Research Article | Open Access

# Positive Solutions of Two-Point Boundary Value Problems for Monge-Ampère Equations

**Academic Editor:**Gennaro Infante

#### Abstract

This paper considers the following boundary value problem: , where is odd. We establish the method of lower and upper solutions for some boundary value problems which generalizes the above equations and using this method we present a necessary and sufficient condition for the existence of positive solutions to the above boundary value problem and some sufficient conditions for the existence of positive solutions.

#### 1. Introduction

In [1], Kutev obtained the existence of a unique nontrivial solution of the following boundary value problem (BVP) where and , , which is related to the following Dirichlet problem of the Monge-Ampère equations in where is the unit ball in and is the Hessian of .

After Kutev’s works, the Monge-Ampère equation has attracted a growing attention in recent years because of its important role in several areas of applied mathematics. For instance, in [2], Hu and Wang obtained some sufficient conditions for the existence and multiplicity of the positive solutions for problem (1) and Dai [3] and Wang [4] discussed the unilateral global bifurcation results for the problem with . And there are some results on the existence and multiplicity and nonexistence of nontrivial radial convex solutions of systems of Monge-Ampère equations (see [5–8]) also. We notice that the nonlinearity is continuous at in the above works. For the case that is singular at , there are fewer results for BVP (1). But some interesting results are presented for BVP (2) in [9–12] where is singular at . We also refer to [7, 8, 13, 14] and references therein for further discussions regarding solutions of the Monge-Ampère equations.

One goal in this paper is to consider the existence of positive solutions under the conditions that and is singular at . Compared to the results in [10], we do not need the monotonicity of .

Our paper is organized as follows. Section 2 lists some lemmas. In Section 3, we obtain a theorem on the upper and lower solutions which is an extension of this method to the class of problem we consider. Section 4 is devoted to our main results on some sufficient conditions for positive solutions of the BVP (1).

#### 2. Preliminaries

Let be the Banach space with

Define a set by It can be easily verified that is indeed a normal cone in (see [2]).

Lemma 1 (see [15, 16]). *Let be a Banach space, and let be a bounded, closed, and convex set. Assume that is a continuous compact operator. Then, has at least one fixed point in .*

*Remark 2. *Obviously, assume that is a Banach space, is a bounded, closed, and convex set and let be a continuous compact operator. Then, has at least one fixed point in also.

Lemma 3. *Assume that function with , , and decreasing in . Then, *

*Proof. *Since is decreasing in , we have which implies Then, for , we have That is, Since and , we have The proof is complete.

*Remark 4. *The idea of the proof of Lemma 3 comes from Lemma in [2].

Now, we list some conditions for convenience: () () is odd.

Throughout this paper, we always assume that condition holds; that is to say, is odd.

#### 3. Upper and Lower Solutions

Consider the following two-boundary value problem: where is continuous and .

Now, we give following definitions.

*Definition 5. *If and then is called a lower solution of the BVP (11).

Similarly, we can define an upper solution of BVP (11).

*Definition 6. *If and then is called an upper solution of BVP (11).

If , define

Theorem 7. *Let , respectively, be lower and upper solutions of BVP (11) with on and for all and **holds. Then, BVP (11) has at least one solution and *

*Proof. *Consider the modified boundary value problem of the form where the function Obviously, if the problem (17) has a solution with , , then problem (11) has a solution , .

For , define an operator A standard argument shows that is continuous and compact. Lemma 1 guarantees that has a fixed point ; that is BVP (17) has a solution .

Now, we claim that which guarantees that is a solution of BVP (11).

Let , .

First, we show that Suppose that . Then, such that , . Let . Obviously, and Let , . Then which implies that is decreasing on . And so for all ; that is, That is,From , we have That is, which means that is increasing on . Then, , . The continuity of at means that , which contradicts (22).

Secondly, we prove that Assume that for some . Since and , there exists a such that with (note )). Then, From the continuity of and in , there exists a such that Integrating both sides of the above inequality from to , we get That is, From , we have And so Then, which contradicts the fact that . Hence, (28) is true.

