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Research article

Double iterative algorithm for solving different constrained solutions of multivariate quadratic matrix equations

  • Received: 05 August 2021 Accepted: 20 October 2021 Published: 03 November 2021
  • MSC : 15A24, 65F45, 65H10

  • Constrained solutions are required in some practical problems. This paper studies the problem of different constrained solutions for a class of multivariate quadratic matrix equations, which has rarely been studied before. First, the quadratic matrix equations are transformed into linear matrix equations by Newton's method. Second, the different constrained solutions of the linear matrix equations are solved by the modified conjugate gradient method, and then the different constrained solutions of the multivariate quadratic matrix equations are obtained. Finally, the convergence of the algorithm is proved. The algorithm can be well performed on the computers, and the effectiveness of the method is verified by numerical examples.

    Citation: Shousheng Zhu. Double iterative algorithm for solving different constrained solutions of multivariate quadratic matrix equations[J]. AIMS Mathematics, 2022, 7(2): 1845-1855. doi: 10.3934/math.2022106

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  • Constrained solutions are required in some practical problems. This paper studies the problem of different constrained solutions for a class of multivariate quadratic matrix equations, which has rarely been studied before. First, the quadratic matrix equations are transformed into linear matrix equations by Newton's method. Second, the different constrained solutions of the linear matrix equations are solved by the modified conjugate gradient method, and then the different constrained solutions of the multivariate quadratic matrix equations are obtained. Finally, the convergence of the algorithm is proved. The algorithm can be well performed on the computers, and the effectiveness of the method is verified by numerical examples.



    Quadratic matrix equation has many different forms, such as AX2+BX+C=O arising in quasi-birth-death processes [1,2] and Riccati equation XCXXEAX+B=O arising in transport theory [3,4,5,6]. There are also some coupled quadratic matrix equations with two or three variables [7,8]. This article will study the general form of these equations. As can be seen, the linear part of these equations can be expressed in the form 3i=1C(l)iXiD(l)i, and the quadratic part of them can be expressed as 3i,j=1XiE(l)ijXj. Therefore, we study the following coupled quadratic matrix equation:

    3i=1C(l)iXiD(l)i+3i,j=1XiE(l)ijXj=S(l)(l=1,2), (1.1)

    where all matrices are n×n real matrices. Each equation in (1.1) consists of three linear terms and nine quadratic terms. Besides, Eq (1.1) have three variables and only two equations, so the solution is not unique.

    As we know, we always need some special kind of solutions in practical applications, such as symmetric solutions are widely used in control theory [8,9] and reflexive solutions which are also called generalized centro-symmetric solutions are used in information theory, linear estimate theory and numerical analysis [10,11]. Liao, Liu, etc. have studied the problem of different constrained solutions of linear matrix equations [12,13]. In this article, we will design a method to obtain different constrained solutions of a class of quadratic matrix equations.

    Many researchers have studied quadratic matrix equations. For example, Bini, Iannazzo and Poloni gave a fast Newton's method for a quadratic matrix equation [4]. Long, Hu and Zhang used an improved Newton's method to solve a quadratic matrix equation [14]. Convergence rate of some iterative methods for quadratic matrix equations arising in transport theory was also described by Guo and Lin [15]. Zhang, Zhu and Liu have studied the constrained solutions of two-variable Riccati matrix equations based on Newton's method and modified conjugate gradient (MCG) method [16,17]. This article will further study the problem of different constrained solutions of coupled quadratic matrix equations with three matrix variables. The algorithm designed in the paper is superior in computing different constrained solutions.

    Notation: Rn×n denotes the set of n×n real matrices. The symbols AT and tr(A) represent the transpose and the trace of the matrix A respectively. AB stands for the Kronecker product of matrices A and B, ¯vec() is an operator that transforms a matrix A into a column vector by vertically stacking the columns of the matrix AT. For example, for the 2×2 matrix

    A=[abcd],

    the vectorization is ¯vec(A)=[a,b,c,d]T. We define an inner product of two matrices A,BRn×n as [A,B]=tr(ATB), then the norm of a matrix A generated by this inner product is Frobenius norm and denoted by A, i.e. A=[A,A].

    Let Ω1 be the set of symmetric matrices. P1,P2Rn×n are said to be symmetric orthogonal matrices if Pi=PTi and P2i=I (i=1,2). XRn×n is said to be a reflexive matrix with respect to P1 if P1XP1=X. Let Ω5 be the set of reflexive matrices. XRn×n is said to be a symmetric reflexive matrix with respect to P2 if XT=X=P2XP2. Let Ω9 be the set of symmetric reflexive matrices. We call (X1,X2,X3) a constrained matrix in Ω159 when X1Ω1, X2Ω5 and X3Ω9. Besides, if the symmetric orthogonal matrices P1 and P2 are changed, we will get different constrained matrices in Ω159.

