Mathematical Problems in Engineering

Mathematical Problems in Engineering / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 163585 | https://doi.org/10.1155/2011/163585

Aijing Liu, Guoliang Chen, "On the Hermitian Positive Definite Solutions of Nonlinear Matrix Equation ", Mathematical Problems in Engineering, vol. 2011, Article ID 163585, 18 pages, 2011. https://doi.org/10.1155/2011/163585

On the Hermitian Positive Definite Solutions of Nonlinear Matrix Equation 𝑋 𝑠 + 𝐴 βˆ— 𝑋 βˆ’ 𝑑 𝟏 𝐴 + 𝐡 βˆ— 𝑋 βˆ’ 𝑑 𝟐 𝐡 = 𝑄

Academic Editor: Mohammad Younis
Received18 Apr 2011
Accepted11 Jul 2011
Published05 Oct 2011

Abstract

Nonlinear matrix equation 𝑋𝑠+π΄βˆ—π‘‹βˆ’π‘‘1𝐴+π΅βˆ—π‘‹βˆ’π‘‘2𝐡=𝑄 has many applications in engineering; control theory; dynamic programming; ladder networks; stochastic filtering; statistics and so forth. In this paper, the Hermitian positive definite solutions of nonlinear matrix equation 𝑋𝑠+π΄βˆ—π‘‹βˆ’π‘‘1𝐴+π΅βˆ—π‘‹βˆ’π‘‘2𝐡=𝑄 are considered, where 𝑄 is a Hermitian positive definite matrix, 𝐴, 𝐡 are nonsingular complex matrices, 𝑠 is a positive number, and 0<𝑑𝑖≀1, 𝑖=1,2. Necessary and sufficient conditions for the existence of Hermitian positive definite solutions are derived. A sufficient condition for the existence of a unique Hermitian positive definite solution is given. In addition, some necessary conditions and sufficient conditions for the existence of Hermitian positive definite solutions are presented. Finally, an iterative method is proposed to compute the maximal Hermitian positive definite solution, and numerical example is given to show the efficiency of the proposed iterative method.

1. Introduction

We consider the nonlinear matrix equation𝑋𝑠+π΄βˆ—π‘‹βˆ’π‘‘1𝐴+π΅βˆ—π‘‹βˆ’π‘‘2𝐡=𝑄,(1.1) where 𝑄 is an 𝑛×𝑛 Hermitian positive definite matrix, 𝐴,𝐡 are 𝑛×𝑛 nonsingular complex matrices, 𝑠 is a positive number, and 0<𝑑𝑖≀1, 𝑖=1,2. Here π΄βˆ— stands for the conjugate transpose of the matrix 𝐴.

Nonlinear matrix equations with the form of (1.1) have many applications in engineering; control theory; dynamic programming; ladder networks; stochastic filtering; statistics and so forth. The solutions of practical interest are their Hermitian positive definite (HPD) solutions. The existence of HPD solutions of (1.1) has been investigated in some special cases. Long et al. [1] studied (1.1) when 𝑠=1, 𝑑1=𝑑2=1. In addition, there have been many papers considering the Hermitian positive solutions of𝑋𝑠+π΄βˆ—π‘‹βˆ’π‘‘π΄=𝑄.(1.2) For instance, the authors [2–5] studied (1.2) when 𝑠=1, 𝑑=1. In Hasanov [6, 7], the authors investigated (1.2) when 𝑠=1, π‘‘βˆˆ(0,1]. Then Peng et al. [8] proposed iterative methods for the extremal positive definite solutions of (1.2) for 𝑠=1 with two cases: 0<𝑑≀1 and 𝑑β‰₯1. Cai and Chen [9, 10] studied (1.2) with two cases: 𝑠 and 𝑑 are positive integers, and 𝑠β‰₯1, 0<𝑑≀1 or 0<𝑠≀1, 𝑑β‰₯1 respectively.

In this paper, we study the HPD solutions of (1.1). The paper is organized as follows. In Section 2, we derive necessary and sufficient conditions for the existence of HPD solutions of (1.1) and give a sufficient condition for the existence of a unique HPD solution of (1.1). We also present some necessary conditions and sufficient conditions for the existence of HPD solutions of (1.1). Then in Section 3, we propose an iterative method for obtaining the maximal HPD solution of (1.1). We give a numerical example in Section 4 to show the efficiency of the proposed iterative method.

We start with some notations which we use throughout this paper. The symbol Cπ‘šΓ—π‘› denotes the set of π‘šΓ—π‘› complex matrices. We write 𝐷>0(𝐷β‰₯0) if the matrix 𝐷 is positive definite(semidefinite). If π·βˆ’πΈ is positive definite(semidefinite), then we write 𝐷>𝐸(𝐷β‰₯𝐸). We use πœ†1(𝐷) and πœ†π‘›(𝐷) to denote the maximal and minimal eigenvalues of a matrix 𝐷. We use ‖𝐷‖ and ‖𝐷‖𝐹 to denote the spectral and Frobenius norm of a matrix 𝐷, and we also use ‖𝑏‖ to denote 𝑙2-norm of a vector 𝑏. We use 𝑋𝑆 and 𝑋𝐿 to denote the minimal and maximal HPD solution of (1.1), that is, for any HPD solution 𝑋 of (1.1), then 𝑋𝑆≀𝑋≀𝑋𝐿. The symbol 𝐼 denotes the 𝑛×𝑛 identity matrix. The symbol 𝜌(𝐷) denotes the spectral radius of 𝐷. Let [𝐷,𝐸]={π‘‹βˆ£π·β‰€π‘‹β‰€πΈ} and (𝐷,𝐸)={π‘‹βˆ£π·<𝑋<𝐸}. For matrices 𝐷=(𝑑1,𝑑2,…,𝑑𝑛)=(𝑑𝑖𝑗) and 𝐸, π·βŠ—πΈ=(𝑑𝑖𝑗𝐸) is a Kronecker product and vec(𝐷) is a vector defined by vec(𝐷)=(𝑑𝑇1,𝑑𝑇2,…,𝑑𝑇𝑛)𝑇.

2. Solvability Conditions and Properties of the HPD Solutions

In this section, we will derive the necessary and sufficient conditions for (1.1) to have an HPD solution and give a sufficient condition for the existence of a unique HPD solution of (1.1). We also will present some necessary conditions and sufficient conditions for the existence of Hermitian positive definite solutions of (1.1).

Lemma 2.1 (see [11]). If 𝐷β‰₯𝐸>0(or 𝐷>𝐸>0), then 𝐷𝑝β‰₯𝐸𝑝>0(or 𝐷𝑝>𝐸𝑝>0) for all π‘βˆˆ(0,1], and 𝐸𝑝β‰₯𝐷𝑝>0(or 𝐸𝑝>𝐷𝑝>0) for all π‘βˆˆ[βˆ’1,0).

Lemma 2.2 (see [12]). Let 𝐷 and 𝐸 be positive operators on a Hilbert space such that 0<π‘š1𝐼≀𝐷≀𝑀1𝐼, 0<π‘š2𝐼≀𝐸≀𝑀2𝐼, and 0<𝐷≀𝐸. Then π·π‘žβ‰€ξ‚΅π‘€1π‘š1ξ‚Άπ‘žβˆ’1πΈπ‘ž,π·π‘žβ‰€ξ‚΅π‘€2π‘š2ξ‚Άπ‘žβˆ’1πΈπ‘ž(2.1) hold for any π‘žβ‰₯1.

Lemma 2.3 (see [13]). Let 𝑓(π‘₯)=π‘₯𝑑(πœβˆ’π‘₯𝑠), 𝜁>0, π‘₯β‰₯0. Then(1)𝑓 is increasing on [0,((𝑑/(𝑠+𝑑))𝜁)1/𝑠] and decreasing on [((𝑑/(𝑠+𝑑))𝜁)1/𝑠,+∞);(2)𝑓max=𝑓(((𝑑/(𝑠+𝑑))𝜁)1/𝑠)=(𝑠/(𝑠+𝑑))(𝑑/(𝑠+𝑑))𝑑/π‘ πœ(𝑑/𝑠)+1.

Lemma 2.4 (see [14]). If 𝐷 and 𝐸 are Hermitian matrices of the same order with 𝐸>0, then 𝐷𝐸𝐷+πΈβˆ’1β‰₯2𝐷.

