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This article is part of the series A Tribute to Professor Ivan Kiguradze.

Open Access Research

Positive solutions for classes of multi-parameter fourth-order impulsive differential equations with one-dimensional singular p-Laplacian

Xuemei Zhang1* and Meiqiang Feng2

Author Affiliations

1 Department of Mathematics and Physics, North China Electric Power University, Beijing, 102206, Republic of China

2 School of Applied Science, Beijing Information Science and Technology University, Beijing, 100192, Republic of China

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Boundary Value Problems 2014, 2014:112  doi:10.1186/1687-2770-2014-112

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2014/1/112


Received:29 January 2014
Accepted:29 April 2014
Published:13 May 2014

© 2014 Zhang and Feng; 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 credited.

Abstract

The authors consider the following impulsive differential equations involving the one-dimensional singular p-Laplacian: ( ϕ p ( y ( t ) ) ) = λ ω ( t ) f ( t , y ( t ) ) , t J , t t k , k = 1 , 2 , , m , Δ y | t = t k = μ I k ( t k , y ( t k ) ) , k = 1 , 2 , , m , a y ( 0 ) b y ( 0 ) = 0 1 h ( s ) y ( s ) d s , a y ( 1 ) + b y ( 1 ) = 0 1 h ( s ) y ( s ) d s , ϕ p ( y ( 0 ) ) = ϕ p ( y ( 1 ) ) = 0 1 h ( t ) ϕ p ( y ( t ) ) d t , where λ > 0 and μ > 0 are two parameters. Several new and more general existence and multiplicity results are derived in terms of different values of λ > 0 and μ > 0 . In this case, our results cover equations without impulsive effects and are compared with some recent results.

Keywords:
multi-parameter; impulsive differential equations; one-dimensional singular p-Laplacian; positive solution; cone and partial ordering

1 Introduction

The theory and applications of the fourth-order ordinary differential equation are emerging as an important area of investigation; it is often referred to as the beam equation. In [1], Sun and Wang pointed out that it is necessary and important to consider various fourth-order boundary value problems (BVPs for short) according to different forms of supporting. Owing to its importance in engineering, physics, and material mechanics, fourth-order BVPs have attracted much attention from many authors; see, for example [2-29] and the references therein.

Very recently, Zhang and Liu [30] studied the following fourth-order four-point boundary value problem without impulsive effect:

{ ( ϕ p ( x ( t ) ) ) = w ( t ) f ( t , x ( t ) ) , t [ 0 , 1 ] , x ( 0 ) = 0 , x ( 1 ) = a x ( ξ ) , x ( 0 ) = 0 , x ( 1 ) = b x ( η ) ,

where 0 < ξ , η < 1 , 0 a < b < 1 . By using the upper and lower solution method, fixed point theorems, and the properties of the Green’s function G ( t , s ) and H ( t , s ) , the authors give sufficient conditions for the existence of one positive solution.

In this paper, we investigate the existence of positive solutions of fourth-order impulsive differential equations with two parameters

{ ( ϕ p ( y ( t ) ) ) = λ ω ( t ) f ( t , y ( t ) ) , t J , t t k , k = 1 , 2 , , m , Δ y | t = t k = μ I k ( t k , y ( t k ) ) , k = 1 , 2 , , m , a y ( 0 ) b y ( 0 ) = 0 1 g ( s ) y ( s ) d s , a y ( 1 ) + b y ( 1 ) = 0 1 g ( s ) y ( s ) d s , ϕ p ( y ( 0 ) ) = ϕ p ( y ( 1 ) ) = 0 1 h ( s ) ϕ p ( y ( s ) ) d s , (1.1)

where λ > 0 and μ > 0 are two parameters, a , b > 0 , J = [ 0 , 1 ] , ϕ p ( s ) is a p-Laplace operator, i.e., ϕ p ( s ) = | s | p 2 s , p > 1 , ( ϕ p ) 1 = ϕ q , 1 p + 1 q = 1 , ω is a nonnegative measurable function on ( 0 , 1 ) , ω 0 on any open subinterval in ( 0 , 1 ) which may be singular at t = 0 and/or t = 1 , t k ( k = 1 , 2 , , m ) (where m is fixed positive integer) are fixed points with 0 = t 0 < t 1 < t 2 < < t k < < t m < t m + 1 = 1 , Δ y | t = t k = y ( t k + ) x ( t k ) , where y ( t k + ) and y ( t k ) represent the right-hand limit and left-hand limit of y ( t ) at t = t k , respectively. In addition, ω, f, I k , g, and h satisfy

(H1) ω L loc 1 ( 0 , 1 ) ;

(H2) f C ( [ 0 , 1 ] × [ 0 , + ) , [ 0 , + ) ) with f ( t , y ) > 0 for all t and y > 0 ;

(H3) I k C ( [ 0 , 1 ] × [ 0 , + ) , [ 0 , + ) ) with I k ( t , y ) > 0 ( k = 1 , 2 , , n ) for all t and y > 0 ;

(H4) g , h L 1 [ 0 , 1 ] are nonnegative and ξ [ 0 , a ) , ν [ 0 , 1 ) , where

ξ = 0 1 g ( t ) d t , ν = 0 1 h ( t ) d t . (1.2)

Some special cases of (1.1) have been investigated. For example, Bai and Wang [14] studied the existence of multiple solutions of problem (1.1) with p = 2 , I k = 0 , k = 1 , 2 , , m and ω 1 for t J . By using a fixed point theorem and degree theory, the authors proved the existence of one or two positive solutions of problem (1.1).

