#### Abstract

We discuss a third-order multipoint boundary value problem under some appropriate resonance conditions. By using the coincidence degree theory, we establish the existence result of solutions. The emphasis here is that the dimension of the linear operator is equal to two. Our results supplement other results.

#### 1. Introduction

In this paper, we are concerned with the following third-order ordinary differential equation: with the boundary conditions where is a Carathéodory function, , , , , , and .

In recent years, many authors have paid much attention to the existence of solutions for multipoint boundary value problems at resonance: we refer the readers to see [1–11]. If the linear equation with boundary conditions (2) has nontrivial solutions, that is, , the BVP (1)-(2) is called a resonance problem. In [5–11], the authors all discussed the case that . In [2, 3], the authors established the existence results for resonance boundary value problems with the case of . However, we will show that some conditions such as assumed in [2, 3] are not necessary. We establish existence of some solutions for BVP (1)-(2) by using the coincidence degree theory of Mawhin [12] at resonance.

According to the constant , the BVP (1)-(2) is divided into the following five resonance cases:); (); (); (); ().

Du et al. [1] studied the existence results of BVP (1)-(2) under the resonance conditions () and (), that is, , but they did not discuss the other three cases. In this paper, under the resonance conditions, (), (), or (), we could imply ; thus we supplement the results in [1].

The layout of this paper is as follows. In Section 2, we briefly present some notations and an abstract existence result due to Mawhin. In Section 3, we study BVP (1)-(2) under the condition () and obtain some existence results. In Section 4, we give an example of the existence results in Section 3.

#### 2. Preliminary

Now, we briefly recall some notations and an abstract existence result by Mawhin [12].

Let , be real Banach spaces and let be a linear operator which is a Fredholm map of index zero and let , be continuous projectors such that , and , . It follows that : is invertible; we denote the inverse of that map by . Let be an open bounded subset of such that ; the map is said to be compact on if the map is bounded and is compact. For more details we refer the reader to the lecture of Mawhin [12].

The theorem we use in this paper is Theorem IV.13 of [12].

Theorem 1. *Let be a Fredholm map of index zero and let be compact on . Assume that the following conditions are satisfied. *(i)* for every .*(ii)* for every .*(iii)*, , where is a continuous projector as above with = .**Then the abstract equation has at least one solution in .*

In the following, we will use the classical spaces , , and . For , we use the norms and , denote the norm in by , and define the Sobolev space as

Let , , and define the linear operator as , where We define as Then BVP (1)-(2) can be written as .

#### 3. Existence Results

Lemma 2. *If the condition () holds, then there exist , , such that
**
where .*

*Proof. *We prove that, for any , there exists , such that .

If else, one has , any ; that is,
Since
thus , .

It is clear that
which is a contradiction to the condition .

Set
Then is a finite set.

If else, there exists a monotone sequence , , , such that
From , we get
Thus

So it is a contradiction. Thus the Lemma is proved.

Lemma 3. *Let hold and ; then : is a Fredholm map of index zero. Furthermore, the linear continuous projector operator can be defined by
**
where
**
And the linear operator : can be written by
**
Furthermore
*

*Proof. *It is clear that , , .

Now we show that
The equation
has a solution satisfying (2) if and only if

In fact, if (19) has a solution such that (2), then from (19) we have
According to the condition , we obtain
On the other hand, if (20) holds, let
where is an arbitrary constant; then is a solution of (19) and (2). Hence (18) holds.

Set
Then we define
It is clear that .

Again from
One has
Thus the operator is a projector.

Now we show that . If , from , we have
Because of
, which yields . On the other hand, if , from and the definition of , so ; thus . Hence, .

For , from , , , we have . And if , from , there exist constants , such that .

From , we obtain
In view of
therefore (30) has a unique solution , which implies . So we have . Since , thus is a Fredholm map of index zero.

Let be defined by
Then, the generalized inverse can be written by

In fact, for , we have
and for , we know

Taking note that , , thus .

It is clear that .

Theorem 4. *Let the condition hold and . Assume the following. * * There exist functions , such that
where , .* * There exists a constant such that for , if or for all , then
* * There exists a constant such that for , if or , then either
or
*

Then BVP (1)-(2) has at least one solution in .

*Proof. *We divide the proof into the following steps.*Step **1.* The set is bounded.

For , since , so , Im ; hence

From (), there exist such that , . , and are absolutely continuous for all , and
which imply

From (), we obtain
So there exists a constant such that ; that is, the set is bounded.*Step **2.* The set is bounded.

For , implies that , , , . From , we get . From (), then ; that is, the set is bounded.*Step **3.* The set , is bounded.

For any , we define the linear isomorphism by
where

Set

For any , we obtain
On account of
therefore, we have

If , then . If and or , from the above equality and (38), one has
which contradicts ; thus . So the set is bounded.*Step **4.* If (39) holds, similar to the above argument, we can prove that the set
is bounded too, where *J* is defined in (44).

Now, we will prove that all conditions of Theorem 1 are satisfied.

Let be an open bounded subset of such that . By the ArzeláAscoli theorem, we can prove that is compact, so is compact on .

Then by the above argument, we have(i) for every ;(ii) Im for every (iii)let .

According to the above argument in Steps 3 and 4, we know for every . Thus, by using the homotopy property of degree, we have
Then by Theorem 1, has at least one solution in dom; that is, BVP (1)-(2) has at least one solution in .

#### 4. Example

*Example 1. *We consider the following boundary value problem:

Let Then the condition () holds.

From Lemma 2, one has . By Lemma 3, we define

Since , then , , .

If , , and , one has Then BVP (53) satisfies Theorem 4. So it has at least one solution in .

*Remark 2. *By using a similar method as employed in the above proof, we could obtain some similar results under the condition () or (), then we omit them.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding to the publication of this paper.

#### Acknowledgments

This paper is sponsored by the Natural Science Foundation of China (11071205, 11101349, 61201431), the Natural Science Foundation of Jiangsu Province, and PAPD of Jiangsu Higher Education Institutions.