Lemma 2.5 (see [15]). If 0<πœƒβ‰€1, and 𝐷 and 𝐸 are positive definite matrices of the same order with 𝐷,𝐸β‰₯𝑏𝐼>0, then β€–π·πœƒβˆ’πΈπœƒβ€–β‰€πœƒπ‘πœƒβˆ’1β€–π·βˆ’πΈβ€– and β€–π·βˆ’πœƒβˆ’πΈβˆ’πœƒβ€–β‰€πœƒπ‘βˆ’(πœƒ+1)β€–π·βˆ’πΈβ€–. Here β€–β‹…β€– stands for one kind of matrix norm.

Lemma 2.6 (see [5]). Let 𝐷 and 𝐸 be two arbitrary compatible matrices. Then 𝜌(π·βˆ—πΈβˆ’πΈβˆ—π·)β‰€πœŒ(π·βˆ—π·+πΈβˆ—πΈ).

Theorem 2.7. Equation (1.1) has an HPD solution if and only if 𝐴,𝐡 can factor as 𝐿𝐴=βˆ—πΏξ€Έπ‘‘1/2𝑠𝑁1𝐿,𝐡=βˆ—πΏξ€Έπ‘‘2/2𝑠𝑁2,(2.2) where 𝐿 is a nonsingular matrix and ξ‚΅πΏπ‘„βˆ’1/2𝑁1π‘„βˆ’1/2𝑁2Qβˆ’1/2ξ‚Ά is column orthonormal.

Proof. If (1.1) has an HPD solution, then 𝑋𝑠>0. Let 𝑋𝑠=πΏβˆ—πΏ be the Cholesky factorization, where 𝐿 is a nonsingular matrix. Then (1.1) can be rewritten as π‘„βˆ’1/2πΏβˆ—πΏπ‘„βˆ’1/2+π‘„βˆ’1/2π΄βˆ—ξ€·πΏβˆ—πΏξ€Έβˆ’π‘‘1/2π‘ ξ€·πΏβˆ—πΏξ€Έβˆ’π‘‘1/2π‘ π΄π‘„βˆ’1/2+π‘„βˆ’1/2π΅βˆ—ξ€·πΏβˆ—πΏξ€Έβˆ’π‘‘2/2π‘ ξ€·πΏβˆ—πΏξ€Έβˆ’π‘‘2/2π‘ π΅π‘„βˆ’1/2=𝐼.(2.3) Let 𝑁1=(πΏβˆ—πΏ)βˆ’π‘‘1/2𝑠𝐴,𝑁2=(πΏβˆ—πΏ)βˆ’π‘‘2/2𝑠𝐡, then 𝐴=(πΏβˆ—πΏ)𝑑1/2𝑠𝑁1, 𝐡=(πΏβˆ—πΏ)𝑑2/2𝑠𝑁2. Moreover, (2.3) turns into π‘„βˆ’1/2πΏβˆ—πΏπ‘„βˆ’1/2+π‘„βˆ’1/2π‘βˆ—1𝑁1π‘„βˆ’1/2+π‘„βˆ’1/2π‘βˆ—2𝑁2π‘„βˆ’1/2=𝐼,(2.4) that is, βŽ›βŽœβŽœβŽœβŽœβŽπΏπ‘„βˆ’1/2𝑁1π‘„βˆ’1/2𝑁2π‘„βˆ’1/2βŽžβŽŸβŽŸβŽŸβŽŸβŽ βˆ—βŽ›βŽœβŽœβŽœβŽœβŽπΏπ‘„βˆ’1/2𝑁1π‘„βˆ’1/2𝑁2π‘„βˆ’1/2⎞⎟⎟⎟⎟⎠=𝐼,(2.5) which means that ξ‚΅πΏπ‘„βˆ’1/2𝑁1π‘„βˆ’1/2𝑁2π‘„βˆ’1/2ξ‚Ά is column orthonormal.
Conversely, if 𝐴,𝐡 have the decompositions as (2.2), let 𝑋=(πΏβˆ—πΏ)1/𝑠, then 𝑋 is an HPD matrix, and it follows from (2.2) and (2.4) that 𝑋𝑠+π΄βˆ—π‘‹βˆ’π‘‘1𝐴+π΅βˆ—π‘‹βˆ’π‘‘2𝐡=πΏβˆ—πΏ+π‘βˆ—1𝑁1+π‘βˆ—2𝑁2=𝑄1/2ξ€·π‘„βˆ’1/2πΏβˆ—πΏπ‘„βˆ’1/2+π‘„βˆ’1/2π‘βˆ—1𝑁1π‘„βˆ’1/2+π‘„βˆ’1/2π‘βˆ—2𝑁2π‘„βˆ’1/2𝑄1/2=𝑄.(2.6) Hence (1.1) has an HPD solution.

Theorem 2.8. Equation (1.1) has an HPD solution if and only if there exist a unitary matrix π‘‰βˆˆC𝑛×𝑛, a column-orthonormal matrix ξ‚€π‘ˆ=π‘ˆ1π‘ˆ2ξ‚βˆˆC2𝑛×𝑛(in which π‘ˆ1,π‘ˆ2∈C𝑛×𝑛), and diagonal matrices 𝐢>0 and 𝑆β‰₯0 with 𝐢2+𝑆2=𝐼 such that 𝑄𝐴=1/2π‘‰βˆ—πΆ2𝑉𝑄1/2𝑑1/2π‘ π‘ˆ1𝑆𝑉𝑄1/2,𝑄𝐡=1/2π‘‰βˆ—πΆ2𝑉𝑄1/2𝑑2/2π‘ π‘ˆ2𝑆𝑉𝑄1/2.(2.7)

Proof. If (1.1) has an HPD solution, we have by Theorem 2.7 that the matrix ξ‚΅πΏπ‘„βˆ’1/2𝑁1π‘„βˆ’1/2𝑁2π‘„βˆ’1/2ξ‚Ά is column orthonormal. According to the CS decomposition theorem (Theorem  3.8 in [16]), there exist unitary matrices 𝑃=𝑃100𝑃2ξ‚βˆˆC3𝑛×3𝑛 (in which 𝑃1∈C𝑛×𝑛,𝑃2∈C2𝑛×2𝑛), π‘‰βˆˆC𝑛×𝑛, such that βŽ›βŽœβŽœβŽπ‘ƒ100𝑃2βŽžβŽŸβŽŸβŽ βŽ›βŽœβŽœβŽœβŽπΏπ‘„βˆ’1/2𝑁1π‘„βˆ’1/2𝑁2π‘„βˆ’1/2βŽžβŽŸβŽŸβŽŸβŽ π‘‰βˆ—=βŽ›βŽœβŽœβŽπΆπ‘†0⎞⎟⎟⎠,(2.8) where 𝐢=diag(cosπœƒ1,…,cosπœƒπ‘›),𝑆=diag(sinπœƒ1,…,sinπœƒπ‘›), and 0β‰€πœƒ1β‰€β‹―β‰€πœƒπ‘›β‰€πœ‹/2. Thus the diagonal matrices 𝐢,𝑆β‰₯0 and 𝐢2+𝑆2=𝐼. Furthermore, noting that 𝐿 is nonsingular, by (2.8), we have 𝐢=𝑃1πΏπ‘„βˆ’1/2π‘‰βˆ—>0,(2.9)𝑃2βŽ›βŽœβŽœβŽπ‘1𝑁2βŽžβŽŸβŽŸβŽ π‘„βˆ’1/2π‘‰βˆ—=βŽ›βŽœβŽœβŽπ‘†0⎞⎟⎟⎠.(2.10)
Equation (2.10) is equivalent to 𝑁1𝑁2=π‘ƒβˆ—2𝑆0𝑉𝑄1/2. Let π‘ƒβˆ—2 be partitioned as π‘ƒβˆ—2=ξ‚€π‘ˆ1π‘ˆ3π‘ˆ2π‘ˆ4, in which π‘ˆπ‘–βˆˆC𝑛×𝑛, 𝑖=1,2,3,4, then we haveβŽ›βŽœβŽœβŽπ‘1𝑁2⎞⎟⎟⎠=βŽ›βŽœβŽœβŽπ‘ˆ1π‘ˆ3π‘ˆ2π‘ˆ4βŽžβŽŸβŽŸβŽ βŽ›βŽœβŽœβŽπ‘†0βŽžβŽŸβŽŸβŽ π‘‰π‘„1/2=βŽ›βŽœβŽœβŽπ‘ˆ1𝑆𝑉𝑄1/2π‘ˆ2𝑆𝑉𝑄1/2⎞⎟⎟⎠,(2.11) from which it follows that 𝑁1=π‘ˆ1𝑆𝑉𝑄1/2, 𝑁2=π‘ˆ2𝑆𝑉𝑄1/2. By (2.9), we have 𝐿=π‘ƒβˆ—1𝐢𝑉𝑄1/2. Then by (2.2), we have 𝐿𝐴=βˆ—πΏξ€Έπ‘‘1/2𝑠𝑁1=𝑄1/2π‘‰βˆ—πΆ2𝑉𝑄1/2𝑑1/2π‘ π‘ˆ1𝑆𝑉𝑄1/2,𝐿𝐡=βˆ—πΏξ€Έπ‘‘2/2𝑠𝑁2=𝑄1/2π‘‰βˆ—πΆ2𝑉𝑄1/2𝑑2/2π‘ π‘ˆ2𝑆𝑉𝑄1/2.(2.12)
Conversely, assume that 𝐴,𝐡 have the decomposition (2.7). Let 𝑋=(𝑄1/2π‘‰βˆ—πΆ2𝑉𝑄1/2)1/𝑠, which is an HPD matrix. Then it is easy to verify that 𝑋 is an HPD solution of (1.1).

