Abstract

We focus on the following elliptic system with critical Sobolev exponents:  ; ; , where , either or , and critical Sobolev exponents and . Conditions on potential functions lead to no compact embedding. Relying on concentration-compactness principle, mountain pass lemma, and genus theory, the existence of solutions to the elliptic system with or will be established.

1. Introduction and Main Result

We aim at the existence of nontrivial weak solutions for the following elliptic system: where , and either or .

Discussion regarding critical growth has been dominated research in various branches of mathematical physics. Here, we quote a few literatures on critical Sobolev exponent and their applications; see Brezis and Nirenberg [1], Wu [2–4], and Garcia-Azorero et al. [5] and the references therein.

By applying variational arguments, Yin and Yang in [6] proved the existence of at least one or two positive solutions to the following system: where are sign-changing functions.

Subsequently, at least one nontrivial solution was established in the work [7]: via mountain pass lemma. More results can be found in [7–10].

In papers cited above, the authors are fond of supposing that potential functions satisfy such conditions:()Functions and satisfy with some constant . Moreover, for every , , where “meas” denotes the Lebesgue measure in .(), , as .

We point out that one of the above conditions has been used in many papers to guarantee the compact embedding for each . Nevertheless, throughout the paper, we are going to weaken the conditions and give assumptions as follows:()Let nonnegative function be homogeneous with , where and either or . Furthermore, there exists such that and for all ()Functions and satisfy with some constant . In addition, for every , , where “meas” denotes the Lebesgue measure in .()Function is positive and with , . Moreover, .()Function is positive and with , .

Condition used in our paper is weaker than which leads to no compact embedding of . As far as the authors are aware, there is less literature dealing with such potential functions. Moreover, a more generalized function appeared in the paper.

In order to state our main result, let us recall some Sobolev spaces and norms. Let , be subspaces of , with the normsObviously, the norm of reflexive Banach space is given byDefineCorresponding functional of (1) defined in isThe main results are enunciated as follows.

Theorem 1. Suppose , , and with hold; then there exists such that if , system (1) has at least a nontrivial solution.

Theorem 2. Suppose , , and with hold; then there exists such that if , problem (1) has infinitely many solutions with , , and as .

This work is structured as follows. Variational framework will be set up and some useful lemmas will be derived in Section 2. The proofs of Theorems 1 and 2 are established in Sections 3 and 4, respectively.

2. Preliminaries

The main purpose of this section is to present that functional satisfies condition. Besides, some useful technical lemmas will be listed.

Lemma 3. Assume that – hold; if is a sequence, then sequence is bounded in .

Proof. Let , in as .
Case  1. When , we can choose such thatCase  2. When , we can choose such thatwhere ; the above inequalities imply the sequence is bounded in and the proof is finished.

As the consequence of Lemma 3, there exist a constant , , and sequence such that

Lemma 4. Let , , and be satisfied with . Suppose sequence is bounded in ; then there exists such that strongly in .

Proof. Since is bounded in , there exist and such thatThus, we have that strongly in , where and .
Step  1. We want to show that, for , there exists such that We can fix and choose and , , whereDenote ; then for , and . What is more, there exist and such that for by condition and is bounded by condition ; then, for every , there exists such that for whereBy (15) and , we observe thatTherefore, we getand by Fatou’s lemmaThus, from (13) and (20), we can get that strongly in .
Step  2. For , let , . By (19), we have thatBy Fatou’s lemma, we can see thatFor each measure subset , we have thatThus, one can get that is uniformly integral and is bounded. Then, we have thatTherefore, from (22) and (24), we can obtainSimilar argument can be applied to get that strongly in . Thus, one can easily get that .

Lemma 5. Assume that and hold with ; if satisfies (12), then

Proof. By and , we get thatwhere and for any . Moreover, we obtain from Hölder inequality thatwhere . By Fatou’s lemma, we haveThen, from (28) and (29), we can deduce . Similarly, we can get . Thus, the proof is completed.

Denote Then the fact in implies as . Similarly, the fact that in implies that , whereNote thatwhereIf , this implies and according to the following lemma.

Lemma 6 (see [11]). There exists constant such that, for all ,where denotes the usual scalar product in .

In the following, we will claim

Based on concentration-compactness principle [12] and (12), it follows thatwhere is the Dirac measure at and , are nonnegative bounded measures. Analogously, we get and . Concentration at of is defined byAnalogously, the concentration at of is given by and .

Step 1. For , we define as follows:One can see that is bounded in ; we obtain . Namely,The fact that produces . In the light of Lions’ principle, we acquireMoreoverAt the same time, we haveFrom the aforementioned equations, it follows thatcombining with (36), we obtain

Step 2. By choosing a suitable cut-off function such thatwe set ; then is bounded in and .
In a similar manner, we gainNote thatwhich deducesNext we will prove and are impossible.

Case 1. If we fix , there exists such that ; then one can getLet ; we gainwhich gives rise to a contradiction, so for all . With the similar argument, we can get .

Case 2. If we fix , one can seeif we can obtainotherwise,moreover,We can easily get that there exists being small such that when , , the right side is positive, which gives rise to a contradiction.
Up to now, we have shown that In view of (12) and the Brezis-Lieb Lemma [13], we acquire As a result of the above, it yields , which implies the functional of (1) fulfils condition.

3. Proof of Theorem 1

The purpose of this section is to give proof of Theorem 1 which mainly relies on mountain pass lemma as follows.

Lemma 7 (see [14]). Let be a real Banach space, , . If ()there exist constants such that for all with ,()there is , , such that ,()the functional satisfies PS condition: every sequence , such that for some , and in as , is relatively compact in , then possesses a critical value which can be characterized as where

In the following, we want to show thatWe suppose and we can get the other case with similar arguments. Let ; then we get the following lemma.

Lemma 8. Under hypotheses , , and with , there exists such that if , we have .

Proof. Forwith , we can easily get as , with large enough, and with sufficiently small. So for some ; namely,We can have ; that is, By direct computation , we can get .
MeanwhileDenote , ; it can achieve its maximum at thus we get .

Proof of Theorem 1. Obviously, functional satisfies conditions – of the Lemma 7 we can define , where denotes the class of all continuous paths joining the origin to . Thus, mountain pass lemma can be applied to conclude that has a critical point with corresponding critical value . So, system (1) has at least one nontrivial solution.

4. Proof of Theorem 2

In this section, we want to apply genus theory to get the main conclusion of Theorem 2. Let be a Banach space and be the class of closed and symmetric subsets with respect to the origin of . For , genus is given by Since with , , thus we can have where .

Moreover, The corresponding functional of (1) turns intowhere and .

Definewe can observe that there exists being small such that if , there exists such that becomes if , respectively.

We set up the truncated functional [15]where and is a nonincreasing function such that for and for . Obviously, is even and satisfies with .

Lemma 9. Assume , , and with hold. Given , there exists such that

Proof. Fix , and suppose is -dimensional subspace of . We can take , with . Thus, for , we havewe can choose and small enough such that . Let ; then, . Hence, .

Lemma 10. Denote and set then, if ,

Proof. For convenience, we denote . Suppose ; we observe that and is a compact set. If , there exists a symmetric and closed set such that . According to deformation lemma, one can get an odd homeomorphism such thatBased on the definition , there exists such that andHence we gainTherefore, and it is in contradiction to (76). Lemma 10 and Kajikiya’s [16] symmetric mountain pass lemma give the proof of Theorem 2.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This work is supported by National Natural Science Foundation of China (11701252) and Project of Shandong Province Higher Educational Science and Technology Program (J16LI03).