Let be the unique, radially symmetric and positive ground state solution of equation (E 0) in . Recall the facts (or see Bahri-Li [10], Bahri-Lions [11], Gidas-Ni-Nirenberg [12] and Kwong [13]):
-
(i)
for some and ;
-
(ii)
for any , there exist positive numbers , and such that for all
and
By Lien-Tzeng-Wang [14], then
(4.1)
For , we define
Clearly, .
First of all, we want to prove that
Lemma 4.1 For and , then
Moreover, we have that
Proof Part I: Since is continuous in H, , and is uniformly bounded in H for any and , then there exists such that for and any ,
From (A1), we have that . Then
It follows that there exists such that for and any ,
From now on, we only need to show that
Since
and by (4.1), then
For , by (4.2), we have that
Since
then
that is, for and ,
Part II: By Lemma 2.4, there is a number such that , where . Hence, from the result of Part I, we have that for and ,
□
Proof of Theorem 1.1 By Lemma 3.2, there exists a (PS)
-sequence in for . Since for and , by Lemma 3.3, there exist a subsequence and such that strongly in H. It is easy to check that is a nontrivial solution of () and . Since and , by Lemma 2.6, we may assume that , . Applying the maximum principle, and in Ω. □
Choosing such that
where and , define and . Suppose for some . Let be given by
where , for and for .
For each , we define
By Lemma 2.4, there exists such that for each . Then we have the following result.
Lemma 4.2 There exists such that if , then for each .
Proof Since
there exists such that
□
We need the following lemmas to prove that for sufficiently small ε, λ, μ.
Lemma 4.3 .
Proof From Part I of Lemma 4.1, we obtain uniformly in i. Similarly to Lemma 2.4, there is a sequence such that and
Let be a minimizing sequence of for . It follows that and
We may assume that and as , where . By the definition of , then . We can deduce that , that is, . □
Lemma 4.4 There exists a number such that if and , then for any .
Proof On the contrary, there exist the sequences and such that , as and for all . It is easy to check that is bounded in H. Suppose that as . Since
then
which is a contradiction. Thus, as . Similarly to the concentration-compactness principle (see Lions [15, 16] or Wang [[6], Lemma 2.16]), then there exist a constant and a sequence such that
(4.3)
where and for some . Let . Then there are a subsequence and such that and weakly in . Using the similar computation of Lemma 2.4, there is a sequence such that and
We deduce that a subsequence satisfies . Then there are a subsequence and such that and weakly in . By (4.3), then and . Applying Ekeland’s variational principle, there exists a (PS)
-sequence for and . Similarly to the proof of Lemma 3.3, there exist a subsequence and such that , strongly in and . Now, we want to show that there exists a subsequence such that .
-
(i)
Claim that the sequence is bounded in . On the contrary, assume that , then
which is a contradiction.
-
(ii)
Claim that On the contrary, assume that , that is, . Then use the argument of (i) to obtain that
which is a contradiction.
Since and , strongly in , we have that
which is a contradiction.
Hence, there exists such that if and , then for any . □
Lemma 4.5 If and , then there exists a number such that for any and .
Proof Using the similar computation in Lemma 2.4, we obtain that there is the unique positive number
such that . We want to show that there exists such that if , then for some constant (independent of u and v). First, for ,
Since , then
(4.4)
and
(4.5)
Moreover, we have that
where . It follows that there exists such that for
(4.6)
Hence, by (4.4), (4.5) and (4.6), for some constant (independent of u and v) for . Now, we obtain that
From the above inequality, we deduce that for any and ,
Hence, there exists such that for ,
By Lemma 4.4, we obtain
or for any and . □
Since , then by Lemma 4.3,
(4.7)
By Lemmas 4.1, 4.2 and (4.7), for any () and ,
(4.8)
Applying above Lemma 4.5, we get that
(4.9)
For each , by (4.8) and (4.9), we obtain that
It follows that
Then applying Ekeland’s variational principle and using the standard computation, we have the following lemma.
Lemma 4.6 For each , there is a -sequence in H for .
Proof See Cao-Zhou [8]. □
Proof of Theorem 1.2 For any and , by Lemma 4.6, there is a -sequence for where . By (4.8), we obtain that
Since satisfies the (PS)
γ
-condition for , then has at least k critical points in for any and . Set and . Replace the terms and of the functional by and , respectively. It follows that () has k nonnegative solutions. Applying the maximum principle, () admits at least k positive solutions. □