    The paper is organized as follows: First, we use Newton's method to convert the quadratic matrix equations into linear matrix equations. Second, MCG method [10,13,16,18] is applied to solve the derived linear matrix equations. Finally, numerical examples are presented to support the theoretical results of this paper.

    As a matter of convenience, we first introduce some notations.

    X=[X1X2X3],X(k)=[X(k)1X(k)2X(k)3],X=[X1X2X3].

    Y, Y(k) and Y are defined in the same way as X, X(k) and X respectively. Then let

    ψ(l)(X)=3i=1C(l)iXiD(l)i+3i,j=1XiE(l)ijXjS(l),
    ϕ(l)X(Y)=3i=1C(l)iYiD(l)i+3i,j=1XiE(l)ijYj+3i,j=1YiE(l)ijXj,

    we can obtain

    ψ(l)(X+Y)=ψ(l)(X)+ϕ(l)X(Y)+3i,j=1YiE(l)ijYj(l=1,2), (2.1)

    where ϕ(l)X(Y): Rn×nRn×n is the Fréchet derivative of ψ(l)(X) at X in the direction Y [1].

    Lemma 2.1. Finding the solution (X1,X2,X3)Ω159 of (1.1) can be transformed into finding the corrected value (Y1,Y2,Y3)Ω159 of ψ(l)(X+Y)=0 (l=1,2). We linearize and solve, to find (Y1,Y2,Y3)Ω159 from the coupled linear matrix equation

    ϕ(l)X(Y)=ψ(l)(X)(l=1,2). (2.2)

    Proof. Supposing that the approximate solution (X1,X2,X3)Ω159 of Eq (1.1) has been obtained. Let Xi=Xi+Yi (i=1,2,3), then finding (X1,X2,X3)Ω159 of (1.1) is transformed into finding the corrected value (Y1,Y2,Y3)Ω159 from

    ψ(l)(X+Y)=O(l=1,2). (2.3)

    The Eq (2.3) is quadratic equations about Yi. As is known, when the norm of Yi is small enough, the quadratic parts 3i,j=1YiE(l)ijYj about Yi in (2.1) can be discarded according to Newton's method. In this way, we can get a linear approximation

    ψ(l)(X+Y)ψ(l)(X)+ϕ(l)X(Y).

    Therefore, finding the solution (X1,X2,X3)Ω159 of (1.1) is transformed into finding (Y1,Y2,Y3)Ω159 from ψ(l)(X)+ϕ(l)X(Y)=O (l=1,2), that is, to solve (2.2).

    According to [14], Newton's method (algorithm 1) is introduced to find constrained solutions in Ω159 of (1.1). Let

    ψ(X)=[ψ(1)(X)ψ(2)(X)],ϕX(Y)=[ϕ(1)X(Y)ϕ(2)X(Y)].
    Algorithm 1: : Newton's method solves the solution X of Eq (1.1)
    Step 1. Choose an initial matrix (X(1)1,X(1)2,X(1)3)Ω159 and set k:=1.
    Step 2. If ψ(X(k))=O, stop, else, solve for (Y(k)1,Y(k)2,Y(k)3)Ω159 from
    ϕX(k)(Y(k))=ψ(X(k)).            (2.4)
    When (2.4) hasn't constrained solutions in Ω159, solve for (Y(k)1,Y(k)2,Y(k)3)Ω159, such that
    ϕX(k)(Y(k))+ψ(X(k))=min.            (2.5)
    Step 3. Compute X(k+1)=X(k)+Y(k), set k:=k+1 and go to step 2.

     | Show Table
    DownLoad: CSV

    The convergent properties about Newton's method can be obtained as follows according to [14] (The proof is similar to Lemma 2.1 in [14]).

    Theorem 2.1. Assume that the real matrix X is a simple root of (1.1), i.e. ψ(X)=O and ϕX(Y) is regular. Then if the starting matrix X(1) is chosen sufficiently close to the solution X, the sequence {X(k)} generated by Newton's method converges quadratically to the solution X.

