#### Abstract

Let be an integer, and let. We discuss the spectrum of discrete linear second-order eigenvalue problems , , where is a parameter, changes sign and on . At last, as an application of this spectrum result, we show the existence of sign-changing solutions of discrete nonlinear second-order problems by using bifurcate technique.

#### 1. Introduction

Let be an integer, . Let us consider the spectrum of the discrete second-order linear eigenvalue problem where is a parameter, and changes sign on ; that is, satisfies the following.

(H0) There exists a proper subset , such that Let be the number of elements in . Then .

In [1], Ince studied the second-order linear eigenvalue problem where is continuous and changes sign. He obtained the following result.

Theorem A. *Problem (4) has an infinite sequence of simple eigenvalues
**
and the eigenfunction corresponding to has exactly simple zeros in (0, 1).*

This result has been extended to one-dimensional Laplacian operator by Anane et al. [2] and to the high-dimensional case by Hess and Kato [3], Bongsoo and Brown [4], and Afrouzi and Brown [5]. Meanwhile, these spectrum results have been used to deal with several nonlinear problems; see, for example, [4–7] and the references therein.

For the discrete case, Atkinson [8] studied the discrete linear eigenvalue problems and obtained that (6) and (7) have exactly real eigenvalues, which can be ordered as . Here , , and is some fixed real number.

In 1995, Jirari [9] studied (6) with the more general boundary conditions where . He got that (6) and (8) have real eigenvalues, which can be ordered as .

However, these two results do not give any information of the eigenfunctions of the linear eigenvalue problems (6) and (7) or (6) and (8).

In 1991, Kelley and Peterson [24] investigated the line eigenvalue problems where on is defined and real valued on and on . They proved the following.

Theorem B. *Problem (9) has exactly real and simple eigenvalues , , which satisfies
**
and the eigenfunction corresponding to has exactly simple generalized zeros.*

Furthermore, when , Agarwal et al. [10] generalized the results of Theorem B to the dynamic equations on time scales with Sturm-Liouville boundary condition. Moreover, under the assumption that the weight functions are not changing sign, several important results on linear Hamiltonian difference systems have also been established by Shi and Chen [11], Bohner [12], and the references therein.

However, there are few results on the spectrum of discrete second-order linear eigenvalue problems when changes its sign on . In 2008, Shi and Yan [13] discussed the spectral theory of left definite difference operators when may change its sign. However, they provided no information about the sign of the eigenvalues and no information about the corresponding eigenfunctions. Recently, Ma et al. [14] obtained that (1) and (2) have two principal eigenvalues and they studied some corresponding discrete nonlinear problems.

It is the purpose of this paper to establish the discrete analogue of Theorem A for the discrete problems (1) and (2). More precisely, we will prove the following.

Theorem 1. *Let be the number of elements in . Let (H0) hold and let . Then*(a)*if , then (1) and (2) have real and simple eigenvalues, which can be ordered as follows:
*(b)*every eigenfunction corresponding to the eigenvalue has exactly simple generalized zeros.*

The rest of the paper is arranged as follows. In Section 2, some preliminaries will be given including Lagrange-type identities. In Section 3, we develop a new method to count the number of negative and positive eigenvalues of (1) and (2), which enable us to prove Theorem 1. Finally in Section 4, we apply our spectrum theory and the Rabinowitz's bifurcation theorem to consider the existence of sign-changing solutions of discrete nonlinear problems where is a real parameter, changes its sign, on , and is continuous.

*Remark 2. *There is also much literature dealing with difference equations similar to (12) subject to various boundary value conditions. We refer to [15–22] and the references therein. However, the weight in these papers.

#### 2. Preliminaries

Recall that . Let , . Then is a Banach space under the norm . Let . Then is a Banach space under the norm .

*Definition 3 (see [22]). *Suppose that a function . If , then is a zero of . If and for some , then is a simple zero of . If for some , then has a node at the point
The nodes and simple zeros of are called the simple generalized zeros of .

To find the eigenvalue of (1) and (2), we rewrite (1) as follows: From (15), it can be seen that if is an eigenvalue of (1) and (2), then . Without loss of generality, we suppose that Further, from (15), (2), and (16), for each , is precisely a polynomial of degree of , we denote it by .

Lemma 4 (Lagrange-type identities). *For ,
*

*Proof. *We write (15) for the two arguments in full, giving
Multiplying, respectively, by and subtracting, we have
and putting and recalling that , we derive (17) with . Induction over then yields (17) from (19) in the general case.

Corollary 5. *For ,
*

*Proof. *Dividing (17) by and making for fixed , then we get the desired result.

Corollary 6. *For , and complex ,
*

*Proof. *Set in (17). Then (21) is obtained.

