Abstract
In this paper, we prove the multiplicity of nontrivial solutions for a class of fractional-order elliptic equation with magnetic field. Under appropriate assumptions, firstly, we prove that the system has at least two different solutions by applying the mountain pass theorem and Ekeland’s variational principle. Secondly, we prove that these two solutions converge to the two solutions of the limit problem. Finally, we prove the existence of infinitely many solutions for the system and its limit problems, respectively.
1. Introduction
In this paper, we consider the multiplicity of nontrivial solutions of the following concave-convex elliptic equation involving variable-order nonlinear fractional magnetic Schrödinger equation: where; is a continuous function; is a bounded subset in with for all ; is the variable-order fractional magnetic Laplace operator; the potential with ; is a parameter; magnetic field with ; are two bounded nonnegative measurable function; ; and . In [1], the fractional magnetic Laplacian has been defined as for . In [2], the variable-order fractional magnetic Laplace is defined as follows: for each , along any . Inspired by them, we define the variable-order fractional magnetic Laplacian as follows: for each ,
Since is a function, magnetic field with , we see that operator is variable order fractional magnetic Laplace operator. Especially, when , reduce to the usual fractional magnetic Laplace operator. When reduce to the usual fractional Laplace operator. Very recently, for , , and ; in [3], under appropriate assumptions, the authors obtained the multiplicity and concentration of the positive solution of the following indefinite semilinear elliptic equations involving concave-convex nonlinearities by the variational method:
For , and ; in [4], the authors obtained the existence, multiplicity, and concentration of nontrivial solutions for the following indefinite fractional elliptic equation by using Nehari manifold decomposition:
When and the authors in [2] give some sufficient conditions to ensure the existence of two different weak solutions and use the variational method and the mountain pass theorem to obtain the two weak solutions of problem (12) which converge to two solutions of its limit problems and the existence of infinitely many solutions to its limit problem:
For , in [1], the authors study the existence of solutions for the following equation on the whole space by using the method of Nehari manifold decomposition,and obtain some sufficient conditions for the existence of nontrivial solutions of the following equation:
In recent years, with the continuous deepening of research, the fractional magnetic problem has attracted extensive attention of researchers. More and more researchers have studied the solvability of the fractional magnetic problem (see [5–8]). We know that the fractional magnetic Laplacian operator is introduced in literature [9]; comes from magnetic Laplacian . However, as far as we know, up to now, few papers have studied the existence and multiplicity of solutions for the variable-order fractional magnetic Schrödinger equation. Therefore, motivated by the above literature, we are interested in the existence and multiplicity of solutions to problem (1) with variable growth. As far as we know, this is the first time to study the existence and multiplicity of nontrivial solutions for the variable-order nonlinear fractional magnetic Schrödinger equation with variable exponents.
It is worth noting that in this paper, we not only prove that there exist two different nontrivial solutions to problem (1), but we also show that the two nontrivial solutions of problem (1) converge to two solutions of the limit problem for problem (1). The novelty of this paper is that, compared with [1], we write the original fractional magnetic Schrödinger equation without variable growth to a variable-order fractional magnetic Schrödinger equation with variable growth. In addition, compared with [2], we write the variable order fractional Schrödinger equation to a variable order fractional magnetic Schrödinger equation.
Inspired by the above works, we assume that and are continuous functions satisfying the following:
(S1): .
(S2): is symmetric, that is, for all .
(V1): is a nonempty bounded domain and .
(V2): there exists a nonempty open domain such that , for all .
(V3): is a continuous function on and .
(V4): there exists a constant such that or all , where is the Hilbert space related to magnetic field (see Section 2).
For the variable exponents , we assume that and satisfy the following assumption:
(H1): for all .
(H2): for all .
In addition, we assume that , satisfy the following assumption:
(H3): are bounded nonnegative measurable function such that , on open interval and
Based on the hypothesis , we can give the following definition of weak solutions for problem (1).
Definition 1. We say that is a weak solution of problem (1), if for any , where will be defined in Section 2.
Now, we will describe the first main result as follows.
Theorem 2. Assume that hold. Let . Then, the problem (1) allows at least two different solutions for all .
Theorem 3. Let and be two solutions obtained in Theorem 2. Then, and in as , where are two nontrivial solutions of the following problem:
Remark 4. In general, if is a continuous function, magnetic field with , then the variable-order fractional magnetic Laplacian can be defined as follows: for each , along any .
