Research

# Global structure of positive solutions for three-point boundary value problems

Jia-Ping Gu1, Liang-Gen Hu1 and Huai-Nian Zhang2*

Author Affiliations

1 Department of Mathematics, Ningbo University, Ningbo, 315211, P.R. China

2 Department of Mathematics and Physics, Beijing Institute of Petrochemical Technology, Beijing, 102617, P.R. China

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Boundary Value Problems 2013, 2013:174  doi:10.1186/1687-2770-2013-174

 Received: 29 April 2013 Accepted: 8 July 2013 Published: 25 July 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

In this paper, we are concerned with the three-point boundary value problem for second-order differential equations

where , , and ; and , satisfies for . The existence of the continuum of a positive solution is established by utilizing the Leray-Schauder global continuation principle. Furthermore, the interval of α about the nonexistence of a positive solution is also given.

MSC: 34B10, 34B18, 34G20.

##### Keywords:
positive solution; global continuous theorem; continuum; differential equation

### 1 Introduction

In this paper, we consider the following three-point boundary value problem for second-order differential equations:

(1.1)

where , , and ; and , satisfies for .

The existence and multiplicity of positive solutions for multi-point boundary value problems have been studied by several authors and many nice results have been obtained; see, for example, [1-6] and the references therein for more information on this problem. The multi-point boundary conditions of ordinary differential equations arose in different areas of applied mathematics and physics. In addition, they are often used to model many physical phenomena which include gas diffusion through porous media, nonlinear diffusion generated by nonlinear sources, chemically reacting systems, infectious diseases as well as concentration in chemical or biological problems. In all these problems, only positive solutions are very meaningful.

In 2009, Sun et al.[1] studied the three-point boundary value problem

(1.2)

where is a parameter, , , and . Based on Krein-Rutmann theorems and the fixed point index theory, they not only established the criteria of the existence and multiplicity of a positive solution, but also obtained the parameter μ in relation with the nonlinear term f and the first eigenvalue of the linear operator.

On the other hand, we note that the nice results in [1] only gave the existence and multiplicity of positive solutions, and if the parameter α is regarded as a variable, then an interesting problem as to what happens to the global structure of positive solutions of (1.2) was not considered. However, this relationship is very useful for computing the numerical solution of (1.2) as it can be used to guide the numerical work. For example, the global bifurcation of solutions for second-order differential equations has been extensively studied in the literature, see [4,7,8].

Motivated by this, in this paper, we consider the three-point boundary value problem for second-order differential equations (1.1) and make use of the Leray-Schauder global continuation theorem in the frame of techniques nicely employed by Ma and Thompson [4] and convex analysis technique. We consider two cases , and , , and establish the existence of continuum of positive solutions, where and . Moreover, the interval of parameter α about the nonexistence of positive solutions is also given. Our main results extend and improve the corresponding results [1,3,4]. In contrast to [[1], Theorem 3.1 and Theorem 3.2], we obtain the global structure and behavior of positive solutions, where the parameter α is regarded as a variable.

The rest of this paper is arranged as follows. In Section 2, we give Green’s function and some lemmas. In Section 3, we consider the case , , and give the existence of the continuum of positive solutions and the interval of parameter α about the nonexistence of positive solutions. In Section 4, we study the case , , and give the existence of global continuum of positive solutions.

### 2 Preliminaries and lemmas

Let denote the Banach space of a continuous function with the maximum norm

Define a set by

then P is a cone.

We assume that

(H0) , , and .

Lemma 2.1 (see [[1], Lemma 2.1])

Suppose that condition (H0) holds and. Then the following linear differential equation

has a unique solution

whereis defined by

(2.1)

For the sake of convenience, we list the following hypotheses:

(H1) .

(H2) satisfies for .

(H3) , (superlinear).

(H4) , (sublinear).

Lemma 2.2Assume that (H0) holds. Letwithforand letube a solution of

(2.2)

Thenfor. Moreover, iffor some, thenfor all.

Proof We only show that if for some , then for all .

If , then we have from [[3], Lemma 2] that the results hold.

Next, we consider the case . If it is not true, then there exists some such that

(2.3)

We separate the proof into two cases: Case I: and Case II: .

Case I. If , then . Since u is concave down in , we obtain that and for all . Set

then we find from the boundary conditions in (2.2) that

This together with the concavity of u leads to

Case II. Consider the case .

(1) If , then and the concavity of u imply that , . Hence, we get that , and (since , we have that leads to . This contradicts (2.3)). Again, since u is concave, we have

Consequently, we obtain that contradicts the condition .

(2) If , then, adopting the same proof as in Case I, we get a contradiction.