Theorem 2.9. If (1.1) has an HPD solution 𝑋, then π‘‹βˆˆ(𝑀,𝑁), where 1𝑀=2ξƒ©ξ‚΅πœ‡1𝜈1ξ‚Ά(1βˆ’π‘‘1)/𝑑1ξ€·π΄π‘„βˆ’1π΄βˆ—ξ€Έ1/𝑑1+ξ‚΅πœ‡2𝜈2ξ‚Ά(1βˆ’π‘‘2)/𝑑2ξ€·π΅π‘„βˆ’1π΅βˆ—ξ€Έ1/𝑑2ξƒͺ,𝑁=π‘„βˆ’π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄βˆ’π΅βˆ—π‘„βˆ’π‘‘2/𝑠𝐡1/𝑠,(2.13) in which πœ‡1 and 𝜈1 are the minimal and maximal eigenvalues of π΄π‘„βˆ’1π΄βˆ— respectively, πœ‡2 and 𝜈2 are the minimal and maximal eigenvalues of π΅π‘„βˆ’1π΅βˆ—, respectively.

Proof. Let 𝑋 be an HPD solution of (1.1), then it follows from 0<𝑋𝑠<𝑄 and Lemma 2.1 that π‘‹βˆ’π‘‘π‘–>π‘„βˆ’π‘‘π‘–/𝑠, 𝑖=1,2. Hence 𝑋𝑠=π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘1π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘2𝐡<π‘„βˆ’π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄βˆ’π΅βˆ—π‘„βˆ’π‘‘2/𝑠𝐡.(2.14) Thus we have 𝑋<π‘„βˆ’π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄βˆ’π΅βˆ—π‘„βˆ’π‘‘2/𝑠𝐡1/𝑠=𝑁.(2.15) On the other hand, from π΄βˆ—π‘‹βˆ’π‘‘1𝐴<𝑄, it follows that π‘„βˆ’1/2π΄βˆ—π‘‹βˆ’π‘‘1/2π‘‹βˆ’π‘‘1/2π΄π‘„βˆ’1/2𝑋<𝐼,βˆ’π‘‘1/2π΄π‘„βˆ’1π΄βˆ—π‘‹βˆ’π‘‘1/2<𝐼,π΄π‘„βˆ’1π΄βˆ—<𝑋𝑑1.(2.16) Let πœ‡1 and 𝜈1 be the minimal and maximal eigenvalues of π΄π‘„βˆ’1π΄βˆ—, respectively. Since 1/𝑑1β‰₯1, and πœ‡1πΌβ‰€π΄π‘„βˆ’1π΄βˆ—β‰€πœˆ1𝐼, by Lemma 2.2, we get (πœ‡1/𝜈1)(1βˆ’π‘‘1)/𝑑1(π΄π‘„βˆ’1π΄βˆ—)1/𝑑1<𝑋.
Similarly, we have (πœ‡2/𝜈2)(1βˆ’π‘‘2)/𝑑2(π΅π‘„βˆ’1π΅βˆ—)1/𝑑2<𝑋, in which πœ‡2 and 𝜈2 are the minimal and maximal eigenvalues of π΅π‘„βˆ’1π΅βˆ—, respectively.
Hence we have 𝑋>1/2((πœ‡1/𝜈1)(1βˆ’π‘‘1)/𝑑1(π΄π‘„βˆ’1π΄βˆ—)1/𝑑1+(πœ‡2/𝜈2)(1βˆ’π‘‘2)/𝑑2(π΅π‘„βˆ’1π΅βˆ—)1/𝑑2)=𝑀.

Theorem 2.10. If π΄βˆ—π‘‹βˆ’π‘‘1𝐴+π΅βˆ—π‘‹βˆ’π‘‘2π΅β‰€π‘„βˆ’π‘€π‘  for all π‘‹βˆˆ[𝑀,𝑄1/𝑠], and 1𝑝=𝑠𝑑1πœ†βˆ’(𝑠+𝑑1)𝑛𝑀‖𝐴‖2𝐹+𝑑2πœ†βˆ’(𝑠+𝑑2)𝑛𝑀‖𝐡‖2𝐹<1,(2.17) where 𝑀 is defined by (2.13), then (1.1) has a unique HPD solution.

