The matrix bounds and fixed-point iteration for the solution of the discrete algebraic Riccati equation

The matrix bounds and fixed-point iteration for the solution of the discrete algebraic Riccati... Abstract In this paper, by constructing the equivalent form of the discrete algebraic Riccati equation (DARE), applying the eigenvalue inequalities of matrix’s sum and product, we provide new lower and upper matrix bounds for the solution of the DARE, which improve some of the recent results. Further, using the derived matrix bounds and a fixed point theorem, we discuss the existence and uniqueness for the solution of the DARE. Finally, we give corresponding numerical examples to demonstrate the effectiveness of our results. 1. Introduction and preliminaries In many control analysis and design problems, particularly in the field of optimal control, many problems reduce to solve the discrete algebraic Riccati equation (DARE) (Lancaster & Rodman, 1995)–(Ionexcu et al., 1999) \begin{equation} P = A^{T}PA -A^{T}PB(I + B^{T}PB)^{-1}B^{T}PA + Q, \end{equation} (1) where A ∈ Rn×n, B ∈ Rn×m, Q ∈ Rn×n is a symmetric positive definite matrix, and the matrix P ∈ Rn×n is the symmetric positive definite solution of the DARE (1). To guarantee the existence of the stabilizing solution of the DARE, it shall be assumed that (A, B) is a stabilizable pair and (A, Q) is a detectable pair. If the pair (A, B) is not stabilizable, this equation cannot have a positive definite solution. Further, when $${\sigma ^{2}_{1}}(A)<1$$, then the eigenvalues of the matrix A satisfy |λ| < 1, which means that A is stable in the discrete-time sense. Thus, (A, B) is automatically stabilzable, but the converse implication is not true (Callier & Desoer, 1991). In many practical control problems, to study the DARE (1), we usually assume that Q is positive definite. In addition, there are many scholars to research this equation on the assumption that Q is positive definite (Komaroff, 1994)–(Lee, 2004). In this paper, we consider the case where Q is positive definite. The DARE (1) has wide application in linear system. Consider the following linear discrete system described by the state space equation: \begin{equation} \begin{cases} x(t+1)=Ax(t)+Bu(t),\\ x(0)=x_{0}, \end{cases} \end{equation} (2) where t = 0, 1, ⋯, N − 1, x(t) ∈ Rn is a state variable, u(t) ∈ Rm is an input variable, A, B are defined in (1). For the system (2), choosing the Hamiltonian function, applying the minimum principle, we can obtain the optimal control $$ u^{*}(t)=-B^{T}A^{-T}(P-Q)x^{*}(t)$$ such that the quadratic performance index of (2) $$ \tilde{\!\!J}=\tfrac{1}{2}x^{T}(N)Px(N)+\tfrac{1}{2}\sum_{t=0}^{N-1}\left[ x^{T}(t)Qx(t)+u^{T}(t)u(t) \right]$$ attains a minimum, where P is the symmetric positive definite solution of the DARE (1). Applying the matrix identity $$ (X^{-1}+YZ)^{-1}=X-XY(I+ZXY)^{-1}ZX$$ from Kailath (1980) with X, Y, Z are square nonsingular matrices, the DARE (1) can be written as \begin{equation} P=A^{T}(P^{-1}+BB^{T})^{-1}A+Q, \end{equation} (3) which shows that (3) is equivalent to (1). In many control design problems, it is significant to investigate the DARE (1) in consideration of its importance (Lancaster & Rodman, 1995)–(Ionexcu et al., 1999). Recently, solving this equation becomes a heated topic. There are few iterative algorithms for deriving the solution of this equation (Komaroff, 1994). In 1994, by using the monotonicity of matrix bounds of the DARE, Komaroff showed the iteration algorithm if this equation possesses positive definite solution. In 1984, Arnold and Laub presented the generalized eigenproblem formulation and a Newtontype iterative refinement procedure for the solution of general forms of algebraic Riccati equations arising in discrete time application (Arnold & Laub, 1984). From Komaroff (1994), we know that the computation of the positive definite solution of this equation is of some difficulty especially when the dimension of the matrices is high. The closer the initial estimate is to the actual solution, the less computer time is expected to be used in the solution algorithm. Thus, there are many works for obtaining the solution bounds, including matrix bounds (Komaroff, 1994)–(Lee & Hsien, 1999), (Komaroff, 1992)–(Zhang & Liu, 2015), summation bounds (Komaroff & Shahian, 1992), eigenvalue bounds (Lee, 2004), (Liu et al., 2010), trace bounds (Kim et al., 1993), (Liu & Zhang, 2009), eigenproblem algorithms and software (Arnold & Laub, 1984), of the DARE in recent years. It is known to us that, amongst these bounds, the matrix bounds are the most general. Further, Scherer in 1999 showed lower bounds of matrix norm in $$H_{2}/H_{\infty }$$ problem (Scherer, 1999). However, this is a numerical bound, not a matrix bound. Sometimes, contrary to numerical bounds, the matrix bounds can reveal the internal structure of relative systems better. Moreover, some necessary and sufficient conditions are established for the constraint generalized Sylvester matrix equations to have a common solution (Wang et al., 2016). In addition, Wang and He (Wang & He, 2014) study some systems of coupled generalized Sylvester matrix equations and present some necessary and sufficient conditions for the solvability to these systems. The paper is organized as follows. In Section 2, we give some notations and several basic preliminary lemmas. In Sections 3, by constructing the equivalent form of the DARE, applying the eigenvalue inequalities of matrix’s sum and product, we provide new lower and upper matrix bounds for the solution of the DARE, which improve some of the recent results. In Section 4, we discuss the existence and uniqueness for the solution of the DARE. In Section 5, we give corresponding numerical examples to demonstrate the effectiveness of the provided results. In Section 6, we offer some concluding remarks. 2. Basic preliminaries Let us introduce some notation and lemmas first. Let Rn×m and N+ denote the set of n × m real matrices and the set of positive integers. For X = (xij) ∈ Rm×n, suppose XT, ||X|| denote the transpose and the spectral norm of X, respectively. If X ∈ Rn×n, X−1 denotes the inverse of X. Suppose X ∈ Rn×n is an arbitrary symmetric matrix, we assume that the eigenvalues of X are arranged so that λ1(X) ≥ λ2(X) ≥ ⋯ ≥ λn(X). For X, Y ∈ Rn×n, suppose the singular values of X are arranged so that σ1(X) ≥ σ2(X) ≥ ⋯ ≥ σn(X). The inequality X > (≥) 0 means that X is a symmetric positive (semi-) definite matrix and X > (≥) Y means X − Y is a symmetric positive (semi-) definite matrix. The identity matrix with appropriate dimensions is represented by I. Lemma 1 (Kim et al., 1993). Let P be the positive definite solution of the DARE (1), then P has the lower matrix bound \begin{equation} P\geq A^{T}(I+QBB^{T})^{-1}QA+Q\equiv P^{*}_{l}. \end{equation} (4) Lemma 2 (Horn & Johnson, 1986, pp. 300, Theorem 5.6.15). Let Xi ∈ Rn×n, then the series $$\sum \limits\nolimits _{i=1}^{\infty} {X_{i}}$$ converges if there is a matrix norm ∥.∥ on Rn×n such that the numerical series $$\sum \limits\nolimits _{i=1}^{\infty} {\|X_{i}\|}$$ converges. Lemma 3 (Marshall & Olkin, 1979, chapter 9, A. 1. a). Let X, Y ∈ Rn×n, for i = 1, 2, ⋯, n, then $$ \lambda_{i}(XY)=\lambda_{i}(YX).$$ Lemma 4 (Marshall & Olkin, 1979, pp. 248, H.1.g). Let X, Y ∈ Rn×n ≥ 0, then $$ \sum\limits_{i=1}^{n}\lambda_{i}(XY)\leq\sum\limits_{i=1}^{n}\lambda_{i}(X)\lambda_{i}(Y).$$ Lemma 5 (Marshall & Olkin, 1979, pp. 242, G.1.a). If X, Y ∈ Rn×n are symmetric matrices and 1 ≤ i1 ≤ ⋯ ≤ it ≤ n, then $$ \sum\limits_{j=1}^{t}\lambda_{i_{j}}(X+Y)\leq \sum\limits_{j=1}^{t}\lambda_{i_{j}}(X)+\sum\limits_{i=1}^{t}\lambda_{i}(Y),\quad t=1,\cdots,n.$$ Lemma 6 (Berinde, 2009, Theorem B). Let (X, d) be a complete metric space and F : X → X be a strict contraction, i.e., a map satisfying $$ d(Fx, Fy)\leq ad(x,y),\quad\ \mbox{for all} \quad\ x, y \in X,$$ where 0 ≤ a < 1 is constant. Then F has a unique fixed point in X. A norm space becomes a metric space if the distance d(x, y) between x and y is ||x − y||. Hence, we get the following conclusion from Lemma 6. Lemma 7 Let (B, ||.||) be a real Banach space, and $$\Omega\subset$$B be a convex closed and bounded subset and F : $$\Omega\,\rightarrow\,\Omega$$ be a contraction map, i.e., a map satisfying $$ ||F(P_{1})-F(P_{2})||\leq p ||P_{1}-P_{2}||, \quad\mbox{for all}\quad \ P_{1}, P_{2} \in \Omega,$$ where 0 ≤ p < 1. Then F has a unique fixed point in $$\Omega$$. Lemma 8 (Zhao et al., 2009, Lemma 2.1). If X, Y ∈ Rn×n, and 0 < X ≤ Y, then $$ ||X||\leq ||Y||.$$ 3. New lower and upper matrix bounds for the DARE In this section, we first establish a new lower bound for the solution of the DARE (1) which improves (4). Theorem 1 Let P be the positive definite solution of the DARE (1), then P has the lower matrix bound \begin{equation} P \geq A^{T}\left\{\left[A^{T}(Q^{-1}+BB^{T})^{-1}A+Q\right]^{-1}+BB^{T}\right\}^{-1}A+Q\equiv P_{l}. \end{equation} (5) Proof. By (3), obviously, P ≥ Q, i.e., \begin{equation} P^{-1}\leq Q^{-1}. \end{equation} (6) Substituting (3) into the right-hand side of (3) shows that \begin{equation} P=A^{T}(P^{-1}+BB^{T})^{-1}A+Q=A^{T}\left\{\left[A^{T}(P^{-1}+BB^{T})^{-1}A+Q\right]^{-1}+BB^{T}\right\}^{-1}A+Q. \end{equation} (7) Substituting (6) into (7) yields $$ P\geq A^{T}\left\{\left[A^{T}(Q^{-1}+BB^{T})^{-1}A+Q\right]^{-1}+BB^{T}\right\}^{-1}A+Q.