We investigate the existence of multiple solutions of (1.1) when 
. We define a functional on
by
Then the functional
is well defined in
and the solutions of (1.4) coincide with the critical points of
. Now we investigate the property of functional
.
Lemma 4.1 (cf. [7]).
is continuous and Frechet differentiable at each
with
We will use a variational reduction method to apply the mountain pass theorem.
Let
be the two-dimensional subspace of
. Both of them have the same eigenvalue
. Then
for
. Let
be the orthogonal complement of
in
. Let
denote
onto
and
denote
onto
. Then every element
is expressed by
where
,
.
Lemma 4.2.
Let
, and
. Let
be given. Then we have that there exists a unique solution
of equation
Let
. Then
satisfies a uniform Lipschitz continuous on
with respect to the
norm (also the norm
).
Proof.
Choose
and let
Then (4.4) can be written as
Since
is a self-adjoint, compact, linear map from
into itself, the eigenvalues of
in
are
, where
or
. Therefore,
is
. Since
the right-hand side of (4.5) defines a Lipschitz mapping because for fixed
maps into itself. By the contraction mapping principle, there exists a unique
(also
) for fixed
. Since
is bounded from
to
there exists a unique solution
of (4.4) for given
.
Let
Then
. If
and
for any
,
, then
Hence
Since 
Therefore,
is continuous on
with respect to norm
(also, to
).
Lemma 4.3.
If
is defined by
, then
is a continuous Frechet derivative
with respect to
and
If
is a critical point of
, then
is a solution of (1.4) and conversely every solution of (1.4) is of this form.
Proof.
Let
and set
. If
, then from (4.4)
Since 
Let
be the two subspaces of
defined as follows:
Given
and considering the function
:
defined by
the function
has continuous partial Fréchet derivatives
and
with respect to its first and second variables given by
Therefore, let
with
and
. Then by Lemma 4.2
If
and
, then
Since
for any
and
, it is easy to know that
And
It follows that
Therefore,
is strictly convex with respect to the second variable.
Similarly, using the fact that
for any
, if
and
are in
and
, then
where
. Therefore,
is strictly concave with respect to the first variable. From (4.17), it follows that
with equality if and only if
.
Since
is strictly concave (convex) with respect to its first (second) variable, [9, Theorem
] implies that
is
with respect to
and
Suppose that there exists
such that
. From (4.24), it follows that
for all
. Then by Lemma 4.2, it follows that
for any
. Therefore,
is a solution of (1.4).
Conversely, if
is a solution of (1.4) and
, then
for any
.
Lemma 4.4.
Let
, and
. Then there exists a small open neighborhood
of 0 in
such that
is a strict local minimum of
.
Proof.
For
, and
, problem (1.4) has a trivial solution
. Thus we have
. Since the subspace
is orthogonal complement of subspace
, we get
and
. Furthermore,
is the unique solution of (4.4) in
for
. The trivial solution
is of the form
and
, where
is an identity map on
,
is continuous, it follows that there exists a small open neighborhood
of 0 in
such that if
then
,
. By Lemma 4.2,
is the solution of (4.5) for any
. Therefore, if
, then for
we have
. Thus
If
, then
. Therefore, in
,
where
. It follows that
is a strict local point of minimum of
.
Proposition 4.5.
If
, then the equation
admits only the trivial solution
in
.
Proof.
is invariant under
and under the map
. So the spectrum
of
restricted to
contains
in
. The spectrum
of
restricted to
contains
in
. From the symmetry theorem in [10], any solution
of this equation satisfies
. This nontrivial periodic solution is periodic with periodic
. This shows that there is no nontrivial solution of 
Lemma 4.6.
Let
and
. Then the functional
, defined on
, satisfies the Palais-Smale condition.
Proof.
Let
be a Palais-Smale sequence that is
is bounded and
in
. Since
is two-dimensional, it is enough to prove that
is bounded in
.
Let
be the solution of (1.4) with
where
. So
By contradiction, we suppose that
, also
. Dividing by
and taking
, we get
Since
, we get
weakly in
. Since
is a compact operator, passing to a subsequence, we get
strongly in
. Taking the limit of both sides of (4.28), it follows that
with
. This contradicts to the fact that for
the following equation
has only the trivial solution by Proposition 4.5. Hence
is bounded in
.
We now define the functional on
, for
,
The critical points of
coincide with solutions of the equation
The above equation (
) has only the trivial solution and hence
has only one critical point
.
Given
, let
be the unique solution of the equation
where
. Let us define the reduced functional
on
by
. We note that we can obtain the same results as Lemmas 4.1 and 4.2 when we replace
and
by
and
. We also note that, for
has only the critical point
.
Lemma 4.7.
Let
,
, and
. Then we have
for all
with
.
The proof of the lemma can be found in [1].
Lemma 4.8.
Let
,
, and
. Then we have
for all
(certainly for also the norm
).
Proof.
Suppose that it is not true that
Then there exists a sequence
in
and a constant
such that
Given
, let
be the unique solution of the equation
Let
. Then
. By dividing
, we have
By Lemma 4.2,
is Lipschitz continuous on
. So the sequence
is bounded in
. Since
(
), it follows that
and
are bounded in
. Since
is a compact operator, there is a subsequence of
converging to some
in
, denoted by itself. Since
is a two-dimensional space, assume that sequence
converges to
with
. Therefore, we can get that the sequence
converges to an element
in
.
On the other hand, since
, dividing this inequality by
, we get
By Lemma 4.2, it follows that for any 
If we set
in (4.40) and divide by
, then we obtain
Let
be arbitrary. Dividing (4.40) by
and letting
, we obtain
where
Then (4.42) can be written in the form
for all
. Put
. Letting
in (4.41), we obtain
where we have used (4.42). Hence
Letting
in (4.39), we obtain
Since
, this contradicts to the fact that
for all
. This proves that
.
Now we state the main result in this paper.
Theorem 4.9.
Let
,
, and
. Then there exist at least three solutions of the equation
two of which are nontrivial solutions.
Proof.
We remark that
is the trivial solution of problem (1.4). Then
is a critical point of functional
. Next we want to find others critical points of
which are corresponding to the solutions of problem (1.4).
By Lemma 4.4, there exists a small open neighborhood
of 0 in
such that
is a strict local point of minimum of
. Since
from Lemma 4.8 and
is a two-dimensional space, there exists a critical point
of
such that
Let
be an open neighborhood of
in
such that
. Since
, we can choose
such that
. Since
satisfies the Palais-Smale condition, by the Mountain Pass Theorem (Theorem 3.3), there is a critical value
where 
If
, then there exists a critical point
of
at level
such that
, 0 ( since
and
). Therefore, in case
, the functional
has also at least 3 critical points
.
If
, then define
where
. Hence,
That is
. By contradiction, assume
. Use the functional
for the deformation theorem (Theorem 4.9) and taking
. We choose
such that
. From the deformation theorem (Theorem 3.2),
and
which is a contradiction. Therefore, there exists a critical point
of
at level
such that
, 0, which means that (1.4) has at least three critical points. Since
, these two critical points coincide with two nontrivial period solutions of problem (1.4).