In this paper, by using variational methods, we study the existence and concentration of ground state solutions for the following fractional Schrödinger equation
where , ϵ is a positive parameter, , stands for the fractional Laplacian, f is a continuous function with subcritical growth, is a -periodic function and satisfies some appropriate assumptions.
In this paper, we are interested in the following fractional Schrödinger equation
where , is a parameter, , stands for the fractional Laplacian operator, f is a continuous function with subcritical growth, the potential is a periodic function verifying
where denotes the spectrum of in . satisfies the following condition
When , the problem (
P
ϵ
) is a classical semilinear elliptic equation. In recent years, the existence and concentration of nontrival solutions for semilinear elliptic equations involving in different nonlinearities and potential functions have been widely studied, such as [1,2,5,6,17,20,25,26,32] and their references. Due to our scope, we would like to mention some works related of the Schrödinger equations with periodic data. Coti Zelati and Rabinowitz [11] considered this kind of problems in the definite case . For the indefinite case, we main consider the situation that 0 lies in the spectrum gap of , thus the space which was spanned by the eigenfunctions corresponding of negative eigenvalue of the operator in is infinite dimensional, it is easy to see that zero is no longer a local minimum point of the corresponding energy functional, thus the usual mountain pass geometry does not work and the classical linking theorem or fountain theorem can not be used to this problem. To overcome this difficulty, there are some different methods have been proposed in the last twenty years. Li and Szulkin [18] first developed a new generalized linking theorem, after that, Szulkin and Weth [31] studied the ground state solutions based on the generalized Nehari manifold or called Nehari–Pankov manifold [23]
they constructed a new minimax characterization such that an indefinite problem can be reduced to a definite problem. Meanwhile, Liu [19] proved the existence of ground state solutions by generalized linking theorem under a more weakly monotone condition, and he used super-quadratic condition to replace the standard Ambrosetti–Rabinowitz condition. Recently, Alves and Germano [1] studied a class of semilinear equations
where the function and satisfy the conditions (
V
), (
A
), is a parameter, f is continuous and satisfies some assumptions, the authors considered the existence and concentration of ground states in subcritical case and critical case, respectively.
In the past several decades, the fractional Schrödinger equations has been widely studied. We refer to Molica Bisci, Rădulescu and Servadei [22] who carefully studied continuous and compact embeddings, trace theory, extension, and regularity in fractional Sobolev space. For the strongly indefinite problem, Fang and Ji [12], Pu et al. [24] used different methods to prove the existence of ground state solutions with different assumptions on the nonlinearity. For more research related fractional PDEs, we refer to [3,4,7,9,10,13–16,21,22,26–28,33,34] and the references therein.
Inspired by [1 ,31], we would like to study the existence and concentration of ground state solutions for the problem (
P
ϵ
). Since the fractional Laplacian operator is a nonlocal operator, our problem becomes more difficult and the -estimate of ground state solutions is more complicated than the local equations.
From now on, we assume the following hypotheses on the nonlinearity f:
;
for , is called the fractional critical exponent;
is strictly increasing on ;
There exists such that
where
We are now in position to give the main result of this paper.
Assume that (
V
), (
A
) and–hold. Then, there existssuch that the problem (
P
ϵ
) has a ground state solution for each. Furthermore, ifdenotes a global maximum point of, then
This paper is organized as follows. In Section 2, we present some preliminaries. In Section 3, we prove some results about the autonomous problems. In Section 4, the existence of ground state solutions for the problem (
P
ϵ
) is proved. The last Section is devoted to prove the concentration result in Theorem 1.1.
Preliminaries
In this section, we introduce the setting framework of the fractional Sobolev space. The fractional Laplace operator can be given by the following singular integral
where is the Cauchy principal value and is a normalization constant, or it also can be defined on the Schwartz class of function through the Fourier transform
Next, we define the fractional Sobolev space as follow
endowed with the natural norm
The variational functional associated with the problem (
P
ϵ
) is
It’s routine to show that .
Let’s define an operator
Since is a continuous periodic function, thus is bounded in . Then it is easy to prove that L is a self-adjoint operator. Let be the square root of the absolute value of L on , then we choose the inner product
with the norm
We define a bilinear map with the following form
There exists an orthogonal decomposition such that and for each . Let and for all . Thus, the functional (2.2) is equivalent to the following form
which also belongs to , and it’s obviously that the critical points of the functional correspond to weak solutions of the problem (
P
ϵ
).
