In this paper, we consider the quasilinear parabolic-elliptic-elliptic attraction-repulsion system
{ut=∇⋅(D(u)∇u)−χ∇⋅(u∇v)+ξ∇⋅(u∇w),x∈Ω,t>0,0=Δv−μ1(t)+f1(u),x∈Ω,t>0,0=Δw−μ2(t)+f2(u),x∈Ω,t>0
under homogeneous Neumann boundary conditions in a smooth bounded domain Ω⊂Rn, n≥2. The nonlinear diffusivity D and nonlinear signal productions f1,f2 are supposed to extend the prototypes
D(s)=(1+s)m−1, f1(s)=(1+s)γ1, f2(s)=(1+s)γ2, s≥0,γ1,γ2>0,m∈R.
We proved that if γ1>γ2 and 1+γ1−m>2n, then the solution with initial mass concentrating enough in a small ball centered at origin will blow up in finite time. However, the system admits a global bounded classical solution for suitable smooth initial datum when γ2<1+γ1<2n+m.
Citation: Ruxi Cao, Zhongping Li. Blow-up and boundedness in quasilinear attraction-repulsion systems with nonlinear signal production[J]. Mathematical Biosciences and Engineering, 2023, 20(3): 5243-5267. doi: 10.3934/mbe.2023243
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In this paper, we consider the quasilinear parabolic-elliptic-elliptic attraction-repulsion system
{ut=∇⋅(D(u)∇u)−χ∇⋅(u∇v)+ξ∇⋅(u∇w),x∈Ω,t>0,0=Δv−μ1(t)+f1(u),x∈Ω,t>0,0=Δw−μ2(t)+f2(u),x∈Ω,t>0
under homogeneous Neumann boundary conditions in a smooth bounded domain Ω⊂Rn, n≥2. The nonlinear diffusivity D and nonlinear signal productions f1,f2 are supposed to extend the prototypes
D(s)=(1+s)m−1, f1(s)=(1+s)γ1, f2(s)=(1+s)γ2, s≥0,γ1,γ2>0,m∈R.
We proved that if γ1>γ2 and 1+γ1−m>2n, then the solution with initial mass concentrating enough in a small ball centered at origin will blow up in finite time. However, the system admits a global bounded classical solution for suitable smooth initial datum when γ2<1+γ1<2n+m.
Since the 1960s, the rapid development of high-speed rail has made it a very important means of transportation. However, the vibration will be caused because of the contact between the wheels of the train and the train tracks during the operation of the high-speed train. Therefore, the analytical vibration model can be mathematically summarized as a quadratic palindromic eigenvalue problem (QPEP) (see [1,2])
(λ2A1+λA0+A⊤1)x=0, |
with Ai∈Rn×n, i=0,1 and A⊤0=A0. The eigenvalues λ, the corresponding eigenvectors x are relevant to the vibration frequencies and the shapes of the vibration, respectively. Many scholars have put forward many effective methods to solve QPEP [3,4,5,6,7]. In addition, under mild assumptions, the quadratic palindromic eigenvalue problem can be converted to the following linear palindromic eigenvalue problem (see [8])
Ax=λA⊤x, | (1) |
with A∈Rn×n is a given matrix, λ∈C and nonzero vectors x∈Cn are the wanted eigenvalues and eigenvectors of the vibration model. We can obtain 1λx⊤A⊤=x⊤A by transposing the equation (1). Thus, λ and 1λ always come in pairs. Many methods have been proposed to solve the palindromic eigenvalue problem such as URV-decomposition based structured method [9], QR-like algorithm [10], structure-preserving methods [11], and palindromic doubling algorithm [12].
On the other hand, the modal data obtained by the mathematical model are often evidently different from the relevant experimental ones because of the complexity of the structure and inevitable factors of the actual model. Therefore, the coefficient matrices need to be modified so that the updated model satisfies the dynamic equation and closely matches the experimental data. Al-Ammari [13] considered the inverse quadratic palindromic eigenvalue problem. Batzke and Mehl [14] studied the inverse eigenvalue problem for T-palindromic matrix polynomials excluding the case that both +1 and −1 are eigenvalues. Zhao et al. [15] updated ∗-palindromic quadratic systems with no spill-over. However, the linear inverse palindromic eigenvalue problem has not been extensively considered in recent years.
