1. Introduction
In recent years, many practical problems have been solved successfully through in-depth research on nonlinear problems of fractional Laplace equations. For example, there have been studies in finance [1] , fluid dynamics [2] , quantum mechanics [3] , physics [4] , materials science [5] , crystal dislocation problems [6] , semi-permeable thin film problems [7] , soft film problems [8] , and very small surface problems [9] , soft film problems [8] and very small surface problems [9] have been studied. With the continuous expansion of the fields involved and the deepening of the problem research, people continue to put forward new problems and explore solutions. In this paper, we study the following semilinear fractional elliptic equation 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. And
,
, where
and
are defined in (1.9) and (1.12),
(1.4)
is the fractional Sobolev critical exponent. Next, let us mention some illuminating work (1.1) related to this problem. In paper [10] , Bartsch Thomas and Sébastien de Valeriola analyzed the case when the relevant function has no lower bound on the
unit sphere, and finally proved the existence of the solution. The nonlinear eigenvalue problems studied are as follows
(1.5)
where
,
, and the function g is superlinear and subcritical,
. In this paper, we study the following fractional nonlinear eigenvalue problem of the form:
(1.6)
where
,
, for all
, with
,
if
and
if
. In this article, we set
. For all
, if
, set that
, and if
, we set that
. And set
is a
-functional
(1.7)
is a
-functional, setting
(1.8)
Using Ekeland’s
-variational principle, we prove the existence of the Palais-Smale sequence.
In this paper, the norm of fractional Sobolev space
is defined
(1.9)
and define
, for
, and
, set
is a continuous map,
(1.10)
endowed the norm on X by
(1.11)
and the corresponding inner product is
Consider the following fractional critical Sobolev space
is defined by
(1.12)
with the norm
(1.13)
where
is the completeness of
. For
, we let
(1.14)
for any
, the embedded
↪
is continuous, exist for the best fractional critical Sobolev constant
(1.15)
In this article, we list all the conditions below on
:
(H1)
is continuous and odd.
(H2)
, satisfying.
such that
Our main result is shown in the following.
Theorem 1.1. For
, under the hypotheses (H1) and (H2) Equation (1.1) admits a couple
of weak solutions such that
and
.
2. Preliminary Lemmas
Lemma 2.1. ( [11] ) For future reference note that from (H1) and (H2) it immediately follows that, for all
Lemma 2.2. Fractional-Gagliardo-Nirenberg-Sobolev inequality (see [12] ): for every
and
, there exists a constant
depending on N, p and s such that
It is equivalent to
with
.
Lemma 2.3. ( [11] ) If (H1) and (H2), and let
be arbitrary but fixed. Then we get:
1)
and
as
2)
and
as
.
Proof. Since
, we get
and through the derivation, we get
Because of
and
, we get for
Thus, we get that
as
. Similarly, when
, it can be obtained by calculation
as
.¨
Lemma 2.4. If (H1) and (H2), there exists
such that
with
Proof. Now we’re going to prove that
, there exists
,
, such that
and
. Then, for
small enough
By lemma 1.1, lemma 1.3 and
, with any
, we get
for
small enough,
thus
Next we are going to prove that
, for
, as
, we get
¨
Lemma 2.5. If (H1) and (H2), there exist
such that
1)
2)
3)
4)
Setting
with
We have
Proof. In order to facilitate readers to read better, we have written down the proof process.
First note that the existence of
is insured by Lemmas 2.1 and 2.2. Now define
with
By we have
Moreover
Therefore, if
holds, our result proves successful. This follows directly from the observation that: for
, there exists
such that
Indeed, the setting
we have, for all
and it suffices to set
.¨
Of course, if
, we get
, it is proved that ended.
Lemma 2.6. ( [10] ) If (H1) and (H2), for a sequence
, such that
Thus, there exists a sequence
such that:
1)
2)
3)
, i.e.
for all
Lemma 2.7. If we fix n, there exists a Palais-Smale sequence
for
at the level
satisfying
For the proof we recall the stretched function from [10]
Now we define
where
and
By lemma 2.7, there exists a Palais-Smale sequence
for
, that is, satisfying
and
are bounded. There exists
such that
and
as
. By lemma 2.6,
there exists
of the form
,
, such that
. Let
From lemma 2.6, we get that as
,
with
(16)
where
. Thus using the fact that
is also bounded, we see that there exists a constant
independent of n such that
From (H2) we have
As a result of
. And then since
is bounded, we can get
After sorting it out, we can get
And
.
Now the lower bound
(see (H2)), proves that
is bounded and consequently
also.
Lemma 2.8. If (H1) and (H2), there exists a sequence
such that:
1)
2)
and
are bounded in
3)
, for all
.
Proof. We claim
is a Palais-Smale sequence of the type we are looking for. Clearly
and is bounded in E. Point (1) is trivial since
. Now let
. We have
Thus, setting
where
.
We see that
If
, from the definition of
, we can get
, and then by taking the derivative, we set
, so
is equal to
, that is,
, then we have
. And by Point (3) of Proposition 2.2, if
, as n is sufficiently large we get that
By Point (2) of Proposition 2.2, for
large since
Let the minimizing sequence
, substitute into the formula. Notice that the particular choice of the minimizing sequence
is used here.
¨
3. Proof of Theorem 2.1.
Lemma 3.1. let
be the PS sequence obtained in Lemma, there exists
such that, up to a subsequence:
and
Proof. By
is bounded in
, there exists
such that
weakly in
. If
converge strongly to
, by Lemma 2.7,
is a critical point for F restricted to
. By
and (H2) there exists
such that
, we get
(3.1)
By
and
, such that
is bounded and strictly greater than zero, that is, there exists
such that, up to a subsequence
(3.2)
otherwise, if
, then we get
, as
, contradiction with
.
Similarly, from (H2), we conclude that
is also bounded and by (17) and (18), we get
by calculation as
, we can be expressed as the existence of
, such that
(3.3)
¨
Lemma 3.2.
in
.
Proof. Next, we’re going to prove that
in
, by calculation we get
with
Up to a subsequence, by the definition of S(a), lemma and (3.1), we have that
is bounded away from zero for
. Thus, there exists
such that
.¨
Lemma 3.3.
in
.
Proof. The next thing we have to prove is the compactness result for
,
in the subspace
.
Next, the function
is weakly continuous. Then for any weakly convergent sequence
, when
, we obtain
using (H2) and (2.1). Now from the compactness of the inclusion
for
if
or
if
(see [13] ), we see that
and
.
From the previous steps 1 to 3, we obtain that there is
. By
, and we deduce that
in
. And by
in
, so we prove the Theorem 1.1.¨