Abstract

Let be a time scale and , . We study the nonlinear fourth-order eigenvalue problem on , , , and obtain the existence and nonexistence of positive solutions when and , respectively, for some . The main tools to prove the existence results are the Schauder fixed point theorem and the upper and lower solution method.

1. Introduction

The deformation of an elastic beam with one end fixed and the other end free can be described by the nonlinear fourth-order dynamic eigenvalue problem on where is a time scale, is a parameter, , and .

Nonlinear dynamic eigenvalue problems of the above type have been studied by some authors, but most of them study only second-order dynamic equations. In 2000, Chyan and Henderson [1] obtained the existence of at least one positive solution for some to second-order case of the dynamic equation in problem (1.1) under conjugate boundary value condition and right focal boundary value condition, respectively. Anderson [2] discussed the same second-order dynamic equation under the Sturm-Liouville boundary value condition and directly generalized the result of [1]. Erbe et al. [3] then studied the general second-order Sturm-Liouville dynamic boundary value problem and obtained the existence, nonexistence, and multiplicity results of positive solutions. In 2005, Li and Liu [4] further studied the dependence of positive solutions on the parameter for the second-order dynamic equation under the right focal boundary value condition. Luo and Ma [5] in 2006 were concerned with the existence and multiplicity of nodal solutions and obtained eigenvalue intervals of the nonlinear second-order dynamic eigenvalue problem under conjugate boundary value condition by using bifurcation methods. In 2007, Sun et al. [6] obtained some sufficient conditions for the nonexistence and existence of at least one or two positive solutions for the -Laplacian three-point dynamic eigenvalue problem with mixed derivatives by using the Krasnosel’skii’s fixed point theorem in a cone. In 2009, Luo [7] derived the eigenvalue intervals in which there exist positive solutions of a singular second-order multipoint dynamic eigenvalue problem with mixed derivatives by making use of the fixed point index theory.

As for the nonlinear higher-order dynamic eigenvalue problems, few papers can be found in the literature to the best of our knowledge. L. Kong and Q. Kong [8], and Boey and Wong [9] discussed the even-order dynamic eigenvalue problem and the right focal eigenvalue problem, respectively, but their problems do not contain (1.1). Particularly for fourth-order problems and special case , Wang and Sun [10] studied the existence of positive solutions for dynamic equations under nonhomogeneous boundary-value conditions describing an elastic beam that is simply supported at its two ends. And both Karaca [11] and Pang and Bai [12] obtained the existence of a solution for two classes of fourth-order four-point problems on time scales by the Leray-Schauder fixed point theorem and the upper and lower solution method, respectively, but the problems they studied are different to (1.1).

This paper studies the relationship between the existence and nonexistence of positive solutions and the value of parameter . We find the existence of a such that problem (1.1) has positive solutions for and no positive solutions for .

The rest of this paper is organized as follows: in Section 2, we firstly introduce the time scales concepts and notations and present some basic properties on time scales which are needed later. Next, Section 3 gives some preliminary results relevant to our discussion, and Section 4 is devoted to establish our main theorems.

2. Introduction to Time Scales

The calculus theory on time scales, which unifies continuous and discrete analysis, is now still an active area of research. We refer the reader to [1316] and the references therein for introduction on this theory. For the convenience of readers, we present some necessary definitions and results here.

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 be the graininess function. And which are derived from the time scale is 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 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 . Similarly, is called the th -derivative of at . We also define the function .

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

Lemma 2.3. If , , is continuous and on , then

Proof. From [15, Theorems  1.28 (Viii) and  1.29], it is clear.

Lemma 2.4 (See [14, Theorem  1.16]). If the -derivative of exists at , then

Define the Banach space 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

Throughout this paper, we assume that both exist and . So there exists a number such that

We also make the following assumptions: is continuous and on ; is continuous. is nondecreasing in , nonincreasing in and .

Set . Then problem (1.1) is equivalent to the system According to [14, Corollary  4.84 and Theorem  4.70], the Green’s function of problems is of the same form and the solution of system (3.3) is Therefore, the solution of problem (1.1) is

Lemma 3.1. Green’s function (3.5) is of the following properties:

Proof. We here only give the proof of (3.10), and the others can be obtained easily. We divide the proof into the following four cases.
Case 1 (). We have
Case 2 (). We have
Case 3 (). We have
Case 4 (). We have

Define

Lemma 3.2. For , one has where , , .

Proof. Firstly, we show that (3.18) holds.
For , we have from that Combining this with we have .
Secondly, we show that (3.19) holds.
For , we have from that Combining this with we have .

Define Then is a Banach space under the norm . Set Then . Since , we have that is completely continuous.

