Abstract
In this paper, we are concerned with the threepoint boundary value problem for secondorder differential equations
where , , and ; and , satisfies for . The existence of the continuum of a positive solution is established by utilizing the LeraySchauder 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 equation1 Introduction
In this paper, we consider the following threepoint boundary value problem for secondorder differential equations:
where , , and ; and , satisfies for .
The existence and multiplicity of positive solutions for multipoint boundary value problems have been studied by several authors and many nice results have been obtained; see, for example, [16] and the references therein for more information on this problem. The multipoint 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 threepoint boundary value problem
where is a parameter, , , and . Based on KreinRutmann 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 secondorder differential equations has been extensively studied in the literature, see [4,7,8].
Motivated by this, in this paper, we consider the threepoint boundary value problem for secondorder differential equations (1.1) and make use of the LeraySchauder 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
Lemma 2.1 (see [[1], Lemma 2.1])
Suppose that condition (H0) holds and. Then the following linear differential equation
has a unique solution
For the sake of convenience, we list the following hypotheses:
Lemma 2.2Assume that (H0) holds. Letwithforand letube a solution of
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
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
This contradicts the hypothesis .
(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
leads to
This is a contradiction.
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
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:
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
Using the LeraySchauder 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;
Then the set
has a continuum ℒ of solutions in, which connects the setwith the set.
3 The superlinear case
From (H0)(H2) and Lemma 2.2, we get that
From (H0)(H2) and Lemma 2.2, we have that
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,
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
Proof First we claim that if , then there exists a positive number such that
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
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
In light of Lemma 2.3, there exists , , such that
Applying the NewtonLeibniz formula, we find
Consequently, combining (3.5) and (3.6), we conclude that
for some constant independent of n. Utilizing the AscoliArzela 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
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
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
Take , where . (H3) implies that there exists a constant such that
Since , we find a sufficiently large such that
This together with (3.9) and (3.10) yields
Put for . Hence, we have from (3.11) that
On the other hand, multiplying (1.1) by ψ and integrating by parts, we find
leads to
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
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
and let
Step 2. We show that there exists satisfying
If it is not true, then there exists such that
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
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
If it is not true, then we have
for some . Then there exists a sequence such that
and
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
This is a contradiction.
(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
a contradiction.
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
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.,
Therefore, we know from equation (3.19) that and is concave down in . Now, we consider two cases. Case I: ; Case II: .
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.,
If , then and imply that , . This contradicts (3.20).
If
then the concavity of u implies that
Thus
contradicts (3.21).
For , we have . Since and , we get from Taylor’s expansion that
where . Substituting (3.23) into (3.22), we find that
From the boundary condition of (3.19), it follows that
leads to
This is a contradiction.
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,
Lemma 4.2Assume that (H0)(H2) and (H4) hold. Letandbe the constants as in Lemma 4.1. Then there exists a positive numberwithsuch that
Proof First, we claim that if , then there exists a positive number such that
Suppose this fails, that is, there exists a sequence with such that
From conditions (H0)(H2) and Lemma 2.2, it follows that
Choose , where . In light of (H4), we get that there exists a constant such 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
Define the nondecreasing function by
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
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 ,
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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