# The Feasibility of Homotopy Continuation Method for a Nonlinear Matrix Equation

Authors: Mengran Wang; Jing Li
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

In this paper, we discuss the feasibility of homotopy continuation method for the nonlinear matrix equationsX+∑i=1sBi∗X−1Bi+∑i=s+1mBi∗XtiBi=I with 0<ti<1. This iterative method does not depend on a good initial approximation to the solution of matrix equation.


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## Body

## 1. Introduction

In this paper, we consider the Hermitian positive definite (HPD) solutions of the nonlinear matrix equation:(1)X+∑i=1sBi∗X−1Bi+∑i=s+1mBi∗XtiBi=I,0<ti<1,where Bii=1,2,…,m are n×n complex nonsingular matrices, I is an n×n identity matrix, and s≥1,m≥2 are positive integers. Here, Bi∗ denotes the conjugate transpose of the matrix Bi.The nonlinear matrix equations of (1) or some special cases are applicable to many fields such as nanoresearch, ladder networks, dynamic programming, control theory, stochastic filtering, and statistics [1–8].Equation (1) is recognized as playing an important role in solving a system of linear equations. For example, in many physical calculations, one must solve the system of linear equation Mx=f, where x and f are column vectors, and(2)M=I0⋯0A10I⋯0A2⋮⋮⋱⋮⋮00⋯IAmA1∗A2∗⋯Am∗Iarises in a finite difference approximation to an elliptic partial differential equation (for more information, refer to [2]). We can rewrite M as M=M˜+D, where(3)M˜=X⋯00⋯0A1⋮⋱⋮⋮⋮⋮0⋯X0⋯0As0⋯0Xps+1⋯0As+1⋮⋱⋮⋮⋱⋮⋮0⋯00⋯XpmAmA1∗⋯As∗As+1∗⋯Am∗I,D˜=I−X⋯00⋯00⋮⋱⋮⋮⋮⋮0⋯I−X0⋯000⋯0I−Xps+1⋯00⋮⋱⋮⋮⋱⋮⋮0⋯00⋯I−Xpm00⋯00⋯00.M
˜ can be factored as M˜=M1˜M2˜ if and only if X is a solution of equation X+∑i=1sAi∗X−1Ai+∑i=s+1mAi∗XpiAi=I, where(4)M1˜=I⋯00⋯00⋮⋱⋮⋮⋮⋮0⋯I0⋯000⋯0I⋯00⋮⋱⋮⋮⋱⋮⋮0⋯00⋯I0A1∗X−1⋯As∗X−1As+1∗Xps+1⋯Am∗XpmI,M2˜=X⋯00⋯0A1⋮⋱⋮⋮⋮⋮0⋯X0⋯0As0⋯0Xps+1⋯0As+1⋮⋱⋮⋮⋱⋮⋮0⋯00⋯XpmAm0⋯00⋯0X.Some special cases of matrix equation (1) and related matrix equations were studied on the solvability, numerical methods, and perturbation analysis by many scholars: X+A∗X−1A=P [9–12]; Xs+A∗X−qA=I [13–17]; X+A∗XqA=Q [18]; X+∑i=1mAi∗X−1Ai=I [19, 20]; Xr+∑i=1mAi∗XδiAi=I [21, 22]; X+∑i=1mAi∗X−qAi=Q [23]; X+∑i=1mBi∗XtiBi=I [24]; and Xs±A∗FXA=Q [25–29]. In addition, many homotopy approaches oriented toward engineering applications were investigated by many scholars: the homotopy analysis method (HAM) [30], the Optimal Homotopy Asymptotic Method (OHAM) [31], the Optimal Homotopy Perturbation Method (OHPM) [32], the homotopy asymptotic method [33], and so on.According to our knowledge, the matrix equation (1) has not been treated explicitly in the literatures. The reason is that (1) does not always have a unique Hermitian positive definite solution. It is hard to find sufficient conditions for the existence of a unique Hermitian positive definite solution because the map FX=I−∑i=1sX−1BiBi∗−∑i=s+1mBi∗XtiBi,0<ti<1 is not monotonic. There are two difficulties for discussing the solvability and iterative method for the matrix equation (1). One is how to find a suitable set and some reasonable restrictions on the coefficient matrices ensuring this equation has a unique Hermitian positive definite solution in this set. The other one is how to find a reasonable expression of FX−FY, which was important for discussing the feasibility of the homotopy continuation iterative method (see [34], for more details), which was not dependent on a good initial approximation to the solution of matrix equation.In this paper, firstly, we derive necessary and sufficient conditions for the existence of Hermitian positive definite solutions to equation (1) in Section 3. Then, discuss the homotopy continuation methods for obtaining the unique Hermitian positive definite solution in Section 4.The following notations are used throughout this paper. We denote byCn×n, ℋn×n, and Un×n the set of all n×n complex matrices, Hermitian matrices, and unitary matrices, respectively. For A=a1,…,an=aij∈Cn×n and a matrix B, A⊗B=aijB is a Kronecker product, and vecA is a vector defined by vecA=a1T,…,anTT. The symbol · stands for the spectral norm. We denote by λiM the eigenvalues of M, by λ1M and λnM the maximal and minimal eigenvalues of M, respectively. For X,Y∈ℋn×n, we write X≥YX>Y, if X−Y is a Hermitian positive semidefinite (definite) matrix. For A,B∈ℋn×n, the sets A,B and A,B are defined by A,B=X∈ℋn×nA≤X≤B and A,B=X∈ℋn×nA<X≤B.


