In this section, we will establish the variational structure relative to problem (1.1) and prove the main results via Morse theory.

Denote and

Equipped with the inner product and norm as follows:

is linearly homeomorphic to . Throughout this paper, we always identify with .

Define the operator by and denote , , where is the identity operator. Set

then has the decomposition . In the rest of this paper, the expression for always means , .

Remark 3.1.

From the discussion in [1, Section 2], we see that , , for and if is even or if is odd.

Define a family of functionals , by

where , . Then, the Fréchet derivative of at , denoted by , can be described as (see [1])

where and

Remark 3.2.

From (3.5) with , we know by computation(or see [1]) that is a critical point of if and only if is a -periodic solution of problem (1.1). Moreover, is differentiable and

where is the derivative of with respect to .

Let , and consist of satisfying

Remark 3.3.

is the solution space of the system , , where

Thus, since possesses of non-degenerate order submatrixes.

Lemma 3.4.

If , and satisfies (3.8), where and , then either or .

Proof.

Setting and , respectively, in (3.8), we have

Comparing the above two equalities, we get

which, by , , implies that

On the other hand, by the definition of and , we have

where . There are two cases to be considered.

Case 1.

for . Then by (3.12), and for , that is, .

Case 2.

There exists such that . By (3.13), we have

If , then which, by (3.13), implies that for , that is, . If , then . This, by (3.12), implies and . Thus, by (3.13), for , that is . The proof is complete.

Set and . The following Lemmas 3.5–3.7 benefit from [4].

Lemma 3.5.

Assume that (f_{1}) and (f_{2}) hold. Let and satisfy and
as . Then,

Proof.

From (f_{2}), we have

where the limitation is uniformly in . It follows that for any , there exists such that

Thus, there exists such that

By the assumption on , we have . It follows from (3.5) that

which, combining with (3.18), implies that

By using, Holder inequality on the above two summations, we get

which leads to

Note that is arbitrarily small, we get (3.15), and the proof is complete.

Lemma 3.6.

Under the conditions of Lemma 3.5, one further has

Proof.

Since , , and are invariant with respect to , we have

If, for the contradiction, (3.23) is false, then there is a subsequence of , called again, and a number , such that , . Then,

where .

By the fact that and are two consecutive eigenvalues of with corresponding eigenspace and , we have and then, the function is strictly decreasing on with as . Besides, . So, by (3.25),

This contradict to (3.15) and the proof is complete.

Lemma 3.7.

Under the assumption of Lemma 3.5, there exists a subsequence of , still called , such that

Proof.

Since as , we can assume (by passing to a subsequence if necessary) that

Thus, (3.16) implies

which implies that there exists a subsequence of , still called , and , such that

Let , then , and, by Lemma 3.6, there is a convergent subsequence of , call it again, such that

To prove (3.27), we only need to show that or . For every , we have as , that is,

If as for , then we can rewrite (3.32) as

Letting in (3.33) and using (3.30) and (3.31), we get

Since for , by setting for , we rewrite (3.34) as

Obviously, if , (3.35) still holds. By Lemma 3.4, or and the proof is complete.

Lemma 3.8.

Assume that and hold. Let and satisfy and as . Then, there exists a subsequence of , still called , such that

Proof.

As that in the above proof, we can assume that satisfies (3.28). Noticing that (f_{3}) implies (f_{2}) and by Lemma 3.7, we have two cases to be considered.

Case 1.

as . We have as and

If , then and are bounded for and . It follows that as for and

By (f_{3}(i)), there exist and such that and for and . Then, for , and ,

Choose such that for and . It follows that

where . Since is a finite dimensional vector space and possesses another norm defined by , , which is equivalent to , there exists a positive constant such that , . Thus, by (3.37)–(3.40),

Obviously, if , the above inequality still holds.

Case 2.

as . By using , we can show that in the same way. The proof is complete.

In the rest of this section, we will use the facts ()–() stated in Section 2 to complete the proofs.

Lemma 3.9.

Let satisfy (f_{1}) and (f_{3}). Then, for every , satisfies the (PS) condition and

Proof.

First we have the following claim:

Claim 1.

For any sequences andifas, then is bounded.

In fact, if is unbounded, there exists a subsequence, still called , such that as . By Lemma 3.8, there exists a subsequence, still called , such that or .

On the other hand, as, that is

Note that , , it follows that . This contradiction proves Claim 1.

Setting in Claim 1, we see that satisfies (PS) condition. Now, we start to prove (3.42). Define a functional as

Claim 2.

There exist such that

In fact, if Claim 2 is not true, there exists and such that and as , which contradict Claim 1.

Noticing that , we set . Then, implies . Consider the flow generated by

The chain rule for differentiation reads . Thus,

and , , which implies that , . Then, the flow is well defined on and is a homeomorphism of to and (see [11])

On the other hand,

Note that is the unique critical point of with Morse index (see Remark 3.1) and nullity . Then, by (2.b), (2.f) and (3.48), we have

The proof is completed.

Proof of Theorem 1.3.

By lemma 3.9, we get (3.42) which, by , implies that there exists with

Since , we have . Denote by and the Morse index and nullity of . By , we get .

Denote . Then, from (3.7) and Remark 3.3, we see that .

In Case (i), is a local minimum of , hence, by (),

which, by comparing with (3.51), implies that . Besides, since . Assume, for the contradiction, that is the unique nontrivial critical point of , then . If or , we have, by (),

from which, () reads , a contradiction.

If , then and . Since , we have . Thus, () with reads , also a contradiction.

In Case (ii), is a local maximum of , hence, by (),

which, by comparing with (3.51), implies that . Besides, since . Assume, for the contradiction, that is the unique nontrivial critical point of , then . If or , then (3.53) holds, from which, ) reads

a contradiction. If , then and, by (),

Note that , we have . Thus, () with and with reads

respectively, which implies that . Then, () reads (3.55), also a contradiction. The proof is complete.

Proof of Theorem 1.4.

As above, there exists with the Morse index , and nullity satisfying , , and (3.51) holds.

On the other hand, is a nondegenerate critical point of with Morse index, denoted by . Thus, and since , which, by comparing with (3.51), implies that .

Assume for the contradiction, that is unique nontrivial critical point of , then . If or , then (3.53) holds and () reads the contradicition .

Now, we consider the case where we have and with (3.56). Since , we know that either or . If , () with reads contradiction . If , by similar argument, we can get (3.57). Thus and () reads the contradiction . The proof is complete.

The proof of the following lemma is similar to that of ([12]) and is omitted.

Lemma 3.10.

Let satisfy
or . Then has a local linking at with respect to the decomposition , where (or , respectively).

Proof of Theorem 1.5.

Now . Thus, is a degenerate critical point of . Let and denote the Morse index and nullity of 0. By Lemma 3.10 and (), we have

where or corresponding to the case () or the case (), respectively. The rest of the proof is similar and is omitted. The proof is complete.