In this section, we show the proofs of our Theorems 1 and 2. Before we come to the proof of Theorem 1, we first recall the following lemma in .
Lemma 4 Let , and . Then there exists such that .
Proof of Theorem 1 Using the hypotheses () and (), it follows from (3) and (4) that
Let , we can find and such that for all
From Lemma 4, we obtain the result that there exists such that
Therefore, one has
Fix ; noticing that , it implies from (8) that there exists small enough such that . Thus we deduce that
By applying the Ekeland’s variational principle in , we obtain the result that there exists a sequence of I.
By the expression of , we can choose such that for all . It follows from and Lemma 3 that I satisfies the condition. Therefore, one has a subsequence still denoted by and such that in and
which implies that is a solution of problem (1). After a direct calculation, we derive , which implies . Since , we have . Applying the Harnack inequality , we see that is a positive solution of problem (1). The proof of Theorem 1 is completed. □
Lemma 5 Let , and . Assume that (), (), (), (), and () hold, then there exists , such that for any , we can find such that .
Proof For convenience, we consider the functional defined by
for all . According to () and (), we can choose such a cut-off function that for , for , and , where is a positive constant. Define
According to () and (), similar to the calculation of , we have the following estimate (as )
where . Define
for all . From (9), we have . Note that and for , so is attained for some . By
Therefore, we deduce from (9) that
By the expression of , we can choose such that for all . Using the definitions of I and , from () and (), we have
for all and . It follows that there exist and such that
for all and . Moreover, using the definitions of I and , it follows from () and (10) that
Let , it follows that
By the above two inequalities, for any , we have
Hence, we can choose such that for all
Therefore, for all and , we have
Set . Let , and , we deduce from (11) and (13) that
Proof of Theorem 2 Choose , from the proof of Theorem 1, we have already seen that problem (1) for any has a positive solution with . Now we only need to find the second positive solution of problem (1). According to (), we can see that (4) and (7) hold. It follows from () and Lemma 4 that there exists such that
According to (4), we have
which implies that
Hence, there exists a positive number such that and for any . It implies from (7) that the functional I has the mountain pass geometry. Define
From Lemma 5, we have . Applying Lemma 3, we know that I satisfies the condition. By the Mountain Pass Theorem , we obtain the result that problem (1) has the second solution with . After a direct calculation, we derive
which implies that . Hence we have . Since , we have . By the Harnack inequality, we obtain the result that is the second positive solution of problem (1). The proof of Theorem 2 is completed. □