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This article has been retracted as it is found to contain a substantial amount of material from the paper "Eigenvalue comparisons for boundary value problems for second order difference equations," authored by Jun Ji and Bo Yang and it is published in "Journal of Mathematical Analysis and Applications" in 2006.

Journal of Applied Mathematics
Volume 2012, Article ID 486230, 10 pages
http://dx.doi.org/10.1155/2012/486230
Research Article

Eigenvalue Comparisons for Second-Order Linear Equations with Boundary Value Conditions on Time Scales

School of Mathematical Sciences, University of Jinan, Jinan, Shandong 250022, China

Received 29 January 2012; Revised 22 March 2012; Accepted 22 March 2012

Academic Editor: Kai Diethelm

Copyright © 2012 Chao Zhang and Shurong Sun. 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.

Abstract

This paper studies the eigenvalue comparisons for second-order linear equations with boundary conditions on time scales. Using results from matrix algebras, the existence and comparison results concerning eigenvalues are obtained.

1. Introduction

In this paper, we consider the eigenvalue problems for the following second-order linear equations:𝑟(𝑡)𝑦Δ(𝑡)Δ+𝜆(1)𝑝(𝑡)𝑦𝜎[](𝑡)=0,𝑡𝜌(𝑎),𝜌(𝑏)𝕋,(1.1)𝑟(𝑡)𝑦Δ(𝑡)Δ+𝜆(2)𝑞(𝑡)𝑦𝜎[](𝑡)=0,𝑡𝜌(𝑎),𝜌(𝑏)𝕋,(1.2) with the boundary conditions𝑦(𝜌(𝑎))𝜏𝑦(𝑎)=𝑦(𝜎(𝑏))𝛿𝑦(𝑏)=0,(1.3) where 𝜆(1) and 𝜆(2) are parameters, 𝜎(𝑡) and 𝜌(𝑡) are the forward and backward jump operators, 𝑦Δ is the delta derivative, 𝑦𝜎(𝑡)=𝑦(𝜎(𝑡)), and [𝜌(𝑎),𝜌(𝑏)]𝕋 is a finite isolated time scale; the discrete interval is given by[]𝜌(𝑎),𝜌(𝑏)𝕋=𝜌(𝑎),𝑎,𝜎(𝑎),𝜎2.(𝑎),,𝜌(𝑏)(1.4)

We assume throughout this paper that(H1)𝑟Δ, 𝑝, and 𝑞 are real-valued functions on [𝜌(𝑎),𝜌(𝑏)]𝕋, 𝑝0(0), 𝑞0(0) on [𝜌(𝑎),𝜌(𝑏)]𝕋 and 𝑟>0 on [𝜌(𝑎),𝑏]𝕋;(H2)𝜏,𝛿[0,1).

First we briefly recall some existing results of eigenvalues comparisons for differential and difference equations. In 1973, Travis [1] considered the eigenvalue problem for boundary value problems of higher-order differential equations. He employed the theory of 𝑢0-positive linear operator on a Banach space with a cone of nonnegative elements to obtain comparison results for the smallest eigenvalues. A representative set of references for these works would be Davis et al. [2], Diaz and Peterson [3], Hankerson and Henderson [4], Hankerson and Peterson [57], Henderson and Prasad [8], and Kaufmann [9]. However, in all the above papers, the comparison results are for the smallest eigenvalues only. The main purpose of this paper is to establish the comparison theorems for all the eigenvalues of (1.1) with (1.3) and (1.2) with (1.3).

Like the eigenvalue comparison for the boundary value problems of linear equations, this type of comparison of eigenvalues in matrix algebra is known as Weyl’s inequality [10, Corllary 6.5.]: If 𝐴,𝐵 are Hermitian matrices, that is, 𝐴=𝐴, where 𝐴 is the conjugate transpose of 𝐴 and 𝐴𝐵 is positive semidefinite, then 𝜆𝑖(𝐴)𝜆𝑖(𝐵), where 𝜆𝑖(𝐴) and 𝜆𝑖(𝐵) are all eigenvalues of 𝐴 and 𝐵. Associated with this conclusion is spectral order of operators. The spectral order has proved to be useful for solving several open problems of spectral theory and has been studied in the context of von Neumann algebras, matrix algebras, and so forth in [1015]. Recently, Hamhalter [15] studied the spectral order in a more general setting of Jordan operator algebras, which is a generalization of the result due to Kato [13]. And as a preparatory material, he extended Olson’s characterization of the spectral order to JBW algebras [14]. Since the boundary value problems (1.1), (1.3) and (1.2), (1.3) can be rewritten into matrix equations, we employ some results from matrix algebras to establish the comparison theorems for the eigenvalues of (1.1), (1.3) and (1.2), (1.3).

