The Numbers of Positive Solutions by the Lusternik-Schnirelmann Category for a Quasilinear Elliptic System Critical with Hardy Terms

Abstract

In this paper, we study the quasilinear elliptic system with Sobolev critical exponent involving both concave-convex and Hardy terms in bounded domains. By employing the technique introduced by Benci and Cerami (1991), we obtain at least distinct positive solutions.

1. Introduction and Main Result

In this paper, we are concerned with the multiplicity of positive solutions of the following critical problem:where is a smooth bounded domain of , , , , is the critical Sobolev exponent, where is the best Hardy constant, and the parameter , , we assume that and where the weight functions and satisfy the following conditions:   with , where and

And the function satisfies the following conditions: , such that  , where   are strictly increasing functions about and for all ,   with   for some constant

Remark 1. We deduce form the conditions , , and that the functional is of class andwhere , and such that and
Moreover, there exists such thatThe proof is almost the same as that in Chu and Tang [1].

Recently, many papers have studied the multiplicity of positive solutions by way of fibering method and the notions of topological indices category for different semilinear, quasilinear, and nonlocal problems involving a critical exponent and concave and convex nonlinearities (see [2–4]). Our goal here is to give a new result for this system by linking the number of positive solutions with the topology of the domain More precisely with the Category index, let us note is the least number of closed and contractible sets in which cover Our main result is the following.

Theorem 2. Let and Suppose that satisfies and the functions , satisfy the condition Then, there exists such that if for each , problem (1) has at least distinct positive solutions.

This paper is composed of four sections. In Section 2, we give some results for the ehari manifold associated of the energy functional and fibering maps. In Section 3, we will build homotopies between and certain sublevel set of the energy functional associated with (1). Finally we prove the result in Section 4.

2. The ehari Manifold Associated with the Energy Functional and Fibering Maps

Let the Sobolev space with the usual norm: Also, the standard norm of the space is Moreover, a pair of functions is said be to a weak solution of problem (1) if for all

We know that looking for weak solutions of (1) is like looking for the critical points of the associated functionalwhere

By the above Remark 1, the functional is well defined on the space and is of class .

Therefore, the solutions of (1) correspond to critical points of Let us denote by the ehari manifold related to , given byNamely,Notice that the functional is not bounded below on the total space for that we consider the functional on the ehari manifold.

Define Let , and by easy calculation we have

Lemma 3. The functional is bounded below on the ehari manifold

Proof. Let , and applying the Hölder inequality and the Sobolev embedding theorem, Young inequality, and Condition we haveand we deduce Thus, is coercive and bounded below on

Now, we split the ehari manifold into three parts, namely, Then, we have the following results.

Lemma 4. Let be a local minimizer of and Then is a critical point of

Proof. The proof is standard; you can see [4].

Lemma 5. There exists such that for all such that then

Proof. Suppose the contrary; that is, there exist with , but Let ; we haveandBy and applying the Minkowski inequality and the Sobolev embedding theorem, we have soCombining (15) and (18), we have thenBy (12) we have thenWe deduct from (20) and (22) that which is a contradiction.

So, we have , and we define

Lemma 6. (i) For some and for so, there exists sequence of for
(ii) If , then there exists a sequence of for

Proof. You find the same proof in the following reference [5].

DenoteWe define a cut-off function such that for , for , , and For , let From Li Wang, Qiaoling Wei, and Dongsheng Kang [6], we havewhere , and verifying , this

Lemma 7.

Proof. Set and and , where , , and Then by , the definition of , and (27), we have We use the following relation: For the reverse inequality, the application of the mountain pass theorem gives us a Palais-Smale sequence for at level and from here we can show that is bounded in using standard arguments. SinceAssuming that , we find From definition (25) of , we get thenSince implies , we deduce from (36) thatThen from (31) and (37) we obtain

Next we prove that satisfies the Palais-Smale condition under some level. Before, we need the following lemma.

