Abstract

We study the effect of the coefficient of the critical nonlinearity on the number of positive solutions for semilinear elliptic systems. Under suitable assumptions for , , and , we should prove that for sufficiently small , there are at least positive solutions of the semilinear elliptic systems , , where is a bounded domain, , , and for .

1. Introduction and Main Results

For , , , and , consider the semilinear elliptic systems where , is a bounded domain with smooth boundary .

Let , , and satisfy the following conditions. , , and are positive continuous functions in and . There exist points and some such that are strict maxima and satisfy and as uniformly in .

Recent studies [110] have investigated the elliptic systems with subcritical or critical exponents and have proved the existence of a ground state solution or the existence of at least two positive solutions for these problems. For the case of , , , and , Lin [11] constructs the compact Palais-Smale sequences that are suitably localized in correspondence of maximum points of . Under assumptions - , she has showed that there are at least positive solutions of the problem for sufficiently small . In this paper, we study the problem and complement the results of [11] to the case . Under assumptions - , we should prove that there exist at least positive solutions of the problem for sufficiently small .

Let be the space with the standard norm Associated with the problem , we consider the -functional , for , The weak solution of the problem is the critical point of the functional ; that is, satisfies for any .

Let with the norm , and let be the best Sobolev constant defined by and let then, by [1, Theorem 5], we have where .

Set where .

The main results of this paper are given as follows.

Theorem 1. Assume that holds. If satisfy , then there exists at least one positive ground state solution of the problem .

Theorem 2. Under the assumptions - , and and , there exists a positive number such that for and , the problem has positive solutions.

This paper is organized as follows. In Section 2, we consider the Nehari manifold where Note that contains all nontrivial weak solution of the problem . Using the argument of Tarantello [12, 13], we split into two parts and for . In Section 3, we prove Theorem 1. In Section 4, since satisfies the -condition for , for sufficiently small , , and some restriction on and , we construct the compact Palais-Smale sequences which are suitably localized in correspondence with the maximum points of and which converge to distinct solutions of the problem belonging to . Hence, we prove Theorem 2 (one is the ground state solution belonging to and the others are in ).

2. Nehari Manifold

Throughout this paper, will be assumed. First, we give some notations.

Notations. We make use of the following notations. , , denote Lebesgue spaces; the norm is denoted by for . , endowed with norm . The dual space of a Banach space will be denoted by . and for all and . is said to be nonnegative in if and in . is said to be positive in if and in . is the Lebesgue measure of . is a ball in . denotes as for . means that there exist the constants such that as is small. denotes as . . , will denote various positive constants, the exact values of which are not important.

Let be the functional defined by

We know that is not bounded below on . From the following lemma, we have that is bounded from below on the Nehari manifold defined in (9).

Lemma 3. The energy functional is coercive and bounded below on .

Proof. If , then by (10), the Hölder inequality, and the Sobolev embedding theorem, we get Hence, we have that is coercive and bounded below on .

Define Then, for , We apply the method in [12]; let

By using equality (17), we get that for . Moreover, we have the following results.

Lemma 4. Let be a constant defined as in (8). If , then .

Proof. Assuming the contrary, there exist with such that . Then, by (16) and (17), for , we have Using and both the Hölder and the Sobolev inequalities, we get This implies which is a contradiction.

For each with , we write Then, the following lemma holds.

Lemma 5. Suppose that , and with . Then, there exist unique such that , and

Proof. This is similar to the proof of Hsu [14, Lemma 2.7].

Applying Lemma 4 ( for ), we write and define

The following lemma shows that the minimizers on are usual critical points for .

Lemma 6. For the case when , if is a local minimizer for on , then in .

Proof. See Brown and Zhang [15, theorem 2.3].

Lemma 7. (i) If and , then one has In particular, .
(ii) If and , then one has , in , and for some positive constant .

Proof. (i) Let . By (16) and (17), we have Then, By the definition of , , we deduce that .
(ii) Let ; by (16) and the Hölder and the Sobolev inequalities, we get This implies By (13) and (31), we obtain that , in , and Thus, if , for all , then

3. Existence of a Ground State Solution

First of all, we define the Palais-Smale (denote by (PS)) sequences and (PS)-condition in for as follows.

Definition 8. (i) For , a sequence is a -sequence in for if and strongly in as .
(ii) satisfies the -condition in if any -sequence in for contains a convergent subsequence.

Proof of Theorem 1. Using the same argument as in Wu [16, Proposition 9] or Hsu [14, Proposition 3.3], there exists a minimizing sequence for on such that Since is coercive on (see Lemma 3), we get that is bounded in . Then, there exist a subsequence and such that This implies First, we claim that is a nontrivial solution of . By (34) and (35), it is easy to verify that is a weak solution of . From and (12), we deduce that Let in (37); by (34), (36), and , we get Thus, is a nontrivial solution of . Now, we prove that strongly in and . By (37), if , then In order to prove that , it suffices to recall that , by (39) and applying Fatou's lemma to get This implies that and . Let ; then Brézis-Lieb lemma [17] implies Therefore, strongly in . Since and , by Lemma 6 we may assume that is a nontrivial nonnegative solution of . By an argument of Hsu [18, Lemma 4.2], we can deduce that and in . Finally, from the maximum principle [19], we deduce that is positive in .

Remark 9. and .

Proof. We claim that . On the contrary, assume that ( for ); then by Lemma 5, there exist unique and such that and . In particular, we have . Since there exists such that . By Lemma 5, which is a contradiction. Hence, and .

