Existence and multiplicity of solutions for nonlocal fourth-order elliptic equations with combined nonlinearities

This paper is concerned with the following nonlocal fourth-order elliptic problem: $$\begin{aligned} \textstyle\begin{cases} \Delta ^{2}u-m(\int _{\varOmega } \vert \nabla u \vert ^{2} \,dx)\Delta u=a(x) \vert u \vert ^{s-2}u+f(x,u), \quad x\in \varOmega , \\ u=\Delta u=0,\quad x\in \partial \varOmega , \end{cases}\displaystyle \end{aligned}$$ 

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 by using the mountain pass theorem, the least action principle, and the Ekeland variational principle, the existence and multiplicity results are obtained.

Problem (1.1) is related to the stationary problems associated with Because of the important background, several researchers have considered problem (1.1) by using variational methods when a(x) ≡ 0, with the function m being bounded or unbounded and f having superlinear growth. We refer the readers to [2][3][4][5][6][7][8][9][10][11] and the references therein.
Recently, in [12], Ru et al. considered problem (1.1) with m(t) = a + bt and a more general f such as By using an iterative method based on the mountain pass lemma and truncation method developed by De Figueiredo et al. [13], they proved that the above problem has at least one nontrivial solution.
One of the important conditions in their work is that f (x, t) satisfies the famous Ambrosetti-Rabinowitz type condition, for short, which is called the (AR) condition: (AR condition) there exist Θ > 2 and t 1 > 0, such that 0 < ΘF(x, t, ξ 1 , ξ 2 ) ≤ tf (x, t, ξ 1 , ξ 2 ), ∀|t| ≥ t 1 , x ∈ Ω, (ξ 1 , where F(x, t, ξ 1 , ξ 2 ) = t 0 f (x, s, ξ 1 , ξ 2 ) ds. It is well known that (AR) is a important technical condition to apply the mountain pass theorem. This condition implies that If f (x, u) is asymptotically linear at u = 0 or u = +∞. then f (x, u) does not satisfy the (AR) condition. In [14], A. Bensedik and M. Bouchekif considered second-order elliptic equations of Kirchhoff type with an asymptotically linear potential On the other hand, the classical equation involving a biharmonic operator has been extensively studied using the mountain pass theorem when a(x) ≡ 0 and f (x, u) is asymptotically linear at u = 0 or u = +∞. We refer the reader to [15,16]. In particular, in [17], Pu et al. considered problem (1.2) when a(x) = 0.
Until now, there are few works on problem (1.1) when a(x) = 0 and f (x, u) does not satisfy the (AR) condition. Inspired by these references, in this paper, we discuss the existence and multiplicity of solutions of problem (1.1) when a(x) = 0 and the nonlinearity f is asymptotically linear at u = 0 or u = +∞.

Preliminaries
Assume that the function m(t) satisfies the following conditions: (M) m : R + → R + is continuous, nondecreasing, and there exists m 1 ≥ m 0 > 0 such that Remark In [14] and [18], the function m(t) is assumed that satisfy (M) and there exits t 0 > 0 such that m(t) = m 1 , ∀t > t 0 .
First, we study the nonlinear eigenvalue problem Let (λ k , φ k ) be the eigenvalue and the corresponding eigenfunction of (-, Via some simple computations, we get and so Λ k (k = 1, 2, . . .) are the eigenvalues of the operator L associated to the eigenfunction φ k .
Assume that the eigenfunctions φ k are suitably normalized with respect to the L 2 (Ω) inner product, namely Expression (2.1) can be rewritten as For each eigenvalue λ k being repeated as often as multiplicity, recall that and if (M) holds, then then we know that It is well known that Similarly, we have

Lemma 2.1 Assume that (M) holds, then
then it is clear that and the deduced norm It is well know that u H is equivalent to ( Ω | u| 2 dx) 1 2 . And there exists τ > 0 such that Denote It is obvious that the norms u and u m 0 are equivalent to the norm u H in H. And since m 0 < m 1 , we have Throughout this paper, we denote by C universal positive constants, unless otherwise specified, and By the Sobolev embedding theorem, there is a positive K q such that Specially, when condition (M) holds and q = 2, by Lemma 2.1, then The mountain pass theorem and the Ekeland variational principle are our main tools, which can be found in [19].

Lemma 2.2 Let E be a real Banach space, and I
There is an e ∈ E with e > ρ such that Then I(u) has a critical value c which can be characterized as

Lemma 2.3
Let V be a complete metric space and I : V → R ∪ {+∞} be lower semicontinuous, bounded from below. Let ε > 0 be given and v ∈ V be such that Then there exists u ∈ V such that and for all w = u in V ,

Main results
A function u ∈ H is called a weak solution of (1.1) if holds for any v ∈ H. Let J : H → R be the functional defined by It is easy to see that J ∈ C 1 (H, R) and the critical points of J in H correspond to the weak solutions of problem (1.1). We make the following assumptions.
Our first main result is concluded as the following theorem: Proof It is easy to see, from condition (F 1 ), that f (x, 0) = 0 for x ∈ Ω. So u = 0 is the trivial solution of (1.1). From condition (F 2 ), we can take ε = 1 2 (Λ 1l) > 0, and there exists T > 0 such that for all |t| ≥ T and a.e. x ∈ Ω. By the continuity of F, there exists C > 0 such that On the other hand, from (M) it follows that Then we have which shows that J is coercive. Moreover, conditions (F 1 ) and (F 2 ) imply that J is weakly lower semicontinuous in H. Therefore we get a global minimum u 1 of J. Next, we prove u 1 = 0, so it is a nontrivial solution of (1.1). From condition (F 1 ), there exists C > 0 such that for all |t| small enough and x ∈ Ω. It follows that for all |t| small enough and x ∈ Ω. From condition (A), we can chose v ∈ H such that Ω a(x)|v| s dx > 0.
Then we have Therefore, we get that J(u 1 ) < 0. It is clear that J(0) = 0. Thus, u 1 is a nontrivial solution of (1.1).
Our second result is the following theorem: Before proving Theorem 3.2, we give two lemmas.
Define functionals J ± : H → R as follows: Proof We just prove that J + (u) satisfies the (PS) condition. The proof for J -(u) is similar. Let {u n } ∈ H be a (PS) sequence, namely Firstly, we claim that {u n } is bounded in H. If not, we may assume that u n → +∞ as n → +∞. Let w n = u n u n , then w n = 1. Passing to a subsequence, we may assume that there exists w ∈ H such that By (F 1 ) and (F 2 ), we see that there exist C 1 and C 2 such that Then we claim that w = 0. Otherwise, if w ≡ 0, we know that w n → 0 strongly in L r (Ω). Dividing (3.2) by u n 2 , we have It follows from (3.1) and (3.5) that which is impossible, so w = 0. Let us define Then, for all v ∈ H, we have where w + (x) = max {w(x), 0}. On the other hand, since u n → +∞, we have |u n (x)| = u n |w n (x)| → +∞ for x ∈ Ω 1 . Therefore, by (F 2 ) and the dominated convergence theorem, we get Combining (3.6) and (3.7), we obtain Now, (3.3) implies that, for all v ∈ H, we have Dividing by u n , we get