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

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:𝑟𝑟((𝑡𝑡))𝑦𝑦^Δ_Δ(𝑡)^Δ+𝜆⁽¹⁾𝑝(𝑡)𝑦^𝜎(𝑡)=0,𝑡∈[𝜌(𝑎),𝜌(𝑏)]_𝕋,(1.1)(𝑡)^Δ+𝜆⁽²⁾𝑞(𝑡)𝑦^𝜎(𝑡)=0,𝑡∈[𝜌(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, 𝑦^Δ is the delta derivative, 𝑦^𝜎(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)𝑟^Δ, 𝑝, and 𝑞 are real-valued functions on [𝜌(𝑎),𝜌(𝑏)]_𝕋, 𝑝≥0(≢0), 𝑞≥0(≢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 [5–7], 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 𝐴−𝐵 is positive semidefinite, then 𝜆^𝑖(𝐴)≥𝜆^𝑖(𝐵), where 𝜆^𝑖(𝐴) and 𝜆^𝑖(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 [10–15]. 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 𝕋⊂𝐑 be a nonempty closed subset. Define the forward and backward jump operators 𝜎,𝜌∶𝕋→𝕋 by𝜎(𝑡)=inf{𝑠∈𝕋∶𝑠>𝑡},𝜌(𝑡)=sup{𝑠∈𝕋∶𝑠<𝑡},(2.1) where inf∅=sup𝕋,sup∅=inf𝕋. We put 𝕋^𝑘=𝕋 if 𝕋 is unbounded above and 𝕋^𝑘=𝕋⧵(𝜌(max𝕋),max𝕋] otherwise. The graininess functions 𝜈,𝜇∶𝕋→[0,∞) are defined by𝜇(𝑡)=𝜎(𝑡)−𝑡,𝜈(𝑡)=𝑡−𝜌(𝑡).(2.2) Let 𝑓 be a function defined on 𝕋. 𝑓 is said to be (delta) differentiable at 𝑡∈𝕋^𝑘 provided there exists a constant 𝑎 such that for any 𝜀>0, there is a neighborhood 𝑈 of 𝑡 (i.e., 𝑈=(𝑡−𝛿,𝑡+𝛿)∩𝕋 for some 𝛿>0) with||𝑓(𝜎(𝑡))−𝑓(𝑠)−𝑎(𝜎(𝑡)−𝑠)||≤𝜀||𝜎(𝑡)−𝑠||,∀𝑠∈𝑈.(2.3) In this case, denote 𝑓^Δ(𝑡)∶=𝑎. If 𝑓 is (delta) differentiable for every 𝑡∈𝕋^𝑘, then 𝑓 is said to be (delta) differentiable on 𝕋. If 𝑓 is differentiable at 𝑡∈𝕋^𝑘, then𝑓^Δ(𝑡)=⎧⎪⎪⎨⎪⎪⎩lim^{𝑠→𝑡𝑠∈𝐓}𝑓(𝑡)−𝑓(𝑠)𝑓(𝜎(𝑡))𝑡−−𝑓𝑠(𝑡)if𝜇(𝑡)=0,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,𝑔∶𝕋→𝐑 and 𝑡∈𝕋^𝑘.(i)If 𝑓 and 𝑔 are differentiable at 𝑡, then 𝑓𝑔 is differentiable at 𝑡 and (𝑓𝑔)^Δ(𝑡)=𝑓^𝜎(𝑡)𝑔^Δ(𝑡)+𝑓^Δ(𝑡)𝑔(𝑡)=𝑓^Δ(𝑡)𝑔^𝜎(𝑡)+𝑓(𝑡)𝑔^Δ(𝑡).(2.5)(ii)If 𝑓 and 𝑔 are differentiable at 𝑡, and 𝑓(𝑡)𝑓^𝜎(𝑡)≠0, then 𝑓⁻¹𝑔 is differentiable at 𝑡 and 𝑔𝑓⁻¹^Δ(𝑡)=𝑔^Δ(𝑡)𝑓(𝑡)−𝑔(𝑡)𝑓^Δ(𝑡)(𝑓^𝜎(𝑡)𝑓(𝑡))⁻¹.(2.6)

3. Eigenvalue Comparisons

In the following, we will write 𝑋≥𝑌 if 𝑋 and 𝑌 are symmetric 𝑛×𝑛 matrices and 𝑋−𝑌 is positive semidefinite. A matrix is said to be positive if every component of the matrix is positive. We denote 𝜌(𝑎)=𝜎⁻¹(𝑎),𝑎=𝜎⁰(𝑎),𝜌(𝑏)=𝜎^𝑛−2(𝑎),𝑏=𝜎^𝑛−1(𝑎),𝜇_𝑖=𝜎^𝑖+1(𝑎)−𝜎^𝑖(𝑎), and 𝑟^𝜎𝑖(𝑎)=𝑟(𝜎^𝑖(𝑎)),𝑖=−1,0,1,2,…,𝑛−1.