$$ This completes the proof. Remark 1 We point out that inequality (5) improves inequality (4). Actually, it is obvious that \begin{align*} &A^{T}\left\{\left[A^{T}(Q^{-1}+BB^{T})^{-1}A+Q\right]^{-1}+BB^{T}\right\}^{-1}A+Q\\ &\quad>A^{T}(Q^{-1}+BB^{T})^{-1}A+Q\\ &\quad=A^{T}\left[Q^{-1}(I+QBB^{T})\right]^{-1}A+Q\\ &\quad=A^{T}(I+QBB^{T})^{-1}QA+Q, \end{align*} which means that inequality (5) is stronger than inequality (4). Denote $$\delta ^{*}_{1}\equiv \lambda _{1}(P^{*}_{l})$$ with $$P^{*}_{l}$$ defined by (4). The following result is obtained in Lee (1997): Theorem A (Lee, 1997, Theorem 2). Let P be the positive definite solution of the DARE (1), and $${\sigma ^{2}_{1}}(A)<1+{\sigma ^{2}_{n}}(B)\delta ^{*}_{1}$$, then P has the upper matrix bound \begin{equation} P\leq \frac{\lambda_{1}(Q)}{1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)}A^{T}A+Q\equiv P^{*}_{u}. \end{equation} (8) Next, we provide a new upper matrix bound which improves Theorem A. Theorem 2 Let P be the positive definite solution of the DARE (1). (1) If σn(B) = 0 and $${\sigma ^{2}_{1}}(A)<1$$, then P has the upper matrix bound \begin{equation} P \leq\sum\limits_{m=0}^{\infty}{(A^{T})^{m}QA^{m}}\equiv P_{u}. \end{equation} (9) (2) If σn(B) > 0, then P has the upper matrix bound \begin{equation} P \leq\frac{\eta_{1}}{1+\delta_{1}{\sigma^{2}_{n}}(B)}A^{T}A+Q\equiv P_{u}, \end{equation} (10) where δ1 ≡ λ1(Pl) with Pl defined by (5), and $$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\eta_{1}\equiv \left\{\lambda_{1}(Q){\sigma^{2}_{n}}(B)+{\sigma^{2}_{1}}(A)-1+\left[\left(\lambda_{1}(Q){\sigma^{2}_{n}}(B)+{\sigma^{2}_{1}}(A)-1\right)^{2}+4\lambda_{1}(Q){\sigma^{2}_{n}}(B)\right]^{\frac{1}{2}}\right\}\left/\left(2{\sigma^{2}_{n}}(B)\right)\!.\right .$$ Proof. 1) If σn(B) = 0, by (3), it is obvious that \begin{equation} P=A^{T}(P^{-1}+BB^{T})^{-1}A+Q\leq A^{T}(P^{-1})^{-1}A+Q= A^{T}PA+Q. \end{equation} (11) Substituting (11) into the right hand of (11) yields \begin{align} \nonumber P&\leq A^{T}PA+Q\leq A^{T}(A^{T}PA+Q)A+Q\\ \nonumber &= (A^{T})^{2}PA^{2}+A^{T}QA+Q\\ \nonumber &\leq (A^{T})^{2}(A^{T}PA+Q)A^{2}+A^{T}QA+Q\\ &\leq \cdots\leq\sum\limits_{m=0}^{n}{(A^{T})^{m}QA^{m}} +(A^{T})^{n+1}PA^{n+1}. \end{align} (12) As $${\sigma ^{2}_{1}}(A)<1$$, choosing spectral norm in Lemma 2, from Lemmas 3 and 4, \begin{align*} \sum\limits_{m=0}^{\infty}\sigma_{1}\left[(A^{T})^{m}QA^{m}\right]&= \sum\limits_{m=0}^{\infty}\lambda_{1}\left[(A^{T})^{m}QA^{m}\right]\\ &= \sum\limits_{m=0}^{\infty}\lambda_{1}\left[A^{m}(A^{T})^{m} \right] \leq \sum\limits_{m=0}^{\infty}{\lambda_{1}\left[A^{m}(A^{T})^{m}\right]\lambda_{1}(Q)}\\ &\leq\sum\limits_{m=0}^{\infty}{{\lambda^{m}_{1}}(AA^{T})\lambda_{1}(Q)}\\ &=\lambda_{1}(Q)\sum\limits_{m=0}^{\infty}\left[{\sigma^{2}_{1}}(A)\right]^{m}\\ &=\frac{\lambda_{1}(Q)}{1-{\sigma^{2}_{1}}(A)}. \end{align*} Obviously, $$\sum \limits\nolimits _{m=0}^{\infty }\,\sigma _{1}\left [(A^{T})^{m}QA^{m}\right ]$$ is convergent. By Lemma 2, $$\sum \limits\nolimits _{m=0}^{\infty} {\,(A^{T})^{m}QA^{m}}$$ is convergent. As $$\sum \limits\nolimits _{m=0}^\infty \,\sigma _{1}[(A^T)^mQA^m]$$ is convergent, then $$ \lim\limits_{n\rightarrow\infty}\sigma_{1}[{(A^{T})^{(n+1)}PA^{(n+1)}}]=0. $$ Since $$ \lambda_{1}\left[{(A^{T})^{(n+1)}PA^{(n+1)}}\right]\leq\sigma_{1}\left[{(A^{T})^{(n+1)}PA^{(n+1)}}\right]\!, $$ thus $$ \lim\limits_{n\rightarrow\infty}\lambda_{1}\left[{(A^{T})^{(n+1)}PA^{(n+1)}}\right]=0. $$ Hence, \begin{equation} \lim\limits_{n\rightarrow\infty}{(A^{T})^{(n+1)}PA^{(n+1)}}=0. \end{equation} (13) Let $$n\rightarrow \infty $$ in (12), in terms of (13), $$ P=\lim\limits_{n\rightarrow\infty}P\leq \lim\limits_{n\rightarrow\infty}\sum\limits_{m=0}^{n}{(A^{T})^{k}QA^{k}} +\lim\limits_{n\rightarrow\infty}(A^{T})^{n+1}PA^{n+1}=\sum\limits_{m=0}^{\infty}{(A^{T})^{m}QA^{m}}.$$ 2) If σn(B) > 0, applying Lemma 5 to (3) satisfies \begin{align} \nonumber P&=A^{T}(P^{-1}+BB^{T})^{-1}A+Q\\ \nonumber&\leq \lambda_{1}(P^{-1}+BB^{T})^{-1}A^{T}A+Q \\ \nonumber&=\frac{1}{\lambda_{n}(P^{-1}+BB^{T})}A^{T}A+Q\\ \nonumber&\leq\frac{1}{\lambda_{n}(P^{-1})+{\sigma^{2}_{n}}(B)}A^{T}A+Q\\ \nonumber&=\frac{1}{\frac{1}{\lambda_{1}(P)}+{\sigma^{2}_{n}}(B)}A^{T}A+Q\\ &=\frac{\lambda_{1}(P)}{1+\lambda_{1}(P){\sigma^{2}_{n}}(B)}A^{T}A+Q. \end{align} (14) Introducing Lemma 5 to (14) yields $$ \lambda_{1}(P)\leq \lambda_{1}\left[ \frac{\lambda_{1}(P)}{1+\lambda_{1}(P){\sigma^{2}_{n}}(B)}A^{T}A+Q\right]\leq \frac{\lambda_{1}(P)}{1+\lambda_{1}(P){\sigma^{2}_{n}}(B)}{\sigma^{2}_{1}}(A)+\lambda_{1}(Q),$$ which is equivalent to \begin{equation} {\sigma^{2}_{n}}(B){\lambda^{2}_{1}}(P)-\lambda_{1}(P)\left[\lambda_{1}(Q){\sigma^{2}_{n}}(B)+{\sigma^{2}_{1}}(A)-1\right]-\lambda_{1}(Q)\leq 0. \end{equation} (15) Solving (15) for λ1(P) leads to \begin{equation} \lambda_{1}(P)\!\leq\! \left\{\lambda_{1}(Q){\sigma^{2}_{n}}(B)\!+\!{\sigma^{2}_{1}}(A)\!-\!1\!+\!\left[\left(\lambda_{1}(Q){\sigma^{2}_{n}}(B)\!+\!{\sigma^{2}_{1}}(A)\!-\!1\right)^{2} \!\!+\!4\lambda_{1}(Q){\sigma^{2}_{n}}(B)\right]^{\frac{1}{2}}\right\}\!\left/\!\left(2{\sigma^{2}_{n}}(B)\right)\right.\!=\!\eta_{1}. \end{equation} (16) Substituting (16) into (14) yields \begin{equation} P\leq \frac{\eta_{1}}{1+\lambda_{1}(P){\sigma^{2}_{n}}(B)}A^{T}A+Q. \end{equation} (17) By Theorem 1, it is obvious that \begin{equation} \lambda_{1}(P)\geq \lambda_{1}(P_{l})=\delta_{1}. \end{equation} (18) Substituting (18) into (17) leads to (10). Remark 2 We point out that the condition of Theorem 2 is weaker than that of Theorem A, and the conclusion of Theorem 2 is better than that of Theorem A. (1) Obviously, if σn(B) = 0, the condition of Theorem 2 reduces to Theorem A. Further, we point out that inequality (9) is tighter than inequality (8). Actually, as $${\sigma ^{2}_{1}}(A)<1$$, by Lemmas 4 and 5, \begin{align*} P_{u}&=\sum\limits_{m=0}^{\infty}{(A^{T})^{m}QA^{m}}\\ &=Q+\sum\limits_{m=1}^{\infty}{(A^{T})^{m}QA^{m}}\\ &=Q+A^{T}\left[\sum\limits_{m=0}^{\infty}{(A^{T})^{m}QA^{m}}\right]A\\ &\leq Q+\sum\limits_{m=0}^{\infty}{\lambda_{1}\left[(A^{T})^{m}QA^{m}\right]}A^{T}A\\ &=Q+ \sum\limits_{m=0}^{\infty}\lambda_{1}\left[A^{m}(A^{T})^{m}Q\right]A^{T}A\\ &\leq Q+\sum\limits_{m=0}^{\infty}{{\lambda^{m}_{1}}(AA^{T})\lambda_{1}(Q)}A^{T}A\\ &=Q+\frac{\lambda_{1}(Q)}{1-{\sigma^{2}_{1}}(A)}A^{T}A =P^{*}_{u}. \end{align*} This implies that inequality (9) is tighter than inequality (8). (2) If σn(B) > 0. (i) The condition $${\sigma ^{2}_{1}}(A)<1+{\sigma ^{2}_{n}}(B)\delta ^{*}_{1}$$ of Theorem A is not required in Theorem 2. This shows that the condition of Theorem 2 is weaker than Theorem A. (ii) Further, if $${\sigma ^{2}_{1}}(A)<1+{\sigma ^{2}_{n}}(B)\delta ^{*}_{1}$$, we prove inequality (10) is tighter than inequality (8). Viewing the proof of Theorem A and Theorem 2, it can be seen that Lee (1997) derives inequality (8) mainly by proving inequality $$ \lambda_{1}(P)\leq \lambda_{1}(Q)\frac{1+{\sigma^{2}_{n}}(B)\delta^{*}_{1}}{1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)}=\eta^{*}_{1}, $$ whilst Theorem 2 derives inequality (12) mainly by proving inequality (16). Hence, to prove inequality (10) is tighter than inequality (8) if $${\sigma ^{2}_{1}}(A)<1+{\sigma ^{2}_{n}}(B)\delta ^{*}_{1}$$, it is only required to prove inequality \begin{equation} \eta_{1}\leq\eta^{*}_{1}. \end{equation} (19) Actually, if $${\sigma ^{2}_{1}}(A)<1+{\sigma ^{2}_{n}}(B)\delta ^{*}_{1}$$, by (15) and (16), we have \begin{equation} \frac{{\eta^{2}_{1}}{\sigma^{2}_{n}}(B)-\eta_{1}\lambda_{1}(Q){\sigma^{2}_{n}}(B)-\eta_{1}\left({\sigma^{2}_{1}}(A)-1\right)+ \lambda_{1}(Q)\delta^{*}_{1}{\sigma^{2}_{n}}(B)} {1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)}=\frac{\lambda_{1}(Q)\left(1+ \delta^{*}_{1}{\sigma^{2}_{n}}(B)\right)}{1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)}. \end{equation} (20) By (3), Remark 1 and (16), it is obvious that $$ 0<\delta^{*}_{1}\leq\delta_{1}\leq\eta_{1},\quad 0<\lambda_{1}(Q)\leq \lambda_{1}(P)\leq \eta_{1}.$$ Hence, \begin{equation} {\eta^{2}_{1}}{\sigma^{2}_{n}}(B)-\eta_{1}\lambda_{1}(Q){\sigma^{2}_{n}}(B)\geq \eta_{1}\delta^{*}_{1} {\sigma^{2}_{n}}(B)-\delta^{*}_{1}\lambda_{1}(Q){\sigma^{2}_{n}}(B). \end{equation} (21) Substituting (21) into (20) yields \begin{align*} \eta^{*}_{1}&\geq\left\{ \eta_{1}\delta^{*}_{1} {\sigma^{2}_{n}}(B)-\delta^{*}_{1}\lambda_{1}(Q){\sigma^{2}_{n}}(B)-\eta_{1}({\sigma^{2}_{1}}(A)-1)+ \lambda_{1}(Q)\delta^{*}_{1}{\sigma^{2}_{n}}(B)\right\}\big/ \left[1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)\right]\\ &=\frac{ \eta_{1}\delta^{*}_{1} {\sigma^{2}_{n}}(B)-\eta_{1}\left({\sigma^{2}_{1}}(A)-1\right)} {1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)}=\eta_{1}, \end{align*} i.e., (19), which implies that inequality (10) is tighter than inequality (8). 