Proof. By the definition of 𝑀, we have 𝑀>0. Hence πœ†π‘›(𝑀)>0.
We consider the map 𝐹(𝑋)=(π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘1π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘2𝐡)1/𝑠 and let π‘‹βˆˆΞ©={π‘‹βˆ£π‘€β‰€π‘‹β‰€π‘„1/𝑠}.Obviously, Ξ© is a convex, closed, and bounded set and 𝐹(𝑋) is continuous on Ξ©.
By the hypothesis of the theorem, we have 𝑄1/𝑠β‰₯ξ€·π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘1π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘2𝐡1/𝑠β‰₯ξ‚€π‘„βˆ’π‘„+𝑀𝑠1/𝑠=𝑀,(2.18) that is, 𝑀≀𝐹(𝑋)≀𝑄1/𝑠. Hence 𝐹(Ξ©)βŠ†Ξ©.
For arbitrary 𝑋,π‘ŒβˆˆΞ©, we have π΄βˆ—π‘‹βˆ’π‘‘1𝐴+π΅βˆ—π‘‹βˆ’π‘‘2π΅β‰€π‘„βˆ’π‘€π‘ ,π΄βˆ—π‘Œβˆ’π‘‘1𝐴+π΅βˆ—π‘Œβˆ’π‘‘2π΅β‰€π‘„βˆ’π‘€π‘ .(2.19) Hence 𝐹(𝑋)=π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘1π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘2𝐡1/𝑠β‰₯ξ‚€π‘„βˆ’π‘„+𝑀𝑠1/𝑠=𝑀β‰₯πœ†π‘›ξ‚€π‘€ξ‚ξ€·πΌ,𝐹(π‘Œ)=π‘„βˆ’π΄βˆ—π‘Œβˆ’π‘‘1π΄βˆ’π΅βˆ—π‘Œβˆ’π‘‘2𝐡1/𝑠β‰₯ξ‚€π‘„βˆ’π‘„+𝑀𝑠1/𝑠=𝑀β‰₯πœ†π‘›ξ‚€π‘€ξ‚πΌ.(2.20) From (2.20), it follows that ‖𝐹(𝑋)π‘ βˆ’πΉ(π‘Œ)𝑠‖𝐹=β€–β€–β€–β€–sβˆ’1𝑖=0𝐹(𝑋)𝑖(𝐹(𝑋)βˆ’πΉ(π‘Œ))𝐹(π‘Œ)π‘ βˆ’1βˆ’π‘–β€–β€–β€–β€–πΉ=‖‖‖‖vecπ‘ βˆ’1𝑖=0𝐹(𝑋)𝑖(𝐹(𝑋)βˆ’πΉ(π‘Œ))𝐹(π‘Œ)π‘ βˆ’1βˆ’π‘–ξƒ­β€–β€–β€–β€–=β€–β€–β€–β€–π‘ βˆ’1𝑖=0ξ€Ίvec𝐹(𝑋)𝑖(𝐹(𝑋)βˆ’πΉ(π‘Œ))𝐹(π‘Œ)π‘ βˆ’1βˆ’π‘–ξ€»β€–β€–β€–β€–=β€–β€–β€–β€–π‘ βˆ’1𝑖=0𝐹(π‘Œ)π‘ βˆ’1βˆ’π‘–βŠ—πΉ(𝑋)𝑖vβ€–β€–β€–β€–β‰₯𝑒𝑐(𝐹(𝑋)βˆ’πΉ(π‘Œ))π‘ βˆ’1𝑖=0πœ†π‘›π‘ βˆ’1𝑀‖‖vec(𝐹(𝑋)βˆ’πΉ(π‘Œ))=π‘ πœ†π‘›π‘ βˆ’1𝑀‖𝐹(𝑋)βˆ’πΉ(π‘Œ)‖𝐹.(2.21)
According to the definition of the map 𝐹, we have 𝐹(𝑋)π‘ βˆ’πΉ(π‘Œ)𝑠=ξ€·π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘1π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘2π΅ξ€Έβˆ’ξ€·π‘„βˆ’π΄βˆ—π‘Œβˆ’π‘‘1π΄βˆ’π΅βˆ—π‘Œβˆ’π‘‘2𝐡=π΄βˆ—ξ€·π‘Œβˆ’π‘‘1βˆ’π‘‹βˆ’π‘‘1𝐴+π΅βˆ—ξ€·π‘Œβˆ’π‘‘2βˆ’π‘‹βˆ’π‘‘2𝐡.(2.22) Combining (2.21) and (2.22), we have by Lemma 2.5 that ‖‖𝐹(𝑋)βˆ’πΉ(π‘Œ)𝐹≀1π‘ πœ†π‘›π‘ βˆ’1𝑀‖𝐹(𝑋)π‘ βˆ’πΉ(π‘Œ)𝑠‖𝐹=1π‘ πœ†π‘›π‘ βˆ’1ξ‚€π‘€ξ‚β€–β€–π΄βˆ—ξ€·π‘Œβˆ’π‘‘1βˆ’π‘‹βˆ’π‘‘1𝐴+π΅βˆ—ξ€·π‘Œβˆ’π‘‘2βˆ’π‘‹βˆ’π‘‘2𝐡‖‖𝐹≀1π‘ πœ†π‘›π‘ βˆ’1𝑀‖𝐴‖2πΉβ€–β€–π‘Œβˆ’π‘‘1βˆ’π‘‹βˆ’π‘‘1‖‖𝐹+‖𝐡‖2πΉβ€–β€–π‘Œβˆ’π‘‘2βˆ’π‘‹βˆ’π‘‘2‖‖𝐹≀1π‘ πœ†π‘›π‘ βˆ’1𝑀𝑑1πœ†βˆ’(𝑑1𝑛+1)𝑀‖𝐴‖2𝐹+𝑑2πœ†βˆ’(𝑑2𝑛+1)𝑀‖𝐡‖2πΉξ‚β€–π‘Œβˆ’π‘‹β€–πΉ=1𝑠𝑑1πœ†βˆ’(𝑠+𝑑1)𝑛𝑀‖𝐴‖2𝐹+𝑑2πœ†βˆ’(𝑠+𝑑2)𝑛𝑀‖𝐡‖2πΉξ‚β€–π‘‹βˆ’π‘Œβ€–πΉ=π‘β€–π‘‹βˆ’π‘Œβ€–πΉ.(2.23)
Since 𝑝<1, we know that the map 𝐹(𝑋) is a contraction map in Ξ©. By Banach fixed point theorem, the map 𝐹(𝑋) has a unique fixed point in Ξ© and this shows that (1.1) has a unique HPD solution in [𝑀,𝑄1/𝑠].

Theorem 2.11. If (1.1) has an HPD solution 𝑋, then πœ†π‘›ξ€·π‘„βˆ’1/2π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄π‘„βˆ’1/2+π‘„βˆ’1/2π΅βˆ—π‘„βˆ’π‘‘2/π‘ π΅π‘„βˆ’1/2≀𝑑𝑠+𝑑t/𝑠𝑠𝑠+𝑑,𝑋≀𝛼𝑄1/𝑠,(2.24) where 𝑑=min{𝑑1,𝑑2}, and 𝛼 is a solution of the equation 𝑦𝑑(1βˆ’π‘¦π‘ )=πœ†π‘›ξ€·π‘„βˆ’1/2π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄π‘„βˆ’1/2+π‘„βˆ’1/2π΅βˆ—π‘„βˆ’π‘‘2/π‘ π΅π‘„βˆ’1/2ξ€Έ(2.25) in [(𝑑/(𝑠+𝑑))1/𝑠,1].

Proof. Consider the sequence defined as follows: 𝛼0=1,π›Όπ‘˜+1=ξƒ©πœ†1βˆ’π‘›ξ€·π‘„βˆ’1/2π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄π‘„βˆ’1/2+π‘„βˆ’1/2π΅βˆ—π‘„βˆ’π‘‘2/π‘ π΅π‘„βˆ’1/2ξ€Έπ›Όπ‘‘π‘˜ξƒͺ1/𝑠,π‘˜=0,1,2,….(2.26) Let 𝑋 be an HPD solution of (1.1), then 𝑋=π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘1π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘2𝐡1/𝑠<𝑄1/𝑠=𝛼0𝑄1/𝑠.(2.27) Assuming that 𝑋<π›Όπ‘˜π‘„1/𝑠, then by Lemma 2.1, we have 𝑋𝑠=π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘1π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘2𝐡<π‘„βˆ’π΄βˆ—ξ€·π›Όπ‘˜π‘„1/π‘ ξ€Έβˆ’π‘‘1π΄βˆ’π΅βˆ—ξ€·π›Όπ‘˜π‘„1/π‘ ξ€Έβˆ’π‘‘2𝐡𝐴<π‘„βˆ’βˆ—π‘„βˆ’π‘‘1/𝑠𝐴+π΅βˆ—π‘„βˆ’π‘‘2/π‘ π΅π›Όπ‘‘π‘˜=𝑄1/2ξƒ©π‘„πΌβˆ’βˆ’1/2π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄π‘„βˆ’1/2+π‘„βˆ’1/2π΅βˆ—π‘„βˆ’π‘‘2/π‘ π΅π‘„βˆ’1/2π›Όπ‘‘π‘˜ξƒͺ𝑄1/2≀𝑄1/2ξƒ©πœ†1βˆ’π‘›ξ€·π‘„βˆ’1/2π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄π‘„βˆ’1/2+π‘„βˆ’1/2π΅βˆ—π‘„βˆ’π‘‘2/π‘ π΅π‘„βˆ’1/2ξ€Έπ›Όπ‘‘π‘˜ξƒͺ𝑄1/2=π›Όπ‘ π‘˜+1𝑄.(2.28) Therefore 𝑋<π›Όπ‘˜+1𝑄1/𝑠. Then by the principle of induction, we get 𝑋<π›Όπ‘˜π‘„1/𝑠,π‘˜=0,1,2,….
Noting that the sequence π›Όπ‘˜ is monotonically decreasing and positive, hence π›Όπ‘˜ is convergent. Let limπ‘˜β†’βˆžπ›Όπ‘˜=𝛼, then 𝛼=(1βˆ’πœ†π‘›(π‘„βˆ’1/2π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄π‘„βˆ’1/2+π‘„βˆ’1/2π΅βˆ—π‘„βˆ’π‘‘2/π‘ π΅π‘„βˆ’1/2)/𝛼𝑑)1/𝑠, that is, 𝛼 is a solution of the equation 𝑦𝑑(1βˆ’π‘¦π‘ )=πœ†π‘›(π‘„βˆ’1/2π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄π‘„βˆ’1/2+π‘„βˆ’1/2π΅βˆ—π‘„βˆ’π‘‘2/π‘ π΅π‘„βˆ’1/2).
Consider the function 𝑓(𝑦)=𝑦𝑑(1βˆ’π‘¦π‘ ), since maxπ‘¦βˆˆ[0,1]𝑑=𝑓𝑠+𝑑1/𝑠=𝑑𝑠+𝑑𝑑/𝑠𝑠,𝑠+𝑑(2.29) from which it follows that πœ†π‘›(π‘„βˆ’1/2π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄π‘„βˆ’1/2+π‘„βˆ’1/2π΅βˆ—π‘„βˆ’π‘‘2/π‘ π΅π‘„βˆ’1/2)≀(𝑑/(𝑠+𝑑))𝑑/𝑠(𝑠/(𝑠+𝑑)).
Next we will prove that ξπ›Όβˆˆ[(𝑑/(𝑠+𝑑))1/𝑠,1]. Obviously, 𝛼≀1. On the other hand, for the sequence π›Όπ‘˜, since 𝛼0=1>(𝑑/(𝑠+𝑑))1/𝑠, we may assume that π›Όπ‘˜>(𝑑/(𝑠+𝑑))1/𝑠 without loss of generality. Then π›Όπ‘˜+1=ξƒ©πœ†1βˆ’π‘›ξ€·π‘„βˆ’1/2π΄βˆ—π‘„βˆ’π‘‘1/π‘ π΄π‘„βˆ’1/2+π‘„βˆ’1/2π΅βˆ—π‘„βˆ’π‘‘2/π‘ π΅π‘„βˆ’1/2ξ€Έπ›Όπ‘‘π‘˜ξƒͺ1/𝑠β‰₯11βˆ’π›Όπ‘‘π‘˜ξ‚€π‘‘ξ‚π‘ +𝑑𝑑/𝑠𝑠ξƒͺ𝑠+𝑑1/𝑠>ξ‚΅11βˆ’(𝑑/(𝑠+𝑑))𝑑/𝑠𝑑𝑠+𝑑𝑑/𝑠𝑠𝑠+𝑑1/𝑠=𝑑𝑠+𝑑1/𝑠.(2.30) Hence π›Όπ‘˜>(𝑑/(𝑠+𝑑))1/𝑠, π‘˜=0,1,2,…. So 𝛼=limπ‘˜β†’βˆžπ›Όπ‘˜β‰₯(𝑑/(𝑠+𝑑))1/𝑠.
Consequently, we have ξπ›Όβˆˆ[(𝑑/(𝑠+𝑑))1/𝑠,1].
This completes the proof.