2. Preliminaries and Notations
In this section, we first give the definition of the variable exponential Lebesgue space. Secondly, we define variable-order fractional magnetic Sobolev spaces and prove the compact conditions between them. Finally, we give the variational setting for problem (1) and theorems that will be used later.
In this paper, we use to represent -dimensional Lebesgue measure of a measurable set . In addition, for each , we will use to represent the real part of and to represent the complex conjugate of . Let and be a nonempty set. A measurable function is called a variable exponent, and we define . If is finite, then the exponent is said to be bounded. The variable exponent Lebesgue space is with the Luxemburg norm then is a Banach space, and when is bounded, we have the following relations
For bounded exponent, the dual space can be identified with , where the conjugate exponent is defined by . If , then the variable exponent Lebesgue space is a separable and reflexive. In particular, with the scalar product
By Lemma 3.2.20 of [10] and , we know that in the variable exponent Lebesgue space, Hölder inequality is still valid. For all with , the following inequality holds:
Let be a nonempty open subset of , and let be a measurable function, and there exist two constants such that for all . Set
Equip with the norm
Especially, if , then the space is the usual fractional Sobolev space .
Lemma 8. Let be a smooth bounded subset of and let satisfying and satisfying . Then, there exists such that , for any . That is, the embedding is continuous. Moreover, is compact.
Proof. By Theorem 2.1 of [2], we know that is continuous and compact, there exists such that . then, for any , we have Hence, the embedding is continuous and compact.
Let be a continuous function and . For a function , define and the corresponding norm denoted by . We consider the space of measurable functions such that ; then, is a Hilbert space. Define as the closure of in ; then, is a Hilbert space. Especially, if , then the space is the variable-order fractional Sobolev space ; if and , then the space is the usual fractional Sobolev space .
In order to define weak solutions of problem (1), we introduce the functional space equipping with the scalar product which induces the following norm Hence, generalizes to the variable-order fractional Sobolev space (see [2]) and the magnetic framework the space introduced in [9]. Next, we state and prove some properties of space , which will be useful in the sequel.
Lemma 6. There exists a constant , depending only on and , such that for any . Thus, is an equivalent norm of .
Proof. For any , by Lemma 3.1 in [9], we have the pointwise diamagnetic inequality from which we immediately have
Since is bounded, there exists such that and ; then, we have
Thus, we obtain where and .
In addition,
Combining the above two aspects, we have which implies that is the equivalent norm of a norm .
Lemma 2.3. Let be a bounded subset of . Assume that is a continuous function satisfying and is a continuous function satisfying . If , then is continuous and is compact, that is, there exists such that
Proof. For any , we have where Since , we have
In view of which is bounded, there exists a compact set such that . By Lemma 2.2 of [11], we know that is locally bounded and is compact, Thus, we obtain
Equations (36)–(39) together with Lemma 6, we have which implies that the embedding is continuous. In addition, by Lemma 5, we know that is compact. Therefore, the embedding is compact.
Next, we give the variational setting for problem (1). For , we need the following scalar product and norm:
Let be equipped with the norm (that is, in ). Obviously, for . Set . Moreover, for , we can get
For simplicity, we let . Therefore, by condition ,
Associated with problem (1), we consider the energy functional
In fact, one can verify that is well-defined of class in and for all . Therefore, if is a critical point of , then is a solution of problem (1).
Now we give the theorems that we need later.
Theorem 8 (see [2, 12]). Let be a real infinite dimensional Banach space and a functional satisfying the condition as well as the following three properties: (1), and there exists two constants such that for all with (2) is even(3)For all finite dimensional subspaces , there exists such that for all , where . Then, poses an unbounded sequence of critical values characterized by a minimax argument
Theorem 9 (see [13]). Let be a real Banach space and . Suppose satisfies the (PS) condition, which is even and bounded from below, and . If for any , there exists a -dimensional subspace of and such that , where , then at least one of the following conclusions holds: (i)There exists a sequence of critical points satisfying for all and as (ii)There exists such that for any , there exists a critical point such that and It is easy to verify that is a separable Hilbert space. Let , then . we define
Theorem 10 (see [14], fountain theorem). Suppose that satisfying . Assume that for every , there exist such that
(D1): .
