# Multiple solutions of three-point boundary value problems for second-order impulsive differential equation at resonance

Yulin Zhao1*, Haibo Chen2 and Qiming Zhang1

Author Affiliations

1 School of Science, Hunan University of Technology, Zhuzhou, Hunan, 412007, PR China

2 Department of Mathematics, Central South University, Changsha, Hunan, 410075, PR China

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Boundary Value Problems 2014, 2014:103  doi:10.1186/1687-2770-2014-103

 Received: 20 December 2013 Accepted: 8 April 2014 Published: 7 May 2014

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

In this paper, by using the coincidence degree theory and the upper and lower solutions method, we deal with the existence of multiple solutions to three-point boundary value problems for second-order differential equation with impulses at resonance. An example is given to show the validity of our results.

##### Keywords:
resonance; coincidence degree; upper and lower solutions; impulsive; three-point boundary value problems

### 1 Introduction

The purpose of the present paper is to investigate the following second-order impulsive differential equations:

{ ( ρ ( t ) u ( t ) ) = f ( t , u ( t ) , u ( t ) ) , t J , t t k , Δ u ( t k ) = I k ( t k , u ( t k ) ) , k = 1 , 2 , , m , Δ u ( t k ) = J k ( t k , u ( t k ) ) , (1.1)

together with the boundary conditions:

u ( 0 ) = 0 , u ( 1 ) = u ( η ) , (1.2)

where J = [ 0 , 1 ] , ρ : J ( 0 , + ) is a continuous differentiable function, f : J × R 2 R is continuous, 0 < η < 1 , I k , J k C ( J , R ) for 1 k m , m is a fixed positive integer, 0 = t 0 < t 1 < t 2 < < t m < t m + 1 = 1 , η t k , Δ u ( t k ) = u ( t k + ) u ( t k ) denotes the jump of u ( t ) at t = t k , Δ u ( t k ) = u ( t k + ) u ( t k ) . u ( t k + ) , u ( t k + ) ( u ( t k ) , u ( t k ) ) represent the right limit (left limit) of u ( t ) and u ( t ) at t = t k , respectively.

Impulsive differential equations describe processes which experience a sudden change of their state at certain moments. The theory of impulse differential equations has been a significant development in recent years and played a very important role in modern applied mathematical models of real processes rising in phenomena studied in physics, population dynamics, chemical technology, biotechnology, and economics; see [1-10] and the references therein.

Recently, several authors (see [6,11-14] and the references therein) have studied the existence of nontrivial or positive solutions for second-order three-point boundary value problem of the type

{ u = f ( t , u , u ) , u ( 0 ) = 0 , u ( 1 ) = a u ( η ) . (1.3)

Note that the nonlinear term f depends on u and its derivative u , then the relative problem becomes more complicated. A general method to deal with this difficulty is to add some conditions to restrict the growth of the u term. One condition is the Caratheodory nonlinearity, the other usual condition is Nagumo condition or Nagumo-Winter condition (see [2,9,12,15-20]). When a 1 , the linear operator L u = u is invertible, this is the so-called non-resonance case. Gupta et al. made use of the Leray-Schauder continuation theorem to get the results on the existence of the solution for the problems (1.3) when a 1 in [14]. By using the Leray-Schauder continuation theorem and in the presence of two pairs of upper and lower solutions, Khan and Webb [12] established the existence of at least three solutions for the problem (1.3) when a 1 . The linear operator L u = u is non-invertible when a = 1 , this is the so-called resonance case, and the Leray-Schauder continuation theorem cannot be applied. In [11], by using the coincidence degree theory of Mawhin [21] and some linear or non-linear growth assumptions on f, Feng and Webb obtained the existence of the solution of the problem (1.3) when a = 1 . By applying the nonlinear alternative of Leray-Schauder, Ma [13] have showed the existence of at least one solution for the problem (1.3) when a = 1 .

Recently, using the coincidence degree theory and the concept of autonomous curvature bound set, Liu and Yu [6] have studied the existence of at least one solution for the problem (1.1)-(1.2) when Δ u ( t k ) = I k ( u ( t k ) , u ( t k ) ) , Δ u ( t k ) = J k ( u ( t k ) , u ( t k ) ) .

In the present paper, we assume that there exist n ( n N and n 2 ) pairs of upper and lower solutions for problem (1.1)-(1.2) and the nonlinear f satisfies a Nagumo-like growth condition with respect to u . By considering a suitably modified nonlinearity and applying the coincidence degree method of Mawhin [21], the existence of multiple solutions for the problem (1.1)-(1.2) is given.

### 2 Preliminaries

Let

X = P C 1 ( J ) { u ( 0 ) = 0 , u ( 1 ) = u ( η ) } , Z = P C ( J ) × R 2 m ,

where

P C ( J ) = { u C ( J ) , u ( t )  and  u ( t + )  exist, and  u ( t k ) = u ( t k ) } . P C 1 ( J ) = { u : J R : u ( t )  is continuously differentiable for  t 0 , 1 , t k ; u ( t ) P C 1 ( J ) =  and  u ( t + )  exist, and  u ( t k ) = u ( t k ) } , J = J { t 1 , t 2 , , t m } .

Obviously, X is a Banach space with the following norm:

u X = max { sup t J | u ( t ) | , sup t J | u ( t ) | } .

In the following, we recall the concept of strict upper and lower solutions for problem (1.1)-(1.2).

