On the Existence of Solutions for the Critical Fractional Laplacian Equation in ℝ^𝑁

Abstract

We study existence of solutions for the fractional Laplacian equation in , , with critical exponent , , , where has a potential well and is a lower order perturbation of the critical power . By employing the variational method, we prove the existence of nontrivial solutions for the equation.

1. Introduction

In the last 20 years, the classical nonlinear Schrödinger equation has been extensively studied by many authors [1–10] and the references therein. We just mention some earlier work about it. Brézis and Nirenberg [1] proved that the critical problem with small linear perturbations can provide positive solutions. In [3], Rabinowitz proved the existence of standing wave solutions of nonlinear Schrödinger equations. Making a standing wave ansatz reduces the problem to that of studying a class of semilinear elliptic equations. Floer and Weinstein [10] proved that Schrödinger equation with potential and cubic nonlinearity has standing wave solutions concentrated near each nondegenerate critical point of .

However, a great attention has been focused on the study of problems involving the fractional Laplacian recently. This type of operator seems to have a prevalent role in physical situations such as combustion and dislocations in mechanical systems or in crystals. In addition, these operators arise in modelling diffusion and transport in a highly heterogeneous medium. This type of problems has been studied by many authors [11–18] and the references therein.

Servadei and Valdinoci [11–14] studied the problem where , is an open bounded set of , , with Lipschitz boundary, is a real parameter, and is a fractional critical Sobolev exponent. is defined as follows: Here is a function such that there exists such that and for any . They proved that problem (1) admits a nontrivial solution for any . They also studied the case and , respectively.

Felmer et al. [15] studied the following nonlinear Schrödinger equation with fractional Laplacian: where , , and is superlinear and has subcritical growth with respect to . The fractional Laplacian can be characterized as , where is the Fourier transform. They gave the proof of existence of positive solutions and further analyzed regularity, decay, and symmetry properties of these solutions.

In this paper, we consider the following problem: with critical exponent , , , where has a potential well, where is the fractional Laplace operator, which may be defined as is the usual fractional Sobolev space.

is a lower order perturbation of the critical power . Now we give our main assumptions. In order to find weak solutions of (5), we will assume the following general hypotheses: is a continuous nonnegative function on and satisfying is a Carathéodory function; uniformly in ;there exist , and , , such that sup for any ;there exists such that, for all and , where .

The aim of this paper is to find solutions for (5) by variational methods. For this, we give the weak formulation of (5) by the following problem: This problem represents the Euler-Lagrange equation of the function defined as where is defined as in . Critical points of are weak solutions of (5). We will prove the existence of the critical points of the functional .

It is convenient to define where will be defined in Section 2 and for any and will be defined in Section 2 and for any

Now, we give our results as follows.

Theorem 1. Let , , , and satisfy and ()–(), respectively. Then, (5) possesses at least one nontrivial solution.

We prove Theorem 1 applying the mountain pass theorem to the functional . Although the Palais-Smale sequences might lose compactness in the whole space , we cannot apply the mountain pass theorem directly. In [15], they used a comparison argument to over this difficulty. But we will use allow it to us to over this problem related to the lack of compactness and to show that the Palais-Smale condition holds true in a suitable range related to .

Theorem 2. Let , , , and satisfy ()–(). Then, (5) possesses at least one nontrivial radial symmetric solution.

For the case , the proof of the existence of solutions is similar to the proof of Theorem 1. To prove symmetry of solutions, we consider the subspace of . consists of radial symmetric functions of .

This paper is organized as follows. In Section 2, we give some preliminary results. In Section 3, we prove the geometry and Palais-Smale condition of the functional and finish the proof of Theorem 1. In Section 4, we finish the proof of Theorem 2.

2. Preliminary Results

We consider the fractional Sobolev space equipped with norm We denote a subspace of the fractional Sobolev space by . This subspace is defined as the completion of with respect to the norm

Obviously, is a Hilbert space, with scalar product and is continuously embedded in .

On the other hand, we also consider the subspace of . consists of radial symmetric functions of and has the same norm with , and its norm is denoted by . is continuously embedded in .

The following two lemmas about the fractional Sobolev space are proved in [15].

Lemma 3. Let ; then one has Consequently, the embedding and are continuous. If further and is a bounded domain, then the bound sequence has a convergent subsequence in .

Thanks to Lemma 3, we can define the constants and and get that and .

Lemma 4. Let . Assume that is bounded in and it satisfies where . Then in for .

Lemma 5. (a) If holds true, then the embedding is compact for any .
(b) The embedding is compact for any .

Proof. (a) We give the proof as in [19].
Case . We will show that strongly in , whenever weakly in . Indeed, let be such that . Given , pick such that for all and denote by the ball of radius in . Then we have that weakly in . The compact embedding implies that for some natural number , On the other hand, by our choice of , we have Combining (22) and (23), we obtain that for all .
Case . Using Lemma 3, together with the interpolation inequality (where ), and by the fact that the embedding is compact, we can obtain that the embedding is also compact.
(b) We give the proof as in Corollary  4.7.2 of [20].
For all , for all , let be the largest number of disjoint balls with radius and the centers lie on the same sphere with radius centered at . It is easily seen that as . By definition, for all and for all , If is a bounded sequence in , for all , , we have We may assume that weakly in ; then, by Lemma 3, after a subsequence , it follows that
By (26), (27), and Lemma 4, we have for .