    In algorithm 1, when X(k) is known, then Y(k) needs to be solved. In this section, MCG method will be used to solve Y(k) from Eq (2.2), that is, to solve Eq (2.4) or Eq (2.5). Consider the general form of Eq (2.2)

    3i=17j=1A(l)ijYiB(l)ij=F(l)(l=1,2), (3.1)

    where all matrices in Eq (3.1) are n×n real matrices. Let

    h(Y)=[h(1)(Y)h(2)(Y)],F=[F(1)F(2)],R=Fh(Y)def=[R(1)R(2)],
    p(R)=[p1(R)p2(R)p3(R)],q(Y)=[q1(Y)q2(Y)q3(Y)],

    where

    h(l)(Y)def=h(l)(Y1,Y2,Y3)=3i=17j=1A(l)ijYiB(l)ij(l=1,2),
    pi(R)=2l=17j=1(A(l)ij)TR(l)(B(l)ij)T(i=1,2,3),
    q1(Y)=12(Y1+YT1),q2(Y)=12(Y2+P1Y2P1),
    q3(Y)=14(Y3+YT3+P2(Y3+YT3)P2),

    and P1,P2Rn×n are symmetric orthogonal matrices.

    In order to solve Eq (3.1), the following two questions will be considered.

    Problem 3.1. Assume that (3.1) has constrained solutions, find (Y1,Y2,Y3)Ω159 from (3.1).

    Problem 3.2. Assume that (3.1) hasn't constrained solutions, find (Y1,Y2,Y3)Ω159, such that

    h(Y)F=min. (3.2)

    Based on the MCG method, we establish the following algorithm (algorithm 2) to solve problem 3.1.

    Algorithm 2: : MCG method to solve problem 3.1
    Step 1. Choose an arbitrary initial matrix (Y(1)1,Y(1)2,Y(1)3)Ω159, set k:=1 and compute
    Rk=Fh(Y(k))def=[R(1)kR(2)k],˜Rk=p(Rk)def=[˜R(1)k˜R(2)k˜R(3)k],Zk=q(˜Rk).
    Step 2. If Rk=O, or RkO and Zk=O, stop, else, compute
    αk=Rk2Zk2,Y(k+1)=Y(k)+αkZk.
    Step 3. Compute
    Rk+1=Fh(Y(k+1))def=[R(1)k+1R(2)k+1],˜Rk+1=p(Rk+1)def=[˜R(1)k+1˜R(2)k+1˜R(3)k+1],
    βk+1=Rk+12Rk2,Zk+1=q(˜Rk+1)+βk+1Zk.
    Step 4. Set k:=k+1 and go to step 2.

     | Show Table
    DownLoad: CSV

    From algorithm 2, we can easily see (Y(k)1,Y(k)2,Y(k)3)Ω159 for k=1,2, and have the following convergent properties (The proof is similar to Theorem 2.1 in [10]):

    Theorem 3.1. Assume that Eq (3.1) has constrained solutions in Ω159. Then for an arbitrary initial matrix (Y(1)1,Y(1)2,Y(1)3)Ω159, a solution of problem 3.1 can be obtained by algorithm 2 within finite number of iterations, which is also a constrained solution in Ω159 of (3.1).

    Algorithm 2 will break if RiO and Zi=O, which means that Eq (3.1) hasn't constrained solution in Ω159 according to Theorem 3.1. Therefore, we need to solve problem 3.2, that is, to find constrained least-squares solutions of (3.1).

    We replace the problem of finding least-squares solutions in Ω159 of (3.1) with finding solutions in Ω159 of equivalent linear matrix equations by the Theorem 3.2, and then an iterative algorithm to find constrained least-squares solutions in Ω159 of (3.1) is constructed according to algorithm 2.

    As a matter of convenience, we introduce some notations:

    u(Y)=h(Y1,Y2,12(Y3+P2Y3P2))def=[u(1)(Y)u(2)(Y)],
    v(Y)=h(YT1,P1Y2P1,12(YT3+P2YT3P2))def=[v(1)(Y)v(2)(Y)],
    g(Y)=[g1(Y)g2(Y)g3(Y)],Q=[Q1(F)Q2(F)Q3(F)],

    where

    g1(Y)=p1(u)+pT1(v),g2(Y)=p2(u)+P1p2(v)P1,
    g3(Y)=12(p3(u)+pT3(v)+P2(p3(u)+pT3(v))P2),
    Q1(F)=p1(F)+pT1(F),Q2(F)=p2(F)+P1p2(F)P1,
    Q3(F)=12(p3(F)+pT3(F)+P2(p3(F)+pT3(F))P2),

    u and v are functions of Y. Then, according to [16,17], we have the following theorem.

    Theorem 3.2. Iterative algorithm for solving problem 3.2 can be replaced by finding constrained solutions in Ω159 from

    g(Y)=Q. (3.3)

    Indeed, (3.3) has constrained solutions in Ω159.