Feng [31] considered problem (1.1) with λ = 1 , I k ( t k , y ( t k ) ) = I k ( y ( t k ) ) , ω 1 for t J and μ = 1 . By using a suitably constructed cone and fixed point theory for cones, the author proved the existence results of multiple positive solutions of problem (1.1).

Motivated by the papers mentioned above, we will extend the results of [14,30,31] to problem (1.1). We remark that on impulsive differential equations with a parameter only a few results have been obtained, not to mention impulsive differential equations with two parameters; see, for instance, [32-34]. However, these results only dealt with the case that p = 2 and μ = 1 .

The rest of the paper is organized as follows: in Section 2, we state the main results of problem (1.1). In Section 3, we provide some preliminary results, and the proofs of the main results together with several technical lemmas are given in Section 4.

2 Main results

In this section, we state the main results, including existence and multiplicity of positive solutions for problem (1.1).

We begin by introducing the notation

f 0 = lim sup y 0 + max t J f ( t , y ) ϕ p ( y ) , f = lim sup y max t J f ( t , y ) ϕ p ( y ) , f 0 = lim inf y 0 + min t J f ( t , y ) ϕ p ( y ) , f = lim inf y min t J f ( t , y ) ϕ p ( y ) , I 0 ( k ) = lim sup y 0 + max t J I k ( t , y ) y , I ( k ) = lim sup y max t J I k ( t , y ) y , I 0 ( k ) = lim inf y 0 + min t J I k ( t , y ) y , I ( k ) = lim inf y min t J I k ( t , y ) y , k = 1 , 2 , , m .

We also choose four numbers r, r 1 , r 2 , and R satisfying

0 < r < r 1 < δ r 2 < r 2 < R < + , (2.1)

where δ is defined in (3.20).

Theorem 2.1Assume that (H1)-(H4) hold.

(i) If f = 0 and I = 0 , then there exist λ 0 > 0 and μ 0 > 0 such that, for any λ > λ 0 and μ > μ 0 , problem (1.1) has a positive solution u ( t ) , t J with

δ r u ( t ) 1 δ R , t J . (2.2)

(ii) If f 0 = 0 and I 0 = 0 , then there exist λ 0 > 0 and μ 0 > 0 such that, for any λ > λ 0 and μ > μ 0 , problem (1.1) has a positive solution u ( t ) with

δ r u ( t ) R , t J . (2.3)

(iii) If f 0 = f = I = I 0 = 0 , then there exist λ 0 > 0 and μ 0 > 0 such that, for any λ > λ 0 and μ > μ 0 , problem (1.1) has at least two positive solutions u 1 ( t ) and u 2 ( t ) with

δ r u ( t ) r 1 < δ r 2 u 2 ( t ) R , t J . (2.4)

Theorem 2.2Assume that (H1)-(H4) hold.

(i) If f = + and I = + , then there exist λ ¯ 0 > 0 and μ ¯ 0 > 0 such that, for any 0 < λ < λ ¯ 0 and 0 < μ < μ ¯ 0 , problem (1.1) has a positive solution u ( t ) , t J with property (2.2).

(ii) If f 0 = + and I 0 = + , then there exist λ ¯ 0 > 0 and μ ¯ 0 > 0 such that, for any 0 < λ < λ ¯ 0 and 0 < μ < μ ¯ 0 , problem (1.1) has a positive solution u ( t ) , t J with property (2.3).

(iii) If f 0 = f = I = I 0 = + , then there exist λ ¯ 0 > 0 and μ ¯ 0 > 0 such that, for any 0 < λ < λ ¯ 0 and 0 < μ < μ ¯ 0 , problem (1.1) has at least two positive solutions u 1 ( t ) and u 2 ( t ) with

δ r u ( t ) r 1 < δ r 2 u 2 ( t ) 1 δ R , t J . (2.5)

3 Preliminaries

Let J = J { t 1 , t 2 , , t m } , and

P C 1 [ 0 , 1 ] = { y C [ 0 , 1 ] : y | ( t k , t k + 1 ) C ( t k , t k + 1 ) , y ( t k ) , y ( t k + )  exists , k = 1 , 2 , , m } .

Then P C 1 [ 0 , 1 ] is a real Banach space with norm

y P C 1 = max { y , y } , (3.1)

where y = sup t J | y ( t ) | , y = sup t J | y ( t ) | .

