Theory and Modern Applications

# New upper and lower bounds, the iteration algorithm for the solution of the discrete algebraic Riccati equation

## Abstract

In this paper, applying the properties of matrix Schur complement and matrix inverse, via some matrix equalities and inequalities, we present new lower and upper solution bounds of the discrete algebraic Riccati equation. Then, by the compressed image principle and a matrix norm inequality, we offer an existence uniqueness condition and a fixed point iteration algorithm for the solution of the discrete algebraic Riccati equation. Finally, a corresponding numerical example demonstrates the effectiveness of the developed results.

## 1 Introduction and preliminaries

The discrete algebraic Riccati equation has many applications in the process of obtaining optimal control and determining system stability [1, 2]. For example, consider the following linear discrete system (see [2]):

$$x(t+1)=Ax(t)+Bu(t),$$
(1.1)

where $$x(t)\in\mathbb{R}^{n}$$ is the state variable, $$u(t)\in\mathbb {R}^{m}$$ is the input variable, $$A\in\mathbb{R}^{n\times n}$$ is the system matrix, $$B\in\mathbb{R}^{n\times m}$$ is the input matrix.

The system (1.1) is associated with the linear quadratic optimal control problem of minimizing the quadratic performance index

$$\widetilde{J}=\sum_{t=0}^{\infty} \left ( \textstyle\begin{array}{@{}c@{}} x(t)\\ u(t) \end{array}\displaystyle \right )^{T} \left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} Q&S^{T}\\ S&G \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{@{}c@{}} x(t)\\ u(t) \end{array}\displaystyle \right ),$$

where $$S\in\mathbb{R}^{m\times n}$$, $$G\in \mathbb{R}^{m\times m}$$, $$Q\in\mathbb{R}^{n\times n}$$ are symmetric positive definite matrices, then the problem switches to how to solve the symmetric positive definite solution $$P\in\mathbb{R}^{n\times n}$$ of the following equation:

$$P=A^{T}PA+Q-\bigl(A^{T}PB+S^{T}\bigr) \bigl(G+B^{T}PB\bigr)^{-1}\bigl(B^{T}PA+S \bigr).$$
(1.2)

If $$S=0$$, $$G=I_{m}$$, (1.2) becomes the discrete algebraic Riccati matrix equation

$$P = A^{T}PA -A^{T}PB\bigl(I + B^{T}PB \bigr)^{-1}B^{T}PA + Q.$$
(1.3)

In many practical control problems, to study the discrete algebraic Riccati equation (1.3), we usually assume that Q is positive semi-definite. When considering certain H-infinity problems and LQG problems and so on in some discrete control linear systems, the assumption that Q is positive definite is only required. Certainly, it is difficult to discuss this equation on the assumption that Q is positive semi-definite. Thus, there are many scholars to research this equation on the assumption that Q is positive definite [3â€“6].

As the discrete algebraic Riccati equations (1.2) and (1.3) play an important role in many control analysis and design problems (Bernstein, 2005 [1]; Kojima et al., 2006 [2]), discussing these equations becomes a heated topic. For one thing, it is important to solve these equations (Kojima et al., 2006 [2]). It is difficult to obtain their solutions, and there are few iterative algorithms for getting the solutions of these equations (Komaroff, 1994 [3]). Reference [3] derives the iterative solution algorithm on the assumption that for the discrete algebraic Riccati equations (1.3) there exists a positive definite solution. In certain control problems, if the original system is assumed to be stabilizable, the exact solutions of the discrete algebraic Riccati equations (1.2) and (1.3) are not often required, while a reasonably tight solution bound will only be required such as the stabilization of jump linear systems (Fang and Loparo, 2002 [7]) and the quadratic optimization problem for a class of singularly perturbed stochastic systems (Dragan, 2005 [8]). Hence, in recent years, many researchers paid much attention to studying the discrete algebraic Riccati equations (1.2) and (1.3). Those include a recursive solution (Assimakis et al., 2003 [9]), a matrix inequality (Saberi, 1998 [10]), prediction, estimation, and smoothing error covariance matrices (Assimakis and Adam, 2013 [11]), iterative and algebraic algorithms (Assimakis and Adam, 2014 [12]; Chiang et al., 2010 [13]), matrix bounds (Choi, 2001 [14]; Kim and Park, 1999 [15]; Gao et al., 2002 [16]; Komaroff, 1994, 1992 [3, 17]; Lee and Chang, 1998 [18]; Lee, 1996, 1997, 2003 [19â€“22], Liu and Zhang, 2011, 2012 [23â€“25], Zhang and Liu, 2012 [26, 27]), eigenvalue bounds (Garloff, 1986 [5]; Lee, 2004 [6]), trace bounds (Kim et al., 1993 [28]), summation bounds (Komaroff and Shahian, 1992 [29]), norm bounds (Patel and Toda, 1978 [30]), and perturbation bounds (Hasanov et al., 2004 [31]; Hasanov, 2010 [32]). The most general bounds are the solution bounds, as they can directly offer the other types of bounds mentioned. Thus, based on the above, there are three problems to be considered as follows:

1. (1)

When does there exist a positive definite solution of the discrete algebraic Riccati equation (1.3)?

2. (2)

If the discrete algebraic Riccati equation (1.3) possesses a positive definite solution under certain conditions, in what range is the solution unique? In addition, how do we estimate the upper and lower solution bounds of this equation?

3. (3)

If the discrete algebraic Riccati equation (1.3) possesses a unique positive definite solution in a certain range, how do we design the iterative solution algorithms? Further, we need to discuss the solution bounds affecting the iterative algorithms.

Therefore, in this paper, we present new lower and upper solution bounds of the discrete algebraic Riccati equation (1.3). Then we offer an existence uniqueness condition and a fixed iterative algorithm for the solution of the discrete algebraic Riccati equation (1.3). Finally, a corresponding numerical example demonstrates the effectiveness of the developed results.

