Existence of Infinitely Many High Energy Solutions for a Fourth-Order Kirchhoff Type Elliptic Equation in R&amp;lt;sup&amp;gt;3&amp;lt;/sup&amp;gt;

Ting Xiao,Qiongfen Zhang,Canlin Gan

doi:10.4236/jamp.2020.88120

Abstract

In this paper, we consider the following fourth-order equation of Kirchhoff type where a, b > 0 are constants, 3 p V ∈ C (R3, R); Δ2: = Δ (Δ) is the biharmonic operator. By using Symmetric Mountain Pass Theorem and variational methods, we prove that the above equation admits infinitely many high energy solutions under some sufficient assumptions on V (x). We make some assumptions on the potential V (x) to solve the difficulty of lack of compactness of the Sobolev embedding. Our results improve some related results in the literature.

Highlights

PreliminariesFirst of all, we shall introduce some properties of the weighted Sobolev space E and present the concept of the nontrivial solutions for problem (1.1)
R3 where a, b > 0 are constants, 3 < p < 5, V ∈ C(R3, R); ∆2 := ∆(∆) is the biharmonic operator
By using Symmetric Mountain Pass Theorem and variational methods, we prove that the above equation admits infinitely many high energy solutions under some sufficient assumptions on V (x)

Summary

Preliminaries

First of all, we shall introduce some properties of the weighted Sobolev space E and present the concept of the nontrivial solutions for problem (1.1). Let φ ∈ C1(E, R), accordding to Ekeland’s variational principal, we say that a sequence {un} is a Palais-Smale sequence at level c, if the sequence {un} satisfying φ(un) = c, φ′(un) → 0 has a convergent subsequence. The functional φ meets the (P S)c condition, if any Palais-Smale sequence at level c has a convergent subsequence. If E = Y ⊕ Z, where Y is finite dimensional, and φ satisfies (i) there exist constants ρ, α > 0 such that φ∂Bρ∩Y ≥ α. Assume that (V0) and (V1) hold, any (P S)sequence {un} ⊂ E defined by (2.4) has a convergent subsequence in E. By Lemma 2.2, the (P S)-sequence {un} ⊂ E is bounded. Due to the boundedness of {un}, we have |uk|p−1uk → |u|p−1u, as k → +∞.

Proof of the Main Result

Conclusion