4. On the existence and uniqueness for the solution of the DARE In this section, we discuss the existence and uniqueness for the solution of the DARE (1) based on the derived lower and upper matrix bounds and a fixed point theorem. Theorem 3 Let $$P_{l}\leq P_{u}, p=||A||^{2}\cdot ||P_{l}^{-1}||^{2}\cdot (||P_{u}^{-1}+BB^{T}||)^{-2}$$, where Pl, η1, Pu are defined as in Theorems 1 and 2. (1) If σn(B) = 0 and $${\sigma ^{2}_{1}}(A)<1$$, then the DARE (1) has a unique positive definite solution P0, and Pl ≤ P0 ≤ Pu. (2) If σn(B) > 0, λ1(Pu) ≤ η1, and \begin{equation} 0<p<1, \end{equation} (22) then the DARE (1) has a unique positive definite solution P0, and Pl ≤ P0 ≤ Pu. Proof. (1) When $${\sigma ^{2}_{1}}(A)<1$$, then the eigenvalues of the matrix A satisfy |λ| < 1, which means that A is stable in the discrete-time sense, hence, (A, B) is automatically stabilizable. By the results of Arnold & Laub (1984), there exists a unique positive definite solution of the DARE (1). From Theorems 1 and 2, the positive definite solution is in [Pl, Pu]. (2) (i) If Pl ≤ Pu, suppose the DARE (1) exists a positive definite solution, from Theorems 1 and 2, the positive definite solution is in [Pl, Pu]. (ii) Define the map F(P) = AT(P−1+BBT)−1A + Q and set $$ P\in \Omega=\{P\mid P_{l}\leq P\leq P_{u}\}.$$ It is obvious that $$\Omega$$ is a convex, closed and bounded set and F(P) is continuous on $$\Omega$$. And we consider a norm space ($$\Omega$$, ||.||), with ||.|| is the spectral norm. In a similar way to the proof of Theorems 1 and 2, for P ∈ $$\Omega$$, we have Pl ≤ F(P) ≤ Pu. Thus F($$\Omega$$) ⊆ $$\Omega$$. For arbitrary P1, P2 ∈ $$\Omega$$, \begin{align*} F(P_{1})&=A^{T}\left(P_{1}^{-1}+BB^{T}\right)^{-1}A+Q,\\ F(P_{2})&=A^{T}\left(P_{2}^{-1}+BB^{T}\right)^{-1}A+Q. \end{align*} Consequently, \begin{align*} F(P_{1})-F(P_{2})&=A^{T}\left[\left(P_{1}^{-1}+BB^{T}\right)^{-1}-\left(P_{2}^{-1}+BB^{T}\right)^{-1}\right]A\\ &=A^{T}\left\{\left(P_{1}^{-1}+BB^{T}\right)^{-1}\left[\left(P_{2}^{-1}+BB^{T}\right)-\left(P_{1}^{-1}+BB^{T}\right)\right]\left(P_{2}^{-1}+BB^{T}\right)^{-1}\right\}A\\ &=A^{T}\left(P_{1}^{-1}+BB^{T}\right)^{-1}\left(P_{2}^{-1}-P_{1}^{-1}\right)\left(P_{2}^{-1}+BB^{T}\right)^{-1}A\\ &=A^{T}\left(P_{1}^{-1}+BB^{T}\right)^{-1}P_{1}^{-1}(P_{1}-P_{2})P_{2}^{-1}\left(P_{2}^{-1}+BB^{T}\right)^{-1}A. \end{align*} Because Pl ≤ P ≤ Pu, according to Lemma 8, we have \begin{align*} ||F(P_{1})-F(P_{2})||&=\left|\left| A^{T}\left(P_{1}^{-1}+BB^{T}\right)^{-1}P_{1}^{-1}(P_{1}-P_{2})P_{2}^{-1}\left(P_{2}^{-1}+BB^{T}\right)^{-1}A\right|\right|\\ &\leq ||A^{T}||\!\cdot\!||A||\!\cdot\!\left|\left|P_{1}^{-1}\right|\right|\!\cdot\!\left|\left|P_{2}^{-1}\right|\right|\!\cdot\!\left|\left|\left(P_{1}^{-1}+BB^{T}\right)^{-1}\right|\right|\!\cdot\!\left|\left|(P_{2}^{-1}+BB^{T})^{-1}\right|\right|\!\cdot\!||P_{1}-P_{2}||\\ &\leq ||A^{T}||\!\cdot\!||A||\!\cdot\!\left|\left|P_{l}^{-1}\right|\right|\!\cdot\!\left|\left|P_{l}^{-1}\right|\right|\!\cdot\!\left(\left|\left|P_{u}^{-1}+BB^{T}\right|\right|\right)^{-1}\!\cdot\!\left(\left|\left|P_{u}^{-1}+BB^{T}\right|\right|\right)^{-1}\!\cdot\!||P_{1}-P_{2}||\\ &= ||A||^{2}\!\cdot\!\left|\left|P_{l}^{-1}\right|\right|^{2}\!\cdot\!\left(\left|\left|P_{u}^{-1}+BB^{T}\right|\right|\right)^{-2}\!\cdot\!||P_{1}-P_{2}||\\ &=p ||P_{1}-P_{2}||. \end{align*} As p < 1, thus the map F(P) is a contraction map in $$\Omega$$. In light of Lemma 7, the map F(P) has a unique fixed point in $$\Omega$$. Combining (i) and (ii) shows that (1) has a unique positive definite solution P0, and Pl ≤ P0 ≤ Pu. This completes the proof. Based on Theorem 3, we present a fixed-point iteration for the solution of the DARE (1). For arbitrarily given P(0) ∈ $$\Omega$$ = {P∣Pl ≤ P ≤ Pu}, we construct the matrix sequence \begin{equation} P^{(k+1)}=A^{T}\left[(P^{(k)})^{-1}+BB^{T}\right]^{-1}A+Q,\quad k=0,1,\cdots. \end{equation} (23) Next, we give Theorem 4 to show that P(k) is convergent and converges to the exact solution of the DARE (1). As it is standard for contraction mapping, the proof of Theorem 4 is omitted. Theorem 4 If the conditions of Theorem 3 are satisfied, then the sequence P(k) given by (23) is convergent and converges to the unique positive definite solution P0 of the DARE (1). By Theorem 4, define P(k) as the kth iteration solution of the DARE (1). For the exact solution P0 and the k iteration solution P(k), in a similar way to the proof of (24), for any small ε > 0, we have the error estimation $$ ||P_{0}-P^{(k)}||\leq p^{k-1}||P_{0}-P^{(1)}||\leq p^{k-1}||P_{u}-P_{l}||<\varepsilon.$$ In the case of the precision ε is permitted, we have the following algorithm. The complexity to the algorithm is about o(n3). Algorithm 1. Input: for the DARE (1), given A, B, Q. Output: P(k). Step 1: compute Pl and Pu. Step 2: compute p, if 0 < p < 1, go to Step 3; otherwise, stop. Step 3: set P(0) ∈ $$\Omega$$; = {P∣Pl ≤ P ≤ Pu}. Step 4: compute $$ P^{(k+1)}=A^{T}\left[\left(P^{(k)}\right)^{-1}+BB^{T}\right]^{-1}A+Q,\quad k=0,1,\cdots.$$ Step 5: for any small ε > 0 and k ∈ N+, if pk−1||Pu − Pl|| < ε, stop; otherwise, let k = k + 1, go to Step 4. 5. Numerical examples In this section, consider the following examples and the whole process is carried out on Matlab 7.1. The precision is 10−8. Example 1 σn(B) > 0 Consider the linear discrete system (2) with $$x(0)=\left ( 0 \quad 1 \quad 0 \quad 0 \quad 0 \quad 0 \right )^{T}\!,$$ $$ A=\left( \begin{array}{@{}cccccc@{}} 0.61 &0.11 & 0.22 & -0.12& 0.13& 0\\ -0.42& 0 & 0.31 & 0.14 & -0.23 & -0.15\\ 0.51 & -0.21 & 0.16 & 0 & 0.21 & -0.32\\ 0 & -0.17 & 0.41 & 0.24 & 0.18 & 0.33\\ -0.13 & 0.25 & 0 & -0.34 & 0.19 & 0.26\\ 0.35 & -0.14 & 0.27 & -0.36 & 0 & 0.43 \end{array} \right)\!,\quad B=\left( \begin{array}{@{}cccccc@{}} 1.1 & 0 & 2.1 & 0 & 3.1 & -1.2\\ 0.2 & 3.1 & -1.1 & 0 & 0.3 & 2.2\\ 1.2 & 4.1 & 5.2 & 3.2 & -2.5 & 0\\ 0 & 2.3 & 1.4 & 4.2 & 5.1 & 3.3\\ 2.4 & -1.5 & 0 & 3.4 & 0.4 & 4.3\\ 0.5 & -2.4 & 1.6 & 0 & 3.5 & 2.6 \end{array} \right)\!.$$ Choose the Hamiltonian function as $$ H(t)=\frac{1}{2}x^{T}(t)Qx(t)+\frac{1}{2}u^{T}(t)u(t)+\mu^{T}(t+1)\left[Ax(t)+Bu(t)\right],\quad t=0,1,\cdots,N-1,$$ where $$ Q=\left( \begin{array}{@{}cccccc@{}} 4.1 & -1.2 & 1.1 & 0 & 0.4 & 1.3\\ -1.2 & 6.2 & 2.1 & 0.5 & 1.5 & 0\\ 1.1 & 2.1 & 8.1 & -1.4 & 0 & 0.6\\ 0 & 0.5 & -1.4 & 7.2 & 2.2 & 1.6\\ 0.4 & 1.5 & 0 & 2.2 & 5.4 & -0.7\\ 1.3 & 0 & 0.6 & 1.6 & -0.7 & 4.5 \end{array} \right)\!.$$ By computation, it is evident that $$ {\sigma_{1}^{2}}(A)= 1.1731>1.$$ Hence, many previous results are invalid. However, $$ {\sigma^{2}_{1}}(A)=1.1731<1+{\sigma^{2}_{n}}(B)\delta^{*}_{1}=25.6061.$$ The lower and upper matrix bounds for the solution P of the DARE (1) found by (4) and (8) are \begin{align*} &P^{*}_{l}=\left( \begin{array}{@{}cccccc@{}} 4.1364 & -1.1974 & 1.1051 & -0.0153 & 0.4061 & 1.2978\\ -1.1974 & 6.2350 & 2.0962 & 0.4915 & 1.5068 & -0.0142\\ 1.1051 & 2.0962 & 8.1173 & -1.4056 & -0.0098 & 0.5970\\ -0.0153 & 0.4915 & -1.4056 & 7.2224 & 2.2026 & 1.6018\\ 0.4061 & 1.5068 & -0.0098 & 2.2026 & 5.4124 & -0.6982\\ 1.2978 & -0.0142 & 0.5970 & 1.6018 & -0.6982 & 4.5161 \end{array} \right)\!,\\ &P^{*}_{u}=\left( \begin{array}{@{}cccccc@{}} 4.5051 & -1.2519 & 1.1770 & -0.0914 & 0.5104 & 1.3071\\ -1.2519 & 6.2714 & 2.0500 & 0.4621 & 1.4945 & 0.0068\\ 1.1770 & 2.0500 & 8.2757 & -1.3922 & 0.0276 & 0.6657\\ -0.0914 & 0.4621 & -1.3922 & 7.3439 & 2.1704 & 1.5209\\ 0.5104 & 1.4945 & 0.0276 & 2.1704 & 5.4779 & -0.6675\\ 1.3071 & 0.0068 & 0.6657 & 1.5209 & -0.6675 & 4.7078 \end{array} \right)\!. \end{align*} The lower and upper matrix bounds for the solution P of the DARE (1) found by (5) and (10) are \begin{align*} &P_{l}=\left( \begin{array}{@{}cccccc@{}} 4.1364& -1.1974 & 1.1051 & -0.0153 & 0.4061 & 1.2978\\ -1.1974 & 6.2350 & 2.0963 & 0.4915 & 1.5068 & -0.0142\\ 1.1051 & 2.0963 & 8.1173 & -1.4057 & -0.0098 & 0.5970 \\ -0.0153 & 0.4915 & -1.4057 & 7.2224 & 2.2027 & 1.6018\\ 0.4061 & 1.5068 & -0.0098 & 2.2026 & 5.4124 & -0.6982 \\ 1.2978 & -0.0142 & 0.5970 & 1.6018 & -0.6982 & 4.5161 \end{array} \right)\!,\\ &P_{u}=\left( \begin{array}{@{}cccccc@{}} 4.5032 & -1.2517 & 1.1766 & -0.0909 & 0.5099 & 1.3070\\ -1.2517 & 6.2711 & 2.0503 & 0.4623 & 1.4945 & 0.0068 \\ 1.1767 & 2.0503 & 8.2749 & -1.3923 & 0.0275 & 0.6654\\ -0.0909 & 0.4623 & -1.3923 & 7.3433 & 2.1706 & 1.5213\\ 0.5099 & 1.4945 & 0.0275 & 2.1706 & 5.4776 & -0.6676 \\ 1.3070 & 0.0068 & 0.6654 & 1.5213 & -0.6676 & 4.7069 \end{array} \right)\!. \end{align*} By Theorem 2.1 in Davies et al. (2008), we get $$ P\leq \left( \begin{array}{@{}cccccc@{}} 95.4639 & -18.9688 & 0.6269 & -11.5408 & 18.2038 & -5.5504\\ -18.9688 & 20.7984 & -3.3799 & -4.3408 & 1.4374 & -3.5664\\ 0.6269 & -3.3799 & 44.4184 & 4.8503 & -3.7130 & 22.7596\\ -11.5408 & -4.3408 & 4.8503 & 44.8398 & -2.2524 &-18.6377\\ 18.2038 & 1.4374 & -3.7130 & -2.2524 & 24.3584 & 2.6238\\ -5.5504 & -3.5664 & 22.7596 &-18.6377 & 2.6238 & 69.5446 \end{array} \right)=M.$$ By computation, we get that $$P_{l}-P^{*}_{l}$$, $$P^{*}_{u}-P_{u}$$ and M − Pu are semi-definite. Hence, $$ P^{*}_{l}\leq P_{l},\quad P_{u}\leq P^{*}_{u},\quad P_{u}\leq M.$$ Thus the bounds Pl and Pu are better than $$P^{*}_{l}$$ and $$P^{*}_{u}, M$$, respectively. By computation, it is evident that Pl ≤ Pu, λ1(Pu) ≤ η1, and $$ 0<p=3.6914\times 10^{-5}<1.