Throughout this paper, we use the following symbol conventions. Let $$\mathbb{R}^{n\times m}$$ and $$N^{+}$$ denote the set of $$n\times m$$ real matrices and positive integers. For $$X=(x_{ij})\in\mathbb{R}^{m\times n}$$, let $$X^{T}$$, $$\|X\|$$ denote the transpose and the spectral norm of X, respectively. If $$X\in \mathbb{R}^{n\times n}$$, $$X^{-1}$$ denotes the inverse of X. If $$X,Y\in\mathbb{R}^{n\times n}$$, the inequality $$X>(\geq)\, 0$$ means that X is a symmetric positive (semi-) definite matrix and the inequality $$X>(\geq)\, Y$$ means $$X-Y$$ is a symmetric positive (semi-) definite matrix. Suppose $$X\in\mathbb {R}^{n\times n}$$ is an arbitrary symmetric matrix, we assume that the eigenvalues of X are arranged so that $$\lambda_{1}(X)\geq\lambda_{2}(X)\geq\cdots\geq\lambda _{n}(X)$$. For $$X\in\mathbb{R}^{n\times n}$$, suppose the singular values of X are arranged so that $$\sigma_{1}(X)\geq\sigma_{2}(X)\geq\cdots\geq\sigma_{n}(X)$$. The identity matrix with appropriate dimensions is represented by I.

Let $$N=\{1,2,\ldots, n\}$$. For nonempty index sets $$\alpha,\beta \subseteq N$$ whose elements are both conventionally arranged in increasing order, we denote by $$|\alpha|$$ the cardinality of Î± and by $$\alpha^{c}=N-\alpha$$ the complement of Î± in N. We write $$X(\alpha,\beta)$$ to mean the submatrix of $$X\in R^{n\times n}$$ lying in the rows indexed by Î± and the columns indexed by Î². $$X(\alpha,\alpha)$$ is abbreviated to $$X(\alpha)$$. Assuming that $$X(\alpha)$$ is nonsingular, denote the Schur complement with respect to $$X(\alpha)$$ by

$$X/\alpha=X/X(\alpha)=X\bigl(\alpha^{c}\bigr)-X\bigl( \alpha^{c},\alpha\bigr)\bigl[X(\alpha )\bigr]^{-1}X\bigl( \alpha,\alpha^{c}\bigr).$$

The following lemmas are used to prove the main results.

### Lemma 1.1

(Zhang [33], p.19, Theorem 1.2)

Let $$X\in \mathbb{R}^{n\times n}$$ be partitioned as

$$X=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} E&F\\ G&H \end{array}\displaystyle \right ),$$

and suppose both X and $$E\in\mathbb{R}^{m\times m}$$ are nonsingular matrices. Then $$X/E$$ is nonsingular and

$$X^{-1}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} E^{-1}+E^{-1}FS^{-1}GE^{-1}&-E^{-1}FS^{-1} \\ -S^{-1}GE^{-1}&S^{-1} \end{array}\displaystyle \right ),$$

where $$S=H-GE^{-1}F=X/E$$.

### Lemma 1.2

(Zhang [33], p.25, Case 5)

If the matrix X is defined as Lemma  1.1, choosing $$\alpha=\{1,2,\ldots, m\}$$, then

$$(X/\alpha)^{-1}=X^{-1}\bigl(\alpha^{c}\bigr).$$

### Lemma 1.3

(Lee [20])

Let P be the positive semi-definite solution of the discrete algebraic Riccati equation (1.3), and $$\sigma^{2}_{1}(A)<1+\sigma^{2}_{n}(B)\delta_{1}$$, then P has the upper matrix bound

$$P\leq\frac{\lambda_{1}(Q)}{1+\delta_{1}\sigma^{2}_{n}(B)-\sigma ^{2}_{1}(A)}A^{T}A+Q\equiv\eta,$$
(1.4)

where $$\delta_{1}\equiv\lambda_{1}[A^{T}(I+QBB^{T})^{-1}QA+Q]$$.

### Definition 1.1

(Rudin [34])

The pair $$(X, d)$$ is called a complete metric space if X is a nonempty set whose elements are denoted by $$x, y, \ldots$$ and assume that on the Cartesian product $$X\times X$$ a distance function d is defined satisfying the following conditions:

1. (D0)

$$d(x,y)\geq0$$,

2. (Dl)

$$d(x, y) = 0$$ if and only if $$x=y$$,

3. (D2)

$$d(x,y)=d(y,x)$$,

4. (D3)

$$d(x,y)\leq d(x,z)+d(z,y)$$,

5. (D4)

every d-Cauchy sequence in X is d-convergent, i.e., if $${x_{n}}$$ is a sequence in X such that $$\lim_{n,m\rightarrow\infty} d(x_{n}, x_{m})= 0$$, then there is an $$x\in X$$ with $$\lim_{n\rightarrow\infty} d(x_{n}, x) = 0$$.

### Definition 1.2

(Rudin [34])

The pair $$(X, \|\cdot\|)$$ is called a normed space if X is a real vector space whose elements are denoted by $$x, y, z,\ldots$$ and assume that for every $$x\in X$$ there is associated a nonnegative real number $$\|x\|$$, called the norm of x, in such a way that:

1. (D0)

$$\|x+y\|\leq\|x\|+\|y\|$$,

2. (Dl)

$$\|\alpha x\|=|\alpha|\cdot\|x\|$$ where Î± is a scalar,

3. (D2)

$$\|x\|>0$$ if $$x\neq0$$.

### Lemma 1.4

(Berinde [35], Theorem B)

Let $$(X, d)$$ be a complete metric space and $$F : X \longrightarrow X$$ be a strict contraction, i.e., a map satisfying

$$d(Fx, Fy)\leq ad(x,y), \quad \textit{for all } x, y \in X,$$

where $$0 \leq a < 1$$ is constant. Then F has a unique fixed point in X.

Every normed space may be regarded as a metric space, in which the distance $$d(x,y)$$ between x and y is $$\|x-y\|$$. Hence, we get the following conclusion from Definitions 1.1, 1.2 and Lemma 1.4.