(3) If , then it follows that . In light of and the concavity of u, we get from (2.3) that

and

where and . Hence, we get from the boundary condition of (2.2) that

Consequently, we get from Case I and Case II that the conclusion holds. □

Remark 2.1 If is positive, then we know from the proof in Lemma 2.2 that may only have zero point at and .

Lemma 2.3Let (H0) hold and letbe a function satisfying

and

Then there existssuch that

Proof If , then from [[4], Lemma 3.3] the conclusion holds.

Next, we consider the case . Clearly, it is easy to see from Lemma 2.2 that for .

(1) If , then . This together with the concavity of u yields . Since , there exists such that .

(2) If , then . Consequently, there exists such that

and

The assumption and the concavity of u in imply that

This completes the proof. □

From (2.1), we define an operator as follows:

(2.4)

Assume that (H0)-(H2) hold, then it is easy to verify that is well defined and completely continuous. We note that u is a positive solution of problem (1.1) if and only if on P.

By a positive solution of (1.1) we mean a solution of (1.1) which is positive on .

Denote by the closure of the set

in .

Using the Leray-Schauder global continuation theorem [[8], Theorem 14.C], Ma and Thompson [[4], Lemma 2.2] obtained the following result.

Lemma 2.4LetPbe a cone in a Banach spaceX. Letbe a bounded and open subset inXwith respect to the topology induced byonP. Assume that the operatoris a continuous, compact map satisfying:

(1) the equationhas no solution on;

(2) with.

Then the set

has a continuumof solutions in, which connects the setwith the set.

### 3 The superlinear case

For a given , we let

From (H0)-(H2) and Lemma 2.2, we get that

(3.1)

For any , we denote

From (H0)-(H2) and Lemma 2.2, we have that

(3.2)

Lemma 3.1 [[1], Theorem 3.2]

Let conditions (H0)-(H3) hold. Then there exist two constantsandwithsuch that problem (1.1) has at least one positive solutionwith. Furthermore,

where.

Proof Since and , we take in [[1], Theorem 3.2] and all the conditions in [[1], Theorem 3.2] are satisfied. Therefore, the conclusion holds. □

Lemma 3.2Assume that (H0)-(H3) hold. Letandbe the constants as in Lemma 3.1. Then there exists a positive numberwithsuch that

whereis defined by (3.2).

Proof First we claim that if , then there exists a positive number such that

(3.3)

Suppose this fails, that is, there exists a sequence with such that

We may suppose that for some . For the sake of convenience, we denote . From (H1), (H2) and Lemma 2.2, we see that for . Set , then we get

(3.4)

where

From Remark 2.1, we know that the set has at most two points and . Therefore, from the hypothesis of , it follows that is continuous in and there exists a constant , independent of n, such that

This together with (3.4) yields

(3.5)

In light of Lemma 2.3, there exists ,  , such that

(3.6)

Applying the Newton-Leibniz formula, we find

Consequently, combining (3.5) and (3.6), we conclude that

for some constant independent of n. Utilizing the Ascoli-Arzela theorem, we have that is a relatively compact set on . Assume, taking a subsequence if necessary, that in . Then and in .

On the other hand, from and the fact as , it follows that

(3.7)

uniformly holds for all . From (2.4) and the fact , we get

This together with (3.7) and the Lebesgue dominated convergence theorem, we get

contradicts . Therefore, the claim (3.3) holds.

Next, we prove that if , then there is a positive number such that

(3.8)

Suppose on the contrary that there exists a sequence with such that

We define . From (H0)-(H2), Lemma 2.2 and the concavity of , it follows that

and

(3.9)

Take , where . (H3) implies that there exists a constant such that

(3.10)

Since , we find a sufficiently large such that

This together with (3.9) and (3.10) yields

(3.11)

Put for . Hence, we have from (3.11) that

On the other hand, multiplying (1.1) by ψ and integrating by parts, we find

Then we obtain

This is a contradiction. Consequently, conclusion (3.8) holds.

Combining (3.3) and (3.8), we let . Thus, the result holds. □

Theorem 3.1Assume that conditions (H0)-(H3) hold. Thencontains a continuum which joinswith.

Proof We divide the proof into four steps.

Step 1. We construct a continuum.

For arbitrarily given , let be as in Lemma 3.2. Define a set by

(3.12)

It follows from Lemma 3.1 and the excision property of the fixed point index that

From Lemma 3.2, we know that has no solutions in . Therefore, from Lemma 2.4, there exists a continuum which joins with . Here is defined by (3.2), and and are defined by (3.1).

Let be the closure of the set

(3.13)

and let

Since , we know that .

