In recent years, many people pay attention to the fractional Laplacian. One of the reasons for this comes from the fact that this operator naturally arises in several phenomena like flames propagation and geophysical fluid dynamics, or in mathematical finance. About the Fractional Sobolev space we can refer   . In this work, we consider the problem
where , , is a bounded domain with Lipschitz boundary. as the fractional Laplacian, which defined as
It is worthy to point out that
we can refer  .
For , we can also define the fractional Laplacian as the operator given by the Fourier multiplier ,that is, for
where we denote by the class of all Schwartz functions in .
We introduce the Sobolev space
and the space
endowed with the norm
where , . This space allows us to deal with the problems proposed in a bounded domain , as we need. The pair yields a Hilbert space  . Moreover, it can be seen that
is a continuous operator.
Theorem 1.1. Let be an increasing locally Lipschitz continuous function. Let . Then (1.1) have a unique solution . Moreover,
In this section, we give some basic results of fractional Sobolev space that will be used in the next section.
Definition 2.1 We say that is a weak solution to (1.1) if we have
for any .
Lemma 2.1.  Let and .Then for all we have
where is the constant defined in (1.3).
Proof. Fixed y we change coordinates and apply Plancherel.
Recalling that we obtain
The integral in brackets is of the form , with
where is the Bessel function of the first kind of order , we can refer  .
Recall that . The formula (1.3) for now follows from
for , we can see  .
Lemma 2.2.  For , , there exists a positive constant , for any ,we have
where is called fractional critical Sobolev exponent. In particular, if then
Lemma 2.3. (Egorov’s theorem)  Let be a sequence of functions and f be a function defined on E, with . Assume that a.e. in E. Then for every there exists a measurable subset A of E such that and uniformly on A,as .
Lemma 2.4. (Vitali)  Let be a sequence of functions and f be a function in . Assume that
1) a.e. in ;
2) if E is a measurable subset of , and we have
uniformly with respect n. Where means measure representing E. Then in .
Proof. Fixed , let be a measurable set, we have
Using assumption (2), we know that there exists such that, if , then for any we have
Since there exists such that if , then
In conclusion the second term of the right-hand side of (2.18) is less than . Let us study the first one. We set , and use Egorov’s theorem, there exist and a measurable set such that , and
for any .Choosing in(2.18), we get the result.
Lemma 2.5. (Stampacchia)  Let H be a Hilbert space, is a continuous and linear form in the second variable such that
1) for ,any , we have
2) for a positive constant C, any we have
Lemma 2.6. (Hölder inequality)  Let p and q are dual indicators, stisfies
where , if , and , then the product of the defined function belongs to , and we have
If and only if there is a real constant m that makes the following formula hold
The first unequal sign of (2.24) is established. If f not constant equals 0,then the second unequal sign of (2.24) is established, if and only if there exists a constant , such that
1) if , then .
2) if , then , and when , we have .
3) if , then , and when , we have .
3. Proof of Theorem 1.1
Theorem 3.1. Let be an increasing function, and g is Lipschitz continuous, that is, there exists a positive constant such that for any we have
Let . Then (1.1) exists a unique solution .
Proof. We define the following form on :
Using Hölder inequality and (3.1) we have
that is, a is well defined. By the definition of a, we know that a is continuous and linear in the second variable. If in , then
the last inequality following from Hölder inequality and (3.1), by lemma 2.2
We know that a satisfies lemma 2.4 from (3.2) and (3.10), the result follows from lemma 2.4.
We define the following function, for :
Proof of theorem 1.1: First, we proof the existence of a solution by approximation. Let , By theorem 3.1 we know that there exists be the solution to problems
We use as a test function in (3.12), we get
Then use Hölder inequality on the right-hand side implies
Because g is increasing, then . This means is uniformly bounded. We can deduce there exists weakly in and a.e., since , by (3.13) there exists a positive constant C such that
for every n.
Now we prove in . Since g is continuous in then it is clear that a.e. in . If E is a subset of , for have
combining (3.14), for we have
Using lemma 2.4, we know that in . Then for any
Finally we prove the solution of problem (1.1) is unique. We assume and are two solutions, , we take as a test function
We can deduce from (3.19) and (3.20)
By the monotonicity of g we know
Combining (3.22) and (3.23) we know a.e. in .
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