## 2. Preliminaries

In this section, we present some lemmas that will be needed to develop this paper.Lemma 1.
(see [35]). If A≥B>0 and 0≤γ≤1, then Aγ≥Bγ.Lemma 2.
(see [24], Lemma 3.3). For every Hermitian positive definite matrix X and 0<t<1, it yields that Xt=sintπ/π∫0∞XλI+X−1λt−1dλ.Lemma 3.
(see [36], Theorem 1.9.1). Let A∈Cm×n,B∈Cp×q,C∈Cn×k,andD∈Cq×r. Then,A⊗BC⊗D=AC⊗BD,A⊗B∗=A∗⊗B∗.Lemma 4.
(see [36], Lemma 1.9.1). Let A∈Cl×m,X∈Cm×n,andB∈Cn×k. Then,(5)vecAXB=BT⊗A·vecX.Lemma 5.
(see [37], Theorem 6.19). Let A∈Cm×m and B∈Cn×n with eigenvalues λi and μj,i=1,2,…,m,j=1,2,…,n, respectively. Then, the eigenvalues of A⊗B are λiμj,i=1,2,…,m and j=1,2,…,n.Lemma 6.
(see [24], Lemma 3.2). Suppose that m≥1,0<t<1, and mt/mt+1<x,y<1. Then,(6)0<fx,y,t=1−x1−yxt−ytx−yxt/2yt/2<1m.


## 3. Hermitian Positive Definite Solutions

In this section, some sufficient and necessary conditions for the existence and uniqueness of HPD solution of (1) are derived.Theorem 1.
(1) has a HPD solution if and only if there exist Qi∈Cn×n,i=1,2,…,m, P∈Un×n, and diagonal matrices Γ,Λ>0 such that(7)Bi=P∗Γ1/2QiΛP,i=1,2,…,s,P∗Γ−ti/2QiΛP,i=s+1,s+2,…,m,where Λ2+Γ=I and ∑i=1mQi∗Qi=I. In this case, Y=P∗ΓP is a HPD solution of (1).Proof.
AssumeY is an HPD solution of (1). According to the spectral decomposition theorem, we have that there exists P∈Un×n and a diagonal matrix Γ>0 such that Y=P∗ΓP. Then, (1) can be expressed as(8)P∗ΓP+∑i=1sBi∗P∗Γ−1PBi+∑i=s+1mBi∗P∗ΓtiPBi=I.
Multiplying the left side of (8) by P and the right side by P∗, we obtain(9)∑i=1sPBi∗P∗Γ−1PBiP∗+∑i=s+1mPBi∗P∗ΓtiPBiP∗=I−Γ.
Recall thatBii=1,2,…,m are nonsingular matrices. Then,(10)0<Γ<I.
Therefore, we can rewrite (10) as(11)∑i=1sI−Γ−1/2PBi∗P∗Γ−1PBiP∗I−Γ−1/2+∑i=s+1mI−Γ−1/2PBi∗P∗ΓtiPBiP∗I−Γ−1/2=I.
Let(12)Λ=I−Γ1/2,Qi=Γ−1/2PBiP∗Λ−1,i=1,2,…,s,Γti/2PBiP∗Λ−1,i=s+1,s+2,…,m.
