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Abstract and Applied Analysis
Volume 2014 (2014), Article ID 931217, 8 pages
Solvability of a Third-Order Multipoint Boundary Value Problem at Resonance
1School of Mathematics and Statistics Xuzhou, Jiangsu Normal University, Jiangsu 221116, China
2School of Mathematics and Systems Science, Shandong University of Science and Technology, Qingdao 266510, China
Received 10 December 2013; Accepted 13 January 2014; Published 23 February 2014
Academic Editor: Jifeng Chu
Copyright © 2014 Zengji Du et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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.
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  at resonance.
Du et al.  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 .
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.
Now, we briefly recall some notations and an abstract existence result by Mawhin .
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 .
The theorem we use in this paper is Theorem IV.13 of .
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
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
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
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 .
Example 1. We consider the following boundary value problem:
Let Then the condition () holds.
Since , then , , .
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.
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.
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