Definition 2.1 A function α ( t ) P C 1 ( J ) C 2 ( J ) is said to be a strict lower solution of the problem (1.1)-(1.2) if

( ρ ( t ) α ( t ) ) > f ( t , α ( t ) , α ( t ) ) , t J , (2.1)

Δ α ( t k ) = I k ( t k , α ( t k ) ) , Δ α ( t k ) J k ( t k , α ( t k ) ) , k = 1 , 2 , , m , (2.2)

α ( 0 ) 0 , α ( 1 ) α ( η ) 0 . (2.3)

Similarly, a function β ( t ) P C 1 ( J ) C 2 ( J ) is said to be a strict upper solution of the problem (1.1)-(1.2) if

( ρ ( t ) β ( t ) ) < f ( t , β ( t ) , β ( t ) ) , t J , (2.4)

Δ β ( t k ) = I k ( t k , β ( t k ) ) , Δ β ( t k ) J k ( t k , β ( t k ) ) , k = 1 , 2 , , m , (2.5)

β ( 0 ) 0 , β ( 1 ) β ( η ) 0 . (2.6)

Remark 2.2 Let f : J × R 2 R be continuous, I k , J k C ( J , R ) , and u P C 1 ( J ) C 2 ( J ) is a solution of the problem (1.1)-(1.2), if α ( β ) is a strict lower solution (strict upper solution) for the problem (1.1)-(1.2) with α u ( u β ), then α < u ( u < β ) on ( 0 , 1 ) .

Definition 2.3 Let α be a strict lower solution and β be a strict upper solution for the problem (1.1)-(1.2) satisfying α ( t ) < β ( t ) on J. We say that f : J × R 2 R has property (H) relative to α and β, if there exists a function ψ C 1 ( [ 0 , + ) , ( 0 , + ) ) such that

| f ( t , u , p ) | < ψ ( | p | ) , (2.7)

for all u ( t ) ( β ( t ) , α ( t ) ) ( α ( t ) , β ( t ) ) , t J , and

0 + s θ s + ψ ( s ) d s = + , (2.8)

where 0 θ < + with | ρ ( t ) | θ , t J .

### 3 The key lemmas

Let dom L = C 2 ( J ) X , and

L : dom L Z , u ( ( ρ ( t ) u ( t ) ) , Δ u ( t 1 ) , , Δ u ( t m ) , Δ u ( t 1 ) , , Δ u ( t m ) ) , N : u z , u ( f ( t , u , u ) , I 1 ( t 1 , u ( t 1 ) ) , , I m ( t m , u ( t m ) ) , N : u z , u J 1 ( t 1 , u ( t 1 ) ) , , J m ( t m , u ( t m ) ) ) .

Then the problem (1.1)-(1.2) can be written as

L u = N u , u dom L .

Lemma 3.1Suppose thatLbe defined in the above. ThenLis a Fredholm operator of index zero. Furthermore

Ker ( L ) = { u X : u = c , c R } , (3.1)

and

Im ( L ) = { ( y , a 1 , , a m , b 1 , , b m ) : ( ρ ( t ) u ( t ) ) = y ( t ) , Δ u ( t k ) = a k , Δ u ( t k ) = b k , k = 1 , , m , for some y dom L } = { ( y , a 1 , , a m , b 1 , , b m ) : η 1 1 ρ ( s ) 0 s y ( τ ) d τ d s + η 1 1 ρ ( s ) t k < s ρ ( t k ) b k d s + η < t k < 1 a k = 0 } . (3.2)

Proof Firstly, it is clear that (3.1) holds. Next, we shall prove that (3.2) holds.

The following problem:

{ ( ρ ( t ) u ( t ) ) = y ( t ) , Δ u ( t k ) = a k , Δ u ( t k ) = b k , k = 1 , , m (3.3)

has a solution u ( t ) satisfying u ( 0 ) = 0 and u ( 1 ) = u ( η ) if and only if

η 1 1 ρ ( s ) 0 s y ( τ ) d τ d s + η 1 1 ρ ( s ) t k < s ρ ( t k ) b k d s + η < t k < 1 a k = 0 . (3.4)

In fact, if (3.3) has a solution u ( t ) satisfying u ( 0 ) = 0 , u ( 1 ) = u ( η ) , then from (3.3) we have

u ( t ) = u ( 0 ) + 0 t 1 ρ ( s ) 0 s y ( τ ) d τ d s + 0 t 1 ρ ( s ) t k < s ρ ( t k ) b k d s + t k < t a k .

According to u ( 0 ) = 0 , u ( 1 ) = u ( η ) , we get (3.4).

On the other hand, if (3.4) holds, setting

u ( t ) = c + 0 t 1 ρ ( s ) 0 s y ( τ ) d τ d s + 0 t 1 ρ ( s ) t k < s ρ ( t k ) b k d s + t k < t a k ,

where c is an arbitrary constant, then u ( t ) is a solution of (3.3) with u ( 0 ) = 0 , u ( 1 ) = u ( η ) . Hence (3.2) holds. □

Take the projector Q : Z Z as follows:

Q ( y , a 1 , , a m , b 1 , , b m ) = ( 1 ϕ ( 1 ) ϕ ( η ) [ η 1 1 ρ ( s ) 0 s y ( τ ) d τ d s + η 1 1 ρ ( s ) t k < s ρ ( t k ) b k d s + η < t k < 1 a k ] , 0 , , 0 ) , (3.5)

where ϕ ( t ) = 0 t s ρ ( s ) d s , t ( 0 , 1 ) . For every ( y , a 1 , , a m , b 1 , , b m ) Z , set

z = ( y 1 , a 1 , , a m , b 1 , , b m ) = ( y , a 1 , , a m , b 1 , , b m ) Q ( y , a 1 , , a m , b 1 , , b m ) .

Thus, we obtain

η 1 1 ρ ( s ) 0 s y 1 ( τ ) d τ d s + η 1 1 ρ ( s ) t k < s ρ ( t k ) b k d s + η < t k < 1 a k = [ η 1 1 ρ ( s ) 0 s y ( τ ) d τ d s + η 1 1 ρ ( s ) t k < s ρ ( t k ) b k d s + η < t k < 1 a k ] × [ 1 1 ϕ ( 1 ) ϕ ( η ) η 1 d s ρ ( s ) ] = 0 .