It is easy to verify thatΓ+Λ2=I and(13)Bi=P∗Γ1/2QiΛP,i=1,2,…,s,P∗Γ−ti/2QiΛP,i=s+1,s+2,…,m.By (11), we obtain ∑i=1mQi∗Qi=I.
Conversely, assume there existP∈Un×n, Qi∈Cn×n, ∑i=1mQi∗Qi=I, and diagonal matrices Γ,Λ>0, Λ2+Γ=I such that(14)Bi=P∗Γ1/2QiΛP,i=1,2,…,s,P∗Γ−ti/2QiΛP,i=s+1,s+2,…,m.Let Y=P∗ΓP such that(15)Y+∑i=1sBi∗Y−1Bi+∑i=s+1mBi∗XtiBi=P∗ΓP+∑i=1sP∗Λ∗Qi∗Γ1/2PP∗ΓP−1P∗Γti/2QiΛP+∑i=s+1mP∗Λ∗Qi∗Γ−ti/2PP∗ΓPtiP∗Γ−ti/2QiΛP=P∗ΓP+∑i=1mP∗ΛQi∗QiΛP=P∗Γ+Λ2P=I,which means Y is an HPD solution of (1).Theorem 2.
Ifλ1∑i=1mBi∗Bi<m/m+12, then (1) has a unique HPD solution X on m/m+1I,I.Proof.Step 1.
We will prove that (1) has a HPD solution on m/m+1I,I under the assumption λ1∑i=1mBi∗Bi<m/m+12.
LetΩ=m/m+1I,I. Define(16)FX=I−∑i=1sBi∗X−1Bi−∑i=s+1mBi∗XtiBi,X∈Ω.
Obviously,Ω is a bounded convex closed set and F is continuous on Ω.
Note that0<ti<1. For any X∈Ω, it follows from Lemma 1 that m/m+1tiI≤Xti≤I, which implies(17)I≥FX≥I−∑i=1sm+1mBi∗Bi−∑i=s+1mBi∗Bi≥I−m+1m∑i=1mBi∗Bi.Recall that λ1∑i=1mBi∗Bi<m/m+12, then I≥FX≥m/m+1I. That is, FX⊆Ω. By Brouwer’s fixed point theorem, the map F has a fixed point X∈Ω, which is a HPD solution of (1).Step 2.
We will prove that if (1) has a HPD solution on m/m+1I,I, then the HPD solution is unique.
IfY1 is a HPD solution of (1), according to Lemma 3.1, there exist P1∈Un×n, Qi∈Cn×n,i=1,2,…,m, and diagonal matrices Γ1,Λ1>0 such that(18)Bi=P1∗Γ11/2QiΛ1P1,i=1,2,…,s,P1∗Γ1−ti/2QiΛ1P1,i=s+1,s+2,…,m,where(19)∑i=1mQi∗Qi=I,Λ12+Γ1=I.
In this case,Y1=P1∗Γ1P1, where Γ1=diagλ11,λ12,…,λ1n with λ1j the eigenvalues of Y1. Similarly, if Y2 is a HPD solution of (1), then there exist P2∈Un×n, Ui∈Cn×n,i=1,2,…,m, and diagonal matrices Γ2,Λ2>0 such that(20)Bi=P2∗Γ21/2UiΛ2P2,i=1,2,…,s,P2∗Γ2−ti/2UiΛ2P2,i=s+1,s+2,…,m,where(21)∑i=1mUi∗Ui=I,Λ22+Γ2=I.In this case, Y2=P2∗Γ2P2, where Γ2=diagλ21,λ22,…,λ2n with λ2j the eigenvalues of. According to Y2 in Lemma 2, we have(22)Y1−Y2=∑i=1sBi∗Y2−1−Y1−1Bi+∑i=s+1mBi∗Y2ti−Y1tiBi=∑i=1sBi∗Y2−1Y1−Y2Y1−1Bi+∑i=s+1mBi∗sintiππ∫0∞Y2λI+Y2−1−Y1λI+Y1−1λti−1Bidλ=∑i=1sBi∗Y2−1Y1−Y2Y1−1Bi−∑i=s+1mBi∗sintiππ∫0∞Y1−Y2λI+Y1−1λti−1Bidλ+∑i=s+1mBi∗sintiππ∫0∞Y2λI+Y2−1Y1−Y2λI+Y1−1λti−1Bidλ.By(23)λI+Yi−1=λI+Pi∗ΓiPi−1=Pi−1λI+Γi−1Pi,i=1,2,we have(24)Y1−Y2=∑i=1sP2∗Γ21/2UiΛ2P2∗P2−1Γ2−1P2Y1−Y2P1∗Γ1−1P1P1∗Γ11/2QiΛ1P1−∑i=s+1msintiππ​​∫0∞P2∗Λ2Ui∗Γ2−ti2P2Y1−Y2P1∗λI+Γ1−1Γ1−ti/2QiΛ1P1λti−1dλ+∑i=s+1msintiππ∫0∞P2∗Λ2Ui∗Γ21−ti2λI+Γ2−1P2Y1−Y2P1∗λI+Γ1−1Γ1−ti/2QiΛ1P1λti−1dλ.