$$ By Theorem 3, the DARE (1) has a unique positive definite solution P0, and Pl ≤ P0 ≤ Pu. Let e(k) = pk−1||Pu − Pl|| be the iteration error at the kth iteration, where k denote the iteration step. Here we take ε = 10−8 and $$P^{(0)}=\frac{P_{l}+P_{u}}{2}$$, i.e., $$ P^{(0)}=\left( \begin{array}{@{}cccccc@{}} 4.3198 & -1.2245 & 1.1409 & -0.0531 & 0.4580 & 1.3024\\ -1.2245 & 6.2531 & 2.0733 & 0.4769 & 1.5006 & -0.0037\\ 1.1409 & 2.0733 & 8.1961 & -1.3990 & 0.0089 & 0.6312\\ -0.0531 & 0.4769 & -1.3990 & 7.2829 & 2.1866 & 1.5616\\ 0.4580 & 1.5006 & 0.0089 & 2.1866 & 5.4450 & -0.6829\\ 1.3024 & -0.0037 & 0.6312 & 1.5616 & -0.6829 & 4.6115 \end{array} \right)\!.$$ From Table 1, obviously, $$ p^{2}||P_{u}-P_{l}||=6.2483\times 10^{-10}<10^{-8},$$ Table 1. Iteration errors for Example 3(ε = 10−8) k − 1 e(k) k − 1 e(k) 1 1.6927 × 10−5 2 6.2483 × 10−10 k − 1 e(k) k − 1 e(k) 1 1.6927 × 10−5 2 6.2483 × 10−10 Table 1. Iteration errors for Example 3(ε = 10−8) k − 1 e(k) k − 1 e(k) 1 1.6927 × 10−5 2 6.2483 × 10−10 k − 1 e(k) k − 1 e(k) 1 1.6927 × 10−5 2 6.2483 × 10−10 then Algorithm 1 needs three iteration steps to converge to the iteration solution P(3) of the DARE (1) as follows: $$ P^{(3)}=\left( \begin{array}{@{}cccccc@{}} 4.1364 & -1.1974 & 1.1051 & -0.0153 & 0.4061 & 1.2978\\ -1.1974 & 6.2350 & 2.0963 & 0.4915 & 1.5068 & -0.0142\\ 1.1051 & 2.0963 & 8.1173 & -1.4057 & -0.0098 & 0.5970\\ -0.0153& 0.4915 & -1.4057 & 7.2224 & 2.2026 & 1.6018\\ 0.4061 & 1.5068 & -0.0098 & 2.2026 & 5.4124 & -0.6982\\ 1.2978 & -0.0142 & 0.5970 & 1.6018 & -0.6982 & 4.5161 \end{array} \right)\!.$$ Applying the minimum principle, considering the initial condition, we can have the optimal control $$ u^{*}(t)\!=\!-B^{T}A^{-T}(P\!-\!Q)x^{*}(t)\!=\!\left( \begin{array}{@{}cccccc@{}} -0.0669 & -0.1524 & -0.0205 & 0.0961 & -0.0199 & 0.0701\\ 0.0425 & 0.0085 & -0.0649 & -0.0643 & 0.0350 & 0.0733\\ -0.1511 & 0.0758 & -0.0414 & 0.0546 & -0.0122 & -0.0126\\ 0.0239 & -0.0184 & 0.0634 & -0.0540 & -0.0877 & -0.0294\\ -0.0418 & -0.0060 & -0.0560 & -0.0163 & -0.0066 & -0.0396\\ 0.0676 & 0.0455 & -0.0566 & 0.0457 & 0.0493 & -0.0467 \end{array} \right)x^{*}(t)$$ such that the quadratic performance index of (2) $$ \widetilde{\!\!J}=\frac{1}{2}x^{T}(N)Px(N)+\frac{1}{2}\sum_{t=0}^{N-1}\left[ x^{T}(t)Qx(t)+u^{T}(t)u(t) \right]$$ attains a minimum. Further, the simulation of the optimal states is shown in Fig. 1. It can be seen that when the iteration step k = 3, the linear discrete system (2) is stable by designing optimal control. Example 2 (Lee, 1997), (Kim et al., 1993). σn(B) = 0 Consider the linear discrete system (2) with $$ [A=\left( \begin{array}{@{}ccc@{}} 0.4&0.2& 0.2\\ -0.6&0& 0.1\\ 0& 0 &0.1 \end{array} \right)\!,\quad\ B=\left( \begin{array}{@{}c@{}} 1\\ 0\\ 1 \end{array} \right)\!,\quad x(0)=\left( \begin{array}{@{}c@{}} 1 \\ 0\\ 0 \end{array} \right)\!. $$ Choose the Hamiltonian function as $$ H(t)=\frac{1}{2}x^{T}(t)Qx(t)+\frac{1}{2}u^{T}(t)u(t)+\mu^{T}(t+1)\left[Ax(t)+Bu(t)\right],\quad t=0,1,\cdots,N-1,$$ where $$ Q=\left( \begin{array}{@{}ccc@{}} 3 &1&1\\ 1 & 2 &0\\ 1& 0& 2 \end{array} \right)\!.$$ As $${\sigma ^{2}_{1}}(A)<1,$$ then the lower and upper matrix bounds for the solution P of the DARE (1) found by (4) and (8) are \begin{align*} &P^{*}_{l}=\left( \begin{array}{@{}ccc@{}} 3.595 & 1.02 & 0.93\\ 1.02 & 2.04 & 0.04\\ 0.93 & 0.04 & 2.06 \end{array} \right)\!,\\ &P^{*}_{u}=\left( \begin{array}{@{}ccc@{}} 7.4677 & 1.6873 & 1.1718\\ 1.6873 & 2.3437 & 0.3437\\ 1.1718 & 0.3437 & 2.5155 \end{array} \right)\!. \end{align*} The lower and upper matrix bounds for the solution P of the DARE (1) found by (5) and (9) with m = 4 are \begin{align*} &P_{l}=\left( \begin{array}{@{}ccc@{}} 3.658 & 1.0404 & 0.9378\\ 1.0404 & 2.0476 & 0.0436\\ 0.9378 & 0.0436 & 2.0621 \end{array} \right)\!,\\ &P_{u}=\left( \begin{array}{@{}ccc@{}} 3.8276 & 1.1662 & 1.1164\\ 1.1662 & 2.1531 & 0.1987\\ 1.1164 & 0.1987 & 2.2928 \end{array} \right)\!. \end{align*} By Theorem 2 in Lee (1997), we get $$ P\leq \left( \begin{array}{@{}ccc@{}} 7.4670 & 1.6872 & 1.1718\\ 1.6872 & 2.3436 & 0.3436\\ 1.1718 & 0.3436 & 2.5154 \end{array} \right)=G.$$ By Theorem 2.1 in Davies et al. (2008), we get $$ P\leq \left( \begin{array}{@{}ccc@{}} 7.4670 & 1.6872 & 1.1718\\ 1.6872 & 2.3436 & 0.3436\\ 1.1718 & 0.3436 & 2.5154 \end{array} \right)=M.$$ By computation, we get that $$P_{l}-P^{*}_{l}$$, $$P^{*}_{u}-P_{u},G-P_{u}$$ and M − Pu are semi-definite. Hence, $$ P^{*}_{l}\leq P_{l},\quad P_{u}\leq P^{*}_{u},\quad P_{u}\leq G,\quad P_{u}\leq M.$$ Thus the bounds Pl and Pu are better than $$P^{*}_{l}$$ and $$P^{*}_{u}, G, M$$, respectively. By computation, obviously, Pl ≤ Pu and $$ 0<p=0.0618<1.$$ According to Theorem 3, the DARE (1) has a unique positive definite solution P0, and Pl ≤ P0 ≤ Pu. Let e(k) = pk−1||Pu − Pl|| be the iteration error at the kth iteration, where k denote the iteration step. Here we choose ε = 10−8 and $$P^{(0)}=\frac{P_{l}+P_{u}}{2}$$, i.e., $$ P^{(0)}=\left( \begin{array}{@{}ccc@{}} 3.7428 & 1.1033 & 1.0271\\ 1.1033 & 2.1003 & 0.1211\\ 1.0271 & 0.1211 & 2.1774 \end{array} \right)\!.$$ The relation between iteration step and iteration error is shown in Table 2 and Fig. 2. From Table 2, we find that $$ p^{7}||P_{u}-P_{l}||=1.6767\times 10^{-9}<10^{-8},$$ then Algorithm 1 needs eight iteration steps to converge to the iteration solution P(8) of the DARE (1) as follows: $$ P^{(8)}=\left( \begin{array}{@{}ccc@{}} 3.6590 & 1.0408 & 0.9380\\ 1.0408 & 2.0480 & 0.0439\\ 0.9380 & 0.0439 & 2.0624 \end{array} \right)\!.$$ Applying the minimum principle, considering the initial condition, we can have the optimal control $$ u^{*}(t)=-B^{T}A^{-T}(P-Q)x^{*}(t)=\left( \begin{array}{@{}ccc@{}} -0.1382 & -0.1069 & -0.1545 \end{array} \right)x^{*}(t)$$ such that the quadratic performance index of (2) $$ \widetilde{\!\!J}=\frac{1}{2}x^{T}(N)Px(N)+\frac{1}{2}\sum_{t=0}^{N-1}\left[ x^{T}(t)Qx(t)+u^{T}(t)u(t) \right]$$ attains a minimum. Further, the simulation of the optimal states is shown in Fig. 3. It can be seen that when the iteration step k = 8, the linear discrete system (2) is stable by designing optimal control. Fig. 1. View largeDownload slide The simulation of the optimal states for Example 1. Fig. 1. View largeDownload slide The simulation of the optimal states for Example 1. Table 2. Iteration errors for Example 2 (ε = 10−8) k − 1 e(k) k − 1 e(k) k − 1 e(k) 1 0.0300 4 7.0981× 10−6 7 1.6767× 10−9 2 0.0019 5 4.3878× 10−7 3 1.1483× 10−4 6 2.7124× 10−8 k − 1 e(k) k − 1 e(k) k − 1 e(k) 1 0.0300 4 7.0981× 10−6 7 1.6767× 10−9 2 0.0019 5 4.3878× 10−7 3 1.1483× 10−4 6 2.7124× 10−8 View Large Table 2. Iteration errors for Example 2 (ε = 10−8) k − 1 e(k) k − 1 e(k) k − 1 e(k) 1 0.0300 4 7.0981× 10−6 7 1.6767× 10−9 2 0.0019 5 4.3878× 10−7 3 1.1483× 10−4 6 2.7124× 10−8 k − 1 e(k) k − 1 e(k) k − 1 e(k) 1 0.0300 4 7.0981× 10−6 7 1.6767× 10−9 2 0.0019 5 4.3878× 10−7 3 1.1483× 10−4 6 2.7124× 10−8 View Large Fig. 2. View largeDownload slide The relation between iteration step and iteration error for Example 2. Fig. 2. View largeDownload slide The relation between iteration step and iteration error for Example 2. Fig. 3. View largeDownload slide The simulation of the optimal states for Example 2. Fig. 3. View largeDownload slide The simulation of the optimal states for Example 2. 6. Conclusions In this paper, we have proposed new lower and upper matrix bounds for the solution of the DARE, which improve some of the previous results. Then, we have discussed the existence and uniqueness for the solution of this equation. The corresponding numerical examples have demonstrated the effectiveness of the derived results. Therefore, our future work is required to show more upper and lower matrix bounds, which are less restrictive and more concise, and also possibly tighter. Further, to get less restrictive and more faster iteration algorithms of this equation is our future work as well. Funding The work was supported in part by National Natural Science Foundation of China (11771370, 11571292), the Key Project of National Natural Science Foundation of China (91430213), the Youth Project of Hunan Provincial Natural Science Foundation of China (2017JJ3305) and the Youth Project of Hunan Provincial Education Department of China (16B257). References Arnold , W. F. & Laub , A. J. ( 1984 ) Generalized eigenproblem algorithms and software for algebraic Riccati equations . Proc. of the IEEE , 72 , 1746 -- 1754 . Google Scholar CrossRef Search ADS Berinde , V. ( 2009 ) A common fixed point theorem for compatible quasi contractive self mappings in metric spaces . Appl. Math. Comput. , 213 , 348 -- 354 . Callier , F. M. & Desoer , C. A. ( 1991 ) Linear System Theory . New York Inc : Springer . Google Scholar CrossRef Search ADS Davies , R. , Shi , P. & Wiltshire , R. ( 2008 ) Upper solution bounds of the continuous and discrete coupled algebraic Riccati equations . 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( 1992 ) Lower summation bounds for the discrete Riccati and Lyapunov equations . IEEE Trans. Automat. Control , 37 , 1078 -- 1080 . Google Scholar CrossRef Search ADS Komaroff , N. ( 1994 ) Iterative matrix bounds and computational solutions to the discrete algebraic Riccati equation . IEEE Trans. Automat. Control , 39 , 1676 -- 1678 . Google Scholar CrossRef Search ADS Lancaster , P. & Rodman , L. ( 1995 ) Algebraic Riccati Equations . New York : The Clarendon Press . Lee , C. H. ( 1997 ) Upper matrix bound of the solution for the discrete Riccati equation . IEEE Trans. Automat. Control , 42 , 840 -- 842 . Google Scholar CrossRef Search ADS Lee , C. H. ( 1997 ) Upper and lower bounds of the solutions of the discrete algebraic Riccati and Lyapunov matrix equations . Internat. J. Control , 68 , 579 -- 598 . Google Scholar CrossRef Search ADS Lee , C. H. & Hsien , T. L. ( 1999 ) New solution bounds for the discrete algebraic matrix Riccati equation . Optimal Control Appl. 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The matrix bounds and fixed-point iteration for the solution of the discrete algebraic Riccati equation

