### 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:

together with the boundary conditions:

where
*m* is a fixed positive integer,

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

Note that the nonlinear term *f* depends on *u* and its derivative
*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
*f*, Feng and Webb obtained the existence of the solution of the problem (1.3) when

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

In the present paper, we assume that there exist *n* (
*f* satisfies a Nagumo-like growth condition with respect to

### 2 Preliminaries

Let

where

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

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

**Definition 2.1** A function

Similarly, a function

**Remark 2.2** Let

**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
*J*. We say that
*H*) relative to *α* and *β*, if there exists a function

for all

where

### 3 The key lemmas

Let

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

**Lemma 3.1***Suppose that**L**be defined in the above*. *Then**L**is a Fredholm operator of index zero*. *Furthermore*

*and*

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

The following problem:

has a solution

In fact, if (3.3) has a solution

According to

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

where *c* is an arbitrary constant, then

Take the projector

where

Thus, we obtain

Then
*L* is a Fredholm operator of index zero.

Take
*L* can be written as

Set
*ψ* defined by (2.8), let

and

**Lemma 3.2***Suppose that there exists a constant*
*then*
*is well defined in*
*and positive on this interval*,
*is also well defined and positive in*
*Moreover*, *if*
*then*
*for any*
*for any*

*Proof* We only consider the case

Integrating the above equation over

However, the left side of the above equation equals ∞ by (2.8). We reach a contradiction.
Hence

Define the following sets:

Define the function

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

(i)
*M**is given in Lemma* 3.2);

(ii)

(iii)
*and*

*Then*

*Proof* Consider the following family of equations:

We will show

If not, then there exist some

Case (I). If

Subcase (1). Suppose

If

which is a contradiction.

If

On the other hand,

which is a contradiction.

If

which is a contradiction.

If

Subcase (2). Suppose

Let

Without loss of generality, we suppose that

(i) If

which is a contradiction.

(ii) If

which is a contradiction.

(iii) If

Case (II).

Subcase (3). Suppose

If

which is a contradiction.

If

Similar to subcase (2), we can show that

On the other hand, for

where *E* is the identity mapping.

Note that

From (3.5) and (3.6), we have

and

By the Ascoli-Arzela theorem, it is easy to show that
*N* is *L*-compact on

Define a mapping

Then it is easy to prove that

In fact, for

Thus we get

Since Ker*L* is one dimensional and

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

**Lemma 3.4***Assume that*

(c1) *there exist lower and upper solutions*
*of the problem* (1.1)-(1.2), *respectively*, *with*

(c2)
*is continuous and has property* (*H*) *relative to**α*, *β*;

(c3)
*are continuous for each*
*and satisfy*

*Then*

*Proof* Choose

and let

Define a set

where

Define the auxiliary functions *F* and

where

We then generalize *F* to
*F*,

Moreover, when
*F*,

where

Next we show

It suffices to show that

In fact, let

Case (1). If there exists

which implies

If

If

if

Hence the function

Case (2). If there exists a

Consequently, we get

By the continuity of *f* and

where

where

which is a contradiction to (3.19).

Case (3). If

Thus we have proved that
*J*. Similarly we can show that
*J*. It then follows that
*J*.

We now shall prove that

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

(a) There exists

(b) There exists some

For case (a), we assume

Since

If
*τ* such that

which is a contradiction.

If
*τ* such that

a contradiction.

If
*τ* such that

Hence

Likewise, we can show that case (b) is also impossible, and thus (3.20) holds. Combining
the above results, we see that if

### 4 Main results

We are now in a position to prove our main result on the existence of at least

**Theorem 4.1***Assume that*

(H_{1}) *there exist**n* (
*and*
*pairs of strict lower and upper solutions*
*of the problem* (1.1)-(1.2) *such that*

(H_{2})
*is continuous on*
*and has property* (*H*) *relative to*

(H_{3})
*are continuous for each*
*and satisfy*

*Then BVP* (1.1)-(1.2) *has at least*
*solutions*
*such that*
*and*

*Proof* Choose

and let

Define the set

Also

For

Then by Lemma 3.4, we have

From the additive property of coincidence degree, we obtain

Since

and

For

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:

where

First, we have

Next, it is clear that