Then z Im L . Hence Z = Im L + R . Since Im L R = { 0 } , we have Z = Im L R , which implies dim Ker ( L ) = dom R = c o dim Im L = 1 . Hence L is a Fredholm operator of index zero.

Take P : Z Z , P u = u ( 0 ) . So the generalized inverse K P : Im L dom L Ker P of L can be written as

K P z ( t ) = K P ( y , a 1 , , a m , b 1 , , b m ) = 0 t 1 ρ ( s ) 0 s y ( τ ) d τ d s + 0 t 1 ρ ( s ) t k < s ρ ( t k ) b k d s + t k < t a k . (3.6)

Set δ : = min t J ρ ( t ) > 0 , d 1 . Then for the function ψ defined by (2.8), let h 1 ( u ) be the solution of the following initial value problem:

δ y y + θ y + ψ ( y ) = 0 , y ( 0 ) = d , (3.7)

and h 2 ( u ) be the solution of the following initial value problem:

δ y y θ y ψ ( y ) = 0 , y ( 0 ) = d . (3.8)

Lemma 3.2Suppose that there exists a constant M > 0 , then h 1 ( u ) is well defined in [ 0 , M ] and positive on this interval, h 2 ( u ) is also well defined and positive in [ M , 0 ] . Moreover, if d 1 , then h 1 ( u ) 1 for any u [ 0 , M ] ; h 2 ( u ) 1 for any u [ M , 0 ] .

Proof We only consider the case y = h 1 ( u ) (in the case y = h 2 ( u ) , the proof is similar). Assume that there exists a u [ 0 , M ] such that y ( u ) = h 1 ( u ) = 0 . Let u 0 = inf { u : h 1 ( u ) = 0 , u [ 0 , M ] } . It follows from (3.7) that

y y θ y + ψ ( y ) = δ 1 , y > 0 .

Integrating the above equation over [ 0 , u 0 ] we get (let τ = y ( s ) )

0 u 0 y ( s ) y ( s ) d s θ y ( s ) + ψ ( y ( s ) ) = 0 d τ d τ θ τ + ψ ( τ ) = δ 1 u 0 δ 1 M .

However, the left side of the above equation equals ∞ by (2.8). We reach a contradiction. Hence h 1 ( u ) > 0 for every u 0 . From d 1 , by the continuity of solution of differential equations on the initial values, we obtain h 1 ( u ) 1 for u [ 0 , M ] . The proof is complete. □

Define the following sets:

G = { ( t , u , p ) : t J , | u | < M , | p | < h 1 ( u )  for  u [ 0 , M ] , G =  and  | p | < h 2 ( u )  for  u [ M , 0 ] } , Ω = { u P C 1 ( J ) : ( t , u ( t ) , u ( t ) ) G , t J ; ( t k + , u ( t k + ) , u ( t k + ) ) G , k = 1 , , m } .

Define the function h ( u ) as

h ( u ) = { h 1 ( u ) , u [ 0 , M ] , h 2 ( u ) , u [ M , 0 ] . (3.9)

Lemma 3.3Let Deg denote the coincidence degree. Let the following conditions hold:

(i) f ( t , M , 0 ) < 0 < f ( t , M , 0 ) , t J (Mis given in Lemma 3.2);

(ii) | f ( t , u , p ) | < ψ ( | p | ) , t J , | u | M , p R ;

(iii) I k ( t k , ± M ) = 0 , and J k ( t k , M ) < 0 < J k ( t k , M ) , k = 1 , , m .

Then

Deg [ ( L , N ) , Ω ] = 1 .

Proof Consider the following family of equations:

L u = λ N u , λ ( 0 , 1 ] . (3.10)

We will show

L u λ N u , u Ω , λ ( 0 , 1 ] . (3.11)

If not, then there exist some λ ( 0 , 1 ] and u Ω such that (3.10) holds. Note that u Ω if and only if ( t , u ( t ) , u ( t ) ) G ¯ and either ( t ¯ , u ( t ¯ ) , u ( t ¯ ) ) G for some t ¯ J , or ( t k 0 + , u ( t k 0 + ) , u ( t k 0 + ) ) Ω for some k 0 { 1 , 2 , , m } . There are two possibilities.

Case (I). If ( t ¯ , u ( t ¯ ) , u ( t ¯ ) ) G , t ¯ [ 0 , 1 ] , t ¯ t k + . In this case, | u ( t ¯ ) | = M , or | u ( t ¯ ) | = h ( u ( t ¯ ) ) .

Subcase (1). Suppose | u ( t ¯ ) | = M . Let g ( t ) = 1 2 ( u ( t ) ) 2 1 2 M . Then g ( t ) 0 , t J and g ( t ¯ ) = 0 . When t t ¯ ( 0 , 1 ) , g ( t ) 0 , we get

0 g ( t ¯ 0 ) = u ( t ¯ ) u ( t ¯ ) .

If g ( t ¯ 0 ) = 0 , then u ( t ¯ ) = 0 . From condition (i), we have

0 g ( t ¯ 0 ) = ( u ( t ¯ ) ) 2 + u ( t ¯ ) ρ ( t ¯ ) ( ρ ( t ¯ ) u ( t ¯ ) ) = ± M ρ ( t ¯ ) λ f ( t ¯ , ± M , 0 ) > 0 ,

If g ( t ¯ 0 ) > 0 , then t ¯ = t k 0 for some k 0 { 1 , , m } and g ( t k 0 0 ) > 0 . Thus from (iii), we have

g ( t k 0 + 0 ) = [ u ( t k 0 ) + λ I k 0 ( t k 0 , u ( t k 0 ) ) ] [ u ( t k 0 ) + λ J k 0 ( t k 0 , u ( t k 0 ) ) ] = u ( t k 0 ) u ( t k 0 ) + u ( t k 0 ) λ J k 0 ( t k 0 , u ( t k 0 ) ) + u ( t k 0 ) λ I k 0 ( t k 0 , u ( t k 0 ) ) + λ 2 I k 0 ( t k 0 , u ( t k 0 ) ) J k 0 ( t k 0 , u ( t k 0 ) ) = g ( t k 0 0 ) + [ ± M λ J k 0 ( t k 0 , ± M ) ] > 0 .