Step 2. We show that there exists satisfying

(3.14)

If it is not true, then there exists such that

(3.15)

Taking , it follows that . Let be a given number by Lemma 3.1 and the set

Then we know, from Lemma 3.1 and the excision property of the fixed point index, that

From Lemma 3.2, has no solution in . Again, using Lemma 2.4, we find a continuum which joins with . This contradicts (3.15). Therefore, the conclusion in (3.14) holds.

Step 3. Let ζ be a continuum satisfying (3.14). We claim that

(3.16)

Suppose on the contrary that there exists a sequence such that and . From (H0)-(H2), Lemma 2.2 and the concavity of , it follows that

and

Adopting the same proof as in the second step in Lemma 3.2, we can find a contradiction. Hence, the result in (3.16) holds.

Step 4. Let ζ be a continuum satisfying (3.14). Next we show that

(3.17)

If it is not true, then we have

for some . Then there exists a sequence such that

and

(3.18)

From conditions (H0)-(H2) and Lemma 2.2, we get that on and the graph of is strictly concave down on .

(1) If , then we obtain from the boundary condition of (3.18) that and . From the strict concavity of , it follows that

implies

(2) If , then (since , we know that and the strict concavity of imply on , which is a contradiction). Put

Then

From the strict concavity of , we get that

Consequently, the conclusion in (3.17) holds. □

Remark 3.1 In contrast to [[1], Theorem 3.2], we obtain the global structure and behavior of positive solutions, where the parameter α is regarded as a variable.

Theorem 3.2Suppose that, and. Let condition (H1) andhold, and letbe a solution of

(3.19)

Then problem (3.19) has no positive solutions.

Proof If , then we know that for is a trivial solution of (3.19).

Suppose on the contrary that there exists a positive solution to equation (3.19), i.e.,

(3.20)

Therefore, we know from equation (3.19) that and is concave down in . Now, we consider two cases. Case I: ; Case II: .

(1) Case I. .

Clearly, we know from the boundary conditions of (3.19) that and . Since u is concave down and u is a positive solution of equation (3.19), we obtain that

implies

i.e.,

(2) Case II. .

If , then and imply that , . This contradicts (3.20).

If

(3.21)

then the concavity of u implies that

Thus

(3.22)

For , we obtain that

For , we have . Since and , we get from Taylor’s expansion that

(3.23)

where . Substituting (3.23) into (3.22), we find that

From the boundary condition of (3.19), it follows that

Therefore, we conclude that if , then problem (3.19) has no positive solutions. □

### 4 The sublinear case

Lemma 4.1 [[1], Theorem 3.1]

Let conditions (H0)-(H2) and (H4) hold. Then there exist two constantsandwithsuch that problem (1.1) has at least one positive solutionwith. Furthermore,

where.

Lemma 4.2Assume that (H0)-(H2) and (H4) hold. Letandbe the constants as in Lemma 4.1. Then there exists a positive numberwithsuch that

whereis defined by (3.2).

Proof First, we claim that if , then there exists a positive number such that

(4.1)

Suppose this fails, that is, there exists a sequence with such that

From conditions (H0)-(H2) and Lemma 2.2, it follows that

The concavity of implies

Choose , where . In light of (H4), we get that there exists a constant such that

Since , we get that

implies

Adopting the same proof as in Lemma 3.2, we get a contradiction. Hence, conclusion (4.1) holds.

Now, we show that if , then there exists a positive number such that

(4.2)

Define the nondecreasing function by

Since , it follows that

(4.3)

Suppose that conclusion (4.2) fails, that is, there exists a sequence with . We may assume that . From (H0)-(H2) and Lemma 2.2, we have that

Define , which implies that

Since , we get from (4.3) that

uniformly holds for . Again, applying the proof method as that in Lemma 3.2, we get a contradiction. Consequently, conclusion (4.2) holds.

If we let , then combining (4.1) and (4.2), we have that the result holds. □

Theorem 4.1Let (H0)-(H2) and (H4) hold. Thencontains a continuum which joinswith.

Proof Applying the method as in Theorem 3.1, we find from Lemma 2.2, Lemma 4.1 and Lemma 4.2 that there exists a continuum satisfying

and

Next, we only show that

Suppose on the contrary that there exists a sequence with such that

Define . From conditions (H0)-(H2) and Lemma 2.2, it follows that for any ,

and the concavity of leads to

Take , where . In light of (H4), we get that there exists a constant such that

Using the same proof as in Lemma 3.2, we get a contradiction.

Hence, the conclusion holds. □

Remark 4.1 In contrast to [[1], Theorem 3.1], we obtain the global structure and behavior of positive solutions, where the parameter α is regarded as a variable.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed significantly in writing this paper. All authors read and approved the final manuscript.

### Acknowledgements

The work was supported partly by NSFC (No. 11201248), K.C. Wong Magna Fund of Ningbo University and Ningbo Natural Science Foundation (No. 2012A610031).

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