Ground State Solutions for Schrödinger Problems with Magnetic Fields and Hardy-Sobolev Critical Exponents

Abstract

A Schrödinger equation and system with magnetic fields and Hardy-Sobolev critical exponents are investigated in this paper, and, under proper conditions, the existence of ground state solutions to these two problems is given.

1. Introduction

In the present paper, we consider the following semilinear stationary Schrödinger equation and system with magnetic fields and Hardy-Sobolev critical exponents: where , are magnetic potentials, , with , and is a smooth bounded domain containing the origin as a interior point. Set , and Then ; is a Hilbert space obtained by the closure of with respect to scaler product: where the bar denotes complex conjugation. Here and in the following, means . We regard the range of function as , except for emphasizing, that is, . denotes the space of -integrable functions with respect to the measure , endowed with norm For , denote the norm by Write and for simplicity.

The first existence results for this kind of problems with a nonsingular magnetic potential were established in [1]. Motivated by the seminal work [1], some papers dealt with the Schrödinger equations with nonsingular magnetic field, for example, [2–11] and references therein. For the singular magnetic potential, we refer to [12, 13]. There are a few works about the nonsingular magnetic problems with critical exponents, such as [9, 14, 15] for Sobolev critical exponent and [16] for Hardy critical exponent.

As far as we have known, there is no result about the problems of this type with Hardy-Sobolev critical exponents, especially for the system case. The current paper is mainly motivated by [15] and applies the least energy solution obtained in [17, 18] to show the existence of the ground state solution to the critical system (2) under proper conditions. Precisely, we get the following results.

Theorem 1. Assume that . If or is continuous at , where the is the skew-symmetric matrix with entries , then (1) has a nontrivial ground state solution.

Theorem 2. Assume that and . If or and are both continuous at , then (2) has a nontrivial ground state solution.

The corresponding energy functional of (2) is where , endowed with norm Define and . By nontrivial solutions of (2), we mean . A solution of (2) is called a ground state solution if and . A ground state solution is semitrivial if it is of type or . Similar definitions are verified for single equation (1).

Remark 3. Although it has been studied that the existence of ground state solutions is heavily dependent on the type of equations without magnetic fields, the symmetric and decaying information about ground state solutions of equations with magnetic fields is not known. In this paper, we find that the latter equation and system also indeed have the ground state solutions provided suitable conditions.

2. Preliminaries

Consider and the corresponding energy functional where , endowed with the norm Define and . Then, by [18], we see that, under the condition , (10) has a positive ground state solution , where and are radial symmetric decreasing with the following decay information: That is, is attained by . Define

As proved in [1, 19], for any , the following diamagnetic inequality holds pointwise for almost every :Then, for , we see that belongs to the usual Sobolev space . Moreover, we have the following.

Lemma 4 (see [20]). The embedding is continuous for , and it is compact for , where .

Lemma 5. If weakly in , where , then .

Proof. Obviously, . Lemma 4 ensures that weakly in and a.e. in . Then, weakly in . Since , the dual space of , we see that where the duality product is taken with respect to and . It is easy to see that By Hölder inequality and (17), we have Thus, the proof is completed by (17), (18), and (19).

Remark 6. It is easy to see that is closed and bounded away from , for any , there exists unique such that . follows from Lemma 4. It is standard to see that any sequence of with constraint is a sequence of without constraints and every sequence of is bounded in . Define where . Then it can be proved that .

3. Ground States

In this section, we firstly suppose that Theorem 1 holds, and then we show Theorem 2 below. In fact, the proof of Theorem 1 is similar to and easier than that of Theorem 2 by the application of the ground state solutions to (cf. [17]), so we shall show Theorem 1 briefly as a remark in the end of this section.

By Theorem 1, we see that and have ground state solutions and with energies and Then, we have the following.

Lemma 7. Assume that and . If or and are continuous at , then .

Proof. Firstly, we prove that . There exists satisfying .
(i) If holds, then by [21], there exist such that . Let where is defined by (15) and is a cut-off function with and for . By direct computation, we have By (26)–(32), for any , we deduce that where is a positive constant, and solves Then, (35) yields that , where Since , we see that as , Setting and we have Then, by (33), we deduce that which implies that (ii) If and are continuous at , then setting we have which imply that and . Then, there exists such that where is a positive number. Let and where is a cut-off function such that in and is defined by (15). By (44), we deduce that Hence, replacing with and following the previous proof for case , we may prove that with minor modifications.
Next to prove , recalling (9), choose satisfying Then, . Noting that and , we deduce that where Then, Similarly, .

Proof of Theorem 2. Note that the functional has a mountain pass structure. It follows from the mountain pass theorem (see [22]) that there exists such that By Remark 6, we see that is bounded in . By Lemma 4, we may assume thatwhere . From we get that i.e., in , the dual space of . Setting and , then in and in . Thus, strongly in , a.e. in , up to a subsequence. Brezis-Lieb Lemma guarantees that Noting that , and , by (55) and Lemma 5, we deduce thatand Assume that up to a subsequence. By (56), we see that Letting , we have Case 1 (). By and (60), we see that . Then, we may assume that for large enough. Recalling (56), (57), and in (13), there exists with as such that . Then, by (60), we obtain that which contradicts Lemma 7. Hence, is impossible.
Case 2 ( or ). Without loss of generality, we may assume that . Thus, is a nontrivial solution of . Then, by (60), we deduce that , a contradiction to Lemma 7. Therefore, Case 2 is impossible.
Since Cases 1 and 2 are both impossible, we have . Remark 6 yields that . It follows from (9) and (20) that there exists a Lagrange multiplier such that where Since and we obtain that and, hence, in . That is, is a ground state solution of (2).

Remark 8. It follows from Remark 6 that the ground state solution of (2) is also a mountain pass solution.

Remark 9. Note that we only use the result in Theorem 1 to construct the values of and which are used in Lemma 7 and then in Case 2 at last of the proof of Theorem 2 to exclude the semitrivial solutions as and to system (2), that is to say that the part of proof of Theorem 2 before Case 2 is independent of Theorem 1. So we can repeat the proof above with a little modification to prove Theorem 1. Indeed, the existence of a ground state solution to is ensured in [17], and we only need to exclude the trivial solution of to (1) which is similar to Case 1 in the proof for the system.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This paper is supported by NSFC (11601109), Hainan Association of Science and Technology Plans for Youth Innovation (201503), and Natural Science Fund of Hainan Province (20161001).