On the other hand, g ( t ) 0 , t J and g ( t k 0 + 0 ) = 0 , thus

0 g ( t k 0 + 0 ) = [ u ( t k 0 ) + λ I k 0 ( t k 0 , u ( t k 0 ) ) ] [ u ( t k 0 ) + λ J k 0 ( t k 0 , u ( t k 0 ) ) ] > 0 ,

If t ¯ = 0 , it is easy to see that g ( 0 ) = 0 . Since u ( 0 ) = 0 , we have g ( 0 ) = 0 , thus we can obtain g ( 0 ) 0 . However, from condition (i) we know

0 ρ ( 0 ) g ( 0 ) = u ( 0 ) λ f ( 0 , u ( 0 ) , 0 ) > 0 ,

If t ¯ = 1 , then | u ( 1 ) | = M . This means that u ( 1 ) G . Since u ( η ) = u ( 1 ) , we have u ( η ) G . However, according to the above arguments, we know u ( η ) G , which is a contradiction.

Subcase (2). Suppose | u ( t ¯ ) | = h ( u ( t ¯ ) ) for t ¯ ( t k 0 , t k 0 + 1 ] , k 0 { 0 , 1 , , m } . Obviously, t ¯ 0 . Since | ( ρ ( t ) u ( t ) ) | = | f ( t , u ( t ) , u ( t ) ) | < ψ ( | u ( t ) | ) . Thus we get

ρ ( t ) | u ( t ) | | ρ ( t ) | | u ( t ) | < ψ ( | u ( t ) | ) , | u ( t ) | < 1 δ [ | ρ ( t ) | | u ( t ) | + ψ ( | u ( t ) | ) ] 1 δ [ θ | u ( t ) | + ψ ( | u ( t ) | ) ] . (3.12)

Let

p ( t ) = 1 2 ( u ( t ¯ ) ) 2 1 2 ( h ( u ( t ) ) ) 2 .

Without loss of generality, we suppose that u ( t ¯ ) = h ( u ( t ¯ ) ) > 0 for t ¯ ( t k 0 , t k 0 + 1 ] . Then we have three possibilities.

(i) If u ( t ¯ ) = 0 , then there exists a sufficiently small neighborhood of t ¯ such that u ( t ) > 0 . Thus p ( t ) = 1 2 ( u ( t ) ) 2 1 2 ( h 1 ( u ( t ) ) ) 2 , and p ( t ) has a local maximum value on t ¯ , which implies p ( t ¯ + 0 ) 0 . But in this case, from (3.7), (3.12), and (ii), we have

p ( t ¯ + 0 ) = u ( t ¯ ) [ u ( t ¯ ) h 1 ( u ( t ¯ ) ) h 1 ( u ( t ¯ ) ) ] > u ( t ¯ ) [ 1 δ ( θ u ( t ¯ ) + ψ ( | u ( t ¯ ) | ) ) + θ h 1 ( u ( t ) ) + ψ ( h 1 ( u ( t ) ) ) δ ] = 0 , (3.13)

(ii) If u ( t ¯ ) < 0 , then there exists a sufficiently small neighborhood of t ¯ such that u ( t ) < 0 . Thus p ( t ) = 1 2 ( u ( t ) ) 2 1 2 ( h 2 ( u ( t ) ) ) 2 , and p ( t ) has a local maximum value on t ¯ , which implies p ( t ¯ ) 0 . But in this case, from (3.8), (3.12), and (ii), we have

p ( t ¯ ) = u ( t ¯ ) [ u ( t ¯ ) h 2 ( u ( t ¯ ) ) h 2 ( u ( t ¯ ) ) ] < u ( t ¯ ) [ 1 δ ( θ u ( t ¯ ) + ψ ( | u ( t ¯ ) | ) ) θ h 2 ( u ( t ) ) + ψ ( h 2 ( u ( t ) ) ) δ ] = 0 ,

(iii) If u ( t ¯ ) > 0 , then there exists a sufficiently small neighborhood of t ¯ such that u ( t ) > 0 . Thus p ( t ) = 1 2 ( u ( t ) ) 2 1 2 ( h 1 ( u ( t ) ) ) 2 . By the same argument as in (3.13), we reach a contradiction.

Case (II). ( t k 0 + , u ( t k 0 + ) , u ( t k 0 + ) ) Ω for some k 0 { 1 , 2 , , m } . In this case, | u ( t k 0 + ) | = M , or | u ( t k 0 + ) | = h ( u ( t k 0 + ) ) .

Subcase (3). Suppose | u ( t k 0 + ) | = M for some k 0 { 1 , 2 , , m } . Let g ( t ) be defined in the above. Obviously, we have g ( t ) 0 for t J and g ( t k 0 + 0 ) = 0 . It follows from (iii) that

0 g ( t k 0 + 0 ) = [ u ( t k 0 ) + λ I k 0 ( t k 0 , u ( t k 0 ) ) ] [ u ( t k 0 ) + λ J k 0 ( t k 0 , u ( t k 0 ) ) ] = u ( t k 0 ) u ( t k 0 ) + u ( t k 0 ) λ J k 0 ( t k 0 , u ( t k 0 ) ) .

If g ( t k 0 + 0 ) = 0 , then u ( t k 0 + ) = 0 . Form (3.11) and (i), we have

0 g ( t k 0 + 0 ) = ( u ( t k 0 + ) ) 2 + u ( t k 0 + ) ρ ( t k 0 + ) ( ρ ( t k 0 ) u ( t k 0 + ) ) = ± M ρ ( t k 0 + ) λ f ( t k 0 , ± M , 0 ) > 0 ,

If g ( t k 0 + 0 ) < 0 , together with (iii), we get g ( t k 0 0 ) = u ( t k 0 ) u ( t k 0 ) < 0 . But g ( t k 0 ) = 0 and g ( t ) 0 for t J , which yields g ( t k 0 0 ) > 0 , and we reach a contradiction.

