Caffarelli–Kohn–Nirenberg inequality for biharmonic equations with inhomogeneous term and Rellich potential

. In this article, multiplicity of nontrivial solutions for an inhomogeneous singular biharmonic equation with Rellich potential are studied. Firstly, a negative energy solution of the studied equations is achieved via the Ekeland’s variational principle and Caffarelli–Kohn–Nirenberg inequality. Then by applying Mountain pass theorem lack of Palais–Smale conditions, the second solution with positive energy is also obtained.


Introduction
We investigate multiplicity of solutions for the following singular biharmonic equations with inhomogeneous terms where , p α = 2(N−α) N−4 , 0 α < 4, f (x) ∈ H −2 0 (R N ) is a given function and f (x) ≡ 0, H −2 0 (R N ) denotes the dual of H 2 0 (R N ), the singular term u |x| 4   comes from models in physics.
In the past decades, nonlinear elliptic equations involving biharmonic operator have received much attention due to their wide application to mechanical and physical models such as clamped plates, thin-elastic plates, and in the research of the Paneitz-Branson equation and the Willmore equation (see [11]).Under the framework of nonlinear function analysis, there are many results on qualitative properties, the existence and multiplicity of solutions for biharmonic equations with singular potential (see [1, 7, 9, 12, 14-16, 19-22, 25, 26], and the references Corresponding author.Emails: yuyang_0522@sina.com(Y.Yu), zhaoylch@sina.com(Y.Zhao).therein).At the beginning, Brezis and Nirenberg [4] considered the following problems: in Ω, u = 0, on ∂Ω, (1.2) where Ω ⊂ R N is a bounded smooth domain, and let where 2 * = 2N N−2 as Sobolev critical exponent.They basically proved that S λ is reachable when N and λ satisfy different conditions.Since the seminal work of Brezis and Nirenberg, the study of critical growth in semilinear and quasilinear problem have gradually become a hot subject.On the basis of (1.2), Jannelli [13] studied the following semilinear elliptic equations involving the Hardy terms and critical exponents, and obtained at least a nontrivial solution when N ≥ 3 and Furthermore, Wang and Zhou [24] considered the problem of [13] with ≥ 0. By using the upper and lower solution method and Mountain pass theorem, they proved the given problem has at least two nontrivial solutions.
Tarantello [23] studied the following semilinear elliptic equations involving inhomogeneous perturbation and critical exponential terms: When f is appropriately small, the author proved that problem (1.3) admits at least two solutions by applying the Mountain pass theorem and the Ekeland's variational principle.By applying similar methods as in Ref. [23], Deng and Wang [8] studied the following nonlinear biharmonic problems with inhomogeneous perturbation terms and critical exponential terms: where N ≥ 5 and Ω ⊂ R N is a bounded smooth domain, 2 * = 2N N−4 .They proved that problem (1.4) has at least two solutions when f is appropriately small.Furthermore, they dealt with the non-existence of solutions for the above studied equation under some assumptions on the perturbation term f .By using the strong Maximum principle and the Comparison principle, Ref. [17] discussed the existence and nonexistence results of the following semilinear biharmonic problems with the optimal exponent p: where p > 1, µ > 0, λ > 0 and Ω ⊂ R N (N > 4) is a smooth bounded domain and 0 ∈ Ω. Mousomi Bhakta [2] considered the following elliptic problem with singular terms: when Ω is an open subset of R N (N ≥ 5), some nonexistence of solutions results are obtained by applying Pohozaev identity and Nehari manifold, In addition, they further discussed the existence of positive solutions when α = 0. Through the analysis of the above mentioned studies, a quite natural question to ask is whether the inhomogeneous biharmonic problem (1.1) possesses multiple nontrivial solution in R N ?As far as we know, when α = 0 and Ω = R N in (1.5), the problem (1.5) does not have a solution.Thanks to lack of compactness of the functional energy, the author obtain that the non-existence result of solution in a bounded domain.Therefore, we consider adding a perturbation term to overcome this difficulty and prove that the energy function I of problem (1.1) admits at least two critical points.One is a negative energy solution obtained by using Ekeland's variational method in [10], and other is a positive energy solution achieved by applying Mountain pass theorem in [1] without Palais-Smale (PS) conditions.The main result of this paper is the following theorem.
Then there exists a constant λ * > 0 such that for any λ ∈ (0, λ * ), the problem (1.1) admits at least two nontrivial solutions which one is of negative energy and the other solution with positive energy, if where S µ will be given in (2.3).