### Lemma 1.5

Let $$(X, \|\cdot\|)$$ be a real Banach space, and $$\Omega \subset X$$ be a convex, closed and bounded subset and $$F : \Omega\longrightarrow\Omega$$ be a contraction map, i.e., a map satisfying

$$\bigl\Vert F(P_{1})-F(P_{2})\bigr\Vert \leq p \|P_{1}-P_{2}\|, \quad \textit{for all } P_{1}, P_{2} \in\Omega,$$

where $$0 \leq p < 1$$ is constant. Then F has a unique fixed point in Î©.

### Lemma 1.6

(Zhao et al. [36], Lemma 2.1)

If $$X,Y\in \mathbb{R}^{n\times n}$$, and $$0< X \leq Y$$, then

$$\|X\|\leq\|Y\|.$$

## 2 New solution bounds for the discrete algebraic Riccati equation (1.3)

In this section, we first present a lemma (Lemma 2.1). Then new lower and upper solution bounds are deduced from Lemma 2.1.

### Lemma 2.1

Define the map

$$F(P)=A^{T}\bigl(P^{-1}+BB^{T}\bigr)^{-1}A+Q,$$
(2.1)

then

$$F(P)=Q^{\frac{1}{2}}\biggl[Q^{-\frac{1}{2}}A^{T} \bigl(P^{-1}+BB^{T}-A\bigl(F(P)\bigr)^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q.$$
(2.2)

### Proof

By (2.1), then

$$Q=F(P)-A^{T}\bigl(P^{-1}+BB^{T} \bigr)^{-1}A=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} F(P)&A^{T} \\ A &P^{-1}+BB^{T} \end{array}\displaystyle \right ) \Big/ \alpha^{c}.$$
(2.3)

Applying Lemma 1.2 to (2.3) yields

$$Q=\left [\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} F(P)&A^{T} \\ A &P^{-1}+BB^{T} \end{array}\displaystyle \right )^{-1}( \alpha) \right ]^{-1},$$

which is equivalent to

$$Q^{-1}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} F(P)&A^{T} \\ A &P^{-1}+BB^{T} \end{array}\displaystyle \right )^{-1}(\alpha).$$
(2.4)

Introducing Lemma 1.1 to (2.4) gives

$$Q^{-1}=F(P)^{-1}+F(P)^{-1}A^{T} \bigl[P^{-1}+BB^{T}-A\bigl(F(P)\bigr)^{-1}A^{T} \bigr]^{-1}AF(P)^{-1}.$$
(2.5)

Multiplying (2.5) on the right and on the left by $$F(P)$$ one gets

$$F(P)Q^{-1}F(P)=F(P)+A^{T}\bigl[P^{-1}+BB^{T}-A \bigl(F(P)\bigr)^{-1}A^{T}\bigr]^{-1}A.$$
(2.6)

Define

$$\widetilde{F}(P)=Q^{-\frac{1}{2}}F(P)Q^{-\frac{1}{2}},$$
(2.7)

multiplying (2.6) on the right and on the left by $$Q^{-\frac{1}{2}}$$ leads to

$$\widetilde{F}(P)^{2}-\widetilde{F}(P)+\frac{1}{4}I=Q^{-\frac {1}{2}}A^{T} \bigl[P^{-1}+BB^{T}-A\bigl(F(P)\bigr)^{-1}A^{T} \bigr]^{-1}AQ^{-\frac {1}{2}}+\frac{1}{4}I,$$

from which one infers that

$$\biggl(\widetilde{F}(P)-\frac{1}{2}I\biggr)^{2}=Q^{-\frac {1}{2}}A^{T} \bigl[P^{-1}+BB^{T}-A\bigl(F(P)\bigr)^{-1}A^{T} \bigr]^{-1}AQ^{-\frac {1}{2}}+\frac{1}{4}I.$$
(2.8)

From (2.1), we get

$$F(P)\geq Q,$$
(2.9)

i.e.,

$$\bigl(F(P)\bigr)^{-1}\leq Q^{-1}.$$
(2.10)

In terms of (2.9), it is evident that

$$F(P)\geq Q > \frac{1}{2}Q,$$

or, equivalently

$$\widetilde{F}(P)\geq\frac{1}{2}I.$$
(2.11)

Combining (2.8) with (2.11) shows that

$$\widetilde{F}(P)-\frac{1}{2}I=\biggl[Q^{-\frac {1}{2}}A^{T} \bigl[P^{-1}+BB^{T}-A\bigl(F(P)\bigr)^{-1}A^{T} \bigr]^{-1}AQ^{-\frac {1}{2}}+\frac{1}{4}I\biggr]^{\frac{1}{2}},$$

i.e.,

$$\widetilde{F}(P)=\biggl[Q^{-\frac {1}{2}}A^{T}\bigl[P^{-1}+BB^{T}-A \bigl(F(P)\bigr)^{-1}A^{T}\bigr]^{-1}AQ^{-\frac {1}{2}}+ \frac{1}{4}I\biggr]^{\frac{1}{2}}+\frac{1}{2}I.$$
(2.12)

Substituting (2.7) into (2.12) gives

$$F(P)=Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T}\bigl(P^{-1}+BB^{T}-A \bigl(F(P)\bigr)^{-1}A^{T}\bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+\frac{1}{2}Q,$$

which completes the proof.â€ƒâ–¡

### Theorem 2.1

Let P be the positive definite solution of the discrete algebraic Riccati equation (1.3), then P has the lower matrix bound

$$P \geq Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T} \bigl(Q^{-1}+BB^{T}\bigr)^{-1}AQ^{-\frac{1}{2}}+ \frac{1}{4}I\biggr]^{\frac {1}{2}}Q^{\frac{1}{2}}+\frac{1}{2}Q \equiv P_{1}.$$
(2.13)

### Proof

From (1.3), we have

$$P = A^{T}P\bigl[P^{-1} -B\bigl(I + B^{T}PB \bigr)^{-1}B^{T}\bigr]PA + Q.$$
(2.14)