A function y P C 1 [ 0 , 1 ] C 4 ( J ) with φ p ( y ) C 2 ( 0 , 1 ) is called a solution of problem (1.1) if it satisfies (1.1).

We shall reduce problem (1.1) to an integral equation. To this goal, firstly by means of the transformation

ϕ p ( y ( t ) ) = x ( t ) , (3.2)

we convert problem (1.1) into

{ x ( t ) + λ ω ( t ) f ( t , y ( t ) ) = 0 , t J , x ( 0 ) = x ( 1 ) = 0 1 h ( t ) x ( t ) d t (3.3)

and

{ y ( t ) = ϕ q ( x ( t ) ) , t J , t t k , Δ y | t = t k = μ I k ( t k , y ( t k ) ) , k = 1 , 2 , , m , a y ( 0 ) b y ( 0 ) = 0 1 g ( s ) y ( s ) d s , a y ( 1 ) + b y ( 1 ) = 0 1 g ( s ) y ( s ) d s . (3.4)

Lemma 3.1If (H1), (H2), and (H4) hold, then problem (3.3) has a unique solutionxgiven by

x ( t ) = λ 0 1 H ( t , s ) ω ( s ) f ( s , y ( s ) ) d s , (3.5)

where

H ( t , s ) = G ( t , s ) + 1 1 ν 0 1 G ( s , τ ) h ( τ ) d τ , (3.6)

G ( t , s ) = { t ( 1 s ) , 0 t s 1 , s ( 1 t ) , 0 s t 1 . (3.7)

Proof The proof of Lemma 3.1 is similar to that of Lemma 2.1 in [31]. □

Write e ( t ) = t ( 1 t ) . Then from (3.6) and (3.7), we can prove that H ( t , s ) and G ( t , s ) have the following properties.

Proposition 3.1If (H4) holds, then we have

H ( t , s ) > 0 , G ( t , s ) > 0 , t , s ( 0 , 1 ) , (3.8)

H ( t , s ) 0 , G ( t , s ) 0 , t , s J , (3.9)

e ( t ) e ( s ) G ( t , s ) G ( t , t ) = t ( 1 t ) = e ( t ) e ¯ = max t [ 0 , 1 ] e ( t ) = 1 4 , t , s J , (3.10)

ρ e ( s ) H ( t , s ) γ s ( 1 s ) = γ e ( s ) 1 4 γ , t , s J , (3.11)

where

γ = 1 1 ν , ρ = 0 1 e ( τ ) h ( τ ) d τ 1 ν . (3.12)

Remark 3.1 From (3.6) and (3.11), we obtain

ρ e ( s ) H ( s , s ) γ s ( 1 s ) = γ e ( s ) 1 4 γ , s J .

Lemma 3.2If (H1), (H3), and (H4) hold, then problem (3.4) has a unique solutionyandycan be expressed in the form

y ( t ) = 0 1 H 1 ( t , s ) ϕ q ( x ( s ) ) d s + μ k = 1 m H 1 ( t , t k ) I k ( t k , y ( t k ) ) , (3.13)

where

H 1 ( t , s ) = G 1 ( t , s ) + 1 a ξ 0 1 G 1 ( s , τ ) g ( τ ) d τ , (3.14)

G 1 ( t , s ) = 1 d { ( b + a s ) ( b + a ( 1 t ) ) , if  0 s t 1 , ( b + a t ) ( b + a ( 1 s ) ) , if  0 t s 1 , d = a ( 2 b + a ) . (3.15)

Proof The proof of Lemma 3.2 is similar to that of Lemma 2.2 in [31]. □

From (3.14) and (3.15), we can prove that H 1 ( t , s ) and G 1 ( t , s ) have the following properties.

Proposition 3.2If (H4) holds, then we have

H 1 ( t , s ) > 0 , G 1 ( t , s ) > 0 , t , s J ; (3.16)

1 d b 2 G 1 ( t , s ) G 1 ( s , s ) 1 d ( b + a ) 2 , t , s J , (3.17)

ρ 1 H 1 ( t , s ) H 1 ( s , s ) ρ 2 , t , s J , (3.18)

where

ρ 1 = b 2 γ 1 a + 2 b , ρ 2 = γ 1 ( b + a ) 2 a + 2 b , γ 1 = 1 a ξ .

Suppose that y is a solution of problem (1.1). Then from Lemma 3.1 and Lemma 3.2, we have

y ( t ) = 0 1 H 1 ( t , s ) ϕ q ( λ 0 1 H ( s , τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) d s + μ k = 1 m H 1 ( t , t k ) I k ( t k , y ( t k ) ) .

Define a cone in P C 1 [ 0 , 1 ] by

K = { y P C 1 [ 0 , 1 ] : y 0 , y ( t ) δ y P C 1 , t J } , (3.19)

where

δ = ρ 1 ρ q 1 ρ 2 γ q 1 . (3.20)

It is easy to see K is a closed convex cone of P C 1 [ 0 , 1 ] .