The following two propositions will be used frequently in the below.
Assume that the sequenceis bounded in, and for each, it satisfiesTheninfor all.
An autonomous problem
In this section, we consider the following autonomous problem
where , , and the conditions (
V
) and – hold. The energy functional associated with the problem (
P
λ
) is
or equivalently
By the assumptions –, belongs to , and the critical points of functional correspond to the weak solutions of the problem (
P
λ
).
In order to get the ground state solutions of the problem (
P
λ
), we shall apply the method of Nehari manifold developed by Szulkin and Weth [31]. Following the same variational framework as in [31], we have the similar results in fractional Sobolev space.
First of all, we define the ground state energy of functional
where is Nehari–Pankov manifold associated with
Moreover, for each , we introduce two important sets which have been studied in [31],
and
where is a subspace in and is a convex subset in . Both of them do not depend on λ, which just rely on the operator , this is a crucial point in the following proofs.
(2.3) shows that is a symmetric bilinear map. Then we can use the similar method in [31, Proposition 2.3] and condition to get this conclusion. □
There existssuch that, where;
for all.
(a) For any , then we have . By , we know that as . Hence if small enough, we have . Then we prove . For each , there exists such that , then . By Lemma 3.1, , thus .
(b) For every , by the definition of c, we have
where by , then we obtain that . □
Ifis a compact subset and.. Otherwise, from, we have the following properties for,
as;
for all.
Then, for the following functionalthere existssuch thatonfor every.
is coercive onfor every.
Suppose by contradiction, assume that there exists a sequence such that for some as . Define the sequence , from Lemma 3.2, we get , hence and for each . There exists a sequence , and such that
If not, from Proposition 2.2, we get that in for . Then for each by , and
which is impossible because s is arbitrary, so (3.3) holds.
Next, we define and . It’s easy to prove that if each element , then . By (3.3), we obtain that . Because , hence we have a.e. in as . By the Fatou’s lemma,
where means the Lebesgue measure of the set . Thereby,
which is absurd. □
For every, the sethas a unique element, and it is the global maximum of.
From Lemma 3.1, we know that . Let and . By Lemma 3.2(a), for small enough, and for by Lemma 3.3, which means that . Since is a convex set, and the functional is weakly supper semi-continuous on , by [30, Theorem 1.2], we can deduce that is attained, and there exists such that . It shows that is a critical point of , so it satisfies for all , which means that . Therefore, . □
From the last lemma, we know that for each , there exists a unique pair , and such that and
. Furthermore, u is a critical point forif and only ifis a critical point forand.
The problem (
P
λ
) has a ground state solution for every.
Next we start studying the property of the following function,
where is the ground state energy for . Define is a ground state solution of (
P
λ
), then and , and from Lemma 3.5, it can be defined by
The functionis decreasing and continuous.
First we consider the monotonicity of , assume that , are the ground state solution of functional and , respectively. Suppose the function is not decreasing, without loss of generality, we assume that such that . By (3.4), there exist and such that
Hence
But from the above assumption, we have
together with and condition , we conclude that
From , , we know if and as . We deduce that a.e. in . Thus , which is impossible because . Then we show that the function is strictly decreasing.
Next we shall show that the function is continuous. To achieve this goal, we want to prove by two cases, an increasing sequence and a decreasing sequence. Because for any sequences always can be divided by an increasing sequence and a decreasing sequence, thus if this two cases both hold, then we could know that the function is continuous.
The increasing case: Choosing the sequence with . From the above result, we have , we need to prove that . Let be the ground state solution of (
P
λ
). For each , there exist and such that
Furthermore,
then for each and . By Lemma (3.3), there exists such that
From (3.5), , namely
Together with (3.6) and (3.7), we get that , which implies that the sequence is bounded in , therefore we obtain that is also bounded,
together with for all , we obtain .
The decreasing case: Choosing the sequence , then . Let be the ground state solution of for each , then there exist and such that
Since is coercive on by Lemma 3.4, it is easy to prove that the sequence is bounded.
Next we start to prove that . First, since is bounded in , there exist , and such that the following inequality is true for each ,
If not, by Proposition 2.2, in for . Then by and , we get , then
which means that , contradicting the equality by Lemma 3.2. Thus (3.8) holds.
