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:
(2.1)
where all ,
Consequently, we have
(2.2)
Let
(2.3)
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
(2.4)
Note that the right-hand side of (2.4) converges to in . Hence is uniformly absolutely continuous. Hence, by Vitali’s theorem,
(2.5)
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
(2.7)
and
(2.8)
Indeed, by (), we know that there exists a positive constant c such that
(2.9)
Thus, for , by (2.9), we have
(2.10)
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
(2.12)
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
(2.13)
Moreover, by the continuity of f, there is a constant such that
(2.14)
By (2.13) and (2.14), for any , we have
(2.15)
as and . This shows
(2.16)
Hence
(2.17)
By (2.12) and (2.17), we see that
for all , i.e.,
(2.18)
for all .
Set
Then, by (2.18), one can obtain
where I is the unit matrix. For any j, take and , where . An easy computation shows that
(2.19)
and
(2.20)
Hence, by , we get that .
Let , where , . A proof similar to (2.16) shows that
(2.21)
and
(2.22)
where , , . Thus
(2.23)
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
(2.24)
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 ([9])
Let , , , , , , and . Then S and ∂Q link.
Lemma 2.2 ([9])
Suppose satisfies the (PS) condition and
() , where and , are bounded self-adjoint,
() is compact,
() there exist a subspace and sets , and constants such that
-
(i)
and ,
-
(ii)
Q is bounded and ,
-
(iii)
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
and
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 [9], () 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
Hence
(2.25)
By (2.25) and Lemma 1.4, for any , we have
(2.26)
Choose . Since , there is small such that . Then, for , (2.26) implies that . Consequently, I satisfies ()(i) with .
Set and
(2.27)
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
(2.28)
for all . Thus, for , by the Hölder inequality (note that ) and orthogonality, we get that
(2.29)
where are constants. Now, choose large and such that
(2.30)
and
(2.31)
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
(2.32)
where are constants.
Let , where , . Then, for large k, by (), Lemma 1.4 and the Hölder inequality, we get that
(2.33)
This implies that
(2.34)
Similarly, one can easily see that
(2.35)
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
and
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 .