Define an operator T λ μ : K P C 1 [ 0 , 1 ] by

( T λ μ y ) ( t ) = 0 1 H 1 ( t , s ) ϕ q ( λ 0 1 H ( s , τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) d s + μ k = 1 m H 1 ( t , t k ) I k ( t k , y ( t k ) ) . (3.21)

From (3.21), we know that y P C 1 [ 0 , 1 ] is a solution of problem (1.1) if and only if y is a fixed point of operator T λ μ .

Lemma 3.3Suppose that (H1)-(H4) hold. Then T λ μ ( K ) K and T λ μ : K K is completely continuous.

Proof The proof of Lemma 3.3 is similar to that of Lemma 2.4 in [31]. □

To obtain positive solutions of problem (1.1), the following fixed point theorem in cones is fundamental, which can be found in [[35], p.94].

Lemma 3.4LetPbe a cone in a real Banach spaceE. Assume Ω 1 , Ω 2 are bounded open sets inEwith 0 Ω 1 , Ω ¯ 1 Ω 2 . If

A : P ( Ω ¯ 2 Ω 1 ) P

is completely continuous such that either

(a) A x x , x P Ω 1 and A x x , x P Ω 2 , or

(b) A x x , x P Ω 1 and A x x , x P Ω 2 ,

thenAhas at least one fixed point in P ( Ω ¯ 2 Ω 1 ) .

Remark 3.2 To make the reader clear what Ω ¯ 2 , Ω 2 , Ω 1 , and Ω 2 Ω ¯ 1 mean, we give typical examples of Ω 1 and Ω 2 , e.g.,

Ω 1 = { x C [ a , b ] : x < r } , Ω 2 = { x C [ a , b ] : x < R }

with 0 < r < R , where x = sup t [ a , b ] | x ( t ) | .

4 Proofs of the main results

For convenience we introduce the following notation:

η = φ q ( 0 1 ω ( s ) d s ) , η = φ q ( t 1 t m ω ( s ) d s )

and

Ω r = { y K : y P C 1 < r } , Ω r = { y K : y P C 1 = r } ,

where r > 0 is a constant.

Proof of Theorem 2.1 Part (i). Noticing that f ( t , y ) > 0 , I k ( t , y ) > 0 ( k = 1 , 2 , , m ) for all t and y > 0 , we can define

m r = min t J , δ r y r { f ( t , y ) } > 0 , m = min { m k , k = 1 , 2 , , m } > 0 ,

where r > 0 , and

m k = min t J , δ r y r { I k ( t , y ) } , k = 1 , 2 , , m .

Let

λ 0 ( 1 2 ρ 1 η r ) p 1 [ ρ m r t 1 ( 1 t m ) ] 1 , μ 0 1 2 m ρ 1 m r .

Then, for u K Ω r and λ > λ 0 , μ > μ 0 , we have

( T λ μ y ) ( t ) = 0 1 H 1 ( t , s ) ϕ q ( λ 0 1 H ( s , τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) d s + μ k = 1 m H 1 ( t , t k ) I k ( t k , y ( t k ) ) ρ 1 ρ q 1 φ q ( λ 0 1 e ( τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) + μ ρ 1 k = 1 m I k ( t k , y ( t k ) ) ρ 1 ρ q 1 φ q ( λ 0 1 e ( τ ) ω ( τ ) m r d τ ) + μ ρ 1 k = 1 m m = ρ 1 ρ q 1 m r q 1 λ q 1 φ q ( 0 1 e ( τ ) ω ( τ ) d τ ) + μ m ρ 1 m ρ 1 ρ q 1 m r q 1 λ q 1 φ q ( t 1 t m e ( τ ) ω ( τ ) d τ ) + μ m ρ 1 m ρ 1 ρ q 1 m r q 1 λ q 1 [ t 1 ( 1 t m ) ] q 1 φ q ( t 1 t m ω ( τ ) d τ ) + μ m ρ 1 m > ρ 1 ρ q 1 m r q 1 λ 0 q 1 [ t 1 ( 1 t m ) ] q 1 φ q ( t 1 t m ω ( τ ) d τ ) + μ 0 m ρ 1 m = ρ 1 ρ q 1 m r q 1 λ 0 q 1 [ t 1 ( 1 t m ) ] q 1 η + μ 0 m ρ 1 m 1 2 r + 1 2 r = r = y P C 1 ,

which implies that

T λ μ y P C 1 > y P C 1 , y K Ω r , λ > λ 0  and  μ > μ 0 . (4.1)

If f = 0 , I = 0 , then there exist l 1 > 0 , l 2 > 0 , and R > r > 0 such that

f ( t , y ) < l 1 φ p ( y ) , I k ( t , y ) < l 2 y , t J , y R , k = 1 , 2 , , m ,

where l 1 satisfies

2 max { ρ 2 , a ( a + b ) } η φ q ( 1 4 γ λ l 1 ) 1 , (4.2)

l 2 satisfies

2 max { ρ 2 , a ( a + b ) } m μ l 2 1 . (4.3)

