Abstract

In this paper, we study the existence of periodic solutions to nonlinear fully third-order differential equation where is continuous and -periodic in . By using the topological transversality method together with the barrier strip technique, we obtain new existence results of periodic solutions to the above equation without growth restrictions on the nonlinearity. Meanwhile, as applications, an example is given to demonstrate our results.

1. Introduction

In this paper, we consider the existence of periodic solutions for nonlinear fully third-order differential equationwhere is continuous and -periodic with respect to .

The third-order periodic problem arises in many areas of applied mathematics and physics, and so it has been extensively studied by many authors via various methods, for instance, see [1–22] and the references therein. Among a substantial number of works, we mention that the upper and lower solutions method is used in [5, 6, 15, 16], Leray–Schauder continuation theorem is used in [1, 10, 20, 21], Leray–Schauder degree theory or the Schauder-fixed-point theorem is used in [10, 12, 13, 18], Mawhin coincidence degree theory is used in [2, 3, 8, 17], and fixed-point theorem in cone or fixed-point index theory is used in [7, 9, 11, 19, 22]. However, to the best of our knowledge, there is no work that refers to periodic solutions of equation (1) using the topological transversality method.

Recently, Kelevedjiev and Todorov [23] have used the topological transversality method and barrier strip technique to study various third-order two-point boundary value problems. But, they did not consider the third-order periodic boundary value problem.

Motivated and inspired by the aforementioned works, the aim of this paper is to establish new existence results of periodic solutions to equation (1) by using the topological transversality method together with barrier strip technique. It is worth mentioning that our results do not need any growth restrictions on the nonlinearity. In addition, compared with the corresponding ones in the known literature, the barrier strip technique we use to estimate a prior bounds of periodic solutions is essentially new.

This work is organized as follows. In Section 2, we first introduce some notations and lemmas and then estimate a prior bounds of periodic solutions of equation (1) by using barrier strip technique. Finally, by using the topological transversality method, we establish the existence results of periodic solutions to equation (1). As applications of our main results, an example is given in the last section.

2. Main Results

Throughout this section, the following assumptions are used: : there exists such that, for any -periodic function , : there exists such that, for any -periodic function , : there exists such that : there are constants with and such that does not change its sign for and for , respectively, hereand .

Let us introduce some notations of the topological transversality method for the convenience of the reader. Let be a convex subset of a Banach space and be open in . Denote by the set of compact operators which are fixed-point-free on . We say that is essential if every operator in which agrees with on has a fixed point in .

The next two lemmas can be found in [24].

Lemma 1. If and is a constant operator, for then is essential.

Lemma 2. Let(i) be essential(ii) be a compact homotopy, , and for and Then, is essential, and therefore, it has a fixed point in .

We note that the existence of -periodic solutions for equation (1) is equivalent to the solvability of the following third-order periodic boundary value problem (for short PBVP):in .

We first consider the family of following PBVPs:where .

A priori bounds for the solutions of PBVP (7), (8) are presented in the following lemmas.

Lemma 3. Suppose that (or ) and hold. Let be a solution of PBVPs (7) and (8) for some . Then,

Proof. At first, we show that (10) holds. Indeed, suppose on the contrary that there exist such that . We may assume that . Let be such thatWithout loss of generality, we assume that ; then, and . It follows from condition thatwhich is a contradiction. This means that (10) holds.
Next, we prove that (9) holds. Indeed, integrating the equation in (7) from 0 to , we obtain thatThis together with the condition (or ) implies that there exists such that . It follows from (10) that, for ,which means that (9) holds. This completes the proof of the lemma.

Lemma 4. Suppose that (or ), , and hold. Let be a solution of PBVPs (7) and (8) for some . Then,

Proof. We estimate by using the barrier strip technique. From condition , does not change its sign for and for , respectively. For the sake of certainty, without loss of generality, we assume thatNotice that , and from Rolle’s mean value theorem, there exists such that . LetWe now assert that the sets and are empty. We shall complete the proof in two steps.

Step 1. Show that . Suppose on the contrary that there exist some . Then, , and . Since is continuous on , there exist such thatand so . Consequently, from assumption (16), we haveand thus,which contradicts (18). This implies that .

Step 2. Prove that . By contradiction, assume that there exist some . Then, , and . We now assert thatIndeed, if , then from the fact , it follows that Notice that , and it follows from the continuity of on that there exist such thatHence, from Lemma 3, we haveand so from (16), it follows thatTherefore,which contradicts (22). This means that (21) holds.
Thus, from (21) and the continuity of on , there exist such thatand thus, . It follows from assumption (16) thatand hence,which contradicts (26). This implies that .
Therefore, by the facts that and the continuity of on , we obtainIn particular, . Notice , and we haveWe now letNotice that, from (30), using the similar arguments on and , we can show thatHence,From this, together with (29), it follows thatwhich implies thatThis completes the proof of the lemma.
Now, denote the Banach space equipped with the norm . SetThen, is a closed and convex subset of , and is an open subset of .
We now give two lemmas which will be used in the proof of our main theorem.

Lemma 5. Suppose that holds. Let the operator be defined by

Then, is essential.

Proof. Define byThen, and for . Thus, it follows from Lemma 1 that is essential. Meanwhile, by a standard argument, it is easy to show that is compact.
We now show thatObviously, for all . Suppose that for some and . Then, on , and soHence, from , we can deduce that , which contradicts . This implies that (39) holds. Therefore, from Lemma 2, is essential. This completes the proof of the lemma.

Lemma 6. Suppose that holds. Let the operator be defined by

Then, is essential.

Proof. The proof is similar to the proof of Lemma 5 and hence is omitted.

Theorem 1. Suppose that , , and hold. Then, equation (1) has at least one -periodic solution satisfying (9), (10), and (15).

Proof. At first, we define operator bywhereIt is easy to check that, for each , is the unique solution of the following boundary value problem:Furthermore, by a standard argument, it is easy to show that the operator is completely continuous.
We now define operator bySuppose that is a fixed point of . Then,It follows that , , andand thus, by (46),Set for . It is easy to see that is a solution of PBVP (6), and validity of (9), (10), and (15) now follows from Lemmas 3 and 4. Therefore, to prove the existence of solutions of PBVP (6) satisfying (9), (10), and (15), it is sufficient to show that the operator has at least one fixed point. Since and is essential by Lemma 5, for the existence of a fixed point of , it is sufficient to verify (i) and (ii) of Lemma 2. Notice that operator is completely continuous, then operator is continuous, and also is compact in . Thus, (i) of Lemma 2 is satisfied. Let for some and . If then which has been proved in the proof of Lemma 5. Let . Then,Hence,and soSet for . We can see that is a solution of PBVPs (7) and (8) with . It follows from Lemmas 4 and 5 thatSince , the first inequality of (52) yieldsand thus,Hence, , and so (ii) of Lemma 2 is verified. This completes the proof of the theorem.

Theorem 2. Suppose that , , and hold. Then, equation (1) has at least one -periodic solution satisfying (9), (10), and (15).

Proof. The proof is similar as that for Theorem 1 except thatand are used in place of and , respectively, and hence is omitted. This completes the proof of the theorem.

3. An Example

In this section, we give an example to demonstrate the applications of the our results.

Example 1. Consider the nonlinear third-order differential equationLetThen, for any -periodic function , we haveand so condition holds. Notice thatuniformly in ; then, there exists such thatthat is, condition holds. In addition, sinceuniformly in , where , it follows that condition holds. Hence, by Theorem 1, third-order differential equation (56) has at least one -periodic solution.

Data Availability

There are no data in this paper.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.