Let(25)W=P2Y1−Y2P1∗.Then, (24) can be expressed as(26)W=∑i=1sΛ2Ui∗Γ2−1/2WΓ1−1/2QiΛ1−∑i=s+1msintiππ∫0∞Λ2Ui∗Γ2−ti/2WλI+Γ1−1Γ1−ti/2QiΛ1λti−1dλ+∑i=s+1msintiππ∫0∞Λ2Ui∗Γ21−ti/2λI+Γ2−1WλI+Γ1−1Γ1−ti/2QiΛ1λti−1dλ.
By (26), Lemmas 3 and 4, we have that(27)vecW=∑i=1sΓ1−1/2QiΛ1T⊗Λ2Ui∗Γ2−1/2·vecW−∑i=s+1msintiππ∫0∞λI+Γ1−1Γ1−ti/2QiΛ1T⊗Λ2Ui∗Γ2−ti/2λti−1dλvecW+∑i=s+1msintiππ∫0∞λI+Γ1−1Γ1−ti/2QiΛ1T⊗Λ2Ui∗Γ21−ti/2λI+Γ2−1λti−1dλvecW=∑i=1sΛ1⊗Λ2QiT⊗Ui∗Γ11/2⊗Γ21/2·vecW−∑i=s+1msintiππΛ1⊗Λ2QiT⊗Ui∗Γ1−ti/2⊗Γ2−ti/2∫0∞λI+Γ1−1⊗Iλti−1dλvecW+∑i=s+1msintiππΛ1⊗Λ2QiT⊗Ui∗Γ1−ti/2⊗Γ21−ti/2∫0∞λI+Γ1−1⊗λI+Γ2−1λti−1dλvecW.Assume that(28)Λ1=diagσ11,σ12,…,σ1n,Λ2=diagσ21,σ22,…,σ2n.
It follows from (10), (19), and (21) that(29)0<σ1j=1−λ1j<1,0<σ2j=1−λ2j<1,j=1,2,…,n.
Let(30)B=Λ1⊗Λ2,Ji=QiT⊗Ui∗,E=Γ11/2⊗Γ21/2i=1,2,…,mCi=Γ1−ti/2⊗Γ2−ti/2sintiππ∫0∞λI+Γ1−1⊗Iλti−1dλ,i=s+1,s+2,…,m,Di=Γ1−ti/2⊗Γ21−ti/2sintiππ∫0∞λI+Γ1−1⊗λI+Γ2−1λti−1dλ,i=s+1,s+2,…,m.
Then, (27) can be expressed as(31)vecW+B∑i=s+1mJiCi−Di−∑i=1sJiE·vecW=0.
By Lemma5, we have(32)B=Λ1⊗Λ2=diagσ1l⋅σ2jn2×n2,l,j=1,2,…,n,Ci=Γ1−ti/2⊗Γ2−ti/2⋅sintiππ∫0∞λI+Γ1−1⊗Iλti−1dλ,=diagλ1l−ti/2·λ2j−ti/2⋅sintiππ∫0∞λ+λ1l−1λti−1dλn2×n2=diagλ1l−ti/2⋅λ2j−ti/2·λ1lti−1n2×n2=diagλ1lti/2−1⋅λ2j−ti/2n2×n2,i=s+1,s+2,…,m,l,j=1,2,…,n,Di=Γ1−ti/2⊗Γ21−ti/2⋅sintiππ∫0∞λI+Γ1−1⊗λI+Γ2−1λti−1dλ=diagλ1l−ti/2·λ2j1−ti/2⋅sintiππ∫0∞λ+λ1l−1λ+λ2j−1λti−1dλn2×n2=diagλ1l−ti/2⋅λ2j1−ti/2λ1lti−1−λ2jti−1λ2j−λ1ln2×n2,i=s+1,s+2,…,m,l,j=1,2,…,n,E=Γ1−1/2⊗Γ2−1/2=diagλ1l−1/2⋅λ2j−1/2n2×n2,l,j=1,2,…,n.