Theorem 2.12. If (1.1) has an HPD solution, then (𝜌(𝐴))2≀𝑠𝑠+𝑑1𝑑1𝑠+𝑑1𝑑1/𝑠(𝜌(𝑄))(𝑑1/𝑠)+1,(2.31)(𝜌(𝐡))2≀𝑠𝑠+𝑑2𝑑2𝑠+𝑑2𝑑2/𝑠(𝜌(𝑄))(𝑑2/𝑠)+1.(2.32)

Proof. For any eigenvalue πœ†(𝐴) of 𝐴, let π‘₯ be a corresponding eigenvector. Multiplying left side of (1.1) by π‘₯βˆ— and right side by π‘₯, we have π‘₯βˆ—π‘‹π‘ π‘₯+π‘₯βˆ—π΄βˆ—π‘‹βˆ’π‘‘1𝐴π‘₯+π‘₯βˆ—π΅βˆ—π‘‹βˆ’π‘‘2𝐡π‘₯=π‘₯βˆ—π‘„π‘₯,(2.33) which yields π‘₯βˆ—π‘‹π‘ ||||π‘₯+πœ†(𝐴)2π‘₯βˆ—π‘‹βˆ’π‘‘1π‘₯+π‘₯βˆ—π΅βˆ—π‘‹βˆ’π‘‘2𝐡π‘₯=π‘₯βˆ—π‘„π‘₯.(2.34) Since 𝑋>0, there exists an unitary matrix π‘ˆ such that 𝑋=π‘ˆβˆ—Ξ›π‘ˆ, where Ξ›=diag(πœ‚1,…,πœ‚π‘›)>0. Then (2.34) turns into the following form: π‘₯βˆ—π‘ˆβˆ—Ξ›π‘ ||||π‘ˆπ‘₯+πœ†(𝐴)2π‘₯βˆ—Uβˆ—Ξ›βˆ’π‘‘1π‘ˆπ‘₯≀π‘₯βˆ—π‘„π‘₯.(2.35) Let 𝑦=(𝑦1,𝑦2,…,𝑦𝑛)𝑇=π‘ˆπ‘₯, then (2.35) reduces to π‘¦βˆ—Ξ›π‘ ||||𝑦+πœ†(𝐴)2π‘¦βˆ—Ξ›βˆ’π‘‘1π‘¦β‰€π‘¦βˆ—π‘ˆπ‘„π‘ˆβˆ—π‘¦,(2.36) from which we obtain ||πœ†||(𝐴)2β‰€π‘¦βˆ—ξ€·π‘ˆπ‘„π‘ˆβˆ—βˆ’Ξ›π‘ ξ€Έπ‘¦π‘¦βˆ—Ξ›βˆ’π‘‘1π‘¦β‰€π‘¦βˆ—ξ€·πœ†1(𝑄)πΌβˆ’Ξ›π‘ ξ€Έπ‘¦π‘¦βˆ—Ξ›βˆ’π‘‘1𝑦=βˆ‘π‘›π‘–=1𝑦2π‘–ξ€·πœ†1(𝑄)βˆ’πœ‚π‘ π‘–ξ€Έβˆ‘π‘›π‘–=1𝑦2π‘–πœ‚βˆ’π‘‘1𝑖.(2.37) Form Lemma 2.3, we know that πœ‚π‘‘1π‘–ξ€·πœ†1(𝑄)βˆ’πœ‚π‘ π‘–ξ€Έβ‰€π‘ π‘ +𝑑1𝑑1𝑠+𝑑1𝑑1/π‘ πœ†(𝑑11/𝑠)+1(𝑄),(2.38) that is, ξ€·πœ†1(𝑄)βˆ’πœ‚π‘ π‘–ξ€Έβ‰€π‘ π‘ +𝑑1𝑑1𝑠+𝑑1𝑑1/π‘ πœ†(𝑑11/𝑠)+1(𝑄)πœ‚βˆ’π‘‘1𝑖.(2.39) Noting that 𝑦≠0, we get 𝑛𝑖=1𝑦2π‘–ξ€·πœ†1(𝑄)βˆ’πœ‚π‘ π‘–ξ€Έβ‰€π‘ π‘ +𝑑1𝑑1𝑠+𝑑1𝑑1/π‘ πœ†(𝑑11/𝑠)+1(𝑄)𝑛𝑖=1𝑦2π‘–πœ‚βˆ’π‘‘1𝑖.(2.40) Consequently, ||πœ†||(𝐴)2β‰€βˆ‘π‘›π‘–=1𝑦2π‘–ξ€·πœ†1(𝑄)βˆ’πœ‚π‘ π‘–ξ€Έβˆ‘π‘›π‘–=1𝑦2π‘–πœ‚βˆ’π‘‘1𝑖≀𝑠𝑠+𝑑1𝑑1𝑠+𝑑1𝑑1/π‘ πœ†(𝑑11/𝑠)+1(𝑄).(2.41)
Then (𝜌(𝐴))2≀(𝑠/(𝑠+𝑑1))(𝑑1/(𝑠+𝑑1))𝑑1/π‘ πœ†(𝑑11/𝑠)+1(𝑄).
Since 𝑄>0, clearly denote πœ†1(𝑄)=𝜌(𝑄), and the last inequality implies directly (2.31).
The proof of (2.32) is similar to that of (2.31), thus it is omitted here.