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 form−𝐷+𝜆⁽¹⁾𝑃𝑦=0,(3.1) where𝐷∶=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝𝒜−+0ℬℬℬ−−+ℬ𝒞𝒞𝒞+−𝑟0𝒞^𝜎2⋯⋯000000(𝑎)𝜇₂⋯0⋯0⋯0⋯⋯⋯𝑟_𝜎𝑛−4⋯0⋯0⋯0(𝑎)𝜇_𝑛−4000000⋯⋯−0𝒟+𝒟𝒟−−+𝒟ℰℰℰ+(1−−0𝛿ℰ)𝑟_𝜎𝑛−1(𝑎)𝜇_𝑛−1𝑃=diag𝜇₋₁⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠𝑝𝜎⁻¹(𝑎),𝜇₀𝑝𝜎⁰(𝑎),…,𝜇_𝑛−3𝑝𝜎^𝑛−3(𝑎),𝜇_𝑛−2𝑝𝜎^𝑛−2(𝑎),(3.2)where 𝒜 donates  (1−𝜏)𝑟^𝜎−1(𝑎)/𝜇₋₁, ℬdonates𝑟^𝜎0(𝑎)/𝜇₀, 𝒞 donates 𝑟^𝜎(𝑎)/𝜇₁, 𝒟 donates 𝑟^𝜎𝑛−3(𝑎)/𝜇_𝑛−3, and ℰ donates 𝑟^𝜎𝑛−2(𝑎)/𝜇_𝑛−2. 𝑦=𝑦𝜎⁰(𝑎),𝑦(𝜎(𝑎)),…,𝑦𝜎^𝑛−2(𝑎),𝑦𝜎^𝑛−1(𝑎)^𝑇.(3.3) And the problem (1.2), (1.3) is equivalent to the equation −𝐷+𝜆⁽²⁾𝑄𝑦=0,(3.4) where𝑄=diag𝜇₋₁𝑞𝜎⁻¹(𝑎),𝜇₀𝑞𝜎⁰(𝑎),…,𝜇_𝑛−3𝑞𝜎^𝑛−3(𝑎),𝜇_𝑛−2𝑞𝜎^𝑛−2(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 𝑒_𝑖 be the 𝑖th column of the identity matrix 𝐼 of order 𝑛 and𝐷₁∶=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝𝒜−+0ℬℬℬ−−+ℬ𝒞𝒞𝒞+−𝑟0𝒞^𝜎2⋯⋯000000(𝑎)𝜇₂⋯0⋯0⋯0⋯⋯⋯𝑟_𝜎𝑛−4⋯0⋯0⋯0(𝑎)𝜇_𝑛−4000000⋯⋯−0𝒟+𝒟𝒟−−+𝒟ℰℰ−ℰ0ℰ⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠(3.6) Define 𝑃_𝑖=𝐼+𝑒_𝑖−1𝑒^𝑇𝑖. It is easily seen that𝐷=𝐷₁+𝑒_𝑛(1−𝛿)𝑟^𝜎𝑛−1(𝑎)𝜇_𝑛−1𝑃𝑒^𝑇𝑛,(3.7)₂𝑃₃⋯𝑃_𝑛𝐷₁𝑃^𝑇𝑛⋯𝑃^𝑇3𝑃^𝑇2=diag(1−𝜏)𝑟^𝜎−1(𝑎)𝜇₋₁,𝑟^𝜎0(𝑎)𝜇₀,…,𝑟^𝜎𝑛−3(𝑎)𝜇_𝑛−3,𝑟^𝜎𝑛−2(𝑎)𝜇_𝑛−2.