#### 3. Spectrum of (1) and (2)

Lemma 7. *For , if is a solution of
**
satisfying , , and on , then
*

Moreover, if , then . If , then .

*Proof. *It is easy to see that . Now, multiplying (22) by , then we get that
which implies the desired result.

Now, we prove some oscillatory properties.

Lemma 8. *For , the polynomial has precisely real and simple zeros.*

*Proof. *This proof is divided into two steps.*Step **1 (each zero of ** is real)*. Suppose on the contrary that has a complex zero ; then . Furthermore,
On the other hand, if is a zero of , then is an eigenvalue of the linear eigenvalue problem
Now, by Lemma 7 and Corollary 6, we get that the above determinant does not equal zero, which is a contradiction. Hence, the zeros of are all real for .*Step **2 (all of the zeros of ** are simple)*. Suppose on the contrary that has a multiple zeros , necessarily real. Then and . Moreover,
However, by Lemma 7 and Corollary 5, we get that the above determinant does not equal zero, a contradiction.

Lemma 9 (see [14]). *Let . Then (1) and (2) have two principal eigenvalues and the corresponding eigenfunctions do not change their sign.*

From Lemma 8, it follows that the spectra of (1) and (2) consist of real eigenvalues. Furthermore, by Lemma 9, there exists such that such real eigenvalues can be ordered as follows: Define by Then we get the following.

Lemma 10. *Let . Then , where denotes the identity operator.*

*Proof. *Suppose that and are two eigenfunctions corresponding to . Then and there exists a constant such that . Now, by the recurrence relation (15), we get that for .

By Lemmas 8–10, all of the eigenvalues of (1) and (2) are real and simple. Now, for fixed , let us investigate the number of sign-changing times of the following sequence, So far, has only been defined for integral values of . We extend it to a continuous function , , specifically, for , to be a linear function of .

Lemma 11. *For fixed real , the zeros of , are simple.*

*Proof. *Suppose that is a zero of . Now, the proof can be divided into two cases.*Case **1*. If , then by virtue of (15), we get that .*Case **2*. If for some . Then exists at and is not zero by the definition of .

Using the same method used in [1, Page 102], we may prove the following.

Lemma 12. *Let be the zeros of . Then is a continuous function of .*

Lemma 13. *For , the zeros, , of , is a decreasing function of . For , is an increasing function of for .*

*Proof. *Let the zero occur in . Since is linear in , the location of this zero is given by
conversely, as given by this equation will actually be a zero of if and if so the given falls in the interval . If we differentiate the right of (31), with respect to , the result is found to be
and this does not equal zero by Lemma 7 and Corollary 5. Furthermore, for , and for , . This combines with the continuity of , and we get the desired result.

Now, we can set up the oscillatory characterization of the eigenvalues of (1) and (2).

Lemma 14. *Let . The sequence
**
exhibits for no changes of sign; for , exactly changes sign; for , exactly changes sign; for , exactly changes sign.*

*Proof. *To prove the times of changes of sign of (33), it is equivalent to find the number of zeros of , . We only deal with the case that ; the case is similar.

By Lemma 12, for , there exist functions , which satisfy and are all decreasing function of ; moreover, for fixed , there are the zeros of . Since and , it follows that for .

For , by Lemma 8, we get that does not have a zero in .

Let be arbitrary. Since is continuous and decreasing, will intersect with at (, ). Moreover, and , which implies that . Thus, for , (33) changes its sign exactly one time.

Now, we claim that for the same , and have no common zero for each .

Suppose the contrary, then there exists such that
Let
Then, for , is the th zero of .*Case **1*. If there exists such that , then by the definition of , we obtain that the signs of and are opposite. Without loss of generality, suppose that and .

Now, consider the variation of when varies. Take sufficiently small, by the continuity of with respect to , for , , and also , since is a continuous function of . However, for , is the th zero of , which implies that , , a contradiction.*Case **2*. If there exists such that , then we consider the sign of and , and we can get the similar contradiction as in Case 1.

This claim implies that, for the same do not intersect with each other. Thus, for , there are functions: , which intersect with in at different points; that is, for , has exactly zeros.

This completes the proof.

Lemma 15. *Problems (1) and (2) have exactly positive eigenvalues and exactly negative eigenvalues.*

*Proof. *First we show that changes its sign exactly times.

Let us consider the following ordered polynomials:
where is a polynomial of degree precisely of .*Observation **1. *Consider
where NSC is the number of sign changes of
*Observation **2*. For , denoted by , the number of the elements in the set
Then . Now
if is large enough. Since
changes sign exactly times, it follows that
changes sign exactly times as is large enough; that is,
This together with (37) implies that . So, changes its sign exactly times.

Next, using the result of first step and Lemma 14, it follows that .

*Proof of Theorem 1. *From Lemma 7–Lemma 15, the results of Theorem 1 hold.