This paper is organized as follows. Section 2 introduces some basic concepts and a fundamental theory about time scales, which will be used in Section 3. By some results from matrix algebras and time scales, the existence and comparison theorems of eigenvalues of boundary value problems (1.1), (1.3) and (1.2), (1.3) are obtained, which will be given in Section 3.

2. Preliminaries

In this section, some basic concepts and some fundamental results on time scales are introduced.

Let 𝕋𝐑 be a nonempty closed subset. Define the forward and backward jump operators 𝜎,𝜌𝕋𝕋 by𝜎(𝑡)=inf{𝑠𝕋𝑠>𝑡},𝜌(𝑡)=sup{𝑠𝕋𝑠<𝑡},(2.1) where inf=sup𝕋,sup=inf𝕋. We put 𝕋𝑘=𝕋 if 𝕋 is unbounded above and 𝕋𝑘=𝕋(𝜌(max𝕋),max𝕋] otherwise. The graininess functions 𝜈,𝜇𝕋[0,) are defined by𝜇(𝑡)=𝜎(𝑡)𝑡,𝜈(𝑡)=𝑡𝜌(𝑡).(2.2) Let 𝑓 be a function defined on 𝕋. 𝑓 is said to be (delta) differentiable at 𝑡𝕋𝑘 provided there exists a constant 𝑎 such that for any 𝜀>0, there is a neighborhood 𝑈 of 𝑡 (i.e., 𝑈=(𝑡𝛿,𝑡+𝛿)𝕋 for some 𝛿>0) with||||||||𝑓(𝜎(𝑡))𝑓(𝑠)𝑎(𝜎(𝑡)𝑠)𝜀𝜎(𝑡)𝑠,𝑠𝑈.(2.3) In this case, denote 𝑓Δ(𝑡)=𝑎. If 𝑓 is (delta) differentiable for every 𝑡𝕋𝑘, then 𝑓 is said to be (delta) differentiable on 𝕋. If 𝑓 is differentiable at 𝑡𝕋𝑘, then𝑓Δ(𝑡)=lim𝑠𝑡𝑠𝐓𝑓(𝑡)𝑓(𝑠)𝑡𝑠if𝜇(𝑡)=0,𝑓(𝜎(𝑡))𝑓(𝑡)𝜇(𝑡)if𝜇(𝑡)>0.(2.4)

For convenience, we introduce the following results ([16, Chapter 1], [17, Chapter 1], and [18, Lemma 1]), which are useful in this paper.

Lemma 2.1. Let 𝑓,𝑔𝕋𝐑 and 𝑡𝕋𝑘.(i)If 𝑓 and 𝑔 are differentiable at 𝑡, then 𝑓𝑔 is differentiable at 𝑡 and (𝑓𝑔)Δ(𝑡)=𝑓𝜎(𝑡)𝑔Δ(𝑡)+𝑓Δ(𝑡)𝑔(𝑡)=𝑓Δ(𝑡)𝑔𝜎(𝑡)+𝑓(𝑡)𝑔Δ(𝑡).(2.5)(ii)If 𝑓 and 𝑔 are differentiable at 𝑡, and 𝑓(𝑡)𝑓𝜎(𝑡)0, then 𝑓1𝑔 is differentiable at 𝑡 and 𝑔𝑓1Δ𝑔(𝑡)=Δ(𝑡)𝑓(𝑡)𝑔(𝑡)𝑓Δ(𝑡)(𝑓𝜎(𝑡)𝑓(𝑡))1.(2.6)

3. Eigenvalue Comparisons

In the following, we will write 𝑋𝑌 if 𝑋 and 𝑌 are symmetric 𝑛×𝑛 matrices and 𝑋𝑌 is positive semidefinite. A matrix is said to be positive if every component of the matrix is positive. We denote 𝜌(𝑎)=𝜎1(𝑎),𝑎=𝜎0(𝑎),𝜌(𝑏)=𝜎𝑛2(𝑎),𝑏=𝜎𝑛1(𝑎),𝜇𝑖=𝜎𝑖+1(𝑎)𝜎𝑖(𝑎), and 𝑟𝜎𝑖(𝑎)=𝑟(𝜎𝑖(𝑎)),𝑖=1,0,1,2,,𝑛1.