(3.8) It follows from (H₁),(H₂), and (3.8) that𝐷₁=𝑃^𝑛−1⋯𝑃³⁻¹𝑃²⁻¹diag(1−𝜏)𝑟^𝜎−1(𝑎)𝜇₋₁,𝑟^𝜎0(𝑎)𝜇₀,…,𝑟^𝜎𝑛−3(𝑎)𝜇_𝑛−3,𝑟^𝜎𝑛−2(𝑎)𝜇_𝑛−2𝑃^2−𝑇𝐷𝑃^3−𝑇⋯𝑃^𝑛−𝑇,(3.9)₁₋₁=𝑃^𝑇𝑛⋯𝑃^𝑇3𝑃^𝑇2diag𝜇⁻¹(1−𝜏)𝑟^𝜎−1(𝑎),𝑟𝜇^𝜎0(⁰𝑎),…,𝑟𝜇^𝜎𝑛−3(^𝑛−3𝑎),𝑟𝜇^𝜎𝑛−2(^𝑛−2𝑎)𝑃²𝑃₃⋯𝑃_𝑛.(3.10)

For any 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑛)^𝑇, we have𝑥^∗𝐷𝑥=𝑥^∗𝐷₁𝑥+(1−𝛿)𝑟^𝜎𝑛−1(𝑎)𝜇_𝑛−1=𝑃_2−𝑇𝑥^∗𝑒_𝑛𝑒^𝑇𝑛𝑥𝑃^3−𝑇⋯𝑃^𝑛−𝑇𝑥^∗diag(1−𝜏)𝑟^𝜎−1(𝑎)𝜇₋₁,𝑟^𝜎0(𝑎)𝜇₀,…,𝑟^𝜎𝑛−3(𝑎)𝜇_𝑛−3,𝑟^𝜎𝑛−2(𝑎)×𝑃_2−𝑇𝜇_𝑛−2𝑃^3−𝑇⋯𝑃^𝑛−𝑇𝑥+(1−𝛿)𝑟^𝜎𝑛−1(𝑎)𝜇_𝑛−1||𝑒^𝑇𝑛𝑥||²≥0.(3.11) Moreover, 𝑥^∗𝐷𝑥=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)𝑦^∗𝑃𝑦>0,(ii)𝜆⁽¹⁾ is real and positive. If 𝜌≠𝜆⁽¹⁾ is an eigenvalue of the boundary value problem (1.1), (1.3) and 𝑥 is a corresponding eigenvector, then 𝑥^∗𝑃𝑦=0.

Proof. (i) It follows from (H₁) and (3.2) that 𝑦^∗𝑃𝑦≥0. Assume the contrary that 𝑦^∗𝑃𝑦=0, we have 𝑦^∗𝐷𝑦=𝜆⁽¹⁾𝑦^∗𝑃𝑦=0. Since 𝐷 is positive definite, then 𝑦=0, which is a contradiction.
(ii) We can write 𝜆⁽¹⁾𝑦^∗𝑃𝑦=𝑦^∗𝜆⁽¹⁾𝑃𝑦=𝑦^∗𝐷𝑦=(𝐷𝑦)^∗𝑦=𝜆⁽¹⁾𝑃𝑦^∗𝑦=𝜆⁽¹⁾𝑦^∗𝑃^∗𝑦=𝜆⁽¹⁾𝑦^∗𝑃𝑦,(3.12) which implies 𝜆⁽¹⁾=𝜆⁽¹⁾, that is, 𝜆 is real. Since 𝐷 is positive definite and 𝑦^∗𝑃𝑦>0, we have 𝜆⁽¹⁾=𝑦^∗𝐷𝑦/𝑦^∗𝑃𝑦>0.
