Positive definite solutions and perturbation analysis of a class of nonlinear matrix equations

Abstract

In this paper, we consider a class of nonlinear matrix equation of the type $X+\sum _{i=1}^mA_i^{*}X^{-q}A_i-\sum _{j=1}^nB_{j}^{*}X^{-r}B_j=Q$, where $0<q,\,r\le 1$ and Q is positive definite. Based on the Schauder fixed point theorem and Bhaskar–Lakshmikantham coupled fixed point theorem, we derive some sufficient conditions for the existence and uniqueness of the positive definite solution to such equations. An iterative method is provided to compute the unique positive definite solution. A perturbation estimation and the explicit expression of Rice condition number of the unique positive definite solution are also established. The theoretical results are illustrated by numerical examples.

1 Introduction

In this paper, we consider the following nonlinear matrix equation

$$\begin{aligned} X+\sum _{i=1}^mA_i^{*}X^{-q}A_i-\sum _{j=1}^nB_{j}^{*}X^{-r}B_j=Q \end{aligned}$$

(1.1)

where $0<q,\,r\le 1$, and $A_i, B_j,\,i=1,2,\ldots ,m,\,j=1,2,\ldots ,n$ and Q are $n\times n$ complex matrices with Q Hermitian positive definite. $A^{*}$ is the conjugate transpose of a matrix A. This type of nonlinear matrix equations arises in many pratical problems. When $m=n=1$ and $q=r=1$, the Eq. (1.1) reduces to be a special stochastic algebraic Riccati equation arising in stochastic control theory, which can be stated as follows. Some stochastic linear quadratic (LQ)control problems lead to computing the positive definite solution of the following stochastic algebraic Riccati equation [26]:

$$\begin{aligned} \begin{array}{rcl}&{}&{}C^*XC- X+S+\Pi _1(X)-\left( L+C^*XP+\Pi _{12}(X)\right) \\ &{}&{}\left( R+P^*XP+\Pi _2(X)\right) ^+\left( L+C^*XP+\Pi _{12}(X)\right) ^*=0 \end{array} \end{aligned}$$

(1.2)

where $Z^+$ stands for the Moore-Penrose inverse of a matrix Z and C, P, S, R and L are given matrices of size $n\times n, n\times m, n\times n, m\times m$ and $n\times m$, respectively, such that

$$\begin{aligned} T=\left( \begin{array}{cc}S&{}L\\ L^*&{}R\end{array}\right) . \end{aligned}$$

is a Hermitian matrix, and the operator

$$\begin{aligned} \Pi (X)=\left( \begin{array}{cc}\Pi _1(X)&{}\Pi _{12}(X)\\ \Pi _{12}(X)^*&{}\Pi _{2}(X)\end{array}\right) \end{aligned}$$

is positive, i.e., $X\ge 0$ implies $\Pi (X)\ge 0$. Consider the following case: C is the identity matrix, P is an $n\times n$ nonsingular matrix, L is the zero matrix and $\Pi _{12}(X)=\Pi _2(X)=0,\ \Pi _1(X)=(R+P^*XP)^{-1}$, where $R+P^*XP$ is positive definite for all positive semidefinite matrices X. Meanwhile, the stochastic rational Riccati Equation (1.2) has the form

$$\begin{aligned} S+\left( R+P^*XP\right) ^{-1}-XP\left( R+P^*XP\right) ^{-1}P^*X=0 \end{aligned}$$

(1.3)

Set $Y=R+P^*XP$, then $P^{-*}(Y-R)=XP$. Thus, we have

$$\begin{aligned} S+Y^{-1}-P^{-*}\left( Y-R\right) Y^{-1}\left( Y-R\right) P^{-1}=0 \end{aligned}$$

which impies that

$$\begin{aligned} Y+R^*Y^{-1}R-P^*Y^{-1}P=2R+P^{*}SP. \end{aligned}$$

Denote by $Q=2R+P^{*}SP$, $A=R$, $B=P$, then Eq. (1.3) can be equivalently written as Eq. (1.1). Therefore, Eq. (1.1) is a special stochastic Riccati equation (1.2). When $r=1$ and $A_i=0$ for all $i=1,2,\ldots ,m$, Eq.(1.1) becomes a sepcial case of nonlinear matrix equation $X=Q+B^*(\hat{X}-C)^{-1}B$ with $C=0$ and $B=diag(B_1,\ldots ,B_n)$ which plays an important role in connection with certain optimal interpolation problems, see for detail [8, 19, 21]. Moreover, the special case when all $A_i=0\,(i=1,2,\ldots ,m)$ or all $B_j=0\,(j=1,2,\ldots ,n)$ are also problems of practical importance and have many applications in control theory, ladder networks, dynamic programming, statistics, stochastic filtering, nano research and etc., see for instance [3, 5, 8–10, 13, 15–17, 23, 25] and the references therein.

In the last few years there has been a constantly increasing interest in developing the theory, numerical solutions, and perturbation analysis for the definite solutions to the nonlinear matrix equations of the form (1.1) in several special cases. For instance, [25] considered the existence and uniqueness of positive definite solutions for the matrix equation $X+\sum _{i=1}^mA_i^*X^{-q}A_i=Q$. Duan [6] and Lim [15] proved, respectively, that the nonlinear matrix equation $X-\sum _{i=1}^mA_i^*X^{-\delta _i}A_i=Q$ always has a unique positive definite solution. Applying matrix differentiation, [8] considered perturbation analysis for the unique positive definite solution of $X-\sum _{i=1}^mA_i^*X^{-1}A_i=Q$. Moreover, positive definite solutions of nonlinear matrix equations $X\pm A^*X^{-q}A=Q$ with $0<q<1$ were extensively investigated [9, 10, 13, 16, 17, 23]. Similar nonlinear matrix equations such as $X^s\pm A^*X^{-t}A=Q$ [11, 24], $X+A^H\bar{X}^{-1}A=I$ [14] and $X^r+\sum _{i=1}^mA_i^*X^{\delta _i}A_i=I$ [22] have been investigated by many authors.

Recently, Mahar Berzig [1] considered positive definite solution to the linear matrix equation $X+\sum _{i=1}^mA_i^*XA_i-\sum _{i=1}^mB_i^*XB_i=Q$. Then [2, 7] considered the existence and uniqueness of the positive definite solution to the nonlinear matrix equation $X+A^*X^{-1}A-B^*X^{-1}B=I$ which is a special case of Eq. (1.1) with $m=n=1$, $q=r=1$ and $Q=I$. However, as far as we know, there is few literature considering the general nonlinear matrix equations $X+\sum _{i=}^mA_i^{*}X^{-q}A_i-\sum _{j=1}^nB_j^{*}X^{-r}B_j=Q$ where $0<q,r\le 1$, m, n are positive integers, and Q is positive definite.

Motivated by this, we consider in this work positive definite solution of the general case of Eq. (1.1) with m, n positive integer numbers and $0<q, r\le 1$. Based on the Lakshmikantham-Bhaskar fixed point theorem, we derive in the second section some sufficient conditions for the existence and uniqueness of positive definite solution to Eq. (1.1) and propose an iterative method to compute the unique positive definite solution. In the third section, we consider the perturbation analysis of nonlinear matrix equations of the form (1.1). An estimation bound of the unique positive definite solution which is sharp and easy to calculate is derived and an explicit expression of the Rice condition number of the unique positive definite solution is also obtained. Theoretical results are illustrated by several numerical examples in Sect. 4.

Throughout this paper, we denote by $C^{n\times n}, H^{n\times n}$ and $H^+(n)$ the sets of all $n\times n$ complex matrices, all $n\times n$ Hermitian matrices and all positive definite Hermitian matrices, respectively. The notation $A\ge 0(A>0)$ means that A is Hermitian positive semidefinite (positive definite). We denote by $\sigma _{1}(A)$ and $\sigma _{n}(A)$ the maximal and minimal singular values of A, respectively. Similarly, $\lambda _{1}(A)$ and $\lambda _{n}(A)$ stand for the maximal and the minimal eigenvalues of A, respectively. For $A, B\in H^{n\times n}$, we write $A\ge B (A>B)$ if $A-B\ge 0 (>0)$ and let

$$\begin{aligned}{}[A,B]=\{X| A\le X\le B\}, \quad [A,B)=\{X| A\le X<B\}. \end{aligned}$$

The symbol $\mathrm{tr}(A)$ denotes the trace of the matrix A and $\Vert \cdot \Vert _{\mathrm{tr}}$ denotes the trace norm which is defined by

$$\begin{aligned} \Vert A\Vert _{\mathrm{tr}}=\sum _{i=1}^{n}\sigma _{i}(A) \end{aligned}$$

where $\sigma _{i}(A), i=1,2,\ldots ,n$ are all singular values of the matrix A. It’s not difficult to verify that $\Vert \cdot \Vert _{\mathrm{tr}}$ is unitary invariant and $\Vert A\Vert _{\mathrm{tr}}=\mathrm{trace}(A)$ if A is Hermitian positive semidefinite. Unless otherwise noted, the symbol $\Vert \cdot \Vert$ stands for the spectral norm(i.e., $\Vert A\Vert =\sqrt{\rho (AA^{*})}=\sigma _{1}(A)$). It’s clear that for any positive definite matrix Q, $\Vert Q\Vert =\lambda _1(Q)$ and $\Vert Q^{-1}\Vert ^{-1}=\lambda _{n}(Q)$.

2 Positive definite solutions

In this section, we provide several sufficient conditions for Eq. (1.1) to have positive definite solutions and also we propose an iterative method for obtaining the positive definite solution.

We start with several lemmas which we need to prove our main results:

Lemma 2.1

[3] If $A>B>0$ (or $A\ge B>0$), then $A^{r}>B^{r}$ (or $A^{r}\ge B^{r}$) for all $r\in (0, 1]$, and $A^{r}<B^{r}$ (or $0<A^{r}\le B^{r}$) for all $r\in [-1, 0)$.

Lemma 2.2

[3] Let f be an operator monotone function on $(0,\infty )$ and let A, B be two positive operators bounded below by a, i.e., $A>aI$ and $B>aI$ for a positive number a. If there exists $f'(a)$, then for every unitary invariant norm $\Vert \cdot \Vert$, we have

$$\begin{aligned} \Vert f(A)-f(B)\Vert <|f'(a)|\cdot \Vert A-B\Vert . \end{aligned}$$

Lemma 2.3

[5] Let $A\ge 0$ and $B\ge 0$ be $n\times n$ matrices. Then

$$\begin{aligned} 0\le \mathrm{tr}(AB)\le \Vert A\Vert \mathrm{tr}(B). \end{aligned}$$

Lemma 2.4

[25] Suppose that $\sum _{i=1}^m\sigma _1^2(Q^{-q/2}A_iQ^{-1/2})<\frac{q^q}{(q+1)^{q+1}}$. Then the nonlinear matrix equation $X+\sum _{i=1}^mA_i^{*}X^{-q}A_i=Q$ has a positive definite solution in $[\frac{q}{q+1}Q,Q]$, where $0<q\le 1$ and Q is positive definite.

Let $(X,\le )$ be a partially ordered set and $F:X\times X \rightarrow X$ be a given mapping. We call that F has the mixed monotone property if F(x, y) is increasing in x and decreasing in y, that is,

$$\begin{aligned} F(x_1,y_1)\le F(x_2,y_2) \end{aligned}$$

for arbitrary$x_1,x_2,y_1, y_2\in X$ with $x_1\le x_2$ and $y_2\le y_1$.

We say that (x, y) is a coupled fixed point of F if $x=F(x,y)$ and $y=F(y,x)$.