Let

$$X=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} P^{-1}&B \\ B^{T} &I+B^{T}PB \end{array}\displaystyle \right ),$$

according to Lemmas 1.1 and 1.2, (2.14) changes to

\begin{aligned} \begin{aligned}[b] P&=A^{T}P\left [\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} P^{-1}&B \\ B^{T} &I+B^{T}PB \end{array}\displaystyle \right ) \Big/\alpha^{c} \right ]PA+Q \\ &=A^{T}P\left [\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} P^{-1}&B \\ B^{T} &I+B^{T}PB \end{array}\displaystyle \right )^{-1}(\alpha) \right ]^{-1}PA+Q \\ &=A^{T}P\left [\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} P+PB(X/\alpha)^{-1}B^{T}P&-PB(X/\alpha)^{-1} \\ -(X/\alpha)^{-1}B^{T}P &(X/\alpha)^{-1} \end{array}\displaystyle \right ) (\alpha) \right ]^{-1}PA+Q. \end{aligned} \end{aligned}
(2.15)

Since

$$X/\alpha=I+B^{T}PB-B^{T}\bigl(P^{-1} \bigr)^{-1}B=I,$$

(2.15) becomes

$$P=A^{T}P\bigl(P+PBB^{T}P\bigr)^{-1}PA+Q=A^{T} \bigl(P^{-1}+BB^{T}\bigr)^{-1}A+Q.$$
(2.16)

By (2.16), choosing $$F(P)=P$$ in Lemma 2.1, we have

$$P=Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T}\bigl(P^{-1}+BB^{T}-AP^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q.$$
(2.17)

From (2.16), we get

$$P\geq Q,$$

i.e.,

$$P^{-1}\leq Q^{-1}.$$
(2.18)

In the light of (2.18), (2.17) turns to

\begin{aligned} P \geq& Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T} \bigl(P^{-1}+BB^{T}\bigr)^{-1}AQ^{-\frac{1}{2}} + \frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+\frac{1}{2}Q \\ \geq& Q^{\frac{1}{2}}\biggl[Q^{-\frac{1}{2}}A^{T} \bigl(Q^{-1}+BB^{T}\bigr)^{-1}AQ^{-\frac{1}{2}} + \frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+\frac{1}{2}Q. \end{aligned}

This completes the proof.â€ƒâ–¡

Furthermore, we propose new lower and upper solution bounds on the basis of Lemma 2.1.

### Theorem 2.2

Let P be the positive definite solution of the discrete algebraic Riccati equation (1.3), and $$\sigma ^{2}_{1}(A)<1+\sigma^{2}_{n}(B)\delta_{1}$$, then P has the lower and upper solution bounds

$$\widetilde{P}_{1}\leq P\leq\widehat{P}_{1},$$
(2.19)

where

$$\widetilde{P}_{1}\equiv Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T} \bigl(Q^{-1}+BB^{T}-A\eta^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q$$

and

$$\widehat{P}_{1}\equiv Q^{\frac{1}{2}}\biggl[Q^{-\frac{1}{2}}A^{T} \bigl(\eta ^{-1}+BB^{T}-AQ^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q,$$

$$\delta_{1}$$ and Î· are defined as in Lemma  1.3.

### Proof

Applying Lemma 1.3 and (2.18) to (2.17), it is simple to see that

\begin{aligned} P =&Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T}\bigl(P^{-1}+BB^{T}-AP^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q \\ \geq& Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T} \bigl(Q^{-1}+BB^{T}-AP^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q \\ \geq& Q^{\frac{1}{2}}\biggl[Q^{-\frac{1}{2}}A^{T} \bigl(Q^{-1}+BB^{T}-A\eta ^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q \end{aligned}

and

\begin{aligned} P =&Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T}\bigl(P^{-1}+BB^{T}-AP^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q \\ \leq& Q^{\frac{1}{2}}\biggl[Q^{-\frac{1}{2}}A^{T}\bigl(\eta ^{-1}+BB^{T}-AP^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q \\ \leq& Q^{\frac{1}{2}}\biggl[Q^{-\frac{1}{2}}A^{T}\bigl(\eta ^{-1}+BB^{T}-AQ^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q. \end{aligned}

This completes the proof.â€ƒâ–¡

### Remark 2.1

For one thing, we point out that $$\widetilde {P}_{1}\geq P_{1} \geq Q$$. Indeed,

\begin{aligned} \widetilde{P}_{1} =&Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T} \bigl(Q^{-1}+BB^{T}-A\eta^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q \\ \geq& Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T} \bigl(Q^{-1}+BB^{T}\bigr)^{-1}AQ^{-\frac{1}{2}} + \frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+\frac{1}{2}Q=P_{1} \end{aligned}

and

\begin{aligned} P_{1} =&Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T} \bigl(Q^{-1}+BB^{T}\bigr)^{-1}AQ^{-\frac{1}{2}} + \frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+\frac{1}{2}Q \\ \geq& Q^{\frac{1}{2}}\biggl(0+\frac{1}{4}I\biggr)^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q=Q, \end{aligned}

which implies that $$\widetilde{P}_{1}\geq P_{1} \geq Q$$.

### Remark 2.2

$$P_{1}$$ and Î· are two different upper bounds. It is difficult to compare them for the same measure. Further, in Section 5, we offer a numerical example which shows that $$P_{1}$$ is better than Î· (i.e., $$P_{1}\leq\eta$$) in a certain case. It is hard to prove that this result holds for the general case in the theory.

## 3 On the solution existence uniqueness of the discrete algebraic Riccati equation (1.3)

In this section, we present a new existence uniqueness condition for the solution of the discrete algebraic Riccati equation (1.3).

### Theorem 3.1

Let $$\sigma^{2}_{1}(A)<1+\sigma^{2}_{n}(B)\delta _{1}$$, $$A^{T}(\widehat{P}_{1}^{-1}+BB^{T})^{-1}A+Q\leq\eta$$, and

$$0< p< 1$$
(3.1)

with

$$p=\|A\|^{2}\cdot\bigl\Vert \widetilde{P}_{1}^{-1} \bigr\Vert ^{2}\cdot\bigl(\bigl\Vert \widehat {P}_{1}^{-1}+BB^{T} \bigr\Vert \bigr)^{-2},$$

where $$\delta_{1}$$, Î· are defined as in Lemma  1.3, $$\widetilde{P}_{1}$$, $$\widehat{P}_{1}$$ are defined by (2.19). Then the discrete algebraic Riccati equation (1.3) has a unique positive definite solution $$P_{0}$$, and $$\widetilde{P}_{1}\leq P_{0}\leq\widehat{P}_{1}$$.

