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

Keyu Zhang12*, Jiafa Xu1 and Wei Dong3

Author Affiliations

1 School of Mathematics, Shandong University, Jinan, Shandong, 250100, China

2 Department of Mathematics, Qilu Normal University, Jinan, Shandong, 250013, China

3 Department of Mathematics, Hebei University of Engineering, Handan, Hebei, 056038, China

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Boundary Value Problems 2013, 2013:120  doi:10.1186/1687-2770-2013-120


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2013/1/120


Received:31 January 2013
Accepted:15 April 2013
Published:10 May 2013

© 2013 Zhang et al.; licensee Springer

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 ,

where J = [ 0 , 1 ] , f C ( [ 0 , 1 ] × R + , R + ) , I k C ( R + , R + ) ( R + : = [ 0 , ) ). 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 . (1.1)

Here J = [ 0 , 1 ] , f C ( [ 0 , 1 ] × R + , R + ) , I k C ( R + , R + ) . Let 0 < t 1 < < t m < 1 be fixed, Δ y | t = t k = y ( t k + ) y ( t k ) , where y ( t k + ) and y ( t k ) denote the right and left limit of y ( t ) at t = t k , 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 , k = 1 , 2 , , m , and p = 1 , problem (1.1) reduces to the differential equation y ( 4 ) ( t ) = f ( t , y ( t ) ) subject to boundary value conditions y ( 0 ) = y ( 1 ) = y ( 0 ) = y ( 1 ) = 0 , which can be used to model the deflection of elastic beams simply supported at the endpoints [9-11]. 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 [12-26] 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 , (1.2)

where w L 1 [ 0 , 1 ] is nonnegative, symmetric on the interval [ 0 , 1 ] (i.e., w ( 1 t ) = w ( t ) for t [ 0 , 1 ] ), f C ( [ 0 , 1 ] × R + , R + ) , f ( 1 t , u ) = f ( t , u ) for all ( t , x ) [ 0 , 1 ] × R + , and g , h L 1 [ 0 , 1 ] are nonnegative, symmetric on [ 0 , 1 ] . 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, [21-29] 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 [21-29].

2 Preliminaries

Let E : = C [ 0 , 1 ] , u : = sup t [ 0 , 1 ] | u ( t ) | . Then ( E , ) is a real Banach space. Let J : = J { t 1 , t 2 , , t m } 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 }

with the norm y P C = max { y , y } . Then ( P C [ 0 , 1 ] , P C ) is also a Banach space.

A function y P C [ 0 , 1 ] C 4 ( J ) is called a solution of (1.1) if it satisfies the differential equation

( | y | p 1 y ) = f ( t , y ) , t J ,

and the function y satisfies the conditions Δ y | t = t k = y ( t k + ) y ( t k ) = I k ( y ( t k ) ) , and the Dirichlet boundary conditions y ( 0 ) = y ( 1 ) = y ( 0 ) = y ( 1 ) = 0 .

Lemma 2.1 (see [21])

Ifyis 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 ) , (2.1)

thenyis a solution of (1.1), where G ( t , s ) = min { t , s } min { 1 s , 1 t } , t , s [ 0 , 1 ] . Note that if f C ( [ 0 , 1 ] × R + , R + ) , I k C ( R + , R + ) , then A : C [ 0 , 1 ] C [ 0 , 1 ] is a completely continuous operator, and the existence of positive solutions for (1.1) is equivalent to that of positive fixed points ofA.

Remark 2.1 By (2.1), we easily find y is concave on [ 0 , 1 ] . Indeed,

y ( t ) = ( 0 1 G ( t , s ) f ( s , y ( s ) ) d s ) 1 p 0

implies y is concave on [ 0 , 1 ] . Furthermore, y ( t k ) = 0 ( k = 1 , 2 , , m ) leads to y ( t ) 0 , t [ 0 , 1 ] .

Let P be a cone in C [ 0 , 1 ] which is defined as

P : = { y C [ 0 , 1 ] : y ( t ) t ( 1 t ) y , t J } .

In what follows, we prove that A ( P ) P .

Lemma 2.2 A ( P ) P .

