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Mathematical Problems in Engineering
Volume 2014, Article ID 610926, 10 pages
http://dx.doi.org/10.1155/2014/610926
Research Article

Existence of Positive Solutions of One-Dimensional Prescribed Mean Curvature Equation

Department of Mathematics, Northwest Normal University, Lanzhou 730070, China

Received 23 October 2013; Accepted 28 February 2014; Published 13 April 2014

Academic Editor: Yuji Liu

Copyright © 2014 Ruyun Ma and Lingfang Jiang. 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

We consider the existence of positive solutions of one-dimensional prescribed mean curvature equation , , , , where is a parameter, and is continuous. Further, when satisfies , , we obtain the exact number of positive solutions. The main results are based upon quadrature method.

1. Introduction and Preliminaries

In this paper, we are interested in the existence of positive solution of one-dimensional prescribed mean curvature equation where is a parameter and satisfies , .

The existence of positive solutions of such type of prescribed mean curvature equations both in one and in higher dimension has been discussed in the last decades by several authors (see Habets and Omari [1, 2], Li and Liu [3], Pan [4], Obersnel and Omari [5, 6], and others [712]) in connection with various configurations of nonlinearities.

Habets and Omari [1] considered the case that the nonlinearity ; they proved that if , then there exist and with such that (1) has exactly one positive solution for , exactly two solutions for , and no positive solution for ; if , then there exists with such that (1) has exactly one positive solution for and no positive solution for ; if , then there exist and such that (1) has no positive solution for and exactly one positive solution for .

Li and Liu [3] studied the case that the nonlinearity , ; they proved that if , then (1) has at most one solution for any ; there exist such that (1) has no positive solution for and exactly one positive solution for ; if , then (1) has at most two solutions for any ; there exists such that (1) has exactly one positive solution for and no positive solution for .

Inspired by [1, 3], we naturally consider (1) with nonlinearity satisfying

Denote , .

Consider the following assumptions:(H1) is continuous, and , ;(H2), for , ;(H3)for fixed , for ;(H4)for fixed , , .

We will prove the following.

Theorem 1. Assume that (H1)–(H3) hold, and, if , then the following conclusions hold:(1)for any , (1) has at most one solution;(2)there exists such that (1) has no solution for and has exactly one solution for .

Theorem 2. Assume that (H1), (H2), and (H4) hold, and, if  , then the following conclusions hold:(1)for any , (1) has at most two solutions;(2)there exists such that (1) has exactly one solution for and has no solution for .

Remark 3. If and , then (2) reduces to , which has been considered by Habets and Omari [1]; for and , which has been studied by Li and Liu [3], and the case has been considered by Zhang and Feng [7].

Remark 4. The problem concerning exact number of solutions of semilinear equation has been discussed by several authors with and the related problems (see [1316]).

2. Quadrature Technique

Let be a solution of problem (1); then it is well known that takes its maximum at , is symmetric with respect to , for , and for . Hence, problem (1) is equivalent to the following problem defined on : Let . If is a solution of (4) with , then is a solution of the following problem defined on : From satisfying and , we see that Therefore, Then Integrating (9) from to leads to By substituting in (7), we get that and .

Let be defined on by For simplicity, we denote It follows that

Lemma 5. has continuous derivatives up to the second order on with respect to and

Proof. From Habets and Omari [1], it is easy to obtain the results.

Lemma 6. Let (H1) and (H2) hold; is strictly decreasing on with respect to , and

Proof. Since we have Then It follows that , .
From the fact that , we have , and ; it follows from the assumption (H1) that Then is strictly decreasing on .

Lemma 7. Let (H1)–(H3) hold, if and . Then has the following properties:(i); (ii); (iii), for .

Proof . (i) From (H2), therefore, Hence, from , then .
(ii) Since , for , thus
(iii) From Lemma 5, we know that , so where .
From the assumption (H3), we have that , ; thus .

Lemma 8. Assume that (H1) and (H2) hold, and let . Then has the following properties:(i)for , is continuous in ;(ii)for , , .

Proof. (i) Applying , we may prove the following (26), (29) and (30) hold. For simplicity, we denote by .
If , then , , and from , we get So Since , we have from (25) that If , then , and , , and, from the fact that , we get that if , then And, if , then therefore, Or
It follows from (26), (29), and (30); we get that is continuous in and (i) holds. (ii) From (25) and the fact that , the following holds.
If , then thus, if , then from (31), we obtain If , then for and for , and, from the fact that , we get the following.
If , then and, if , then Then It follows from (26), (29), (33), (36), and Lemma 6 that and .

Lemma 9. Let (H1), (H2), and (H4) hold; assume that . Then has the following properties, for fixed (i);(ii);(iii); (iv), for .

Proof. (i) Since , we choose ; thus from (H2), we get therefore, Hence, from , we have .
(ii) From Lemma 5, we have that according to the assumption (H2), we have and, from the assumption (H4), that is, , , we have ; thus Thus Therefore, .
(iii) It is proved in the same way as in Lemma 7(ii).
(iv) From Lemma 5, we get that From (H1) where , using the assumption (H4) and the fact that , we arrive at Therefore, , for .

Lemma 10. Assume that . Denote . Then(i) is strictly decreasing in ;(ii) is continuous in ;(iii), .

Proof. (i) From Lemma 9, we get Let . From the definition of and Lemma 6, we obtain And, for and , It follows that, for , thus, Hence, , which implies that is strictly decreasing in .
(ii) Let . Since , there exists such that, for and , It follows that , for , . Further, Define Thus, if , then and, if , then It follows that the singular integral in (53) and (54) converges uniformly in , and hence and are continuous in . By (i), we have We will see that the equalities hold in (55). If this does not hold, then we may assume, for instance, Choose an increasing sequence satisfying and then choose such that Since , we may assume that there exists such that ; by the continuity of and , we have or which contradicts (56). Therefore, . Similarly, we have . So is continuous at .
(iii) Note that , for any , , and , and, from (55), we have Hence, .
Fix and . For , the definition of and the fact that imply .
Since , Hence, .

3. The Proofs of the Main Results

Proof of Theorem 1. Similar to the proof of Theorem 1.1 [3], it follows from Lemmas 7 and 8 that the results are easy to prove.

Proof of Theorem 2. Similar to the proof of Theorem 1.2 [3], it follows from Lemmas 8, 9, and 10 that the results are easy to prove.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This paper was supported by the NSFC (no. 11361054), the SRFDP (no. 20126203110004), and the Gansu Provincial National Science Foundation of China (no. 1208RJZA258).

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