The proof of our main result in this section is based on the following two fixed point theorems.

Lemma 2.5

(Schauder fixed point theorem) Let S be a nonempty, compact, convex subset of a normed vector space. Every continuous function $f:S\rightarrow S$ mapping S into itself has a fixed point.

Lemma 2.6

(Bhaskar–Laksmikantam’s coupled fixed point theorem [4]) Let $(X,\le )$ be a partially ordered set and d be a metric on X. Let the map $F:X\times X\rightarrow X$ be continuous and mixed monotone on X. Assume that there exists a $\delta \in [0,1)$ with

$$\begin{aligned} d(F(x,y),F(u,v))\le \frac{\delta }{2}[d(x,u)+d(y,v)] \end{aligned}$$

for all $x\ge u$ and $y\le v$. Suppose also that

• (1) there exist $x_{0},y_{0}\in X$ such that $x_{0}\le F(x_{0},y_{0})$ and $y_{0}\ge F(y_{0},x_{0})$;

• (2) every pair of elements in X has a lower bound and an upper bound, that is, for every $(x,y)\in (X\times X)$, there exist $z_{1}$ and $z_{2}$ such that $x,y\le z_{1}$ and $x,y\ge z_{2}$.

Then there exists a unique $\bar{x}\in X$ such that $\bar{x}=F(\bar{x},\bar{x})$. Moreover, the sequences $\{x_{k}\}$ and $\{y_{k}\}$ generated by $x_{k+1}=F(x_k,y_k)$ and $y_{k+1}=F(y_k,x_k)$ converge to $\bar{x}$, with the following estimate

$$\begin{aligned} \mathrm{max}\left\{ d(x_k,\bar{x}),d(y_k,\bar{x})\right\} \le \frac{\delta ^k}{1-\delta }\mathrm{max}\left\{ d(x_1,x_0),d(y_1,y_0)\right\} . \end{aligned}$$

Theorem 2.1

If $\sum _{i=1}^m\sigma _1^2(Q^{-q/2}A_iQ^{-1/2})<\frac{q^q}{(q+1)^{q+1}}$, then Eq. (1.1) has a positive definite solution $\bar{X}\ge \frac{q}{q+1}Q$.

Proof

Suppose $\sum _{i=1}^m\sigma _1^2(Q^{-q/2}A_iQ^{-1/2})<\frac{q^q}{(q+1)^{q+1}}$, then there exists a positive definite $\tilde{X}\in [\frac{q}{q+1}Q, Q]$ such that $\tilde{X}+\sum _{i=1}^mA_i^{*}\tilde{X}^{-q}A_i=Q$ according to Lemma 2.4. Then

$$\begin{aligned} \tilde{X}=Q-\sum _{i=1}^mA_i^{*}\tilde{X}^{-q}A_i\le Q-\sum _{i=1}^mA_i^{*}\tilde{X}^{-q}A_i+\sum _{j=1}^nB_j^{*}\tilde{X}^{-r}B_j\le Q+\sum _{j=1}^nB_j^{*}\tilde{X}^{-r}B_j. \end{aligned}$$

Denote $D=\{X| \tilde{X}\le X\le Q+\sum _{j=1}^nB_j^{*}\tilde{X}^{-r}B_j\}$. Consider the map

$$\begin{aligned} G(X)=Q-\sum _{i=1}^mA_i^{*}X^{-q}A_i+\sum _{j=1}^nB_j^{*}X^{-r}B_j, \ \ X\in D. \end{aligned}$$

Clearly, D is a bounded convex set in $H^+(n)$ and G is continuous on D. Then we have for arbitrary $X\in D$,

$$\begin{aligned} G(X)= & {} Q-\sum _{i=1}^mA_i^{*}X^{-q}A_i+\sum _{j=1}^nB_j^{*}X^{-r}B_j\ge Q-\sum _{i=1}^mA_i^{*}X^{-q}A_i\\\ge & {} Q-\sum _{i=1}^mA_i^{*}\tilde{X}^{-q}A_i=\tilde{X}, \end{aligned}$$

and

$$\begin{aligned} G(X)= & {} Q-\sum _{i=1}^mA_i^{*}X^{-q}A_i+\sum _{j=1}^nB_j^{*}X^{-r}B_j\le Q+\sum _{j=1}^nB_j^{*}X^{-r}B_j\\\le & {} Q+\sum _{j=1}^nB_j^{*}\tilde{X}^{-r}B_j. \end{aligned}$$

which implies $G(D)\subseteq D$. Applying Schauder fixed point theorem (Lemma 2.5), G has a fixed point $\bar{X}$ in D, which is a positive definite solution to Eq. (1.1) and clearly $\bar{X}\ge \tilde{X}\ge \frac{q}{q+1}Q$. $\square$

Remark 2.1

From the proof of Theorem 2.1, if $\sum _{i=1}^m\sigma _1^2(Q^{-q/2}A_iQ^{-1/2})<\frac{q^q}{(q+1)^{q+1}}$, then Eq. (1.1) has a positive definite solution $\bar{X}$ satisfying $\frac{q}{q+1}Q\le \tilde{X}\le \bar{X}\le Q+\sum _{j=1}^nB_j^{*}\tilde{X}^{-r}B_j\le Q+\frac{(q+1)^r}{q^r}\sum _{j=1}^nB_j^{*}Q^{-r}B_j$ in which the second inequality holds from Lemma 2.1 ($\tilde{X}\ge \frac{q}{q+1}Q$ gives $\tilde{X}^{-r}\le \frac{(q+1)^r}{q^r}Q^{-r}$.)

Theorem 2.2

Suppose that

$$\begin{aligned} \sum _{i=1}^m\Vert A_i^*A_i\Vert <\frac{1}{2}\frac{q^q}{\Vert (q+1)Q^{-1}\Vert ^{q+1}},\ \ \sum _{j=1}^n\Vert B_j^*B_j\Vert <\frac{1}{2r}\frac{q^{r+1}}{\Vert (q+1)Q^{-1}\Vert ^{r+1}}. \end{aligned}$$

(2.1)

Then Eq. (1.1) has a unique positive definite solution $\bar{X}\in [\frac{q}{q+1}Q, +\infty )$ and the sequences $\{X_k\}$ and $Y_k$ defined by

$$\begin{aligned} \left\{ \begin{array}{l} X_0=\frac{q}{q+1}Q,\ \ Y_{0}=\frac{2r(q+1)+q}{2r(q+1)}Q, \\ \\ X_{k+1}=Q-\sum _{i=1}^mA_i^{*}X_{k}^{-q}A_i+\sum _{j=1}^nB_j^*Y_k^{-r}B_j,\\ \\ Y_{k+1}=Q-\sum _{i=1}^mA_i^{*}Y_{k}^{-q}A_i+\sum _{j=1}^nB_j^*X_k^{-r}B_j, \ \ k=0,1,2,\ldots \end{array}\right. \end{aligned}$$

(2.2)

converges to $\bar{X}$ and the error estimation is given by

$$\begin{aligned} \mathrm{max}\{\Vert X_k-X\Vert _{\mathrm{tr}}, \ \Vert Y_k-Y\Vert _{\mathrm{tr}}\}\le \frac{\delta ^k}{1-\delta }\mathrm{max}\left\{ \Vert X_1-X_0\Vert _{\mathrm{tr}}, \ \Vert Y_1-Y_0\Vert _{\mathrm{tr}}\right\} , \end{aligned}$$

(2.3)

where

$$\begin{aligned} \delta =2\cdot \mathrm{max}\left\{ \frac{\Vert (q+1)Q^{-1}\Vert ^{q+1}}{q^q}\sum _{i=1}^m\Vert A_i^{*}A_i\Vert ,\ \ \frac{r\cdot \Vert (q+1)Q^{-1}\Vert ^{r+1}}{q^{r+1}}\sum _{j=1}^n\Vert B_j^{*}B_j\Vert \right\} . \end{aligned}$$

Proof

Consider the following two-variable map

$$\begin{aligned} F(X,Y)=Q-\sum _{i=1}^mA_i^*X^{-q}A_i+\sum _{j=1}^nB_j^{*}Y^{-r}B_j, \ \ X,Y\in \Omega , \end{aligned}$$

where $\Omega =\{Z\in H^+(n): Z\ge \frac{q}{q+1}Q\}$. Obviously, F is continuous on $\Omega \times \Omega$.

We divide our proof into five aspects:

• (1) For arbitrary $X,Y\in \Omega$, combining (2.1) with lemma 2.1, we obtain that

  $$\begin{aligned} F(X,Y)\ge & {} Q-\sum _{i=1}^mA_i^*X^{-q}A_i\ge Q-\frac{(q+1)^q}{q^q}\sum _{i=1}^mA_i^*Q^{-q}A_i\\\ge & {} Q-\frac{1}{2}\frac{1}{q+1}Q\ge \frac{q}{q+1}Q, \end{aligned}$$

  in which the third inequality holds since

  $$\begin{aligned} \frac{(q+1)^q}{q^q}\sum _{i=1}^mA_i^*Q^{-q}A_i\le & {} \frac{(q+1)^q}{q^q}\Vert Q^{-1}\Vert ^{q}\sum _{i=1}^mA_i^*A_i \le \frac{1}{2}\frac{1}{(q+1)\Vert Q^{-1}\Vert }I\\= & {} \frac{1}{2}\frac{1}{(q+1)}\lambda _n(Q)I\le \frac{1}{2}\frac{1}{q+1}Q. \end{aligned}$$

  Thus, $F(X,Y)\in \Omega$ which implies that $F: \Omega \times \Omega \rightarrow \Omega$.

• (2) For arbitrary $X_1,X_2,Y_1,Y_2\in \Omega$ with $X_1\le X_2$ and $Y_1\ge Y_2$, we have

  $$\begin{aligned} F(X_1,Y_1)= & {} Q-\sum _{i=1}^mA_i^{*}X_1^{-q}A_i+\sum _{j=1}^nB_j^*Y_1^{-r}B_j\le Q\\&-\sum _{i=1}^mA_i^{*}X_2^{-q}A_i+\sum _{j=1}^nB_j^*Y_2^{-r}B_j=F\left( X_2,Y_2\right) \end{aligned}$$

  which implies that F is mixed monotone on $\Omega$.