It follows that(33)Ci−Di=diagλ1lti/2−1·λ2j−ti/2−λ1l−ti/2⋅λ2j1−ti/2λ1lti−1−λ2jti−1λ2j−λ1ln2×n2=diagλ1lti−λ2jtiλ1l−λ2jλ1lti/2λ2jti/2n2×n2,i=s+1,s+2,…,m.
Note thatB is nonsingular. Multiplying the left side of (31) by B−1, we have(34)B−1vecW+∑i=s+1mJiCi−Di−∑i=1sJiE⋅vecW=I+∑i=s+1mJiCi−DiB−∑i=1sJiEBB−1·vecW=0.
A combination of (30) and Lemma 3 gives(35)Ji∗Ji=QiT⊗Ui∗∗QiT⊗Ui∗=Q¯i⊗UiQiT⊗Ui∗=Q¯iQiT⊗UiUi∗=QiQi∗¯⊗UiUi∗.
It follows (19), (21), and Lemma 5 that 0<Ji≤1. Therefore,(36)∑i=s+1mJiCi−DiB≤∑i=s+1mCi−DiB,∑i=1sJiEB≤∑i=1sEB.
By the hypothesis of the theorem, we havem/m+1I<Y1,Y2<I, which implies that m/m+1<λ1l,λ2j<1;l,j=1,2,…,n. Note that mti/mti+1<m/m+1;i=1,2,…,m. Then, mti/mti+1<λ1l,λ2j<1;l,j=1,2,…,n;i=1,2,…,m. Therefore, it follows (32) and (33) that(37)Ci−DiB=maxl,j1−λ1l1−λ2jλ2jti−λ1ltiλ2j−λ1lλ2jti/2λ1lti/2=maxl,jfλ1l,λ2j,ti,where fx,y,t is defined in Lemma 6. It is easy to verify that(38)EB=maxl,j1−λ1l1−λ2jλ2j1/2λ1l1/2<1m.
A combination of Lemma6, (36)–(38) gives that(39)∑i=s+1mJiCi−DiB−∑i=1sJiEiB<m−sm+sm=1,which implies I+∑i=s+1mJiCi−DiB−∑i=1sJiEB is nonsingular. It follows (34) that vecW=0. Recall that W=P2Y1−Y2P1∗. Therefore, Y1=Y2, which means the HPD solution on m/m+1I,I of (1) is unique.


## 4. The Homotopy Continuation Iterative Method

In this section, by means of the homotopy continuation iterative method (see [34], for more details), we derive a numerical iterative process for solving the matrix equation (1).Define the nonlinear mapF:Pn×n⊂Cn×n⟶Cn×n by(40)FX=I−∑i=1sBi∗X−1Bi−∑i=s+1mBi∗XtiBi.Consider the homotopy H:0,1×Pn×n⟶Pn×n:(41)Ht,X=I−t∑i=1sBi∗X−1Bi−t∑i=s+1mBi∗XtiBi−X.Then, at t=0, the solution of Ht,X=0 is a known matrix I, while at t=1, the solution X of Ht,X=0 also solves FX=X. To discuss the numerical method for solving the homotopy equation Ht,X=0, we rewrite the homotopy equation Ht,X=0 as the following fixed point form.Assume thatG:0,1×Pn×n⊂0,1×Cn×n⟶Pn×n is a map such that(42)Xt=Gt,Xt,t∈0,1,where X:0,1⟶Pn×n denotes the solution of Ht,X=0. Then, for each t, we can consider the iterative process:(43)Xn+1=Gt,Xn.Since for a fixedt, this process will converge to Xt only for starting values near that point; to overcome the local convergence of iterative process, we consider the following numerical continuation process.A partition ofJ=0,1:(44)0=t0<t1<⋯<tN=1,and a sequence of integers jk, k=1,…,N−1, is chosen with the property that the points(45)Xk,j+1=Gtk,Xk,j,j=0,…,jk−1,k=1,…,N−1,Xk+1,0=Xk,jk,X1,0=X0,are well-defined and such that(46)XN,j+1=G1,XN,j,converges to X1 as j⟶∞.The main idea is to choose partition (44) so that Xtk lies in some domain of attraction Dtk+1, for each k, 1≤k≤N. Then, if Xk,0∈Dtk+1, the sequence generated by (43) for t=tk must produce an iterate Xk,jk∈Dtk+1, which in turn can be taken as the starting point Xk+1,0=Xk,jk for the next iteration involving tk+1. Thus, with X1,0=X0 as initial point, the entire process can be carried out until finally tk=tN=1 is reached. For t=1, XN,0=XN,jN−1 is then in D1 which ensures that (45) converges to X1 as j⟶∞.To discuss the feasibility of the abovementioned numerical continuation process, we will use the following definition and lemmas which can be found in [34].Definition 1.