    Proof. When (Y1,Y2,Y3)Ω159, we have Y1=YT1, Y2=P1Y2P1 and Y3=YT3=P2Y3P2. Therefore, solving problem 3.2 is equivalent to solving (Y1,Y2,Y3)Ω159 from

    [u(Y)v(Y)][FF]=min. (3.4)

    Now we have to prove that solving the problem (3.4) is equivalent to finding constrained solutions in Ω159 of (3.3). We let multiply operation prior to Kronecker product operation between matrices. Let

    G(l)i=7j=1A(l)ij(B(l)ij)T,
    H(l)im=7j=1A(l)ijPm(PmB(l)ij)T,(i=1,2,3;l,m=1,2),
    M=[G(1)1G(1)212(G(1)3+H(1)32)G(2)1G(2)212(G(2)3+H(2)32)G(1)1Tn,nH(1)2112(G(1)3+H(1)32)Tn,nG(2)1Tn,nH(2)2112(G(2)3+H(2)32)Tn,n],
    y=[¯vec(Y1)¯vec(Y2)¯vec(Y3)],f=[¯vec(F)¯vec(F)],

    where Tn,n denotes a commutation matrix such that Tn,n¯vec(An×n)=¯vec(AT) [19] and let Tn,n only work on ¯vec. Then applying ¯vec to the following equations:

    {u(Y)=F,v(Y)=F, (3.5)

    we can get the equivalent equation: My=f. Besides, MTMy=MTf, the normal equation of My=f, is the vectorization of (3.3). Therefore, the least-squares solution of My=f is also a solution of MTMy=MTf, and the vectorization of the solution of (3.3). So the solution of (3.4) is also a solution of (3.3), and vice versa.

    Above all, iterative algorithm for solving problem 3.2 can be replaced by finding constrained solutions in Ω159 of (3.3).

    As we all know, normal equations always have solutions, and the vectorization of Eq (3.3) is a normal equation, so Eq (3.3) also has solutions. Suppose ˜Y=(˜YT1,˜YT2,˜YT3)T (whether ˜YΩ159 or not) is a solution of (3.3), then g(˜Y)=Q. Let Yi=qi(˜Y) (i=1,2,3), then (Y1,Y2,Y3)Ω159 and g(Y)=Q. Hence, (3.3) has constrained solutions in Ω159.

    We use the MCG method to find constrained solutions in Ω159 of (3.3) by algorithm 3, which is also a process to solve the problem 3.2.

    Algorithm 3: : MCG method to solve problem 3.2
    Step 1. Choose an arbitrary initial matrix (Y(1)1,Y(1)2,Y(1)3)Ω159, set k:=1 and compute
    Rk=Qg(Y(k)),˜Rk=g(Rk),Zk=˜Rk.
    Step 2. If Rk=O, stop, else, compute
    αk=Rk2Zk2,Y(k+1)=Y(k)+αkZk.
    Step 3. Compute
    Rk+1=Qg(Y(k+1)),˜Rk+1=g(Rk+1),
    βk+1=Rk+12Rk2,Zk+1=˜Rk+1+βk+1Zk.
    Step 4. Set k:=k+1 and go to step 2.

     | Show Table
    DownLoad: CSV

    From algorithm 3, we can see that (Y(k)1,Y(k)2,Y(k)3)Ω159 for k=1,2, and have the following convergent properties (The proof is similar to Theorem 2 in [13]).

    Theorem 3.3. For an arbitrary initial matrix (Y(1)1,Y(1)2,Y(1)3)Ω159, a solution of problem 3.2 can be obtained by algorithm 3 within finite number of iterations, and it is also a constrained least-squares solution in Ω159 of (3.1).

    In this section, we design two computation programmes to find constrained solutions of (1.1). Then two numerical examples are given to illustrate the proposed results. All computations are performed using MATLAB. Because of the influence of roundoff errors, we regard a matrix A as zero matrix if A107.

    Let n be the order of the matrix Xi, k,k1,k2 be the iteration numbers of algorithm 1, algorithm 2 and algorithm 3 respectively, and t be the computation time (seconds).

    Programme 1.

    (1) Choose an initial matrix (X(1)1,X(1)2,X(1)3)Ω159 and set k:=1.

    (2) If ψ(X(k))=O, stop, else, solve for (Y(k)1,Y(k)2,Y(k)3)Ω159 from (2.4) using algorithm 2. When algorithm 2 breaks, that is (2.4) hasn't constrained solution in Ω159, solve for (Y(k)1,Y(k)2,Y(k)3)Ω159 from (2.5) using algorithm 3.