• (3) For arbitrary $X,Y,U,V\in \Omega$ with $X\ge U\ge \frac{q}{q+1}Q\ge \frac{q}{q+1}\lambda _{n}(Q)I$ and $\frac{q}{q+1}\lambda _{n}(Q)I\le \frac{q}{q+1}Q\le Y\le V$, we have by Lemma 2.3 that

  $$\begin{aligned} \Vert F(X,Y)-F(U,V)\Vert _{\mathrm{tr}}= & {} \Vert \sum _{i=1}^mA_i^{*}\left( U^{-q}-X^{-q}\right) A_i\\&+\sum _{j=1}^nB_j^{*}\left( Y^{-r}-V^{-r}\right) B_j\Vert _{\mathrm{tr}}\\\le & {} \Vert \sum _{i=1}^mA_i^{*}\left( U^{-q}-X^{-q}\right) A_i\Vert _{\mathrm{tr}}\\&+\Vert \sum _{j=1}^nB_j^{*}\left( Y^{-r}-V^{-r}\right) B_j\Vert _{\mathrm{tr}}\\= & {} \sum _{i=1}^m\mathrm{tr}\left( A_i^*\left( U^{-q}-X^{-q}\right) A_i\right) \\&+\sum _{j=1}^n\mathrm{tr}\left( B_j^*\left( Y^{-r}-V^{-r}\right) B_j\right) \\\le & {} \sum _{i=1}^m\Vert A_i^{*}A_i\Vert \cdot \Vert U^{-q}-X^{-q}\Vert _{\mathrm{tr}}\\&+\sum _{j=1}^n\Vert B_j^{*}B_j\Vert \cdot \Vert Y^{-r}-V^{-r}\Vert _{\mathrm{tr}}. \end{aligned}$$

  Using Lemma 2.2, we obtain that

  $$\begin{aligned} \Vert U^{-q}-X^{-q}\Vert _{\mathrm{tr}}\le & {} q\cdot \frac{(q+1)^{q+1}}{q^{q+1}}\cdot \frac{1}{\lambda _{n}^{q+1}(Q)}\Vert U-X\Vert _{\mathrm{tr}}\\= & {} \frac{(q+1)^{q+1}}{q^q}\cdot \frac{1}{\lambda _{n}^{q+1}(Q)}\Vert U-X\Vert _{\mathrm{tr}}, \end{aligned}$$

  Similarly,

  $$\begin{aligned} \Vert Y^{-r}-V^{-r}\Vert _{\mathrm{tr}}\le r\cdot \frac{(q+1)^{r+1}}{q^{r+1}}\cdot \frac{1}{\lambda _{n}^{r+1}(Q)}\Vert Y-V\Vert _{\mathrm{tr}}, \end{aligned}$$

  It is clear that $\Vert Q^{-1}\Vert =\sigma _1(Q^{-1})=\lambda _1(Q^{-1})=\frac{1}{\lambda _n(Q)}$ since Q is positive definite. Then we have

  $$\begin{aligned} \begin{array}{rcl} \Vert F(X,Y)-F(U,V)\Vert _{\mathrm{tr}}&{}\le &{}\frac{(q+1)^{q+1}}{q^q}\cdot \frac{1}{\lambda _{n}^{q+1}(Q)}\cdot \sum _{i=1}^m\Vert A_i^{*}A_i\Vert \cdot \Vert U-X\Vert _{\mathrm{tr}}\\ \\ &{}&{}+r\cdot \frac{(q+1)^{r+1}}{q^{r+1}}\cdot \frac{1}{\lambda _{n}^{r+1}(Q)}\cdot \sum _{j=1}^n\Vert B_j^{*}B_j\Vert \cdot \Vert Y-V\Vert _{\mathrm{tr}}\\ \\ &{}=&{}\frac{\Vert (q+1)Q^{-1}\Vert ^{q+1}}{q^q}\sum _{i=1}^m\Vert A_i^{*}A_i\Vert \cdot \Vert U-X\Vert _{\mathrm{tr}}\\ \\ &{}&{}+r\cdot \frac{\Vert (q+1)Q^{-1}\Vert ^{r+1}}{q^{r+1}}\sum _{j=1}^n\Vert B_j^{*}B_j\Vert \cdot \Vert Y-V\Vert _{\mathrm{tr}}\\ \\ &{}\le &{}\delta \cdot \frac{1}{2}\cdot (\Vert U-X\Vert _{\mathrm{tr}}+\Vert Y-V\Vert _{\mathrm{tr}})\end{array} \end{aligned}$$

  where

  $$\begin{aligned}&0<\delta =2\cdot \mathrm{max}\{\frac{\Vert (q+1)Q^{-1}\Vert ^{q+1}}{q^q}\\&\quad \sum _{i=1}^m\Vert A_i^{*}A_i\Vert ,\ \ \frac{r\cdot \Vert (q+1)Q^{-1}\Vert ^{r+1}}{q^{r+1}}\sum _{j=1}^n\Vert B_j^{*}B_j\Vert \}<1. \end{aligned}$$

• (4) Let $X_0=\frac{q}{q+1}Q,\ Y_0=\frac{2r(q+1)+q}{2r(q+1)}Q$. Then we have by Lemma 2.1 that

  $$\begin{aligned} F(X_0,\ Y_0)= & {} Q-\sum _{i=1}^mA_i^*X_0^{-q}A_i+\sum _{j=1}^nB_j^*Y_0^{-r}B_j\\ \\= & {} Q-\frac{(q+1)^q}{q^q}\sum _{i=1}^mA_i^*Q^{-q}A_i+\sum _{j=1}^nB_j^*Y_0^{-r}B_j\\ \\\ge & {} Q-\frac{(q+1)^q}{q^q}\Vert Q^{-q}\Vert \sum _{i=1}^m \Vert A_i^*A_i\Vert \cdot I\\ \\\ge & {} Q-\frac{(q+1)^q}{q^q}\Vert Q^{-1}\Vert ^q\cdot \frac{1}{2}\frac{q^q}{\Vert (q+1)Q^{-1}\Vert ^{q+1}}I\\ \\= & {} Q-\frac{1}{2(q+1)}\lambda _{n}(Q)I\\ \\\ge & {} Q-\frac{1}{q+1}\lambda _{n}(Q)I\\ \\\ge & {} Q-\frac{1}{q+1}Q\\ \\= & {} \frac{q}{q+1}Q=X_0, \end{aligned}$$

  and similarly,

  $$\begin{aligned} F(Y_0,X_0)= & {} Q-\sum _{i=1}^mA_i^*Y_0^{-q}A_i+\sum _{j=1}^nB_j^*X_0^{-r}B_j\\ \\\le & {} Q+\frac{(q+1)^r}{q^r}\sum _{j=1}^nB_j^*Q^{-r}B_j\\ \\\le & {} Q+\frac{(q+1)^r}{q^r}\cdot \Vert Q^{-1}\Vert ^r\sum _{j=1}^n\Vert B_j^*B_j\Vert \cdot I\\ \\\le & {} Q+\frac{(q+1)^r}{q^r}\Vert Q^{-1}\Vert ^r\cdot \frac{q^{r+1}}{2r(q+1)^{r+1}\Vert Q^{-1}\Vert ^{r+1}}I\\ \\= & {} Q+\frac{1}{2r}\frac{q}{q+1}\lambda _{n}(Q)I\\ \\\le & {} \frac{2r(q+1)+q}{2r(q+1)}Q\\ \\= & {} Y_0. \end{aligned}$$

• (5) For arbitrary pair $X,Y\in H^+(n)$, it’s well known that

  $$\begin{aligned} \mathrm{min}\{\lambda _{n}(X),\ \ \lambda _{n}(Y)\}I\le X,Y\le \mathrm{max}\{\lambda _1(X),\ \lambda _1(Y)\}I \end{aligned}$$

  which means that each pair has both lower bound and upper bound.

$\square$

Combining the above 5 aspects with Lemma 2.6, there exists a unique $\bar{X}\in \Omega$ such that $\bar{X}=F(\bar{X}, \ \bar{X})$. Moreover, the sequences $\{X_k\}$ and $Y_k$ generated by (2.2) converge to $\bar{X}$, with the estimation (2.3).

Remark 2.2

In case $m=n=1$, $q=r=1$ and $Q=I$ in Theorem 2.2, we obtain Theorem 2.3 in [7].

3 Perturbation analysis

Consider the perturbed matrix equation

$$\begin{aligned} \tilde{X}+\sum _{i=1}^m\tilde{A}_i^{*}\tilde{X}^{-q}\tilde{A}_i -\sum _{j=1}^n\tilde{B}_j^{*}\tilde{X}^{-r}\tilde{B}_j=\tilde{Q} \end{aligned}$$

(3.1)

where $\tilde{A}_i, i=1,2,\ldots ,m; \tilde{B}_j,j=1,\ldots ,n$ and $\tilde{Q}$ are the slightly perturbed matrices of the matrices $A_i, i=1,\ldots ,m; B_j,j=1,\ldots ,n$ and Q, respectively. In this section, we show that if $\Vert \tilde{A}_i-A_i\Vert ,i=1,\ldots ,m;\ \Vert \tilde{B}_j-B_j\Vert ,j=1,\ldots ,n.$ and $\Vert \tilde{Q}-Q\Vert$ are sufficiently small, then the unique positive definite solution $\tilde{X}$ to the perturbed matrix equation (3.1) exists. We derive a perturbation estimate for the unique positive definite solution X and give an explicit expression of the Rice condition number of X.

Denote $\Delta Q=\tilde{Q}-Q, \ \Delta A_i=\tilde{A}_i-A_i, i=1,2,\ldots ,m; \ \Delta B_j=\tilde{B}_j-B_j,j=1,2,\ldots ,n,\ \Delta X=\tilde{X}-X.$

Theorem 3.1

Let

$$\begin{aligned}&\mathrm{(i)}\ \ \theta _1:=\frac{q^{q}}{(q+1)^{q+1}} -2\sum _{i=1}^m\Vert A_i\Vert ^{2}\Vert Q^{-1}\Vert ^{q+1}>0,\ \ \theta _2:=\frac{q^{r+1}}{r(q+1)^{r+1}}\nonumber \\&\quad -2\sum _{j=1}^n\Vert B_j\Vert ^{2}\Vert Q^{-1}\Vert ^{r+1}>0, \end{aligned}$$

(3.2)

$$\begin{aligned}&\mathrm{(ii)}\ \ \Vert \Delta Q\Vert \le \frac{1}{\Vert Q^{-1}\Vert }\cdot \min \left\{ 1-\root q+1 \of {1-\theta _1}, \ 1-\root r+1 \of {1-\theta _2}\right\} , \end{aligned}$$

(3.3)

$$\begin{aligned}&\mathrm{(iii)}\ \ \sum _{i=1}^m\left( \Vert \tilde{A}_i\Vert ^{2}-\Vert A_i\Vert ^{2}\right) < \frac{(q+1)^{q+1}-q^{q}}{2(q+1)^{q+1}\Vert Q^{-1}\Vert ^{q+1}} \theta _1,\ \sum _{j=1}^n\left( \Vert \tilde{B}_j\Vert ^{2}-\Vert B_j\Vert ^{2}\right) \nonumber \\&\quad < \frac{(q+1)^{r+1}-q^{r+1}/r}{2(q+1)^{r+1}\Vert Q^{-1}\Vert ^{r+1}} \theta _2. \end{aligned}$$

(3.4)

Then nonlinear matrix equations $X+\sum _{i=1}^mA_i^{*}X^{-q}A_i-\sum _{j=1}^nB_j^*X^{-r}B_j=Q$ and $\tilde{X}+\sum _{i=1}^m\tilde{A}_i^{*}\tilde{X}^{-q}\tilde{A}_i-\sum _{j=1}^n\tilde{B}_j^{*}\tilde{X}^{-r}\tilde{B}_j=\tilde{Q}$ have unique positive definite solutions X and $\tilde{X},$ respectively. Moreover, we have

$$\begin{aligned} \begin{array}{rcl} \Vert \Delta X\Vert &{}\le &{}\frac{1}{\xi } \cdot \left[ \Vert \Delta Q\Vert +2\sum _{i=1}^m \left( \Vert X^{-q}A_i\Vert \cdot \Vert \Delta A_i\Vert +\Vert X^{-q}\Vert \cdot \Vert \Delta A_i\Vert ^{2}\right) \right. \\ \\ &{}&{}+\left. 2\sum _{j=1}^n\left( \Vert X^{-r}B_j\Vert \cdot \Vert \Delta B_j\Vert +\Vert X^{-r}\Vert \cdot \Vert \Delta B_j\Vert ^{2}\right) \right] ,\end{array} \end{aligned}$$

(3.5)

where

$$\begin{aligned} \xi =1-qb^{-(q+1)}\sum _{i=1}^m\Vert \tilde{A}_i\Vert ^{2}-rb^{-(r+1)}\sum _{j=1}^n\Vert \tilde{B}_j\Vert ^{2},\ \ b=\frac{q}{q+1}\mathrm{min}\left\{ \lambda _{n}(Q),\lambda _{n}(\tilde{Q})\right\} . \end{aligned}$$

Proof

Since $\theta _1>0, \ \theta _2>0$, we know from Theorem 2.2 that Eq. (1.1) has unique positive definite solution $X\ge \frac{q}{q+1}Q$. Notice that $\theta _1<1$ and

$$\begin{aligned} \Vert \tilde{Q}^{-1}\Vert \le \Vert Q^{-1}\Vert +\Vert Q^{-1}\Vert \cdot \Vert \Delta Q\Vert \cdot \Vert \tilde{Q}^{-1}\Vert \le \Vert Q^{-1}\Vert +\left( 1-\root q+1 \of {1-\theta _1}\right) \Vert \tilde{Q}^{-1}\Vert , \end{aligned}$$

which gives

$$\begin{aligned} \Vert \tilde{Q}^{-1}\Vert ^{q+1}\le \frac{\Vert Q^{-1}\Vert ^{q+1}}{1-\theta _1}. \end{aligned}$$

Similarly, $\Vert \tilde{Q}^{-1}\Vert ^{r+1}\le \frac{\Vert Q^{-1}\Vert ^{r+1}}{1-\theta _2}$.

