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

Authors: Chao Zhang; Shurong Sun
Journal: Journal of Applied Mathematics
 (2012)

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## 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.


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## Body

## 1. Introduction

In this paper, we consider the eigenvalue problems for the following second-order linear equations:(1.1)(r(t)yΔ(t))Δ+λ(1)p(t)yσ(t)=0,t∈[ρ(a),ρ(b)]T,(1.2)(r(t)yΔ(t))Δ+λ(2)q(t)yσ(t)=0,t∈[ρ(a),ρ(b)]T,
with the boundary conditions(1.3)y(ρ(a))-τy(a)=y(σ(b))-δy(b)=0,
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(1.4)[ρ(a),ρ(b)]T:={ρ(a),a,σ(a),σ2(a),…,ρ(b)}.We assume throughout this paper that(H1)
rΔ, 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 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 [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. Section2 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.Let𝕋⊂R be a nonempty closed subset. Define the forward and backward jump operators σ,ρ:𝕋→𝕋 by(2.1)σ(t)=inf⁡{s∈T:s>t},ρ(t)=sup⁡{s∈T:s<t},
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,μ:𝕋→[0,∞) are defined by(2.2)μ(t)=σ(t)-t,ν(t)=t-ρ(t).
Let f be a function defined on 𝕋. f is said to be (delta) differentiable at t∈𝕋k provided there exists a constant a such that for any ɛ>0, there is a neighborhood U of t (i.e., U=(t-δ,t+δ)∩𝕋 for some δ>0) with(2.3)|f(σ(t))-f(s)-a(σ(t)-s)|≤ɛ|σ(t)-s|,∀s∈U.
In this case, denote fΔ(t):=a. If f is (delta) differentiable for every t∈𝕋k, then f is said to be (delta) differentiable on 𝕋. If f is differentiable at t∈𝕋k, then(2.4)fΔ(t)={lim⁡s→ts∈Tf(t)-f(s)t-sifμ(t)=0,f(σ(t))-f(t)μ(t)ifμ(t)>0.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.
Letf,g:𝕋→R and t∈𝕋k.(i)
Iff and g are differentiable at t, then fg is differentiable at t and
(2.5)(fg)Δ(t)=fσ(t)gΔ(t)+fΔ(t)g(t)=fΔ(t)gσ(t)+f(t)gΔ(t).(ii)
Iff and g are differentiable at t, and f(t)fσ(t)≠0, then f-1g is differentiable at t and
(2.6)(gf-1)Δ(t)=(gΔ(t)f(t)-g(t)fΔ(t))(fσ(t)f(t))-1.


