_HTML.gif, a 4 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq89_HTML.gif satisfying
a 3 p p κ 1 + π 3 2 σ p a 4 p κ 2 k = 1 m sin ( π t k ) > π 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equl_HTML.gif
such that
f ( t , y ) a 3 y p c , I k ( y ) a 4 y c , t [ 0 , 1 ] , y 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ6_HTML.gif
(3.2)
(H4) There is a ρ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq79_HTML.gif such that σ ρ y ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq90_HTML.gif and 0 t 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq81_HTML.gif imply
f ( t , y ) ξ p ρ p , I k ( y ) ξ k ρ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equm_HTML.gif
where ξ , ξ k 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq91_HTML.gif satisfy
ξ + k = 1 m ξ k > 0 , ξ t 1 t m G ( 1 2 , s ) ( 0 1 G ( s , τ ) d τ ) 1 p d s + k = 1 m G ( 1 2 , t k ) ξ k > 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equn_HTML.gif
(H5) There exist 0 < r 0 < ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq83_HTML.gif and b 1 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq92_HTML.gif, b 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq93_HTML.gif satisfying
b 1 2 + b 2 2 0 , b 1 p p κ 3 + π 2 b 2 p κ 4 k = 1 m sin ( π t k ) 0 1 ( t ( 1 t ) ) p sin ( π t ) d t < π 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equo_HTML.gif
such that
f ( t , y ) b 1 y p , I k ( y ) b 2 y , t [ 0 , 1 ] , 0 < y < r 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ7_HTML.gif
(3.3)
(H6) There exist c > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq87_HTML.gif and b 3 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq94_HTML.gif, b 4 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq95_HTML.gif satisfying
b 3 2 + b 4 2 0 , b 3 p p κ 5 + π 2 b 4 p κ 6 k = 1 m sin ( π t k ) 0 1 ( t ( 1 t ) ) p sin ( π t ) d t < π 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equp_HTML.gif
such that
f ( t , y ) b 3 y p + c , I k ( y ) b 4 y + c , t [ 0 , 1 ] , y 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ8_HTML.gif
(3.4)

Theorem 3.1 Suppose that (H1)-(H3) are satisfied. Then (1.1) has at least two positive solutions.

Proof If y B ρ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq50_HTML.gif, it follows from (H1) that
A y 0 1 G ( s , s ) ( 0 1 G ( s , τ ) f ( τ , y ( τ ) ) d τ ) 1 p d s + k = 1 m G ( t k , t k ) I k ( y ( t k ) ) ρ ( η 0 1 G ( s , s ) ( 0 1 G ( s , τ ) d τ ) 1 p d s + k = 1 m G ( t k , t k ) η k ) < ρ = y . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equq_HTML.gif
Now Lemma 2.3 yields
i ( A , B ρ P , P ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ9_HTML.gif
(3.5)
Let r ( 0 , r 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq96_HTML.gif. Then for y B r P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq97_HTML.gif, we find
y ( t ) t ( 1 t ) y σ r , t [ t 1 , t m ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ10_HTML.gif
(3.6)
where σ = min t [ t 1 , t m ] t ( 1 t ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq98_HTML.gif. Let M 1 : = { y P : y = A y + λ ψ  for some  λ 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq99_HTML.gif, where ψ ( t ) = sin ( π t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq100_HTML.gif. Next, from (H2), we prove M 1 { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq101_HTML.gif. Indeed, y M 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq102_HTML.gif implies y ( t ) ( A y ) ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq103_HTML.gif. Lemma 2.6, together with this, leads to
y p ( t ) [ 0 1 G ( t , s ) ( 0 1 G ( s , τ ) f ( τ , y ( τ ) ) d τ ) 1 p d s + k = 1 m G ( t , t k ) I k ( y ( t k ) ) ] p κ 1 [ 0 1 G ( t , s ) ( 0 1 G ( s , τ ) f ( τ , y ( τ ) ) d τ ) 1 p d s ] p + κ 1 [ k = 1 m G ( t , t k ) I k ( y ( t k ) ) ] p κ 1 0 1 0 1 G ( t , s ) G ( s , τ ) f p p ( τ , y ( τ ) ) d τ d s + κ 2 k = 1 m G ( t , t k ) I k p ( y ( t k ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ11_HTML.gif
(3.7)
Multiply both sides of the above by sin ( π t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq104_HTML.gif and integrate over [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq24_HTML.gif and use (2.2) to obtain
0 1 y p ( t ) sin ( π t ) d t κ 1 0 1 sin ( π t ) 0 1 0 1 G ( t , s ) G ( s , τ ) f p p ( τ , y ( τ ) ) d τ d s d t + κ 2 k = 1 m 0 1 sin ( π t ) G ( t , t k ) I k p ( y ( t k ) ) d t κ 1 π 4 0 1 f p p ( t , y ( t ) ) sin ( π t ) d t + κ 2 π 2 k = 1 m I k p ( y ( t k ) ) sin ( π t k ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ12_HTML.gif
(3.8)
Combining this and (3.1), we get
0 1 y p ( t ) sin ( π t ) d t a 1 p p κ 1 π 4 0 1 y p ( t ) sin ( π t ) d t + a 2 p κ 2 π 2 k = 1 m y p ( t k ) sin ( π t k ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ13_HTML.gif
(3.9)