Consequently, we have

$$\begin{aligned} \begin{aligned}&\sum _{i=1}^m\Vert \tilde{A}_i\Vert ^{2}\Vert \tilde{Q}^{-1}\Vert ^{q+1} <\frac{\Vert Q^{-1}\Vert ^{q+1}}{1-\theta _1}\left[ \sum _{i=1}^m\Vert A_i\Vert ^{2}+ \frac{(q+1)^{q+1}-q^{q}}{2(q+1)^{q+1}\Vert Q^{-1}\Vert ^{q+1}}\theta _1\right] \\&\quad =\frac{2(q+1)^{q+1}\sum _{i=1}^m\Vert A_i\Vert ^{2}\Vert Q^{-1}\Vert ^{q+1}+[(q+1)^{q+1}-q^{q}]\theta _1}{2(1-\theta _1)(q+1)^{q+1}}\\&\quad =\frac{q^{q}}{2(q+1)^{q+1}}\cdot \frac{2\sum _{i=1}^m\Vert A_i\Vert ^{2}\frac{(q+1)^{q+1}}{q^{q}}\Vert Q^{-1}\Vert ^{q+1} +[\frac{(q+1)^{q+1}}{q^{q}}-1]\theta _1}{1-\theta _1}\\&\quad =\frac{q^{q}}{2(q+1)^{q+1}},\end{aligned} \end{aligned}$$

(3.6)

and similarly

$$\begin{aligned} \begin{aligned}&\sum _{j=1}^n\Vert \tilde{B}_j\Vert ^{2}\Vert \tilde{Q}^{-1}\Vert ^{r+1} <\frac{\Vert Q^{-1}\Vert ^{r+1}}{1-\theta _2}\left[ \sum _{j=1}^n\Vert B_j\Vert ^{2}+ \frac{(q+1)^{r+1}-q^{r+1}/r}{2(q+1)^{r+1}\Vert Q^{-1}\Vert ^{r+1}}\theta _2\right] \\&\quad =\frac{2(q+1)^{r+1}\sum _{j=1}^n\Vert B_j\Vert ^{2}\Vert Q^{-1}\Vert ^{r+1}+[(q+1)^{r+1}-q^{r+1}/r]\theta _2}{2(1-\theta _2)(q+1)^{r+1}}\\&\quad =\frac{q^{r+1}}{2r(q+1)^{r+1}}\cdot \frac{2\sum _{j=1}^n\Vert B_j\Vert ^{2}\cdot \frac{r(q+1)^{r+1}}{q^{r+1}}\Vert Q^{-1}\Vert ^{r+1} +[\frac{r(q+1)^{r+1}}{q^{r+1}}-1]\theta _2}{1-\theta _2}\\&\quad =\frac{q^{r+1}}{2r(q+1)^{r+1}},\end{aligned} \end{aligned}$$

(3.7)

Applying Theorem 2.2, we obtain that the perturbed matrix equation (3.1) has unique positive definite solution $\tilde{X}>\frac{q}{q+1}\tilde{Q}$.

In the following, we show the estimate (3.5):

Since $X >\frac{q}{q+1}\lambda _{n}(Q)I$ and $\tilde{X}>\frac{q}{q+1}\lambda _{n}(\tilde{Q})I$. Let $b=\frac{q}{q+1}\mathrm{min}\{\lambda _{n}(Q), \lambda _{n}(\tilde{Q})\}$. Then $X,\tilde{X}>bI$ and consequently,

$$\begin{aligned} \Vert X^{-q}-\tilde{X}^{-q}\Vert \le qb^{-(q+1)}\Vert \Delta X\Vert ,\ \text{ and } \Vert X^{-r}-\tilde{X}^{-r}\Vert \le rb^{-(r+1)}\Vert \Delta X\Vert \end{aligned}$$

from Lemma 2.2.

Since $X+\sum _{i=1}^mA_i^{*}X^{-q}A_i-\sum _{j=1}^nB_j^*X^{-r}B_j=Q$ and $\tilde{X}+\sum _{i=1}^m\tilde{A}_i^{*}\tilde{X}^{-q}\tilde{A}_i-\sum _{j=1}^n\tilde{B}_j^*\tilde{X}^{-r}\tilde{B}_j=\tilde{Q}$, then

$$\begin{aligned} \tilde{X}-X+\sum _{i=1}^m\tilde{A}_i^{*}\tilde{X}^{-q}\tilde{A}_i-\sum _{i=1}^mA_i^*X^{-q}A_i -\sum _{j=1}^n\tilde{B}_j^{*}\tilde{X}^{-r}\tilde{B}_j+\sum _{j=1}^nB_j^*X^{-r}B_j=\tilde{Q}-Q, \end{aligned}$$

i.e.,

$$\begin{aligned} \Delta X= & {} \Delta Q+\sum _{i=1}^m\tilde{A}_i^{*}\left( X^{-q}-\tilde{X}^{-q}\right) \tilde{A}_i +\sum _{i=1}^m\left[ A_i^{*}X^{-q}A_i-\tilde{A}_i^{*}X^{-q}\tilde{A}_i\right] \\&-\sum _{j=1}^n\tilde{B}_j^{*}\left( X^{-r}-\tilde{X}^{-r}\right) \tilde{B}_j +\sum _{j=1}^n\left[ -B_j^{*}X^{-r}B_j+\tilde{B}_j^{*}X^{-r}\tilde{B}_j\right] \\= & {} \Delta Q+\sum _{i=1}^m\tilde{A}_i^{*}\left( X^{-q}-\tilde{X}^{-q}\right) \tilde{A}_i\\&-\sum _{i=1}^m\left[ \Delta A_i^{*}X^{-q}A_i+\Delta A_i^{*}X^{-q}\Delta A_i+A_i^{*}X^{-q}\Delta A_i\right] \\&+\sum _{j=1}^n\tilde{B}_j^{*}\left( \tilde{X}^{-r}-X^{-r}\right) \tilde{B}_j\\&+\sum _{j=1}^n\left[ \Delta B_j^{*}X^{-r}B_j+\Delta B_j^{*}X^{-r}\Delta B_j+B_j^{*}X^{-r}\Delta B_j\right] . \end{aligned}$$

Hence

$$\begin{aligned} \Vert \Delta X\Vert\le & {} \Vert \Delta Q\Vert +\Vert X^{-q}-\tilde{X}^{-q}\Vert \sum _{i=1}^m\Vert \tilde{A}_i\Vert ^{2}+\sum _{i=1}^m \left( \Vert \Delta A_i^{*}X^{-q}\Delta A_i\Vert +2\Vert \Delta A_i^{*}X^{-q}A_i\Vert \right) \\&+\Vert X^{-r}-\tilde{X}^{-r}\Vert \sum _{j=1}^n\Vert \tilde{B}_j\Vert ^{2}+ \sum _{j=1}^n\left( \Vert \Delta B_j^{*}X^{-r}\Delta B_j\Vert +2\Vert \Delta B_j^{*}X^{-r}B_j\Vert \right) \\\le & {} \Vert \Delta Q\Vert +qb^{-(q+1)}\Vert \Delta X\Vert \sum _{i=1}^m\Vert \tilde{A}_i\Vert ^{2}\\&+\sum _{i=1}^m\left( 2\Vert X^{-q}A_i\Vert \cdot \Vert \Delta A_i\Vert +\Vert X^{-q}\Vert \cdot \Vert \Delta A_i\Vert ^{2}\right) \\&+rb^{-(r+1)}\Vert \Delta X\Vert \sum _{j=1}^n\Vert \tilde{B}_j\Vert ^{2}+\sum _{j=1}^n\left( 2\Vert X^{-r}B_j\Vert \cdot \Vert \Delta B_j\Vert +\Vert X^{-r}\Vert \cdot \Vert \Delta B_j\Vert ^{2}\right) . \end{aligned}$$

Denote $\xi =1-qb^{-(q+1)}\sum _{i=1}^m\Vert \tilde{A}_i\Vert ^{2}-rb^{-(r+1)}\sum _{j=1}^n\Vert \tilde{B}_j\Vert ^{2}$. Notice from the proof of (3.6) and (3.7) that

$$\begin{aligned} \Vert Q^{-1}\Vert ^{q+1}\cdot \sum _{i=1}^m\Vert \tilde{A}_i\Vert ^{2}< & {} \frac{\Vert Q^{-1}\Vert ^{q+1}}{1-\theta _1}\left[ \sum _{i=1}^m\Vert A_i\Vert ^{2}+ \frac{(q+1)^{q+1}-q^{q}}{2(q+1)^{q+1}\Vert Q^{-1}\Vert ^{q+1}}\theta _1\right] \\= & {} \frac{q^{q}}{2(q+1)^{q+1}} \end{aligned}$$

and

$$\begin{aligned} \Vert Q^{-1}\Vert ^{r+1}\cdot \sum _{j=1}^n\Vert \tilde{B}_j\Vert ^{2}< & {} \frac{\Vert Q^{-1}\Vert ^{r+1}}{1-\theta _2}\left[ \sum _{j=1}^n\Vert B_j\Vert ^{2}+ \frac{(q+1)^{r+1}-q^{r+1}/r}{2(q+1)^{r+1}\Vert Q^{-1}\Vert ^{r+1}}\theta _2\right] \\= & {} \frac{q^{r+1}}{2r(q+1)^{r+1}}. \end{aligned}$$

which implies that if $b=\frac{q}{q+1}\lambda _n(Q)=\frac{q}{q+1}\frac{1}{\Vert Q^{-1}\Vert }$, then

$$\begin{aligned} \begin{array}{rcl}\xi &{}=&{}1-qb^{-(q+1)}\sum _{i=1}^m\Vert \tilde{A}_i\Vert ^{2} -rb^{-(r+1)}\sum _{j=1}^n\Vert \tilde{B}_j\Vert ^{2}\\ \\ &{}=&{}1-q\frac{(q+1)^{q+1}}{q^{q+1}}\Vert Q^{-1}\Vert ^{q+1}\sum _{i=1}^m\Vert \tilde{A}_i\Vert ^{2} -r\frac{(q+1)^{r+1}}{q^{r+1}}\Vert Q^{-1}\Vert ^{r+1}\sum _{j=1}^n\Vert \tilde{B}_j\Vert ^{2}\\ \\ &{}>&{} 1-q\frac{(q+1)^{q+1}}{q^{q+1}}\frac{q^{q}}{2(q+1)^{q+1}} -r\frac{(q+1)^{r+1}}{q^{r+1}}\cdot \frac{1}{2r}\cdot \frac{q^{r+1}}{(q+1)^{r+1}}=0,\end{array} \end{aligned}$$