### Proof

(1) If $$\sigma^{2}_{1}(A)<1+\sigma^{2}_{n}(B)\delta_{1}$$, using (1.4) and the inequalities

$$\eta^{-1} \leq Q ^{-1}, \qquad -AQ ^{-1}A^{T} \leq -A\eta^{-1}A^{T},$$

we get $$\widetilde{P}_{1}\leq\widehat{P}_{1}$$. Suppose the discrete algebraic Riccati equation (1.3) possesses a positive definite solution, from Theorem 2.2, then the positive definite solution is in $$[\widetilde{P}_{1},\widehat{P}_{1}]$$, i.e., $$\widetilde {P}_{1} \leq P \leq\widehat{P}_{1}$$.

(2) Define the map $$F(P)=A^{T}(P^{-1}+BB^{T})^{-1}A+Q$$ and set

$$P\in\Omega=\{P\mid\widetilde{P}_{1}\leq P\leq\widehat{P}_{1} \}.$$

It is obvious that Î© is a convex, closed, and bounded set and $$F(P)$$ is continuous on Î©. We consider a Banach space $$(\Omega, \|\cdot\|)$$, where $$\|\cdot\|$$ is the spectral norm.

For one thing, since $$P\leq\widehat{P}_{1}$$, $$A^{T}(\widehat {P}_{1}^{-1}+BB^{T})^{-1}A+Q\leq\eta$$,

$$F(P)=A^{T}\bigl(P^{-1}+BB^{T}\bigr)^{-1}A+Q \leq A^{T}\bigl(\widehat {P}_{1}^{-1}+BB^{T} \bigr)^{-1}A+Q\leq\eta.$$

For another, since $$\widetilde{P}_{1}\leq P$$,

$$F(P)=A^{T}\bigl(P^{-1}+BB^{T}\bigr)^{-1}A+Q \geq F(\widetilde {P}_{1})=A^{T}\bigl(\widetilde{P}_{1}^{-1}+BB^{T} \bigr)^{-1}A+Q\geq Q.$$

Thus, by Lemma 2.1, we obtain

\begin{aligned} F(P) =&Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T} \bigl(P^{-1}+BB^{T}-A\bigl(F(P)\bigr)^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q \\ \geq& Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T} \bigl(Q^{-1}+BB^{T}-A\bigl(F(P)\bigr)^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q \\ \geq& Q^{\frac{1}{2}}\biggl[Q^{-\frac{1}{2}}A^{T} \bigl(Q^{-1}+BB^{T}-A\eta ^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q=\widetilde{P}_{1} \end{aligned}

and

\begin{aligned} \begin{aligned} F(P)&=Q^{\frac{1}{2}}\biggl[Q^{-\frac {1}{2}}A^{T} \bigl(P^{-1}+BB^{T}-A\bigl(F(P)\bigr)^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q \\ &\leq Q^{\frac{1}{2}}\biggl[Q^{-\frac{1}{2}}A^{T}\bigl(\eta ^{-1}+BB^{T}-A\bigl(F(P)\bigr)^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q \\ &\leq Q^{\frac{1}{2}}\biggl[Q^{-\frac{1}{2}}A^{T}\bigl(\eta ^{-1}+BB^{T}-AQ^{-1}A^{T} \bigr)^{-1}AQ^{-\frac{1}{2}} +\frac{1}{4}I\biggr]^{\frac{1}{2}}Q^{\frac{1}{2}}+ \frac{1}{2}Q= \widehat{P}_{1}. \end{aligned} \end{aligned}

Consequently, for $$P\in\Omega$$, we get $$\widetilde{P}_{1}\leq F(P)\leq \widehat{P}_{1}$$.

Thus $$F(\Omega)\subseteq\Omega$$.

For arbitrary $$P_{1},P_{2}\in\Omega$$,

\begin{aligned}& F(P_{1})=A^{T}\bigl(P_{1}^{-1}+BB^{T} \bigr)^{-1}A+Q, \\& F(P_{2})=A^{T}\bigl(P_{2}^{-1}+BB^{T} \bigr)^{-1}A+Q. \end{aligned}

Consequently,

\begin{aligned} F(P_{1})-F(P_{2}) =&A^{T}\bigl[ \bigl(P_{1}^{-1}+BB^{T}\bigr)^{-1}- \bigl(P_{2}^{-1}+BB^{T}\bigr)^{-1}\bigr]A \\ =&A^{T}\bigl\{ \bigl(P_{1}^{-1}+BB^{T} \bigr)^{-1}\bigl[\bigl(P_{2}^{-1}+BB^{T} \bigr)-\bigl(P_{1}^{-1}+BB^{T}\bigr)\bigr] \bigl(P_{2}^{-1}+BB^{T}\bigr)^{-1}\bigr\} A \\ =&A^{T}\bigl(P_{1}^{-1}+BB^{T} \bigr)^{-1}\bigl(P_{2}^{-1}-P_{1}^{-1} \bigr) \bigl(P_{2}^{-1}+BB^{T}\bigr)^{-1}A \\ =&A^{T}\bigl(P_{1}^{-1}+BB^{T} \bigr)^{-1}P_{1}^{-1}(P_{1}-P_{2})P_{2}^{-1} \bigl(P_{2}^{-1}+BB^{T}\bigr)^{-1}A. \end{aligned}

