Ground State Solutions for a Nonlocal System in Fractional Orlicz-Sobolev Spaces

El-Houari, Hamza; Moussa, Hicham; Chadli, Lalla Saâdia

doi:https://doi.org/10.1155/2022/3849217

International Journal of Differential Equations

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Research Article | Open Access

Volume 2022 | Article ID 3849217 | https://doi.org/10.1155/2022/3849217

Ground State Solutions for a Nonlocal System in Fractional Orlicz-Sobolev Spaces

Hamza El-Houari,¹Hicham Moussa,¹and Lalla Saâdia Chadli¹

Academic Editor: Sining Zheng

Received19 Oct 2021

Revised18 Jan 2022

Accepted16 Mar 2022

Published20 May 2022

Abstract

We consider an elliptic system driven by the fractional -Laplacian operator, with Dirichlet boundary conditions type. By using the Nehari manifold approach, we get a nontrivial ground state solution on fractional Orlicz–Sobolev spaces.

1. Introduction

We consider the following fractional elliptic systemwhere is a bounded open subset of with Lipschitz boundary , , and is the nonlocal fractional -Laplacian operator of elliptic type introduced in [1] and is defined asfor all , where is the principal value and are of class and satisfies the following conditions: is strictly increasing.

For example, when we choose and for , then satisfies and . In this case, the operator (2) is named fractional -Laplacian operator and reads asand the system (1) reduces to the fractional -Laplacian system studied in [2] and given as

The existence of solutions for systems like (3) has also received a wide range of interests. For this, we find that, in the literature, many researchers have studied this type of system using some important methods, such as variational method, Nehari manifold and fibering method, and three critical points theorem (see for instance [3–6]). This kind of operator can be used for many purposes, such as phase transition phenomena, population dynamics, and continuum mechanics (see for example [7–9]). It is natural to think about what results can be obtained when the -Laplace operator is changed to a fractional -Laplacian operator. From what we know is, there is a number of results obtained about the fractional Orlicz–Sobolev spaces, such as a nonlocal Kirchhoff problem (see [10]):where is the Kirchhoff function and is an function, i.e., is a convex function such that

In a similar way, Fernández Bonder and Salort in [11] studied the following problem with a nonlinear nonhomogeneous reaction term:where is a Carathéodory function. Therefore, in [12], we have investigated the following type of fractional -Kirchhoff systemwhere the functional is defined by

To the best of our knowledge, this is one of the first works devoted to the studies of the existence of a solution to the nonlocal problem (1) by using the suitable Nehari manifold method. The main difficulty in this work arises from the complicated method which we are applying to find the solution. Furthermore, the nonhomogeneities of the operator (2).

To state the outcome, we assume the following condition on the function : :where . We can notice that by applying arguments as in [13], we arrive at which implies the condition :

: there exists such thatand satisfies that : is a function such that for all ,: x .

: the functionsare increasing on and , respectively.

This work is structured as follows. In Section 2, we recall briefly some properties on Orlicz and fractional Orlicz–Sobolev spaces. Section 3 is related to specify the assumptions on the data and to show the existing results of the problem (1) and its proof based on Nehari manifolds.

2. Some Preliminary Results

The reader is referred to [1, 14–17] for more details on Orlicz and fractional Orlicz-Sobolev space.

We point out that an function . If for a certain constant ,

The relationship-related and is given by , where is the complementary to . We assume that

Then, we introduced the conjugate function , given by the following expression of its inverse in :

Let and be two functions. The notation means that, for each ,

The Orlicz space is defined as follows:

The usual norm on is .

We recall that the Hölder inequality holds

And the Young inequality reads as follows:

After this, we list a few inequalities that will be used for our proofs. The proof is provided in [18].

Lemma 1. Let and , and is an function; then, these assertions are equivalent as follows:

Lemma 2. If is an function satisfies (22), then we have

Lemma 3. Let be the complement of and , , where and . If is an function and (22) holds, then satisfies

Lemma 4. We have , i.e., .

Remark 1. By Lemma 1, Lemma 3, and (22), we show that .
We now look at the definition of the fractional Orlicz–Sobolev spaces (see[1]), and we set some properties on these spaces. The fractional Orlicz–Sobolev spaces defined as follows:where in and . This space is equipped with the normwhere is the Gagliardo seminorm, defined byTo deal with the problem under consideration, we chooseIn these spaces, the generalized Poincaré inequality reads as follows (see [19]):where is a positive constant. Then, is a Banach space whose norm is equivalent to . Also it is a separable (resp. reflexive) space if and only if (resp. and ). Furthermore, if and is convex, then the space is uniformly convex Theorem 1 (see [1]).

