Theorem 2.1 Assume that f satisfies (), () and (). Then (1.1) possesses at least one 4r-periodic solution.
Proof Let denote the usual normal orthogonal bases in and set
where is the set of all positive integers. Then . For any , it has a Fourier expansion as follows:
where all ,
Consequently, we have
Then (2.2) and (2.3) show that and are two equivalent norms on E. Henceforth we use the norm as the norm for E. And the spaces M, N are mutually orthogonal with respect to the associated inner product.
First, we prove that satisfies (1.2) in E.
Let be any sequence which converges to some u weakly in E. By the compactness of the embedding , we assume that
Thus, a.e. for all . Since satisfies (), there exist positive constants and such that
Note that the right-hand side of (2.4) converges to in . Hence is uniformly absolutely continuous. Hence, by Vitali’s theorem,
Moreover, since L is a bounded self-adjoint linear operator on E,
According to (2.5) and (2.6), we get that is weak-to-weak continuous.
Next, we prove
Indeed, by (), we know that there exists a positive constant c such that
Thus, for , by (2.9), we have
where is a given constant. Since , (2.10) implies (2.7). The proof of (2.8) is similar. In fact, when , by (2.9), we get that
This implies (2.8).
Note that (2.10) implies
The combination of (2.8) and (2.11) implies that there is such that (1.3) holds with , . By Lemma 1.1 we know that there are a sequence and a constant c such that
Finally, we show that the sequence is bounded in E. To do this, assume that , and write . Then . From Lemma 1.4, there is a renamed subsequence such that
By (), for any , there exists a constant such that
Moreover, by the continuity of f, there is a constant such that
By (2.13) and (2.14), for any , we have
as and . This shows
By (2.12) and (2.17), we see that
for all , i.e.,
for all .
Then, by (2.18), one can obtain
where I is the unit matrix. For any j, take and , where . An easy computation shows that
Hence, by , we get that .
Let , where , . A proof similar to (2.16) shows that
where , , . Thus
On the other hand, (2.12) implies
which contradicts (2.23). Thus must be bounded. Consequently, there is a renamed subsequence of such that in E. Hence, by the weak-to-weak continuity of , we have
Now, the combination of (2.12) and (2.24) implies that . This completes the proof. □
Remark 2.1 Let , and . Then satisfies all the conditions of Theorem 2.1.
In order to give our another result, we still need the following preliminaries.
Let , be the projectors of E onto M, N associated with the given splitting of E. Set
Recall that K is continuous and maps bounded sets to relatively compact sets since K is compact. Let with , a given subspace of E. Then ∂Q will refer to the boundary of Q in .
Definition 2.1 We say S and ∂Q link if whenever and for all , then for all .
Lemma 2.1 ()
Let , , , , , , and . Then S and ∂Q link.
Lemma 2.2 ()
Suppose satisfies the (PS) condition and
() , where and , are bounded self-adjoint,
() is compact,
() there exist a subspace and sets , and constants such that
Q is bounded and ,
S and ∂Q link.
Then I possesses a critical value .
The following is our another main result.
Theorem 2.2 Assume that f satisfies (), () and the following conditions:
() for all ,
() as ,
() there exist constants , and such that
Then (1.1) possesses at least one nonconstant 4r-periodic solution.
Proof We will show that I satisfies the hypotheses of Lemma 2.2. This will lead to a nonconstant 4r-periodic solution of (1.1). We divide the proof of Theorem 2.2 into the following three parts.
First, we prove that I satisfies () and () of Lemma 2.2.
Note that , and L is bounded self-adjoint on E. We see that I satisfies () of Lemma 2.2 with , and
By Proposition B.37 in , () implies that is compact. Hence () holds.
Next, we show that I satisfies () of Lemma 2.2.
By (), for any , there is such that
whenever . By (), there is a constant such that
By (2.25) and Lemma 1.4, for any , we have
Choose . Since , there is small such that . Then, for , (2.26) implies that . Consequently, I satisfies ()(i) with .
where and are free constants for the moment. Define . Then and S and ∂Q link by Lemma 2.1.
By (), there are constants such that
for all . Thus, for , by the Hölder inequality (note that ) and orthogonality, we get that
where are constants. Now, choose large and such that
By (2.27), (2.29), (2.30), (2.31) and (), one can easily check that . Hence I satisfies ()(ii).
To sum up, I satisfies () of Lemma 2.2.
Finally, we check that I satisfies the (PS) condition. Let be a sequence such that and as . Then, for large k, by () and (2.28), we have
where are constants.
Let , where , . Then, for large k, by (), Lemma 1.4 and the Hölder inequality, we get that
This implies that
Similarly, one can easily see that
By (2.32), (2.34) and (2.35), there is a constant such that
which implies that is bounded in E.
By the compactness of , going if necessary to a subsequence, we can assume that
Let , where and . Then
as . Similarly, we have as . Hence in E. Hence I satisfies the (PS) condition.
Therefore, Theorem 2.2 follows from Lemma 2.2. □
Remark 2.2 Let and . Then satisfies all the conditions of Theorem 2.2 with and .