Research Article | Open Access
Yanmin Niu, Baoqiang Yan, "Exact Number of Positive Solutions for a Class of Two-Point Boundary Value Problems", Journal of Applied Mathematics, vol. 2013, Article ID 589483, 8 pages, 2013. https://doi.org/10.1155/2013/589483
Exact Number of Positive Solutions for a Class of Two-Point Boundary Value Problems
This paper considers the existence of positive solutions of the boundary value problems and , where and is a positive parameter. Using a time-map approach, we obtain the exact number of positive solutions in different cases.
1. Introduction and Main Results
The study of multiplicity results to boundary value problem where is a positive parameter, is very interesting because of its applications. As we know, when , the boundary value problem where and are fixed given numbers and is a parameter, comes from the elliptic problem with , which was raised by Ambrosetti et al. in .
Cheng  investigated the following two-point boundary value problem: where is a positive parameter and and got the exact number of positive solutions.
Now, in this paper we consider the more general case where is a positive parameter, , and .
Define , where satisfies
For and , let be given byThe main results of this paper are as follows.
Theorem 1. If , (5) has exactly one positive solution for any .
Theorem 2. If , (5) has exactly one positive solution for and none for .
Theorem 3. If , (5) has exactly one positive solution for and none for .
Theorem 4. If , (5) has exactly one positive solution for and none for .
Theorem 5. Assume that . Define where and . One has the following.(1)If , (5) has exactly one positive solution for and none for .(2)If , there exists such that (5) has exactly two positive solutions for , exactly one for or , and none for .
We assume throughout this section that and Denote that , where is given by (6). For and , let Define a function as It is clear that . Now we claim that for and if and only if .
Let and ; then . Immediately, we get that . In order to judge the sign of , we just judge the sign of . Since , , and , we can get the following two results.(1)For , the function has a stable point . When , we have , and when , we have .(2)For , we have on . Combining and with the above two results, we obtain the monotony of .(3)For , the function has a stable point on . When , we have and when , we have .(4)For , we have that on .(5)For or , we have that on .Then from , we infer that for .
We consider the integration . It is clear that is a flaw point. Since on , we consider the integration .
Using Lagrange theorem, we obtain that where are constants.
Proof. (1) Assume that and is a positive solution of (5). Let satisfy . It follows from that
This implies that and if . And combine to obtain
Then, we have that
It follows that
From (19) and we have that . With (17) and (18) we obtain the result (1) of this theorem.
(2) Since and is a positive solution of the boundary value problem we have that is a positive solution of (5).
Lemma 7. is differentiable on , and where
Lemma 8. Consider the following: where is given by (7).
Proof. From and the Lebesgue theorem, we have that On the other hand, from , we can obtain that This completes the proof of Lemma 8.
Lemma 9. Consider the following:
Proof. From we have thatFrom Lemma 7 and the Lebesgue theorem, we can obtain the results of this lemma.
3. The Proof of Theorem 5
Lemma 11. For and ,
Condition implies that
With and (44), we have that for .
It follows that Combining we have the results of this lemma.
Lemma 12. For ,
Lemma 13. has continuous derivatives up to second order on as follows: where .
Lemma 14. There exist and such that
Proof. From (41) and , we have that
And since , let satisfy
From , we have that and
On the other hand, from (10), (38), and Lemma 11, we have that
for , .
This means that for , . It follows from Lemma 13 and (58) that Now, from (59)–(62) we have (55), where
Lemma 15. If , then for . If , then there exists , such that for and for .
Proof. It follows from Lemma 14 that if and then By (40), (42), and (51) we can obtain that With (10) and (41), (66) implies that If , then from Lemma 12 and (64)–(67) we have the results of this lemma, immediately. If , then by Lemmas 12 and 14 we have that for near , and so for near . Thus, it follows from (64)–(67) that for .
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Copyright © 2013 Yanmin Niu and Baoqiang Yan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.