Nontrivial Solutions of a Fully Fourth-Order Periodic Boundary Value Problem
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.
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  motivated our study.
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 ). 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 ). 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
Theorem 5. Assume that there exist , , and a nonnegative function . Suppose further that (H0) satisfies the Carathéodory conditions;(H1)if , , , , then (H2)if , , , , then where , , , fulfil (14). Then PBVP (1), (2) has at least one solution such that
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.
Theorem 6. Assume that all conditions in Theorem 5 hold with the exception of (H1), 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.
This work was supported by the National Natural Science Foundation of China (11201008).
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