and if $b=\frac{q}{q+1}\lambda _n(\tilde{Q})=\frac{q}{q+1}\frac{1}{\Vert \tilde{Q}^{-1}\Vert }$, then

$$\begin{aligned} \begin{array}{rcl}\xi &{}=&{}1-qb^{-(q+1)}\sum _{i=1}^m\Vert \tilde{A}_i\Vert ^{2} -rb^{-(r+1)}\sum _{j=1}^n\Vert \tilde{B}_j\Vert ^{2}\\ \\ &{}=&{}1-q\frac{(q+1)^{q+1}}{q^{q+1}}\Vert \tilde{Q}^{-1}\Vert ^{q+1}\sum _{i=1}^m\Vert \tilde{A}_i\Vert ^{2} -r\frac{(q+1)^{r+1}}{q^{r+1}}\Vert \tilde{Q}^{-1}\Vert ^{r+1}\sum _{j=1}^n\Vert \tilde{B}_j\Vert ^{2}\\ \\ &{}>&{} 1-q\frac{(q+1)^{q+1}}{q^{q+1}}\frac{q^{q}}{2(q+1)^{q+1}} -r\frac{(q+1)^{r+1}}{q^{r+1}}\cdot \frac{1}{2r}\cdot \frac{q^{r+1}}{(q+1)^{r+1}}\\ \\ &{}=&{}0.\end{array} \end{aligned}$$

Thus we obtain that

$$\begin{aligned} \Vert \Delta X\Vert\le & {} \frac{1}{\xi } \cdot \left[ \Vert \Delta Q\Vert +\sum _{i=1}^m\left( 2\Vert X^{-q}A_i\Vert \cdot \Vert \Delta A_i\Vert +\Vert X^{-q}\Vert \cdot \Vert \Delta A_i\Vert ^{2}\right) \right. \\&+\left. \sum _{j=1}^n \left( 2\Vert X^{-r}B_j\Vert \cdot \Vert \Delta B_j\Vert +\Vert X^{-r}\Vert \cdot \Vert \Delta B_j\Vert ^{2}\right) \right] . \end{aligned}$$

$\square$

By the theory of condition number developed by Rice [20], we give in the following an explicit expression of the condition number of the unique positive definite solution X in case $0<q, r<1$.

3.1 The complex case

Lemma 3.1

[13] Let X be any $n\times n$ positive definite matrix, $0<q<1$. Then

$$\begin{aligned} \mathrm{(i)}\ \ X^{-q}= & {} \frac{\sin \mathrm{q}\pi }{\pi }\int _{0}^{\infty }(\lambda I+X)^{-1}\lambda ^{-q}d\lambda ,\\ \mathrm{(ii)}\ \ X^{-q}= & {} \frac{\sin q\pi }{q\pi }\int _{0}^{\infty } (\lambda I+X)^{-1}X(\lambda I+X)^{-1}\lambda ^{-q}d\lambda . \end{aligned}$$

From Theorem 3.1, we see that if $\Vert \Delta A_i\Vert , i=1,2,\ldots ,m,$ $\Vert \Delta B_j\Vert , j=1,2,\ldots ,n$ and $\Vert \Delta Q\Vert$ are sufficiently small, then the perturbed matrix equation (3.1) has a unique positive definite solution $\tilde{X}$. Subtracting (1.1) from (3.1) gives rise to

$$\begin{aligned} \Delta X+\sum _{i=1}^m\left( \tilde{A}_i^{*}\tilde{X}^{-q}\tilde{A}_i-A_i^{*}X^{-q}A_i\right) +\sum _{j=1}^n\left( -\tilde{B}_j^{*}\tilde{X}^{-r}\tilde{B}_j+B_j^{*}X^{-r}B_j\right) =\Delta Q, \end{aligned}$$

$\mathrm{i.e.,}$

$$\begin{aligned}&\Delta X+\sum _{i=1}^m\left[ A_i^{*}\left( \tilde{X}^{-q}-X^{-q}\right) A_i+\tilde{A}_i^{*}\left( \tilde{X}^{-q}-X^{-q}\right) \Delta A_i+\Delta A_i^{*}\left( \tilde{X}^{-q}-X^{-q}\right) A_i\right] \nonumber \\&\quad \quad +\sum _{j=1}^n \left[ -B_j^{*}\left( \tilde{X}^{-r}-X^{-r}\right) B_j-\tilde{B}_j^{*}\left( \tilde{X}^{-r}-X^{-r}\right) \Delta B_j-\Delta B_j^{*}\left( \tilde{X}^{-r}-X^{-r}\right) B_j\right] \nonumber \\&\quad =\Delta Q-\sum _{i=1}^m\left( \Delta A_i^{*}X^{-q}A_i+\Delta A_i^{*}X^{-q}\Delta A_i+A_i^{*}X^{-q}\Delta A_i\right) \nonumber \\&\quad \quad +\sum _{j=1}^n\left( \Delta B_j^{*}X^{-r}B_j+\Delta B_j^{*}X^{-r}\Delta B_j+B_j^{*}X^{-r}\Delta B_j\right) , \end{aligned}$$

(3.8)

Notice that

$$\begin{aligned} \left( \lambda I+X+\Delta X\right) ^{-1}-(\lambda I+X)^{-1}= & {} - \left( \lambda I+X+\Delta X\right) ^{-1}\Delta X (\lambda I+X)^{-1}\\= & {} -(\lambda I+X)^{-1}\Delta X\left( \lambda I+X+\Delta X\right) ^{-1}. \end{aligned}$$

Using Lemma 3.1, we have

$$\begin{aligned}&\begin{array}{rcl}\tilde{X}^{-q}-X^{-q}&{}=&{}\frac{\sin q\pi }{\pi }\int _{0}^{\infty } [(\lambda I+X+\Delta X)^{-1}-(\lambda I+X)^{-1}]\lambda ^{-q}d\lambda \\ &{}=&{}\frac{\sin q\pi }{\pi }\int _{0}^{\infty }-(\lambda I+X)^{-1}\Delta X(\lambda I+X+\Delta X)^{-1}\lambda ^{-q}d\lambda \\ &{}=&{}\frac{\sin q\pi }{\pi }\int _{0}^{\infty }-(\lambda I+X)^{-1}\Delta X(\lambda I+X)^{-1}\lambda ^{-q}d\lambda \\ &{}&{}+\frac{\sin q\pi }{\pi }\int _{0}^{\infty } (\lambda I+X)^{-1}\Delta X(\lambda I+X+\Delta X)^{-1}\Delta X(\lambda I+X)^{-1}\lambda ^{-q}d\lambda \end{array},\nonumber \\&\begin{array}{rcl} \tilde{X}^{-r}-X^{-r}&{}=&{}\frac{\sin r\pi }{\pi }\int _{0}^{\infty } [(\lambda I+X+\Delta X)^{-1}-(\lambda I+X)^{-1}]\lambda ^{-r}d\lambda \\ \\ &{}=&{}\frac{\sin r\pi }{\pi }\int _{0}^{\infty }-(\lambda I+X)^{-1}\Delta X(\lambda I+X+\Delta X)^{-1}\lambda ^{-r}d\lambda \\ \\ &{}=&{}\frac{\sin r\pi }{\pi }\int _{0}^{\infty }-(\lambda I+X)^{-1}\Delta X(\lambda I+X)^{-1}\lambda ^{-r}d\lambda \\ &{}&{}+\frac{\sin r\pi }{\pi }\int _{0}^{\infty } (\lambda I+X)^{-1}\Delta X(\lambda I+X+\Delta X)^{-1}\Delta X(\lambda I+X)^{-1}\lambda ^{-r}d\lambda \end{array}. \end{aligned}$$

(3.9)

Combining (3.9) with (3.8), we obtain that

$$\begin{aligned} \begin{array}{rcl} &{}&{}\Delta X-\frac{\sin q\pi }{\pi }\sum _{i=1}^m\int _{0}^{\infty }A_i^{*} (\lambda I+X)^{-1}\Delta X(\lambda I+X)^{-1}\lambda ^{-q}A_id\lambda \\ \\ &{}&{}\quad \quad +\frac{\sin r\pi }{\pi }\sum _{j=1}^n\int _{0}^{\infty }B_j^{*} (\lambda I+X)^{-1}\Delta X(\lambda I+X)^{-1}\lambda ^{-r}B_jd\lambda \\ \\ &{}&{}\quad =E+h(\Delta X),\end{array} \end{aligned}$$

(3.10)

where

$$\begin{aligned} E= & {} \Delta Q-\sum _{i=1}^m\left( \Delta A_i^{*}X^{-q}A_i+\Delta A_i^{*}X^{-q}\Delta A_i+A_i^{*}X^{-q}\Delta A_i\right) \\&+\sum _{j=1}^n\left( \Delta B_j^{*}X^{-r}B_j+\Delta B_j^{*}X^{-r}\Delta B_j+B_j^{*}X^{-r}\Delta B_j\right) ,\\ h\left( \Delta X\right)= & {} -\frac{\sin q\pi }{\pi }\sum _{i=1}^m A_i^{*}\int _{0}^{\infty } \left( \lambda I+X\right) ^{-1}\Delta X\left( \lambda I+X+\Delta X\right) ^{-1}\\&\times \,\Delta X\left( \lambda I+X\right) ^{-1}\lambda ^{-q}d\lambda A_i\\&+\frac{\sin q\pi }{\pi }\sum _{i=1}^m\tilde{A}_i^{*}\int _{0}^{\infty }\left( \lambda I+X\right) ^{-1} \Delta X\left( \lambda I+X+\Delta X\right) ^{-1}\lambda ^{-q}d\lambda \Delta A_i\\&+\frac{\sin q\pi }{\pi }\sum _{i=1}^m\Delta A_i^{*}\int _{0}^{\infty } \left( \lambda I+X\right) ^{-1}\Delta X\left( \lambda I+X+\Delta X\right) ^{-1}\lambda ^{-q}d\lambda A_i\\&+\frac{\sin r\pi }{\pi }\sum _{j=1}^nB_j^{*}\int _{0}^{\infty } \left( \lambda I+X\right) ^{-1}\Delta X\left( \lambda I+X+\Delta X\right) ^{-1}\\&\times \,\Delta X\left( \lambda I+X\right) ^{-1}\lambda ^{-r}d\lambda B_j\\&-\frac{\sin r\pi }{\pi }\sum _{j=1}^n\tilde{B}_j^{*}\int _{0}^{\infty }\left( \lambda I+X\right) ^{-1} \Delta X\left( \lambda I+X+\Delta X\right) ^{-1}\lambda ^{-r}d\lambda \Delta B_j\\ \\&-\frac{\sin r\pi }{\pi }\sum _{j=1}^n\Delta B_j^{*}\int _{0}^{\infty } \left( \lambda I+X\right) ^{-1}\Delta X\left( \lambda I+X+\Delta X\right) ^{-1}\lambda ^{-r}d\lambda B_j. \end{aligned}$$

Lemma 3.2

Let

$$\begin{aligned} \sum _{i=1}^m\Vert A_i^*A_i\Vert <\frac{q^{q}}{2\Vert (q+1)Q^{-1}\Vert ^{q+1}}, \ \ \sum _{j=1}^n\Vert B_j^*B_j\Vert <\frac{q^{r+1}}{2r\Vert (q+1)Q^{-1}\Vert ^{r+1}}. \end{aligned}$$

Then the linear operator $\mathbf{L }: H^{n\times n}\rightarrow H^{n\times n}$ defined by

$$\begin{aligned} \begin{array}{rcl}\mathbf{L } W&{}=&{}W-\frac{\sin q\pi }{\pi }\sum _{i=1}^m\int _{0}^{\infty }A_i^{*} \left( \lambda I+X\right) ^{-1}W(\lambda I+X)^{-1}A_i\lambda ^{-q}d\lambda \\ \\ &{}&{}\ \ \ \ +\frac{\sin r\pi }{\pi }\sum _{j=1}^n\int _{0}^{\infty }B_j^{*} (\lambda I+X)^{-1}W\left( \lambda I+X\right) ^{-1}B_j\lambda ^{-r}d\lambda \end{array} \end{aligned}$$

(3.11)

is invertible.