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Abstract

Abstract In this paper, by constructing the equivalent form of the discrete algebraic Riccati equation (DARE), applying the eigenvalue inequalities of matrix’s sum and product, we provide new lower and upper matrix bounds for the solution of the DARE, which improve some of the recent results. Further, using the derived matrix bounds and a fixed point theorem, we discuss the existence and uniqueness for the solution of the DARE. Finally, we give corresponding numerical examples to demonstrate the effectiveness of our results. 1. Introduction and preliminaries In many control analysis and design problems, particularly in the field of optimal control, many problems reduce to solve the discrete algebraic Riccati equation (DARE) (Lancaster & Rodman, 1995)–(Ionexcu et al., 1999) \begin{equation} P = A^{T}PA -A^{T}PB(I + B^{T}PB)^{-1}B^{T}PA + Q, \end{equation} (1) where A ∈ Rn×n, B ∈ Rn×m, Q ∈ Rn×n is a symmetric positive definite matrix, and the matrix P ∈ Rn×n is the symmetric positive definite solution of the DARE (1). To guarantee the existence of the stabilizing solution of the DARE, it shall be assumed that (A, B) is a stabilizable pair and (A, Q) is a detectable pair. If the pair (A, B) is not stabilizable, this equation cannot have a positive definite solution. Further, when $${\sigma ^{2}_{1}}(A)<1$$, then the eigenvalues of the matrix A satisfy |λ| < 1, which means that A is stable in the discrete-time sense. Thus, (A, B) is automatically stabilzable, but the converse implication is not true (Callier & Desoer, 1991). In many practical control problems, to study the DARE (1), we usually assume that Q is positive definite. In addition, there are many scholars to research this equation on the assumption that Q is positive definite (Komaroff, 1994)–(Lee, 2004). In this paper, we consider the case where Q is positive definite. The DARE (1) has wide application in linear system. Consider the following linear discrete system described by the state space equation: \begin{equation} \begin{cases} x(t+1)=Ax(t)+Bu(t),\\ x(0)=x_{0}, \end{cases} \end{equation} (2) where t = 0, 1, ⋯, N − 1, x(t) ∈ Rn is a state variable, u(t) ∈ Rm is an input variable, A, B are defined in (1). For the system (2), choosing the Hamiltonian function, applying the minimum principle, we can obtain the optimal control $$ u^{*}(t)=-B^{T}A^{-T}(P-Q)x^{*}(t)$$ such that the quadratic performance index of (2) $$ \tilde{\!\!J}=\tfrac{1}{2}x^{T}(N)Px(N)+\tfrac{1}{2}\sum_{t=0}^{N-1}\left[ x^{T}(t)Qx(t)+u^{T}(t)u(t) \right]$$ attains a minimum, where P is the symmetric positive definite solution of the DARE (1). Applying the matrix identity $$ (X^{-1}+YZ)^{-1}=X-XY(I+ZXY)^{-1}ZX$$ from Kailath (1980) with X, Y, Z are square nonsingular matrices, the DARE (1) can be written as \begin{equation} P=A^{T}(P^{-1}+BB^{T})^{-1}A+Q, \end{equation} (3) which shows that (3) is equivalent to (1). In many control design problems, it is significant to investigate the DARE (1) in consideration of its importance (Lancaster & Rodman, 1995)–(Ionexcu et al., 1999). Recently, solving this equation becomes a heated topic. There are few iterative algorithms for deriving the solution of this equation (Komaroff, 1994). In 1994, by using the monotonicity of matrix bounds of the DARE, Komaroff showed the iteration algorithm if this equation possesses positive definite solution. In 1984, Arnold and Laub presented the generalized eigenproblem formulation and a Newtontype iterative refinement procedure for the solution of general forms of algebraic Riccati equations arising in discrete time application (Arnold & Laub, 1984). From Komaroff (1994), we know that the computation of the positive definite solution of this equation is of some difficulty especially when the dimension of the matrices is high. The closer the initial estimate is to the actual solution, the less computer time is expected to be used in the solution algorithm. Thus, there are many works for obtaining the solution bounds, including matrix bounds (Komaroff, 1994)–(Lee & Hsien, 1999), (Komaroff, 1992)–(Zhang & Liu, 2015), summation bounds (Komaroff & Shahian, 1992), eigenvalue bounds (Lee, 2004), (Liu et al., 2010), trace bounds (Kim et al., 1993), (Liu & Zhang, 2009), eigenproblem algorithms and software (Arnold & Laub, 1984), of the DARE in recent years. It is known to us that, amongst these bounds, the matrix bounds are the most general. Further, Scherer in 1999 showed lower bounds of matrix norm in $$H_{2}/H_{\infty }$$ problem (Scherer, 1999). However, this is a numerical bound, not a matrix bound. Sometimes, contrary to numerical bounds, the matrix bounds can reveal the internal structure of relative systems better. Moreover, some necessary and sufficient conditions are established for the constraint generalized Sylvester matrix equations to have a common solution (Wang et al., 2016). In addition, Wang and He (Wang & He, 2014) study some systems of coupled generalized Sylvester matrix equations and present some necessary and sufficient conditions for the solvability to these systems. The paper is organized as follows. In Section 2, we give some notations and several basic preliminary lemmas. In Sections 3, by constructing the equivalent form of the DARE, applying the eigenvalue inequalities of matrix’s sum and product, we provide new lower and upper matrix bounds for the solution of the DARE, which improve some of the recent results. In Section 4, we discuss the existence and uniqueness for the solution of the DARE. In Section 5, we give corresponding numerical examples to demonstrate the effectiveness of the provided results. In Section 6, we offer some concluding remarks. 2. Basic preliminaries Let us introduce some notation and lemmas first. Let Rn×m and N+ denote the set of n × m real matrices and the set of positive integers. For X = (xij) ∈ Rm×n, suppose XT, ||X|| denote the transpose and the spectral norm of X, respectively. If X ∈ Rn×n, X−1 denotes the inverse of X. Suppose X ∈ Rn×n is an arbitrary symmetric matrix, we assume that the eigenvalues of X are arranged so that λ1(X) ≥ λ2(X) ≥ ⋯ ≥ λn(X). For X, Y ∈ Rn×n, suppose the singular values of X are arranged so that σ1(X) ≥ σ2(X) ≥ ⋯ ≥ σn(X). The inequality X > (≥) 0 means that X is a symmetric positive (semi-) definite matrix and X > (≥) Y means X − Y is a symmetric positive (semi-) definite matrix. The identity matrix with appropriate dimensions is represented by I. Lemma 1 (Kim et al., 1993). Let P be the positive definite solution of the DARE (1), then P has the lower matrix bound \begin{equation} P\geq A^{T}(I+QBB^{T})^{-1}QA+Q\equiv P^{*}_{l}. \end{equation} (4) Lemma 2 (Horn & Johnson, 1986, pp. 300, Theorem 5.6.15). Let Xi ∈ Rn×n, then the series $$\sum \limits\nolimits _{i=1}^{\infty} {X_{i}}$$ converges if there is a matrix norm ∥.∥ on Rn×n such that the numerical series $$\sum \limits\nolimits _{i=1}^{\infty} {\|X_{i}\|}$$ converges. Lemma 3 (Marshall & Olkin, 1979, chapter 9, A. 1. a). Let X, Y ∈ Rn×n, for i = 1, 2, ⋯, n, then $$ \lambda_{i}(XY)=\lambda_{i}(YX).$$ Lemma 4 (Marshall & Olkin, 1979, pp. 248, H.1.g). Let X, Y ∈ Rn×n ≥ 0, then $$ \sum\limits_{i=1}^{n}\lambda_{i}(XY)\leq\sum\limits_{i=1}^{n}\lambda_{i}(X)\lambda_{i}(Y).$$ Lemma 5 (Marshall & Olkin, 1979, pp. 242, G.1.a). If X, Y ∈ Rn×n are symmetric matrices and 1 ≤ i1 ≤ ⋯ ≤ it ≤ n, then $$ \sum\limits_{j=1}^{t}\lambda_{i_{j}}(X+Y)\leq \sum\limits_{j=1}^{t}\lambda_{i_{j}}(X)+\sum\limits_{i=1}^{t}\lambda_{i}(Y),\quad t=1,\cdots,n.$$ Lemma 6 (Berinde, 2009, Theorem B). Let (X, d) be a complete metric space and F : X → X be a strict contraction, i.e., a map satisfying $$ d(Fx, Fy)\leq ad(x,y),\quad\ \mbox{for all} \quad\ x, y \in X,$$ where 0 ≤ a < 1 is constant. Then F has a unique fixed point in X. A norm space becomes a metric space if the distance d(x, y) between x and y is ||x − y||. Hence, we get the following conclusion from Lemma 6. Lemma 7 Let (B, ||.||) be a real Banach space, and $$\Omega\subset$$B be a convex closed and bounded subset and F : $$\Omega\,\rightarrow\,\Omega$$ be a contraction map, i.e., a map satisfying $$ ||F(P_{1})-F(P_{2})||\leq p ||P_{1}-P_{2}||, \quad\mbox{for all}\quad \ P_{1}, P_{2} \in \Omega,$$ where 0 ≤ p < 1. Then F has a unique fixed point in $$\Omega$$. Lemma 8 (Zhao et al., 2009, Lemma 2.1). If X, Y ∈ Rn×n, and 0 < X ≤ Y, then $$ ||X||\leq ||Y||.