It follows from Lemma 2.1(ii), (2.4), and (1.4) that the boundary value problem (1.1), (1.3) can be written in the form𝐷+𝜆(1)𝑃𝑦=0,(3.1) where𝑟𝐷=𝒜+0000+𝒞𝒞0000𝒞𝒞+𝜎2(𝑎)𝜇2𝑟000000𝜎𝑛4(𝑎)𝜇𝑛4+𝒟𝒟0000𝒟𝒟+0000+(1𝛿)𝑟𝜎𝑛1(𝑎)𝜇𝑛1𝜇𝑃=diag1𝑝𝜎1(𝑎),𝜇0𝑝𝜎0(𝑎),,𝜇𝑛3𝑝𝜎𝑛3(𝑎),𝜇𝑛2𝑝𝜎𝑛2(,𝑎)(3.2)where 𝒜 donates  (1𝜏)𝑟𝜎1(𝑎)/𝜇1, donates𝑟𝜎0(𝑎)/𝜇0, 𝒞 donates 𝑟𝜎(𝑎)/𝜇1, 𝒟 donates 𝑟𝜎𝑛3(𝑎)/𝜇𝑛3, and donates 𝑟𝜎𝑛2(𝑎)/𝜇𝑛2. 𝑦𝜎𝑦=0(𝜎𝑎),𝑦(𝜎(𝑎)),,𝑦𝑛2(𝜎𝑎),𝑦𝑛1(𝑎)𝑇.(3.3) And the problem (1.2), (1.3) is equivalent to the equation 𝐷+𝜆(2)𝑄𝑦=0,(3.4) where𝜇𝑄=diag1𝑞𝜎1(𝑎),𝜇0𝑞𝜎0(𝑎),,𝜇𝑛3𝑞𝜎𝑛3(𝑎),𝜇𝑛2𝑞𝜎𝑛2.(𝑎)(3.5) Since the solutions of (1.1), (1.3) can be written into the form of vectors, then the nontrivial solution corresponding to 𝜆 is called an eigenvector.

Let 𝑒𝑖 be the 𝑖th column of the identity matrix 𝐼 of order 𝑛 and𝐷1𝑟=𝒜+0000+𝒞𝒞0000𝒞𝒞+𝜎2(𝑎)𝜇2𝑟000000𝜎𝑛4(𝑎)𝜇𝑛4+𝒟𝒟0000𝒟𝒟+0000(3.6) Define 𝑃𝑖=𝐼+𝑒𝑖1𝑒𝑇𝑖. It is easily seen that𝐷=𝐷1+𝑒𝑛𝑟(1𝛿)𝜎𝑛1(𝑎)𝜇𝑛1𝑒𝑇𝑛,𝑃(3.7)2𝑃3𝑃𝑛𝐷1𝑃𝑇𝑛𝑃𝑇3𝑃𝑇2𝑟=diag(1𝜏)𝜎1(𝑎)𝜇1,𝑟𝜎0(𝑎)𝜇0𝑟,,𝜎𝑛3(𝑎)𝜇𝑛3,𝑟𝜎𝑛2(𝑎)𝜇𝑛2.(3.8) It follows from (H1),(H2), and (3.8) that𝐷1=𝑃𝑛1𝑃31𝑃21𝑟diag(1𝜏)𝜎1(𝑎)𝜇1,𝑟𝜎0(𝑎)𝜇0𝑟,,𝜎𝑛3(𝑎)𝜇𝑛3,𝑟𝜎𝑛2(𝑎)𝜇𝑛2𝑃2𝑇𝑃3𝑇𝑃𝑛𝑇𝐷,(3.9)11=𝑃𝑇𝑛𝑃𝑇3𝑃𝑇2𝜇diag1(1𝜏)𝑟𝜎1,𝜇(𝑎)0𝑟𝜎0𝜇(𝑎),,𝑛3𝑟𝜎𝑛3,𝜇(𝑎)𝑛2𝑟𝜎𝑛2𝑃(𝑎)2𝑃3𝑃𝑛.(3.10)

