Lemma 4.1
-
(i)
If , then the functional J has a (PS)
-sequence .
-
(ii)
If , then the functional J has a -sequence .
Proof The proof is similar to that of [21] and is omitted. □
Lemma 4.2 Suppose that . Then J has a minimizer such that is a positive solution of (1.1) and .
Proof By Lemma 4.1(i), there exists a (PS)
-sequence of J such that
(4.1)
Since J is coercive on (see Lemma 3.1), we get that is bounded in E. Passing to a subsequence (still denoted by ), we can assume that there exists such that
(4.2)
which implies that
(4.3)
First, we claim that is a solution of (1.1). By (4.1) and (4.2), it is easy to see that is a solution of (1.1). Furthermore, from and (3.3), we deduce that
(4.4)
Taking in (4.4), by (4.1), (4.2) and the fact , we get
Therefore, is a nontrivial solution of (1.1).
Next, we prove that strongly in E and . Noting and applying the Fatou lemma, we have
Therefore, and . Set . By the Brézis-Lieb lemma [17], we get
Then standard argument shows that strongly in E. Moreover, we have . Otherwise, if , then by Lemma 3.5 there exist unique such that and . Since
there exists such that . By Lemma 3.5 we get that
which is a contradiction. Since and , by Lemma 3.2 we may assume that is a nontrivial nonnegative solution of (1.1).
In particular , . Indeed, without loss of generality, we may assume that . Then as is a nontrivial nonnegative solution of
by the standard regularity theory, we have in Ω and
Moreover, we may choose such that
Now,
and so by Lemma 3.6 there is unique such that . Moreover,
and
This implies
which is a contradiction.
Finally, from the maximum principle [22] we deduce that in Ω and is thus a positive solution of (1.1). □
Let be defined as in (1.4) and set , where is a cut-off function:
The following results are already known.
Lemma 4.3 [4]
As we have the following estimates:
Lemma 4.4 [11]
Suppose that (ℋ) holds, is defined as in (1.6) and are the minimizers of defined as in (1.4). Then and has the minimizers , where .
Lemma 4.5 Under the assumptions of Theorem 1.2, there exist and such that for all there holds
(4.8)
In particular, for all .
Proof For all , define the functions and
Note that and as t is closed to 0. Thus, is attained at some finite with . Furthermore, , where and are the positive constants independent of ε.
Choose small enough such that for all . Set . Then for all and , which implies that there exists satisfying , for all . Note that
(4.9)
From (4.9) and Lemmas 4.3, 4.4 it follows that
Consequently,
and
where we have used the assumption .
Therefore we can choose , such that
The definition of in Lemma 2.1 implies that
Note that
Taking ε small enough, there exists such that for all ,
(4.10)
Choose . Then for all there holds
(4.11)
Finally, we prove that for all . Recall that . By Lemma 3.5, the definition of and (4.11), we can deduce that there exists such that and
The proof is thus complete by taking . □
Lemma 4.6 Set . Then for all , problem (1.1) has a positive solution such that and .
Proof By Lemma 4.1, there exists a -sequence of J for all . From Lemmas 2.3, 3.4 and 4.5, it follows that and J satisfies the condition for all . Since J is coercive on , we get that is bounded in E. Therefore, there exist a subsequence (still denoted by ) and such that strongly in E and for all . Since and , by Lemma 3.2 we may assume that is a nontrivial nonnegative solution of (1.1). Moreover, by (3.7) and , we get
This implies that and . From the strong maximum principle [22] it follows that is a positive solution of (1.1). □
Proof of Theorems 1.1 and 1.2. By Lemma 4.2, we obtain that (1.1) has a positive solution for all . On the other hand, from Lemma 4.6, we can get the second positive solution for all . Since , this implies that and are distinct. □