(D2): as .
(D3): satisfies condition for every .
Then, has an unbounded sequence of critical values which have the form
where .
Theorem 11 (see [15], dual fountain theorem). Suppose that satisfying . Assume that for every , there exist such that
(D4): .
(D5): .
(D6): as .
(D7): satisfies condition for every .
Then, has a sequence of negative critical values converging to .
3. Proof of Theorem 1
In this part, we first recall that definitions of functional satisfies the condition and condition in at the level and use the usual mountain pass theorem (see [2]) to find a sequence in . Second, we show that functional satisfies the condition in at the level . Finally, we give the proof of problem (1).
Definition 12 (see [2]). Let and . The functional satisfies the condition if any sequence such that admits a strongly convergent subsequence in .
Definition 13 (see [16]). Let and . The functional satisfies the condition (with respect to ) if any sequence such that , admits a strongly convergent subsequence in .
Remark 14. From Remark 2.1 in [16], we get that the condition means the condition.
Theorem 15 (Theorem 3.1, [2]). Let be a real Banach space and with Suppose that (i)there exist and such that for each subject to (ii)there exists with such that Define . Then, and there exists a sequence .
In order to obtain our main results by using the mountain pass theorem, we first prove that satisfies the mountain pass geometry (i) and (ii).
Lemma 16. Assume that (S1), (V1)–(V4), and (H1)–(H3) hold. Then, for each , there exists and such that
Proof. In view of (42) and the fractional Sobolev inequality, for each , one has where are two constants of embedding from variable-order fractional Sobolev space to and , respectively. Making use of (43) and (50), we obtain that for each with .
Set
Define as follows where
As long as that is, that is, we can easily show that for , we have
By it is easy to derive that
By letting and we can easily get
Lemma 17. Suppose that , and hold. Then, there exists such that , where is given by Lemma 16.
Proof. Notice that is a bounded nonnegative measurable function such that on open interval ; then, we can select such that For all , combining (43) with (44), we have
Since , then there exists large enough such that and . By letting , we can easily reach the conclusion.
Define where is a restriction of on and
Observe that for all Obviously, is independent of . From the proofs of Lemma 16 and Lemma 17, we can easily derive that satisfies the mountain pass geometry. Since for all , we have for all . Evidently, for any . Consequently, there exists such that being . Then, for all . In view of Lemma 16, Lemma 17, and Theorem 15, it is easy to get that for all , there exists such that
Lemma 18. Assume that (S1),(V1)–(V4) and (H1)–(H3) hold. Then, satisfies the condition in for all and .
Proof. Assume that be a sequence in with ; then, , It follows from (42) and (43) and the Hölder inequality that On the contrary, we suppose that is not bounded in . Then, there exists a subsequence still denoted by such that as . Then, it follows from (69) that which leads to a contradiction. Hence, is bounded in for all . Therefore, there exist a subsequence of still denoted by and in such that where . The next step is to show that in . By Lemma 7, we can get in . Thus,
Making use of Hölder inequality, we can obtain
Combining (H3), (71), and (72), we have
Similarly, we have
We notice that in and ; then, we obtain
Therefore, which means that
It follows from (43) that
Proof of Theorem 2. In view of Lemma 16, Lemma 17, and Theorem 15, we can easily infer that for all , there exists a sequence for on . It derives from Lemma 18 and for all that there exists a subsequence of still denoted by and such that in . Furthermore, and is a solution of problem (1).
The next step is to prove that system (1) has another solution. Set where and is given by Lemma 16. Then, for all . For this purpose, we first prove there exists such that for all sufficiently small. Let such that . Making use of the hypothesis and (43), we obtain that for small enough,
Consequently, there exists such that for all sufficiently small.
Applying Lemma 16 and the Ekeland variational principle to , there exists a sequence such that for all . Now we prove that for enough large. On the contrary, we suppose that for infinitely many . Without loss of generality, we can suppose that for . It follows from Lemma 16 that
Combining with (82), we obtain which is contradictive with . Next, we prove in as . Let where small enough such that for fixed large. Then, which implies that . Hence, by using (50), we get that is,
Set ; we obtain for each fixed large. Similarly, choosing and small enough and repeating the process above, we can easily get that for each fixed large.