In what follows, we will distinguish three cases.

Case 1. a 1 p p κ 1 = π 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq105_HTML.gif. By (H2), we know a 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq106_HTML.gif. (3.9) implies
a 2 p κ 2 π 2 k = 1 m y p ( t k ) sin ( π t k ) 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equr_HTML.gif

Therefore, y ( t k ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq38_HTML.gif ( k = 1 , 2 , , m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq12_HTML.gif), and then y ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq39_HTML.gif, t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq40_HTML.gif by Remark 2.1, which contradicts y B r P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq97_HTML.gif.

Case 2. a 1 p p κ 1 > π 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq107_HTML.gif. Equation (3.9) implies
( a 1 p p κ 1 π 4 1 ) 0 1 y p ( t ) sin ( π t ) d t + a 2 p κ 2 π 2 k = 1 m y p ( t k ) sin ( π t k ) 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equs_HTML.gif

and thus y ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq108_HTML.gif, t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq40_HTML.gif, which also contradicts y B r P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq97_HTML.gif.

Case 3. a 1 p p κ 1 < π 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq109_HTML.gif. Since 0 1 y p ( t ) sin ( π t ) d t 2 r p π https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq110_HTML.gif, we have by (3.6) and (3.9),
2 [ π 4 a 1 p p κ 1 ] r p π [ π 4 a 1 p p κ 1 ] 0 1 y p ( t ) sin ( π t ) π 2 σ p r p a 2 p κ 2 k = 1 m sin ( π t k ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equt_HTML.gif
Therefore,
a 1 p p κ 1 + π 3 2 σ p a 2 p κ 2 k = 1 m sin ( π t k ) π 4 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equu_HTML.gif
which contradicts (H2). So, we have y A y λ ψ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq111_HTML.gif for all y B r P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq112_HTML.gif and λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq113_HTML.gif. Now, by virtue of Lemma 2.4, we obtain
i ( A , B r P , P ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ14_HTML.gif
(3.10)
On the other hand, by (H3), we prove M 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq114_HTML.gif is bounded in P. By (3.2) together with (3.8), we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ15_HTML.gif
(3.11)

where c 1 : = 2 κ 1 c p p π 5 + κ 2 c p π 2 k = 1 m sin ( π t k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq115_HTML.gif. Now we distinguish the following two cases.