For any 𝑥=(𝑥1,𝑥2,,𝑥𝑛)𝑇, we have𝑥𝐷𝑥=𝑥𝐷1𝑟𝑥+(1𝛿)𝜎𝑛1(𝑎)𝜇𝑛1𝑥𝑒𝑛𝑒𝑇𝑛𝑥=𝑃2𝑇𝑃3𝑇𝑃𝑛𝑇𝑥𝑟diag(1𝜏)𝜎1(𝑎)𝜇1,𝑟𝜎0(𝑎)𝜇0𝑟,,𝜎𝑛3(𝑎)𝜇𝑛3,𝑟𝜎𝑛2(𝑎)𝜇𝑛2×𝑃2𝑇𝑃3𝑇𝑃𝑛𝑇𝑥+𝑟(1𝛿)𝜎𝑛1(𝑎)𝜇𝑛1||𝑒𝑇𝑛𝑥||20.(3.11) Moreover, 𝑥𝐷𝑥=0 implies 𝑥=0. Hence, the matrix 𝐷 is positive definite.

Lemma 3.1. If 𝜆(1) is an eigenvalue of the boundary value problem (1.1), (1.3) and 𝑦 is a corresponding eigenvector, then(i)𝑦𝑃𝑦>0,(ii)𝜆(1) is real and positive. If 𝜌𝜆(1) is an eigenvalue of the boundary value problem (1.1), (1.3) and 𝑥 is a corresponding eigenvector, then 𝑥𝑃𝑦=0.

Proof. (i) It follows from (H1) and (3.2) that 𝑦𝑃𝑦0. Assume the contrary that 𝑦𝑃𝑦=0, we have 𝑦𝐷𝑦=𝜆(1)𝑦𝑃𝑦=0. Since 𝐷 is positive definite, then 𝑦=0, which is a contradiction.
(ii) We can write 𝜆(1)𝑦𝑃𝑦=𝑦𝜆(1)𝑃𝑦=𝑦𝐷𝑦=(𝐷𝑦)𝜆𝑦=(1)𝑃𝑦𝑦=𝜆(1)𝑦𝑃𝑦=𝜆(1)𝑦𝑃𝑦,(3.12) which implies 𝜆(1)=𝜆(1), that is, 𝜆 is real. Since 𝐷 is positive definite and 𝑦𝑃𝑦>0, we have 𝜆(1)=𝑦𝐷𝑦/𝑦𝑃𝑦>0.
If 𝜌𝑃𝑥=𝐷𝑥 and 𝜌𝜆(1), then 𝜆(1)𝑥𝜌𝑃𝑦=𝜆(1)𝑥𝑃𝑦𝜌𝑥𝑃𝑦=𝑥𝜆(1)𝑃𝑦(𝜌𝑃𝑥)𝑦=𝑥𝐷𝑦(𝐷𝑥)𝑦=0.(3.13) Hence, 𝑥𝑃𝑦=0. This completes the proof.

Lemma 3.2. If 𝜆(1) is an eigenvalue of the boundary value problem (1.1), (1.3), then 1/𝜆(1) is an eigenvalue of 𝐷1/2𝑃𝐷1/2. If 𝛼 is a positive eigenvalue of 𝐷1/2𝑃𝐷1/2, then 1/𝛼 is an eigenvalue of (1.1), (1.3), respectively.

Proof. If 𝜆(1) is an eigenvalue of the boundary value problem (1.1), (1.3) and 𝑦 is a corresponding eigenvector, then 𝜆(1)>0 and 𝜆(1)𝑃𝑦=𝐷𝑦. Therefore, 𝜆(1)𝑃𝑦=𝐷1/2𝐷1/2𝐷𝑦,1/2𝑃𝐷1/2𝐷1/2𝑦=1𝜆(1)𝐷1/2𝑦.(3.14)
With a similar argument, one can get that if 𝛼 is a positive eigenvalue of 𝐷1/2𝑃𝐷1/2, then 1/𝛼 is an eigenvalue of (1.1), (1.3). This completes proof.