Similar to subcase (2), we can show that | u ( t k 0 + ) | = h ( u ( t k 0 + ) ) is also impossible. Combining the results of case (I) and case (II) we obtain (3.11).

On the other hand, for λ ( 0 , 1 ] , it follows from [21] that (3.10) is equivalent to the following family of operator equations:

u = P u + Q N u + λ K P ( E Q ) N u , (3.14)

where E is the identity mapping.

Note that Ker L = R , and

Ω Ker L = { c R : ( t , c , p ) G , t J ; ( t k + , c , p ) G ,  for some  k = 1 , , m } .

From (3.5) and (3.6), we have

Q N u = ( 1 ϕ ( 1 ) ϕ ( η ) [ η 1 1 ρ ( s ) 0 s f ( τ , u ( τ ) , u ( τ ) ) d τ d s + η 1 1 ρ ( s ) t k < s ρ ( t k ) J k ( t k , u ( t k ) ) d s + η < t k < 1 I k ( t k , u ( t k ) ) ] , 0 , , 0 ) ,

and

K P ( E Q ) N u = 1 ϕ ( 1 ) ϕ ( η ) [ 0 t 1 ρ ( s ) 0 s f ( τ , u ( τ ) , u ( τ ) ) d τ d s + 0 t 1 ρ ( s ) t k < s ρ ( t k ) J k ( t k , u ( t k ) ) d s + t k < t I k ( t k , u ( t k ) ) ] ϕ ( t ) ϕ ( 1 ) ϕ ( η ) [ η 1 1 ρ ( s ) 0 s f ( τ , u ( τ ) , u ( τ ) ) d τ d s + η 1 1 ρ ( s ) t k < s ρ ( t k ) J k ( t k , u ( t k ) ) d s + η < t k < 1 I k ( t k , u ( t k ) ) ] .

By the Ascoli-Arzela theorem, it is easy to show that Q N ( Ω ¯ ) is bounded and K P ( E Q ) N : Ω ¯ X is compact. Thus N is L-compact on Ω ¯ .

Define a mapping H : Ω ¯ × [ 0 , 1 ] X ,

H ( u , λ ) = P u + Q N u + λ K P ( E Q ) N u , ( u , λ ) Ω ¯ × [ 0 , 1 ] .

Then it is easy to prove that H ( u , λ ) is completely continuous and we claim that

u = H ( u , λ ) , u Ω , 0 λ 1 . (3.15)

In fact, for 0 < λ 1 , it follows from (3.10) and (3.14) that (3.15) holds. For λ = 0 , if there exists a u ¯ Ω such that u ¯ = H ( u ¯ , 0 ) , that is u ¯ = P u ¯ + Q N u ¯ , in this case, Q N u ¯ = 0 , u ¯ Ker L , hence u ¯ = M or u ¯ = M . However, it follows from (iii) that

Q N ( c ) = 1 ϕ ( 1 ) ϕ ( η ) [ η 1 1 ρ ( s ) 0 s f ( τ , c , 0 ) d τ d s + η 1 1 ρ ( s ) t k < s ρ ( t k ) J k ( t k , c ) d s + η < t k < 1 I k ( t k , c ) ] = 1 ϕ ( 1 ) ϕ ( η ) η 1 1 ρ ( s ) 0 s f ( τ , c , 0 ) d τ d s + η 1 1 ρ ( s ) t k < s ρ ( t k ) J k ( t k , c ) d s , c R .

Thus we get Q N ( M ) > 0 and Q N ( M ) < 0 , which contradicts u = P u + Q N u for u Ker L . Therefore (3.15) holds. As follows from [21] and by using the invariance of Leray-Schauder degree under homotopy, we obtain

Deg ( E H ( , 1 ) , Ω , 0 ) = Deg ( E H ( , 0 ) , Ω , 0 ) = Deg B ( ( E P Q N ) | Ker L Ω ¯ , Ker L Ω , 0 ) = Deg B ( ( Q N ) | Ker L Ω ¯ , Ker L Ω , 0 ) .

Since KerL is one dimensional and Q N ( M ) > 0 , Q N ( M ) < 0 , we get

Deg B ( ( Q N ) | Ker L Ω ¯ , Ker L Ω , 0 ) = 1 .

From the property of coincidence degree we proved Lemma 3.3. □

Lemma 3.4Assume that

(c1) there exist lower and upper solutions α ( t ) , β ( t ) of the problem (1.1)-(1.2), respectively, with α ( t ) < β ( t ) ;

(c2) f : J × R 2 R is continuous and has property (H) relative toα, β;

(c3) I k , J k are continuous for each k = 1 , 2 , , m , and satisfy

I k ( t k , α ( t k ) ) = I k ( t k , β ( t k ) ) = 0 , and J k ( t k , α ( t k ) ) < 0 < J k ( t k , β ( t k ) ) .

Then Deg [ ( L , N ) , Ω α β ] = 1 .

Proof Choose M > 0 large enough for u , p R , u ( β ( t ) , α ( t ) ) ( α ( t ) , β ( t ) ) , t J , such that

f ( t , β ( t ) , 0 ) + M β ( t ) > 0 , M > β ( t ) , t J , f ( t , α ( t ) , 0 ) M α ( t ) < 0 , M < α ( t ) , t J , J k ( t k , α ( t k ) ) M α ( t k ) < 0 < J k ( t k , β ( t k ) ) + M β ( t k ) , k = 1 , , m ,

and let h i ( u ) ( i = 1 , 2 ) be defined in (3.7) and (3.8), then it follows from Lemma 3.2 that we can choose d > 0 large enough such that

min [ 0 , M ] h 1 ( u ) > max { max t J | β ( t ) | , max t J | α ( t ) | } , min [ M , 0 ] h 2 ( u ) > max { max t J | β ( t ) | , max t J | α ( t ) | } .