Lemma 3.3. Suppose and hold, and is a solution of problem (1.1), then

Proof. By , (3.7), (3.8) and the fact that , are nonnegative functions, (3.26) holds. For , from (3.7), (3.26), (3.2) and (3.12), we have Therefor (3.27) holds.
For , from (3.7), (3.11), and (3.12), we have Thus (3.28), (3.29), and (3.30) hold.

At the end of this section, we state a lemma of the upper and lower solution method, which is needed for some proofs in next section.

Lemma 3.4 (See [17, Theorem  3.3.8]). Let be a cone with nonempty interior in space , and a completely continuous and increasing operator. Suppose the following conditions hold:
(i)there exist , such that ;(ii)there exist and a constant , such that ;(iii) is in the interior of , and there exists , such that . Then has at least one fixed point in .

4. The Main Result

Our main result is the following existence theorem.

Theorem 4.1. Suppose and hold, and either or () holds. Here;).Then there exists , such that problem (1.1) has at least one positive solution for , and has no positive solution for .

To prove Theorem 4.1, first, we show that problem (1.1) has positive solutions for some small enough.

Theorem 4.2. Suppose and hold. Then there exists , such that problem (1.1) has at least one positive solution for .

Proof. The fixed point of defined in (3.25) is the solution of problem (1.1), so it will be enough to find the fixed point of .
Set Let For , we have Set If , then Consequently, for , Then that is, . By the Schauder fixed point theorem, has at least one fixed point in satisfying . From Lemma 3.3, is a positive solution of problem (1.1).

Next, we show that there exist no positive solution for some large enough.

Theorem 4.3. Suppose that and , hold, and either or holds. Then problem (1.1) has no positive solution for .

Proof. Suppose is a solution to problem (1.1) for some . We divide our discussions into two cases.
Case 1 (, and hold). By , for a fixed , there is such that If with , then from (3.30) and (3.28), we have Further by (4.8), If with , then from (3.27), there is with Set , we have from (4.10) and (4.11) that Combining this with (3.7), (4.13), (3.29), (3.12), and (3.2), we have for , If we choose such that then which is a contradiction.
Case 2 (, , and hold). Similar to Case 1, by , there is also an , such that Thus for , we have from (3.7), (4.17), (3.28), and (3.12) that If we take such that then which is a contradiction.
Therefore, problem (1.1) has no positive solution for .

Define the set.

Theorem 4.4. One has

Proof. Let be defined as Theorem 4.2. For , the result holds from Theorem 4.2. So, we discuss the case that .
For , three cases will be discussed.;;.Cases  (1) and  (3) are clear from Theorem 4.2 and the assumption , respectively. Now, we deal with Case  (2).
Define Then is a cone with nonempty interior in . For , let be defined as (3.25). Then and is an increasing operator from . Set and as positive solutions of problem (1.1) at and , respectively. Then So is an upper solution of the operator and is a lower solution. If , then there exists a positive solution satisfying for by the upper and lower solution method. If , we verify the conditions of Lemma 3.4.
Clearly, the condition (i) in Lemma 3.4 holds for .
For , we have from (3.10), (3.12), (3.11), and (3.2) that Then and the condition (ii) of Lemma 3.4 is satisfied for .
From (3.26) and (3.27), we have that is in the interior of . From (3.8), there is This implies for and the condition (iii) of Lemma 3.4 is satisfied.
By Lemmas 3.4 and 3.3, we get a positive solution in . That is, .

Now, we give the proof of Theorem 4.1.

Proof of Theorem 4.1. From Theorems 4.2 and 4.3, is bounded. Thus, we can define . Firstly, we show that .
Choose a sequence , which belongs to a compact subinterval in , and . Then there exists , for , Let satisfy If is uniformly bounded, then there exists such that and Consequently, By Lemma 3.3, is a positive solution of problem (1.1), and .
Next, we will prove that is uniformly bounded and the discussions will be divided into two cases.
Case 1 (, , and hold). From , there exists , where satisfies Suppose on the contrary that is unbounded. Then which implies that there exists , for . From (3.30) and (3.28), we have Subsequently, for , , by (4.32), (3.29), (3.12), (4.28), (3.2), and (4.33). This is a contradiction.
Case 2 (, , and hold). From , there exists , where satisfies Suppose on the contrary that is unbounded. Then similar to Case 1, there exists such that for . Thus for , , we have by (4.37), (3.28), (3.12), (4.28) and (4.38). This is also a contradiction.
According to the definition of and , and Theorem 4.4, we complete the proof.

Acknowledgments

The authors would like to thank editor and the referees for carefully reading this paper and suggesting many valuable comments. This paper is supported by China Postdoctoral Science Foundation Funded Projects (no. 201104602 and no. 20100481239) and General Project for Scientific Research of Liaoning Educational Committee (no. L2011200).