Research Article | Open Access

Hua Luo, "Bifurcation from Interval and Positive Solutions of a Nonlinear Second-Order Dynamic Boundary Value Problem on Time Scales", *Abstract and Applied Analysis*, vol. 2012, Article ID 316080, 15 pages, 2012. https://doi.org/10.1155/2012/316080

# Bifurcation from Interval and Positive Solutions of a Nonlinear Second-Order Dynamic Boundary Value Problem on Time Scales

**Academic Editor:**Paul Eloe

#### Abstract

Let be a time scale with . We give a global description of the branches of positive solutions to the nonlinear boundary value problem of second-order dynamic equation on a time scale , , which is not necessarily linearizable. Our approaches are based on topological degree theory and global bifurcation techniques.

#### 1. Introduction

Let be a time scale with , we consider the existence of positive solutions, in this paper, for a nonlinear boundary value problem of second-order dynamic equation on a time scale as follows: Research for the existence of solutions to the dynamic boundary value problem is rapidly growing in recent years. A great many existence results of positive solutions have been established for problem (1.1), see [1–5] and the references therein. The main tool used by them is the fixed point theorem in cones, and the key conditions in these papers do not depend on the first eigenvalue, , of the following linear problem: and the corresponding existence conditions are not optimal.

In 2006, for , , , with for , Luo and Ma [6] obtained the existence of at least one positive solution to problem: under the condition (H) if either or , where and is the first eigenvalue of the linear problem: The approaches adopted by Luo and Ma [6] are based on global bifurcation techniques. They obtained the existence of at least one positive solution by considering the branches of solutions, which bifurcate from one point. The key conditions in [6] depend on the first eigenvalue of the corresponding linear problem and the condition (H) is optimal!

In this paper, we will use the following assumptions. (A1) is continuous and there exist functions , such that for some functions with uniformly for , and for some functions with uniformly for . (A2) for . (A3) There exists a function such that

Obviously, (A1) means that is not necessarily linearizable at and . We consider the existence of positive solutions of problem (1.1) in this paper by using bifurcation techniques. The difference from [6] is that the branches of positive solutions under consideration now bifurcate from not one point, but an interval. Our main idea is from [7], in which they considered positive solutions of fourth-order boundary value problems for differential equations. The main tool we will use is the following global bifurcation theorem for problems which is not necessarily linearizable.

Theorem A (Rabinowitz, [8]). *Let be a real reflexive Banach space. Let be completely continuous such that , for all . Let be such that is an isolated solution of the equation
**
for , and , where , are not bifurcation points of (1.9). Furthermore, assuming that
**
where is an isolating neighborhood of the nontrivial solution, and denote the degree of on with respect to . Let
**
Then there exists a connected component of containing , and either*(i)* is unbounded, or*(ii)*.*

The rest of the paper is organized as follows. In Section 2, we firstly introduce the time scales concepts and notations that we will use in this paper. Next, Section 3 states some notations and proves some necessary preliminary results, and Section 4 studies the bifurcation from the trivial solution for a nonlinear problem which is not necessarily linearizable and then establishes our main result.

#### 2. Introduction for Time Scales

A time scale is a nonempty closed subset of , assuming that has the topology that it inherits from the standard topology on . Define the forward and backward jump operators by Here, we put , . Let which is derived from the time scale be and . Define interval on by .

*Definition 2.1. * If is a function and , then the *-derivative* of at the point is defined to be the number (provided that it exists) with the property that for each , there is a neighborhood of such that
for all . The function is called -differentiable on if exists for all .

* The second **-derivative of ** at *, if it exists, is defined to be . We also define the function and .

*Definition 2.2. * If holds on , we define the Cauchy -integral by

Lemma 2.3 (see [2, Theorems 2.7 and 2.8]). * Assume , then
**
Furthermore, if , is a continuous function on , then
*

Define the Banach space (denoted by ) to be the set of continuous functions with the norm For , we define the Banach space to be the set of the th -differential functions for which with the norm where

#### 3. Preliminaries and Necessary Lemmas

Assuming that , then from [9, Theorem 2.9], linear problem has a unique principal eigenvalue , with a corresponding positive eigenfunction.

Let , , and
We will work essentially in the Banach space with the norm
where
By a *positive solution* of problem (1.1), we mean is a solution of (1.1) with in and .

Lemma 3.1. *For , we have
*

*Proof. *By , we have that
and so
Therefore,

Define the linear operator , with Then is a closed operator, and is completely continuous, see [10, Lemma 3.7].

Let be the closure of the set of positive solutions to the problem We extend the function to a continuous function defined on by Then on . For , the arbitrary solution to the eigenvalue problem satisfies that on , and consequently, the graph of is concave down on . This together with the boundary conditions imply that Thus, (3.14) implies that is a nonnegative solution of problem (3.13), and the closure of the set of nontrivial solutions of (3.13) in is exactly .