$$ 3. New lower and upper matrix bounds for the DARE In this section, we first establish a new lower bound for the solution of the DARE (1) which improves (4). Theorem 1 Let P be the positive definite solution of the DARE (1), then P has the lower matrix bound \begin{equation} P \geq A^{T}\left\{\left[A^{T}(Q^{-1}+BB^{T})^{-1}A+Q\right]^{-1}+BB^{T}\right\}^{-1}A+Q\equiv P_{l}. \end{equation} (5) Proof. By (3), obviously, P ≥ Q, i.e., \begin{equation} P^{-1}\leq Q^{-1}. \end{equation} (6) Substituting (3) into the right-hand side of (3) shows that \begin{equation} P=A^{T}(P^{-1}+BB^{T})^{-1}A+Q=A^{T}\left\{\left[A^{T}(P^{-1}+BB^{T})^{-1}A+Q\right]^{-1}+BB^{T}\right\}^{-1}A+Q. \end{equation} (7) Substituting (6) into (7) yields $$ P\geq A^{T}\left\{\left[A^{T}(Q^{-1}+BB^{T})^{-1}A+Q\right]^{-1}+BB^{T}\right\}^{-1}A+Q.$$ This completes the proof. Remark 1 We point out that inequality (5) improves inequality (4). Actually, it is obvious that \begin{align*} &A^{T}\left\{\left[A^{T}(Q^{-1}+BB^{T})^{-1}A+Q\right]^{-1}+BB^{T}\right\}^{-1}A+Q\\ &\quad>A^{T}(Q^{-1}+BB^{T})^{-1}A+Q\\ &\quad=A^{T}\left[Q^{-1}(I+QBB^{T})\right]^{-1}A+Q\\ &\quad=A^{T}(I+QBB^{T})^{-1}QA+Q, \end{align*} which means that inequality (5) is stronger than inequality (4). Denote $$\delta ^{*}_{1}\equiv \lambda _{1}(P^{*}_{l})$$ with $$P^{*}_{l}$$ defined by (4). The following result is obtained in Lee (1997): Theorem A (Lee, 1997, Theorem 2). Let P be the positive definite solution of the DARE (1), and $${\sigma ^{2}_{1}}(A)<1+{\sigma ^{2}_{n}}(B)\delta ^{*}_{1}$$, then P has the upper matrix bound \begin{equation} P\leq \frac{\lambda_{1}(Q)}{1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)}A^{T}A+Q\equiv P^{*}_{u}. \end{equation} (8) Next, we provide a new upper matrix bound which improves Theorem A. Theorem 2 Let P be the positive definite solution of the DARE (1). (1) If σn(B) = 0 and $${\sigma ^{2}_{1}}(A)<1$$, then P has the upper matrix bound \begin{equation} P \leq\sum\limits_{m=0}^{\infty}{(A^{T})^{m}QA^{m}}\equiv P_{u}. \end{equation} (9) (2) If σn(B) > 0, then P has the upper matrix bound \begin{equation} P \leq\frac{\eta_{1}}{1+\delta_{1}{\sigma^{2}_{n}}(B)}A^{T}A+Q\equiv P_{u}, \end{equation} (10) where δ1 ≡ λ1(Pl) with Pl defined by (5), and $$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\eta_{1}\equiv \left\{\lambda_{1}(Q){\sigma^{2}_{n}}(B)+{\sigma^{2}_{1}}(A)-1+\left[\left(\lambda_{1}(Q){\sigma^{2}_{n}}(B)+{\sigma^{2}_{1}}(A)-1\right)^{2}+4\lambda_{1}(Q){\sigma^{2}_{n}}(B)\right]^{\frac{1}{2}}\right\}\left/\left(2{\sigma^{2}_{n}}(B)\right)\!.\right .$$ Proof. 1) If σn(B) = 0, by (3), it is obvious that \begin{equation} P=A^{T}(P^{-1}+BB^{T})^{-1}A+Q\leq A^{T}(P^{-1})^{-1}A+Q= A^{T}PA+Q. \end{equation} (11) Substituting (11) into the right hand of (11) yields \begin{align} \nonumber P&\leq A^{T}PA+Q\leq A^{T}(A^{T}PA+Q)A+Q\\ \nonumber &= (A^{T})^{2}PA^{2}+A^{T}QA+Q\\ \nonumber &\leq (A^{T})^{2}(A^{T}PA+Q)A^{2}+A^{T}QA+Q\\ &\leq \cdots\leq\sum\limits_{m=0}^{n}{(A^{T})^{m}QA^{m}} +(A^{T})^{n+1}PA^{n+1}. \end{align} (12) As $${\sigma ^{2}_{1}}(A)<1$$, choosing spectral norm in Lemma 2, from Lemmas 3 and 4, \begin{align*} \sum\limits_{m=0}^{\infty}\sigma_{1}\left[(A^{T})^{m}QA^{m}\right]&= \sum\limits_{m=0}^{\infty}\lambda_{1}\left[(A^{T})^{m}QA^{m}\right]\\ &= \sum\limits_{m=0}^{\infty}\lambda_{1}\left[A^{m}(A^{T})^{m} \right] \leq \sum\limits_{m=0}^{\infty}{\lambda_{1}\left[A^{m}(A^{T})^{m}\right]\lambda_{1}(Q)}\\ &\leq\sum\limits_{m=0}^{\infty}{{\lambda^{m}_{1}}(AA^{T})\lambda_{1}(Q)}\\ &=\lambda_{1}(Q)\sum\limits_{m=0}^{\infty}\left[{\sigma^{2}_{1}}(A)\right]^{m}\\ &=\frac{\lambda_{1}(Q)}{1-{\sigma^{2}_{1}}(A)}. \end{align*} Obviously, $$\sum \limits\nolimits _{m=0}^{\infty }\,\sigma _{1}\left [(A^{T})^{m}QA^{m}\right ]$$ is convergent. By Lemma 2, $$\sum \limits\nolimits _{m=0}^{\infty} {\,(A^{T})^{m}QA^{m}}$$ is convergent. As $$\sum \limits\nolimits _{m=0}^\infty \,\sigma _{1}[(A^T)^mQA^m]$$ is convergent, then $$ \lim\limits_{n\rightarrow\infty}\sigma_{1}[{(A^{T})^{(n+1)}PA^{(n+1)}}]=0. $$ Since $$ \lambda_{1}\left[{(A^{T})^{(n+1)}PA^{(n+1)}}\right]\leq\sigma_{1}\left[{(A^{T})^{(n+1)}PA^{(n+1)}}\right]\!, $$ thus $$ \lim\limits_{n\rightarrow\infty}\lambda_{1}\left[{(A^{T})^{(n+1)}PA^{(n+1)}}\right]=0. $$ Hence, \begin{equation} \lim\limits_{n\rightarrow\infty}{(A^{T})^{(n+1)}PA^{(n+1)}}=0. \end{equation} (13) Let $$n\rightarrow \infty $$ in (12), in terms of (13), $$ P=\lim\limits_{n\rightarrow\infty}P\leq \lim\limits_{n\rightarrow\infty}\sum\limits_{m=0}^{n}{(A^{T})^{k}QA^{k}} +\lim\limits_{n\rightarrow\infty}(A^{T})^{n+1}PA^{n+1}=\sum\limits_{m=0}^{\infty}{(A^{T})^{m}QA^{m}}.$$ 2) If σn(B) > 0, applying Lemma 5 to (3) satisfies \begin{align} \nonumber P&=A^{T}(P^{-1}+BB^{T})^{-1}A+Q\\ \nonumber&\leq \lambda_{1}(P^{-1}+BB^{T})^{-1}A^{T}A+Q \\ \nonumber&=\frac{1}{\lambda_{n}(P^{-1}+BB^{T})}A^{T}A+Q\\ \nonumber&\leq\frac{1}{\lambda_{n}(P^{-1})+{\sigma^{2}_{n}}(B)}A^{T}A+Q\\ \nonumber&=\frac{1}{\frac{1}{\lambda_{1}(P)}+{\sigma^{2}_{n}}(B)}A^{T}A+Q\\ &=\frac{\lambda_{1}(P)}{1+\lambda_{1}(P){\sigma^{2}_{n}}(B)}A^{T}A+Q. \end{align} (14) Introducing Lemma 5 to (14) yields $$ \lambda_{1}(P)\leq \lambda_{1}\left[ \frac{\lambda_{1}(P)}{1+\lambda_{1}(P){\sigma^{2}_{n}}(B)}A^{T}A+Q\right]\leq \frac{\lambda_{1}(P)}{1+\lambda_{1}(P){\sigma^{2}_{n}}(B)}{\sigma^{2}_{1}}(A)+\lambda_{1}(Q),$$ which is equivalent to \begin{equation} {\sigma^{2}_{n}}(B){\lambda^{2}_{1}}(P)-\lambda_{1}(P)\left[\lambda_{1}(Q){\sigma^{2}_{n}}(B)+{\sigma^{2}_{1}}(A)-1\right]-\lambda_{1}(Q)\leq 0. \end{equation} (15) Solving (15) for λ1(P) leads to \begin{equation} \lambda_{1}(P)\!\leq\! \left\{\lambda_{1}(Q){\sigma^{2}_{n}}(B)\!+\!{\sigma^{2}_{1}}(A)\!-\!1\!+\!\left[\left(\lambda_{1}(Q){\sigma^{2}_{n}}(B)\!+\!{\sigma^{2}_{1}}(A)\!-\!1\right)^{2} \!\!+\!4\lambda_{1}(Q){\sigma^{2}_{n}}(B)\right]^{\frac{1}{2}}\right\}\!\left/\!\left(2{\sigma^{2}_{n}}(B)\right)\right.\!=\!\eta_{1}. \end{equation} (16) Substituting (16) into (14) yields \begin{equation} P\leq \frac{\eta_{1}}{1+\lambda_{1}(P){\sigma^{2}_{n}}(B)}A^{T}A+Q. \end{equation} (17) By Theorem 1, it is obvious that \begin{equation} \lambda_{1}(P)\geq \lambda_{1}(P_{l})=\delta_{1}. \end{equation} (18) Substituting (18) into (17) leads to (10). Remark 2 We point out that the condition of Theorem 2 is weaker than that of Theorem A, and the conclusion of Theorem 2 is better than that of Theorem A. (1) Obviously, if σn(B) = 0, the condition of Theorem 2 reduces to Theorem A. Further, we point out that inequality (9) is tighter than inequality (8). Actually, as $${\sigma ^{2}_{1}}(A)<1$$, by Lemmas 4 and 5, \begin{align*} P_{u}&=\sum\limits_{m=0}^{\infty}{(A^{T})^{m}QA^{m}}\\ &=Q+\sum\limits_{m=1}^{\infty}{(A^{T})^{m}QA^{m}}\\ &=Q+A^{T}\left[\sum\limits_{m=0}^{\infty}{(A^{T})^{m}QA^{m}}\right]A\\ &\leq Q+\sum\limits_{m=0}^{\infty}{\lambda_{1}\left[(A^{T})^{m}QA^{m}\right]}A^{T}A\\ &=Q+ \sum\limits_{m=0}^{\infty}\lambda_{1}\left[A^{m}(A^{T})^{m}Q\right]A^{T}A\\ &\leq Q+\sum\limits_{m=0}^{\infty}{{\lambda^{m}_{1}}(AA^{T})\lambda_{1}(Q)}A^{T}A\\ &=Q+\frac{\lambda_{1}(Q)}{1-{\sigma^{2}_{1}}(A)}A^{T}A =P^{*}_{u}. \end{align*} This implies that inequality (9) is tighter than inequality (8). (2) If σn(B) > 0. (i) The condition $${\sigma ^{2}_{1}}(A)<1+{\sigma ^{2}_{n}}(B)\delta ^{*}_{1}$$ of Theorem A is not required in Theorem 2. This shows that the condition of Theorem 2 is weaker than Theorem A. (ii) Further, if $${\sigma ^{2}_{1}}(A)<1+{\sigma ^{2}_{n}}(B)\delta ^{*}_{1}$$, we prove inequality (10) is tighter than inequality (8). Viewing the proof of Theorem A and Theorem 2, it can be seen that Lee (1997) derives inequality (8) mainly by proving inequality $$ \lambda_{1}(P)\leq \lambda_{1}(Q)\frac{1+{\sigma^{2}_{n}}(B)\delta^{*}_{1}}{1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)}=\eta^{*}_{1}, $$ whilst Theorem 2 derives inequality (12) mainly by proving inequality (16). Hence, to prove inequality (10) is tighter than inequality (8) if $${\sigma ^{2}_{1}}(A)<1+{\sigma ^{2}_{n}}(B)\delta ^{*}_{1}$$, it is only required to prove inequality \begin{equation} \eta_{1}\leq\eta^{*}_{1}. \end{equation} (19) Actually, if $${\sigma ^{2}_{1}}(A)<1+{\sigma ^{2}_{n}}(B)\delta ^{*}_{1}$$, by (15) and (16), we have \begin{equation} \frac{{\eta^{2}_{1}}{\sigma^{2}_{n}}(B)-\eta_{1}\lambda_{1}(Q){\sigma^{2}_{n}}(B)-\eta_{1}\left({\sigma^{2}_{1}}(A)-1\right)+ \lambda_{1}(Q)\delta^{*}_{1}{\sigma^{2}_{n}}(B)} {1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)}=\frac{\lambda_{1}(Q)\left(1+ \delta^{*}_{1}{\sigma^{2}_{n}}(B)\right)}{1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)}. \end{equation} (20) By (3), Remark 1 and (16), it is obvious that $$ 0<\delta^{*}_{1}\leq\delta_{1}\leq\eta_{1},\quad 0<\lambda_{1}(Q)\leq \lambda_{1}(P)\leq \eta_{1}.$$ Hence, \begin{equation} {\eta^{2}_{1}}{\sigma^{2}_{n}}(B)-\eta_{1}\lambda_{1}(Q){\sigma^{2}_{n}}(B)\geq \eta_{1}\delta^{*}_{1} {\sigma^{2}_{n}}(B)-\delta^{*}_{1}\lambda_{1}(Q){\sigma^{2}_{n}}(B). \end{equation} (21) Substituting (21) into (20) yields \begin{align*} \eta^{*}_{1}&\geq\left\{ \eta_{1}\delta^{*}_{1} {\sigma^{2}_{n}}(B)-\delta^{*}_{1}\lambda_{1}(Q){\sigma^{2}_{n}}(B)-\eta_{1}({\sigma^{2}_{1}}(A)-1)+ \lambda_{1}(Q)\delta^{*}_{1}{\sigma^{2}_{n}}(B)\right\}\big/ \left[1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)\right]\\ &=\frac{ \eta_{1}\delta^{*}_{1} {\sigma^{2}_{n}}(B)-\eta_{1}\left({\sigma^{2}_{1}}(A)-1\right)} {1+\delta^{*}_{1}{\sigma^{2}_{n}}(B)-{\sigma^{2}_{1}}(A)}=\eta_{1}, \end{align*} i.e., (19), which implies that inequality (10) is tighter than inequality (8). 