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
.

Then

. If

and

for any

,

, then

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

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).