    (3) Compute X(k+1)=X(k)+Y(k), set k:=k+1 and go to step 2.

    Programme 2.

    (1) Choose an initial matrix (X(1)1,X(1)2,X(1)3)Ω159 and set k:=1.

    (2) If ψ(X(k))=O, stop, else, solve for (Y(k)1,Y(k)2,Y(k)3)Ω159 from (2.5) using algorithm 3. Especially, when (2.4) has constrained solutions in Ω159, the constrained least-squares solutions in Ω159 are also its constrained solutions in Ω159.

    (3) Compute X(k+1)=X(k)+Y(k), set k:=k+1 and go to step 2.

    Example 4.1. Consider (1.1) with the following parameters:

    P1=[010100001],P2=[010100001],
    C(l)i=C+l×ones(3),D(l)i=(C(l)i)T,E(l)ij=uiuTi,
    S(l)=3i=1C(l)iXiD(l)i+3i,j=1XiE(l)ijXj(i,j=1,2,3;l=1,2),

    where

    C=[100011101],X1=[100.50100.502],
    X2=[100.5010.5002],X3=[100.25010.250.250.252],
    u1=(1,1,0)T,u2=(0,1,1)T,u3=(0,0,1)T.

    We can easily see that (1.1) has the constrained solution (X1,X2,X3)Ω159. By applying programmes 1 and 2 with the initial matrix X(1)i=eye(3), Y(1)i=zeros(3)(i=1,2,3), we obtain the constrained solution in Ω159 of (1.1) as follows:

    X(5)1=[1.00000.00000.50000.00001.00000.00000.50000.00002.0000],
    X(5)2=[1.00000.00000.50000.00001.00000.50000.00000.00002.0000],
    X(5)3=[1.00000.00000.25000.00001.00000.25000.25000.25002.0000].

    The iteration numbers and computation time are listed in Table 1.

    Table 1.  The iteration numbers and computation time of programmes 1 and 2.
    Results n k k1 k2 t
    Programme 1 3 5 97 0 0.4521
    Programme 2 3 5 0 184 3.4310

     | Show Table
    DownLoad: CSV

    From the results in Table 1, we see that programme 1 is more effective when the derived linear matrix equations are always have constrained solutions in Ω159.

    Example 4.2. Consider (1.1) with the following parameters:

    C(l)i=i×C+l×ones(3),D(l)i=(C(l)i)T,
    E(l)ij=uiuTi(i,j=1,2,3;l=1,2),
    S(1)=[18.532293650.54.5435.537.5],S(2)=[2.568118183.529.52217.522.5],

    where

    C=[100011101],
    u1=(1,1,0)T,u2=(0,1,1)T,u3=(0,0,1)T.

    By applying programmes 1 and 2 with the initial matrix X(1)i=ones(3), Y(1)i=zeros(3) (i=1,2,3), P1 and P2 are identity matrices, we obtain a special constrained solution (now, X1 and X3 are symmetric matrices, X2 is a general matrix) in Ω159 of (1.1) as follows:

    X(4)1=[1.13521.52491.59241.52491.01880.99041.59240.99041.1223],
    X(4)2=[0.93320.84111.01521.84110.97300.95492.01520.95491.0135],
    X(4)3=[0.97691.62451.41501.62451.01281.02991.41501.02990.9482].

    When X(1)i=eye(3) (i=1,2,3), others remain unchanged, we obtain another special constrained solution in Ω159 of (1.1) as follows:

    X(6)1=[1.81471.88332.46921.88331.10891.02272.46921.02272.6495],
    X(6)2=[1.25810.07730.61450.92270.68730.58801.61450.58800.2754],
    X(6)3=[0.34662.22761.12612.22761.35421.00981.12611.00981.1756].

    Thus it can be seen that if the constrained solution of Eq (1.1) is not unique, we can get different constrained solutions in Ω159 when choosing different initial matrices.

    In this paper, an iterative algorithm is studied to find different constrained solutions. By using the proposed algorithm, we compute a set of different constrained solution in Ω159 of multivariate quadratic matrix equations. The provided examples illustrate the effectiveness of the new iterative algorithm.

    There are still some results we can obtain by changing the initial matrices X(1)i and Y(1)i, the direction matrix Zk in algorithm 2 and Eq (3.3) in algorithm 3. In this way, we can get other kind of constrained solutions, which are not only interesting but also valuable. It remains to study in our further work.

    This research was funded by Doctoral Fund Project of Shandong Jianzhu University grant number X18091Z0101.

    The author declares no conflict of interest.



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