(see [34]). If a partition (44) exists so that with some sequence of integers jk, the entire process (45)−(46) is well defined so that (46) converges to X1, and then the numerical continuation process (45)−(46) is called feasible.Definition 2.
(see [34]). Let G:D⊂Rn⟶Rn be a given mapping. Then, any nonempty set D0⊂D is a domain of attraction of the iterative process:(47)xn+1=Gxn,n=0,1,…,with respect to the point x∗, if for any x0∈D0, we have xn⊂D and limn⟶∞xn=x∗.
Ifx∗∈intD0 for some domain of attraction D0, then x∗ is a point of attraction of (47).Lemma 7.
(see [34]). Let G:D⊂Rn⟶Rn be Fréchet differentiable at the fixed point x∗∈intD of G. If ρG′x∗<1, then x∗ is a point of attraction of (47) and, more precisely, there is an open ball Sx∗,r with center x∗ and radius r>0 which is a domain of attraction of (47) with respect to x∗. Here, ρ⋅ denotes the spectral radius of G′x∗.Lemma 8.
(see [34]). Let G:0,1×D⊂0,1×Rn⟶Rn, where D is open and assume that x:0,1⟶D is continuous and satisfies xt=Gt,xt. Let G have a Fréchet derivative with respect to x at t,xt for every t∈0,1. If Gxt,xt is continuous on 0,1×D and ρGxt,xt<1 for all t∈0,1, then the numerical continuation process (45)−(46) is feasible.In what follows, we derive a sufficient condition for the existence of a unique HPD solution of the homotopy equationHt,X=0 for all t∈0,1.Theorem 3.
Ifλ∑i=1mBi∗Bi<m/m+12, then for arbitrary t∈0,1, the homotopy equation Ht,X=0 has a unique HPD solution on m/m+1I,I.Proof.
Sincet>0, then the homotopy equation Ht,X=0 can be rewritten as(48)X+∑i=1stBi∗X−1tBi+∑i=s+1mtBi∗XtitBi=I.By the hypothesis of the theorem, we have(49)λ1∑i=1mtBi∗tBi=λ1∑i=1mtBi∗Bi≤λ1∑i=1mBi∗Bi<mm+12.It follows from Theorem 2 that the homotopy equation Ht,X=0 has a unique HPD solution on m/m+1I,I.
In the next theorem, the local convergence of the iterative process (43) is obtained.Theorem 4.
IfFX=X has a unique HPD solution X∗ on m/m+1I,I, then there exists an open ball NX∗,δ with center X∗ and radius δ>0 such that, for any starting value X0∈NX∗,δ, Xn=FXn−1 converges to X∗ as n⟶∞.Proof.
Recall thatFX=I−∑i=1sBi∗X−1Bi−∑i=s+1mBi∗XtiBi. According to Lemma 2, for any h∈Pn×n, we have(50)FX∗+h−FX∗=∑i=1sBi∗X∗−1−X∗+h−1Bi+∑i=s+1mBi∗X∗ti−X∗+htiBi=∑i=1sBi∗X∗+h−1hX∗−1Bi−∑i=s+1mBi∗sintiππ∫0∞hλI+X∗+h−1λti−1Bidλ+∑i=s+1mBi∗sintiππ∫0∞X∗λI+X∗−1hλI+X∗+h−1λti−1Bidλ.
By the definition of Fréchet derivative, we obtain(51)F′X∗h=∑i=1sBi∗X∗−1hX∗−1Bi−∑i=s+1mBi∗sintiππ∫0∞hλI+X∗−1λti−1Bidλ+∑i=s+1mBi∗sintiππ∫0∞X∗λI+X∗−1hλI+X∗−1λti−1Bidλ.
Letλ be any eigenvalue of F′X∗. Then, there exists a nonzero matrix h∗ such that(52)F′X∗h∗=λh∗.