Let α = R δ . Thus, when y K Ω α we have

y ( t ) δ y P C 1 = δ α = R , t J ,

and then we get

( T λ μ y ) ( t ) = 0 1 H 1 ( t , s ) ϕ q ( λ 0 1 H ( s , τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) d s ( T λ μ y ) ( t ) = + μ k = 1 m H 1 ( t , t k ) I k ( t k , y ( t k ) ) ( T λ μ y ) ( t ) ρ 2 ( 1 4 λ γ ) q 1 φ q ( 0 1 ω ( τ ) f ( τ , y ( τ ) ) d τ ) + μ ρ 2 k = 1 m I k ( t k , y ( t k ) ) ( T λ μ y ) ( t ) ρ 2 ( 1 4 λ γ ) q 1 φ q ( 0 1 ω ( τ ) l 1 ϕ p ( y ( τ ) ) d τ ) + μ ρ 2 k = 1 m l 2 y ( t k ) ( T λ μ y ) ( t ) ρ 2 ( 1 4 λ γ ) q 1 φ q ( 0 1 ω ( τ ) l 1 ϕ p ( y P C 1 ) d τ ) + μ ρ 2 k = 1 m l 2 y P C 1 ( T λ μ y ) ( t ) ρ 2 ( 1 4 λ γ ) q 1 l 1 q 1 y P C 1 φ q ( 0 1 ω ( τ ) d τ ) + μ ρ 2 m l 2 y P C 1 ( T λ μ y ) ( t ) = ρ 2 ( 1 4 λ γ ) q 1 l 1 q 1 y P C 1 η + μ ρ 2 m l 2 y P C 1 ( T λ μ y ) ( t ) 1 2 y P C 1 + 1 2 y P C 1 = y P C 1 , (4.4)

| ( T λ μ y ) ( t ) | 0 1 | H 1 t ( t , s ) | ϕ q ( λ 0 1 H ( s , τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) d s | ( T λ μ y ) ( t ) | + μ k = 1 m | H 1 t ( t , t k ) | I k ( t k , y ( t k ) ) | ( T λ μ y ) ( t ) | a ( b + a ) ( 1 4 λ γ ) q 1 φ q ( 0 1 ω ( τ ) f ( τ , y ( τ ) ) d τ ) + μ a ( b + a ) k = 1 m I k ( t k , y ( t k ) ) | ( T λ μ y ) ( t ) | a ( b + a ) ( 1 4 λ γ ) q 1 φ q ( 0 1 ω ( τ ) l 1 ϕ p ( y ( τ ) ) d τ ) + μ a ( b + a ) k = 1 m l 2 y ( t k ) | ( T λ μ y ) ( t ) | a ( b + a ) ( 1 4 λ γ ) q 1 φ q ( 0 1 ω ( τ ) l 1 ϕ p ( y P C 1 ) d τ ) | ( T λ μ y ) ( t ) | + μ a ( b + a ) k = 1 m l 2 y P C 1 | ( T λ μ y ) ( t ) | a ( b + a ) ( 1 4 λ γ ) q 1 l 1 q 1 y P C 1 η + μ a ( b + a ) m l 2 y P C 1 | ( T λ μ y ) ( t ) | 1 2 y P C 1 + 1 2 y P C 1 = y P C 1 , (4.5)

where

H 1 t ( t , s ) = G 1 t ( t , s ) = { a ( b + a s ) , if  0 s t 1 , a ( b + a ( 1 s ) ) , if  0 t s 1

and

max t , s J , t s | H 1 t ( t , s ) | = max t , s J , t s | G 1 t ( t , s ) | = a ( b + a ) .

It follows from (4.4) and (4.5) that

T λ μ y P C 1 y P C 1 , y K Ω α . (4.6)

Applying (b) of Lemma 3.4 to (4.1) and (4.6) shows that T λ μ has a fixed point y K ( Ω ¯ α Ω r ) with r y P C 1 α = 1 δ R . Hence, since for y K we have y ( t ) δ y P C 1 , t J , it follows that (2.2) holds. This gives the proof of part (i).

Part (ii). Noticing that f ( t , y ) > 0 , I k ( t , y ) > 0 ( k = 1 , 2 , , m ) for all t and y > 0 , we can define

m R = min t J , δ R y R { f ( t , y ) } > 0 , m = min { m k , k = 1 , 2 , , m } > 0 ,

where R > 0 , and

m k = min t J , δ R y R { I k ( t , y ) } , k = 1 , 2 , , m .

Let

λ 0 ( 1 2 ρ 1 η R ) p 1 [ ρ m R t 1 ( 1 t m ) ] 1 , μ 0 1 2 m ρ 1 m R .