Since $$\widetilde{P}_{1}\leq P\leq\widehat{P}_{1}$$, according to Lemma 1.6, we have

\begin{aligned} \bigl\Vert F(P_{1})-F(P_{2})\bigr\Vert =&\bigl\Vert A^{T}\bigl(P_{1}^{-1}+BB^{T} \bigr)^{-1}P_{1}^{-1}(P_{1}-P_{2})P_{2}^{-1} \bigl(P_{2}^{-1}+BB^{T}\bigr)^{-1}A\bigr\Vert \\ \leq&\bigl\Vert A^{T}\bigr\Vert \cdot \Vert A\Vert \cdot\bigl\Vert P_{1}^{-1}\bigr\Vert \cdot \bigl\Vert P_{2}^{-1}\bigr\Vert \cdot\bigl\Vert \bigl(P_{1}^{-1}+BB^{T}\bigr)^{-1}\bigr\Vert \\ &{}\cdot \bigl\Vert \bigl(P_{2}^{-1}+BB^{T} \bigr)^{-1}\bigr\Vert \cdot \Vert P_{1}-P_{2} \Vert \\ \leq&\bigl\Vert A^{T}\bigr\Vert \cdot \Vert A\Vert \cdot\bigl\Vert \widetilde{P}_{1}^{-1}\bigr\Vert \cdot\bigl\Vert \widetilde {P}_{1}^{-1}\bigr\Vert \cdot\bigl(\bigl\Vert \widehat{P}_{1}^{-1}+BB^{T}\bigr\Vert \bigr)^{-1} \\ &{}\cdot\bigl(\bigl\Vert \widehat {P}_{1}^{-1}+BB^{T} \bigr\Vert \bigr)^{-1}\cdot \Vert P_{1}-P_{2} \Vert \\ =& \Vert A\Vert ^{2}\cdot\bigl\Vert \widetilde{P}_{1}^{-1} \bigr\Vert ^{2}\cdot \bigl(\bigl\Vert \widehat{P}_{1}^{-1}+BB^{T} \bigr\Vert \bigr)^{-2}\cdot \Vert P_{1}-P_{2} \Vert \\ =&p \Vert P_{1}-P_{2}\Vert . \end{aligned}
(3.2)

As $$p<1$$, thus the map $$F(P)$$ is a contraction map in Î©. In the light of Lemma 1.5, the map $$F(P)$$ has a unique fixed point in Î©.

Combing (1) and (2) shows that (1.3) has a unique positive definite solution $$P_{0}$$, and $$\widetilde{P}_{1}\leq P_{0}\leq\widehat {P}_{1}$$.â€ƒâ–¡

## 4 A fixed point iteration algorithm for the solution of the discrete algebraic Riccati equation (1.3)

In this section, based on Theorem 3.1, we present a fixed point iteration algorithm for the solution of the discrete algebraic Riccati equation (1.3).

For arbitrary given $$P^{(0)}\in\Omega=\{P\mid\widetilde{P}_{1}\leq P\leq\widehat{P}_{1}\}$$, we construct the matrix sequence

$$P^{(k+1)}=A^{T}\bigl[\bigl(P^{(k)} \bigr)^{-1}+BB^{T}\bigr]^{-1}A+Q, \quad k=0,1, \ldots.$$
(4.1)

Next, we prove $$P^{(k)}$$ is convergent and converges to the exact solution of the discrete algebraic Riccati equation (1.3).

### Theorem 4.1

If the conditions of Theorem  3.1 are satisfied, then the sequence $$P^{(k)}$$ given by (4.1) is convergent and converges to the unique positive definite solution $$P_{0}$$ of the discrete algebraic Riccati equation (1.3).

### Proof

For $$P^{(0)}\in\Omega$$, in terms of (4.1), for $$k\in N^{+}$$, we obtain a sequence $$\{P^{(k)}\}$$, then

\begin{aligned} P^{(k)}-P^{(k-1)} =&A^{T}\bigl\{ \bigl[ \bigl(P^{(k-1)}\bigr)^{-1}+BB^{T}\bigr]^{-1}- \bigl[\bigl(P^{(k-2)}\bigr)^{-1}+BB^{T} \bigr]^{-1}\bigr\} A \\ =&A^{T}\bigl\{ \bigl[\bigl(P^{(k-1)}\bigr)^{-1}+BB^{T} \bigr]^{-1}\bigl\{ \bigl[\bigl(P^{(k-2)}\bigr)^{-1}+BB^{T} \bigr]- \bigl[\bigl(P^{(k-1)}\bigr)^{-1}+BB^{T}\bigr] \bigr\} \\ &{}\times\bigl[\bigl(P^{(k-2)}\bigr)^{-1}+BB^{T} \bigr]^{-1}\bigr\} A \\ =&A^{T}\bigl[\bigl(P^{(k-1)}\bigr)^{-1}+BB^{T} \bigr]^{-1}\bigl[\bigl(P^{(k-2)}\bigr)^{-1}- \bigl(P^{(k-1)}\bigr)^{-1}\bigr] \bigl[\bigl(P^{(k-2)} \bigr)^{-1}+BB^{T}\bigr]^{-1}A \\ =&A^{T}\bigl[\bigl(P^{(k-1)}\bigr)^{-1}+BB^{T} \bigr]^{-1}\bigl(P^{(k-1)}\bigr)^{-1} \bigl(P^{(k-1)}-P^{(k-2)}\bigr) \bigl(P^{(k-2)} \bigr)^{-1} \\ &{}\times\bigl[\bigl(P^{(k-2)}\bigr)^{-1}+BB^{T} \bigr]^{-1}A. \end{aligned}

In a similar way to the proof of (3.2), we obtain

$$\bigl\Vert P^{(k)}-P^{(k-1)}\bigr\Vert \leq p\bigl\Vert P^{(k-1)}-P^{(k-2)}\bigr\Vert .$$

Applying mathematical induction, it is easy to get

$$\bigl\Vert P^{(k)}-P^{(k-1)}\bigr\Vert \leq p^{k-1} \bigl\Vert P^{(1)}-P^{(0)}\bigr\Vert .$$
(4.2)

Because $$0< p<1$$, thus $$\lim_{k\rightarrow\infty}p^{k}=0$$. Then, $$\forall \varepsilon>0$$, $$\exists N_{1}\in N^{+}$$, $$\forall k> N_{1}$$, we get

$$p^{k}< \frac{\varepsilon(1-p)}{\|P^{(1)}-P^{(0)}\|}.$$
(4.3)