4. Existence of Solutions

Throughout this section, - will be assumed. First of all, we want to show that satisfies the -condition in for , where is defined in the following lemma.

Lemma 10. If is a -sequence for with weakly in , then and there exists a constant such that .

Proof. Let and . If is a -sequence for with weakly in , it is easy to check that in . Then, we get ; that is, . Thus, by (13), the Hölder, the Young, and the Sobolev inequalities, we have where .

Lemma 11. If is a -sequence for , then is bounded in .

Proof. See Hsu and Lin [8, Lemma 2.3].

Recall that and let where is given in Lemma 10.

Lemma 12. satisfies the -condition in for .

Proof. Let be a -sequence for with . Write . We know from Lemma 11 that is bounded in , and then weakly up to a subsequence; is a critical point of . Furthermore, we may assume that , weakly in and , strongly in for all , and , a.e. on . Hence, we have that and Let and . Then, we obtain and by an argument of Han [20, Lemma 2.1], Since , in and (47)–(49), we deduce that Hence, we may assume that Assume that ; by the definition of , and (52), we obtain which implies that . In addition, from Lemma 10, (50), and (52), we get which is a contradiction. Hence, ; that is, strongly in .

From assumption , we can choose such that and , where and for .

Define where , .

Then, we have the following separation result.

Lemma 13. If and for , then .

Proof. For any satisfying , we get which implies that Hence, from (58), we obtain which is a contradiction.

For , we set and define

Recall that the best Sobolev constant is defined as It is well known that is a minimizer of , and . Fix a maximum point of ( ). Let be a cut-off function such that , , and for , for . We define where and .

From now on, we assume that and .

Lemma 14. There exist , , such that for and , one has where is the positive constant given in Lemma 12.
In particular, for all .

Proof. It is well known that (or see Brézis and Nirenberg [21], Cheng and Ma [22, Lemma 3.2], Struwe [23], and Willem [24, Lemma 1.46]) as , For , and ,
Set . By Lemma 5, there exists such that for . Furthermore, Hence, there exists for any which implies and then First, we consider the functional defined by
Step I. Show that .
According to condition , we conclude that From (66), (74), , and , we can deduce that Using (67) and (75), then Since by (7) and (76), we conclude that
Step II. Let be the positive constant given in Lemma 10. We can choose such that for all , we have Since is continuous in , , and is uniformly bounded in for any (see (67)), then there exists (independent of ) such that for any , According to condition , and . Applying the results of Step I and (68), we have that for and , where and .
Therefore, we can choose and such that
This implies that There exist , such that for all and , we have Thus, we can choose . Then, for all , there holds
Step III. For and , by Lemma 7, (72), and (85), we get

To proceed, we need to quote the concentration-compactness principle (see [24, 25]) about the case of systems.

Lemma 15. Let be a sequence such that Then, it follows that Moreover, if and , then and concentrate at a single point.

Proof. See Han [20, Lemma 2.2].

Lemma 16. For any , there exist such that

Proof. Fix . Assume the contrary. There then exists a sequence with as such that as . Consequently, there exists a sequence such that as , It then follows easily that is uniformly bounded in , and since and are continuous on , we obtain From (90), and by the Hölder and the Sobolev inequalities, we can fix such that Thus, up to a subsequence, we infer that Furthermore, by , we deduce which implies On the other hand, we have, as , Hence, together with (96), we get and then from (95), we also have Therefore, Set ; then, we have . Moreover, by (94),(98), and (100), we get Thus, up to a subsequence, we may assume that Since is bounded, from (101) and Lemma 15, we deduce that If and , we deduce that which is a contradiction.
Thus, or . If , from (103)–(105), we get and . Then, which means that is achieved by . It is impossible since cannot be achieved on any bounded domain . Hence, Then, on , and from (103), (104), we easily have . By Lemma 15, we conclude that such that Observe that ; which implies that by the definition of . On the other hand, from (95) and (101), we get which is impossible, because is not a constant function by condition .

Throughout this section, take ; and are as in Lemmas 14 and 16. Using the idea of Tarantello [12], we have the following results. For , , we define

Lemma 17. For each and , there exist and a differentiable function such that , for all and for all .

Proof. For , define a function by Then, and According to the implicit function theorem, there exist and a differentiable function such that ; which is equivalent to that is, for all . Furthermore, by the continuity of the functions and , we have that still holds if is sufficiently small. This implies that .

Proposition 18. If , then there exists a -sequence in for .

Proof. If denotes the closure of , at first we note that for all . It then follows from Lemmas 14 and 16, that Hence, Now, we fix . Applying the Ekeland variational principle [26], there exists a minimizing sequence such that Using (119), we may assume that for sufficiently large. Applying Lemma 17 with , we obtain the function for some such that for all . Let and ; we set and . Since , we deduce from (121) that By the mean-value theorem, we obtain Therefore, Now, we observe that , and consequently we get from (125) that Then, we write the pervious inequality in the following form: We can find a constant independent of such that For a fixed , let in (127). Using the fact that we obtain This implies

Now, we complete the proof of Theorem 2. By Lemmas 12, 14 and Proposition 18, for all , there exists a sequence and , , such that Moreover, , and by Lemma 7 (ii), we get , , in , Thus, is a nontrivial solution of the problem and for . Set and . Replace the terms and of the functional by and , respectively. It then follows that is a nonnegative solution of the problem . Applying the maximum principle [19], is a positive solution of the problem . Since , where is a positive solution of equation as in Theorem 1. From Lemma 13, we conclude that are disjoint for . This implies that and are distinct positive solutions of the problem .