Case 1. a 3 p p κ 1 π 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq116_HTML.gif. (H3) implies
( a 3 p p κ 1 π 4 ) 0 1 t p ( 1 t ) p sin ( π t ) d t + π 2 σ p a 4 p κ 2 k = 1 m sin ( π t k ) > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equv_HTML.gif
Combining this and (3.11), we have
( a 3 p p κ 1 π 4 ) 0 1 y p ( t ) sin ( π t ) d t + π 2 a 4 p κ 2 k = 1 m y p ( t k ) sin ( π t k ) π 4 c 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equw_HTML.gif
Therefore,
y p π 4 c 1 ( a 3 p p κ 1 π 4 ) 0 1 t p ( 1 t ) p sin ( π t ) d t + π 2 σ p a 4 p κ 2 k = 1 m sin ( π t k ) : = N 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equx_HTML.gif
Case 2. a 3 p p κ 1 < π 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq117_HTML.gif. (3.11) implies
2 [ π 4 a 3 p p κ 1 ] y p π + π 4 c 1 [ π 4 a 3 p p κ 1 ] 0 1 y p ( t ) sin ( π t ) d t + π 4 c 1 π 2 a 4 p κ 2 k = 1 m y p ( t k ) sin ( π t k ) π 2 σ p a 4 p y p κ 2 k = 1 m sin ( π t k ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equy_HTML.gif
and thus
y p π 5 c 1 2 a 3 p p κ 1 + π 3 σ p a 4 p κ 2 k = 1 m sin ( π t k ) 2 π 4 : = N 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equz_HTML.gif
Therefore, we obtain the boundedness of M 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq114_HTML.gif, as claimed. Taking R > sup { ρ , N 1 p , N 2 p } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq118_HTML.gif, we have y A y λ ψ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq111_HTML.gif for all y B R P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq119_HTML.gif and λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq113_HTML.gif. Now, by virtue of Lemma 2.4, we obtain
i ( A , B R P , P ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ16_HTML.gif
(3.12)
Combining (3.5), (3.10), and (3.12), we arrive at
i ( A , ( B R B ¯ ρ ) P , P ) = 0 1 = 1 , i ( A , ( B ρ B ¯ r ) P , P ) = 1 0 = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equaa_HTML.gif

Now A has at least two fixed points, one on ( B R B ¯ ρ ) P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq120_HTML.gif and the other on ( B ρ B ¯ r ) P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq121_HTML.gif. Hence (1.1) has at least two positive solutions. The proof is completed. □

Theorem 3.2 Suppose that (H4)-(H6) are satisfied. Then (1.1) has at least two positive solutions.

