# The Solutions of Second-Order Linear Matrix Equations on Time Scales

Authors: Kefeng Li; Chao Zhang
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

This paper studies the solutions of second-order linear matrix equations on time scales. Firstly, the necessary and sufficient conditions for the existence of a solution of characteristic equation are introduced; then two diverse solutions of characteristic equation are applied to express general solution of the matrix equations on time scales.


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## Body

## 1. Introduction

In this paper, we consider the solutions for the following second-order linear matrix equations:(1)[P(t)XΔ(t)]Δ+QXΔ(t)+R(t)Xσ(t)=0,t∈𝕋,
where P(t)=Pμ(t), R(t)=R/μ(t), P,Q,R,X(t)∈Rm×m, XΔ is the delta derivative, σ(t) is the forward jump operators, Xσ(t)∶=X(σ(t)), μ(t) is the graininess function, and 𝕋 is an infinite isolated time scale, which is given by
(2)𝕋∶={t0,t1,t2,…,tn,…}.As a tool for establishing a unified framework for continuous and discrete analysis, a theory of dynamic equations on measure chains was introduced by Hilger in his Ph.D. thesis [1] in 1988. In many cases, it is necessary to study a special case of measure chains-time scales. In the last decade the investigation of dynamic systems on time scales has involved much interesting, including quite a few fields, such as the theory of calculus, the oscillation of the dynamic systems, the eigenvalue problems and boundary value problems, and partial differential equations on time scales, and so forth [2–5]. Up to now, there are few results about matrix equations on time scales. In 1998, Agarwal and Bohner [6] studied the quadratic functionals for second order matrix equations on time scales; in 2002, Erbe and Peterson [7] obtained oscillation criteria for a second-order self-adjoint matrix differential equation on time scales in terms of the eigenvalues of the coefficient matrices and the graininess function. The theory of dynamic systems on time scales is of very important theoretical significance and has a wide range of applications.Based on the related literatures, the researches of solutions of matrix difference or differential equations have few results. In 1999, Barkatou [8] proposed an algorithm for the rational solutions of special matrix difference equations and discussed their applications; in 2003, Freiling and Hochhaus [9] investigated some properties of the solution for rational matrix difference equations; in 2004, Xu and Zhang [10] studied the representation of general solution for second order homogeneous matrix difference equations; in 2011, Wu and Zhou [11] obtained the particular solutions of one kind of second order matrix differential equations. Since the continuous case and the discrete case are two special cases of time scales, so, we study the solutions of second order linear matrix equations on time scales.This paper is organized as follows. Section2 introduces some basic concepts and fundamental theory about time scales. By using the characteristic equation of the matrix equation the solutions of (1) 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
(3)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
(4)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 that 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
(5)|f(σ(t))-f(s)-a(σ(t)-s)|≤ε|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
(6)fΔ(t)={lims∈𝕋s→tf(t)-f(s)t-s,ifμ(t)=0,f(σ(t))-f(t)μ(t),ifμ(t)>0.For convenience, we introduce the following results ([3, Lemma 1] and [4, Chapter 1]), which are useful in this paper.Lemma 1.
Letf,g:𝕋→R and t∈𝕋k. If f and g are differentiable at t, then fg is differentiable at t and
(7)(fg)Δ(t)=fσ(t)gΔ(t)+fΔ(t)g(t)=fΔ(t)gσ(t)+f(t)gΔ(t).