Lemma 3.3. For any 1𝑖,𝑗𝑛, define 𝛾=min{𝑖,𝑗}. We have(i)𝑒𝑇𝑖𝐷11𝑒𝑗=𝜇1/(1𝜏)𝑟𝜎1(𝑎)+𝛾2𝑘=0(𝜇𝑘/𝑟𝜎𝑘(𝑎));(ii)𝑒𝑇𝑖𝐷1𝑒𝑗 = ((𝜇1/(1𝜏)𝑟𝜎1(𝑎))+𝛾2𝑘=0(𝜇𝑘/𝑟𝜎𝑘(𝑎)))((𝜇𝑛1/(1𝛿)𝑟𝜎𝑛1(𝑎))+(𝜇1/(1𝜏)𝑟𝜎1(𝑎))+𝑛2𝑘=0(𝜇𝑘/𝑟𝜎𝑘(𝑎)))((𝜇1/(1𝜏)𝑟𝜎1(𝑎))+𝑖2𝑘=0(𝜇𝑘/𝑟𝜎𝑘(𝑎)))((𝜇1/(1𝜏)𝑟𝜎1(𝑎))+𝑗2𝑘=0(𝜇𝑘/𝑟𝜎𝑘(𝑎)))/(𝜇𝑛1/(1𝛿)𝑟𝜎𝑛1(𝑎)) + (𝜇1/(1𝜏)𝑟𝜎1(𝑎))+𝑛2𝑘=0(𝜇𝑘/𝑟𝜎𝑘(𝑎)).

Proof. It is easy to see that 𝑃𝑖𝑒𝑗=𝑒𝑗 if 𝑖𝑗, while 𝑃𝑖𝑒𝑗=𝑒𝑗1+𝑒𝑗 if 𝑖=𝑗. Hence, 𝑃2𝑃3𝑃𝑛𝑒𝑗=𝑒1+𝑒2++𝑒𝑗.(3.15)(i)It is seen from (3.10) and (3.15) that 𝑒𝑇𝑖𝐷11𝑒𝑗=𝑃2𝑃3𝑃𝑛𝑒𝑖𝑇𝜇diag1(1𝜏)𝑟𝜎1,𝜇(𝑎)0𝑟𝜎0𝜇(𝑎),,𝑛3𝑟𝜎𝑛3,𝜇(𝑎)𝑛2𝑟𝜎𝑛2𝑃(𝑎)2𝑃3𝑃𝑛𝑒𝑗=𝑒1+𝑒2++𝑒𝑖𝑇𝜇diag1(1𝜏)𝑟𝜎1,𝜇(𝑎)0𝑟𝜎0𝜇(𝑎),,𝑛3𝑟𝜎𝑛3,𝜇(𝑎)𝑛2𝑟𝜎𝑛2×𝑒(𝑎)1+𝑒2++𝑒𝑗=𝜇1(1𝜏)𝑟𝜎1+(𝑎)𝛾2𝑘=0𝜇𝑘𝑟𝜎𝑘.(𝑎)(3.16)(ii)It follows from (3.7) and the Sherman-Morrison updating formula [19] that 𝐷1=𝐷11𝐷11𝑒𝑛𝑒𝑇𝑛𝐷11𝜇𝑛1/(1𝛿)𝑟𝜎𝑛1(𝑎)+𝑒𝑇𝑛𝐷11𝑒𝑛,(3.17) leading to 𝑒𝑇𝑖𝐷1𝑒𝑗=𝑒𝑇𝑖𝐷11𝑒𝑗𝑒𝑇𝑖𝐷11𝑒𝑛𝑒𝑇𝑛𝐷11𝑒𝑗𝜇𝑛1/(1𝛿)𝑟𝜎𝑛1(𝑎)+𝑒𝑇𝑛𝐷11𝑒𝑛,(3.18) which, together with (i), further implies the result (ii). This completes the proof.