If 𝜌𝑃𝑥=𝐷𝑥 and 𝜌≠𝜆⁽¹⁾, then 𝜆⁽¹⁾−𝜌𝑥^∗𝑃𝑦=𝜆⁽¹⁾𝑥^∗𝑃𝑦−𝜌𝑥^∗𝑃𝑦=𝑥^∗𝜆⁽¹⁾𝑃𝑦−(𝜌𝑃𝑥)^∗𝑦=𝑥^∗𝐷𝑦−(𝐷𝑥)^∗𝑦=0.(3.13) Hence, 𝑥^∗𝑃𝑦=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 𝐷^−1/2𝑃𝐷^−1/2. If 𝛼 is a positive eigenvalue of 𝐷^−1/2𝑃𝐷^−1/2, 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, 𝐷𝜆⁽¹⁾𝑃𝑦=𝐷^1/2𝐷^1/2𝑦,_−1/2𝑃𝐷^−1/2𝐷^1/2𝑦=1𝜆⁽¹⁾𝐷^1/2𝑦.(3.14)
With a similar argument, one can get that if 𝛼 is a positive eigenvalue of 𝐷^−1/2𝑃𝐷^−1/2, then 1/𝛼 is an eigenvalue of (1.1), (1.3). This completes proof.

Lemma 3.3. For any 1≤𝑖,𝑗≤𝑛, define 𝛾=min{𝑖,𝑗}. We have(i)𝑒^𝑇𝑖𝐷¹⁻¹𝑒_𝑗=𝜇₋₁/(1−𝜏)𝑟^𝜎−1(𝑎)+∑^{𝛾−2𝑘=0}(𝜇_𝑘/𝑟^𝜎𝑘(𝑎));(ii)𝑒^𝑇𝑖𝐷⁻¹𝑒_𝑗 = ((𝜇₋₁/(1−𝜏)𝑟^𝜎−1(𝑎))+∑^{𝛾−2𝑘=0}(𝜇_𝑘/𝑟^𝜎𝑘(𝑎)))((𝜇_𝑛−1/(1−𝛿)𝑟^𝜎𝑛−1(𝑎))+(𝜇₋₁/(1−𝜏)𝑟^𝜎−1(𝑎))+∑^{𝑛−2𝑘=0}(𝜇_𝑘/𝑟^𝜎𝑘(𝑎))) − ((𝜇₋₁/(1−𝜏)𝑟^𝜎−1(𝑎))+∑^{𝑖−2𝑘=0}(𝜇_𝑘/𝑟^𝜎𝑘(𝑎)))((𝜇₋₁/(1−𝜏)𝑟^𝜎−1(𝑎))+∑^{𝑗−2𝑘=0}(𝜇_𝑘/𝑟^𝜎𝑘(𝑎)))/(𝜇_𝑛−1/(1−𝛿)𝑟^𝜎𝑛−1(𝑎)) + (𝜇₋₁/(1−𝜏)𝑟^𝜎−1(𝑎))+∑^{𝑛−2𝑘=0}(𝜇_𝑘/𝑟^𝜎𝑘(𝑎)).

