In this paper, we focus our attention on the following problem:
where is a bounded domain in , , and , , , and with here denotes the fractional Laplace operator defined, up to a normalization factor, by
The aim of this paper is to study the existence of solutions, we will see that if , then by concentration-compactness principle, together with mini-max arguments, we can prove the existence of solutions for (1.1). We now summarize the main result of the paper.
Theorem 1.1. Let , and with . Moreover, is bounded on . Then
1) For any , there exists , then for any , (1.1) has a consequence of weak solutions .
2) For any , there exist , then for any , (1.1) has a consequence of weak solutions .
We denote by the usual fractional Sobolev space endowed with the so-called Gagliardo norm
Then we defined
endowed with the norm
we refer to  for a general definition of and its properties.
Observe that by [  , Proposition 3.6] we have the following identity
In this work, the Sobolev constant is given by (can be seen in [  , theorem 7.58])
2. Statements of the Result
We will use a variational approach to find a solution of (1.1). Firstly, we will associate a suitable functional to our problem, the Euler-Lagrange functional related to problem (1) is given by defined as follow
To proof that J satisfy the Palais Smale condition at level c, we need the following lemma.
Lemma 2.1  Letting be a regular function that satisfies that for some
Let be a bilinear form defined by
then, for every , there exist positive constant and , such that for , one has
and . (2.5)
To establish the next auxiliary result we consider a radial, nonincreasing cut-off function
Lemma 2.2.  Letting be a uniformly bounded in and the function defined in (2.6). Then,
Lemma 2.3.  With the same assumptions of Lemma 2.8 we have that
where B is defined in (2.4).
Lemma 2.4.  (Minimax principle) Assume that , and is a family of nonempty subset of X, denote
If the following conditions holds:
1) c is a finite real number;
2) there exists an , such that is invariant with respect to the family of mappings;
that is, for any , there holds
Then, E possesses a sequence at level c define as (6.1.1); Furthermore, if E satisfies the condition (or the condition at level c), then c is a critical value of E.
3. Proof of Theorem 1.1
Firstly, recalling that J is said to satisfy the Palais Smale condition at level c if any sequence such that and has a convergent subsequence.
Lemma 3.1. The sequence for J is bounded.
Proof. Note that satisfies
where as . Choose as test function in (3.2), we get that
therefore, by (3.1) and (3.2), we have
which yields the boundeness of in ,since .
If , then for , similar to the proof of , we get
Which also yields the boundedness of sequence .
Lemma 3.2. Assume that . Then
1) For any , there exists , such that for any , then J satisfies .
2) For any there exists such that for any , then J satisfies .
Proof. By Lemma3.1 is bounded in , up to a subsequence, we get that
, . (3.5)
Following  it is easy to prove that could also be the -norm. Applying [  , Theorem1.5], we have that the exist an index. Set a sequence of point and two sequences of nonnegative real numbers , such that
in the sense of measures, with
for every (3.8)
here denotes the Dirac Delta at , while is the constant given in (1.7), we consider a nonincreasing cut-off function satisfying
Set taking the derivative of (1.6), for any . We obtain that
Then, taking as a test function in
by (3.10), we have
therefore, by (3.5) (3.6) and (3.7) we get
Since is regular function with compact support, it is easy to see that it satisfies the hypothesis of Lemma 2.1, by Lemma 2.2 and Lemma 2.3 applied to the sequence , it follows that the left hand side of (3.13) goes to zero. We obtain that
Clearly, if , we get ; if , by (3.8), we get or .
suppose that , we know that
according to the embedded theorem, we have
This yields that
Thus, if , we get that
However, if is given, we can choose so small for every that last term on the right-hand side above is greater than 0 which is contradiction when
is the same as greater than 0. We see that cannot occur if or are choose properly. Thus . As consequence, we obtain that in , that is . This implies convergence of in . Finally using the continuity of the inverse operator . We obtain strong convergence of in . #
Next, by using the classical concept and properties of the genus, we construct a min-max class of the critical point.
For a Banach space X, We define the set
For , define
If there is no mapping as above for any , there . we refer to  for the properties of the genus.
Proposition 3.3.  Let ,
1) If there exists an odd map , then ;
2) If , then ;
4) If S is a sphere centered at the origin in , then ;
5) If A is compact, there exists a symmetric Neighborhood N of A, such that .
According Holder inequality, we get that
We define the function
Then it is easy to see that given , there exists so small that for every , there exists such that for , for . and . Analogously, for given , we can choose with the property that as above for each . Clearly, .
As in  , Let be a nonincreasing function such that if and . if . Set , we make the following truncation of the function J:
It is clear that and is bounded from below.
Lemma 3.4.  1) For any and or any and , if , then and .
2) For any , there exists such that if and then satisfies .
3) For any ,there exists such that if and then satisfies .
Lemma 3.5. Denote . Then for any , there is such that .
Proof. Denote by the closure of with the respect to norm , in . Extending functions in
by 0 outside . Let be a m-dimensional subspace of . For any . We write with and . From the assumptions of , it is easy to see for every with that there exists such that
For . Since all the norms are equivalent, we get
Therefore for given and . we can choose sufficiently small so that .#
Let . Then , Hence by proposition 3.3 (2) and (4) .
We denote and let
because and is bounded from below.
Proposition 3.6. Let be as in Lemma 3.5 (2) and (3). Then all given by (3.25) are critical values of and as .
Proof. Denote . Then by Lemma 3.4 (2) and (3), if , is compact. It is clear that . By (3.26) . Hence . Moreover, since satisfied, it follows from a standard argument (see  ) that all are critical values of . Now, we claim that . If because is compact and , it follows from Proposition 3.3 (5) that and there exists such that . By the deformation Lemma  , there exists and an odd homeomorphism such that
Since is increasing anad converges to . there exists such that
And exists a such that
By Proposition 3.3 (3), we obtain
By Proposition 3.3 (1), we obtain
consequently, from (3.28), we get
on the other hand, by (3.27) and (3.29)
which implies that
this contradicts to (3.32).Hence . #
By (1) of Lemma 3.4 if . This and Proposition 3.6 give Theorem1.1.
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