Existence of Multiple Solutions for a Class of Biharmonic Equations

and obtained at least three nontrivial solutions by using the minimax method and Morse theory. Problem (1) has been studied extensively in recent years; we refer the reader to [5–11] and the references therein. In [12], the author showed that the problem (1) admits at least three (or four or five) nontrivial solutions by using the minimax method and Mountain Pass Theory. However, to the best of author’s knowledge, there have been very few results dealingwith (1) using a symmetricMountain PassTheorem.This paper will make some contribution in the research field. In this paper, we study the problem (1) by a symmetric Mountain Pass Theorem at the resonant and nonresonant case and obtain infinitely many solutions of the equation. Consider eigenvalue problem

This fourth-order semilinear elliptic problem can be considered as an analogue of a class of second-order problems which have been studied by many authors.A main tool of seeking solutions of the problem is the Mountain Pass Theorem (see [1][2][3]).In [4], Pei studied the following problem: and obtained at least three nontrivial solutions by using the minimax method and Morse theory.Problem (1) has been studied extensively in recent years; we refer the reader to [5][6][7][8][9][10][11] and the references therein.In [12], the author showed that the problem (1) admits at least three (or four or five) nontrivial solutions by using the minimax method and Mountain Pass Theory.However, to the best of author's knowledge, there have been very few results dealing with (1) using a symmetric Mountain Pass Theorem.This paper will make some contribution in the research field.In this paper, we study the problem (1) by a symmetric Mountain Pass Theorem at the resonant and nonresonant case and obtain infinitely many solutions of the equation.

Lemma 2. Let 𝑋
Proof.It is similar to the proof of Lemma 2.5 in [13].
The main results of this paper are as follows.

Proofs of Theorems
Proof of Theorem 5. (i) Assume that {  } ⊂  is a () sequence, that is, We claim that {  } is bounded.Assume as a contradiction that Without loss of generality, we assume From (15)  (17) Next we consider the two possible cases: (a)V ̸ = 0, (b)V = 0.In case (a), from ( 2 ) we derive For In case (b), we have We can easily see that V ̸ ≡ 0. In fact, if V ≡ 0, then as ,  → ∞.Hence   →  in .We verify Φ  () satisfies () condition.
Proof of Theorem 6.Similar to the proof of Theorem 5(i), we have We can easily see that V ̸ ≡ 0. In fact, if V ≡ 0, then where lim