Preliminaries
This section will mainly give some preparation to the proof of Theorem 1.1.
Due to the fact that the space Note that µ < μ and by the following Rellich inequality [18] is optimal, then we can show that the norm From [3], we have the following Caffarelli-Kohn-Nirenberg (CKN) inequality where the constant C(N, α) > 0. For each µ with 0 < µ < µ, the best Sobolev constant S µ can be given by where S µ is achieved in R N .By applying (2.1) and (2.2), we know S µ > 0.
To obtain our results, the energy function I of problem (1.1) can be defined by 2), it is easy to obtain that the energy function Proof.For n sufficiently large, there hold where o n (1) means that for n → ∞, o n (1) → 0 .Thus, there holds which means that {u n } ∞ n=1 is a bounded sequence in H 2 0 (R N ).Up to a subsequence if necessary, there holds (2.7) Thus, it is easy to obtain that Lemma 2.2.For some c ∈ R, let {u n } ∞ n=1 in H 2 0 (R N ) be a (PS) c sequence for the energy functional I, that is to say n=1 , we know that the sequence {u n } ∞ n=1 possesses a weak convergent subsequence, still denoted by {u n } ∞ n=1 , then we can get that u n u 0 in H 2 0 (R N ), and u n → u 0 a.e. in R N , as n → ∞.Denote w n = u n − u 0 , then we have w n 0, as n → +∞.On the basis of Brezis-Lieb Lemma (see [5]), we could obtain that Therefore, there holds And It follows from Lemma 2.1 that I (u 0 ) = 0, combining with (2.8) we can infer that (1).
In this situation, we may assume that Suppose ξ > 0, together with the definition of S µ , we have ξ ≥ S pα pα −2 µ .Furthermore, by (2.8), we obtain This ends the proof of Lemma 2.2.

Proof of Theorem 1.1
In this section, we first take advantage of some analytical skills and functional idea to prove that the functional I can admit a local minimizer, which is a nontrivial negative energy solution.After that we show the existence of a nontrivial solution with positive energy via using Mountain pass theorem without (PS) condition.
We now show that there exists a nontrivial solution with negative solution.
On account of the continuity of f on R N and combining with f ≡ 0, we can choose φ ∈ C 0 (R N \ {0}) such that R N f (x)φdx > 0. Then for t > 0 sufficiently small with tφ µ < η 0 , there holds Therefore, we have According to the complete metric space B η 0 with respect to the norm of H 2 0 (R N ), then applying the Ekeland's variational principle to I(u) in B η 0 yields that there exist a (PS) which is a contradiction.Then the above proof yields that u * is a critical point of the functional I satisfying c 1 = I(u * ) < 0. Furthermore, it follows from (2.3) and (3.2) that Thus, we can deduce that the problem (1.1) possesses a nontrivial solution u * with negative energy.
Proof.From Theorem 2.1 of [2], we know that there is a nontrivial nonnegative solution as λ = 0 for problem(1.1),and then denote it as z(x).Next, we may choose u We now claim that there holds Indeed, the inequality (3.6) holds obviously if the function f (x) ≥ 0 or f (x) ≤ 0 for each x ∈ R N .Now if there is a point x 0 ∈ R N satisfying f (x 0 ) > 0, then by the continuity of the function f , we can deduce that there exists an open neighborhood B(x 0 , τ) ⊂ R N of x 0 , τ > 0, such that the function f (x) > 0 for all x ∈ B(x 0 , τ).Therefore, one can deduce from the definition of z(x − x 0 ), that To prove the inequality (3.5), we discuss the functions g and g defined by Obviously, there holds for every λ ∈ (0, Λ 2 ).Thus from the continuity of function g, there exists some t 1 > 0 sufficiently small, such that for all t ∈ (0, t 1 ).
For another thing, by the definition of g there holds This together with the definition of g, we have Then from (3.6), one has For all λ ∈ (0, Λ 3 ), we conclude that and this ends the proof.

Lemma 2 . 1 .
Assume that the sequence{u n } ∞ n=1 in H 2 0 (R N ) be a (PS) c sequence for the energy function I of problem (1.1) at level c ∈ R. Then u n u in H 2 0 (R N ) and I (u) = 0.