In the following, we use to denote the norm of for any . Let be a Hilbert space with the inner product and norm given respectively by
It is well known that W is continuously embedded in for . We define an operator by
Since M is an antisymmetric constant matrix, Γ is self-adjoint on W. Moreover, we denote by χ the self-adjoint extension of the operator with the domain .
Let , the domain of . We define respectively on E the inner product and the norm
where denotes the inner product in .
By a similar proof of Lemma 3.1 in [15], we can prove that if conditions (L1) and (L2) hold, then
(2.1)
Therefore, it is easy to prove that the spectrum has a sequence of eigenvalues (counted with their multiplicities)
and the corresponding system of eigenfunctions () forms an orthogonal basis in .
By (J1), we may let
Then one has the orthogonal decomposition
with respect to the inner product . Now, we introduce respectively on E the following new inner product and norm:
(2.2)
where with and . Clearly, the norms and are equivalent (see [4]), and the decomposition is also orthogonal with respect to both inner products and . Hence, by (J1), E with equivalent norms, besides, we have
(2.3)
and
(2.4)
where a and b are defined in (J1).
For problem (1.1), we consider the following functional:
Then I can be rewritten as
Let . In view of the assumptions of H, we know and the derivatives are given by
for any with and . By the discussion of [24], the (weak) solutions of system (1.1) are the critical points of the functional . Moreover, it is easy to verify that if is a solution of (1.1), then and as (see Lemma 3.1 in [25]).
The following abstract critical point theorem plays an important role in proving our main result. Let E be a Hilbert space with the norm and have an orthogonal decomposition , is a closed and separable subspace. There exists a norm satisfying for all and inducing a topology equivalent to the weak topology of N on a bounded subset of N. For with , , we define . Particularly, if is -bounded and , then weakly in N, strongly in , weakly in E (cf.[26]).
Let , with . Let and . For , let
with , . We define
For , define
where denotes various finite-dimensional subspaces of E, since .
The variant weak linking theorem is as follows.
Lemma 2.1 ([26])
The family of-functionalshas the form
where. Assume that
-
(a)
, , ;
-
(b)
as;
-
(c)
is-upper semicontinuous, is weakly sequentially continuous onE. Moreover, maps bounded sets to bounded sets;
-
(d)
, .
Then, for almost all, there exists a sequencesuch that
where.
In order to apply Lemma 2.1, we shall prove a few lemmas. We pick such that . For , we consider
It is easy to see that satisfies condition (a) in Lemma 2.1. To see (c), if and , then and in E, a.e. on ℝ, going to a subsequence if necessary. It follows from the weak lower semicontinuity of the norm, Fatou’s lemma and the fact for all and by (1.5) in (H4) that
Thus we get . It implies that is -upper semicontinuous. is weakly sequentially continuous on E due to [27].
Lemma 2.2Under assumptions of Theorem 1.1, then
Proof By the definition of and (H4), we have
which is due to . □
Therefore, Lemma 2.2 implies that condition (b) holds. To continue the discussion, we still need to verify condition (d), that is, the following two lemmas.
Lemma 2.3Under assumptions of Theorem 1.1, there are two positive constantssuch that
Proof By (H1), (H3), (2.4) and the Sobolev embedding theorem, for all ,
where C is a positive constant. It implies the conclusion if we take sufficiently small. □
Lemma 2.4Under assumptions of Theorem 1.1, then there is ansuch that
where.
Proof Suppose by contradiction that there exist , and such that . If , then by (H2) and (2.3), we have
Therefore, and . Let , then
It follows from and the definition of I that
(2.5)
There are renamed subsequences such that , , and there is a renamed subsequence such that in E and a.e. on ℝ.
We claim that
(2.6)
Case 1. If . Let be the subset of ℝ where , then for all we have . It follows from (H4) and that
Case 2. If , then by (H4) and , we have
Therefore, Cases 1 and 2 imply that (2.6) holds. Therefore, by (2.5), (2.6) and the facts , , we have
that is, . Thus, . It follows from (H4) that
which contradicts (2.5). The proof is finished. □
Therefore, Lemmas 2.3 and 2.4 imply that condition (d) of Lemma 2.1 holds. Applying Lemma 2.1, we soon obtain the following fact.
Lemma 2.5Under assumptions of Theorem 1.1, for almost all, there exists a sequencesuch that
where the definition ofis given in Lemma 2.1.
Lemma 2.6Under assumptions of Theorem 1.1, for almost all, there exists asuch that
Proof Let be the sequence obtained in Lemma 2.5. Since is bounded, we can assume in E and a.e. on ℝ. By (H1), (H3), (2.1) and Theorem A.4 in [27], we have
(2.7)
and
(2.8)
By Lemma 2.5 and the fact that is weakly sequentially continuous, we have
That is, . By Lemma 2.5, we have
It follows from (2.7), (2.8) and the fact that
The proof is finished. □
Applying Lemma 2.6, we soon obtain the following fact.
Lemma 2.7Under assumptions of Theorem 1.1, for every, there are sequencesandwithsuch that
Lemma 2.8Under assumptions of Theorem 1.1, then
where, , and the constantdoes not depend onu, w, r.
Proof This follows from (H5) if we take and . □
Lemma 2.9The sequences given in Lemma 2.7 are bounded.
Proof Write , where . Suppose that
Let , then , , and . Thus in E and a.e. on ℝ, after passing to a subsequence.
Case 1. If . Let be the subset of ℝ where . Then and on . It follows from (H4) and that
which together with Lemmas 2.3 and 2.7 and in for all (by (2.1)) implies that
It is a contradiction.
Case 2. If . We claim that there is a constant C independent of and such that
(2.9)
Since
it follows from the definition of I that
(2.10)
Take in (2.10), then it follows from Lemma 2.8 that
Thus (2.9) holds.
Let be a fixed constant and take
Therefore, (2.9) implies that
It follows from and Lemma 2.7 that
(2.11)
Note that Lemmas 2.3 and 2.7 and (H4) imply that
It follows from the fact as due to that
(2.12)
for all sufficiently large n. We take , by (2.12) and , we have
(2.13)
for all sufficiently large n. By (H1) and (H3), we have
(2.14)
For all sufficiently large n, by (2.13) and (2.14), it follows from and in for all (by (2.1)) that
This implies that as , contrary to (2.11).
Therefore, are bounded. The proof is finished. □