Theorem 2.13. If 𝑄≀𝐼 and (1.1) has an HPD solution, then πœŒξ‚€π΄π‘ /𝑑1+ξ€·π΄βˆ—ξ€Έπ‘ /𝑑1ξ‚ξ‚€π΄β‰€πœŒ(𝑄),πœŒπ‘ /𝑑1βˆ’ξ€·π΄βˆ—ξ€Έπ‘ /𝑑1ξ‚β‰€πœŒ(𝑄),(2.42)πœŒξ‚€π΅π‘ /𝑑2+ξ€·π΅βˆ—ξ€Έπ‘ /𝑑2ξ‚ξ‚€π΅β‰€πœŒ(𝑄),πœŒπ‘ /𝑑2βˆ’ξ€·π΅βˆ—ξ€Έπ‘ /𝑑2ξ‚β‰€πœŒ(𝑄).(2.43)

Proof. If (1.1) has an HPD solution, we have by Theorem 2.7 that 𝐿𝐴=βˆ—πΏξ€Έπ‘‘1/2𝑠𝑁1𝐿,𝐡=βˆ—πΏξ€Έπ‘‘2/2𝑠𝑁2,(2.44) and the matrix ξ‚΅πΏπ‘„βˆ’1/2𝑁1π‘„βˆ’1/2𝑁2π‘„βˆ’1/2ξ‚Ά is column orthonormal. From which we have πΏβˆ—πΏ+π‘βˆ—1𝑁1+π‘βˆ—2𝑁2=𝑄.(2.45) Hence, ξ‚€π΄π‘„βˆ’π‘ /𝑑1+ξ€·π΄βˆ—ξ€Έπ‘ /𝑑1=πΏβˆ—πΏ+π‘βˆ—1𝑁1+π‘βˆ—2𝑁2βˆ’ξ€·πΏβˆ—πΏξ€Έ1/2𝑁𝑠/𝑑11βˆ’ξ€·π‘βˆ—1𝑠/𝑑1ξ€·πΏβˆ—πΏξ€Έ1/2=ξ‚€ξ€·πΏβˆ—πΏξ€Έ1/2βˆ’π‘π‘ /𝑑11ξ‚βˆ—ξ‚€ξ€·πΏβˆ—πΏξ€Έ1/2βˆ’π‘π‘ /𝑑11+π‘βˆ—2𝑁2+ξ‚€π‘βˆ—1𝑁1βˆ’ξ€·π‘βˆ—1𝑠/𝑑1𝑁𝑠/𝑑11β‰₯0.(2.46)
Similarly, we have 𝑄+(𝐴𝑠/𝑑1+(π΄βˆ—)𝑠/𝑑1)β‰₯0.
Thus, βˆ’π‘„β‰€(𝐴𝑠/𝑑1+(π΄βˆ—)𝑠/𝑑1)≀𝑄. Hence 𝜌(𝐴𝑠/𝑑1+(π΄βˆ—)𝑠/𝑑1)β‰€πœŒ(𝑄).
On the other hand, by Lemma 2.6 and (2.2), we get πœŒξ‚€π΄π‘ /𝑑1βˆ’ξ€·π΄βˆ—ξ€Έπ‘ /𝑑1𝐿=πœŒβˆ—πΏξ€Έ1/2𝑁𝑠/𝑑11βˆ’ξ€·π‘βˆ—1𝑠/𝑑1ξ€·πΏβˆ—πΏξ€Έ1/2ξ‚ξ‚€πΏβ‰€πœŒβˆ—ξ€·π‘πΏ+βˆ—1𝑠/𝑑1𝑁𝑠/𝑑11ξ‚ξ€·πΏβ‰€πœŒβˆ—πΏ+π‘βˆ—1𝑁1ξ€Έβ‰€πœŒ(𝑄).(2.47) The proof of (2.43) is similar to that of (2.42).

If 𝑑1=𝑑2, we denote 𝑑=𝑑1=𝑑2. Then (1.1) turns into𝑋𝑠+π΄βˆ—π‘‹βˆ’π‘‘π΄+π΅βˆ—π‘‹βˆ’π‘‘π΅=𝑄.(2.48) Consider the following equations:π‘₯𝑠+π‘‘βˆ’πœ†π‘›(𝑄)π‘₯𝑑+πœ†1ξ€·π΄βˆ—π΄ξ€Έ+πœ†1ξ€·π΅βˆ—π΅ξ€Έ=0,(2.49)π‘₯𝑠+π‘‘βˆ’πœ†1(𝑄)π‘₯𝑑+πœ†π‘›ξ€·π΄βˆ—π΄ξ€Έ+πœ†π‘›ξ€·π΅βˆ—π΅ξ€Έ=0.(2.50)

We assume that 𝐴,𝐡, and 𝑄 satisfyπœ†1ξ€·π΄βˆ—π΄ξ€Έ+πœ†1ξ€·π΅βˆ—π΅ξ€Έ<π‘ πœ‰π‘ +π‘‘π‘‘βˆ—πœ†π‘›(𝑄),(2.51) where πœ‰βˆ—=((𝑑/(𝑠+𝑑))πœ†π‘›(𝑄))1/𝑠. By (2.51) and Lemma 2.3, we know that (2.49) has two positive real roots 𝛼2<𝛽1. We also get that (2.50) has two positive real roots 𝛼1<𝛽2. It is easy to prove that 0<𝛼1≀𝛼2<πœ‰βˆ—<𝛽1≀𝛽2<πœ†11/𝑠(𝑄).(2.52)

We define matrix sets as follows: πœ‘1=𝑋=π‘‹βˆ—βˆ£0<𝑋<𝛼1𝐼,πœ‘2=𝑋=π‘‹βˆ—βˆ£π›Ό1𝐼≀𝑋≀𝛼2𝐼,πœ‘3=𝑋=π‘‹βˆ—βˆ£π›Ό2𝐼<𝑋<𝛽1𝐼,πœ‘4=𝑋=π‘‹βˆ—βˆ£π›½1𝐼≀𝑋≀𝛽2𝐼,πœ‘5=𝑋=π‘‹βˆ—βˆ£π›½2𝐼<𝑋<πœ†11/𝑠.(𝑄)𝐼(2.53)

Theorem 2.14. Suppose that 𝐴,𝐡, and 𝑄 satisfy (2.51), that is, πœ†1ξ€·π΄βˆ—π΄ξ€Έ+πœ†1ξ€·π΅βˆ—π΅ξ€Έ<𝑠𝑑𝑠+𝑑𝑠+𝑑𝑑/π‘ πœ†π‘›(𝑑/𝑠)+1(𝑄).(2.54) Then(i) Equation (2.48) has a unique HPD solution in πœ‘4;(ii) Equation (2.48) has no HPD solution in πœ‘1,πœ‘3,πœ‘5.