Proof

It suffices to show that for any matrix $V\in H^{n\times n}$, the following equation

$$\begin{aligned} \mathbf{L }W=V \end{aligned}$$

(3.12)

has a unique solution. Define the operator $\mathbf{G }: H^{n\times n}\rightarrow H^{n\times n}$ by

$$\begin{aligned}&\mathbf{G }Z=\frac{\sin q\pi }{\pi }\sum _{i=1}^m\int _{0}^{\infty }X^{-1/2}A_i^{*} (\lambda I+X)^{-1}X^{1/2}ZX^{1/2}(\lambda I+X)^{-1}A_iX^{-1/2}\lambda ^{-q}d\lambda \\&\quad -\frac{\sin r\pi }{\pi }\sum _{j=1}^n\int _{0}^{\infty }X^{-1/2}B_j^{*} (\lambda I+X)^{-1}X^{1/2}ZX^{1/2}(\lambda I+X)^{-1}B_jX^{-1/2}\lambda ^{-r}d\lambda . \end{aligned}$$

Let $Y=X^{-1/2}WX^{-1/2}$. Thus (3.12) is equivalent to

$$\begin{aligned} Y-\mathbf G Y=X^{-1/2}VX^{-1/2}. \end{aligned}$$

(3.13)

Notice that $\Vert X^{-1}\Vert <\frac{q+1}{q}\Vert Q^{-1}\Vert$. According to Lemma 3.1 (ii), we have

$$\begin{aligned} \Vert \mathbf G Y\Vert\le & {} \Vert \frac{\sin q\pi }{\pi }\sum _{i=1}^m\int _{0}^{\infty }X^{-1/2}A_i^{*} (\lambda I+X)^{-1}X^{1/2}YX^{1/2}(\lambda I+X)^{-1}A_iX^{-1/2}\lambda ^{-q}d\lambda \Vert \\&+\Vert \frac{\sin r\pi }{\pi }\sum _{j=1}^n\int _{0}^{\infty }X^{-1/2}B_j^{*} (\lambda I+X)^{-1}X^{1/2}YX^{1/2}(\lambda I+X)^{-1}B_jX^{-1/2}\lambda ^{-r}d\lambda \Vert \\\le & {} \Vert \frac{\sin q\pi }{\pi }\sum _{i=1}^m\int _{0}^{\infty }X^{-1/2}A_i^{*} (\lambda I+X)^{-1}X(\lambda I+X)^{-1}A_iX^{-1/2}\lambda ^{-q}d\lambda \Vert \cdot \Vert Y\Vert \\&+\Vert \frac{\sin r\pi }{\pi }\sum _{j=1}^n\int _{0}^{\infty }X^{-1/2}B_j^{*} (\lambda I+X)^{-1}X(\lambda I+X)^{-1}B_jX^{-1/2}\lambda ^{-r}d\lambda \Vert \cdot \Vert Y\Vert \\= & {} q\Vert Y\Vert \cdot \sum _{i=1}^m\Vert X^{-1/2}A_i^{*}X^{-q}A_iX^{-1/2}\Vert +r\Vert Y\Vert \cdot \sum _{j=1}^n\Vert X^{-1/2}B_j^{*}X^{-r}B_jX^{-1/2}\Vert \\\le & {} q\Vert Y\Vert \cdot \sum _{i=1}^m\Vert X^{-1/2}A_i^{*}X^{-q/2}\Vert ^{2} +r\Vert Y\Vert \cdot \sum _{j=1}^n\Vert X^{-1/2}B_j^{*}X^{-r/2}\Vert ^{2}\\\le & {} q\Vert Y\Vert \cdot \sum _{i=1}^m\Vert A_i\Vert ^{2}\Vert X^{-1}\Vert ^{q+1} +r\Vert Y\Vert \cdot \sum _{j=1}^n\Vert B_j\Vert ^{2}\Vert X^{-1}\Vert ^{r+1}\\\le & {} q\Vert Y\Vert \cdot \frac{(q+1)^{q+1}}{q^{q+1}}\cdot \sum _{i=1}^m\Vert A_i\Vert ^{2}\Vert Q^{-1}\Vert ^{q+1}\\&+r\Vert Y\Vert \cdot \frac{(q+1)^{r+1}}{q^{r+1}}\cdot \sum _{j=1}^n\Vert B_j\Vert ^{2}\Vert Q^{-1}\Vert ^{r+1}\\< & {} \Vert Y\Vert . \end{aligned}$$

Then $\Vert \mathbf G \Vert <1$ which implies that $\mathbf I -\mathbf G$ is invertible. Therefore, for any matrix $V\in H^{n\times n}$, Eq. (3.13) has a unique solution Y. Thus equation (3.12) has a unique solution W for any $V\in H^{n\times n}$ which implies that the operator $\mathbf L$ is invertible. The proof is then completed. $\square$

Let $C_i=X^{-q}A_i, i=1,2,\ldots ,m; \ D_j=X^{-r}B_j, j=1,2,\ldots ,n$. We can rewrite (3.10) as

$$\begin{aligned} \begin{array}{rcl} \Delta X&{}=&{}\mathbf L ^{-1}\left[ \Delta Q-\sum _{i=1}^m \left( C_i^{*}\Delta A_i+\Delta A_i^{*}C_i\right) +\sum _{j=1}^n\left( D_j^{*}\Delta B_j+\Delta B_j^{*}D_j\right) \right] \\ \\ &{}&{}-\mathbf L ^{-1}\left( \sum _{i=1}^m\Delta A_i^{*}X^{-q}\Delta A_i+\sum _{j=1}^n\Delta B_j^{*}X^{-r}\Delta B_j\right) \\ \\ &{}&{}+\mathbf L ^{-1}(h(\Delta X)).\end{array} \end{aligned}$$

Then we have

$$\begin{aligned} \begin{array}{rcl} \Delta X&{}=&{}\mathbf L ^{-1}\left( \Delta Q-\sum _{i=1}^m \left( C_i^{*}\Delta A_i+\Delta A_i^{*}C_i\right) +\sum _{j=1}^n\left( D_j^{*}\Delta B_j+\Delta B_j^{*}D_j\right) \right) \\ \\ &{}&{}+\textit{O}\left( \Vert \left( \Delta A_1,\cdots ,\Delta A_m,\Delta B_1,\cdots ,\Delta B_n, \Delta Q\right) \Vert _{F}^{2}\right) ,\end{array} \end{aligned}$$

(3.14)

$(\Delta A_1,\cdots ,\Delta A_m,\Delta B_1,\cdots ,\Delta B_n, \Delta Q)\rightarrow 0.$

By Rice’s condition number theory [20], we define the condition number of the unique positive definite solution X of Eq. (1.1) as follows:

$$\begin{aligned} \textit{C}(X)=\lim _{\delta \rightarrow 0} \sup _{\Vert \left( \frac{\Delta A_1}{\mu _1},\cdots ,\frac{\Delta A_m}{\mu _m}, \frac{\Delta B_1}{\eta _1},\ldots ,\frac{\Delta B_n}{\eta _n}, \frac{\Delta Q}{\rho }\right) \Vert _{F} \le \delta }\frac{\Vert \Delta X\Vert _{F}}{\xi \delta }, \end{aligned}$$

(3.15)

where $\xi , \mu _1,\cdots ,\mu _m, \eta _1,\cdots ,\eta _n, \rho$ are positive parameters. Taking $\xi =\mu _1=\cdots =\mu _m=\eta _1=\cdots =\eta _n=\rho =1$ in (3.15) gives the absolute condition number $\textit{C}_{\mathrm{abs}} (X)$ and taking $\xi =\Vert X\Vert _{F}, \mu _i=\Vert A_i\Vert _{F}, i=1,\ldots ,m, \eta _j=\Vert B_j\Vert _{F}, j=1,\ldots ,n, \rho =\Vert Q\Vert _{F}$ gives the relative condition number $\textit{C}_{\mathrm{rel}}(X)$.

Substituting (3.14) into (3.15), we get

$$\begin{aligned} \textit{C}(X)= & {} \frac{1}{\xi }\max _{\mathop {\Delta A_i, \Delta B_j\in C^{n \times n},\Delta Q\in H^{n\times n}}\limits ^{\left( \frac{\Delta A_1}{\mu _1},\cdots ,\frac{\Delta A_m}{\mu _m}, \frac{\Delta B_1}{\eta _1},\ldots ,\frac{\Delta B_n}{\eta _n}, \frac{\Delta Q}{\rho }\right) \ne 0}} \\&\times \frac{\Vert \mathbf L ^{-1}\left( \Delta Q-\sum _{i=1}^m\left( C_i^{*}\Delta A_i+\Delta A_i^{*}C_i\right) +\sum _{j=1}^n\left( D_j^{*}\Delta B_j+\Delta B_j^{*}D_j\right) \right) \Vert _{F}}{\Vert \left( \frac{\Delta A_1}{\mu _1},\cdots ,\frac{\Delta A_m}{\mu _m}, \frac{\Delta B_1}{\eta _1},\ldots ,\frac{\Delta B_n}{\eta _n}, \frac{\Delta Q}{\rho }\right) \Vert _{F}}\\= & {} \frac{1}{\xi }\max _{\mathop {K_i, F_j\in C^{n \times n},H\in H^{n\times n}}\limits ^{\left( K_1,\cdots ,K_m, F_1,\ldots ,F_n, H\right) \ne 0}} \\&\times \frac{\Vert \mathbf L ^{-1}\left( \rho H-\sum _{i=1}^m\mu _i\left( C_i^{*}K_i+K_i^{*}C_i\right) + \sum _{j=1}^n\eta _j\left( D_j^{*}F_j+F_j^{*}D_j\right) \right) \Vert _{F}}{\Vert \left( K_1,\cdots ,K_m, F_1,\ldots ,F_n, H\right) \Vert _{F}}\\= & {} \frac{1}{\xi }\max _{\mathop {K_i, F_j\in C^{n \times n},H\in H^{n\times n}}\limits ^{\left( K_1,\cdots ,K_m, F_1,\ldots ,F_n, H\right) \ne 0}} \\&\times \frac{\Vert \mathbf L ^{-1}\left( \rho H-\sum _{i=1}^m\mu _i\left( C_i^{*}K_i+K_i^{*}C_i\right) +\sum _{j=1}^n \eta _j\left( D_j^{*}F_j+F_j^{*}D_j\right) \right) \Vert _{F}}{\Vert \left( -K_1,\cdots ,-K_m, F_1,\ldots ,F_n, H\right) \Vert _{F}}\\= & {} \frac{1}{\xi }\max _{\mathop {E_i, F_j\in C^{n \times n},H\in H^{n\times n}}\limits ^{\left( E_1,\cdots ,E_m, F_1,\ldots ,F_n, H\right) \ne 0}} \\&\times \frac{\Vert \mathbf L ^{-1}\left( \rho H+\sum _{i=1}^m\mu _i\left( C_i^{*}E_i+E_i^{*}C_i\right) + \sum _{j=1}^n\eta _j\left( D_j^{*}F_i+F_j^{*}D_j\right) \right) \Vert _{F}}{\Vert \left( E_1,\cdots ,E_m, F_1,\ldots ,F_n, H\right) \Vert _{F}}. \end{aligned}$$