Proof If y B ρ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq50_HTML.gif, then we find
y ( t ) t ( 1 t ) y = σ ρ , t [ t 1 , t m ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ17_HTML.gif
(3.13)
By (H4),
( A y ) ( 1 2 ) t 1 t m G ( 1 2 , s ) ( 0 1 G ( s , τ ) f ( τ , y ( τ ) ) d τ ) 1 p d s + k = 1 m G ( 1 2 , t k ) I k ( y ( t k ) ) ρ ( ξ t 1 t m G ( 1 2 , s ) ( 0 1 G ( s , τ ) d τ ) 1 p d s + k = 1 m G ( 1 2 , t k ) ξ k ) > ρ = y , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equab_HTML.gif
so that
A y > y , y B ρ P . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equac_HTML.gif
Now Lemma 2.3 yields
i ( A , B ρ P , P ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ18_HTML.gif
(3.14)
Let r ( 0 , r 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq96_HTML.gif. Then for y B r P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq97_HTML.gif, we find
y ( t ) t ( 1 t ) y = t ( 1 t ) r , t [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ19_HTML.gif
(3.15)
Let M 2 : = { y P : y = λ A y  for some  λ [ 0 , 1 ] } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq122_HTML.gif. Next, from (H5), we prove M 2 = { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq123_HTML.gif. Indeed, if y M 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq124_HTML.gif, we have
y p ( t ) ( A y ) p ( t ) = [ 0 1 G ( t , s ) ( 0 1 G ( s , τ ) f ( τ , y ( τ ) ) d τ ) 1 p d s + k = 1 m G ( t , t k ) I k ( y ( t k ) ) ] p κ 3 0 1 0 1 G ( t , s ) G ( s , τ ) f p p ( τ , y ( τ ) ) d τ d s + κ 4 k = 1 m G ( t , t k ) I k p ( y ( t k ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equad_HTML.gif
Multiply both sides of the above by sin ( π t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq104_HTML.gif and integrate over [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq24_HTML.gif and use (2.2) to obtain
0 1 y p ( t ) sin ( π t ) d t κ 3 π 4 0 1 f p p ( t , y ( t ) ) sin ( π t ) d t + κ 4 π 2 k = 1 m I k p ( y ( t k ) ) sin ( π t k ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ20_HTML.gif
(3.16)
Combining this and (3.3), we have
0 1 y p ( t ) sin ( π t ) d t b 1 p p κ 3 π 4 0 1 y p ( t ) sin ( π t ) d t + b 2 p κ 4 π 2 k = 1 m y p ( t k ) sin ( π t k ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equae_HTML.gif
Consequently,
r p ( π 4 b 1 p p κ 3 ) 0 1 ( t ( 1 t ) ) p sin ( π t ) d t ( π 4 b 1 p p κ 3 ) 0 1 y p ( t ) sin ( π t ) d t r p π 2 b 2 p κ 4 k = 1 m sin ( π t k ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equaf_HTML.gif
which contradicts (H5). This implies M 2 = { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq125_HTML.gif, and thus y λ A y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq61_HTML.gif for all y B r P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq97_HTML.gif and λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq126_HTML.gif. Now Lemma 2.5 yields
i ( A , B r P , P ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ21_HTML.gif
(3.17)
On the other hand, by (H6), we prove M 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq127_HTML.gif is bounded in P. By (3.4) together with (3.16), we obtain
0 1 y p ( t ) sin ( π t ) d t κ 3 π 4 0 1 ( b 3 y p ( t ) + c ) p p sin ( π t ) d t + κ 4 π 2 k = 1 m ( b 4 y ( t k ) + c ) p sin ( π t k ) b 3 p p κ 5 π 4 0 1 y p ( t ) sin ( π t ) d t + b 4 p κ 6 π 2 k = 1 m y p ( t k ) sin ( π t k ) + c 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equag_HTML.gif
where c 2 : = 2 κ 5 c p p π 5 + κ 6 c p π 2 k = 1 m sin ( π t k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq128_HTML.gif. Therefore,
y p ( π 4 b 3 p p κ 5 ) 0 1 ( t ( 1 t ) ) p sin ( π t ) d t ( π 4 b 3 p p κ 5 ) 0 1 y p ( t ) sin ( π t ) d t y p π 2 b 4 p κ 6 k = 1 m sin ( π t k ) + π 4 c 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equah_HTML.gif
namely,
y p π 4 c 2 ( π 4 b 3 p p κ 5 ) 0 1 ( t ( 1 t ) ) p sin ( π t ) d t π 2 b 4 p κ 6 k = 1 m sin ( π t k ) : = N 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equai_HTML.gif
This proves the boundedness of M 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq127_HTML.gif, as required. Choosing R > N 2 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq129_HTML.gif and R > ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq130_HTML.gif, we have y λ A y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq61_HTML.gif for all y B R P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq119_HTML.gif and λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq131_HTML.gif. Now Lemma 2.5 yields
i ( A , B R P , P ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ22_HTML.gif
(3.18)
Combining (3.14), (3.17), and (3.18), we obtain
i ( A , ( B R B ¯ ρ ) P , P ) = 1 0 = 1 , i ( A , ( B ρ B ¯ r ) P , P ) = 0 1 = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equaj_HTML.gif

Hence A has at least two fixed points, one on ( B R B ¯ ρ ) P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq132_HTML.gif and the other on ( B ρ B ¯ r ) P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq133_HTML.gif, and thus (1.1) has at least two positive solutions. The proof is completed. □

4 An example

Let us consider the problem
{ ( | y | p 1 y ) = y α + y β , t J , 0 < α < p < β , Δ y | t = t k = c k y ( t k ) , c k 0 , k = 1 , 2 , , m , y ( 0 ) = y ( 1 ) = y ( 0 ) = y ( 1 ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equ23_HTML.gif
(4.1)
Taking ρ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq134_HTML.gif in (H1), k = 1 m G ( t k , t k ) c k < 2 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq135_HTML.gif, and η > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq136_HTML.gif is chosen such that 2 < η < 2 6 1 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq137_HTML.gif. Set f ( t , y ) = y α + y β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq138_HTML.gif, 0 < α < p < β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq139_HTML.gif, η k = c k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq140_HTML.gif. Therefore, f ( t , y ) ρ α + ρ β = 2 < η p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq141_HTML.gif, I k ( y ) = c k y c k ρ = η k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq142_HTML.gif, and
η + k = 1 m η k > 0 , η 0 1 G ( s , s ) ( 0 1 G ( s , τ ) d τ ) 1 p d s + k = 1 m G ( t k , t k ) η k < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equak_HTML.gif
As a result, (H1) holds. On the other hand, by simple computation, we have
lim inf y 0 + min t [ 0 , 1 ] f ( t , y ) y p = + , lim inf y + min t [ 0 , 1 ] f ( t , y ) y p = + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_Equal_HTML.gif
Therefore,
  1. (i)