Proof. Consider the map 𝐺(𝑋)=(π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘π΅)1/𝑠, which is continuous on πœ‘4. Obviously, πœ‘4 is a convex, closed, and bounded set. If π‘‹βˆˆπœ‘4, πœ†π‘ 1(𝐺(𝑋))=πœ†1(𝐺(𝑋)𝑠)=πœ†1ξ€·π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘π΅ξ€Έβ‰€πœ†1πœ†(𝑄)βˆ’π‘›ξ€·π΄βˆ—π΄ξ€Έ+πœ†π‘›ξ€·π΅βˆ—π΅ξ€Έπœ†π‘‘1(𝑋)β‰€πœ†1πœ†(𝑄)βˆ’π‘›ξ€·π΄βˆ—π΄ξ€Έ+πœ†π‘›ξ€·π΅βˆ—π΅ξ€Έπ›½π‘‘2=𝛽𝑠2.(2.55) Hence, we have πœ†1(𝐺(𝑋))<𝛽2. One has πœ†π‘ π‘›(𝐺(𝑋))=πœ†π‘›(𝐺(𝑋)𝑠)=πœ†π‘›ξ€·π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘π΅ξ€Έβ‰₯πœ†π‘›πœ†(𝑄)βˆ’1ξ€·π΄βˆ—π΄ξ€Έ+πœ†1ξ€·π΅βˆ—π΅ξ€Έπœ†π‘‘π‘›(𝑋)β‰₯πœ†π‘›πœ†(𝑄)βˆ’1ξ€·π΄βˆ—π΄ξ€Έ+πœ†1ξ€·π΅βˆ—π΅ξ€Έπ›½π‘‘1=𝛽𝑠1.(2.56) Hence, we have πœ†π‘›(𝐺(𝑋))<𝛽1.
Thus, 𝐺(𝑋) maps πœ‘4 into itself.
For arbitrary 𝑋,π‘Œβˆˆπœ‘4, similar to (2.21) and (2.22), we have ‖𝐺(𝑋)π‘ βˆ’πΊ(π‘Œ)𝑠‖𝐹β‰₯𝑠𝛽1π‘ βˆ’1‖‖𝐺(𝑋)βˆ’πΊ(π‘Œ)𝐹,𝐺(𝑋)π‘ βˆ’πΊ(π‘Œ)𝑠=π΄βˆ—ξ€·π‘Œβˆ’π‘‘βˆ’π‘‹βˆ’π‘‘ξ€Έπ΄+π΅βˆ—ξ€·π‘Œβˆ’π‘‘βˆ’π‘‹βˆ’π‘‘ξ€Έπ΅.(2.57) Combining (2.57), we have by Lemma 2.5 and (2.49) ‖‖𝐺(𝑋)βˆ’πΊ(π‘Œ)𝐹≀1𝑠𝛽1π‘ βˆ’1‖𝐺(𝑋)π‘ βˆ’πΊ(π‘Œ)𝑠‖𝐹=1𝑠𝛽1π‘ βˆ’1β€–β€–π΄βˆ—ξ€·π‘Œβˆ’π‘‘βˆ’π‘‹βˆ’π‘‘ξ€Έπ΄+π΅βˆ—ξ€·π‘Œβˆ’π‘‘βˆ’π‘‹βˆ’π‘‘ξ€Έπ΅β€–β€–πΉβ‰€1𝑠𝛽1π‘ βˆ’1‖𝐴‖22+‖𝐡‖22ξ€Έβ€–β€–π‘Œβˆ’π‘‘βˆ’π‘‹βˆ’π‘‘β€–β€–πΉβ‰€1𝑠𝛽1π‘ βˆ’1ξ€·πœ†1ξ€·π΄βˆ—π΄ξ€Έ+πœ†1ξ€·π΅βˆ—π΅ξ€Έξ€Έπ‘‘π›½1βˆ’(𝑑+1)β€–π‘Œβˆ’π‘‹β€–πΉ=π‘‘π‘ πœ†1ξ€·π΄βˆ—π΄ξ€Έ+πœ†1ξ€·π΅βˆ—π΅ξ€Έπ›½1𝑠+π‘‘β€–π‘‹βˆ’π‘Œβ€–πΉ=π‘‘π‘ ξ‚΅πœ†π‘›(𝑄)𝛽𝑠1ξ‚Άβˆ’1β€–π‘‹βˆ’π‘Œβ€–πΉ<β€–π‘‹βˆ’π‘Œβ€–πΉ.(2.58)
Thus, we know that the map 𝐺(𝑋) is a contraction map in πœ‘4. By Banach fixed point theorem, the map 𝐺(𝑋) has a unique fixed point in πœ‘4 and this shows that (2.48) has a unique HPD solution in πœ‘4.
Assume 𝑋 is the HPD solution of (2.48), then πœ†π‘ 1(𝑋)=πœ†1(𝑋𝑠)=πœ†1ξ€·π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘π΅ξ€Έβ‰€πœ†1πœ†(𝑄)βˆ’π‘›ξ€·π΄βˆ—π΄ξ€Έ+πœ†π‘›ξ€·π΅βˆ—π΅ξ€Έπœ†π‘‘1,(𝑋)(2.59) that is, πœ†1𝑠+𝑑(𝑋)βˆ’πœ†1(𝑄)πœ†π‘‘1(𝑋)+πœ†π‘›(π΄βˆ—π΄)+πœ†π‘›(π΅βˆ—π΅)≀0. So, 𝛼1β‰€πœ†1(𝑋)≀𝛽2, thus (2.48) has no HPD solution in πœ‘1,πœ‘5. πœ†π‘ π‘›(𝑋)=πœ†π‘›(𝑋𝑠)=πœ†π‘›ξ€·π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘π΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘π΅ξ€Έβ‰₯πœ†π‘›πœ†(𝑄)βˆ’1ξ€·π΄βˆ—π΄ξ€Έ+πœ†1ξ€·π΅βˆ—π΅ξ€Έπœ†π‘‘π‘›,(𝑋)(2.60) that is, πœ†π‘›π‘ +𝑑(𝑋)βˆ’πœ†π‘›(𝑄)πœ†π‘‘π‘›(𝑋)+πœ†1(π΄βˆ—π΄)+πœ†1(π΅βˆ—π΅)β‰₯0. So, πœ†π‘›(𝑋)≀𝛼2 or πœ†π‘›(𝑋)β‰₯𝛽1, thus (2.48) has no HPD solution in πœ‘3.
This completes the proof.

3. Iterative Method for the Maximal HPD Solution

In this section, we consider the iterative method for obtaining the maximal HPD solution 𝑋𝐿 of (1.1). We propose the following algorithm which avoids calculating matrix inversion in the process of iteration.

Algorithm 1. Step 1. Input initial matrices: 𝑋0=𝛾𝑄1/𝑠,π‘Œ0=𝛾+1𝑄2π›Ύβˆ’1/𝑠,(3.1) where π›Ύβˆˆ(𝛼,1), and 𝛼 is defined in Theorem 2.11.Step 2. For π‘˜=0,1,2,…, compute π‘Œπ‘˜+1=π‘Œπ‘˜ξ€·2πΌβˆ’π‘‹π‘˜π‘Œπ‘˜ξ€Έ,π‘‹π‘˜+1=ξ€·π‘„βˆ’π΄βˆ—π‘Œπ‘‘1π‘˜+1π΄βˆ’π΅βˆ—π‘Œπ‘‘2π‘˜+1𝐡1/𝑠.(3.2)

Theorem 3.1. If (1.1) has an HPD solution, then it has the maximal one 𝑋𝐿. Moreover, to the sequences π‘‹π‘˜ and π‘Œπ‘˜ generated by Algorithm 1, one has 𝑋0>𝑋1>𝑋2>…,limπ‘˜β†’βˆžπ‘‹π‘˜=𝑋𝐿;π‘Œ0<π‘Œ1<π‘Œ2<β‹―,limπ‘˜β†’βˆžπ‘Œπ‘˜=π‘‹πΏβˆ’1.(3.3)