## 3. Eigenvalue Comparisons

In the following, we will writeX≥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)=σn-2(a),b=σn-1(a),μi=σi+1(a)-σi(a), and rσi(a)=r(σi(a)),i=-1,0,1,2,…,n-1.It follows from Lemma2.1(ii), (2.4), and (1.4) that the boundary value problem (1.1), (1.3) can be written in the form(3.1)(-D+λ(1)P)y=0,
where(3.2)D≔(A+B-B0⋯000-BB+C-C⋯0000-CC+rσ2(a)μ2⋯000⋯⋯⋯⋯⋯⋯⋯000⋯rσn-4(a)μn-4+D-D0000⋯-DD+E-E000⋯0-EE+(1-δ)rσn-1(a)μn-1)P=diag⁡(μ-1p(σ-1(a)),μ0p(σ0(a)),…,μn-3p(σn-3(a)),μn-2p(σn-2(a))),where 𝒜 donates  (1-τ)rσ-1(a)/μ-1, ℬdonatesrσ0(a)/μ0, 𝒞 donates rσ(a)/μ1, 𝒟 donates rσn-3(a)/μn-3, and  ℰ donates rσn-2(a)/μn-2. (3.3)y=(y(σ0(a)),y(σ(a)),…,y(σn-2(a)),y(σn-1(a)))T.
And the problem (1.2), (1.3) is equivalent to the equation (3.4)(-D+λ(2)Q)y=0,
where(3.5)Q=diag⁡(μ-1q(σ-1(a)),μ0q(σ0(a)),…,μn-3q(σn-3(a)),μn-2q(σn-2(a))).
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.Letei be the ith column of the identity matrix I of order n and(3.6)D1:=(A+B-B0⋯000-BB+C-C⋯0000-CC+rσ2(a)μ2⋯000⋯⋯⋯⋯⋯⋯⋯000⋯rσn-4(a)μn-4+D-D0000⋯-DD+E-E000⋯0-EE)
Define Pi=I+ei-1eiT. It is easily seen that(3.7)D=D1+en(1-δ)rσn-1(a)μn-1enT,(3.8)P2P3⋯PnD1PnT⋯P3TP2T=diag⁡((1-τ)rσ-1(a)μ-1,rσ0(a)μ0,…,rσn-3(a)μn-3,rσn-2(a)μn-2).
It follows from (H1),(H2), and (3.8) that(3.9)D1=Pn-1⋯P3-1P2-1diag⁡((1-τ)rσ-1(a)μ-1,rσ0(a)μ0,…,rσn-3(a)μn-3,rσn-2(a)μn-2)P2-TP3-T⋯Pn-T,(3.10)D1-1=PnT⋯P3TP2Tdiag⁡(μ-1(1-τ)rσ-1(a),μ0rσ0(a),…,μn-3rσn-3(a),μn-2rσn-2(a))P2P3⋯Pn.For anyx=(x1,x2,…,xn)T, we have(3.11)x*Dx=x*D1x+(1-δ)rσn-1(a)μn-1x*enenTx=(P2-TP3-T⋯Pn-Tx)*diag⁡((1-τ)rσ-1(a)μ-1,rσ0(a)μ0,…,rσn-3(a)μn-3,rσn-2(a)μn-2)×(P2-TP3-T⋯Pn-Tx)+(1-δ)rσn-1(a)μn-1|enTx|2≥0.
Moreover, x*Dx=0 implies x=0. Hence, the matrix D is positive definite.Lemma 3.1.
Ifλ(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(3.12)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,
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.
IfρPx=Dx and ρ≠λ(1), then
(3.13)(λ(1)-ρ)x*Py=λ(1)x*Py-ρx*Py=x*(λ(1)Py)-(ρPx)*y=x*Dy-(Dx)*y=0.
Hence, x*Py=0. This completes the proof.Lemma 3.2.
Ifλ(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.
Ifλ(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,
(3.14)λ(1)Py=D1/2D1/2y,D-1/2PD-1/2(D1/2y)=1λ(1)(D1/2y).
With a similar argument, one can get that ifα 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 any1≤i,j≤n, define γ=min{i,j}. We have(i)
eiTD1-1ej=μ-1/(1-τ)rσ-1(a)+∑k=0γ-2(μk/rσk(a));(ii)
eiTD-1ej = ((μ-1/(1-τ)rσ-1(a))+∑k=0γ-2(μk/rσk(a)))((μn-1/(1-δ)rσn-1(a))+(μ-1/(1-τ)rσ-1(a))+∑k=0n-2(μk/rσk(a))) − ((μ-1/(1-τ)rσ-1(a))+∑k=0i-2(μk/rσk(a)))((μ-1/(1-τ)rσ-1(a))+∑k=0j-2(μk/rσk(a)))/(μn-1/(1-δ)rσn-1(a)) + (μ-1/(1-τ)rσ-1(a))+∑k=0n-2(μk/rσk(a)).Proof.
It is easy to see thatPiej=ej if i≠j, while Piej=ej-1+ej if i=j. Hence,
(3.15)P2P3⋯Pnej=e1+e2+⋯+ej.(i)
It is seen from (3.10) and (3.15) that
(3.16)eiTD1-1ej=(P2P3⋯Pnei)Tdiag⁡(μ-1(1-τ)rσ-1(a),μ0rσ0(a),…,μn-3rσn-3(a),μn-2rσn-2(a))(P2P3⋯Pnej)=(e1+e2+⋯+ei)Tdiag⁡(μ-1(1-τ)rσ-1(a),μ0rσ0(a),…,μn-3rσn-3(a),μn-2rσn-2(a))×(e1+e2+⋯+ej)=μ-1(1-τ)rσ-1(a)+∑k=0γ-2μkrσk(a).(ii)
It follows from (3.7) and the Sherman-Morrison updating formula [19] that
(3.17)D-1=D1-1-D1-1enenTD1-1(μn-1/(1-δ)rσn-1(a))+enTD1-1en,
leading to
(3.18)eiTD-1ej=eiTD1-1ej-eiTD1-1enenTD1-1ej(μn-1/(1-δ)rσn-1(a))+enTD1-1en,
which, together with (i), further implies the result (ii). This completes the proof.Theorem 3.4.
(i) Ifλ(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) Ifλ1(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 eithery(a)=0 or y(b)=0. By the boundary condition (1.3), we can easily deduce a contradiction y(t)≡0.
(ii) It follows fromD-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.
Ifp(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.
Letp(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),σm-2(a)]𝕋 and p(t)>0 for all t∈[σm-1(a),ρ(b)]𝕋; we can write D-1P as follows:
(3.19)D-1P=(0V0Z),
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 yZ>0 such that ZyZ=(1/λ1(1))yZ. Let yV=λ1(1)VyZ and y=(yVT,yZT)T. Obviously, we have
(3.20)D-1Py=1λ1(1)y,wherey>0.
This completes the proof.Lemma 3.5.
Ifλ(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.
Letx≠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
(3.21)(-D+λ(1)P)z=x(a)(-D+λ(1)P)y-y(a)(-D+λ(1)P)x=0,
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.
LetN≥1 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 D-1/2PD-1/2.Proof.
Suppose thatα1≥α2≥⋯≥αn≥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
(3.22)CTD-1/2PD-1/2C=diag⁡(α1α2⋯αn),
therefore, we have that
(3.23)rank⁡(P)=rank⁡(CTD-1/2PD-1/2C)=rank⁡(diag⁡(α1α2⋯αn))
indicating that the number of positive αi is the same as that of positive number in P which is equal to N.
Suppose thatαi0=αi0+1>0 for some i0 where 1≤i0≤N-1. Observe that CTD-1/2PD-1/2Cei=αiei in view of (3.22), which further implies that
(3.24)D(D-1/2Cei)=1αiP(D-1/2Cei)i=i0,i0+1.
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/αi∣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.
Letj 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 1≤i≤k.Proof.
It follows from Lemma3.6 that
(3.25)α1=1λ1(1)>⋯>αj=1λj(1)>0,αj+1=⋯=αn=0,β1=1λ1(2)>⋯>βk=1λk(2)>0,βk+1=⋯=αn=0
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
(3.26)D-1/2PD-1/2≥D-1/2QD-1/2.
By Weyl’s inequality and (3.26), we have
(3.27)αi≥βi1≤i≤n.
Finally, it is easily seen from (3.25) and (3.27) that
(3.28)1λi(1)≥1λi(2)1≤i≤k,
implying that λi(1)≤λi(2) for 1≤i≤k. This completes the proof.



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