Consequently, $$\forall k> N_{1}$$, $$\forall v\in N^{+}$$, using the triangular inequality, (4.2), and (4.3), we obtain

\begin{aligned} \bigl\Vert P^{(k+v)}-P^{(k)}\bigr\Vert \leq& \bigl\Vert P^{(k+v)}-P^{(k+v-1)}\bigr\Vert +\bigl\Vert P^{(k+v-1)}-P^{(k+v-2)} \bigr\Vert +\cdots +\bigl\Vert P^{(k+1)}-P^{(k)}\bigr\Vert \\ \leq&\bigl(p^{k+v-1}+p^{k+v-2}+\cdots+p^{k}\bigr)\bigl\Vert P^{(1)}-P^{(0)}\bigr\Vert \\ =&\frac{p^{k}-p^{k+v}}{1-p}\bigl\Vert P^{(1)}-P^{(0)}\bigr\Vert \\ < &\frac{p^{k}}{1-p}\bigl\Vert P^{(1)}-P^{(0)}\bigr\Vert \\ < &\varepsilon. \end{aligned}

Hence $$P^{(k)}$$ is a Cauchy sequence in Î©, then $$P^{(k)}$$ is convergent. Denote $$\lim_{k\rightarrow\infty}P^{(k)}=P_{0}$$. Let $$k\rightarrow\infty$$ in (4.1), then

$$P_{0}=A^{T}\bigl(P_{0}^{-1}+BB^{T} \bigr)^{-1}A+Q.$$

This shows that $$P_{0}$$ is a solution of (1.3). According to Theorem 3.1, $$P_{0}$$ is the unique positive definite solution of the discrete algebraic Riccati equation (1.3). This completes the proof.â€ƒâ–¡

By Theorem 4.1, define $$P^{(k)}$$ as the kth iterative solution of the discrete algebraic Riccati equation (1.3). For the iterative solution $$P^{(k+1)}$$ and $$P^{(k)}$$, by (4.2), for any small $$\varepsilon>0$$, we have the error estimation

$$\bigl\Vert P^{(k+1)}-P^{(k)}\bigr\Vert \leq p^{k} \bigl\Vert P^{(1)}-P^{(0)}\bigr\Vert \leq p^{k} \Vert \widehat {P}_{1}-\widetilde{P}_{1}\Vert < \varepsilon.$$

In the case of the precision Îµ is permitted, we have the following algorithm.

### Algorithm 4.1

Input::

For the discrete algebraic Riccati equation (1.3), given A, B, Q.

Output::

$$P^{(k)}$$.

Step 1::

Compute $$\widetilde{P}_{1}$$ and $$\widehat{P}_{1}$$.

Step 2::

Compute p, if $$0 < p < 1$$, go to Step 3; otherwise, stop.

Step 3::

Set $$P^{(0)}\in\Omega=\{P\mid\widetilde{P}_{1}\leq P\leq \widehat{P}_{1}\}$$.

Step 4::

Compute

$$P^{(k+1)}=A^{T}\bigl[\bigl(P^{(k)} \bigr)^{-1}+BB^{T}\bigr]^{-1}A+Q, \quad k=0,1, \ldots.$$
Step 5::

For any small $$\varepsilon>0$$ and $$k\in N^{+}$$, if $$\|P^{(k+1)}-P^{(k)}\|<\varepsilon$$, stop; otherwise, let $$k=k+1$$, go to Step 4.

## 5 A numerical example

In this section, we demonstrate the effectiveness of our results by the following real application example. In many engineering fields such as solid mechanics, quantum mechanics, parameter identification and automatic control, we often need to study the system stability and optimal control of linear discrete systems. Sometimes, these problems reduce to discussing the symmetric positive definite solution of the corresponding discrete algebraic Riccati matrix equations (1.2) and (1.3). The whole process of the example is carried out on Matlab 7.1 and the precision is 10âˆ’8.

### Example 5.1

Consider the following linear discrete system (see [2]):

$$x(t+1)=Ax(t)+Bu(t),$$
(5.1)

where $$x(t)\in\mathbb{R}^{n}$$ is the state variable, $$u(t)\in\mathbb {R}^{m}$$ is the input variable, $$A\in\mathbb{R}^{n\times n}$$ is the system matrix, $$B\in\mathbb{R}^{n\times m}$$ is the input matrix. Here, we choose

\begin{aligned}& A=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.02 & 0.05 & 0.1\\ 0.03 & 0.04 & 0\\ 0.06 & -0.1 & 0.05 \end{array}\displaystyle \right ), \\& B=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.1 & 0 & 0.2\\ 0.3 & 0.4 & 0\\ 0 & 0.15 & 0.16 \end{array}\displaystyle \right ). \end{aligned}

The quadratic performance index of (5.1) is

$$J=\frac{1}{2}\sum_{t=0}^{\infty} \bigl[ x^{T}(t)Qx(t)+u^{T}(t)u(t) \bigr],$$
(5.2)

where

$$Q=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.6 & 0.12 & 0.1\\ 0.12 & 0.4 & 0.05\\ 0.1 & 0.05 & 0.3 \end{array}\displaystyle \right )$$

is symmetric positive definite.

In order to seek the optimal control such that the quadratic performance index (5.2) attains the minimum, we use the minimum principle. Choose the Hamiltonian function as

$$H(t)=\frac{1}{2}x^{T}(t)Qx(t)+\frac{1}{2}u^{T}(t)u(t)+ \chi ^{T}(t+1)\bigl[Ax(t)+Bu(t)\bigr].$$

Then we can obtain the canonical equation

$$\chi(t)=\frac{\partial H(t)}{\partial x(t)}=Qx(t)+A^{T}\chi(t+1),$$
(5.3)

and the control equation

$$\frac{\partial H(t)}{\partial u(t)}=u(t)+B^{T}\chi(t+1)=0,$$

i.e.,

$$u(t)=-B^{T}\chi(t+1).$$
(5.4)

Suppose

$$\chi(t)=Px(t).$$
(5.5)

Substituting (5.5) into the canonical equation (5.3) yields

$$Px(t)=Qx(t)+A^{T}Px(t+1).$$
(5.6)

Using the state equation (5.1) and the control equation (5.4), we have

$$x(t+1)=Ax(t)+Bu(t)=Ax(t)-BB^{T}Px(t+1),$$

then

$$x(t+1)= \bigl(I+BB^{T}P \bigr)^{-1}Ax(t).$$
(5.7)

Substituting (5.7) into (5.6), eliminating $$x(t)$$ on both sides of the equality, applying the Sherman-Morrison-Woodbury equality

$$(I+ST)^{-1}=I-S(I+TS)^{-1}T,$$
(5.8)

we get the discrete algebraic Riccati equation (1.3),

$$P = A^{T}PA -A^{T}PB\bigl(I + B^{T}PB \bigr)^{-1}B^{T}PA + Q.$$

In terms of (5.4) and (5.5), by (5.8), we obtain the optimal control

$$u(t)=- \bigl(I+B^{T}PB \bigr)^{-1}B^{T}PAx(t)$$
(5.9)

such that the quadratic performance index (5.2) attains a minimum, where P is the symmetric positive definite solution of the discrete algebraic Riccati equation (1.3).