#### 4. Application

As an application, we consider the existence of sign-changing solutions of the discrete nonlinear boundary value problems (12), (13).

In this section we suppose that(H1) with , ; for , for ;(H2)there exist such that

for simplicity, we give some notations at first.

For , , let denote the set of functions in such that(1) has exactly simple generalized zeros in ;(2).

Define . They are disjoint in . Finally, let and let .

Theorem 16. *Suppose that (H0), (H1), and (H2) hold. Assume that . Then*(i)*if
* *problems (12) and (13) have at least four sign-changing solutions , , , and ;*(ii)*if
* *problems (12) and (13) have at least four sign-changing solutions , , , and .**Moreover, for , there also exist at least two sign-changing solutions and ; meanwhile, for , there also exist at least two sign-changing solutions and .*

Theorem 17. *Suppose that (H0), (H1), and (H2) hold. Assume that . Then*(i)*if
* *problems (12) and (13) have at least four sign-changing solutions , , , and ;*(ii)*if
* *problems (12) and (13) have at least four sign-changing solutions , , , and .**Moreover, for , there exist at least two sign-changing solutions and ; meanwhile, for , there also exist at least two sign-changing solutions and .*

If condition (H2) is replaced by(2),

then we obtain the following result.

Theorem 18. *Let (H0), (H1), and (2) hold. Then (12) and (13) have a sign-changing solution in , if and only if . Moreover, for , there exist at least two solutions and , and there also exist at least two solutions and for .*

Recall that ; then Let be such that Clearly Let us consider as a bifurcation problem from the trivial solution .

Equation (52) can be converted to the equivalent equation Further we note that for near in , since where and

Lemma 19. *Suppose that is a nontrivial solution of (12) and (13); then there exists such that
*

*Proof. *Suppose on the contrary that for every
Then there exists such that
Since on , we may assume that
On the other hand, it follows from (52) and that
which implies that
However, by (59) and the fact , we get
which contradicts (58).

Now, the results of Rabinowitz [23] for (52) can be stated as follows.(i)For each integer , there exists a continuum of solutions to (52) joining to infinity in . Moreover, .(ii)For each integer , there exists a continuum of solutions to (52) joining to infinity in . Moreover, .

Lemma 20. *Suppose that (H0), (H1), and (H2) hold. Then for ,
*

*Proof. *Suppose on the contrary that there exists such that
By (H0), (H1), and (H2), there exists such that is strictly increasing in for . Then
and since ,
Subtracting, we get
Let , and applying boundary value problem (12) and (13), we have

Let , such that
Let be the Green function of the boundary value problem
From , , applying Theorem 6.8 and Corollary 6.7 of [24], we have and for .

Problem (70) is equivalent to
By using the positivity of and , we have
that is, . This contradicts (64).

There exists such that
Note that in this case , so we can choose such that is strictly decreasing in for . Then
and since ,
Subtracting, we get
Let , and applying boundary value condition (13), we have
Similar to the above proof, we have . This contradicts (73).

In the following we will investigate other sign-changing solutions of problems (1) and (2).

Let be such that Clearly Let us consider as a bifurcation problem from infinity. We note that (80) is equivalent to (12) and (13).

Now, the results of Rabinowitz [25] for (80) can be stated as follows.(i)For each integer , there exists a continuum of solutions to (80) meeting .(ii)For each integer , there exists a continuum of solutions to (80) meeting .

Lemma 21. *Suppose that (H0), (H1), and (H2) hold. Then for , we have
*

*Proof. *It is similar to the proof of Lemma 20, so we omit it.

Lemma 22. *Suppose (H0), (H1), and (H2) hold. Then
*

*Proof. *Firstly, we will prove . Take as an interval such that and is a neighborhood of whose projection on lies in and whose projection on is bounded away from . Then by [25, Theorem 1.6 and Corollary 1.8], we have that either(1) is bounded in in which case meets; or(2) is unbounded.

Moreover if (2) occurs and has a bounded projection on , then meets where .

Obviously Lemma 21 implies that (1) does not occur. So is unbounded.

Lemma 19 guarantees that is a component of solutions of (80) in which meets . Therefore there is no such that also meets . Otherwise, there will exist such that has a multiple zero point ; that is, and . However this contradicts with Lemma 19, and consequently is unbounded. Thus
and similarly we have

*Proof of Theorems 16, 17, and 18. *From Lemmas 19–22 we have already completed the proof of Theorems 16 and 17. We note that if , then and which imply that the results of Theorem 18 hold.

#### Conflict of Interests

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

#### Acknowledgments

This work is supported by NSFC (nos. 11061030, 11326127, and 11361054), SRFDP (no. 20126203110004), and Gansu Provincial National Science Foundation of China (no. 1208RJZA258).