In [11, Lemma  6], Servadei and Valdinoci proved the following result.

Lemma 6. Assume that satisfies ()–(). Then, for any , there exists such that a.e. , and for any and so, as a consequence, where is defined as in () and .

In [17, Lemma  4], Servadei and Valdinoci proved the following result.

Lemma 7. Assume that satisfies ()–(). Then, there exist two positive measurable functions and such that a.e. , and for any , where is defined as in (), , and , .

3. The Proof of Theorem 1

In this section we study the mountain pass geometry and Palais-Smale condition in a suitable energy range and finish the proof of Theorem 1. We consider the functional where is defined as in ().

Then and critical points of are solutions of The Fréchet derivative of is

Proposition 8. Let , , , and satisfy and ()–(), respectively. Then, there exist and such that for any with it results that .

Proof. Let be a function in . By Lemma 3 and (30), we get that, for any ,
Choosing such that , by (35), it easily follows that for suitable positive constants , , and .
Now, let be such that . Since , we can choose sufficiently small, so that
Hence, Proposition 8 is proved.

Proposition 9. There exists with a.e. in , such that

Proof. By definition of , we have that there exists such that for any . We have a.e. in , or else we can take . Indeed, by triangle inequality, a.e. , , and so Thus, It follows that there exists such that thanks to (39). Since , so we can choose such that
Hence, Proposition 9 is proved.

Proposition 10. Let , , , and satisfy and ()–(), respectively. Then, there exists such that a.e. in , , and , where and are given in Proposition 8.

In particular, we can construct as follows: with as in (38) and large enough.

Proof. We fix such that . By (31), we get where , are constant. Since and , passing to the limit as , we get that , so that the assertion follows taking , with sufficiently large.
In particular, we can take , then with large enough.

We easily see that , with given in Proposition 8. Now, set where with in Proposition 9.

In [11, the proof of Theorem  1], Servadei and Valdinoci proved the following result.

Proposition 11. The constant is given in (47) such that where is given in Proposition 8 and is defined in formula (12).

By [6, Theorem  2.2], we have a sequence in such that as .

Proposition 12. There exists such that, up to a subsequence, as .

Proof. We proceed by steps.
Step  1. The sequence is bounded in .
Proof. For any by (50) and (51) it easily follows that there exists such that As a consequence of (52), we have By and , we have and We obtain that is bounded in by , (53), and (54).
Step  2. Problem (10) admits a solution .
Proof. By Step  1 and since is a reflexive space, up to a subsequence, still denoted by , there exists such that weakly in ; that is, as . By , (53), and (54), we have that is bounded in . Since is a reflexive space, up to a subsequence, as , while by Lemma 5(a), up to a subsequence, as , for any . By (56) and since is bounded in , we have as . By the proof of Lemma 6 [11, Lemma  6], we get
Moreover, we have, by taking , By (61), being bounded in , and Lemma 5(a), we obtain Since is a reflexive space, we get as . It is easily seen that as and so, in particular, as . Since (51) holds true, for any ,
Passing to the limit in this expression as and taking into account (55), (57), (59), and (65), we get for any ; that is, is a solution of problem (10).
Step  3. The following equality holds true:
Proof. By Step  2, taking as a test function in (10), we have so that The last inequality follows from assumption .
Now, we conclude the proof of Proposition 12.
We write ; then, weakly in . Moreover, since (58) holds true, by the Brézis-Lieb Lemma, we get Then,
By and , we get By the definition of , we have and so . Either or . If , the proof is complete. Assuming that , we obtain, from (49), (68), and (72), which is a contradiction. Thus and as . This ends the proof of Proposition 12.

We have finished the proof of Theorem 1 by Propositions 8, 10, and 12 and the mountain pass theorem.

4. The Proof of Theorem 2

In this section we consider the case , study the mountain pass geometry and Palais-Smale condition in a suitable energy range, and finish the proof of Theorem 2. We consider the functional where is defined as in . Then and critical points of are solutions of The Fréchet derivative of is for any .

Similar to the proof of Theorem 1, we have the following conclusions.

Proposition 13. Let , , and satisfy ()–(). Then, there exist and such that for any with it results that .

Proposition 14. There exists with a.e. in , such that

Proposition 15. Let , , and satisfy ()–(). Then, there exists such that a.e. in , , and , where and are given in Proposition 13.

In particular, we can construct as follows: with as in (81) and large enough.

We easily see that , with given in Proposition 13. Now, set where with in Proposition 13.

Proposition 16. The constant is given in (83) such that where is given in Proposition 13 and is defined in formula (13).

By [6, Theorem  2.2], we have a sequence in such that as .

Proposition 17. There exists such that, up to a subsequence, as .

We have finished the proof of Theorem 2 by Propositions 13, 15, and 17 and the mountain pass theorem.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The work is supported by the National Nature Science Foundation of China (11271331).