Define a set Ω α β as

Ω α β = { u P C 1 ( J ) | ( t , u , p ) : α ( t ) < u < β ( t ) , | p | < h ( u ) , t J } ,

where h ( u ) is given in (3.9).

Define the auxiliary functions F and I ¯ k , J ¯ k as follows:

F ( t , u , p ) = f ( t , n ( t , u ) , q ( t , p ) ) + u ( t ) n ( t , u ) , t J , I ¯ k ( t k , u ( t k ) ) = I k ( t k , n ( t k , u ( t k ) ) ) , k { 1 , , m } , J ¯ k ( t k , u ( t k ) ) = J k ( t k , n ( t k , u ( t k ) ) ) + u ( t k ) n ( t k , u ( t k ) ) , k { 1 , , m } ,

where

n ( t , u ( t ) ) = { β ( t ) , β ( t ) < u M , u ( t ) , α ( t ) u β ( t ) , α ( t ) , M u < α ( t ) , q ( t , p ) = { h ( u ) , p > h ( u ) , | u | M , p , | p | h ( u ) , | u | M , h ( u ) , p < h ( u ) , | u | M .

We then generalize F to J × R 2 and I ¯ k , J ¯ k : J × R R . It is easy to see that F, I ¯ k , J ¯ k are continuous and satisfy

F ( t , M , 0 ) < 0 < F ( t , M , 0 ) , I ¯ k ( t k , ± M ) = 0 , and J ¯ k ( t k , M ) < 0 < J ¯ k ( t k , + M ) , k { 1 , , m } .

Moreover, when | u | M , F, I ¯ k , J ¯ k are bounded. It follows from Lemma 3.3 that

Deg [ ( L , N ¯ ) , Ω ] = 1 ,

where

N ¯ u = ( F ( t , u , u ) , I ¯ 1 ( t 1 , u ( t 1 ) ) , , I ¯ m ( t m , u ( t m ) ) , J ¯ 1 ( t 1 , u ( t 1 ) ) , , J ¯ m ( t m , u ( t m ) ) ) .

Next we show

Deg [ ( L , N ¯ ) , Ω α β ] = 1 . (3.16)

It suffices to show that

L u = N ¯ u , u Ω ¯ Ω α β . (3.17)

In fact, let u Ω ¯ such that L u = N ¯ u and assume that

max t J { u ( t ) β ( t ) } = u ( τ ) β ( τ ) 0 .

Case (1). If there exists τ ( 0 , 1 ) , τ t k such that the function y ( t ) = u ( t ) β ( t ) attains its maximum value y ( τ ) 0 , which implies y ( τ ) = 0 and y ( τ ) 0 . But on the other hand,

ρ ( τ ) y ( τ ) = ( ρ ( τ ) u ( τ ) ) ( ρ ( τ ) β ( τ ) ) = F ( τ , β ( τ ) , u ( τ ) ) + u ( τ ) β ( τ ) ( ρ ( τ ) β ( τ ) ) = f ( τ , β ( τ ) , β ( τ ) ) + u ( τ ) β ( τ ) ( ρ ( τ ) β ( τ ) ) > 0 , (3.18)

which implies y ( τ ) > 0 . We reach a contradiction.

If τ = 0 , then y ( 0 ) 0 , y ( 0 ) 0 . On the other hand, since u ( 0 ) = 0 and β ( t ) is a strict upper solution of problem (1.1)-(1.2), we see that y ( 0 ) = u ( 0 ) β ( 0 ) 0 . Thus y ( 0 ) = 0 . Note that y ( t ) assumes the maximum value at t = 0 , there exists τ 0 ( 0 , 1 ) , τ 0 t k such that y ( τ 0 ) 0 and y ( τ 0 ) 0 . By the same argument as in (3.18) where τ = τ 0 , we reach a contradiction.

If τ = 1 , then y ( 1 ) = u ( 1 ) β ( 1 ) 0 . According to (2.6), we get

y ( 1 ) = u ( 1 ) β ( 1 ) u ( 1 ) β ( η ) = u ( η ) β ( η ) = y ( η ) ,

if y ( η ) > y ( 1 ) , which is a contradiction. Thus y ( η ) = y ( 1 ) , which implies that y ( t ) also attains its maximum value at t = η . By the same argument as in (3.18) where τ = η , we reach a contradiction.

Hence the function y ( t ) cannot have any nonnegative maximum value on the interval ( 0 , 1 ) , t t k for k { 0 , 1 , , m } .

Case (2). If there exists a τ J such that y ( τ ) = u ( τ ) β ( τ ) = ε 0 , from case (1), we get τ = t k for some k = 1 , 2 , , m . Hence n ( t k , u ( t k ) ) = β ( t k ) , β ( t k ) u ( t k ) , and consequently we have y ( t k + ) = y ( t k ) = ε , which implies y ( t k + ) 0 , because

u ( t k + ) u ( t k ) = J ¯ k ( t k , u ( t k ) ) = J k ( t k , β ( t k ) ) + u ( t k ) β ( t k ) J k ( t k , β ( t k ) ) β ( t k + ) β ( t k ) .

Consequently, we get u ( t k + ) = β ( t k + ) or y ( t k + ) = 0 and D + y ( t k + ) 0 .

