Research Article | Open Access
Huxiao Luo, Shengjun Li, Xianhua Tang, "Nontrivial Solution for the Fractional -Laplacian Equations via Perturbation Methods", Advances in Mathematical Physics, vol. 2017, Article ID 5317213, 9 pages, 2017. https://doi.org/10.1155/2017/5317213
Nontrivial Solution for the Fractional -Laplacian Equations via Perturbation Methods
We study the existence of nontrivial solution of the following equation without compactness: , where , , is the fractional -Laplacian, and the subcritical -superlinear term is 1-periodic in for . Our main difficulty is that the weak limit of (PS) sequence is not always the weak solution of fractional -Laplacian type equation. To overcome this difficulty, by adding coercive potential term and using mountain pass theorem, we get the weak solution of perturbation equations. And we prove that as . Finally, by using vanishing lemma and periodic condition, we get that is a nontrivial solution of fractional -Laplacian equation.
This article is concerned with the fractional -Laplacian equationswhere , , and satisfies the following conditions. , is 1-periodic in for , and uniformly in for some , where ; as uniformly for ; if , and there exists and such that where .
The fractional -Laplacian is defined on smooth functions by This definition is consistent, up to a normalization constant depending on and , with the usual definition of the linear fractional Laplacian operator when . There is, currently, a rapidly growing literature on problems involving these nonlocal operators. This type of problem arises in many different applications, such as continuum mechanics, phase transition phenomena, population dynamics, and game theory, as they are the typical outcome of stochastically stabilization of Lévy processes; see [1–9] and the references therein. The literature on nonlocal operators and their applications is very interesting and quite large; we refer the interested reader to [4, 10–21] and the references therein. For the basic properties of fractional Sobolev spaces, we refer the interested reader to [22, 23].
The main purpose of this paper is to consider the existence of nontrivial solutions for equation (1). Our main difficulty is that the weak limit of (PS) sequence is not always the weak solution of (1). To overcome this problem, we apply the perturbation method [22, 24–26]. First, we consider the perturbation equation by adding coercive potential termwhere is a parameter and satisfies the following conditions:, ., .
And we prove that the energy functional of (5) has the geometry of the mountain pass theorem that it satisfies the Cerami condition and finally that the obtained solutions have the uniform bounds. Finally, we verify that as and is the nontrivial solution of (1).
Now, we give the main result of this article.
Theorem 1. Suppose that – hold. Then (1) possesses at least a nontrivial solution.
Remark 2. In order to get our result, there are mainly three difficulties. (i)The working space has not compactness.(ii)The classical AR condition for the nonlinearity is not satisfied.(iii)If is a Palais Smale sequence of (see Section 2) and converges weakly to , one can not obtain that is a weak solution of the fractional -Laplacian type equation (1).
Notation 1. In this paper we make use of the following notation: (i) is the usual norm of the space .(ii), and , denote positive (possibly different) constants.(iii)We denote the weak convergence in and its by “” and the strong convergence by “”.(iv) denote being infinitely small (possibly different) when .
2. Variational Framework
Before stating this section, we define the Gagliardo seminorm by where is a measurable function. On one hand, we define fractional Sobolev space by endowed with the norm where Moreover, (1) is variational and its solutions are the critical points of the functional defined in by From , it is easy to check that is well defined on and , and
On the other hand, we consider the fractional Sobolev space endowed with the norm We also need the following inner norm: and let Obviously, we have the two norms and are equivalent. Next, the following lemma discusses the continuous and compact embedding for for all . For the proof of the lemma, it was proved in  in the case . For the general case, the proof is similar. We give it here for readers convenience.
Lemma 3. Assume that and hold. Then is continuously embedded in for all . Moreover, can be compactly embedded into for all .
Proof. Let be a bounded sequence of such that in . Then, by Theorem in , in for . We claim thatTo prove (16), we only need to show that, for any , there exists such that Set then Now choose such that ; then we have Since is bounded and condition holds, we may choose , large enough such that and are small enough. Hence, , we have from which (16) follows.
To prove the lemma for general exponent , we use an interpolation argument. Let in , we have just proved that in . That is, as . Moreover, because the embedding is continuous and is bounded in , we also have Since , there is a number such that Then by Hölder inequality This implies in .
From Lemma 3, there exists such that where denotes the usual norm in for all .
Next, we define the energy functional on by We also need the following inner norm: by Lemma 3, we have that the norms and are also equivalent. By (), (), the energy functional is well defined and of class . Moreover, the derivative of is for all .
In what follows, we give the vanishing lemma which is introduced by Lion.
Lemma 4 (see ). Assume is a bounded sequence in which satisfies for some . Then
3. Proofs of the Main Result
The proof of Theorem 1 is divided into several lemmas. We show that the functional has the geometry of the mountain pass theorem that it satisfies the Cerami condition and finally that the obtained solutions have the uniform bounds.
Lemma 5. Suppose that , , – are satisfied. Then there exists , , such that for fixed , where and are independent of .
Proof. For any , it follows from and that there exists such that where , and then For , let So, from the Sobolev inequality, one has So one has, for , since . Hence, by fixing and letting be small enough, it is easy to see that there is such that this lemma holds.
Lemma 6. Suppose that , , – are satisfied. then there exists with such that for fixed , where is given by Lemma 5.
Proof. Using , we obtain there exists such that Next, for we have This implies for all . Since is arbitrary, by the above inequality, we get Consequently, as . Hence, let be big enough and ; then we have ; we complete the proof.
Definition 7. We say that satisfies Cerami condition in , if, for any sequence such that as , there exists a convergent subsequence of .
Lemma 8. Suppose that , and – are satisfied. Then the functional satisfies Cerami condition.
Proof. Let be a sequence in so thatWe shall prove that contains a convergent subsequence.
(i) We claim that is bounded in . Observe that for largeArguing indirectly, assume by contradiction that ; then . Set ; then . By Lemma 3, one has for . Observe that from (42) and it follows that Set for By , for all , and as . For let Since if , one has and It follows from (43) that Invoking , set and . Since one sees . Fix arbitrarily . Using (49), as uniformly in , which implies by Hölder inequality that as uniformly in . Using (49) again, for any fix , as .
Let . By there is such that for all ; consequently, for all . By , (43), (51), and Hölder inequality we can take large so that for all . Note that there is independent of such that for . By (52) there is such that for all . Now the combination of (53), (55), and (62) implies that for which contradicts (45). Hence is bounded in .
(ii) By (i), we can conclude that is bounded in . Going if necessary to a subsequence, we can assume that in . From Lemma 3, we have in for all . By the boundedness of in , we have By Hölder inequality and the above inequality we also have Similarly, By , , for , there exists such that Then using Hölder inequality we have Since is bounded in and is arbitrarily small, we haveBy (62) andwe havewhere we have used the following elementary inequality: where the constant is independent from the variable and . Recall that , as ; it is clear thatFrom (64), (66), we have as . Therefore, satisfies Cerami condition.