SinceX∗ is the unique HPD solution of FX=X, then by Theorem 1, there exists P∈Un×n, Qi∈Cn×n,i=1,2,…,m, and diagonal matrices Γ,Λ>0 such that(53)Bi=P∗Γ1/2QiΛP,i=1,2,…,s,P∗Γ−ti/2QiΛP,i=s+1,s+2,…,m,where(54)∑i=1mQi∗Qi=I,Γ+Λ2=I.
In this case,X∗=P∗ΓP, where Γ=diagλ1,λ2,…,λn with λj the eigenvalues of X∗. Therefore, (52) can be rewritten as(55)F′X∗h∗=∑i=1sP∗ΛQi∗Γ−1/2Ph∗P∗Γ−1/2QiΛP−∑i=s+1msintiππ∫0∞P∗ΛQi∗Γ−ti/2Ph∗P∗λI+Γ−1λti−1Γ−ti/2QiΛPdλ+∑i=s+1msintiππ∫0∞P∗ΛQi∗Γ1−ti/2λI+Γ−1Ph∗P∗λI+Γ−1λti−1Γ−ti/2QiΛPdλ=λh∗.
Letz=Ph∗P∗. It follows that(56)∑i=1sΛQi∗Γ−1/2zΓ−1/2QiΛ−∑i=s+1msintiππ∫0∞ΛQi∗Γ−ti/2zλI+Γ−1λti−1Γ−ti/2QiΛdλ+∑i=s+1msintiππ∫0∞ΛQi∗Γ1−ti/2λI+Γ−1zλI+Γ−1λti−1Γ−ti/2QiΛdλ=λz.
Define the operatorℒ:Cn×n⟶Cn×n by(57)ℒz=∑i=1sΛQi∗Γ−1/2zΓ−1/2QiΛ−∑i=s+1msintiππ∫0∞ΛQi∗Γ−ti/2zλI+Γ−1Γ−ti/2QiΛλti−1dλ+∑i=s+1msintiππ∫0∞ΛQi∗Γ1−ti/2λI+Γ−1zλI+Γ−1Γ−ti/2QiΛλti−1dλ.
Then,(58)ℒz=λz.
Using (58) and Lemmas 3 and 4, we can rewrite (57) as(59)vecℒz=∑i=1sΓ−1/2QiΛT⊗ΛUi∗Γ−1/2⋅vecz−∑i=s+1msintiππ∫0∞λI+Γ−1Γ−ti/2QiΛT⊗ΛQi∗Γ−ti/2λti−1dλ⋅vecz+∑i=s+1msintiππ∫0∞λI+Γ−1Γ−ti/2QiΛT⊗ΛQi∗Γ1−ti/2λI+Γ−1λti−1dλ⋅vecz=∑i=1sΛ⊗ΛQiT⊗Ui∗Γ1/2⊗Γ1/2⋅vecz−∑i=s+1msintiππΛ⊗ΛQiT⊗Qi∗Γ−ti/2⊗Γ−ti/2∫0∞λI+Γ−1⊗Iλti−1dλ⋅vecz+∑i=s+1msintiππΛ⊗ΛQiT⊗Qi∗Γ−ti/2⊗Γ1−ti/2∫0∞λI+Γ−1⊗λI+Γ−1λti−1dλvecz​=λvecz.
Let(60)B=Λ⊗Λ,Ji=QiT⊗Qi∗,Ci=Γ−ti/2⊗Γ−ti/2·sintiππ∫0∞λI+Γ−1⊗Iλti−1dλ,i=s+1,s+2,…,m,Di=Γ−ti/2⊗Γ1−ti/2·sintiππ∫0∞λI+Γ−1⊗λI+Γ−1λti−1dλ,i=s+1,s+2,…,m,E=Γ−1/2⊗Γ−1/2=diagλl−1/2·λj−1/2n2×n2;l,j=1,2,…,n.
Then, (59) can be rewritten as(61)vecℒz=∑i=s+1mBJiDi−Ci+∑i=1sBJiE⋅vecz=λ⋅vecz.
According to (36), (55), (57), (59), and (61), we have that(62)ρF′X∗=maxλ=ρ∑i=s+1mBJiDi−Ci+∑i=1sBJiE≤∑i=s+1mBJiDi−Ci+∑i=1sBJiE≤∑i=s+1mBCi−Di+∑i=1sBE.
Assume that(63)Λ=diagσ1,σ2,…,σn.
By (54), we have(64)0<σj=1−λj,j=1,2,…,n.