Let , we have that is bounded, then up to a subsequence, in . Fixing , then is bounded and the sequence does not weakly converge to zero in . Thus, by Lemma 3.3, there exists such that for every ,
and let , we have
Hence, by (3.9) and (3.10), we derive that for every , then . Namely the sequence is bounded in , which implies that is bounded, thus,
This means that
and together with for each , we obtain . We complete the proof. □
Existence of ground state solution for equation
In this section, we start to prove the existence of ground state solutions for problem (
P
ϵ
). Denote the energy functional associated with the problem (
P
ϵ
)
or the following equivalent form
and the corresponding Nehari manifold is
Without loss of generality, we assume that
Next, we define the ground state energy of on Nehari manifold . Moreover, the similar process in Lemma 3.5 can be used in , then for each , there is only one point in , hence, there exists a unique pair and such that
The ground state energy of can be given by
For convenience, we denote some notes , and .
To begin with, we show a relation between and in order to help us finish the below proofs.
.
Let as , and for each , thus
Let be the ground state energy of the functional for . From the last section, there exists a unique pair with and such that
By the definition of , we obtain
hence, together with (4.3), we can deduce that for each and .
Next, we know that the autonomous problem with has a ground state solution from Section 3, there exist , such that the following equation holds,
By Lemma 3.3, we know that the sequence is bounded in . Then we assume that and in and satisfies
Therefore, by the weakly lower semicontinuous of the norm, and the Fatou’s lemma,
so we obtain . □
From the last inequalities, we found that , then both and are the elements of . But there is only one point in , so and , .
There existssuch thatfor any.
From assumption (
A
), we know , then by Proposition 3.8. Because , choosing small enough, we can obtain the result. □
The functionalis coercive onfor each.
Suppose by contradiction, assume that there exists a sequence satisfying
where h is a fixed number in . Let , from Lemma 3.2, we know that , and then . So there exists a sequence , and satisfying the following inequality for each ,
If not, we can obtain in for all by Proposition 2.1, then by and , for each . Hence,
because m is arbitrary but h has been fixed, it’s impossible, thus (4.5) holds.
Now we define and , we have and . Up to a subsequence if necessary, in by (4.5). Hence a.e. in , and a.e. in as .
Then from the Fatou’s lemma and Lemma 3.2,
where is the Lebesgue measure of the set .
Let , then
obtain a contradiction. The proof is completed. □
Under the assumptions (
V
), (
A
) and–, the problem (
P
ϵ
) has a ground state solution for each, wherewas given by Corollary
4.3
.
From Lemma 3.7 and using the Ekeland variational principle in [31], there exists a sequence for such that
The sequence is bounded in since is coercive on , we need to show that in . By the definition of , let be a sequence converging to zero, we have
Hence
it follows that , and is bounded in , then from Proposition 2.2, there exists a sequence , and such that
First, we prove that the sequence is bounded in , suppose by contradiction, assume that is unbounded and as . Define , then by (4.6). Choosing the test function , we obtain
Passing to the limit , we get
By the density of in , we obtained
This implies that v is a nontrival solution of and , by the Fatou’s lemma,
It follows that , this yields a contradiction since for any by Corollary 4.3. Hence is bounded, we can select such that for all and
we deduce that . Repeating the method in (4.7) and (4.8), we can obtain that is a nontrival solution for (
P
ϵ
), thus .
On the other hand, by the Fatou’s lemma and , we have
Hence , it follows that w is a ground state solution of (
P
ϵ
). We complete the proof. □
The concentration result of ground state solutions
In this section, we start to prove the concentration result of the maximum points of the solutions which was given by the last section. For convenience, we denote , , , and , where is a sequence in such that as . From the argument in Lemma 4.1, we have for all . Define is the set of the maximum points of . We need to show that if is the maximum point of for every , then
where . In other words, we need to prove that for the sequence as , up to a subsequence if necessary, .
The following lemmas are important tools to prove the main result of this section.
The sequenceis bounded in.
It is easy to check this point by Proposition 4.4. □
There exist,andsuch thatFurthermore, the sequenceis bounded inand,.
First of all, since is bounded, we want to show that there exist , and such that
Suppose by contradiction, we obtain that in for by Proposition 2.2. By (
A
), and ,
therefore, , which is impossible because by Lemma 3.2. Define the sequence , then for some subsequence .
