Existence of Solutions for Fractional Klein-Gordon-Maxwell Systems ()
1. Introduction
In recent years, by studying the nonlinear problems related to fractional Laplacian, many practical problems have been solved. For example, in the financial market problem, phase transformation problem, anomalous diffusion problem, crystal dislocation problem, semi-permeable film problem, soft film problem, minimal surface problem (see [1] and references for more details). As it involves more and more fields, the research on the problem is more and more in-depth, and people keep putting forward new problems at the same time, also keep producing new ways to solve the problem. In this paper, we study the following fractional Klein-Gordon-Maxwell system on
(1.1)
where
is a fixed constant and
is the fractional Laplacian operator, defined as
(1.2)
where
is a constant, dependent on s can be expressed as
(1.3)
and P. V. stands the principal value.
,
, where
and
are defined in (1.9) and (1.11),
is the fractional Sobolev critical exponent. Next, let us mention some illuminating work (1.1) related to this problem. In [2] , the critical Klein-Gordon-Maxwell system with external potential is not only studied when the potential well is steep,
(1.4)
where
and
are positive parameters,
, where
and
satisfy the following hypotheses:
(V)
and there exists
such that the set
is bounded;
(V’) the set
is non-empty and has smooth boundary with
;
(
)
and
;
(
)
, where
. Moreover, there exist
,
and
such that
for
.
The existence of the solution and the phenomenon of concentration are proved by using the penalized technique and the elliptic estimation. In addition, the existence of the solution is proved when the potential well is not steep, that is to investigate whether the problem has a solution without any restrictions on
and
, that is, consider the following problem
(1.5)
In [3] , when the nonlinearity exhibits critical growth, the existence of a positive ground state solution to the problem is proved by the Nehari method,
(1.6)
where
,
,
with
,
, and
are functions, where
satisfies some of the following hypotheses:
(
) V is periodic in
;
(
) There exists
, such that
,
and
,
and V* > 0 are studied respectively in both cases is the ground state of existence. Benci and Fortunato first studied the following system in [4] ,
(1.7)
They proved infinitely many radially symmetric solutions using the variational method when
and for sub-critical exponents p satisfying
. Based on the nonlinear Klein-Gordon field and electrostatic field of the relationship between research, many researchers on the system of the existence, nonexistence and diversity some results are obtained. In [5] , the existence of nontrivial solutions is investigated separately for different
cases by means of the Ekeland’s variational principle and the mountain pass theorem. Carriao, Cunha and Miyagaki in [6] such as periodic potential
to replace the constant
, considered the critical problem of the existence of the ground state solutions accordingly. After this, more attention was paid to the following Klein-Gordon-Maxwell system
(1.8)
Inspired by the above literature, the existence of a non-trivial solution of system (1.1) will be discussed in this paper. To illustrate our results, we set the potential functions
and
satisfy the following assumptions:
(V1)
,
for all
and
;
(V2) there exist
and
such that
for
;
(f1)
and
;
(f2) there exist
and
such that
for
;
(f3) the function
is increasing for
.
Notations:
In this paper, the norm of fractional Sobolev space
is defined
(1.9)
and define
, endowed the norm on X by
(1.10)
and the corresponding inner product is
Therefore,
is equivalent to the usual norm on
, where
is referred to in (V1). Consider the following fractional critical Sobolev space
is defined by
(1.11)
with the norm
(1.12)
where
is the completeness of
. For
, we let
(1.13)
and
for
, where
. For any
, the embedded
↪
is continuous, exist for the best fractional critical Sobolev constant
(1.14)
for any
,
(1.15)
For this paper, taking C uniformly represents all normal numbers. The main research results can be summarized as follows:
Theorem 1.1. If (V1)-(V2), and (f1)-(f3) hold with
, then there exists
such that for
, problem (1) admits a nontrivial solution
.
2. Preliminary Lemmas
By (V1), there exist
such that
for
. Moreover, there exists
such that
for
. Then
(2.1)
By the embedding
↪
is continuous, so in the same way, we get
(2.2)
The purpose of this paper is to find the solution of (1.1). To this end, we give the weak formula of (1.1) by the following questions:
(2.3)
The relevant functional can be defined by (1.1):
(2.4)
We take the derivative of that and we get
for any
, we have
(2.5)
Next up, we define
, where
, the function
defined as
(2.6)
and
(2.7)
Critical points of
are weak solutions of (1.1). We will prove the existence of the critical points of the functional
.
Lemma 2.1. ( [7] ) If (f1) and (f3) is true, then
1)
, where
;
2)
is increasing for
.
Lemma 2.2. Let assume
be a Carathéodory function verifying conditions (f1), we get that for any
, there exists
such that
(2.8)
(2.9)
Let
, we also derive that for any
, there exists
such that
(2.10)
Lemma 2.3. ( [8] ) For any
, there exists a unique
satisfying
And the map
is continuously differentiable and for any
,
1)
on
;
2)
and
,
where
and
are positive constants.
Lemma 2.4. ( [9] ) Let
and
. Then, the following estimates hold true:
(2.11)
(2.12)
and
(2.13)
as
, for some positive constant
depending on s.
Lemma 2.5. ( [10] ) If
in X, then, up to subsequences,
in
as
.
3. The Proof of Theorem 1.1
Lemma 3.1. Let
. Define
where
. The minimum
is given by a non-negative function
. Moreover,
.
