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

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Abstract

The authors consider the following impulsive differential equations involving the one-dimensional singular p-Laplacian: ( ϕ p ( y ( t ) ) ) =λω(t)f(t,y(t)), tJ, t t k , k=1,2,,m, Δ y | t = t k =μ I k ( t k ,y( t k )), k=1,2,,m, ay(0)b y (0)= 0 1 h(s)y(s)ds, ay(1)+b y (1)= 0 1 h(s)y(s)ds, ϕ p ( y (0))= ϕ p ( y (1))= 0 1 h(t) ϕ p ( y (t))dt, 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.

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 [229] 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, 0a<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), ω0 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) fC([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)dt,ν= 0 1 h(t)dt.
(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 ω1 for tJ. 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 )), ω1 for tJ 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, [3234]. 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 2.1 Assume that (H1)-(H4) hold.

  1. (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), tJ with

    δru(t) 1 δ R,tJ.
    (2.2)
  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

    δru(t)R,tJ.
    (2.3)
  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

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

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

  1. (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), tJ with property (2.2).

  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), tJ with property (2.3).

  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

    δru(t) r 1 <δ r 2 u 2 (t) 1 δ R,tJ.
    (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 yP 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 3.1 If (H1), (H2), and (H4) hold, then problem (3.3) has a unique solution x given by

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

where

H(t,s)=G(t,s)+ 1 1 ν 0 1 G(s,τ)h(τ)dτ,
(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(1t). Then from (3.6) and (3.7), we can prove that H(t,s) and G(t,s) have the following properties.

Proposition 3.1 If (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,sJ,
(3.9)
e(t)e(s)G(t,s)G(t,t)=t(1t)=e(t) e ¯ = max t [ 0 , 1 ] e(t)= 1 4 ,t,sJ,
(3.10)
ρe(s)H(t,s)γs(1s)=γe(s) 1 4 γ,t,sJ,
(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

ρe(s)H(s,s)γs(1s)=γe(s) 1 4 γ,sJ.

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

y(t)= 0 1 H 1 (t,s) ϕ q ( x ( s ) ) ds+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)dτ,
(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 3.2 If (H4) holds, then we have

H 1 (t,s)>0, G 1 (t,s)>0,t,sJ;
(3.16)
1 d b 2 G 1 (t,s) G 1 (s,s) 1 d ( b + a ) 2 ,t,sJ,
(3.17)
ρ 1 H 1 (t,s) H 1 (s,s) ρ 2 ,t,sJ,
(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 ) ds+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 :KP 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 yP 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 3.3 Suppose that (H1)-(H4) hold. Then T λ μ (K)K and T λ μ :KK 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 3.4 Let P be a cone in a real Banach space E. Assume Ω 1 , Ω 2 are bounded open sets in E with 0 Ω 1 , Ω ¯ 1 Ω 2 . If

A:P( Ω ¯ 2 Ω 1 )P

is completely continuous such that either

  1. (a)

    Axx, xP Ω 1 and Axx, xP Ω 2 , or

  2. (b)

    Axx, xP Ω 1 and Axx, xP Ω 2 ,

then A has 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 uK 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 ,yK 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,tJ,yR,k=1,2,,m,

where l 1 satisfies

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

l 2 satisfies

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

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

y(t)δ y P C 1 =δα=R,tJ,

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 ,yK 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 yK( Ω ¯ α Ω r ) with r y P C 1 α= 1 δ R. Hence, since for yK we have y(t)δ y P C 1 , tJ, 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 yK 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 ,yK 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(tJ,0yr,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 ,yK 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 yK( Ω ¯ R Ω r ) with r y P C 1 R. Hence, since for yK we have y(t)δ y P C 1 for tJ, 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 ,yK 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 ,yK Ω r
(4.10)

and

T λ μ y P C 1 < y P C 1 ,yK 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 yK 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 ,yK 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(tJ,yR,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

2μ ρ 1 m l 4 δ1.
(4.16)

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

y(t)δ y P C 1 =δα=R,tJ,

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 ,yK 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 yK( Ω ¯ α Ω r ) with r y P C 1 α= 1 δ R. Hence, since for yK we have y(t)δ y P C 1 , tJ, 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 yK 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 ,yK Ω R .
(4.19)

It follows from (4.18) and (4.19) that

T λ μ y P C 1 < y P C 1 ,yK 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(tJ,0yr,k=1,2,,m),

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

Therefore, for yK 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 ,yK 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 yK( Ω ¯ R Ω r ) with r y P C 1 R. Hence, since for yK we have y(t)δ y P C 1 , tJ, 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 ,yK 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 ,yK Ω r
(4.23)

and

T λ μ y P C 1 > y P C 1 ,yK 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.

  1. (i)

    Two parameters λ>0 and μ>0 are considered.

  2. (ii)

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

  3. (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.

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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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Correspondence to Xuemei Zhang.

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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.

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Keywords

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