Abstract

In this paper, we study the following fractional Schrödinger equation in , in with and . By using the constrained variational method, we show the existence of solutions with prescribed norm for this problem.

1. Introduction

This paper concerns with the following fractional Schrödinger problem: where , , and . Here, the fractional Laplacian in is defined by where stands for the Cauchy principal value and is a normalization constant.

In the present paper, the motivation for studying such equations comes from mathematical physics: searching for the form of standing wave of the evolution equation leads to looking for solutions of (1). Here, is the imaginary unit and . This class of Schrödinger-type equations is of particular interest in fractional quantum mechanics for the study of particles on stochastic fields modelled by Lévy processes. A path integral over the Lévy flight paths and a fractional Schrödinger equation of fractional quantum mechanics are formulated by Laskin [1, 2] from the idea of Feynman and Hibbs’s path integrals.

On the other hand, problem (1) has attracted considerable attention in the recent period. Part of the motivation is to consider as a fixed parameter and then to search for a solution solving (1). In this direction, mainly by variational methods, many researches have been devoted to the study of the existence, multiplicity, uniqueness, regularity, and asymptotic decay properties of the solutions to fractional Schrödinger equation (1). For this information, we can refer to [311] and the references therein. Besides, some more complicated fractional equations and systems were also studied, and indeed, some interesting results were obtained. Nearly, Mingqi et al. [12] investigated a critical Schrödinger-Kirchhoff type systems driven by nonlocal integrodifferential operators and by applying the mountain pass theorem and Ekeland’s variational principle; the authors obtained the existence and asymptotic behavior of solutions for this system under some suitable assumptions. Later, in [13], the same authors as in [12] studied a diffusion model of Kirchhoff-type. Under some appropriate conditions, by employing the Galerkin method, the local existence of nonnegative solutions was obtained, and then by virtue of a differential inequality technique, they proved that the local nonnegative solutions blow up in finite time with arbitrary negative initial energy and suitable initial values. Moreover, in [14], Mingqi et al. concerned with a class of fractional Kirchhoff-type problems with the Trudinger-Moser nonlinearity. By applying minimax techniques combined with the fractional the Trudinger-Moser inequality, they found the existence of a ground state solution with positive energy and the existence of nonnegative solutions with negative energy by using Ekeland’s variational principle. In [15], the three authors considered a fractional Choquard-Kirchhoff-type problem involving an external magnetic potential and a critical nonlinearity and established a fractional version of the concentration-compactness principle with magnetic field, and then together with the mountain pass theorem, they verified the existence of nontrivial radial solutions in nondegenerate and degenerate cases. Furthermore, Mingqi et al. [16] concerned the Schrödinger-Kirchhoff-type problems involving the fractional -Laplacian and critical exponent. By using the concentration-compactness principle in fractional Sobolev spaces, they showed the existence of pairs of solutions for any , and by applying Krasnoselskii’s genus theory, they also got the existence of infinitely many solutions under some suitable conditions for the parameter. For more information on this direction, one can refer to [1724] and the references therein.

In the present paper, inspired by the fact that physicists are often interested in normalized solutions, we look for solutions in having a prescribed norm to equation (1). Such types of problems were studied extensively in recent years for the classical Schrödinger equations with the standard Laplacian operator. We refer the interested reader to [2531] and to the references therein. But up to our knowledge, not much is obtained for the existence of normalized solutions of equation (1) in with a fractional Laplacian operator. So, in this paper, the aim is to get the normalized solutions of equation (1). Here, we give the definition of prescribed - norm solutions. For fixed , if is a solution of problem (1) such that we call it a prescribed - norm solution. Naturally, a prescribed - norm solution of (1) can be a constrained critical point of the functional on the sphere in , where

Note that for any , is a well-defined and functional. Set

It is standard that if is a minimizer of (7), then is a solution of (1) with prescribed - norm with the constraint being the Lagrange multiplier. However, it is worth mentioning that dealing with this kind of problem, one has to face the main difficulty concerning with the lack of compactness of the minimizing sequence . In fact, we will encounter two possible bad scenarios that and with . In order to avoid the possible cases and to get that the infimum is obtained, we prove an important lemma (Lemma 6) that guarantees the compactness of minimizing sequence. As a consequence of this lemma, setting we can get our main result as follows:

Theorem 1. If , then has a minimizer if and only if .
In particular, there is a prescribed - norm solution of (1) with the constraint . But, when , has no minimizer for any .

Remark 2. In fact, for any , we can infer that . To see this, letting be arbitrary and considering for any , we find and Hence, as and the conclusion is as follows:
Finally, we give the following notations which can be used in this paper: (i) is the usual Sobolev space endowed with the standard norm(ii)Denote (iii) is the norm of the Lebesgue space for (iv)Denote by various positive constants which may vary from one line to another and which are not important for the analysis of the problem

This paper is organized as follows: In Section 2, we will give some preliminary results which are crucial to prove our main result. And then the proof of our main result is given in Section 3.

2. Preliminaries

In this part, we give some important results. First, similar to the classical Gagliardo-Nirenberg inequality to the Laplacian operator, we introduce the fractional version of Gagliardo-Nirenberg inequality as follows:

Lemma 3 (see [32]). Let , ,, , and . We havewith andIn particular, when , one haswith .

In [33], the authors have established the Pohozaev identity for the fractional Laplacian operator.

Applying the Pohozaev identity, we have the following:

Lemma 4. If is a critical point of on , then , where

Proof. Define the functional energy corresponding to (1) as Then any critical point of satisfies the Pohozaev identity for (1) (see [33]), that is, On the other hand, if is a critical point of restricted to , there is a Lagrange multiplier such that So, for any , we have

Furthermore, if we now know that by (18), we find . As a result,

Using Lemma 3, the estimate (13) leads to the following fact:

Lemma 5. If , then for any , functional is bounded from below and coercive on .

