Research Article | Open Access

Haitong Li, Minghe Pei, Libo Wang, "Nontrivial Solutions of a Fully Fourth-Order Periodic Boundary Value Problem", *Journal of Applied Mathematics*, vol. 2014, Article ID 895862, 7 pages, 2014. https://doi.org/10.1155/2014/895862

# Nontrivial Solutions of a Fully Fourth-Order Periodic Boundary Value Problem

**Academic Editor:**Daqing Jiang

#### Abstract

We investigate the solvability of a fully fourth-order periodic boundary value problem of the form where satisfies Carathéodory conditions. By using the coincidence degree theory, the existence of nontrivial solutions is obtained. Meanwhile, as applications, some examples are given to illustrate our results.

#### 1. Introduction

In this paper, we consider a fully nonlinear fourth-order periodic boundary value problem of the form subject to the boundary conditions where satisfies Carathéodory conditions; that is, (i)for a.e. , the function is continuous;(ii)for every , the function is measurable;(iii)for each , there is a real valued function such that for a.e. and .

It is well known that fourth-order periodic boundary value problems are important research topics which arise in a variety of different areas, such as nonlinear oscillations, fluid mechanical, and nonlinear elastic mechanical phenomena, and thus have been extensively studied; for instance, see [1–30] and references therein. However, most of the works in the above-mentioned references allow only having , or , , in the right-hand side nonlinear function ; see [2–11, 13, 15–18, 20–30]. The works on the fully nonlinear cases of which contains explicitly and every derivative of up to order three have been quite rarely seen; see [1, 12, 14, 19].

The aim of this paper is to establish the existence of solutions and nontrivial solutions for the fully nonlinear fourth-order PBVP (1), (2). Our main tool is the coincidence degree theory. The paper [31] motivated our study.

#### 2. Preliminary

In this section, we present some lemmas which are needed for our main results.

At first, we will briefly recall some notations that are needed for our discussion.

Let , be real Banach spaces. A linear mapping will be called a* Fredholm mapping of index zero* if the following two conditions hold:(i) is a closed subspace of ;(ii).

Let be a Fredholm mapping of index zero; then there exist continuous projectors and such that
so that
It follows that is invertible. We denote the inverse of that map by . Let be an open bounded subset of such that ; the map will be called -*compact* on , if and are compact.

Lemma 1 (see [32]). *Let be a linear Fredholm mapping of index zero and let be an open bounded set. Let be -compact on and let be -completely continuous such that*(i)*;*(ii)*for every ,
* *and assume that . Then equation
* *has at least one solution in .*

Lemma 2 (see [33]). *Let be a linear Fredholm mapping of index zero and let be an open bounded set. Let be -compact on and the coincidence degree is well defined. If there exists with such that
**
then .*

In the following, we take Banach space with the norm , and . Define a linear map by where and is the usual Sobolev space. It is easy to see that is a Fredholm mapping of index zero. Also define a nonlinear map by

Define two projects and as follows:

Let be Green function for the homogeneous BVP Then can be given by Hence the map is continuous. We note that if satisfying Carathéodory conditions, then is bounded and continuous by Lebesgue's dominated convergence theorem. Furthermore, is -compact on every bounded set .

#### 3. Main Results

For and we put

In order to introduce our main theorem, we need some lemmas.

Lemma 3. *Let be a nonnegative function and . Let , , , , and fulfil (14). Then for any , the inequalities
**
imply
*

*Proof. *Since , there exists such that
We will show that
By contradiction, assume that there exists such that
Then there exists such that
There are two cases to consider.*Case **1*. Consider on . In this case, integrating (16) from to and using (15) and (21) we infer that
Thus , which contradicts (20).*Case **2*. Consider on . Similar to Case 1, we have
Thus, , which contradicts (20). Therefore (19) is true. Furthermore, from the fact it follows that .

Finally, we show that
Suppose on the contrary that there exists satisfying
Then there exists such that
There are two cases to consider.*Case 1* Consider on . Similar to Case 1, we have , which contradicts (25).*Case 2* Consider on . Similar to Case 2, one has , which also contradicts (25). Hence (24) is true.

