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 (f1) and (f2) hold. Let
and
satisfy
and
as
. Then,
Proof.
From (f2), 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 (f3) implies (f2) 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 (f3(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 (f1) and (f3). Then, for every
,
satisfies the (PS) condition and
Proof.
First we have the following claim:
Claim 1.
For any sequences
and
if
as
, 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.