    There exist 0 < r 0 < ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq83_HTML.gif and a 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq143_HTML.gif, a 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq144_HTML.gif such that (H2) holds.

     
  2. (ii)

    There exist c > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq87_HTML.gif and a 3 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq145_HTML.gif, a 4 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-120/MediaObjects/13661_2013_Article_343_IEq146_HTML.gif such that (H3) holds.

     

Consequently, the problem (4.1) has at least two positive solutions by Theorem 3.1.

Declarations

Acknowledgements

Research supported by the NNSF-China (10971046), Shandong and Hebei Provincial Natural Science Foundation (ZR2012AQ007, A2012402036), GIIFSDU (yzc12063), IIFSDU (2012TS020) and the Project of Shandong Province Higher Educational Science and Technology Program (J09LA55).

Authors’ Affiliations

(1)
School of Mathematics, Shandong University
(2)
Department of Mathematics, Qilu Normal University
(3)
Department of Mathematics, Hebei University of Engineering

References

  1. Bernis F: Compactness of the support in convex and non-convex fourth order elasticity problem. Nonlinear Anal. 1982, 6: 1221-1243. 10.1016/0362-546X(82)90032-3MathSciNetView Article
  2. Zill D, Cullen M: Differential Equations with Boundary Value Problems. 5th edition. Brooks/Cole, Pacific Grove; 2001.
  3. Myers T, Charpin J: A mathematical model for atmospheric ice accretion and water flow on a cold surface. Int. J. Heat Mass Transf. 2004, 47: 5483-5500. 10.1016/j.ijheatmasstransfer.2004.06.037View Article
  4. Myers T, Charpin J, Chapman S: The flow and solidification of thin fluid film on an arbitrary three-dimensional surface. Phys. Fluids 2002, 12: 2788-2803.MathSciNetView Article
  5. Halpern D, Jensen O, Grotberg J: A theoretic study of surfactant and liquid delivery into the lungs. J. Appl. Physiol. 1998, 85: 333-352.
  6. Meméli F, Sapiro G, Thompson P: Implicit brain imaging. Hum. Brain Mapp. 2004, 23: 179-188.
  7. Toga A: Brain Warping. Academic Press, New York; 1998.
  8. Hofer M, Pottmann H: Energy-minimizing splines in manifolds. ACM Trans. Graph. 2004, 23: 284-293. 10.1145/1015706.1015716View Article
  9. Li Y: Existence and multiplicity positive solutions for fourth-order boundary value problems. Acta Math. Appl. Sin. 2003, 26: 109-116. (in Chinese)MathSciNet
  10. O’Regan D: Fourth (and higher) order singular boundary value problems. Nonlinear Anal. 1990, 14: 1001-1038. 10.1016/0362-546X(90)90066-PMathSciNetView Article
  11. O’Regan D: Solvability of some fourth (and higher) order singular boundary value problems. J. Math. Anal. Appl. 1991, 161: 78-116. 10.1016/0022-247X(91)90363-5MathSciNetView Article
  12. Graef J, Kong L: Necessary and sufficient conditions for the existence of symmetric positive solutions of singular boundary value problems. J. Math. Anal. Appl. 2007, 331: 1467-1484. 10.1016/j.jmaa.2006.09.046MathSciNetView Article
  13. Graef J, Kong L: Necessary and sufficient conditions for the existence of symmetric positive solutions of multi-point boundary value problems. Nonlinear Anal. 2008, 68: 1529-1552. 10.1016/j.na.2006.12.037MathSciNetView Article
  14. Li J, Shen J: Existence of three positive solutions for boundary value problems with p -Laplacian. J. Math. Anal. Appl. 2005, 311: 457-465. 10.1016/j.jmaa.2005.02.054MathSciNetView Article