In summary, from (19) and (24) it follows that estimate (17) holds. This completes the proof of the lemma.

Lemma 4. *Let and let be a nonnegative function. Let , , , , and fulfil (14). Then for any function the inequalities
**
imply
*

*Proof. *For every , from (28), there exist such that
Integrating (27) by (14) and (30) we get
This completes the proof of the lemma.

Now, we apply Lemma 1 to establish the existence results of solutions for the fourth-order PBVP (1), (2).

Theorem 5. *Assume that there exist , , and a nonnegative function . Suppose further that *(H_{0})* satisfies the Carathéodory conditions;*(H_{1})*if , , , , then
*(H_{2})*if , , , , then
* *where , , , fulfil (14). Then PBVP (1), (2) has at least one solution such that
*

*Proof. *Let
Then iff there exist some such that

Now, we show that
where , . To do this, we assume that is the solution of the following periodic boundary value problem:
Integrating the equation as above on , we obtain
Thus, by Wirtinger inequality,
Hence from (38) it follows that
If , then, from , the following contradiction holds:
Therefore, .

Finally, we show that, for every ,
To do this, let and let be a solution of the following PBVP:
Then . In fact, let
Then, by (33),
for a.e. . Applying Lemma 3, we obtain
Thus according to (32), we have
for a.e. . It follows from Lemma 4 that
Thus . This implies that condition (ii) of Lemma 1 is valid.

In summary, all conditions of Lemma 1 are satisfied. Therefore the conclusion of Theorem 5 holds. This completes the proof of the theorem.

Next, we establish the existence result of nontrivial solutions for the fourth-order PBVP (1), (2) by means of Lemma 2.

Theorem 6. *Assume that all conditions in Theorem 5 hold with the exception of (H _{1}), which is replaced by the following:*

*there exists a constant such that if , , , , then*

*and if , , , , then*

*Then PBVP (1), (2) has at least one nontrivial solution satisfying (34).*

*Proof. *From the proof of Theorem 5 and Lemma 1, it follows that has a solution in
and . Without loss of generality, we assume that and . We also assume that for all .

Now we assert that
In fact, suppose that there exist and such that
Applying to both sides of above equality, it follows that
that is,
Notice that and ; we have
Consequently, from assumption one has
This together with (56) it follows that
which is a contradiction. This implies that
Thus from Lemma 2 it follows that
Hence
Therefore has a solution in ; that is, PBVP (1), (2) has at least one nontrivial solution. This completes the proof of the theorem.

Finally, we give some examples to illustrate our results.

*Example 7. *Consider the fourth-order periodic boundary value problem
where is a parameter, is nonnegative, and
is a constant.

Let
Then satisfies Carathéodory conditions. Taking any , then , , , and are well defined by (14).

Now, we assert that all conditions of Theorem 5 are satisfied when
In fact, without loss of generality, we can assume . In this case, we choose . It is easy to see that, for every ,
and, for every ,
Hence condition of Theorem 5 is satisfied. In addition, for , , , , we have
Therefore, condition of Theorem 5 is also satisfied. Hence, from Theorem 5, the fourth-order PBVP (63) has at least a solution , provided

*Example 8. *Consider the fourth-order periodic boundary value problem
where is a parameter, , is odd, is even, and is a nonnegative function.

Let
Then satisfies Carathéodory conditions. We choose ; then is well defined by (14).

Now, we assert that satisfies all conditions of Theorem 6 when
In fact, without loss of generality, we can assume that . Choose and . Then it is easy to see that, for every ,
and, for every ,
On the other hand, for every ,
In summary, all conditions of Theorem 6 are satisfied. Therefore, from Theorem 6, the fourth-order PBVP (71) has at least one nontrivial solution , provided .

#### Conflict of Interests

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

#### Acknowledgment

This work was supported by the National Natural Science Foundation of China (11201008).

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Copyright © 2014 Haitong Li 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.