Theorem 3.4. (i) If 𝜆(1) is an eigenvalue of the boundary value problem (1.1), (1.3) and 𝑦0 is a corresponding eigenvector, then 𝑦(𝑎)0 and 𝑦(𝑏)0.
(ii) If 𝜆1(1)>0 is the smallest eigenvalue of the boundary value problem (1.1), (1.3), then there exists a positive eigenvector 𝑦>0 corresponding to 𝜆1(1).

Proof. (i) Assume the contrary that either 𝑦(𝑎)=0 or 𝑦(𝑏)=0. By the boundary condition (1.3), we can easily deduce a contradiction 𝑦(𝑡)0.
(ii) It follows from 𝐷1𝑃𝑦=(1/𝜆1(1))𝑦 that 1/𝜆1(1) is the maximum eigenvalue of 𝐷1𝑃 and the 𝑦 is an eigenvector corresponding to 1/𝜆1(1). By Lemma 3.3(ii), we have that all the elements of 𝐷1 are positive, then 𝐷1 is a positive matrix. Since 𝑝(𝑡)0 for all 𝑡[𝜌(𝑎),𝜌(𝑏)]𝕋, hence, the following discussions are divided into two cases.
Case 1. If 𝑝(𝑡)>0 for all 𝑡[𝜌(𝑎),𝜌(𝑏)]𝕋, then we obtain that the matrix 𝐷1𝑃 is positive and therefore, the result follows from the Perron-Forbenius theorem [20].Case 2. Let 𝑝(𝑡)=0 for some 𝑡[𝜌(𝑎),𝜌(𝑏)]𝕋. Without loss of generality, we assume that 𝑝(𝑡)=0 for all 𝑡[𝜌(𝑎),𝜎𝑚2(𝑎)]𝕋 and 𝑝(𝑡)>0 for all 𝑡[𝜎𝑚1(𝑎),𝜌(b)]𝕋; we can write 𝐷1𝑃 as follows: 𝐷1𝑃=0𝑉0𝑍,(3.19) where 𝑉 is an 𝑚×(𝑛𝑚) matrix and 𝑍 is an (𝑛𝑚)×(𝑛𝑚) matrix. Both 𝑉 and 𝑍 are positive matrices. 1/𝜆1(1) is also the maximum eigenvalue of 𝑍. Applying the Perron-Forbenius theorem to the positive matrix 𝑍, there exists a positive vector 𝑦𝑍>0 such that 𝑍𝑦𝑍=(1/𝜆1(1))𝑦𝑍. Let 𝑦𝑉=𝜆1(1)𝑉𝑦𝑍 and 𝑦=(𝑦𝑇𝑉,𝑦𝑇𝑍)𝑇. Obviously, we have 𝐷11𝑃𝑦=𝜆1(1)𝑦,where𝑦>0.(3.20) This completes the proof.

Lemma 3.5. If 𝜆(1) is an eigenvalue of the boundary value problem (1.1), (1.3), then the dimension of the null space of (𝐷+𝜆(1)𝑃) is 1.

Proof. Let 𝑥0 and 𝑦0 be any two eigenvectors of the boundary value problem (1.1), (1.3) corresponding to 𝜆(1) and define 𝑧=𝑥(𝑎)𝑦𝑦(𝑎)𝑥. Obviously, we have 𝐷+𝜆(1)𝑃𝑧=𝑥(𝑎)𝐷+𝜆(1)𝑃𝑦𝑦(𝑎)𝐷+𝜆(1)𝑃𝑥=0,(3.21) which, together with 𝑧(𝑎)=0, indicates that 𝑧=0, that is, 𝑥(𝑎)𝑦=𝑦(𝑎)𝑥. Therefore, 𝑥 and 𝑦 are linearly dependent. So the dimension of the null space of (𝐷+𝜆(1)𝑃) is 1. This completes the proof.

Lemma 3.6. Let 𝑁1 be the number of positive elements in the set {𝑝(𝑡)𝑡[𝜌(𝑎),𝜌(𝑏)]𝕋}. Then there are 𝑁 distinct eigenvalues 𝜆𝑖(1)(𝑖=1,2,,𝑁) of the boundary value problem (1.1), (1.3) and 𝛼𝑖=1/𝜆𝑖(1)(𝑖=1,2,,𝑁) are the only positive eigenvalues of 𝐷1/2𝑃𝐷1/2.