Proof We easily see that t ( 1 t ) G ( s , s ) G ( t , s ) G ( s , s ) , t , s [ 0 , 1 ] . 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 ) ) .

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 ) ) ] .

Therefore, ( A y ) ( t ) t ( 1 t ) A y , for any t [ 0 , 1 ] , as required. This completes the proof. □

We denote B ρ : = { u E : u < ρ } for ρ > 0 in the sequel.

Lemma 2.3 (see [30])

Suppose A : P P is a completely continuous operator and has no fixed points on B ρ P .

1. If A y y for all y B ρ P , then i ( A , B ρ P , P ) = 1 , whereiis fixed point index onP.

2. If A y y for all y B ρ P , then i ( A , B ρ P , P ) = 0 .

Lemma 2.4 (see [30])

If A : B ¯ ρ P P is a completely continuous operator. If there exists y 0 P { 0 } such that y A y λ y 0 , λ 0 , y B ρ P , then i ( A , B ρ P , P ) = 0 .

Lemma 2.5 (see [30])

If 0 B ρ and A : B ¯ ρ P P is a completely continuous operator. If y λ A y , y B ρ P , 0 λ 1 , then i ( A , B ρ P , P ) = 1 .

Lemma 2.6Let ψ ( t ) : = sin ( π t ) . 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 ) . (2.2)

Lemma 2.7 (Jensen’s inequalities)

Let θ > 0 , n 1 , a i 0 ( i = 1 , 2 , , n ), and φ C ( [ 0 , 1 ] , R + ) . Then

3 Main results

Let p : = max { 1 , p } , p : = min { 1 , p } , κ 1 : = 2 p 1 , κ 2 : = 2 m ( p 1 ) , κ 3 : = 2 p 1 , κ 4 : = 2 m ( p 1 ) , κ 5 : = 2 p p + p 2 , κ 6 : = 2 ( m + 1 ) ( p 1 ) . We now list our hypotheses.

(H1) There is a ρ > 0 such that 0 y < ρ and 0 t 1 imply

f ( t , y ) η p ρ p , I k ( y ) η k ρ ,

where η , η k 0 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 .

(H2) There exist 0 < r 0 < ρ and a 1 0 , a 2 0 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

such that

f ( t , y ) a 1 y p , I k ( y ) a 2 y , t [ 0 , 1 ] , 0 < y < r 0 , (3.1)

where σ : = min t [ t 1 , t m ] t ( 1 t ) > 0 .

(H3) There exist c > 0 and a 3 0 , a 4 0 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

such that

f ( t , y ) a 3 y p c , I k ( y ) a 4 y c , t [ 0 , 1 ] , y 0 . (3.2)

(H4) There is a ρ > 0 such that σ ρ y ρ and 0 t 1 imply

f ( t , y ) ξ p ρ p , I k ( y ) ξ k ρ ,

where ξ , ξ k 0 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 .

(H5) There exist 0 < r 0 < ρ and b 1 0 , b 2 0 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

such that

f ( t , y ) b 1 y p , I k ( y ) b 2 y , t [ 0 , 1 ] , 0 < y < r 0 . (3.3)

(H6) There exist c > 0 and b 3 0 , b 4 0 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

such that

f ( t , y ) b 3 y p + c , I k ( y ) b 4 y + c , t [ 0 , 1 ] , y 0 . (3.4)

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

Proof If y B ρ P , 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 .

Now Lemma 2.3 yields

i ( A , B ρ P , P ) = 1 . (3.5)

Let r ( 0 , r 0 ) . Then for y B r P , we find

y ( t ) t ( 1 t ) y σ r , t [ t 1 , t m ] , (3.6)

where σ = min t [ t 1 , t m ] t ( 1 t ) > 0 . Let M 1 : = { y P : y = A y + λ ψ  for some  λ 0 } , where ψ ( t ) = sin ( π t ) . Next, from (H2), we prove M 1 { 0 } . Indeed, y M 1 implies y ( t ) ( A y ) ( t ) . 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 ) ) . (3.7)

Multiply both sides of the above by sin ( π t ) and integrate over [ 0 , 1 ] 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 ) . (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 ) . (3.9)

In what follows, we will distinguish three cases.