Proof. It is easy to see that 𝑃_𝑖𝑒_𝑗=𝑒_𝑗 if 𝑖≠𝑗, while 𝑃_𝑖𝑒_𝑗=𝑒_𝑗−1+𝑒_𝑗 if 𝑖=𝑗. Hence, 𝑃₂𝑃₃⋯𝑃_𝑛𝑒_𝑗=𝑒₁+𝑒₂+⋯+𝑒_𝑗.(3.15)(i)It is seen from (3.10) and (3.15) that 𝑒^𝑇𝑖𝐷¹⁻¹𝑒_𝑗=𝑃₂𝑃₃⋯𝑃_𝑛𝑒_𝑖^𝑇diag𝜇⁻¹(1−𝜏)𝑟^𝜎−1(𝑎),𝑟𝜇^𝜎0(⁰𝑎),…,𝑟𝜇^𝜎𝑛−3(^𝑛−3𝑎),𝑟𝜇^𝜎𝑛−2(^𝑛−2𝑎)𝑃²=𝑒₁𝑃₃⋯𝑃_𝑛𝑒_𝑗+𝑒₂+⋯+𝑒_𝑖^𝑇diag𝜇⁻¹×𝑒(1−𝜏)𝑟^𝜎−1(𝑎),𝑟𝜇^𝜎0(⁰𝑎),…,𝑟𝜇^𝜎𝑛−3(^𝑛−3𝑎),𝑟𝜇^𝜎𝑛−2(^𝑛−2𝑎)₁=𝜇+𝑒₂₋₁+⋯+𝑒_𝑗(1−𝜏)𝑟^𝜎−1(𝑎)+^𝛾−2_𝑘=0𝜇_𝑘𝑟^𝜎𝑘(𝑎).(3.16)(ii)It follows from (3.7) and the Sherman-Morrison updating formula [19] that 𝐷⁻¹=𝐷¹⁻¹−𝐷¹⁻¹𝑒_𝑛𝑒^𝑇𝑛𝐷¹⁻¹𝜇_𝑛−1/(1−𝛿)𝑟^𝜎𝑛−1(𝑎)+𝑒^𝑇𝑛𝐷¹⁻¹𝑒_𝑛,(3.17) leading to 𝑒^𝑇𝑖𝐷⁻¹𝑒_𝑗=𝑒^𝑇𝑖𝐷¹⁻¹𝑒_𝑗−𝑒^𝑇𝑖𝐷¹⁻¹𝑒_𝑛𝑒^𝑇𝑛𝐷¹⁻¹𝑒_𝑗𝜇_𝑛−1/(1−𝛿)𝑟^𝜎𝑛−1(𝑎)+𝑒^𝑇𝑛𝐷¹⁻¹𝑒_𝑛,(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 𝑦≠0 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 𝐷⁻¹𝑃𝑦=(1/𝜆¹⁽¹⁾)𝑦 that 1/𝜆¹⁽¹⁾ is the maximum eigenvalue of 𝐷⁻¹𝑃 and the 𝑦 is an eigenvector corresponding to 1/𝜆¹⁽¹⁾. By Lemma 3.3(ii), we have that all the elements of 𝐷⁻¹ are positive, then 𝐷⁻¹ 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 𝐷⁻¹𝑃 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),𝜎^𝑚−2(𝑎)]_𝕋 and 𝑝(𝑡)>0 for all 𝑡∈[𝜎^𝑚−1(𝑎),𝜌(b)]_𝕋; we can write 𝐷⁻¹𝑃 as follows: 𝐷⁻¹𝑃=⎛⎜⎜⎝00𝑍𝑉⎞⎟⎟⎠,(3.19) where 𝑉 is an 𝑚×(𝑛−𝑚) matrix and 𝑍 is an (𝑛−𝑚)×(𝑛−𝑚) 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 𝑦_𝑍>0 such that 𝑍𝑦_𝑍=(1/𝜆¹⁽¹⁾)𝑦_𝑍. Let 𝑦_𝑉=𝜆¹⁽¹⁾𝑉𝑦_𝑍 and 𝑦=(𝑦^𝑇𝑉,𝑦^𝑇𝑍)^𝑇. Obviously, we have 𝐷⁻¹𝑃𝑦=1𝜆¹⁽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 𝑥≠0 and 𝑦≠0 be any two eigenvectors of the boundary value problem (1.1), (1.3) corresponding to 𝜆⁽¹⁾ and define 𝑧=𝑥(𝑎)𝑦−𝑦(𝑎)𝑥. Obviously, we have −𝐷+𝜆⁽¹⁾𝑃𝑧=𝑥(𝑎)−𝐷+𝜆⁽¹⁾𝑃𝑦−𝑦(𝑎)−𝐷+𝜆⁽¹⁾𝑃𝑥=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 𝑁≥1 be the number of positive elements in the set {𝑝(𝑡)∣𝑡∈[𝜌(𝑎),𝜌(𝑏)]_𝕋}. Then there are 𝑁 distinct eigenvalues 𝜆^𝑖(1)(𝑖=1,2,…,𝑁) of the boundary value problem (1.1), (1.3) and 𝛼_𝑖=1/𝜆^𝑖(1)(𝑖=1,2,…,𝑁) are the only positive eigenvalues of 𝐷^−1/2𝑃𝐷^−1/2.