Proof. By (f1), we get that for
, there exists
, such that
, for
.
So exist
, such that
, for
. Choose
and
. By (f2) and Lemma 2.3, we get
and
. So we derive that
and
. By the mountain pass theorem in [11] , there is
satisfying
and
.
Step 1:
is bounded in X. Then by lemma 2.1 and
, we get
So
is bounded.
Step 2:
. By lemma 2.3, there exists
and
such that
and
By Lemma 2.3 and (f2),
And by Lemma 2.4, there exists
such that for
,
in consideration of
. Then by the definition of
, we have
. So we have
.
Step 3:
. Let
,
. And
, we get
,
is bounded,
and
, so
(3.1)
We assume
weakly in
. If
, by the Lions Lemma, we have
in
for any
. So
.
Because of
, we assume
, where
. By (3.1) and Lemma 2.3, we have
. Then by the Sobolev embedding theorem
and when we take the limit of both sides, we have
(3.2)
this is in contradiction with
. Thus, we assume that there exists
, such that
. Therefore we deduce that there exists
satisfying
weakly in X, thus,
,
and
. Go through again with Lemma 2.5, we get
and
By Fatou’s lemma
and
, we get
By the definition of
, we get
.
Step 4:
is attained by
. There exists
such that
and
. We put
in there, so we get
, with Lemma 2.2 we get
In addition, by the Sobolev embedding theorem, we have
. Then
(3.3)
So
is bounded and
.
Since
and
, we can assume that
. Similarly, we deduce that there exists
such that
weakly in
, where
is non-negative. By Lemma 2.5, we have
. So by (31) and Fatou’s lemma,
However, by definition of
, we get a contradiction with
. □
Lemma 3.2. ( [2] ) For any
, there exists
such that
Recall that a sequence
is a Ceramisequence sequence for the functional
if
and
as
. We need the following variation of the mountain pass lemma in [12] .
Theorem 3.1. Let X be a real Banach space and assume
satisfies
for some
and
with
. Let
(3.4)
where
. Then there exists a Ceramisequence sequence
for the functional K satisfying
.
Proof of Theorem 1.1. Let
. Since
, Lemma 2.2 and (17), we get there exists
such that
(3.5)
Therefore, there exists
such that
for
. Choose
such that
. By (2.10), there exists
is a constant such that
for
. Because of
, we get that
. Therefore, there exists
such that
. At the same time, and we get
. Let
. By Theorem 3.1, there exists a sequence
is a Ceramisequence sequence for the functional
, such that
and
as
, where
with
.
Step 1:
is bounded. Next we show that
is bounded. Since
. Now, for any
, let
. Then we get
weakly in X and
a.e.
. Since
a.e.
. By (2.10), let
, there exists
such that for
and
, there holds
(3.6)
Thanks to the Young’s inequality, we have
(3.7)
where
. On the basis of choosing
small in (3.7), we get that there exists
such that
(3.8)
By Lemma 2.1, we have that
for
and
. Next, by (3.8) and Lemma 2.3, we get
(3.9)
as
. By
and
, we get a contradiction.
As a result,
. By
, we have
. So
a.e.
, from which we get that
a.e.
. Might as well set
. Let
is an open bounded subset of
defined as
And we can see that the measure of
is positive. By
and
, for
, we get that
as
. Obviously,
for
, from which we have
(3.10)
By (2.10), we set that
for
,
and
. On the basis of choosing
small in (3.7), we get that there exists
such that
for
and
. By
for
and
, we get
(3.11)
Together with (2.2), we get that
(3.12)
By (3.10) and (3.12), we have
(3.13)
However, by the embedding
↪
is continuous,
and Lemma 2.3, we get
which contradict (3.13).
Step 2:
. Let
. By (2.10) and Lemma 2.3, we get that for
, there exists
such that
(3.14)
So there exist a small
and a large
such that
independent of
satisfying
(3.15)
Observe that
(3.16)
Choose
. By Lemma 3.2, there exists
such that
(3.17)
From absolute value inequality, we get
, by (V2) and (3.17), set
, we get that
(3.18)
Then, set
for
, defined as
And
(3.19)
By (f3), we get that
is increasing for
, we deduce that
has a unique critical point which is its maximum value. By
, the critical point is reached, i.e.
, this critical point should be achieved. So
. By (3.16), (3.18) and Lemma 2.3, we get
(3.20)
By
, we set
, there holds
. Then there exists
such that for
(3.21)
Combining with (3.15) and (3.21), we get that
for
and
. By the definition of
, we proved that
.
Now, by step 1
is bounded,
and
0, without loss of generality, we may assume that
a.e. and
weakly in X. Because if
,
weakly in X. If
, by the Lions Lemma, we derive that
in
, where
. Similar to the principle of Lemma 3.1, we launch a contradiction. We’re not going to prove it here. There exists
such that
, so we deduce that there exists
with
satisfying
weakly in X. By
weakly in X and
, we get
and
. Therefore,
(3.22)
From
weakly in X and Lemma 2.2, we have
. So
. By (3.22), we get
(3.23)
So let’s take the limit of both sides, by Fatou’s lemma, we get
(3.24)
This contradiction with Step 2
. Therefore,
weakly in X. By
and Lemma 2.2, we get
.