Case 1. a 1 p p κ 1 = π 4 . By (H2), we know a 2 > 0 . (3.9) implies

a 2 p κ 2 π 2 k = 1 m y p ( t k ) sin ( π t k ) 0 .

Therefore, y ( t k ) = 0 ( k = 1 , 2 , , m ), and then y ( t ) 0 , t [ 0 , 1 ] by Remark 2.1, which contradicts y B r P .

Case 2. a 1 p p κ 1 > π 4 . 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 ,

and thus y ( t ) 0 , t [ 0 , 1 ] , which also contradicts y B r P .

Case 3. a 1 p p κ 1 < π 4 . Since 0 1 y p ( t ) sin ( π t ) d t 2 r p π , 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 ) .

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 ,

which contradicts (H2). So, we have y A y λ ψ for all y B r P and λ 0 . Now, by virtue of Lemma 2.4, we obtain

i ( A , B r P , P ) = 0 . (3.10)

On the other hand, by (H3), we prove M 1 is bounded in P. By (3.2) together with (3.8), we obtain

(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 ) . Now we distinguish the following two cases.

Case 1. a 3 p p κ 1 π 4 . (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 .

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 .

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 .

Case 2. a 3 p p κ 1 < π 4 . (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 ) ,

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 .

Therefore, we obtain the boundedness of M 1 , as claimed. Taking R > sup { ρ , N 1 p , N 2 p } , we have y A y λ ψ for all y B R P and λ 0 . Now, by virtue of Lemma 2.4, we obtain

i ( A , B R P , P ) = 0 . (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 .

Now A has at least two fixed points, one on ( B R B ¯ ρ ) P and the other on ( B ρ B ¯ r ) P . Hence (1.1) has at least two positive solutions. The proof is completed. □

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

Proof If y B ρ P , then we find

y ( t ) t ( 1 t ) y = σ ρ , t [ t 1 , t m ] . (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 ,

so that

A y > y , y B ρ P .

Now Lemma 2.3 yields

i ( A , B ρ P , P ) = 0 . (3.14)

Let r ( 0 , r 0 ) . Then for y B r P , we find

y ( t ) t ( 1 t ) y = t ( 1 t ) r , t [ 0 , 1 ] . (3.15)

Let M 2 : = { y P : y = λ A y  for some  λ [ 0 , 1 ] } . Next, from (H5), we prove M 2 = { 0 } . Indeed, if y M 2 , 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 ) ) .

Multiply both sides of the above by sin ( π t ) and integrate over [ 0 , 1 ] 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 ) . (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 ) .

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 ) ,

which contradicts (H5). This implies M 2 = { 0 } , and thus y λ A y for all y B r P and λ [ 0 , 1 ] . Now Lemma 2.5 yields

i ( A , B r P , P ) = 1 . (3.17)

On the other hand, by (H6), we prove M 2 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 ,

where c 2 : = 2 κ 5 c p p π 5 + κ 6 c p π 2 k = 1 m sin ( π t k ) . 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 ,

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 .

This proves the boundedness of M 2 , as required. Choosing R > N 2 p and R > ρ , we have y λ A y for all y B R P and λ [ 0 , 1 ] . Now Lemma 2.5 yields

i ( A , B R P , P ) = 1 . (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 .

Hence A has at least two fixed points, one on ( B R B ¯ ρ ) P and the other on ( B ρ B ¯ r ) P , 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 . (4.1)

Taking ρ = 1 in (H1), k = 1 m G ( t k , t k ) c k < 2 3 , and η > 0 is chosen such that 2 < η < 2 6 1 p . Set f ( t , y ) = y α + y β , 0 < α < p < β , η k = c k . Therefore, f ( t , y ) ρ α + ρ β = 2 < η p , I k ( y ) = c k y c k ρ = η k , 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 .

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 = + .

Therefore,

(i) There exist 0 < r 0 < ρ and a 1 > 0 , a 2 > 0 such that (H2) holds.

(ii) There exist c > 0 and a 3 > 0 , a 4 > 0 such that (H3) holds.

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

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

KZ and JX obtained the results in a joint research. All the authors read and approved the final manuscript.

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).

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