- Open Access
Nonlinear biharmonic boundary value problem
© Jung and Choi; licensee Springer. 2014
- Received: 28 June 2013
- Accepted: 8 January 2014
- Published: 4 February 2014
We consider the nonlinear biharmonic equation with variable coefficient and polynomial growth nonlinearity and Dirichlet boundary condition. We get two theorems. One theorem says that there exists at least one bounded solution under some condition. The other one says that there exist at least two solutions, one of which is a bounded solution and the other of which has a large norm under some condition. We obtain this result by the variational method, generalized mountain pass geometry and the critical point theory of the associated functional.
MSC:35J20, 35J25, 35Q72.
- biharmonic boundary value problem
- polynomial growth
- variational method
- generalized mountain pass geometry
- critical point theory
where is a continuous function which changes sign in Ω.
We assume that g satisfies the following conditions:
where if ,
(g4) as .
Remark 1.1 The real number ξ in the definition (g3) is not automatically nonnegative. The reason is as follows.
Since and , and . By , we have two cases: one case is that and . The other case is that and . Thus ξ is not nonnegative.
Remark 1.2 We obtain the boundedness of as follows.
Thus we obtain the boundedness of .
- (ii)If , (g2) can be dropped. If , it suffices that
If and , where and is a small number, then (g1)-(g4) are satisfied.
has also infinitely many eigenvalues , and corresponding eigenfunctions , . We note that , and that for .
She showed that if and , then (1.4) has a negative solution. She obtained this result by degree theory. Micheletti and Pistoia  also proved that if and then (1.4) has at least four solutions by the variational linking theorem and Leray-Schauder degree theory.
Since changes sign, the open subsets and are nonempty. Now we can write . Our main results are as follows.
Theorem 1.1 Assume that , g satisfies (g1)-(g4) and is bounded. Then (1.1) has at least one bounded nontrivial solution.
The outline of Theorem 1.1 and Theorem 1.2 is as follows: In Section 2, we prove that the corresponding functional of (1.1), which is introduced in (2.1), is continuous and Fréchet differentiable and satisfies the condition. In Section 3, we prove Theorem 1.1. In Section 4, we prove Theorem 1.2 by the variational method, the generalized mountain pass geometry and the critical point theory.
, for some ,
if and only if ,
which are proved in .
Then . Let be the orthogonal projection on and be the orthogonal projection on .
By (g1) and (g2), I is well defined. By Proposition 2.1, and I is Fréchet differentiable in H.
This implies that and is weakly continuous.
The proof of Proposition 2.1 is the same as that of Appendix B in .
Proposition 2.2 (Palais-Smale condition)
Assume that , , and g satisfies (g1)-(g4). We also assume that is bounded or that there exists an such that . Then satisfies the Palais-Smale condition.
from which the boundedness of follows. Thus converges weakly in H. Since with compact and the weak convergence of imply the strong convergence of and hence condition holds. □
We shall show that satisfies the generalized mountain pass geometrical assumptions.
We recall the generalized mountain pass geometry.
there are constants and a bounded neighborhood of 0 such that , and
there is an and such that if , then .
We have the following generalized mountain pass geometrical assumptions.
there are constants , and a bounded neighborhood of 0 such that , and
there is an and such that if , then , and
there exists such that and .
- (ii)Let be a ball with radius , e be a fixed element in and . Then , , . We note that
- (iii)If we choose such that , in Ω and , then we have
for all . Since , for great enough, is such that and . □
Since , b is bounded: .
for some constants , from which we conclude that is bounded and the proof of Theorem 1.1 is complete. □
for , from which we can conclude that is bounded and the proof of Theorem 1.2(i) is complete.
Next we shall prove Theorem 1.2(ii). We may assume that for all . Let us set , .
for small . Since , there exist great enough for each n and an such that and if and , so the lemma is proved. □
which means that is bounded. This is absurd because of the fact that . Thus we complete the proof. □
This work (Choi) was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (KRF-2013010343).
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