ABSTRACT

We consider the existence of a nontrivial solution for the Dirichlet boundary value problem

We prove an abstract result on the existence of a critical point for the functional on a Hilbert space via the local linking theorem. Different from the works in the literature, the new theorem is constructed under the condition instead of condition.

1. Introduction and Main Results

Consider the Dirichlet boundary value problem

(1)

where and is a bounded domain whose boundary is a smooth manifold.

We assume that, where. In [1], Li and Willem established the existence of a nontrivial solution for problem (1) under the following well-known Ambrosetti-Rabinowitz superlinearity condition: there exists and such that

(AR)

for all and, which has been used extensively in the literature; see [1-4] and the references therein. It is easy to see that condition (AR) does not include some superquadratic nonlinearity like

(G0)

In [5], Qin Jiang and Chunlei Tang completed the Theorem 4 in [1], and obtained the existence of a nontrivial solution for problem (1) under a new superquadratic condition which covered the case of (G0). The conditions are as follows:

(G1), as uniformly on(G2), as uniformly on(G3) There are constants and

such that

for all(G4) There are constants, and

such that

for all andIf 0 is an eigenvalue of (with Dirichlet boundary condition) assume also the condition that:

(G5) There exists such that:

1), for all,; or 2), for all,.

Note that (G4) is also (AR) type condition.

The aim of this paper is to consider the nontrivial solution of problem (1) in a more general sense. Without the Ambrosetti-Rabinowitz superlinearity condition (AR) or (G4), the superlinear problems become more complicated. We do not know in our situations whether the (PS) or sequence are bounded. However, we can check that any Cerami (or) sequence is bounded. The definition of (or) sequence can be found in [6].

We will obtain the same conclusion under the condition instead of condition. So we only need the following conditions instead of (G3) (G4):

(G3') Let satisfying 1) if2) if, where

,.

It is easy to see that function

satisfies conditions of (G1) (G2) (G5) and (G3').

Our main result is the following theorem:

Theorem 1.1. Suppose that satisfies (G1) (G2) (G5) and (G3'). If 0 is an eigenvalue of (with Dirichlet boundary condition). Then problem (1) has at least one nontrivial solution.

Remark 1. There are many functions which are superlinear but it is not necessary to satisfy AmbrosettiRabinowitz condition. For example,

where. Then it is easy to check that (AR) does not hold. On the other hand, in order to verify (AR), it usually is an annoying task to compute the primitive function of and sometimes it is almost impossible. For example,

where.

Remark 2. Our condition is much weaker than (AR) type condition (cf. [6]).

2. Proof of Theorem

Define a functional in the space by

where, , is the space spanned by the eigenvectors corresponding to negative (positive) eigenvalue of.

In this paper, we shall use the following local linking theorem (Lemma 2.1) to prove our Theorem . Let be a real Banach space with and such that, . For every multi-index, let. We know that,. A sequence is admissible if for every there is such that . We say satisfies the condition if every sequence such that is admissible and satisfies

contains a subsequence which converges to a critical point of.

Lemma 2.1. ([6]) Suppose that satisfies the following assumptions:

(f₁) has a local linking at 0(f₂) satisfies condition(f₃) maps bounded sets into bounded sets(f₄) For every, as, on.

Then has at least two critical points.

Proof of Theorem 1. We shall apply Lemma 2.1 to the functional associated with (1), we consider the case where 0 is an eigenvalue of and

(2)

The other case are similar.

1) and maps bounded sets into bounded sets.

Let,. Choose Hilbertian basis for and for, define

Assumption (G3') implies there are constants such that

(3)

so

where.

Hence and maps bounded sets into bounded sets.

In fact,

so (f₃) holds.

2) has a local linking at 0.

It follows from (g2) and (g3) that, for any, there exists, such that

(4)

we obtain, on, for some,

choosing, then,.

Decompose into when ,. Also set . Since is a finitedimensional space, there exists, such that

(5)

First we set and

On, we have, by (5)

hence, by (2)

On, we have also by (5)

hence, by (4)

and for some

Therefore we deduce that

choosing, then, , Let, then (f₁) holds.

3) satisfies condition.

Consider a sequence such that is admissible and

I) is bounded.

For n large, from assumption (g3'), with, for some,

where

So

(6)

Arguing indirectly, assume. Set , Then and for all . In addition, using (6)

hence by Interpolation inequality for norms, for

(7)

where or.

Since,

so

From (7) for some

as, therefore, , a contradiction. Hence sequence is bounded.

II) From (I) we see that is bounded in, going if necessary to a subsequence, we can assume that in. Since, in.

which implies that in. Similarly, in. It follows then that in and.

4) Finally, we claim that, for every,

Indeed, by (g1) we have, there exists such that

so on,

where. □

REFERENCES

NOTES

^*Supported by NSFC(11101237).

^#Corresponding author.