By the continuity of f and β ( t ) is a strict upper solution of problem (1.1)-(1.2), there exists a sequence ζ j R with ζ j > 0 , ζ j 0 , as j such that

D + ρ ( t k ) β ( t k + ) = lim j sup ρ ( t k + ζ j ) β ( t k + ζ j ) ρ ( t k ) β ( t k ) ζ j = lim j ( ρ ( t k + ζ j ) β ( t k + ζ j ) ) < lim j [ f ( t k + ζ j , β ( t k + ζ j ) , β ( t k + ζ j ) ) ] = f ( t k , β ( t k + ) , β ( t k + ) ) , (3.19)

where ζ j ( t k , t k + ζ j ) are from the mean value theorem. As before, we also get

D + ρ ( t k ) u ( t k + ) = F ( t k , u ( t k + ) , u ( t k + ) ) ,

where D + ρ ( t k ) u ( t k + ) = lim j inf ρ ( t k + ζ j ) u ( t k + ζ j ) ρ ( t k ) u ( t k ) ζ j . As a result, we can obtain

D + ρ ( t k ) β ( t k + ) D + ρ ( t k ) u ( t k + ) = f ( t k , β ( t k + ) , β ( t k + ) ) + y ( t k + ) f ( t k , β ( t k + ) , β ( t k + ) )

which is a contradiction to (3.19).

Case (3). If y ( t ) = u ( t ) β ( t ) < ε for all t J , then there must be a k 0 ( 1 k 0 m ) such that y ( t k 0 + ) = u ( t k 0 + ) β ( t k 0 + ) = ε , and y ( t k 0 ) < ε , which implies β ( t k 0 + ) β ( t k 0 ) < u ( t k 0 + ) u ( t k 0 ) . Namely, Δ β ( t k 0 ) < Δ u ( t k 0 ) . However, this is impossible because

Δ u ( t k 0 ) = I ¯ k 0 ( t k 0 , n ( t k 0 , u ( t k 0 ) ) ) = I k 0 ( t k 0 , β ( t k 0 ) ) = Δ β ( t k 0 ) .

Thus we have proved that u ( t ) < β ( t ) on J. Similarly we can show that α ( t ) < u ( t ) on J. It then follows that α ( t ) < u ( t ) < β ( t ) on J.

We now shall prove that

| u ( t ) | < h ( u ( t ) ) for  t [ t k , t k + 1 ] , k = 0 , 1 , , m . (3.20)

Assume that (3.20) cannot hold. Thus there are two possibilities:

(a) There exists τ ( t k 0 , t k 0 + 1 ] for some k 0 { 0 , 1 , , m } such that | u ( τ ) | h ( u ( τ ) ) .

(b) There exists some k 0 { 0 , 1 , , m } such that | u ( t k 0 + ) | h ( u ( t k 0 + ) ) .

For case (a), we assume max t ( t k 0 , t k 0 + 1 ] { u ( t ) h ( u ( t ) ) } = max t ( t k 0 , t k 0 + 1 ] r ( t ) = u ( τ ) h ( u ( τ ) ) 0 .

Since | ( ρ ( t ) u ( t ) ) | = | F ( t , u ( t ) , u ( t ) ) | = | f ( t , u ( t ) , h ( u ( t ) ) ) | < ψ ( h ( u ( t ) ) ) . Thus we have

ρ ( t ) | u ( t ) | | ρ ( t ) | | u ( t ) | < ψ ( h ( u ( t ) ) ) , | u ( t ) | < 1 δ [ | ρ ( t ) | | u ( t ) | + ψ ( h ( u ( t ) ) ) ] 1 δ [ θ | u ( t ) | + ψ ( h ( u ( t ) ) ) ] . (3.21)

If u ( τ ) = 0 , then there exists a sufficiently small neighborhood of τ such that u ( t ) > 0 . Thus r ( t ) = u ( t ) h 1 ( u ( t ) ) and r ( τ + 0 ) 0 . However, it follows from (3.7) and (3.21) that

0 r ( τ + 0 ) = u ( τ ) h 1 ( u ( τ ) ) u ( τ ) > 1 δ [ θ u ( τ ) + ψ ( h 1 ( u ( τ ) ) ) ] + ψ ( h 1 ( u ( τ ) ) ) + θ h 1 ( u ( τ ) ) δ h 1 ( u ( τ ) ) u ( τ ) = ψ ( h 1 ( u ( τ ) ) ) δ h 1 ( u ( τ ) ) [ u ( τ ) h 1 ( u ( τ ) ) ] 0 , (3.22)

If u ( τ ) < 0 , then there exists a sufficiently small neighborhood of τ such that u ( t ) < 0 . Thus r ( t ) = u ( t ) h 2 ( u ( t ) ) and r ( τ ) = 0 . However, it follows from (3.8) and (3.21) that

0 = r ( τ ) = u ( τ ) h 2 ( u ( τ ) ) u ( τ ) < 1 δ [ θ u ( τ ) + ψ ( h 1 ( u ( τ ) ) ) ] ψ ( h 2 ( u ( τ ) ) ) + θ h 2 ( u ( τ ) ) δ h 2 ( u ( τ ) ) u ( τ ) = ψ ( h 2 ( u ( τ ) ) ) δ h 2 ( u ( τ ) ) [ u ( τ ) h 2 ( u ( τ ) ) ] 0 ,

If u ( τ ) > 0 , then there exists a sufficiently small neighborhood of τ such that u ( t ) > 0 . Thus r ( t ) = u ( t ) h 1 ( u ( t ) ) and r ( τ ) = 0 . By the similarly argument as in (3.22), we reach a contradiction.

Hence u ( t ) < h ( u ( t ) ) , t ( t k , t k + 1 ] , k = 0 , 1 , , m . Similarly, we can prove h ( u ( t ) ) < u ( t ) , t ( t k , t k + 1 ] , k = 0 , 1 , , m .

Likewise, we can show that case (b) is also impossible, and thus (3.20) holds. Combining the above results, we see that if u Ω , L u = N ¯ u , then u Ω α β . Hence (3.17) is proved. From the property of coincidence, we know that (3.16) holds. Since in Ω α β , F = f , we have N = N ¯ . The proof is complete. □

### 4 Main results

We are now in a position to prove our main result on the existence of at least 2 n 1 solutions of boundary value problem (1.1)-(1.2).