Obviously, $$\sigma^{2}_{1}(A)<1+\sigma^{2}_{n}(B)\delta_{1}$$, then the upper matrix bound for the solution P of the discrete algebraic Riccati equation (1.3) by (1.4) is

$$\eta=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.60339617 & 0.11736623 & 0.10346548\\ 0.11736623 & 0.40977266 & 0.05000000\\ 0.10346548 & 0.05000000 & 0.30866371 \end{array}\displaystyle \right ),$$

$$\lambda_{1}(\eta)=0.69307554, \qquad \lambda_{2}(\eta)= 0.35445560,\qquad \lambda_{3}(\eta)= 0.27430140.$$

The lower matrix bound for the solution P of the discrete algebraic Riccati equation (1.3) found by (2.13) is

$$P_{1}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.60205607 & 0.11950250 & 0.10293022\\ 0.11950250 & 0.40416982 & 0.05112541\\ 0.10293022 & 0.05112541 & 0.30720800 \end{array}\displaystyle \right ),$$

$$\lambda_{1}(P_{1})=0.69252521,\qquad \lambda_{2}(P_{1})= 0.34803838, \qquad \lambda_{3}(P_{1})=0.27287030.$$

The lower and upper matrix bounds for the solution P of the discrete algebraic Riccati equation (1.3) found by (2.19) are

$$\widetilde{P}_{1}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.60204980 & 0.11953871 & 0.10295826\\ 0.11953871 & 0.40415536 & 0.05121050\\ 0.10295826 & 0.05121050 & 0.30729715 \end{array}\displaystyle \right ),$$

\begin{aligned}& \lambda_{1}(\widetilde{P}_{1})=0.69258489,\qquad \lambda _{2}(\widetilde{P}_{1})= 0.34800103, \qquad \lambda_{3}(\widetilde{P}_{1})=0.27291639, \\& \widehat{P}_{1}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.60209132 & 0.11949876 & 0.10300401\\ 0.11949876 & 0.40422207 & 0.05115031\\ 0.10300401 & 0.05115030 & 0.30738615 \end{array}\displaystyle \right ), \end{aligned}

\begin{aligned}& \lambda_{1}(\widehat{P}_{1})= 0.69261472 ,\qquad \lambda_{2}(\widehat{P}_{1})= 0.34808987, \\& \lambda_{3}(\widehat{P}_{1})=0.27299495. \end{aligned}

Obviously,

$$\widetilde{P}_{1}\geq P_{1}, \qquad \eta\geq \widehat{P}_{1}.$$

This shows that the lower bound $$\widetilde{P}_{1}$$ is better than $$P_{1}$$, and the upper bound $$\widehat{P}_{1}$$ is better than Î·.

By computation, obviously, $$\widetilde{P}_{1}\leq\widehat{P}_{1}$$, $$A^{T}(\widehat{P}_{1}^{-1}+BB^{T})^{-1}A+Q\leq\eta$$, and

$$0< p=0.01643822< 1.$$

According to Theorem 3.1, the discrete algebraic Riccati equation (1.3) has a unique positive definite solution $$P_{0}$$, and $$\widetilde{P}_{1}\leq P_{0}\leq\widehat{P}_{1}$$.

Let $$e(k)=\|P^{(k+1)}-P^{(k)}\|$$ be the iteration error at the kth iteration, k denote the iteration number, where we choose $$\varepsilon=10^{-8}$$ and $$P^{(0)}=\widetilde{P}_{1}$$ and $$P^{(0)}=Q$$, respectively. Then Algorithm 4.1 and Theorem 3.1 of [3] need 4 and 5 iteration steps, respectively, to converge to the iteration solution of the discrete algebraic Riccati equation (1.3) as follows:

$$P=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.60208663 & 0.11950317 & 0.10300002\\ 0.11950317 & 0.40420121 & 0.05116389\\ 0.10300002 & 0.05116389 & 0.30735947 \end{array}\displaystyle \right ).$$

The relation between iteration step and iteration error is shown in Table 1 and Figure 1. Obviously, Table 1 and Figure 1 show that compared to Theorem 3.1 of [3], Algorithm 4.1 has less errors and less iteration steps.

Thus, by (5.9), we can have the optimal control

\begin{aligned} u(t) =&- \bigl(I+B^{T}PB \bigr)^{-1}B^{T}PAx(t) \\ =& \left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.00644824 & 0.00704508 & 0.00957459\\ 0.00909563 & 0.00265056 & 0.00810075\\ 0.00724564 & 0.00089025 & 0.01614435 \end{array}\displaystyle \right )x(t) \end{aligned}

such that the quadratic performance index (5.2) attains a minimum.

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## Acknowledgements

The work was supported in part by National Natural Science Foundation of China (11571292, 11471279), National Natural Science Foundation for Youths of China (11401505), the Key Project of National Natural Science Foundation of China (91430213), the General Project of Hunan Provincial Natural Science Foundation of China (2015JJ2134), and the General Project of Hunan Provincial Education Department of China (15C1320).

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Correspondence to Jianzhou Liu.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Zhang, J., Liu, J. New upper and lower bounds, the iteration algorithm for the solution of the discrete algebraic Riccati equation. Adv Differ Equ 2015, 313 (2015). https://doi.org/10.1186/s13662-015-0649-6