THE GROUND STATE SOLUTIONS FOR CRITICAL FRACTIONAL PROBLEMS WITH STEEP POTENTIAL WELL

In this paper we investigate the existence of ground state solutions for a class of critical fractional problems. Under suitable assumptions of nonlinear terms and parameters, we get the existence of the ground states solutions.


Introduction
In this paper we investigate the existence of ground state solutions for the following critical fractional Schrödinger equation where µ > 0, 2 * s = 6 3−2s is the fractional critical Sobolev exponent, (−∆) s is the fractional laplacian operator defined by a normalization constant as where ).More details on the fractional laplace operator (−∆) s and fractional Sobolev spaces can be referred to [10].We consider the more general form of the equation (1.1) When s = 1, V (x) is steep potential, Clapp and Ding [7] established the existence and multiplicity of solutions when f (x, u) = λu + u 2 * −1 .After Bartsch and Wang in [5] firstly introduced the steep potential for which V (x) = 1 + λa(x), many researchers have done similar researches.For example, T. Bartsch et al. [4] considered the positive solutions for nonlinear Schrödinger equations.L.F. Yin et al. in [20] got existence and concentration of ground state solutions by variational method for f (x, u) =|u| 4 u + λu q−2 u.For some other important results, one can refer to [1,2,6,9,13,15] and the references therein.
When s ∈ (0, 1), X.D. Fang et al. [12] considered the ground state solution and multiple solutions, where V and f (x, u) are periodic, asymptotically linear and satisfying a monotonicity condition.In [21], H. Zhang et al. considered the superlinear fractional Schrödinger equation where V and f are asymptotically periodic in x.F. Patricio et al. in [11] got the positive solution where f (x, u) is superlinear and subcritical growth with respect to u.J. Zhang et al. in [22] studied the critical case and obtained the existence of ground state solutions by establishing Pohožave type identity when s ∈ ( 3 4 , 1).Inspired by the works described above and [3,8,14,18], in this paper, we discuss the fractional critical problem under the following conditions: The following is our main result.
Remark 1.1.Without Ambrosetti-Rabinowitz conditions, the boundedness of Palais-Smale sequence is difficult to get.Moreover, the minimum value of the energy functional is greater than zero, which can not be easily obtained by variational method.In order to overcome these difficulties, we establish the Pohozaev type identity and the Nehari-Pohozaev-Palais-Smale sequence.

Preliminaries and the functional setting
We now collect some preliminary results for the fractional Laplacian.For any s ∈ [ 3 4 , 1), the fractional order Sobolev space: where [u] H s is the so-called Gagliardo seminorm defined as Then the fractional Sobolev space Then H s (R 3 ) is a Hilbert space with the inner product It is well known that the fractional soblev space H s (R 3 ) is continuously embedding into L q where q ∈ [2, 2 * s ], and H s (R 3 ) is compactly embedding into L q loc , where q ∈ [1, 2 * s ).Then we define the best fractional Sobolev constants Moreover, for any µ > 0, we define the following space with the inner product as follows Then the energy functional I : E → R is It is easy to know that I(u) is well defined and the derivative is given by Thus, u is a solution of (1.1) if and only if u is a critical point of I.
Lemma 2.3.If µ > 0, (f 1 ) − (f 3 ) and (a 1 ), (a 3 ) hold, then there is a Proof.From [17], we know that the best Soblev constant can be attained by ) and , and By a direct calculation, there exists a κ 0 such that (2.5) From Lemma 2.2, there exists 0 and (a 3 ), we have Then we get that combine with (2.4) and (2.5), one get that t ε ≤ C, where C is a constant independent of ε.For some constant C 1 > 0, one gets t ε ≥ C 1 > 0 for ε > 0 small.Otherwise, there is a sequence ε n → 0 as n → ∞ such that t εn → 0 as n → ∞, along a subsequence, we get t εn u εn → 0 in E as n → ∞, therefore it is a contradiction.From (f 3 ) and [17], for any M > 0, there is . Hence, for ε small enough, we have Together with (2.5), one has Moreover we get that by (a 2 ), (2.5), (2.6), (2.7) and t ε ≤ C, we consider Lemma 2.4.Assume (f 1 ) − (f 3 ), (a 1 ) and (a 3 ) hold, let {u k } ⊂ H s (R) be a sequence such that thus, we derive that We have known that when K is big enough, it is a contradiction.Thus, u k is bounded.
Proof.By Lemma 2.1, Lemma 2.2 and Lemma 2.4, we get that there is a bounded sequence {u k } ⊂ E, along a subsequence still written as So we derive that I ′ (u) = 0. Suppose that u = 0, then u k ⇀ 0, B r (y) be a bounded ball, we have Next, we will discuss two cases to show that this hypothesis does not hold.
. By Lebesgue dominated convergence theorem, there is Thus We take a subsequence, which is still recorded as {u k }.Then there exists l ≥ 0 such that (2.9) Suppose l = 0, then ∥ u k ∥→ 0 in E, I(u k ) → 0. But we have already known that I(u k ) → c > 0, there is a contradiction.Therefore, l > 0 holds.By I(u k ) → c > 0, (2.8) and (2.9), we get by the definition of S, we derive that where where q ∈ (2, 2 * s ] and uniformly in k.From u k → 0 in L p loc (R 3 ) with q ∈ (2, 2 * s ], (2.11) and (2.12), as r → ∞, we have This is a contradiction.Hence, there exists u ̸ = 0 such that u k ⇀ u in E and I ′ (u) = 0, the proof is finished.By the argument above, we know that N is not empty.Now we claim that m > 0.
We establish the Pohozaev type identity of equation (1.1) as follows Combined with (2.13) (2.14) and (a 2 ), for s ∈ [ 3 4 , 1), u ̸ = 0, we have Thus, we get m > 0. Assume u k ⇀ u weakly in E, similar to the proof of Lemma 2.5, we derive I(u) = m and I ′ (u) = 0 with u ̸ = 0 from I(u k ) → m < 1 3 S 3 2 and I ′ (u k ) = 0, so m is attained by u.The proof is completed.