Lemma 8. Let with and for some    Let be a bounded sequence in , such that weakly in Then as

Proof. (The idea of this proof was borrowed from [7])

Lemma 9. satisfies the condition forwhere is independent on and

Proof. The proof is similar to that of Lemma 2.1 in [8].

Let , with , and putThen the following lemma holds. Its proof is similar to the lemma [4] (or see Tarantello [9]).

Lemma 10. Let , with , so there are unique number positives and such that with and

Lemma 11. For some , and such that , we have

Proof. First, we claim that there exist positive constants independent of such thatLet and We obtainThen, by and (27) we deduct thatthen is bounded above as Using Lemma 10, we have then we can also suppose that is bounded below. By a direct calculation we haveand the constant is a positive. So We have , and there exist such thatLet By the following relation, for and , we have , and we obtain By (51) we have Then, there exists such that , and we haveso by definition we deduct that For the case , so we get the same result.

For the existence of the first solution of our problem (1)

Lemma 12. There exists such that if , then has a minimizer and its satisfies is a positive solution of (1).

Proof. Taking into account the fact that and Lemma 11 we have Hence, for the proof of (i) just use the following Lemmas 11 and 9. Now let be solution of problem (1) such that Moreover, we have In fact, if , by Lemma 10, there are unique , such that and In particular, we have Since there exists such that By Lemma 10which is impossible and by the maximum principle, we deduct that is a positive solution of problem (1).

3. Some Technical Results

Lemma 13. Let and decreasing sequences in for some and converging to , so

Proof. By Lemma 6 there exists a sequence , such thatThere exists a real number sequence satisfying SoSince, by Lemma 10 for all we haveMoreover, implies that and we deduct that and We get that is bounded in
We can now say that is a bounded sequence, if we assume by contradiction that We find , so which implies by (60) that which is a contradiction with (63).

We consider the following lemma. See section 5.3 in [10].

Lemma 14. Suppose that is Banach space and Assume that, for and , (1)F satisfies the condition for ,(2) Then has at least critical points in

Let us consider tow subset of Note that and are homotopically equivalent to for some We may assume We considerRecall that implies that is extension in with outside of Let asWe denote bySimilar to , can be shown to satisfy restricted versions of the same three Lemmas 7, 9, and 11. We considerLet for all , that is,

and the map given by

with radial. For all Then can be rewritten

Remark 15. and Next, we define the map by

Lemma 16. For some such that we have

Proof. We show by the absurd that there exist sequence of , , and such thatand let (up to a subsequence of ). By Remark 15, we have and Standard calculations show that is bounded in and by this we obtainandas We have by Lemmas 13 and 7 and its restricted version for that then by (83), (84) and (85)Now, it is easy to see that the sequence given byverifies For a subsequence of we have By the same way used in [[10],Lemma 1.40] (see also [7] ), we obtain and Since It is easy to confirm that and are equal either to or to Since is independent of and is never achieved except when (see also [11]), so necessarily . Then the measure is concentrated at a single point of ,
and we haveTherefore and this is impossible.

Lemma 17. For some such that , we have

Proof. This classic proof is omitted for brevity. An identical proof can be found in [12], Lemma 14.

4. The Proof of Theorem 2

Denote by the restriction of on

Lemma 18. If , for some , so the functional verifies the Palais-Smale condition for

Proof. By [[10], Proposition 5.12], there exists a sequence . If is a for at level , there exists a sequence such thatwhere Recall that , so
If ,andBy the same argument employed in Lemma 5, we get andand we deduct that This is contradiction.
Moreover we assume that , as Since , so as then Thus,then by Lemma 9 the proof is finished.

Lemma 19. For some such that if , then every critical point of is a critical point of in

Proof. For the proof of this lemma, it is similar to Lemma 18.

Proof of Theorem 2. Applying Lemmas 9 and 12, satisfies condition for all Then, by Lemmas 17 and 14, admits at least critical points in Hence, we deduce from Lemma 19 that has at least critical points in Moreover, , at least critical points in

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.