Proof. Since 𝑋𝐿 is an HPD solution of (1.1), by Theorem 2.11, we have 𝑋𝐿≀𝛼𝑄1/𝑠, thus 𝑋0=𝛾𝑄1/𝑠>𝛼𝑄1/𝑠β‰₯𝑋𝐿,π‘Œ0=𝛾+1𝑄2π›Ύβˆ’1/𝑠<1π›Ύπ‘„βˆ’1/𝑠<1π‘„ξπ›Όβˆ’1/π‘ β‰€π‘‹πΏβˆ’1.(3.4) By Lemmas 2.1 and 2.4, we have π‘Œ1=π‘Œ0ξ€·2πΌβˆ’π‘‹0π‘Œ0ξ€Έ=2π‘Œ0βˆ’π‘Œ0𝑋0π‘Œ0≀𝑋0βˆ’1<π‘‹πΏβˆ’1,π‘Œ1βˆ’π‘Œ0=π‘Œ0βˆ’π‘Œ0𝑋0π‘Œ0=π‘Œ0ξ€·π‘Œ0βˆ’1βˆ’π‘‹0ξ€Έπ‘Œ0=1βˆ’π›Ύ2𝑄4π›Ύβˆ’1/𝑠>0.(3.5) According to Lemma 2.1 and π‘Œ1<π‘‹πΏβˆ’1, we have 𝑋1=ξ€·π‘„βˆ’π΄βˆ—π‘Œπ‘‘11π΄βˆ’π΅βˆ—π‘Œπ‘‘21𝐡1/𝑠>ξ€·π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘1πΏπ΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘2𝐿𝐡1/𝑠=𝑋𝐿,𝑋𝑠1βˆ’π‘‹π‘ 0=βˆ’π΄βˆ—ξ€·π‘Œπ‘‘11βˆ’π‘Œπ‘‘10ξ€Έπ΄βˆ’π΅βˆ—ξ€·π‘Œπ‘‘21βˆ’π‘Œπ‘‘20𝐡<0,(3.6) that is, 𝑋𝑠1<𝑋𝑠0, by Lemma 2.1 again, it follows that 𝑋1<𝑋0.
Hence 𝑋0>𝑋1>𝑋𝐿, and π‘Œ0<π‘Œ1<π‘‹πΏβˆ’1.
Assume that π‘‹π‘˜βˆ’1>π‘‹π‘˜>𝑋𝐿, and π‘Œπ‘˜βˆ’1<π‘Œπ‘˜<π‘‹πΏβˆ’1, we will prove the inequalities π‘‹π‘˜>π‘‹π‘˜+1>𝑋𝐿, and π‘Œπ‘˜<π‘Œπ‘˜+1<π‘‹πΏβˆ’1.
By Lemmas 2.1 and 2.4, we have π‘Œπ‘˜+1=2π‘Œπ‘˜βˆ’π‘Œπ‘˜π‘‹π‘˜π‘Œπ‘˜β‰€π‘‹π‘˜βˆ’1<π‘‹πΏβˆ’1,π‘‹π‘˜+1=ξ€·π‘„βˆ’π΄βˆ—π‘Œπ‘‘1π‘˜+1π΄βˆ’π΅βˆ—π‘Œπ‘‘2π‘˜+1𝐡1/𝑠>ξ€·π‘„βˆ’π΄βˆ—π‘‹βˆ’π‘‘1πΏπ΄βˆ’π΅βˆ—π‘‹βˆ’π‘‘2𝐿𝐡1/𝑠=𝑋𝐿.(3.7) Since π‘Œπ‘˜β‰€π‘‹βˆ’1π‘˜βˆ’1<π‘‹π‘˜βˆ’1, we have π‘Œπ‘˜βˆ’1>π‘‹π‘˜, thus we have by Lemma 2.1 that π‘Œπ‘˜+1βˆ’π‘Œπ‘˜=π‘Œπ‘˜ξ€·π‘Œπ‘˜βˆ’1βˆ’π‘‹π‘˜ξ€Έπ‘Œπ‘˜π‘‹>0,π‘ π‘˜+1βˆ’π‘‹π‘ π‘˜=βˆ’π΄βˆ—ξ€·π‘Œπ‘‘1π‘˜+1βˆ’π‘Œπ‘‘1π‘˜ξ€Έπ΄βˆ’π΅βˆ—ξ€·π‘Œπ‘‘2π‘˜+1βˆ’π‘Œπ‘‘2π‘˜ξ€Έπ΅<0,(3.8) that is, π‘‹π‘ π‘˜+1<π‘‹π‘ π‘˜, by Lemma 2.1 again, it follows that π‘‹π‘˜+1<π‘‹π‘˜.
Hence we have by induction that 𝑋0>𝑋1>𝑋2>β‹―>π‘‹π‘˜>𝑋𝐿,π‘Œ0<π‘Œ1<π‘Œ2<β‹―<π‘Œπ‘˜<π‘‹πΏβˆ’1(3.9) are true for all π‘˜=0,1,2,…, and so limπ‘˜β†’βˆžπ‘‹π‘˜ and limπ‘˜β†’βˆžπ‘Œπ‘˜ exist. Suppose limπ‘˜β†’βˆžπ‘‹π‘˜=𝑋, limπ‘˜β†’βˆžπ‘Œπ‘˜=ξπ‘Œ, taking the limit in the Algorithm 1 leads to ξξπ‘‹π‘Œ=βˆ’1 and 𝑋=(π‘„βˆ’π΄βˆ—ξπ‘‹βˆ’π‘‘1π΄βˆ’π΅βˆ—ξπ‘‹βˆ’π‘‘2𝐡)1/𝑠. Therefore 𝑋 is an HPD solution of (1.1), thus 𝑋≀𝑋𝐿. Moreover, as each π‘‹π‘˜>𝑋𝐿, so 𝑋β‰₯𝑋𝐿, then 𝑋=𝑋𝐿. The theorem is proved.

Theorem 3.2. If (1.1) has an HPD solution and after π‘˜ iterative steps of Algorithm 1, one has β€–πΌβˆ’π‘‹π‘˜π‘Œπ‘˜β€–<πœ€, then β€–β€–π‘‹π‘ π‘˜+π΄βˆ—π‘‹βˆ’π‘‘1π‘˜π΄+π΅βˆ—π‘‹βˆ’π‘‘2π‘˜β€–β€–π΅βˆ’π‘„β‰€πœ€πœ†π‘›βˆ’1𝑀𝑑1πœ†(1βˆ’π‘‘11)/𝑠‖(𝑄)𝐴‖2+𝑑2πœ†(1βˆ’π‘‘21)/𝑠‖(𝑄)𝐡‖2,(3.10) where 𝑀 is defined by (2.13).

Proof. From the proof of Theorem 3.1, we have π‘„βˆ’1/𝑠<((𝛾+1)/2𝛾)π‘„βˆ’1/𝑠<π‘Œπ‘˜<π‘‹π‘˜βˆ’1<π‘‹πΏβˆ’1 for all π‘˜=1,2,…. Thus we have by Theorem 2.9 that π‘„βˆ’1/𝑠<π‘Œπ‘˜<π‘‹π‘˜βˆ’1<π‘€βˆ’1. And this implies πœ†1βˆ’1/𝑠(𝑄)𝐼<π‘Œπ‘˜<π‘‹π‘˜βˆ’1<πœ†π‘›βˆ’1𝑀𝐼.(3.11) Since π‘‹π‘ π‘˜+π΄βˆ—π‘‹βˆ’π‘‘1π‘˜π΄+π΅βˆ—π‘‹βˆ’π‘‘2π‘˜ξ€·π΅βˆ’π‘„=π‘„βˆ’π΄βˆ—π‘Œπ‘‘1π‘˜π΄βˆ’π΅βˆ—π‘Œπ‘‘2π‘˜π΅ξ€Έ+π΄βˆ—π‘‹βˆ’π‘‘1π‘˜π΄+π΅βˆ—π‘‹βˆ’π‘‘2π‘˜π΅βˆ’π‘„=π΄βˆ—ξ€·π‘‹βˆ’π‘‘1π‘˜βˆ’π‘Œπ‘‘1π‘˜ξ€Έπ΄+π΅βˆ—ξ€·π‘‹βˆ’π‘‘2π‘˜βˆ’π‘Œπ‘‘2π‘˜ξ€Έπ΅,(3.12) we have by Lemma 2.5 that β€–β€–π‘‹π‘ π‘˜+π΄βˆ—π‘‹βˆ’π‘‘1π‘˜π΄+π΅βˆ—π‘‹βˆ’π‘‘1π‘˜β€–β€–=β€–β€–π΄π΅βˆ’π‘„βˆ—ξ€·π‘‹βˆ’π‘‘1π‘˜βˆ’π‘Œπ‘‘1π‘˜ξ€Έπ΄+π΅βˆ—ξ€·π‘‹βˆ’π‘‘2π‘˜βˆ’π‘Œπ‘‘2π‘˜ξ€Έπ΅β€–β€–β‰€β€–π΄β€–2β€–β€–π‘‹βˆ’π‘‘1π‘˜βˆ’π‘Œπ‘‘1π‘˜β€–β€–+‖𝐡‖2β€–β€–π‘‹βˆ’π‘‘2π‘˜βˆ’π‘Œπ‘‘2π‘˜β€–β€–β‰€ξ‚€π‘‘1πœ†βˆ’(𝑑11βˆ’1)/𝑠(𝑄)‖𝐴‖2+𝑑2πœ†βˆ’(𝑑21βˆ’1)/𝑠(𝑄)‖𝐡‖2ξ‚β€–β€–π‘‹π‘˜βˆ’1βˆ’π‘Œπ‘˜β€–β€–β‰€ξ‚€π‘‘1πœ†(1βˆ’π‘‘11)/𝑠(𝑄)‖𝐴‖2+𝑑2πœ†(1βˆ’π‘‘21)/𝑠(𝑄)‖𝐡‖2ξ‚β€–β€–π‘‹π‘˜βˆ’1β€–β€–β€–β€–πΌβˆ’π‘‹π‘˜π‘Œπ‘˜β€–β€–β‰€πœ€πœ†π‘›βˆ’1𝑀𝑑1πœ†(1βˆ’π‘‘11)/𝑠‖(𝑄)𝐴‖2+𝑑2πœ†(1βˆ’π‘‘21)/𝑠‖(𝑄)𝐡‖2.(3.13)

4. Numerical Example

In this section, we give a numerical example to illustrate the efficiency of the proposed algorithm. All the tests are performed by MATLAB 7.0 with machine precision around 10βˆ’16. We stop the practical iteration when the residual ‖𝑋sπ‘˜+π΄βˆ—π‘‹βˆ’π‘‘1π‘˜π΄+π΅βˆ—π‘‹βˆ’π‘‘2π‘˜π΅βˆ’π‘„β€–πΉβ‰€1.0π‘’βˆ’010.

Example 4.1. Let 𝑠=5, 𝑑1=0.2, 𝑑2=0.5, and βŽ›βŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽβŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽ βŽ›βŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽβŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽ ,βŽ›βŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽβŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽ .𝐴=2001001