The same argument shows that Consequently, (28) and (36) show that (20) holds. Then, is a solution to BVP (11).

*Remark 8. *The idea of the proof comes from [17, 18].

#### 4. Main Results

Suppose that is defined in Section 2. Now, we list the following condition for convenience: () , continuous and , such that

*Remark 9. *Assume that where , are constants. Let , , . Then, if , we have A similar argument shows that if , we have

We have the following theorems.

Theorem 10. *Suppose that holds and . Then, the necessary and sufficient condition of the existence of the positive solution in for BVP (1) is *

*Proof. **Necessity*. Suppose that is a positive solution. We have Let From and , Lemma 3 implies that and Let be a constant such that , . By (37) and (44), we have That is, Since we have from (46) Hence, (41) holds.*Sufficiency.* Since (41) holds, set Obviously and are decreasing in . Then, Lemma 3 means that . From (41), we know that there exists such that . Consequently, where , .

Let and , where is big enough such that , . Obviously, Since (note (50)) we have (note ) From we can obtain That is, And so (53) is Similarly, we have Differentiating on , we have with , , which implies that is a lower solution to BVP (1).

Similarly, we have with , . Hence, is an upper solution to BVP (1).

Theorem 7 guarantees that BVP (1) has at least one positive solution in .

*Remark 11. *Idea of the proof comes from [19].

Theorem 12. *Assume that holds and the following conditions are true:**there exist and such that **there exists such that **Then, BVP (1) has at least one positive solution . If, moreover, is decreasing, then the solution is unique.*

*Proof. *By assumption (1), there is such that Without restriction, we can assume that is a decreasing sequence and with for all .

Let It is easy to see that Let which guarantees that Set From (63), we have From and condition (2), we choose an such that Let Equation (70) guarantees that Now, Theorem 7 implies that there exists such that Let , , and , . It is easy to prove that Theorem 7 guarantees that there exists such that Proceeding the above proof, we can get a sequence such that Now, we claim that First, we show that Obviously, . Suppose . Let , , and . Obviously, (note ) and , , for all , which implies that For , condition (1) guarantees that Integrate (80) from to to obtain From , we have That is, which means that is decreasing on . The continuity of at means that . This contradicts that . Hence, (78) is true.

Secondly, we show that Suppose that there is a such that . There exists a with and ; that is, . Now, implies that , which together with condition (1) guarantees that The continuity of and at guarantees that there exists a such that Integrate (86) from to to obtain From , we have That is, which means that is decreasing on . This contradicts that . Consequently, (84) is true.

Consider now the pointwise limit as follows: It is clear that which means that For any , from (91) and we have That is, The Ascoli-Arzela theorem guarantees that converges uniformly on and converges uniformly on . Since is arbitrary, we find that It remains only to prove the continuity of at .

Let be given. Take with . Since is continuous at , we can find such that From (91), we obtain that which together with (92) means that is continuous at . From (92) and (96), is positive solution to BVP (1) in .

Now, we show that the uniqueness of follows when is decreasing on .

Suppose that, by contradiction, and are two different solutions. Obviously First, we show that In fact, if , let . Obviously , for all and (notice ). The monotonicity of implies that Integrate (101) from to to obtain From , we have That is, Then, is decreasing on , which together with means that . This is a contradiction. Hence, (100) is true.

Secondly, we show that Suppose that there exists with . Without loss of generality, we suppose that . Let . Equation (100) guarantees that , , and . The monotonicity of implies that The continuity of and at implies that there exists a such that Integrate (107) from to to obtain Now, implies that That is, Then, is decreasing on , which contradicts . Hence, (105) is true.

#### Conflict of Interests

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

#### Acknowledgments

The authors thank the referees for their suggestions. This research is supported by Young Award of Shandong Province (ZR2013AQ008) and the Fund of Science and Technology Plan of Shandong Province (2014GGH201010).

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