Theorem 1 (see [19]). Let be an function , , and regularity with bounded boundary. If , , and hold, then the embedding is continuous, and the embedding is compact for any N function .

Remark 2. The assumption implies that there is such that ; then, by Lemma 4 and Theorem 1, the following embedding are compact. i.e., there exists a constant such thatNow, we defined our working space under the normSo, by the above arguments, is a separable and reflexive Banach space.

3. Nehari Manifolds in Fractional Orlicz–Sobolev Spaces

We observe that the energy functional on corresponding to system (1) iswhere , and is the function defined in (9). Now, let us prove that the energy functional is well defined. To do this, we need to see the following:

Remark 3. By (13) and the fact thatwe show thatwhere .
The fractional -Laplacian operator specified in (2) is indeed defined between and its dual space . Indeed, in [1], Theorem 6.12, the following expression is found:for all . Throughout the document, we note by and (i.e., regular Borel measure on the set ).

Lemma 5. The function is well defined, and it is the , and we havefor all .

Proof. First, we can see thatfor all . It follows from (39) for each that . Next, we prove that . On the one hand, let with strongly in , for ; we have and by Hölder inequality,On the other hand, if in , then in , so by Dominated convergence theorem, there exists a subsequence and a function in such thatAndAgain by dominated convengeance theorem, we obtain thatCombining (43) in (40), we get is continuous. Now, we turn to prove thatBy (32) and (36), we haveThen, is well defined in . Now, by (13) and (32) and the similar argument in [[20] Lemma 3.1,] we see that (44) is holds. This completes the proof.
To find the critical points of , we will minimize the energy functional on the constraint of Nehari manifoldThe main result of this section is as follows:

Theorem 2. We assume that - and - hold true. Then, the system (1) has a nontrivial ground state solution in the way that there is such thatwhere .

Lemma 6 (see [16] Lemma 4.1). The following properties are true:

Lemma 7 (see [21]). We suppose that is convex, in and . Then, . Similary, we can obtain that, in . Therefore, in .

Lemma 8. We suppose that -, (22), and hold true. Then, we have

Proof. Let be sequence in such that . By (13) and Hölder inequality, we haveSince in , it is clear that in and in , then and are bounded. By Remark 2, there exists such thatCombining (50) and (51), then we haveHence, the compact embedding (see Remark 2) make sure that in ; then, for any positive , we can choose such thatFurthermore,for all and for small . Since is an arbitrary one, by joining (52) with (54), we concluded that item helds. With a similar discussion to the one above, we can show that is proven to be true.

Lemma 9. We assume that -, -. Then, there exists a constant such that for each .

Proof. The proof is augmented by contradiction. We suppose that there exists a subsequence denoted by such that , for each integer . Using , the construction of (see (46)), (13), the Hölder inequality, and (32), we havewhere .
According to Lemma 1, Remark 2, and for large enough, last inequality rewritesDividing the last expression by , we get toThe fact that , we get that for a large ,which is a contradiction.

Lemma 10. We assume that - and - hold true. Then, the function given byis and .

Proof. We setSo, by definition of , we have andFurthermore, there is with such thatThus,We recall that by :and as a consequence,Using - and , we haveIndeed, using Young’s inequality, (15) and the fact , we infer thatHence,Now, let us see that is continuous. Let be a sequence such that .
Then,By arguments as above,where . By (64) and (65), we get thatApplying the Lebesgue dominated convergence theorem,Hence, we have which completes the proof of Lemma 10.
At this step, in view of determining the behavior of on , we introduce the fibering map associated to the Nehari manifold given byUsing (38), then the first derivative of the map is given by

Lemma 11. We suppose that - and - hold. Then,

Proof. By equation (36), there exists such thatfor a given , and by Poincaré’s inequality and Lemma 6, we haveUsing Lemma 2.3 in [10] and (23), we infer thatWith the above arguments, we have for all ,We note that , so the last inequality (79) rewrites as follows:Hence,Now, for item (ii), by using (23), we haveMoreover, we can see thatIn fact from the condition and (83), we infer thatCombining (84) in (82), we proved item (ii). Now, for (iii), we first show that, by equation (13), there exists such thatNow, using and (85) in (78), we haveIn fact from Lemma 2.3 in [10] and (23), we infer thatThen,Since , we conclude the result for item (iii). For the last item, we haveUsing (35), we haveWe can easily see thatAs a consequence of , we infer thatIt follows using (91) and (92) and inequalities (90) thatUsing F₂, we conclude thatAt this point, last limit just above and (89), we infer the item (iv).