Next, we prove the sequence is bounded in by contradiction. Suppose there exists a subsequence such that as , then we use the same idea in the proof of Theorem 4.5. Define is a ground state solution of , for any , we have
Passing the limit , and by the density of in , we have
It implies that v is a nontrivial solution of and . By the Fatou’s lemma and Lemma 4.1,
Together with by Proposition 3.8, we get a contradiction. So the sequence is bounded in . Then up to a subsequence, suppose that . Repeating above process, we have
It follows that v is a nontrivial solution of equation and . Similarly, by the above method, we obtained . Hence by Lemma 3.8, while according to condition (
A
), we deduce that . So we obtain and . □
For each, there existsandsuch thatandwhere.
Because is continuous in and as by condition , hence is bounded in , the first inequality holds for some η.
We consider the second inequality in two intervals and .
Because H is a bounded closed subset, by the continuity of f, so is bounded in H, then there exists a number such that and for all .
From conditions , , we can get
where , are some fixed constants. Because and , it’s easy to see that
where .
Multiplying on both sides, we get
Based on the above argument, we can get the desired conclusion. □
The sequencestrongly converges to v in.
First, (5.1) shows that . Using the method in Lemma 5.2 again, we obtain
It implies that v is also a ground state solution of , and
Therefore,
From , we obtain that
From Lemma 5.1, the sequence is bounded in , then up to a subsequence in , then in with , and a.e. in . For some subsequence, and from [8, Theorem 4.9], there exists such that
Using , there exists such that
Therefore, there is a constant such that for each .
Next we define the function
We want to show .
From in with and , we deduce that in for any ,
By Lemma 5.3, there exist positive numbers σ and η such that
and
Let and be the measures of and , respectively. Define . From the Hölder inequality, we have
Let R large enough if necessary, such that the following inequality holds for each ,
Therefore,
Together with (5.2), it follows that
and
Using the similar method, we can prove that
Hence, we have
Since and , we obtain
It means that in for some subsequence. Likewise, we have in . Then from to know that in . We finish the proof. □
.
Suppose by contradiction, from Lemma 5.4, we get , which is a contradiction with . □
To get the -estimate of the solution sequence , we first introduce the Bessel kernel to show the solution of the following equation
where .
Using the result in [13], the solution of (5.3) can be expressed as
where is the Bessel kernel.
Next, we introduce some properties of the Bessel kernel that will be used below, more details can be found in [13,29].
Letbe a solution of, then there existssuch that, for each, andinfor.
Furthermore, there existsverifyingsuch thatandinfor all.
From [13, Theorem 3.4], we can directly obtain that there exists such that for each . By Proposition 2.1, we know that for . For each , there exists such that
Then using the Hölder inequalities, it follows that for and for every .
From Lemma 5.1 and Lemma 5.4, in , hence in by Sobolev embedding. For each , there exists such that
Define , using the Hölder inequalities again,
as in and , in for . We complete the proof. □
The sequencesatisfiesasfor all.
First, for each , we define two sets
and
From (5.4),
From Lemma 5.1, is bounded in , by the continuous embedding in Proposition 2.1, , which implies that is bounded in .
Because , are bounded in , and , where by , we have
where , are some constants.
Define , , , then
where p is from (5.6). It is easy to check
From (5.6), Lemma 5.6(b) and the Hölder inequalities, we have the following estimate for all ,
where and P is a constant.
Next, we start to estimate the second term. From Lemma 5.7, there exists such that and for each . We define
We can check that holds. From Lemma 5.6(c), we have and , combining with the Hölder inequality, we obtain
From Lemma 5.4, we can choose large enough for each σ such that the following inequality holds for all ,
Hence, due to , we have
Then
where Q is a constant related to P and C. For each σ, there exists for each σ such that as for all . Then let σ small enough, we can get the result. We complete the proof. □
Suppose that is a maximum point of for each ,
By the definition, . Then is a maximum point of .
As a direct result of Corollary 5.5, for some , then let η equal to σ in Lemma 5.8, then we have
which implies that the sequence is bounded in , together with by Lemma 5.2, then up to a subsequence, we have
Hence
The proof is completed. □
Footnotes
Acknowledgements
The authors would like to thank the referee for many useful comments which clarify the paper.
C. Ji was partially supported by Natural Science Foundation of Shanghai (20ZR1413900, 18ZR1409100).
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