In this section, let be the Banach space of continuous functions endowed with , and the ordering if for all . Define the cone by
where λ is given as in Lemma 2.5.
For convenience of the reader, we denote
Lemma 3.1
Let
be the operator defined by
Then is completely continuous.
Proof By Lemma 2.5, we have
Thus, . In view of non-negativity and continuity of , and , we find that is continuous.
Let be bounded, i.e., there exists a positive constant such that , for all . Let , then, for , we have
Hence, is uniformly bounded. Further for any and , we have
Hence, . For any and , we have
That is to say, is equicontinuous. By the Arzela-Ascoli theorem, we see that is completely continuous. The proof is completed. □
We are now ready to prove our main results.
Theorem 3.1 Let be nonnegative continuous on . Assume that there exist constants a, b with such that
(H1) , for ;
(H2) , for .
Then the boundary value problem (1.1) has at least two positive solutions and satisfying , and , where λ is given as in Lemma 2.5.
Proof Let be the nonnegative continuous concave functional defined by
Evidently, for each , we have .
It’s easy to see that is completely continuous and . We choose , then
So . Hence, if , then for . Thus for , from assumption (H1), we have
Consequently,
That is,
Therefore, condition (B1) of Lemma 2.7 is satisfied. Now if , then . By assumption (H2), we have
which shows that , that is, for . This shows that condition (B2) of Lemma 2.7 is satisfied. Finally, we show that (B3) of Lemma 2.7 also holds. Assume that with , then by the definition of cone P, we have
So condition (B3) of Lemma 2.7 is satisfied. Thus using Lemma 2.7, T has at least two fixed points. Consequently, the boundary value problem (1.1) has at least two positive solutions and in satisfying , and . The proof is completed. □
Theorem 3.2 Let be nonnegative continuous on . Assume that there exist constants a, b, c with such that
(H3) , for ;
(H4) , for ;
(H5) , for .
Then the boundary value problem (1.1) has at least three positive solutions , and with , , and , where λ is given as in Lemma 2.5.
Proof If , then . By assumption (H5), we have
This shows that . Using the same arguments as in the proof of Lemma 3.1, we can show that is a completely continuous operator. It follows from the conditions (H3) and (H4) in Theorem 3.2 that . Similarly with the proof of Theorem 3.1, we have and
Moreover, for and , we have
So all the conditions of Lemma 2.6 are satisfied. Thus using Lemma 2.6, T has at least three fixed points. So, the boundary value problem (1.1) has at least three positive solutions , and with , , and . The proof is completed. □
Theorem 3.3 Let be nonnegative continuous on . If the following assumptions are satisfied:
(H6) ;
(H7) there exists a constant such that
then the boundary value problem (1.1) has at least two positive solutions and such that .
Proof From Lemma 3.1, we obtain is completely continuous. In view of , there exists such that
where .
Let . Then, for any , we have
which implies for . Hence, Lemma 2.9 implies
On the other hand, since , there exists such that
where .
Let and . Then , for any . By using the method to get (3.1), we obtain
which implies for . Thus, from Lemma 2.9, we have
Finally, let . Then, for any , by (H7), we then get
which implies for . Using Lemma 2.9 again, we get
Note that , by the additivity of fixed-point index and (3.1)-(3.3), we obtain
and
Hence, T has a fixed point in , and has a fixed point in . Clearly, and are positive solutions of the boundary value problem (1.1) and . The proof is completed. □
Theorem 3.4 Let be nonnegative continuous on . If the following assumptions are satisfied:
(H8) ;
(H9) there exists a constant such that
Then the boundary value problem (1.1) has at least two positive solutions and such that .
Proof From Lemma 3.1, we obtain is completely continuous. In view of , there exists such that
where .
Let . Then, for any , we have
which implies for . Hence, Lemma 2.9 implies
Next, since , there exists such that
where . We consider two cases.
Case 1: Suppose that f is bounded, which implies that there exists such that for all and .
Take . Then, for with , we get
Case 2: Suppose that f is unbounded. In view of being continuous, there exist and such that
Then, for with , we obtain
So, in either case, if we always choose , then we have
Thus, from Lemma 2.9, we have
Finally, Let . Then, for any , , by (H9), and we then obtain
which implies for . An application of Lemma 2.9 again shows that
Note that ; by the additivity of fixed-point index and (3.4)-(3.6), we obtain
and
Hence, T has a fixed point in , and it has a fixed point in . Consequently, and are positive solutions of the boundary value problem (1.1) and . The proof is completed. □