Let L be the matrix of the operator $\mathbf L$. Then it is not difficult to see that

$$\begin{aligned} L= & {} I\otimes I-\frac{\sin q\pi }{\pi }\sum _{i=1}^m\int _{0}^{\infty } \left[ \left( \lambda I+X\right) ^{-1}A_i\right] ^{T}\otimes \left[ \left( \lambda I+X\right) ^{-1}A_i\right] ^{*}\lambda ^{-q}d\lambda \\&+\frac{\sin r\pi }{\pi }\sum _{j=1}^n\int _{0}^{\infty }\left[ \left( \lambda I+X\right) ^{-1}B_j\right] ^{T}\otimes \left[ \left( \lambda I+X\right) ^{-1}B_j\right] ^{*}\lambda ^{-r}d\lambda . \end{aligned}$$

Denote by $\beta =\mathrm{vec}(H)=a+kb, w_i=\mathrm{vec}(E_i)=u^{(i)}+kv^{(i)}, i=1,\ldots ,m; \varepsilon _j=\mathrm{vec}(F_j)=p^{(j)}+kq^{(j)}, \ j=1,2,\ldots ,n,$ where k is the imaginary unit satisfying $k^2=-1$. Let

$$\begin{aligned}&g_1^{(i)}=\left( \begin{array}{cc}u^{(i)}\\ v^{(i)}\end{array} \right) , g_2^{(j)}=\left( \begin{array}{cc}p^{(j)}\\ q^{(j)} \end{array}\right) , g_1=\left( \begin{array}{cc}g_1^{(1)}\\ \vdots \\ g_1^{(m)}\end{array} \right) , g_2=\left( \begin{array}{cc}g_2^{(1)}\\ \vdots \\ g_2^{(n)}\end{array}\right) , g=\left( \begin{array}{cc}a\\ b\\ g_1\\ g_2\end{array}\right) ;\\&L^{-1}\left( I\otimes C_i^{*}\right) =L^{-1}\left( I\otimes \left( X^{-q}A_i\right) ^{*}\right) =U_1^{(i)}+k\Omega _1^{(i)},\\&L^{-1}\left( C_i^{T}\otimes I\right) \Pi =L^{-1}\left( \left( X^{-q}A_i\right) ^{T}\otimes I\right) \Pi =U_2^{(i)}+k\Omega _2^{(i)},\\&L^{-1}\left( I\otimes D_i^{*}\right) =L^{-1}\left( I\otimes \left( X^{-r}B_i\right) ^{*}\right) =V_1^{(j)}+k\Phi _1^{(j)},\\&L^{-1}\left( D_i^{T}\otimes I\right) \Pi =L^{-1}\left( \left( X^{-r}B_i\right) ^{T}\otimes I\right) \Pi =V_2^{(j)}+k\Phi _2^{(j)}, \end{aligned}$$

where $U_1^{(i)},U_2^{(i)},V_1^{(j)},V_2^{(j)},\Omega _1^{(i)},\Omega _2^{(i)},\Phi _1^{(j)},\Phi _2^{(j)}\in R^{n^{2}\times n^{2}}, i=1,\ldots ,m, j=1,\ldots ,n,$ and $\Pi$ is the vec-permutation matrix, such that $\mathrm{vec}(K^{T})=\Pi \mathrm{vec}K.$ For $i=1,2,\ldots ,m; j=1,\ldots ,n.$

Denote

$$\begin{aligned} L^{-1}= & {} S+k\Sigma , S,\Sigma \in R^{n^{2}\times n^{2}},\ \ S_{c}=\left[ \begin{array}{cc} S &{}-\Sigma \\ \Sigma &{} S \\ \end{array}\right] ,\\ U_{c}^{(i)}= & {} \left[ \begin{array}{l@{\quad }l} U_1^{(i)}+U_2^{(i)} &{}\Omega _2^{(i)}-\Omega _1^{(i)} \\ \Omega _1^{(i)}+\Omega _2^{(i)} &{} U_1^{(i)}-U_2^{(i)}\\ \end{array}\right] , \ \ V_c^{(j)}=\left[ \begin{array}{l@{\quad }l} V_1^{(j)}+V_2^{(j)} &{}\Phi _2^{(j)}-\Phi _1^{(j)} \\ \Phi _1^{(j)}+\Phi _2^{(j)} &{} V_1^{(j)}-V_2^{(j)}\\ \end{array}\right] . \end{aligned}$$

Then we obtain that

$$\begin{aligned} \textit{C}(X)= & {} \frac{1}{\xi }\max _{\mathop {E_i, F_j\in C^{n \times n},H\in H^{n\times n}}\limits ^{\left( E_1,\cdots ,E_m, F_1,\ldots ,F_n, H\right) \ne 0}} \\&\times \frac{\Vert \mathbf L ^{-1}\left( \rho H+\sum _{i=1}^m\mu _i\left( C_i^{*}E_i+E_i^{*}C_i\right) + \sum _{j=1}^n\eta _j\left( D_j^{*}F_j+F_j^{*}D_j\right) \right) \Vert _{F}}{\Vert \left( E_1,\cdots ,E_m, F_1,\ldots ,F_n, H\,\right) \Vert _{F}}\\= & {} \frac{1}{\xi }\max _{\mathop {E_i, F_j\in C^{n \times n},H\in H^{n\times n}}\limits ^{\left( E_1,\cdots ,E_m, F_1,\ldots ,F_n, H\right) \ne 0}} \\&\times \frac{\Vert \rho L^{-1}\mathrm{vec}(H) +\sum _{i=1}^m\mu _iL^{-1}\mathrm{vec}\left( C_i^{*}E_i+E_i^{*}C_i\right) +\sum _{j=1}^n\eta _jL^{-1}\mathrm{vec}\left( D_j^{*}F_j+F_j^{*}D_j\right) \Vert }{\Vert \mathrm{vec}\left( E_1,\cdots ,E_m, F_1,\ldots ,F_n, H\right) \Vert }\\= & {} \frac{1}{\xi }\max _{\mathop {E_i, F_j\in C^{n \times n},H\in H^{n\times n}}\limits ^{\left( E_1,\cdots ,E_m, F_1,\ldots ,F_n, H\right) \ne 0}} \\&\times \frac{\Vert \rho S_{c}\left( \begin{array}{c} a\\ b\\ \end{array}\right) +\sum _{i=1}^m\mu _i U_c^{(i)}\left( \begin{array}{c} u^{(i)}\\ v^{(i)}\\ \end{array}\right) +\sum _{j=1}^n\eta _jV_c^{(j)}\left( \begin{array}{c} p^{(j)}\\ q^{(j)}\\ \end{array}\right) \Vert }{\Vert \mathrm{vec}\left( E_1,\cdots ,E_m, F_1,\ldots ,F_n, H\right) \Vert }\\= & {} \frac{1}{\xi }\max _{g\ne 0}\frac{\Vert (\rho S_{c}, \mu _1 U_c^{(1)},\cdots ,\mu _m U_c^{(m)},\eta _1 V_c^{(1)},\ldots ,\eta _n V_c^{(n)})g\Vert }{\Vert g\Vert }\\= & {} \frac{1}{\xi }\Vert \left( \rho S_{c}, \mu _1 U_c^{(1)},\cdots ,\mu _m U_c^{(m)},\eta _1 V_c^{(1)},\ldots ,\eta _n V_c^{(n)}\right) \Vert . \end{aligned}$$

Theorem 3.2

Let

$$\begin{aligned} \sum _{i=1}^m\Vert A_i^*A_i\Vert <\frac{1}{2}\frac{q^q}{\Vert (q+1)Q^{-1}\Vert ^{q+1}},\ \ \ \ \sum _{j=1}^n\Vert B_j^*B_j\Vert <\frac{1}{2r}\frac{q^{r+1}}{\Vert (q+1)Q^{-1}\Vert ^{r+1}}. \end{aligned}$$

Then the condition number $\textit{C}(X)$ defined by (3.15) has the following explicit expression

$$\begin{aligned} \textit{C}(X)=\frac{1}{\xi }\Vert \left( \rho S_{c}, \mu _1 U_c^{(1)},\cdots ,\mu _m U_c^{(m)}, \eta _1 V_c^{(1)},\ldots , \eta _n V_c^{(n)}\right) \Vert , \end{aligned}$$

(3.16)

where $S_{c}, U_c^{(i)}, V_c^{(j)}, i=1,2,\ldots ,m; j=1,2,\ldots ,n$ are defined as above.

Remark 3.1

From (3.16), we have the relative condition number

$$\begin{aligned} \textit{C}_{\mathrm{rel}}(X)=\frac{\Vert \left( \Vert Q\Vert _{F} S_{c}, \Vert A_1\Vert _{F}U_{c}^{(1)},\cdots ,\Vert A_m\Vert _{F}U_{c}^{(m)}, \Vert B_1\Vert _F V_c^{(1)},\cdots ,\Vert B_n\Vert _F V_c^{(n)}\right) \Vert }{\Vert X\Vert _{F}}.\nonumber \\ \end{aligned}$$

(3.17)

3.2 The real case

In this section we consider the real case, i.e., all the coefficient matrices $A_i,i=1,2,\ldots ,m; B_j,j=1,2,\ldots ,n,$ and Q of Eq. (1.1) are real. In such a case the corresponding unique positive definite solution X is also real. Similar to Theorem 3.2, we obtain the following theorem.