4. On the existence and uniqueness for the solution of the DARE In this section, we discuss the existence and uniqueness for the solution of the DARE (1) based on the derived lower and upper matrix bounds and a fixed point theorem. Theorem 3 Let $$P_{l}\leq P_{u}, p=||A||^{2}\cdot ||P_{l}^{-1}||^{2}\cdot (||P_{u}^{-1}+BB^{T}||)^{-2}$$, where Pl, η1, Pu are defined as in Theorems 1 and 2. (1) If σn(B) = 0 and $${\sigma ^{2}_{1}}(A)<1$$, then the DARE (1) has a unique positive definite solution P0, and Pl ≤ P0 ≤ Pu. (2) If σn(B) > 0, λ1(Pu) ≤ η1, and \begin{equation} 0<p<1, \end{equation} (22) then the DARE (1) has a unique positive definite solution P0, and Pl ≤ P0 ≤ Pu. Proof. (1) When $${\sigma ^{2}_{1}}(A)<1$$, then the eigenvalues of the matrix A satisfy |λ| < 1, which means that A is stable in the discrete-time sense, hence, (A, B) is automatically stabilizable. By the results of Arnold & Laub (1984), there exists a unique positive definite solution of the DARE (1). From Theorems 1 and 2, the positive definite solution is in [Pl, Pu]. (2) (i) If Pl ≤ Pu, suppose the DARE (1) exists a positive definite solution, from Theorems 1 and 2, the positive definite solution is in [Pl, Pu]. (ii) Define the map F(P) = AT(P−1+BBT)−1A + Q and set $$ P\in \Omega=\{P\mid P_{l}\leq P\leq P_{u}\}.$$ It is obvious that $$\Omega$$ is a convex, closed and bounded set and F(P) is continuous on $$\Omega$$. And we consider a norm space ($$\Omega$$, ||.||), with ||.|| is the spectral norm. In a similar way to the proof of Theorems 1 and 2, for P ∈ $$\Omega$$, we have Pl ≤ F(P) ≤ Pu. Thus F($$\Omega$$) ⊆ $$\Omega$$. For arbitrary P1, P2 ∈ $$\Omega$$, \begin{align*} F(P_{1})&=A^{T}\left(P_{1}^{-1}+BB^{T}\right)^{-1}A+Q,\\ F(P_{2})&=A^{T}\left(P_{2}^{-1}+BB^{T}\right)^{-1}A+Q. \end{align*} Consequently, \begin{align*} F(P_{1})-F(P_{2})&=A^{T}\left[\left(P_{1}^{-1}+BB^{T}\right)^{-1}-\left(P_{2}^{-1}+BB^{T}\right)^{-1}\right]A\\ &=A^{T}\left\{\left(P_{1}^{-1}+BB^{T}\right)^{-1}\left[\left(P_{2}^{-1}+BB^{T}\right)-\left(P_{1}^{-1}+BB^{T}\right)\right]\left(P_{2}^{-1}+BB^{T}\right)^{-1}\right\}A\\ &=A^{T}\left(P_{1}^{-1}+BB^{T}\right)^{-1}\left(P_{2}^{-1}-P_{1}^{-1}\right)\left(P_{2}^{-1}+BB^{T}\right)^{-1}A\\ &=A^{T}\left(P_{1}^{-1}+BB^{T}\right)^{-1}P_{1}^{-1}(P_{1}-P_{2})P_{2}^{-1}\left(P_{2}^{-1}+BB^{T}\right)^{-1}A. \end{align*} Because Pl ≤ P ≤ Pu, according to Lemma 8, we have \begin{align*} ||F(P_{1})-F(P_{2})||&=\left|\left| A^{T}\left(P_{1}^{-1}+BB^{T}\right)^{-1}P_{1}^{-1}(P_{1}-P_{2})P_{2}^{-1}\left(P_{2}^{-1}+BB^{T}\right)^{-1}A\right|\right|\\ &\leq ||A^{T}||\!\cdot\!||A||\!\cdot\!\left|\left|P_{1}^{-1}\right|\right|\!\cdot\!\left|\left|P_{2}^{-1}\right|\right|\!\cdot\!\left|\left|\left(P_{1}^{-1}+BB^{T}\right)^{-1}\right|\right|\!\cdot\!\left|\left|(P_{2}^{-1}+BB^{T})^{-1}\right|\right|\!\cdot\!||P_{1}-P_{2}||\\ &\leq ||A^{T}||\!\cdot\!||A||\!\cdot\!\left|\left|P_{l}^{-1}\right|\right|\!\cdot\!\left|\left|P_{l}^{-1}\right|\right|\!\cdot\!\left(\left|\left|P_{u}^{-1}+BB^{T}\right|\right|\right)^{-1}\!\cdot\!\left(\left|\left|P_{u}^{-1}+BB^{T}\right|\right|\right)^{-1}\!\cdot\!||P_{1}-P_{2}||\\ &= ||A||^{2}\!\cdot\!\left|\left|P_{l}^{-1}\right|\right|^{2}\!\cdot\!\left(\left|\left|P_{u}^{-1}+BB^{T}\right|\right|\right)^{-2}\!\cdot\!||P_{1}-P_{2}||\\ &=p ||P_{1}-P_{2}||. \end{align*} As p < 1, thus the map F(P) is a contraction map in $$\Omega$$. In light of Lemma 7, the map F(P) has a unique fixed point in $$\Omega$$. Combining (i) and (ii) shows that (1) has a unique positive definite solution P0, and Pl ≤ P0 ≤ Pu. This completes the proof. Based on Theorem 3, we present a fixed-point iteration for the solution of the DARE (1). For arbitrarily given P(0) ∈ $$\Omega$$ = {P∣Pl ≤ P ≤ Pu}, we construct the matrix sequence \begin{equation} P^{(k+1)}=A^{T}\left[(P^{(k)})^{-1}+BB^{T}\right]^{-1}A+Q,\quad k=0,1,\cdots. \end{equation} (23) Next, we give Theorem 4 to show that P(k) is convergent and converges to the exact solution of the DARE (1). As it is standard for contraction mapping, the proof of Theorem 4 is omitted. Theorem 4 If the conditions of Theorem 3 are satisfied, then the sequence P(k) given by (23) is convergent and converges to the unique positive definite solution P0 of the DARE (1). By Theorem 4, define P(k) as the kth iteration solution of the DARE (1). For the exact solution P0 and the k iteration solution P(k), in a similar way to the proof of (24), for any small ε > 0, we have the error estimation $$ ||P_{0}-P^{(k)}||\leq p^{k-1}||P_{0}-P^{(1)}||\leq p^{k-1}||P_{u}-P_{l}||<\varepsilon.$$ In the case of the precision ε is permitted, we have the following algorithm. The complexity to the algorithm is about o(n3). Algorithm 1. Input: for the DARE (1), given A, B, Q. Output: P(k). Step 1: compute Pl and Pu. Step 2: compute p, if 0 < p < 1, go to Step 3; otherwise, stop. Step 3: set P(0) ∈ $$\Omega$$; = {P∣Pl ≤ P ≤ Pu}. Step 4: compute $$ P^{(k+1)}=A^{T}\left[\left(P^{(k)}\right)^{-1}+BB^{T}\right]^{-1}A+Q,\quad k=0,1,\cdots.$$ Step 5: for any small ε > 0 and k ∈ N+, if pk−1||Pu − Pl|| < ε, stop; otherwise, let k = k + 1, go to Step 4. 5. Numerical examples In this section, consider the following examples and the whole process is carried out on Matlab 7.1. The precision is 10−8. Example 1 σn(B) > 0 Consider the linear discrete system (2) with $$x(0)=\left ( 0 \quad 1 \quad 0 \quad 0 \quad 0 \quad 0 \right )^{T}\!,$$ $$ A=\left( \begin{array}{@{}cccccc@{}} 0.61 &0.11 & 0.22 & -0.12& 0.13& 0\\ -0.42& 0 & 0.31 & 0.14 & -0.23 & -0.15\\ 0.51 & -0.21 & 0.16 & 0 & 0.21 & -0.32\\ 0 & -0.17 & 0.41 & 0.24 & 0.18 & 0.33\\ -0.13 & 0.25 & 0 & -0.34 & 0.19 & 0.26\\ 0.35 & -0.14 & 0.27 & -0.36 & 0 & 0.43 \end{array} \right)\!,\quad B=\left( \begin{array}{@{}cccccc@{}} 1.1 & 0 & 2.1 & 0 & 3.1 & -1.2\\ 0.2 & 3.1 & -1.1 & 0 & 0.3 & 2.2\\ 1.2 & 4.1 & 5.2 & 3.2 & -2.5 & 0\\ 0 & 2.3 & 1.4 & 4.2 & 5.1 & 3.3\\ 2.4 & -1.5 & 0 & 3.4 & 0.4 & 4.3\\ 0.5 & -2.4 & 1.6 & 0 & 3.5 & 2.6 \end{array} \right)\!.$$ Choose the Hamiltonian function as $$ H(t)=\frac{1}{2}x^{T}(t)Qx(t)+\frac{1}{2}u^{T}(t)u(t)+\mu^{T}(t+1)\left[Ax(t)+Bu(t)\right],\quad t=0,1,\cdots,N-1,$$ where $$ Q=\left( \begin{array}{@{}cccccc@{}} 4.1 & -1.2 & 1.1 & 0 & 0.4 & 1.3\\ -1.2 & 6.2 & 2.1 & 0.5 & 1.5 & 0\\ 1.1 & 2.1 & 8.1 & -1.4 & 0 & 0.6\\ 0 & 0.5 & -1.4 & 7.2 & 2.2 & 1.6\\ 0.4 & 1.5 & 0 & 2.2 & 5.4 & -0.7\\ 1.3 & 0 & 0.6 & 1.6 & -0.7 & 4.5 \end{array} \right)\!.$$ By computation, it is evident that $$ {\sigma_{1}^{2}}(A)= 1.1731>1.$$ Hence, many previous results are invalid. However, $$ {\sigma^{2}_{1}}(A)=1.1731<1+{\sigma^{2}_{n}}(B)\delta^{*}_{1}=25.6061.$$ The lower and upper matrix bounds for the solution P of the DARE (1) found by (4) and (8) are \begin{align*} &P^{*}_{l}=\left( \begin{array}{@{}cccccc@{}} 4.1364 & -1.1974 & 1.1051 & -0.0153 & 0.4061 & 1.2978\\ -1.1974 & 6.2350 & 2.0962 & 0.4915 & 1.5068 & -0.0142\\ 1.1051 & 2.0962 & 8.1173 & -1.4056 & -0.0098 & 0.5970\\ -0.0153 & 0.4915 & -1.4056 & 7.2224 & 2.2026 & 1.6018\\ 0.4061 & 1.5068 & -0.0098 & 2.2026 & 5.4124 & -0.6982\\ 1.2978 & -0.0142 & 0.5970 & 1.6018 & -0.6982 & 4.5161 \end{array} \right)\!,\\ &P^{*}_{u}=\left( \begin{array}{@{}cccccc@{}} 4.5051 & -1.2519 & 1.1770 & -0.0914 & 0.5104 & 1.3071\\ -1.2519 & 6.2714 & 2.0500 & 0.4621 & 1.4945 & 0.0068\\ 1.1770 & 2.0500 & 8.2757 & -1.3922 & 0.0276 & 0.6657\\ -0.0914 & 0.4621 & -1.3922 & 7.3439 & 2.1704 & 1.5209\\ 0.5104 & 1.4945 & 0.0276 & 2.1704 & 5.4779 & -0.6675\\ 1.3071 & 0.0068 & 0.6657 & 1.5209 & -0.6675 & 4.7078 \end{array} \right)\!. \end{align*} The lower and upper matrix bounds for the solution P of the DARE (1) found by (5) and (10) are \begin{align*} &P_{l}=\left( \begin{array}{@{}cccccc@{}} 4.1364& -1.1974 & 1.1051 & -0.0153 & 0.4061 & 1.2978\\ -1.1974 & 6.2350 & 2.0963 & 0.4915 & 1.5068 & -0.0142\\ 1.1051 & 2.0963 & 8.1173 & -1.4057 & -0.0098 & 0.5970 \\ -0.0153 & 0.4915 & -1.4057 & 7.2224 & 2.2027 & 1.6018\\ 0.4061 & 1.5068 & -0.0098 & 2.2026 & 5.4124 & -0.6982 \\ 1.2978 & -0.0142 & 0.5970 & 1.6018 & -0.6982 & 4.5161 \end{array} \right)\!,\\ &P_{u}=\left( \begin{array}{@{}cccccc@{}} 4.5032 & -1.2517 & 1.1766 & -0.0909 & 0.5099 & 1.3070\\ -1.2517 & 6.2711 & 2.0503 & 0.4623 & 1.4945 & 0.0068 \\ 1.1767 & 2.0503 & 8.2749 & -1.3923 & 0.0275 & 0.6654\\ -0.0909 & 0.4623 & -1.3923 & 7.3433 & 2.1706 & 1.5213\\ 0.5099 & 1.4945 & 0.0275 & 2.1706 & 5.4776 & -0.6676 \\ 1.3070 & 0.0068 & 0.6654 & 1.5213 & -0.6676 & 4.7069 \end{array} \right)\!. \end{align*} By Theorem 2.1 in Davies et al. (2008), we get $$ P\leq \left( \begin{array}{@{}cccccc@{}} 95.4639 & -18.9688 & 0.6269 & -11.5408 & 18.2038 & -5.5504\\ -18.9688 & 20.7984 & -3.3799 & -4.3408 & 1.4374 & -3.5664\\ 0.6269 & -3.3799 & 44.4184 & 4.8503 & -3.7130 & 22.7596\\ -11.5408 & -4.3408 & 4.8503 & 44.8398 & -2.2524 &-18.6377\\ 18.2038 & 1.4374 & -3.7130 & -2.2524 & 24.3584 & 2.6238\\ -5.5504 & -3.5664 & 22.7596 &-18.6377 & 2.6238 & 69.5446 \end{array} \right)=M.$$ By computation, we get that $$P_{l}-P^{*}_{l}$$, $$P^{*}_{u}-P_{u}$$ and M − Pu are semi-definite. Hence, $$ P^{*}_{l}\leq P_{l},\quad P_{u}\leq P^{*}_{u},\quad P_{u}\leq M.