Lemma 12. We assume that - and - hold true; then, for each , there is an only such that . Moreover, for each .

Proof. On the one hand, let . By Lemma 11, we have for small enough and for large enough. On the other hand, the map is continuous, and there is at least one number such that , which means that . Now, let us see that there is an only such that . Using (78), we haveIndeed,We remark that implies thatwhere ; then, we haveUsing inequality (98) in (97), we infer thatHence,It can be seen thatAccording to and (101), we getTherefore by (102), we have that is a decreasing function that vanishes once in . So, there exists a unique such that . Therefore, the function has one critical point, namely and . Furthermore, by Lemma 11, it follows that is a maximal point of on and, in fact , implying that . The arguments above also show that , for every . And finally, because if and only if , we conclude that for every . This completes the proof.

Proposition 1. We assume that -- hold true. Then is a -submanifold of , i.e., any critical point of is a critical point of .

Proof. We consider the functional defined bywhereAccording to Lemma 10, we can see that , and by using equation (59), we show thatWe also show thatNow, we setBy using Lemma 10, it follows that . Furthermore, from Lemma 11, we infer that is the global maximum of , and also in the proof of Lemma 12, we see thatUsing the fact that and 0 is a regular value for , the set is a -submanifold of . Now, we assume that is a critical point of . According to the theorem of Lagrange multiplier, there exists a real constant such thatUsing (108), we infer that . Therefore, so that is a free critical point of . This completes the proof.

Lemma 13. We assume that -, - hold. Let be a minimizing sequence of over the Nehari manifold . Then, is bounded in .

Proof. Let be a minimizing sequence that is and . To prove the boundedness of , we argue by contradiction and consider that there exists a subsequence of , always denoted by such that . We approach the problem in two cases.

Case 1. We suppose that and also . Let and . Then, is bounded in separable and reflexive Banach space . By Theorem 1 and Remark 2, there exists a point such that(i) in ; and in a.e in (ii) in and in a.e in We claim that has nonzero Lebesgue measure. Firstly, we assume by the way of contradiction that both and have zero Lebesgue measure, that is in and in . Since , thenLet be a constant. Now, we observe thatNow using , , and the continuous of the function , we infer thatCombining (111), (112), and using Lemma 6, we getPassing to the limit in (113), we getwhich is impossible. Therefore, and has nonzero Lebesgue measure. It is remembered that we are assuming that and . Hence,
On the one hand, employing Lemma 1, we havePassing to the limit above, we haveOn the other hand, applying Fatou’s lemma and the fact that and has nonzero Lebesgue measure, we haveIt is clear that when . Then, by using and last inequalities just above, we concludewhich is impossible.

Case 1. We suppose that or for some and all . Without loss of generality, we assume that and , for some and for all . Let and , then and . By Theorem 1 and Remark 2, there exists a point such that(i) in and in a.e in (ii) in ; and in a.e in By similar argument in Case 1, we prove that has nonzero Lebesgue measure. It is recalled that we are assuming that and . Hence,Applying Lemma 1 and the last equation just above, we getPassing to the limit to the inequalities just above, we haveIn other way, using Fatou’s lemma and the fact that , we haveIt is clear to see thatAccording to , (123), and last inequalities just above, we concludewhich is impossible. Thus, is bounded in . The proof is complete.

Lemma 14. We assume that -, - hold. Then, there exists such that

Proof. Let be a minimizing sequence of over the Nehari manifold . By Lemma 13, there exists such that(i) in and in (ii)We claim that . We assume on the contrary that . The fact that and by using , we haveApplying Lemma 8, we getAs a consequence of Lemma 6, which is contradicting Lemma 9. That proves the claim.
Accordingly, from Lemma 10, we have that is weakly lower continuous. Hence,We recall that . By Lemma 12, there is such that . Hence, .(i)We claim that so that is in .Furthermore, by contrary, we assume that . Then,where , and Now, we can see by using thatAgain by using , we havewhich impliesThen, we infer thatand then the functions and are increasing in and , respectively.
Moreover, the functionsThen, we conclude thatNow, using the weak lower continuity of the functionsand the continuity of the function , we infer that which is impossible. Thus, , and then . This finishes the proof. Now, we have all tools to prove our result.

3.1. Proof of Theorem 2

Let be a minimizing sequence for over . By the proof of Lemma 13, there is such that

Applying Lemma 7, we get

Since Lemma 5 infer that , it follows that . By Lemma 14, and

By Proposition 1 the set is a submanifold of so that is a critical point of . Again Proposition 1 shows that is a critical point of .

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest.

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Copyright © 2022 Hamza El-Houari et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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