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 ,

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 [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 ,
(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, [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 ] , 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])

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 ) ,
(2.1)

then y is 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 of A.

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. 1.

    If A y y for all y B ρ P , then i ( A , B ρ P , P ) = 1 , where i is fixed point index on P.

     
  2. 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 2.6 Let 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
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 } , 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 3.1 Suppose 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
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 ) . 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 3.2 Suppose 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,
  1. (i)

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

     
  2. (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.

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 ArticleGoogle Scholar
  2. Zill D, Cullen M: Differential Equations with Boundary Value Problems. 5th edition. Brooks/Cole, Pacific Grove; 2001.Google Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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.Google Scholar
  6. Meméli F, Sapiro G, Thompson P: Implicit brain imaging. Hum. Brain Mapp. 2004, 23: 179-188.Google Scholar
  7. Toga A: Brain Warping. Academic Press, New York; 1998.Google Scholar
  8. Hofer M, Pottmann H: Energy-minimizing splines in manifolds. ACM Trans. Graph. 2004, 23: 284-293. 10.1145/1015706.1015716View ArticleGoogle Scholar
  9. Li Y: Existence and multiplicity positive solutions for fourth-order boundary value problems. Acta Math. Appl. Sin. 2003, 26: 109-116. (in Chinese)MathSciNetGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 654871Google Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  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 ArticleGoogle Scholar
  30. Guo D, Lakshmikantham V: Nonlinear Problems in Abstract Cones. Academic Press, Orlando; 1988.Google Scholar

Copyright

© Zhang et al.; licensee Springer 2013

This article is published under license to BioMed Central Ltd. 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.