  15. Zhao J, Wang L, Ge W: Necessary and sufficient conditions for the existence of positive solutions of fourth order multi-point boundary value problems. Nonlinear Anal. 2010, 72: 822-835. 10.1016/j.na.2009.07.036MathSciNetView Article
  16. Luo Y, Luo Z: Symmetric positive solutions for nonlinear boundary value problems with ϕ -Laplacian operator. Appl. Math. Lett. 2010, 23: 657-664. 10.1016/j.aml.2010.01.027MathSciNetView Article
  17. Zhang X, Feng M, Ge W: Symmetric positive solutions for p -Laplacian fourth-order differential equations with integral boundary conditions. J. Comput. Appl. Math. 2008, 222: 561-573. 10.1016/j.cam.2007.12.002MathSciNetView Article
  18. Zhao X, Ge W: Successive iteration and positive symmetric solution for a Sturm-Liouville-like four-point boundary value problem with a p -Laplacian operator. Nonlinear Anal. 2009, 71: 5531-5544. 10.1016/j.na.2009.04.060MathSciNetView Article
  19. Yang J, Wei Z: Existence of positive solutions for fourth-order m -point boundary value problems with a one-dimensional p -Laplacian operator. Nonlinear Anal. 2009, 71: 2985-2996. 10.1016/j.na.2009.01.191MathSciNetView Article
  20. Xu J, Yang Z: Positive solutions for a fourth order p -Laplacian boundary value problem. Nonlinear Anal. 2011, 74: 2612-2623. 10.1016/j.na.2010.12.016MathSciNetView Article
  21. Feng M: Multiple positive solutions of fourth-order impulsive differential equations with integral boundary conditions and one-dimensional p -Laplacian. Bound. Value Probl. 2011., 2011: Article ID 654871
  22. Xu J, Kang P, Wei Z: Singular multipoint impulsive boundary value problem with p -Laplacian operator. J. Appl. Math. Comput. 2009, 30: 105-120. 10.1007/s12190-008-0160-2MathSciNetView Article
  23. Zhang X, Ge W: Impulsive boundary value problems involving the one-dimensional p -Laplacian. Nonlinear Anal. 2009, 70: 1692-1701. 10.1016/j.na.2008.02.052MathSciNetView Article
  24. Feng M, Du B, Ge W: Impulsive boundary value problems with integral boundary conditions and one-dimensional p -Laplacian. Nonlinear Anal. 2009, 70: 3119-3126. 10.1016/j.na.2008.04.015MathSciNetView Article
  25. Bai L, Dai B: Three solutions for a p -Laplacian boundary value problem with impulsive effects. Appl. Math. Comput. 2011, 217: 9895-9904. 10.1016/j.amc.2011.03.097MathSciNetView Article
  26. Shi G, Meng X: Monotone iterative for fourth-order p -Laplacian boundary value problems with impulsive effects. Appl. Math. Comput. 2006, 181: 1243-1248. 10.1016/j.amc.2006.02.024MathSciNetView Article
  27. Zhang X, Yang X, Ge W: Positive solutions of n th-order impulsive boundary value problems with integral boundary conditions in Banach spaces. Nonlinear Anal. 2009, 71: 5930-5945. 10.1016/j.na.2009.05.016MathSciNetView Article
  28. Zhang X, Feng M, Ge W: Existence of solutions of boundary value problems with integral boundary conditions for second-order impulsive integro-differential equations in Banach spaces. J. Comput. Appl. Math. 2010, 233: 1915-1926. 10.1016/j.cam.2009.07.060MathSciNetView Article
  29. Lin X, Jiang D: Multiple positive solutions of Dirichlet boundary value problems for second order impulsive differential equations. J. Math. Anal. Appl. 2006, 321: 501-514. 10.1016/j.jmaa.2005.07.076MathSciNetView Article
  30. Guo D, Lakshmikantham V: Nonlinear Problems in Abstract Cones. Academic Press, Orlando; 1988.

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