Fourth-order elliptic problems with critical nonlinearities by a sublinear perturbation

where Ω is a nonempty bounded open subset of the Euclidean space (R, |·|), N > 5, with sufficient smooth boundary, 2∗ = 2N/(N − 4), 1 < q < 2, λ and μ are positive parameters. Bernis, Garcia-Azorero and Peral [3] study a fourth-order problem with a critical growth, which presents several difficulties. Indeed, the Palais–Smale condition, as well as the weak lower semi-continuity of the associated functional, may fail because the Sobolev embedding is not compact. To be precise, consider the problem

Bernis, Garcia-Azorero and Peral [3] study a fourth-order problem with a critical growth, which presents several difficulties. Indeed, the Palais-Smale condition, as well as the weak lower semi-continuity of the associated functional, may fail because the Sobolev embedding is not compact. To be precise, consider the problem ∆ 2 u = |u| 2 * −2 u + µ|u| s−2 u in Ω, where µ > 0 is a parameter. Bernis, Garcia-Azorero and Peral [3] study this problem following the idea of Ambrosetti, Brezis and Cerami [2]. They proved the following result.
Moreover, they also proved that if µ > Λ, the previous problem admits no solution (see [3,Thm. 2.1]). Their proof is combination of topological and variational methods. Precisely, they determine the existence of a first solution by using the method of sub-and super-solutions and then prove that this solution is the minimum of a suitable functional and apply the mountain pass theorem so ensuring the existence of a second solution. For other result of fourth-order problem and variational problem, we refer the reader to [1,5,8,[10][11][12][13][14][15][16] and references therein.
In this paper, we investigate a fourth-order problem with critical growth (P λ ). Our approach is due to Bonanno [4,6]. Using the variational method, we will ensure that problem (P λ ) has one positive solution when the parameters λ and µ are in a suitable interval. Furthermore, when λ = 1, we can get another positive solution, where µ is in a suitable interval, and give the estimate of the parameter µ.
At first, we give the variational framework of this problem. As usual, put X := for all u ∈ X. Obviously, |ξ| 2 * /2 * + µ|ξ| q /q 0 for all ξ ∈ R. By the Sobolev embedding, and by Talenti [17] we obtain Due to (2), by the Hölder inequality it follows that where "|Ω|" denotes the Lebesgue measure of the set Ω and that the embedding X → L s (Ω) is not compact if s = 2 * . http://www.journals.vu.lt/nonlinear-analysis Fix r > 0 and put where c 2 * , c q are given by (2) and (3). Now, we give the first result of this paper.
Next, we obtain the following existence result of two solutions. At the same time, an estimate of parameters is also obtained.

Preliminaries
We present some definitions on differentiability of functionals and refer the reader to [4,Sect. 2]. Let X be a real Banach space. We denote the dual space of X by X * , while ·, · stands for the duality pairing between X * and X. A functional I : It is called continuously Gâteaux differentiable if it is Gâteaux differentiable for any u ∈ X and the functional u → I(u) is a continuous map from X to its dual X * . Let Φ, Ψ : X → R be two continuously Gâteaux differentiable functionals and put Fix r 1 , r 2 ∈ [−∞, +∞] with r 1 < r 2 . We say that the functional I verifies the Palais-Smale condition cut off lower at r 1 and upper at r 2 (in short (PS) [r2] [r1] -condition) if any sequence (u n ) such that has a convergent subsequence.
When we fix r 1 = −∞, that is, Φ(u n ) < r 2 for all n ∈ N, we denote this type of Palais-Smale condition with (PS) [r2] . When, in addition, r 2 = +∞, it is the classical Palais-Smale condition. Now, we recall the following local minimum theorem.
) Let X be a real Banach space, and let Φ, Ψ : X → R be two continuously Gâteaux differentiable functionals such that inf Assume that there are r ∈ R andũ ∈ X with 0 < Φ(ũ) < r such that and, for each

Proof of the main results
Firstly, we establish the following result. Lemma 1. Let Φ and Ψ be the functional defined in (1) and fix r > 0. Then, for each Going to a subsequence if necessary, we can assume u n → u 0 a.e. on Ω.
Taking (i) into account, for a constant c, lim n→∞ I λ (u n ) = c. Moreover, (u n ) is bounded in L 2 * (Ω). Now, we proof our result by many steps.
Step 1. u 0 is a weak solution of problem (P λ ). Since (u n ) is bounded in L 2 * (Ω), we get that (u 2 * −1 n ) is bounded in L 2 * /(2 * −1) (Ω). Indeed, we have Therefore, we get that In fact, since u n → u 0 a.e. x ∈ Ω, we obtain u 2 * −1 a.e. x ∈ Ω, and that, together with the boundedness of (u 2 * −1 One has for all v ∈ X. Therefore, due to (ii), we obtain that for all v ∈ X, that is, u 0 is a weak solution of (P λ ).
Step 2. We prove that Let us consider the nonlinear term So, It follows that for all u ∈ X, u (2r) 1/2 , we obtained Noting (iii) and Φ is sequentially weakly lower semicontinuous, we have That is, Step 3. Let v n = u n − u 0 . We get that In fact, u n 2 = v n + u 0 2 = v n 2 + u 0 2 + 2 v n , u 0 , so, we obtained Moreover, by the Brezis-Lieb lemma (see [7, Thm. 1]) one has Finally, since u → Ω (1/q)|u| q dx is locally Lipschitz in L q (Ω) (see, for example, [9, Thm. 7.2.1]) and u n → u 0 in L q (Ω), we obtained that is, We get (5).
Therefore, seen in the proof of (5) and we get that |u n | q−1 → |u 0 | q−1 in L q/(q−1) (Ω) (see the first step) and u n → u 0 in L q (Ω).
Since u 0 is a weak solution of (P λ ), one has that is, (6).
Conclusion. Finally, we observe that v n 2 is bounded in R. Thus, there is a subsequence, still denoted by v n 2 , which converges to b ∈ R. That is, lim n→∞ v n 2 = b. If b = 0, we have proved the lemma. In this situation, we have lim n→∞ u n − u 0 = 0.
Proof of Theorem 3. Fix µ ∈]0, µ * [. From Theorem 2 there exists a positive solution u µ of (P λ ) such that u µ is a local minimum for the functional where F is the primitive of f (t), and We consider a new problem in Ω, v = ∆v = 0 on ∂Ω.
Clearly, if v µ is a positive weak solution to (7), then w µ = u µ + v µ is a weak solution of (P λ ) such that w µ > u µ > 0. Now, our aim is to prove that (7)  Clearly, nonzero critical points of J are positive weak solutions of (7). Since u µ is a local minimum of I, one has I(u µ + v) − I(u µ ) 0 for all v ∈ X such that v < δ for some δ > 0. So, taking into account that for all v ∈ X (see [3]), we get J(v) 0 for all v ∈ X such that v < δ. That is, 0 is a local minimum of J. By using the same proof in [3], the functional J admits a positive critical point v µ for which w µ = u µ + v µ is the second weak solution of (7), and the proof is complete.