Theorem 4.1Assume that

(H1) there existn ( n N and n 2 ) pairs of strict lower and upper solutions { α i ( t ) } i = 1 n , { β i ( t ) } i = 1 n of the problem (1.1)-(1.2) such that

α 1 ( t ) < β 1 ( t ) < α 2 ( t ) < β 2 ( t ) < < α n ( t ) < β n ( t ) , t J ;

(H2) f : J × R 2 R , f ( t , u , p ) is continuous on J × R 2 and has property (H) relative to α 1 ( t ) , β n ( t ) ;

(H3) I k , J k are continuous for each k = 1 , 2 , , m , and satisfy

I k ( t k , α i ( t k ) ) = I k ( t k , β i ( t k ) ) = 0 , and J k ( t k , α i ( t k ) ) < 0 < J k ( t k , β i ( t k ) ) , i = 1 , 2 , , n .

Then BVP (1.1)-(1.2) has at least 2 n 1 solutions u 1 ( t ) , u 2 ( t ) , , u 2 n 1 ( t ) such that α 1 ( t ) < u 1 ( t ) < β 1 ( t ) , α 2 ( t ) < u 2 ( t ) < β 2 ( t ) , , α n ( t ) < u n ( t ) < β n ( t ) , and

min t J u n + 1 ( t ) < α 2 ( t ) , max t J u n + 1 ( t ) > β 1 ( t ) , , min t J u 2 n 1 ( t ) < α n ( t ) , max t J u 2 n 1 ( t ) > β n 1 ( t ) .

Proof Choose M > 0 large enough such that, for i = 1 , 2 , , n ,

f ( t , β i ( t ) , 0 ) + M β i ( t ) > 0 , M > β i ( t ) , t J , f ( t , α i ( t ) , 0 ) M α i ( t ) < 0 , M < α i ( t ) , t J , J k ( t k , α i ( t k ) ) M α i ( t k ) < 0 < J k ( t k , β i ( t k ) ) + M β i ( t k ) , k = 1 , , m ,

and let h ( u ) be defined in (3.10), then it follows from Lemma 3.4 that we can take d > 0 large enough such that, for all u [ M , M ] ,

min [ 0 , M ] h 1 ( u ) > max { max t J | β i ( t ) | , max t J | α i ( t ) | , i = 1 , 2 , , n } , min [ M , 0 ] h 2 ( u ) > max { max t J | β i ( t ) | , max t J | α i ( t ) | , i = 1 , 2 , , n } .

Define the set G i ( i = 1 , 2 , , n ) as

G i = { u P C 1 ( J ) | ( t , u , p ) : α i ( t ) < u < β i ( t ) , | p | < h ( u ) , t J } .

Also

G n + 1 = { u P C 1 ( J ) | ( t , u , p ) : α 1 ( t ) < u < β n + 1 ( t ) , | p | < h ( u ) , t J } .

For i = 1 , 2 , , n + 1 , we define the sets

Ω i = { u P C 1 ( J ) : ( t , u ( t ) , u ( t ) ) G i , t J ; ( t k + , u ( t k + ) , u ( t k + ) ) G i , k = 1 , , m } .

Then by Lemma 3.4, we have

Deg [ ( L , N ) , Ω i ] = 1 , i = 1 , 2 , , n + 1 .

From the additive property of coincidence degree, we obtain

Deg [ ( L , N ) , Ω n + 1 Ω 1 Ω 2 Ω n ¯ ] = n 1 1 .

Since n 2 is arbitrary, we first deal with the case n = 2 , the above discussion implies that the equation L u = N u , that is, the problem (1.1)-(1.2) has at least one solution in the set Ω 1 , Ω 2 and Ω 3 Ω 1 Ω 2 ¯ respectively. That is, there exist at least three different solutions u 1 ( t ) , u 2 ( t ) and u 3 ( t ) such that

α 1 ( t ) < u 1 ( t ) < β 1 ( t ) , α 2 ( t ) < u 2 ( t ) < β 2 ( t ) , on  J ,

and min t J u 3 ( t ) < α 2 ( t ) , max t J u 3 ( t ) > β 1 ( t ) .

For n = 3 , replacing α 1 , α 2 , β 1 , β 2 by α 2 , α 3 , β 2 , β 3 respectively, then we can obtain another two different solutions u 4 ( t ) , u 5 ( t ) of the problem (1.1)-(1.2) such that

α 3 ( t ) < u 4 ( t ) < β 3 ( t ) , and min t J u 5 ( t ) < α 3 ( t ) , max t J u 5 ( t ) > β 2 ( t ) .

Along this way, we can complete the proof by the induction method. □

We now present an example to illustrate that the assumptions of our theorem can be verified.

Example 1 Consider the following boundary value problem:

{ ( ρ ( t ) u ) = a 2 a sin 2 π u ( t ) cos 4 π u ( t ) + b t 2 ( ρ ( t ) u ) = + 1 2 ( u ) 2 ( 1 + sin 2 u ) , t [ 0 , 1 ] , t t k , Δ u ( t k ) = ( t k u ( t k ) ) cos 2 π u ( t k ) , k = 1 , , m , Δ u ( t k ) = t k 16 t g π u ( t k ) , k = 1 , , m , u ( 0 ) = 0 , u ( 1 ) = u ( η ) , (4.1)

where ρ ( t ) = c + sin π t , c > 1 , 0 < b < a 1 12 , 0 < η < 1 , 0 < t 1 < t 2 < < t m < 1 , η t k .

First, we have ρ ( t ) c 1 > 0 and | ρ ( t ) | = | cos t | 1 , thus θ = 1 , δ = c 1 .

Next, it is clear that { α i } i = 1 n = { 1 4 , 3 4 , , n 5 4 } and