Now, we are going to seek a function u in E such that
We consider the associated functional of (3.1),
By (g1), I is well defined.
We recall the critical point theorem for the indefinite functional (cf. ).
Theorem 3.1 (Critical point theorem for the indefinite functional)
Let E be a real Hilbert space with and . Suppose that satisfies (PS), and that
, where and is bounded and self-adjoint, ,
is compact, and
there exist a subspace and sets , and constants such that
Q is bounded and ,
S and ∂Q link.
Then I possesses a critical value .
The eigenvalues of are , where j is odd and k is even, and . Thus is a compact operator.
It is convenient for the following to rearrange the eigenvalues , where j is odd and k is even, by increasing magnitude: from now on we denote by the sequence of negative eigenvalues, by the sequence of positive ones, so that
We note that each eigenvalue has a finite multiplicity and that , as .
We shall show that the functional I satisfies the geometrical assumptions of the critical point theorem for the indefinite functional.
By the following lemma, and the solutions of (3.1) coincide with the nonzero critical points of .
Lemma 3.1 Assume that g satisfies (g1)-(g3). Then is continuous and Fréchet differentiable in E with the Fréchet derivative
for all . Moreover, if we set
then is continuous with respect to weak convergence, is compact, and
This implies that and is weakly continuous.
For the proof of Lemma 3.1, refer to .
Lemma 3.2 Assume that g satisfies (g1)-(g3). The problem
has only a trivial solution.
Proof Let . Since , . Thus (3.4) has only a trivial solution. □
Lemma 3.3 Assume that g satisfies (g1)-(g3). Then satisfies the Palais-Smale condition: If for a sequence , is bounded from above and as , then has a convergent subsequence.
Proof Let be a sequence with and as . It suffices to show that is bounded. By contradiction, we suppose that as . Let . Then and converges weakly to an element, say w. Then we have
Since is bounded, letting m∞ in (3.5), we have
On the other hand, we have
Letting in (3.7), we have
By (3.6) and (3.8), and . Thus , so converges strongly to w and by (3.8), w is a weak solution of the problem
By Lemma 3.2, , which is absurd for the fact that . Thus is bounded. □
Lemma 3.4 Assume that g satisfies the conditions (g1)-(g3). There exist a subspace and sets , and constants and such that
there are and such that if , then ,
there exists such that and ,
and ∂Q link.
Proof (i) Let . Since is bounded, there exists a constant such that
for . Then there exist a constant and a constant such that if , then .
(ii) Let us choose an element . Let . Then , , . We note that
Thus we have
for . Then there exists such that if , then .
(iii) Let us choose . By (ii), .
(iv) S and Q link at the set . □
Proof of Theorem 1.1 By Lemma 3.1, is continuous and Fréchet differentiable in E. By Lemma 3.3, the functional I satisfies the (PS) condition. We note that . By Lemma 3.4, there are constants , and a bounded neighborhood of 0 such that , and there are and such that if . Let us set
By Theorem 3.1, I possesses a critical value . Moreover, c can be characterized as
Thus we prove that I has at least one nontrivial critical point, we denote by a critical point of I such that . We claim that c is bounded. In fact, we have
and by (g3),
for some constant . Since , c is bounded:
We claim that is bounded. In fact, by contradiction, and, for any , imply that
is not bounded, which is absurd to the fact that is bounded. Thus is bounded, so (1.1) has at least one bounded weak solution, and hence we prove Theorem 1.1. □