Proof. If , from Lemma 3, we have So, Since and , our result follows.

From the above Lemma, we can prove that

Lemma 6. If , then (i)there exists such that for all , (ii)for any such that , admits a minimizer(iii)the functional is continuous about each

Proof. (i)Letting such that with , we seeIf we take , one has Since and , for is large enough, we get and then (i) has been proved. (ii)We will divide it into three steps to show (ii).Step 1. We claim that for any , . To see this, let be a minimizing sequence on for . Since , from (22), we can get On the other hand, applying (24), we obtain that for , where with . Moreover, using , we can deduce that and for all , Thus, combining (28) and (29), for all , we find and from (26), As a result, if , and if , It follows from (31) and (32) that the claim holds.
Step 2. We will show that all the minimizing sequences for have a weak limit, up to translations, different from zero. Let be a minimizing sequence on for . Note that for any sequence , is still a minimizing sequence for . So the proof of this step can be finished if we can prove the existence of a sequence such that the weak limit of is different from zero.
Applying Lion’s lemma, we know that if then in for any , where . Since , we see that Therefore, it follows the compactness of the embedding that the weak limit of the sequence is not the trivial function.
Step 3. Finally, we verify that has a minimizer for . Suppose to be a minimizing sequence on for with . Then by Lemma 5, is bounded in and for any . So there exists such that in and then we can get If we set and , then by the Step 2, . Now, we want to prove that . To see this, we assume that . From (36), we find since . Noting that , (37) tells that which contradicts to (31) and (32) and then .
Since , we have . Hence, if we would verify that in , it remains to show that up to a subsequence. First, by assumption, there is such that So, which implies is bounded and up to a subsequence; there exists with .
On the other hand, we find , Hence, Notice that by the interpolation inequality, we get with and , and then As a result, Using , , (42) and (45), one can get that is a Cauchy sequence in and hence as . (iii)Now, we come to prove that if , then . For any , let such that . Using Lemma 5, we deduce that is bounded in and then and are bounded. So it is easy to find that On the other hand, letting be a minimizing sequence for , we have Hence, from (46) and (47), follows.

3. Proof of the Main Result

To prove our main theorem, we first give the following important results:

Lemma 7. When , there exists such that has no minimizer for all .

Proof. We prove it by contradiction and suppose that there exist with as and such that .

Since for any , , we have and then by Lemma 3, with . Due to and , (48) tells us that

So, from (48) and Lemma 4, we infer that which is impossible since from Lemma 4.

Recall that

We have

Lemma 8. If , then .

Proof. First, it follows from Lemma 3 that for any , if So,

Thus, for any with small enough and then follows.

On the other hand, taking , then for all and

This implies that if is large enough, and our result has been proved.

Lemma 9. When , we have the following: (i)(ii) if (iii) and is strictly decreasing about if (iv)Moreover, when , then and as and if

Proof. (i)We prove it by contradiction. Suppose that , and then from the definition of , for any , we can get . Hence, it follows from Lemma 6 (ii) that has a minimizer for any . But this gives a contradiction with Lemma 7. On the other hand, using Lemma 6 (i), we can infer that and so (i) follows.(ii)First, from the definition of and for any , we know that for . Furthermore, using the continuity of (see Lemma 6 (iii)), we have and (ii) is proved.(iii)By the definition of , we have for . So, by Lemma 6 (ii), admits a minimizer . Setting with with , we can find that which implies that since and . This tells that (iii) holds.(iv)It follows from Lemma 8 that . On the other hand, from the definition of , it is direct to see that for . Now, it remains to show that if . Here, we claim that for any , there exists such that . In fact, suppose that for all . Then for an arbitrary , taking and for , we find and from (55), which tells that for . This contradicts the definition of since and so the claim holds.Now we consider another scaling for all . Then and This yields as , and we get our result.

Proof of Theorem 1. We will divide it into three steps to prove Theorem 1.
Step 1. First, we prove that if , has a minimizer. Set for all . Then and using Lemmas 6 (iii) and 9 (ii), we find .
So applying Lemmas 9 (iii) and 6 (ii), and have a minimizer, denoted by . By Lemma 5, is bounded in .
Now, we claim that as . Suppose that as by contradiction. Since , similar to (49) and (48), we find with , which, combining (58), implies that for is large enough. This contradicts , and then the claim holds.
With the same argument as the proof of (34), there exists and a sequence such that Letting , then is bounded in and there exists such that , and .
Thus, by (60), we can check that . Finally, we come to show is a minimizer of . First, we have and then by Lemma 6 (iii) and 9 (ii), On the other hand, from Lemma 9 (ii), we have and and then is impossible. So by (61) and (62), we can get and is a minimizer of . Now if we assume that , using (55), one has This yields a contradiction with , and thus .
Step 2. We will prove that has a minimizer if and only if if .
Suppose that there is such that has a minimizer. Then from the definition of and (55), we get that and for any . This contradicts the definition of .
On the other hand, if , using Lemmas 9 (iii) and 6 (ii), admits a minimizer.
Step 3. Finally, it follows from the definition of that for any and if , Thus, if we assume that has a minimizer for some , then from (64) and Lemma 9 (iv), we have which completes the Proof of Theorem 1.

Data Availability

Data sharing is not applicable to this article as no data sets were generated or analyzed during the current study.

Conflicts of Interest

The author declares that there are no conflicts of interest.

Acknowledgments

This work was partially supported by the NSFC (No. 11601194) and the Natural Science Foundation of Jiangsu Province of China (BK20180976). The author thanks the referee for the thoughtful reading of the details of the paper and nice suggestions to improve the results.