Proof. Suppose that 𝛼1𝛼2𝛼𝑛0 are all eigenvalues of 𝐷1/2𝑃𝐷1/2. Since 𝐷1/2𝑃𝐷1/2 is real and symmetric that there exists an orthogonal matrix 𝐶 such that 𝐶𝑇𝐷1/2𝑃𝐷1/2𝛼𝐶=diag1𝛼2𝛼𝑛,(3.22) therefore, we have that 𝐶rank(𝑃)=rank𝑇𝐷1/2𝑃𝐷1/2𝐶𝛼=rankdiag1𝛼2𝛼𝑛(3.23) indicating that the number of positive 𝛼𝑖 is the same as that of positive number in 𝑃 which is equal to 𝑁.
Suppose that 𝛼𝑖0=𝛼𝑖0+1>0 for some 𝑖0 where 1𝑖0𝑁1. Observe that 𝐶𝑇𝐷1/2𝑃𝐷1/2𝐶𝑒𝑖=𝛼𝑖𝑒𝑖 in view of (3.22), which further implies that 𝐷𝐷1/2𝐶𝑒𝑖=1𝛼𝑖𝑃𝐷1/2𝐶𝑒𝑖𝑖=𝑖0,𝑖0+1.(3.24) Thus, we have two independent vectors in the null space of (𝐷+𝜆(1)𝑃) for 𝜆(1)=1/𝛼𝑖0, which contradicts Lemma 3.5. Thus, from Lemma 3.2, we see that {𝜆𝑖(1)=1/𝛼𝑖𝑖=1,2,,𝑁} gives the complete set of eigenvalues of the boundary value problem (1.1), (1.3). This completes the proof.

Theorem 3.7. Let 𝑗 be the number of positive elements in the set {𝑝(𝑡)𝑡[𝜌(𝑎),𝜌(𝑏)]𝕋} and 𝑘 the number of positive elements in the set {𝑞(𝑡)𝑡[𝜌(𝑎),𝜌(𝑏)]𝕋}. Let {𝜆1(1)<𝜆2(1)<<𝜆𝑗(1)} be the set of all eigenvalues of the boundary value problem (1.1), (1.3) and {𝜆1(2)<𝜆2(2)<<𝜆𝑘(2)} the set of all eigenvalues of the boundary value problem (1.2), (1.3). If 𝑝(𝑡)𝑞(𝑡) for all 𝑡[𝜌(𝑎),𝜌(𝑏)]𝕋, then 𝜆𝑖(1)𝜆𝑖(2) for 1𝑖𝑘.

Proof. It follows from Lemma 3.6 that 𝛼1=1𝜆1(1)>>𝛼𝑗=1𝜆𝑗(1)>0,𝛼𝑗+1==𝛼𝑛𝛽=0,1=1𝜆1(2)>>𝛽𝑘=1𝜆𝑘(2)>0,𝛽𝑘+1==𝛼𝑛=0(3.25) are the eigenvalues of 𝐷1/2𝑃𝐷1/2 and 𝐷1/2𝑄𝐷1/2, respectively. If 𝑝(𝑡)𝑞(𝑡) for all 𝑡[𝜌(𝑎),𝜌(𝑏)]𝕋, then 𝑃𝑄, implying 𝐷1/2𝑃𝐷1/2𝐷1/2𝑄𝐷1/2.(3.26) By Weyl’s inequality and (3.26), we have 𝛼𝑖𝛽𝑖1𝑖𝑛.(3.27) Finally, it is easily seen from (3.25) and (3.27) that 1𝜆𝑖(1)1𝜆𝑖(2)1𝑖𝑘,(3.28) implying that 𝜆𝑖(1)𝜆𝑖(2) for 1𝑖𝑘. This completes the proof.

Acknowledgments

Many thanks are due to Kai Diethelm (the editor) and the anonymous reviewer(s) for helpful comments and suggestions. This paper was supported by the NNSF of China (Grants nos. 11071143 and 11101241), the NNSF of Shandong Province (Grants nos. ZR2009AL003, ZR2010AL016, and ZR2011AL007), the Scientific Research and Development Project of Shandong Provincial Education Department (J11LA01), and the NSF of University of Jinan (XKY0918).

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