According to Lemma5 and (64), we have(65)B=diagσl⋅σjn2×n2=diag1−λl·1−λjn2×n2,(66)Ci=Γ−ti/2⊗Γ−ti/2·sintiππ∫0∞λI+Γ−1⊗Iλti−1dλ=diagλl−ti/2·λj−ti/2·sintiππ∫0∞λ+λl−1λti−1dλn2×n2=diagλl−ti/2·λj−ti/2·λlti−1n2×n2=diagλlti/2−1·λj−ti/2n2×n2,i=s+1,s+2,…,m,l,j=1,2,…,n,(67)Di=Γ−ti/2⊗Γ1−ti/2⋅sintiππ∫0∞λI+Γ−1⊗λI+Γ−1λti−1dλ=diagλl−ti/2·λj1−ti/2⋅sintiππ∫0∞λ+λl−1λ+λj−1λti−1dλn2×n2=diagλl−ti/2·λj1−ti/2λlti−1−λjti−1λj−λln2×n2,i=s+1,s+2,…,m,l,j=1,2,…,n.
Sincem/m+1I<X∗<I, then mti/mti+1<m/m+1<λl,λj<1,i=1,2,…,m,l,j=1,2,…,n. By (65)–(67), we have that(68)Ci−DiB=maxl,j1−λl1−λjλjti−λltiλj−λlλjti/2λlti/2=maxl,jfλl,λj,tiEiB=maxl,j1−λl1−λjλjti/2λlti/2,where fx,y,t is defined in Lemma 6. Combining Lemma 6 (62) with (68) gives that(69)ρF′X∗≤∑i=s+1mBCi−Di+∑i=1sBE<1.According to Lemma 7, there exists an open ball NX∗,δ with center X∗ and radius δ>0 such that, for any starting value X0∈NX∗,δ, Xn=FXn−1 converges to X∗ as n⟶∞.
In the next theorem, we will prove the numerical continuation process (45)−(46) is feasible.Theorem 5.
Ifλ1∑i=1mBi∗Bi<m/m+12, then the numerical continuation process (45)−(46) is feasible.Proof.
Define the mapG:0,1×Pn×n⟶Pn×n by(70)Gt,Xt=I−t∑i=1sBi∗X−1Bi−t∑i=s+1mBi∗XtiBi.
In the following, we will prove the numerical continuation processes (45)−(46) are feasible.
By Lemma2, for any h∈Pn×n, we have(71)Gt,Xt+h−Gt,Xt,(72)Gt,Xt+h−Gt,Xt=t∑i=1sBi∗X−1−X+h−1Bi+t∑i=s+1mBi∗Xti−X+htiBi=t∑i=1sBi∗X−1hX+h−1Bi−t∑i=s+1mBi∗sintiππ∫0∞hλI+X+h−1λti−1Bidλ+t∑i=s+1mBi∗sintiππ∫0∞XλI+X−1hλI+X+h−1λti−1Bidλ.
By the definition of Fréchet derivative, we obtain(73)G′t,Xth=t∑i=1sBi∗X−1hX−1Bi−t∑i=s+1mBi∗sintiππ∫0∞hλI+X−1λti−1Bidλ+t∑i=s+1mBi∗sintiππ∫0∞XλI+X−1hλI+X−1λti−1Bidλ.
Using the same technique described in Theorem4, we have that(74)ρG′t,Xt<t∑i=1smaxl,j1−λl1−λjλj1/2λl1/2+∑i=s+1mmaxl,j1−λl1−λjλjti−λltiλj−λlλlti/2λjti/2<1,where mti/mti+1<m/m+1<λl,λj<1,l,j=1,2,…,n.
According to Lemma2, the numerical continuation processes (45)−(46) are feasible.


## 5. Conclusions

We have introduced a class of nonlinear matrix equations that is wider than those studied earlier in the literature. We have derived some sufficient and necessary conditions for the existence and uniqueness of HPD solution of (1). In order to apply the homotopy continuation iterative method proposed by Avila [34] to the nonlinear matrix equation (1), we have constructed a related homotopy matrix equation Ht,X=0 and have derived a sufficient condition for the existence of a unique HPD solution of this equation. We have cited the definition and the judgment theorem of feasible for the numerical continuation process (45)−(46), which was proposed in [34]. And then we have derived the condition for feasible of the numerical continuation processes (45)−(46). Furthermore, we obtained the feasible conclusion of the numerical method by verifying the conditions in Theorem 2.5 in [34].



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