Journal of Applied Mathematics

Volume 2013 (2013), Article ID 589483, 8 pages

http://dx.doi.org/10.1155/2013/589483

Research Article

Exact Number of Positive Solutions for a Class of Two-Point Boundary Value Problems

Yanmin Niu and Baoqiang Yan

School of Mathematical Sciences, Shandong Normal University, Jinan 250014, China

Received 25 September 2013; Accepted 4 November 2013

Academic Editor: Yansheng Liu

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.

Abstract.

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 [1].

Under different assumptions on , there are many results for the above problems and elliptic equations (see [2–8]).

In [9, 10], Liu considered the case of and and gave the exact number of solutions and many interesting properties of the solutions.

Cheng [11] 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 by

The 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 .

2. The Proofs of Theorems 1–4

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.

The following Lemma 6 is listed to show that to study the number of positive solutions of (5) is equivalent to study the shape of the map on . Lemmas 7–9 show some properties of on .

Lemma 6. Let be the unique solution of the problem

where . One has the following.

(1)If and is a positive solution of (5), , , and for .(2)If and , is a positive solution of (5) with .

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

Proof. Equation (21) can be obtained by (11), immediately. From

we have that

It follows from (21) that is differentiable on and (22) is true.

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 that

From Lemma 7 and the Lebesgue theorem, we can obtain the results of this lemma.

In the following section, we give the proofs of Theorems 1–5. For convenience, we denote that

Hence from Lemma 7, we have that

Proof of Theorem 1. From , we obtain that

for , . Thus for . By Lemmas 6, 8, and 9, we have the results of this theorem.

Proof of Theorem 2. From , (33), and (34), we have that for . It follows from Lemmas 6, 8, and 9 that the results of Theorem 2 hold.

Proof of Theorem 3. Conditions and imply that . With (33), we obtain that

for , , which means that for . Thus by Lemmas 6, 8, and 9, we have the results of this theorem.

Proof of Theorem 4. From and (33), we have that

for , . Hence for . By Lemmas 6, 8, and 9, we have the results of this theorem.

3. The Proof of Theorem 5

In this section we always assume that . Denote that

where , , and are given by (10), (6), and

Remark 10. From , (6), and (41), it is obvious that , , and .

Lemma 11. For and ,

Proof. Let

Condition implies that

With and (44), we have that for .

It follows that

Combining

we have the results of this lemma.

Lemma 12. For ,

Proof. From Lemma 7 we have that

It follows from Lebesgue theorem that

Finally, implies that

Combing (49) and (50), we complete the proof of this lemma.

Lemma 13. has continuous derivatives up to second order on as follows:

where .

Proof. Equation (51) can be obtained by

and Lemma 7, immediately. From (24) we have that

and so (52) holds.

Lemma 14. There exist and such that

Proof. From (41) and , we have that

And since , let satisfy

Set

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 .

Proof of Theorem 5. From Lemmas 6, 8, 9, and 15, we can obtain the results of Theorem 5.

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