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).