Proof. Suppose that 𝛼₁≥𝛼₂≥⋯≥𝛼_𝑛≥0 are all eigenvalues of 𝐷^−1/2𝑃𝐷^−1/2. Since 𝐷^−1/2𝑃𝐷^−1/2 is real and symmetric that there exists an orthogonal matrix 𝐶 such that 𝐶^𝑇𝐷^−1/2𝑃𝐷^−1/2𝐶=diag𝛼₁𝛼₂⋯𝛼_𝑛,(3.22) therefore, we have that rank(𝑃)=rank𝐶^𝑇𝐷^−1/2𝑃𝐷^−1/2𝐶=rankdiag𝛼₁𝛼₂⋯𝛼_𝑛(3.23) indicating that the number of positive 𝛼_𝑖 is the same as that of positive number in 𝑃 which is equal to 𝑁.
Suppose that 𝛼_𝑖0=𝛼_𝑖0+1>0 for some 𝑖₀ where 1≤𝑖₀≤𝑁−1. Observe that 𝐶^𝑇𝐷^−1/2𝑃𝐷^−1/2𝐶𝑒_𝑖=𝛼_𝑖𝑒_𝑖 in view of (3.22), which further implies that 𝐷𝐷^−1/2𝐶𝑒_𝑖=1𝛼_𝑖𝑃𝐷^−1/2𝐶𝑒_𝑖𝑖=𝑖₀,𝑖₀+1.(3.24) Thus, we have two independent vectors in the null space of (−𝐷+𝜆⁽¹⁾𝑃) for 𝜆⁽¹⁾=1/𝛼_𝑖0, which contradicts Lemma 3.5. Thus, from Lemma 3.2, we see that {𝜆^𝑖(1)=1/𝛼_𝑖∣𝑖=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⁾<⋯<𝜆^𝑗(1)} be the set of all eigenvalues of the boundary value problem (1.1), (1.3) and {𝜆¹⁽²⁾<𝜆²⁽²⁾<⋯<𝜆^𝑘(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 𝜆^𝑖(1)≤𝜆^𝑖(2) for 1≤𝑖≤𝑘.

Proof. It follows from Lemma 3.6 that 𝛼₁=1𝜆¹⁽¹⁾>⋯>𝛼_𝑗=1𝜆^𝑗(1)𝛽>0,𝛼_𝑗+1=⋯=𝛼_𝑛=0,₁=1𝜆¹⁽²⁾>⋯>𝛽_𝑘=1𝜆^𝑘(2)>0,𝛽_𝑘+1=⋯=𝛼_𝑛=0(3.25) are the eigenvalues of 𝐷^−1/2𝑃𝐷^−1/2 and 𝐷^−1/2𝑄𝐷^−1/2, respectively. If 𝑝(𝑡)≥𝑞(𝑡) for all 𝑡∈[𝜌(𝑎),𝜌(𝑏)]_𝕋, then 𝑃≥𝑄, implying 𝐷^−1/2𝑃𝐷^−1/2≥𝐷^−1/2𝑄𝐷^−1/2.(3.26) By Weyl’s inequality and (3.26), we have 𝛼_𝑖≥𝛽_𝑖1≤𝑖≤𝑛.(3.27) Finally, it is easily seen from (3.25) and (3.27) that 1𝜆^𝑖(1)≥1𝜆^𝑖(2)1≤𝑖≤𝑘,(3.28) implying that 𝜆^𝑖(1)≤𝜆^𝑖(2) for 1≤𝑖≤𝑘. 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).