Then, for y K Ω R and λ > λ 0 , μ > μ 0 , we have

( T λ μ y ) ( t ) = 0 1 H 1 ( t , s ) ϕ q ( λ 0 1 H ( s , τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) d s + μ k = 1 m H 1 ( t , t k ) I k ( t k , y ( t k ) ) ρ 1 ρ q 1 φ q ( λ 0 1 e ( τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) + μ ρ 1 k = 1 m I k ( t k , y ( t k ) ) ρ 1 ρ q 1 φ q ( λ 0 1 e ( τ ) ω ( τ ) m R d τ ) + μ ρ 1 k = 1 m m = ρ 1 ρ q 1 m R q 1 λ q 1 φ q ( 0 1 e ( τ ) ω ( τ ) d τ ) + μ m ρ 1 m ρ 1 ρ q 1 m R q 1 λ q 1 φ q ( t 1 t m e ( τ ) ω ( τ ) d τ ) + μ m ρ 1 m ρ 1 ρ q 1 m R q 1 λ q 1 [ t 1 ( 1 t m ) ] q 1 φ q ( t 1 t m ω ( τ ) d τ ) + μ m ρ 1 m > ρ 1 ρ q 1 m R q 1 λ 0 q 1 [ t 1 ( 1 t m ) ] q 1 φ q ( t 1 t m ω ( τ ) d τ ) + μ 0 m ρ 1 m = ρ 1 ρ q 1 m R q 1 λ 0 q 1 [ t 1 ( 1 t m ) ] q 1 η + μ 0 m ρ 1 m 1 2 R + 1 2 R = y P C 1 ,

which implies that

T λ μ y P C 1 > y P C 1 , y K Ω R , λ > λ 0  and  μ > μ 0 . (4.7)

If f 0 = 0 , I 0 = 0 , then there exist l 1 > 0 , l 2 > 0 , and 0 < r < R such that

f ( t , y ) < l 1 φ p ( y ) , I k ( t , y ) < l 2 y ( t J , 0 y r , k = 1 , 2 , , m ) ,

where l 1 and l 2 satisfy (4.2) and (4.3), respectively.

Similar to the proof of (4.6), we can prove that

T λ μ y P C 1 y P C 1 , y K Ω r . (4.8)

Applying (a) of Lemma 3.4 to (4.7) and (4.8) shows that T λ μ has a fixed point y K ( Ω ¯ R Ω r ) with r y P C 1 R . Hence, since for y K we have y ( t ) δ y P C 1 for t J , it follows that (2.3) holds. This gives the proof of part (ii).

Consider part (iii). Choose two numbers r 1 and r 2 satisfying (2.1). By part (i) and part (ii), there exist λ 0 > 0 and μ 0 > 0 such that

T λ μ y P C 1 > y P C 1 , y K Ω r i , i = 1 , 2 . (4.9)

Since f 0 = f = I = I 0 = 0 , from the proof of part (i) and part (ii), it follows that

T λ μ y P C 1 < y P C 1 , y K Ω r (4.10)

and

T λ μ y P C 1 < y P C 1 , y K Ω R . (4.11)

Applying Lemma 3.4 to (4.9)-(4.11) shows that T λ μ has two fixed points y 1 and y 2 such that y 1 K ( Ω ¯ r 1 Ω r ) and y 2 K ( Ω ¯ R Ω r 2 ) . These are the desired distinct positive solutions of problem (1.1) for λ 0 > 0 and μ 0 > 0 satisfying (2.4). Then the result of part (iii) follows. □

Proof of Theorem 2.2 Part (i). Noticing that f ( t , y ) > 0 , I k ( t , y ) > 0 ( k = 1 , 2 , , m ) for all t and y > 0 , we can define

M r = max t J , δ r y r { f ( t , y ) } > 0 , M = max { M k , k = 1 , 2 , , m } > 0 ,

where r > 0 , and

M k = max t J , δ r y r { I k ( t , y ) } , k = 1 , 2 , , m .

Let

λ ¯ 0 4 ( 1 2 max { ρ 2 , a ( a + b ) } η r ) p 1 ( M r γ ) 1 , μ ¯ 0 1 2 max { ρ 2 , a ( a + b ) } m M r .

Then, for y K Ω r and λ < λ ¯ 0 , μ < μ ¯ 0 , we have

( T λ μ y ) ( t ) = 0 1 H 1 ( t , s ) ϕ q ( λ 0 1 H ( s , τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) d s + μ k = 1 m H 1 ( t , t k ) I k ( t k , y ( t k ) ) ρ 2 ( 1 4 γ ) q 1 φ q ( λ 0 1 ω ( τ ) f ( τ , y ( τ ) ) d τ ) + μ ρ 2 k = 1 m I k ( t k , y ( t k ) ) ρ 2 ( 1 4 γ λ ) q 1 φ q ( 0 1 ω ( τ ) M r d τ ) + μ ρ 2 k = 1 m M = ρ 2 ( 1 4 γ λ M r ) q 1 φ q ( 0 1 ω ( τ ) d τ ) + μ ρ 2 m M < ρ 2 ( 1 4 γ λ ¯ 0 M r ) q 1 η + μ ¯ 0 ρ 2 m M 1 2 r + 1 2 r = y P C 1 . (4.12)