$$ Thus the bounds Pl and Pu are better than $$P^{*}_{l}$$ and $$P^{*}_{u}, M$$, respectively. By computation, it is evident that Pl ≤ Pu, λ1(Pu) ≤ η1, and $$ 0<p=3.6914\times 10^{-5}<1.$$ By Theorem 3, the DARE (1) has a unique positive definite solution P0, and Pl ≤ P0 ≤ Pu. Let e(k) = pk−1||Pu − Pl|| be the iteration error at the kth iteration, where k denote the iteration step. Here we take ε = 10−8 and $$P^{(0)}=\frac{P_{l}+P_{u}}{2}$$, i.e., $$ P^{(0)}=\left( \begin{array}{@{}cccccc@{}} 4.3198 & -1.2245 & 1.1409 & -0.0531 & 0.4580 & 1.3024\\ -1.2245 & 6.2531 & 2.0733 & 0.4769 & 1.5006 & -0.0037\\ 1.1409 & 2.0733 & 8.1961 & -1.3990 & 0.0089 & 0.6312\\ -0.0531 & 0.4769 & -1.3990 & 7.2829 & 2.1866 & 1.5616\\ 0.4580 & 1.5006 & 0.0089 & 2.1866 & 5.4450 & -0.6829\\ 1.3024 & -0.0037 & 0.6312 & 1.5616 & -0.6829 & 4.6115 \end{array} \right)\!.$$ From Table 1, obviously, $$ p^{2}||P_{u}-P_{l}||=6.2483\times 10^{-10}<10^{-8},$$ Table 1. Iteration errors for Example 3(ε = 10−8) k − 1 e(k) k − 1 e(k) 1 1.6927 × 10−5 2 6.2483 × 10−10 k − 1 e(k) k − 1 e(k) 1 1.6927 × 10−5 2 6.2483 × 10−10 Table 1. Iteration errors for Example 3(ε = 10−8) k − 1 e(k) k − 1 e(k) 1 1.6927 × 10−5 2 6.2483 × 10−10 k − 1 e(k) k − 1 e(k) 1 1.6927 × 10−5 2 6.2483 × 10−10 then Algorithm 1 needs three iteration steps to converge to the iteration solution P(3) of the DARE (1) as follows: $$ P^{(3)}=\left( \begin{array}{@{}cccccc@{}} 4.1364 & -1.1974 & 1.1051 & -0.0153 & 0.4061 & 1.2978\\ -1.1974 & 6.2350 & 2.0963 & 0.4915 & 1.5068 & -0.0142\\ 1.1051 & 2.0963 & 8.1173 & -1.4057 & -0.0098 & 0.5970\\ -0.0153& 0.4915 & -1.4057 & 7.2224 & 2.2026 & 1.6018\\ 0.4061 & 1.5068 & -0.0098 & 2.2026 & 5.4124 & -0.6982\\ 1.2978 & -0.0142 & 0.5970 & 1.6018 & -0.6982 & 4.5161 \end{array} \right)\!.$$ Applying the minimum principle, considering the initial condition, we can have the optimal control $$ u^{*}(t)\!=\!-B^{T}A^{-T}(P\!-\!Q)x^{*}(t)\!=\!\left( \begin{array}{@{}cccccc@{}} -0.0669 & -0.1524 & -0.0205 & 0.0961 & -0.0199 & 0.0701\\ 0.0425 & 0.0085 & -0.0649 & -0.0643 & 0.0350 & 0.0733\\ -0.1511 & 0.0758 & -0.0414 & 0.0546 & -0.0122 & -0.0126\\ 0.0239 & -0.0184 & 0.0634 & -0.0540 & -0.0877 & -0.0294\\ -0.0418 & -0.0060 & -0.0560 & -0.0163 & -0.0066 & -0.0396\\ 0.0676 & 0.0455 & -0.0566 & 0.0457 & 0.0493 & -0.0467 \end{array} \right)x^{*}(t)$$ such that the quadratic performance index of (2) $$ \widetilde{\!\!J}=\frac{1}{2}x^{T}(N)Px(N)+\frac{1}{2}\sum_{t=0}^{N-1}\left[ x^{T}(t)Qx(t)+u^{T}(t)u(t) \right]$$ attains a minimum. Further, the simulation of the optimal states is shown in Fig. 1. It can be seen that when the iteration step k = 3, the linear discrete system (2) is stable by designing optimal control. Example 2 (Lee, 1997), (Kim et al., 1993). σn(B) = 0 Consider the linear discrete system (2) with $$ [A=\left( \begin{array}{@{}ccc@{}} 0.4&0.2& 0.2\\ -0.6&0& 0.1\\ 0& 0 &0.1 \end{array} \right)\!,\quad\ B=\left( \begin{array}{@{}c@{}} 1\\ 0\\ 1 \end{array} \right)\!,\quad x(0)=\left( \begin{array}{@{}c@{}} 1 \\ 0\\ 0 \end{array} \right)\!. $$ Choose the Hamiltonian function as $$ H(t)=\frac{1}{2}x^{T}(t)Qx(t)+\frac{1}{2}u^{T}(t)u(t)+\mu^{T}(t+1)\left[Ax(t)+Bu(t)\right],\quad t=0,1,\cdots,N-1,$$ where $$ Q=\left( \begin{array}{@{}ccc@{}} 3 &1&1\\ 1 & 2 &0\\ 1& 0& 2 \end{array} \right)\!.$$ As $${\sigma ^{2}_{1}}(A)<1,$$ then the lower and upper matrix bounds for the solution P of the DARE (1) found by (4) and (8) are \begin{align*} &P^{*}_{l}=\left( \begin{array}{@{}ccc@{}} 3.595 & 1.02 & 0.93\\ 1.02 & 2.04 & 0.04\\ 0.93 & 0.04 & 2.06 \end{array} \right)\!,\\ &P^{*}_{u}=\left( \begin{array}{@{}ccc@{}} 7.4677 & 1.6873 & 1.1718\\ 1.6873 & 2.3437 & 0.3437\\ 1.1718 & 0.3437 & 2.5155 \end{array} \right)\!. \end{align*} The lower and upper matrix bounds for the solution P of the DARE (1) found by (5) and (9) with m = 4 are \begin{align*} &P_{l}=\left( \begin{array}{@{}ccc@{}} 3.658 & 1.0404 & 0.9378\\ 1.0404 & 2.0476 & 0.0436\\ 0.9378 & 0.0436 & 2.0621 \end{array} \right)\!,\\ &P_{u}=\left( \begin{array}{@{}ccc@{}} 3.8276 & 1.1662 & 1.1164\\ 1.1662 & 2.1531 & 0.1987\\ 1.1164 & 0.1987 & 2.2928 \end{array} \right)\!. \end{align*} By Theorem 2 in Lee (1997), we get $$ P\leq \left( \begin{array}{@{}ccc@{}} 7.4670 & 1.6872 & 1.1718\\ 1.6872 & 2.3436 & 0.3436\\ 1.1718 & 0.3436 & 2.5154 \end{array} \right)=G.$$ By Theorem 2.1 in Davies et al. (2008), we get $$ P\leq \left( \begin{array}{@{}ccc@{}} 7.4670 & 1.6872 & 1.1718\\ 1.6872 & 2.3436 & 0.3436\\ 1.1718 & 0.3436 & 2.5154 \end{array} \right)=M.$$ By computation, we get that $$P_{l}-P^{*}_{l}$$, $$P^{*}_{u}-P_{u},G-P_{u}$$ and M − Pu are semi-definite. Hence, $$ P^{*}_{l}\leq P_{l},\quad P_{u}\leq P^{*}_{u},\quad P_{u}\leq G,\quad P_{u}\leq M.$$ Thus the bounds Pl and Pu are better than $$P^{*}_{l}$$ and $$P^{*}_{u}, G, M$$, respectively. By computation, obviously, Pl ≤ Pu and $$ 0<p=0.0618<1.$$ According to Theorem 3, the DARE (1) has a unique positive definite solution P0, and Pl ≤ P0 ≤ Pu. Let e(k) = pk−1||Pu − Pl|| be the iteration error at the kth iteration, where k denote the iteration step. Here we choose ε = 10−8 and $$P^{(0)}=\frac{P_{l}+P_{u}}{2}$$, i.e., $$ P^{(0)}=\left( \begin{array}{@{}ccc@{}} 3.7428 & 1.1033 & 1.0271\\ 1.1033 & 2.1003 & 0.1211\\ 1.0271 & 0.1211 & 2.1774 \end{array} \right)\!.$$ The relation between iteration step and iteration error is shown in Table 2 and Fig. 2. From Table 2, we find that $$ p^{7}||P_{u}-P_{l}||=1.6767\times 10^{-9}<10^{-8},$$ then Algorithm 1 needs eight iteration steps to converge to the iteration solution P(8) of the DARE (1) as follows: $$ P^{(8)}=\left( \begin{array}{@{}ccc@{}} 3.6590 & 1.0408 & 0.9380\\ 1.0408 & 2.0480 & 0.0439\\ 0.9380 & 0.0439 & 2.0624 \end{array} \right)\!.$$ Applying the minimum principle, considering the initial condition, we can have the optimal control $$ u^{*}(t)=-B^{T}A^{-T}(P-Q)x^{*}(t)=\left( \begin{array}{@{}ccc@{}} -0.1382 & -0.1069 & -0.1545 \end{array} \right)x^{*}(t)$$ such that the quadratic performance index of (2) $$ \widetilde{\!\!J}=\frac{1}{2}x^{T}(N)Px(N)+\frac{1}{2}\sum_{t=0}^{N-1}\left[ x^{T}(t)Qx(t)+u^{T}(t)u(t) \right]$$ attains a minimum. Further, the simulation of the optimal states is shown in Fig. 3. It can be seen that when the iteration step k = 8, the linear discrete system (2) is stable by designing optimal control. Fig. 1. View largeDownload slide The simulation of the optimal states for Example 1. Fig. 1. View largeDownload slide The simulation of the optimal states for Example 1. Table 2. Iteration errors for Example 2 (ε = 10−8) k − 1 e(k) k − 1 e(k) k − 1 e(k) 1 0.0300 4 7.0981× 10−6 7 1.6767× 10−9 2 0.0019 5 4.3878× 10−7 3 1.1483× 10−4 6 2.7124× 10−8 k − 1 e(k) k − 1 e(k) k − 1 e(k) 1 0.0300 4 7.0981× 10−6 7 1.6767× 10−9 2 0.0019 5 4.3878× 10−7 3 1.1483× 10−4 6 2.7124× 10−8 View Large Table 2. Iteration errors for Example 2 (ε = 10−8) k − 1 e(k) k − 1 e(k) k − 1 e(k) 1 0.0300 4 7.0981× 10−6 7 1.6767× 10−9 2 0.0019 5 4.3878× 10−7 3 1.1483× 10−4 6 2.7124× 10−8 k − 1 e(k) k − 1 e(k) k − 1 e(k) 1 0.0300 4 7.0981× 10−6 7 1.6767× 10−9 2 0.0019 5 4.3878× 10−7 3 1.1483× 10−4 6 2.7124× 10−8 View Large Fig. 2. View largeDownload slide The relation between iteration step and iteration error for Example 2. Fig. 2. View largeDownload slide The relation between iteration step and iteration error for Example 2. Fig. 3. View largeDownload slide The simulation of the optimal states for Example 2. Fig. 3. View largeDownload slide The simulation of the optimal states for Example 2. 6. Conclusions In this paper, we have proposed new lower and upper matrix bounds for the solution of the DARE, which improve some of the previous results. Then, we have discussed the existence and uniqueness for the solution of this equation. The corresponding numerical examples have demonstrated the effectiveness of the derived results. Therefore, our future work is required to show more upper and lower matrix bounds, which are less restrictive and more concise, and also possibly tighter. Further, to get less restrictive and more faster iteration algorithms of this equation is our future work as well. Funding The work was supported in part by National Natural Science Foundation of China (11771370, 11571292), the Key Project of National Natural Science Foundation of China (91430213), the Youth Project of Hunan Provincial Natural Science Foundation of China (2017JJ3305) and the Youth Project of Hunan Provincial Education Department of China (16B257). References Arnold , W. F. & Laub , A. J. ( 1984 ) Generalized eigenproblem algorithms and software for algebraic Riccati equations . Proc. of the IEEE , 72 , 1746 -- 1754 . Google Scholar CrossRef Search ADS Berinde , V. ( 2009 ) A common fixed point theorem for compatible quasi contractive self mappings in metric spaces . Appl. Math. Comput. , 213 , 348 -- 354 . Callier , F. M. & Desoer , C. A. ( 1991 ) Linear System Theory . New York Inc : Springer . Google Scholar CrossRef Search ADS Davies , R. , Shi , P. & Wiltshire , R. ( 2008 ) Upper solution bounds of the continuous and discrete coupled algebraic Riccati equations . 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Published: Feb 1, 2018

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