## 3. Main Results

We assume throughout this paper thatP is an invertible matrix. It follows from Lemma 1, (6), and (2) that the matrix equation (1) can be written in the form
(8)Xσ2(t)-BXσ(t)-AX(t)=0,t∈𝕋,
where Xσ2(t)=Xσ(σ(t)), B=2Em-P-1Q-P-1R, A=P-1Q-Em, and Em is the identity matrix.Definition 2.
The equation(9)F(Λ)=Λ2-BΛ-A=0,
where Λ∈Rm×m, is called the characteristic equation of (8).Definition 3.
The functions(10)F(λ)=λ2-Bλ-A,h(λ)=detF(λ)
are called the eigenmatrix and eigenpolynomial of (8). Such a λ which satisfies h(λ)=0 is called an eigenvalue of F(λ).
In order to study the solutions of matrix equation (8), we now introduce some results about the characteristic equation (9).Theorem 4.
IfAB=BA and there exists C∈Rm×m such that C2=B2+4A, BC=CB, then the characteristic equation (9) has solutions.Proof.
LetΛ1=(1/2)(B+C), Λ2=(1/2)(B-C). Then Λ1 and Λ2 are solutions of (9).Remark 5.
Generally, ifAB=BA and C2=B2+4A, we cannot easily get BC=CB.Example 6.
Let(11)A=(1201414),B=(102-1).
Then
(12)C=(3011+31),BC=(305+231+3-1),CB=(303+231+3-1).
Obviously, BC≠CB.So, Huang and Chen [12] and J. Huang and H. Huang [13] obtained the following results.Definition 7.
Letλ0 be an eigenvalue of F(λ). Then
(13)Vλ0={x∣(λ02-Bλ0-A)x=0,x∈Rm×1}
is called a characteristic subspace of F(λ) corresponding to λ0; the nonzero vector x of Vλ0 is called the eigenvector corresponding to λ0.Theorem 8.
There exist the diagonalizable solutions of characteristic equation (9) if and only if the dimension of the sum of characteristic subspaces Vλi is n; that is, dim{∑i=1hVλi}=m, where λi(i=1,2,…,h) are the distinct eigenvalues of F(λ).Definition 9.
Letλi,i=1,2,…,s, be eigenvalues of F(λ). The extension vectors set of Vλiare as follows:
(14)EVλi={(x0,x1,…,xk)∣(λi2-Bλi-A)x0=0,(λi2-Bλi-A)x1=-(2λiEm-B)x0,k≤ki-1,(λi2-Bλi-A)xk=-(2λiEm-B)xk-1-xk-2Em},
where ki is the multiplicity of λi, (x0,x1,…,xk) is called the component of EVλi.Theorem 10.
The characteristic equation (9) has solutions if and only if there exist some components (x0,x1,…,xk),…,(y0,y1,…,yq) of the set ∑i=1sEVλi in (14), and they are m linearly independent extension vectors.In the following, we discuss the general solutions of the matrix equation (8) by using the solutions of the characteristic equation (9).Theorem 11.
IfΛ1 and Λ2 are two diverse solutions of the characteristic equation (9), which satisfy Λ1Λ2=Λ2Λ1, and Λ1-Λ2 is invertible, then
(15)X(tn)=(∑k=1n‍Λ1k-1Λ2n-k)X(t1)-(∑k=1n-1‍Λ1n-kΛ2k)X(t0)
is the general solution of the matrix equation (8).Proof.
SinceΛ1 and Λ2 are two solutions of (9), then
(16)Λ12-BΛ1-A=0,Λ22-BΛ2-A=0,
which deduce
(17)(Λ1+Λ2)(Λ1-Λ2)=B(Λ1-Λ2).
Hence, by the invertibility of Λ1-Λ2, we have
(18)B=Λ1+Λ2,-A=Λ1Λ2,(Λ1-Λ2)2=B2+4A.
Let C=Λ1-Λ2. It follows from Λ1Λ2=Λ2Λ1 that
(19)AB=BA,AC=CA,BC=CB.
Replacing B and A with (1/2)(B+C)+(1/2)(B-C) and -(1/4)(B+C)(B-C) in matrix equation (8), we obtain
(20)2xσ2(t)-(B-C)Xσ(t)=12(B+C)×[2Xσ(t)-(B-C)X(t)].
Let
(21)Y(t)=2Xσ(t)-(B-C)X(t).
Then (20) can be rewritten as
(22)Yσ(t)=12(B+C)Y(t).
Hence, the general solution of (21) is as follows:
(23)Y(tn)=12n(B+C)nY(t0),
where Y(t0)=2X(t1)-(B-C)X(t0). Putting Y(tn) into (21), we can get the general solution of (8):
(24)x(tn)=12n-1(B-C)n-1X(t1)+∑k=1n-1‍12k(B-C)k-1Y(tn-k)=-12n∑k=1n-1‍(B+C)n-k(B-C)kX(t0)+12n-1∑k=1n‍(B+C)k-1(B-C)n-kX(t1)=(∑k=1n‍Λ1k-1Λ2n-k)X(t1)-(∑k=1n-1‍Λ1n-kΛ2k)X(t0).Theorem 12.
IfΛ1 and Λ2 are two diverse solutions of the characteristic equation (9) and Λ1-Λ2 is invertible, then
(25)X(tn)=Λ1nC1+Λ2nC2,
where C1=(Λ2-Λ1)-1(Λ2X(t0)-X(t1)) and C2=(Λ1-Λ2)-1(Λ1X(t0)-X(t1)), is the general solution of the matrix equation (8).Proof.
Firstly, we prove (25) is a solution of (8). Putting (25) into (8), we have
(26)(Λ1n+2C1+Λ2n+2C2)-B(Λ1n+1C1+Λ2n+1C2)-A(Λ1nC1+Λ2nC2)=(Λ12-BΛ1-A)Λ1nC1+(Λ22-BΛ2-A)×Λ2nC2=0.
By using X(t0)=C1+C2, X(t1)=Λ1C1+Λ2C2, and Λ1-Λ2 is invertible, then we get
(27)C1=(Λ2-Λ1)-1(Λ2X(t0)-X(t1)),C2=(Λ1-Λ2)-1(Λ1X(t0)-X(t1)).
Next, we will prove all the solutions of (8) can be written into the form of (25).
Obviously, byC1=(Λ2-Λ1)-1(Λ2X(t0)-X(t1)) and C2=(Λ1-Λ2)-1(Λ1X(t0)-X(t1)), we have X(t0)=C1+C2, X(t1)Λ1C1+Λ2C2; then X(t0) and X(t1) are in the form of (25). Assume the solution of (8) can be written in the form of (25) for n≤k; that is,
(28)X(tk)=Λ1kC1+Λ2kC2forn≤k.
Then, it follows from (8) and Λ1, Λ2  being solutions of (9) that
(29)x(tk+3)=Xσ2(tk+1)=BXσ(tk+1)+AX(tk+1)=BX(tk+2)+AX(tk+1)=B(Λ1k+2C1+Λ2k+2C2)+A(Λ1k+1C1+Λ2k+1C2)=(BΛ1+A)Λ1k+1C1+(BΛ2+A)Λ2k+1C2=Λ12Λ1k+1C1+Λ22Λ2k+1C2=Λ1k+3C1+Λ2k+3C2.
By using the method of mathematical induction, we have all the solutions of (8) can be written in the form of (25).



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