In short, we have which immediately concludes that in as . Therefore, is a sequence for the functional . Making use of a similar proof as Lemma 18, there exists such that in . Therefore, we obtain a nontrivial solution of problem (1) satisfying
Hence, it is easy to conclude that which completes the proof.
4. Proof of Theorem 3
In this section, we mainly give the proof of Theorem 3. In addition, inspired by [2, 17], we obtain the method to prove Theorem 3.
Proof of Theorem 3. For each sequence such that as , set to be the critical points of obtained in Theorem 2, where . Therefore, one has
In view of (92) and (93), it gains where is independent of . So we can suppose that in and in . Making use of Fatou’s lemma, we can easily obtain
Consequently, a.e. in and .
Next, we will prove that in . Indeed, combining in and , one has
Then,
We notice that
Therefore, we have
On the other hand, the weak lower semicontinuity of norm yields that
To sum up, we can see that
By Proposition 3.32 of [18], we can obtain that in . We notice that for any . Hence,
Since the density of in , we can obtain that is a weak solution of problem (12).
Together with (92), a.e. in and the constants are independent of ; we have which implies that and .
Now we consider the case where ; that is where are two nonnegative constants. Correspondingly, the energy functional : is
Next, we mainly prove the existence of infinitely many solutions to problem (107) by using four different methods.
Theorem 19. Assume that (S1), (S2), (V1)–(V4), and (H1)–(H3) hold. Let . Then, problem (107) has infinitely many solutions.
Proof. Method 1: It is easy to verify that functional is even and satisfies . Furthermore, Lemma 18 shows that functional is bounded from below in and satisfies the condition. For any and , let ; then, for any , one has
We find that there exists such that , since all norms are equivalent on finite dimensional Banach space. Then, by , it gains
Letting small enough, we can obtain . Furthermore, we assert that (ii) of Theorem 9 does not work. In fact, (109) means since as small enough with the given . Thus, by Theorem 9, we get that problem (1) has a sequence of solutions with as . In short, problem (1) has infinitely many solutions for all .
Method 2. To start with, we assert that for any finite dimensional subspace of , there exists such that for all , where Indeed, for each , we can easily get that
We observe that there exists such that , since all norms are equivalent on finite dimensional Banach space . Then, by , it gains
Thus, there exists large enough such that for all , with and . Consequently, the assertion is valid.
From Lemma18, we know that satisfies the condition for any . Obliviously, and is an even functional. In short, it follows from Theorem 8 that there exists an unbounded sequence of solutions of problem (1) for all .
Lemma 19 (see Lemma 4.1, [10]). Let for all . For any , define Then, , as .
Method 3. By Remark 14 and Lemma 18, we know that satisfies the condition for any . To start with, we will prove (D1) is satisfied. It follows from (36) and (43) that
We find that there exists such that , since all norms are equivalent on finite dimensional Banach space . Then,
Then, by , we can easily obtain that (D1) is satisfied for big enough. Next, we will show (D2) is fulfilled. In view of (43) and Lemma 20, for , one has
Choosing . Combing with Lemma 20, we know that as . Thus, one has as and as . In conclusion, (D2) is fulfilled. It is easy to check that satisfying . Thus, by Theorem 10, we can obtain that problem (1) has infinitely many solutions for all .
Method 4. First, we will show that (D4) is fulfilled. By (42) and (43), one has
Since , we can choose small enough such that for all with , hold. Then, Choosing
Combing with Lemma 20, we know that as . Thus, one has as . Hence, there exists such that as . Consequently, for and with , we can obtain that . Next, we will show (D5) is fulfilled. For any , with , we get
We find that there exists such that and , since all norms are equivalent on finite dimensional Banach space . Then, as small enough. Now we check (D6) is fulfilled. It follows from (D4) that for and with , thanks to as . Thus, one has as . Thus, (D6) is also satisfied. In conclusion, by Theorem 11, we can obtain that problem (1) has infinitely many solutions for all .
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that there is no conflict of interests regarding the publication of this paper.
Authors’ Contributions
All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.
Acknowledgments
This work is supported by the Natural Sciences Foundation of Yunnan Province under Grant 2018FE001(-136), 2017zzx199, the National Natural Sciences Foundation of People's Republic of China under Grants 11961078 and 11561072, the Yunnan Province, Young Academic and Technical Leaders Program (2015HB010), the Natural Sciences Foundation of Yunnan Province under Grant 2016FB011