Similar to the proof of (4.5), we can prove

| ( T λ μ y ) ( t ) | < y P C 1 . (4.13)

It follows from (4.12) and (4.13) that

T λ μ y P C 1 < y P C 1 , y K Ω r . (4.14)

If f = , I = , then there exist l 3 > 0 , l 4 > 0 , and R > r > 0 such that

f ( t , y ) > l 3 φ p ( y ) , I k ( t , y ) > l 4 y ( t J , y R , k = 1 , 2 , , m ) ,

where l 3 satisfies

2 ρ 1 ρ q 1 λ q 1 l 3 q 1 δ [ t 1 ( 1 t m ) ] q 1 η 1 , (4.15)

l 4 satisfies

2 μ ρ 1 m l 4 δ 1 . (4.16)

Let α = R δ . Thus, when y K Ω α we have

y ( t ) δ y P C 1 = δ α = R , t J ,

and then we get

( T λ μ y ) ( t ) = 0 1 H 1 ( t , s ) ϕ q ( λ 0 1 H ( s , τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) d s + μ k = 1 m H 1 ( t , t k ) I k ( t k , y ( t k ) ) ρ 1 ρ q 1 φ q ( λ 0 1 e ( τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) + μ ρ 1 k = 1 m I k ( t k , y ( t k ) ) ρ 1 ρ q 1 λ q 1 φ q ( 0 1 e ( τ ) ω ( τ ) l 3 ϕ p ( y ( τ ) ) d τ ) + μ ρ 1 k = 1 m l 4 y ( t k ) ρ 1 ρ q 1 λ q 1 φ q ( 0 1 e ( τ ) ω ( τ ) l 3 ϕ p ( δ y P C 1 ) d τ ) + μ ρ 1 k = 1 m l 4 δ y P C 1 = ρ 1 ρ q 1 λ q 1 l 3 q 1 δ y P C 1 φ q ( 0 1 e ( τ ) ω ( τ ) d τ ) + μ ρ 1 m l 4 δ y P C 1 ρ 1 ρ q 1 λ q 1 l 3 q 1 δ y P C 1 φ q ( t 1 t m e ( τ ) ω ( τ ) d τ ) + μ ρ 1 m l 4 δ y P C 1 ρ 1 ρ q 1 λ q 1 l 3 q 1 δ y P C 1 [ t 1 ( 1 t m ) ] q 1 φ q ( t 1 t m ω ( τ ) d τ ) + μ ρ 1 m l 4 δ y P C 1 > ρ 1 ρ q 1 λ q 1 l 3 q 1 δ y P C 1 [ t 1 ( 1 t m ) ] q 1 η + μ ρ 1 m l 4 δ y P C 1 1 2 α + 1 2 α = α .

This yields

T λ μ y P C 1 y P C 1 , y K Ω α . (4.17)

Applying (b) of Lemma 3.4 to (4.14) and (4.17) shows that T λ μ has a fixed point y K ( Ω ¯ α Ω r ) with r y P C 1 α = 1 δ R . Hence, since for y K we have y ( t ) δ y P C 1 , t J , it follows that (2.2) holds. This gives the proof of part (i).

Part (ii). Noticing that f ( t , y ) > 0 , I k ( t , y ) > 0 ( k = 1 , 2 , , m ) for all t and y > 0 , we can define

M R = max t J , 0 y R { f ( t , y ) } > 0 , M = max { M k , k = 1 , 2 , , m } > 0 ,

where R > 0 , and

M k = max t J , 0 y R { I k ( t , y ) } , k = 1 , 2 , , m .

Let

λ ¯ 0 4 ( R 2 ρ 2 η ) p 1 ( γ M R ) 1 , μ ¯ 0 R 2 ρ 2 m M .

Then, for y K Ω R and λ < λ ¯ 0 , μ < μ ¯ 0 , we have

( T λ μ y ) ( t ) = 0 1 H 1 ( t , s ) ϕ q ( λ 0 1 H ( s , τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) d s + μ k = 1 m H 1 ( t , t k ) I k ( t k , y ( t k ) ) ρ 2 ( 1 4 γ ) q 1 φ q ( λ 0 1 ω ( τ ) f ( τ , y ( τ ) ) d τ ) + μ ρ 2 k = 1 m I k ( t k , y ( t k ) ) ρ 2 ( 1 4 γ λ ) q 1 φ q ( 0 1 ω ( τ ) M R d τ ) + μ ρ 2 k = 1 m M = ρ 2 ( 1 4 γ λ M R ) q 1 φ q ( 0 1 ω ( τ ) d τ ) + μ ρ 2 m M < ρ 2 ( 1 4 γ λ ¯ 0 M R ) q 1 η + μ ¯ 0 ρ 2 m M 1 2 R + 1 2 R = y P C 1 . (4.18)

Similar to the proof of (4.5), we can prove

| ( T λ μ y ) ( t ) | y P C 1 , y K Ω R . (4.19)

It follows from (4.18) and (4.19) that

T λ μ y P C 1 < y P C 1 , y K Ω R . (4.20)

If f 0 = , I 0 = , then there exist l 3 > 0 , l 4 > 0 , and 0 < r < R such that

f ( t , y ) > l 3 φ p ( y ) , I k ( t , y ) > l 4 y ( t J , 0 y r , k = 1 , 2 , , m ) ,

where l 3 and l 4 satisfy (4.15) and (4.16), respectively.