Theorem 3.3

Let $A_i,i=1,2,\ldots ,m; B_j,j=1,2,\ldots ,n,$ and Q be real. Suppose that

$$\begin{aligned} \sum _{i=1}^m\Vert A_i^*A_i\Vert <\frac{1}{2}\frac{q^q}{\Vert (q+1)Q^{-1}\Vert ^{q+1}},\ \ \ \ \sum _{j=1}^n\Vert B_j^*B_j\Vert <\frac{1}{2r}\frac{q^{r+1}}{\Vert (q+1)Q^{-1}\Vert ^{r+1}}. \end{aligned}$$

Then the condition number $\textit{C}(X)$ defined by (3.15) has the explicit expression

$$\begin{aligned} \textit{C}(X)=\frac{1}{\xi }\Vert \left( \rho S_{r}, \mu _1U_r^{(1)},\cdots ,\mu _mU_r^{(m)}, \eta _1V_r^{(1)},\cdots ,\eta _nV_r^{(n)}\right) \Vert , \end{aligned}$$

(3.18)

where

$$\begin{aligned} S_{r}= & {} \left[ I\otimes I-\frac{\sin q\pi }{\pi }\sum _{i=1}^m\int _{0}^{\infty }\left( \left( \lambda I+X\right) ^{-1}A_i\right) ^{T}\otimes \left( \left( \lambda I+X\right) ^{-1}A_i\right) ^{T}\lambda ^{-q}d\lambda \right. \\&+\left. \frac{\sin r\pi }{\pi }\sum _{j=1}^n\int _{0}^{\infty }\left( \left( \lambda I+X\right) ^{-1}B_j\right) ^{T}\otimes \left( \left( \lambda I+X\right) ^{-1}B_j\right) ^{T}\lambda ^{-r}d\lambda \right] ^{-1},\\ U_r^{(i)}= & {} S_{r}\left[ I\otimes \left( A_i^{T}X^{-q}\right) +\left( \left( A_i^{T}X^{-q}\right) \otimes I\right) \Pi \right] ,\quad i=1,2,\ldots ,m;\\ V_r^{(j)}= & {} S_{r} \left[ I\otimes \left( B_j^{T}X^{-r}\right) + \left( \left( B_j^{T}X^{-r}\right) \otimes I\right) \Pi \right] ,\quad j=1,2,\ldots ,n. \end{aligned}$$

Remark 3.2

In the real case the relative condition number is given by

$$\begin{aligned} \textit{C}_{\mathrm{rel}}(X)=\frac{\Vert \left( \Vert Q\Vert _{F} S_{r}, \Vert A_1\Vert _{F}U_r^{(1)}\cdots \Vert A_m\Vert _{F}U_r^{(m)}, \Vert B_1\Vert _{F}V_r^{(1)}\cdots \Vert B_n\Vert _{F}V_r^{(n)}\right) \Vert }{\Vert X\Vert _{F}}.\nonumber \\ \end{aligned}$$

(3.19)

4 Numerical experiments

In this section, several numerical examples are given to illustrate the theoretical results. All the tests are carried out using MATLAB 7.1 with machine precision around $10^{-16}$. The practical stopping criterion used is the residual $\Vert X+\sum _{i=1}^mA_i^{*}X^{-q}A_i-\sum _{j=1}^nB_j^*X^{-r}B_j-Q\Vert <10^{-15}$.

Example 4.1

Consider Eq. (1.1) with the case $q=0.5,r=0.7$, $m=n=2$, and the matrices $A_1, A_2, B_1, B_2$ and Q as follows:

$$\begin{aligned} A_1= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} 0.1 &{} 0.05 &{} 0.05 \\ 0.05 &{} 0.1 &{} 0.05 \\ 0.05 &{} 0.05 &{} 0.1 \end{array} \right) ,\ \ \ \ A2=\left( \begin{array}{c@{\quad }c@{\quad }c} 0.5 &{} -0.02 &{} -0.02 \\ -0.02&{}0.5&{} -0.02 \\ -0.02 &{} -0.02 &{} 0.5 \end{array} \right) ,\\ B_1= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} -0.04 &{} 0.01 &{} -0.02 \\ 0.05 &{} 0.07 &{} -0.013 \\ 0.011 &{} 0.09 &{} 0.06 \end{array} \right) ,\ \ \ \ B_2=\left( \begin{array}{c@{\quad }c@{\quad }c} 0.01 &{} 0.001 &{} 0.01 \\ 0.001&{}0.01&{} 0.001 \\ 0.01 &{} 0.001 &{} 0.01 \end{array} \right) ,\\ Q= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} 2 &{} 0.2 &{} 0.2 \\ 0.2&{} 2 &{} 0.2 \\ 0.2 &{} 0.2 &{} 2 \end{array} \right) . \end{aligned}$$

By computation, $\Vert A_1^*A_1\Vert +\Vert A_2^*A_2\Vert -\frac{1}{2}\cdot \frac{q^q}{\Vert (q+1)Q^{-1}\Vert ^{q+1}} =-0.1544$, $\Vert B_1^*B_1\Vert +\Vert B_2^*B_2\Vert -\frac{1}{2r}\cdot \frac{q^{r+1}}{\Vert (q+1)Q^{-1}\Vert ^{r+1}} =-0.2831$. According to Theorem 2.2, using iteration (2.2) and iterating 15 steps, then we get the unique positive definite solution to Eq. (1.1):

$$\begin{aligned} \bar{X}\approx X_{15}=\left( \begin{array}{c@{\quad }c@{\quad }c} 1.8026 &{} 0.2190 &{} 0.2168\\ 0.2190 &{} 1.8077 &{} 0.2191\\ 0.2168 &{} 0.2191 &{} 1.8025 \end{array} \right) \end{aligned}$$

with the residual $\Vert X_{15}+A_1^{*}X_{15}^{-q}A_1 +A_2^{*}X_{15}^{-q}A_2-B_1^*X_{15}^{-r}B_1-B_2^*X_{15}^{-r}B_2-Q\Vert = 4.4524e-016$ or

$$\begin{aligned} \bar{X}\approx Y_{15}=\left( \begin{array}{c@{\quad }c@{\quad }c} 1.8026 &{} 0.2190 &{} 0.2168\\ 0.2190 &{} 1.8077 &{} 0.2191\\ 0.2168 &{} 0.2191 &{} 1.8025 \end{array} \right) \end{aligned}$$

with the residual $\Vert Y_{15}+A_1^{*}Y_{15}^{-q}A_1 +A_2^{*}Y_{15}^{-q}A_2-B_1^*Y_{15}^{-r}B_1-B_2^*Y_{15}^{-r}B_2-Q\Vert = 4.4593e-016$.

Example 4.2

Let $m=n=1$ and

$$\begin{aligned} A=\frac{\sqrt{3}}{45}\left( \begin{array}{ccccc} 1&{}0&{}0&{}0&{}1\\ -1&{}1&{}0&{}0&{}1\\ -1&{}-1&{}1&{}0&{}1\\ -1&{}-1&{}-1&{}1&{}1\\ -1&{}-1&{}-1&{}-1&{}1 \end{array}\right) ,\, B=\frac{A+A^{*}}{2}, \end{aligned}$$

$q=0.5, r=0.3, \ X=\mathrm{diag}(0.725, 2,3,2,1), \ Q=X+A^{*}X^{-q}A-B^{*}X^{-r}B$.

Consider the perturbed matrix equation

$$\begin{aligned} \tilde{X}+\tilde{A}^{*}_{j}\tilde{X}^{-q}\tilde{A}_{j}-\tilde{B}^{*}_{j}\tilde{X}^{-r}\tilde{B}_{j}=\tilde{Q}_{j}, \end{aligned}$$

where $\epsilon _{j}=0.1^{2j},\ \ \tilde{A}_{j}=A+\epsilon _{j}(I+E),\ \tilde{B}_{j}=B+\epsilon _{j}(I+2E) \ \ \tilde{X}_{j}=X+\epsilon _{j}(I-E), \ \ \tilde{Q}_{j}=\tilde{X}_{j}+\tilde{A}^{*}_{j}\tilde{X}_{j}^{-q}\tilde{A}_{j} -\tilde{B}^{*}_{j}\tilde{X}_{j}^{-q}\tilde{B}_{j},$ with $e=(1,1,1,1,1)$ and $E=e'e$.

Now we compute the perturbation bounds for Eq. (1.1).

By computation, $\theta _1=\frac{q^{q}}{(q+1)^{q+1}}-2\Vert A\Vert ^{2}\Vert Q^{-1}\Vert ^{q+1}=0.3378>0,$ $\theta _2=\frac{q^{r+1}}{r(q+1)^{r+1}}- 2\Vert B\Vert ^{2}\Vert Q^{-1}\Vert ^{r+1}=0.7891>0,$ and $\lambda _{n}(X-\frac{q}{q+1}Q)=0.4821>0$ which implies that X is the unique positive definite solution of Eq. (1.1) by Theorem 2.2. Obviously, $\tilde{X}_{j}$ are positive definite solutions of the perturbed matrix equations $\tilde{X}+\tilde{A}^{*}_{j}\tilde{X}^{-q}\tilde{A}_{j}-\tilde{B}^{*}_{j}\tilde{X}^{-q}\tilde{B}_{j}=\tilde{Q}_{j}$. Moreover, it is not difficult to verify that for each $j=1,2,3,4,5$, the corresponding equations $\tilde{X}+\tilde{A}^{*}_{j}\tilde{X}^{-q}\tilde{A}_{j}-\tilde{B}^{*}_{j}\tilde{X}^{-q}\tilde{B}_{j}=\tilde{Q}_{j}$ and $\tilde{X}_{j}$ satisfy the assumption $\theta _{j1}=\frac{q^{q}}{(q+1)^{q+1}}- 2\Vert \tilde{A}_j\Vert ^{2}\Vert \tilde{Q}_j^{-1}\Vert ^{q+1}>0,$ $\theta _{j2}=\frac{q^{r+1}}{r(q+1)^{r+1}}- 2\Vert \tilde{B}_j\Vert ^{2}\Vert \tilde{Q}_j^{-1}\Vert ^{r+1}>0,$ and the conditions $\lambda _{n}(\tilde{X}_{j}-\frac{q}{q+1}\tilde{Q}_{j})>0$,respectively. Thus by Theorem 2.2, $\tilde{X}_{j}\ (j=1,2, \ldots , 5)$ are the unique positive definite solutions of the corresponding perturbed matrix equations, respectively. We denote $\Delta X^{(j)}=\tilde{X}^{j}-X$. All the conditions of Theorem 3.1 are satisfied for $j=1,2, 3, 4, 5$. The results are given in the following table.

	j = 1	j = 2	j = 3	j = 4	j = 5
True error $\Vert \Delta X^{(j)}\Vert$	0.0400	$4.0000e{-004}$	$4.0000e{-006}$	$4.0000e{-008}$	$4.0000e{-010}$
Our result (3.5)	0.0985	$6.9607e{-004}$	$6.9346e{-006}$	$6.9343e{-008}$	$6.9343e{-010}$

Example 4.3

Consider Eq. (1.1) with $q=r=0.5, m=2, n=1$ and

$$\begin{aligned} A_1=\left( \begin{array}{cc} 0 &{} a_{1}\\ 0.02 &{} 0 \end{array}\right) ,\ \ A_{2}=\left( \begin{array}{cc} 0 &{} a_2\\ 0 &{} 0 \end{array}\right) ,\ \ B=\left( \begin{array}{cc} 0 &{} 0\\ b &{} 0 \end{array}\right) , \ \ Q=\left( \begin{array}{cc} 1.1 &{} 0\\ 0 &{} 1.2 \end{array}\right) , \end{aligned}$$

where $a_{1}=0.25+10^{-k}$, $a_{2}=0.15+10^{-k}$ and $b=0.35+10^{-k}$. Denote $\theta _1=\Vert A_1^*A_1\Vert +\Vert A_2^*A_2\Vert -\frac{1}{2}\frac{q^q}{\Vert (q+1)Q^{-1}\Vert ^{q+1}}$, $\theta _2=\Vert B^*B\Vert -\frac{1}{2r}\frac{q^{r+1}}{\Vert (q+1)Q^{-1}\Vert ^{r+1}}$, For k from 1 to 6, we compute $\theta _1$ and $\theta _2$ to see that the conditions of Theorem 3.3 are always satisfied. Results for $\textit{C}_{\mathrm{rel}}(X)$ by (3.19) with different vales of k are listed below where $\textit{C}_{\mathrm{rel}}(X)$ is the relative condition number of the unique positive definite solution of Eq.(1.1).

k	1	2	3	4	5	6
$\theta _1$	$-0.037027$	$-0.128827$	$-0.136225$	$-0.136947$	$-0.137019$	$-0.137026$
$\theta _2$	$-0.019527$	$-0.092427$	$-0.098826$	$-0.099457$	$-0.099520$	$-0.099526$
$\textit{C}_{\mathrm{rel}}(X_{L})$	1.014454	1.000561	0.999674	0.999590	0.999582	0.999581

The numerical results listed in the second line show that the unique positive definite solution X is well-conditioned in such cases.