Let , and let be the Nemytskii operator associated with the function :

Lemma 3.2. *Let on . Let be such that in , . Then
**
Moreover,
**
whenever . *

Let be the Nemytskii operator associated with the function Then (3.13), with , is equivalent to the operator equation In the following we will apply the Brouwer degree theory, mainly to the mapping , For , let .

Lemma 3.3. * Let be a compact interval with . Then there exists a number with the property
*

*Proof. *Suppose to the contrary that there exist sequences in and in and in , such that for all . By Lemma 3.2, for .

Set . Then from , we have . Since is bounded in , we infer that is relatively compact in , hence (for a subsequence) with in , . Let be defined as . Then is linear. For , .

Now, from condition (A1), we have that
According to
we get

Let and denote the eigenfunctions corresponding to and , respectively. Denote . Then is nondecreasing. From uniformly for , we have
According to and , we have from the first inequality in (3.24) that
Notice that
by Lemma 3.1. Let , by integration by parts (2.5), we have
and consequently
Similarly, we deduce from the second inequality in (3.24) that
Thus, . This contradicts .

*Remark 3.4. *If , then the value of do not contribute to the value of (3.28). Thus we can discuss the integral from to . For and are not defined at , we may define the values of and to be zero at , since the functions and are zero at . The details of the discussion can be found in [9, Page 497].

Corollary 3.5. * For and , . *

* Proof. *Lemma 3.3, applied to the interval , guarantees the existence of such that for
Hence, for any ,
which implies the assertion.

On the other hand, we have the following.

Lemma 3.6. * Suppose . Then there exists such that for all with , for all,
**
where is the positive eigenfunction corresponding to . *

* Proof. *We assume again to the contrary that there exist and a sequence with and in such that for all . As
and in , we can conclude from Lemma 3.2 that for .

Notice that has a unique orthogonal decomposition
with . Since on and , we have from (3.35) that .

Choose such that
By (A1), there exists , such that
Since , there exists , such that
and consequently
Applying (3.35) and (3.39), it follows that
Thus,
This contradicts (3.36).

Corollary 3.7. *For and , . *

* Proof. *Let , where is the number asserted in Lemma 3.6. As is bounded in , there exists such that , for all. By Lemma 3.6,
Hence,

Now, using Theorem A, we may prove the following.

Lemma 3.8. * is a bifurcation interval from the trivial solution for (3.19). There exists an unbounded component of positive solutions of (3.19) which meets . Moreover,
*

*Proof. *For fixed with , set , and . It is easy to check that for , all of the conditions of Theorem A are satisfied. So there exists a connected component of solutions of (3.19) containing , and either(i) is unbounded, or(ii).

By Lemma 3.3, the case (ii) cannot occur. Thus, is unbounded bifurcated from in . Furthermore, we have also from Lemma 3.3 that for any closed interval , if , then the fact that in is impossible. So must be bifurcated from in .

#### 4. The Main Result

We obtain the following main result in this paper.

Theorem 4.1. *Let (A1), (A2), and (A3) hold. Assuming that either
**
or
**
Then problem (1.1) has at least one positive solution.*

* Proof. *It is clear that any solution to (3.19) of the form yields a solution of problem (1.1). We will show that crosses the hyperplane in . To do this, it is enough to show that joins to . Let satisfy
We note that for all since is the only solution to (3.19) for , and .*Case 1* ()*.* In this case, we show that the interval
We divide the proof into two steps.* Step 1.* We show that is bounded.

Since , . From (A3), we have

Let denote the nonnegative eigenfunction corresponding to . From (4.5), by integration by parts formula (2.5), we have
Thus,
* Step 2.* We show that joins to .

From (4.3) and (4.7), we have that . Notice that (3.19) is equivalent to the equation
where is the Green’s function for problem , . So we have from (A1),
We divide the both sides of (4.9) by and set . Since is bounded in , choosing a subsequence and relabelling if necessary, we see that for some with in , and . Similar to the proof of Lemma 3.3, we have that
and , it is easy to verify that
which implies that

Let and denote the nonnegative eigenfunctions corresponding to and , respectively. Then we have from the first inequality in (4.12) that
By integration by parts formula (2.5), we obtain that
and consequently
Similarly, we deduce from the second inequality in (4.12) that
Thus,
So joins to .*Case 2* ()*.* In this case, if is such that
then
and, moreover,

Assuming that is bounded, applying a similar argument to that used in Step 2 of Case 1, after taking a subsequence and relabelling if necessary, it follows that
Again joins to and the result follows.

#### Acknowledgments

This work supported by China Postdoctoral Science Foundation funded Project (no. 201104602 and no. 20100481239), General Project for Scientific Research of Liaoning Educational Committee (no. L2011200), Teaching and Research Project of DUFE (no. YY12012), and the NSFC (no. 71201019).

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#### Copyright

Copyright © 2012 Hua Luo. 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.