Therefore, for y K Ω r , we obtain

( T λ μ y ) ( t ) = 0 1 H 1 ( t , s ) ϕ q ( λ 0 1 H ( s , τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) d s + μ k = 1 m H 1 ( t , t k ) I k ( t k , y ( t k ) ) ρ 1 ρ q 1 φ q ( λ 0 1 e ( τ ) ω ( τ ) f ( τ , y ( τ ) ) d τ ) + μ ρ 1 k = 1 m I k ( t k , y ( t k ) ) ρ 1 ρ q 1 λ q 1 φ q ( 0 1 e ( τ ) ω ( τ ) l 3 ϕ p ( y ( τ ) ) d τ ) + μ ρ 1 k = 1 m l 4 y ( t k ) ρ 1 ρ q 1 λ q 1 φ q ( 0 1 e ( τ ) ω ( τ ) l 3 ϕ p ( δ y P C 1 ) d τ ) + μ ρ 1 k = 1 m l 4 δ y P C 1 = ρ 1 ρ q 1 λ q 1 l 3 q 1 δ y P C 1 φ q ( 0 1 e ( τ ) ω ( τ ) d τ ) + μ ρ 1 m l 4 δ y P C 1 ρ 1 ρ q 1 λ q 1 l 3 q 1 δ y P C 1 φ q ( t 1 t m e ( τ ) ω ( τ ) d τ ) + μ ρ 1 m l 4 δ y P C 1 ρ 1 ρ q 1 λ q 1 l 3 q 1 δ y P C 1 [ t 1 ( 1 t m ) ] q 1 φ q ( t 1 t m ω ( τ ) d τ ) + μ ρ 1 m l 4 δ y P C 1 > ρ 1 ρ q 1 λ q 1 l 3 q 1 δ y P C 1 [ t 1 ( 1 t m ) ] q 1 η + μ ρ 1 m l 4 δ y P C 1 1 2 y P C 1 + 1 2 y P C 1 = y P C 1 .

This yields

T λ μ y P C 1 > y P C 1 , y K Ω r . (4.21)

Applying (a) of Lemma 3.4 to (4.20) and (4.21) shows that T λ μ has a fixed point y K ( Ω ¯ R Ω r ) with r y P C 1 R . Hence, since for y K we have y ( t ) δ y P C 1 , t J , it follows that (2.3) holds. This gives the proof of part (ii).

Consider part (iii). Choose two numbers r 1 and r 2 satisfying (2.1). By part (i) and part (ii), there exist λ ¯ 0 > 0 and μ ¯ 0 > 0 such that

T λ μ y P C 1 < y P C 1 , 0 < λ < λ ¯ 0 , 0 < μ < μ ¯ 0 , y K Ω r i , i = 1 , 2 . (4.22)

Since f 0 = f = I = I 0 = , from the proof of part (i) and part (ii), it follows that

T λ μ y P C 1 > y P C 1 , y K Ω r (4.23)

and

T λ μ y P C 1 > y P C 1 , y K Ω R . (4.24)

Applying Lemma 3.4 to (4.22)-(4.24) shows that T λ μ has two fixed points y 1 and y 2 such that y 1 K ( Ω ¯ r 1 Ω r ) and y 2 K ( Ω ¯ R Ω r 2 ) . These are the desired distinct positive solutions of problem (1.1) for 0 < λ < λ ¯ 0 and 0 < μ < μ ¯ 0 satisfying (2.5). Then the proof of part (iii) is complete. □

Remark 4.1 Comparing with Feng [31], the main features of this paper are as follows.

(i) Two parameters λ > 0 and μ > 0 are considered.

(ii) ω L loc 1 ( 0 , 1 ) , not only ω ( t ) 1 for t J .

(iii) It follows from the proof of Theorem 2.1 that the conditions of Corollary 3.2 in [31] are not the optimal conditions, which guarantee the existence of at least one positive solution for problem (1.1). In fact, if f 0 = , or f = 0 , I ( k ) = 0 , we can prove that problem (1.1) has at least one positive solution, respectively.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

XZ completed the main study and carried out the results of this article. MF checked the proofs and verified the calculation. All the authors read and approved the final manuscript.

Acknowledgements

This work is sponsored by the project NSFC (11301178, 11171032) and the Fundamental Research Funds for the Central Universities (2014MS58). The authors are grateful to anonymous referees for their constructive comments and suggestions, which has greatly improved this paper.

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