In this work, we just consider the linear inverse palindromic eigenvalue problem (IPEP). It can be stated as the following problem:
Problem IPEP. Given a pair of matrices (Λ,X) in the form
Λ=diag{λ1,⋯,λp}∈Cp×p, |
and
X=[x1,⋯,xp]∈Cn×p, |
where diagonal elements of Λ are all distinct, X is of full column rank p, and both Λ and X are closed under complex conjugation in the sense that λ2i=ˉλ2i−1∈C, x2i=ˉx2i−1∈Cn for i=1,⋯,m, and λj∈R, xj∈Rn for j=2m+1,⋯,p, find a real-valued matrix A that satisfy the equation
AX=A⊤XΛ. | (2) |
Namely, each pair (λt,xt), t=1,⋯,p, is an eigenpair of the matrix pencil
P(λ)=Ax−λA⊤x. |
It is known that the mathematical model is a "good" representation of the system, we hope to find a model that is closest to the original model. Therefore, we consider the following best approximation problem:
Problem BAP. Given ˜A∈Rn×n, find ˆA∈SA such that
‖ˆA−˜A‖=minA∈SA‖A−˜A‖, | (3) |
where ‖⋅‖ is the Frobenius norm, and SA is the solution set of Problem IPEP.
In this paper, we will put forward a new direct method to solve Problem IPEP and Problem BAP. By partitioning the matrix Λ and using the QR-decomposition, the expression of the general solution of Problem IPEP is derived. Also, we show that the best approximation solution ˆA of Problem BAP is unique and derive an explicit formula for it.
We first rearrange the matrix Λ as
Λ=[10 00Λ1 000 Λ2]t2s2(k+2l) t 2s2(k+2l), | (4) |
where t+2s+2(k+2l)=p, t=0 or 1,
Λ1=diag{λ1,λ2,⋯,λ2s−1,λ2s},λi∈R, λ−12i−1=λ2i, 1≤i≤s,Λ2=diag{δ1,⋯,δk,δk+1,δk+2,⋯,δk+2l−1,δk+2l}, δj∈C2×2, |
with
δj=[αj+βji00αj−βji], i=√−1, 1≤j≤k+2l,δ−1j=ˉδj, 1≤j≤k,δ−1k+2j−1=δk+2j, 1≤j≤l, |
and the adjustment of the column vectors of X corresponds to those of Λ.
Define Tp as
Tp=diag{It+2s,1√2[1−i1i],⋯,1√2[1−i1i]}∈Cp×p, | (5) |
where i=√−1. It is easy to verify that THpTp=Ip. Using this matrix of (5), we obtain
˜Λ=THpΛTp=[1000Λ1000˜Λ2], | (6) |
˜X=XTp=[xt,⋯,xt+2s,√2yt+2s+1,√2zt+2s+1,⋯,√2yp−1,√2zp−1], | (7) |
where
˜Λ2=diag{[α1β1−β1α1],⋯,[αk+2lβk+2l−βk+2lαk+2l]}≜ |
and \tilde{\Lambda}_2 \in {\bf \mathbb{R}}^{2(k+2l) \times 2(k+2l)} , \tilde{X} \in {\bf \mathbb{R}}^{n \times p} . y_{t+2s+j} and z_{t+2s+j} are, respectively, the real part and imaginary part of the complex vector x_{t+2s+j} for j = 1, 3, \cdots, 2(k+2l)-1 . Using (6) and (7), the matrix equation (2) is equivalent to
\begin{equation} A\tilde{X} = A^\top \tilde{X}\tilde{\Lambda}. \end{equation} | (8) |
Since \text{rank}(X) = \text{rank}(\tilde{X}) = p . Now, let the QR-decomposition of \tilde{X} be
\begin{equation} \tilde{X} = Q\left[ \begin{array}{c} R \\ 0 \\ \end{array} \right], \end{equation} | (9) |
where Q = [Q_1, Q_2]\in \mathbb{R}^{n \times n} is an orthogonal matrix and R \in \mathbb{R}^{p \times p} is nonsingular. Let
\begin{equation} \begin{array}{cc} Q^\top AQ = \left[ \begin{array}{cc} A_{11}& A_{12} \\ A_{21}& A_{22} \\ \end{array}\right]&\begin{array}{c} p \\ n-p \\ \end{array} \\ \begin{array}{cc} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p & \ n-p \\ \end{array}& \end{array}. \end{equation} | (10) |
Using (9) and (10), then the equation of (8) is equivalent to
\begin{eqnarray} && A_{11}R = A_{11}^\top R\tilde{\Lambda}, \end{eqnarray} | (11) |
\begin{eqnarray} && A_{21}R = A_{12}^\top R\tilde{\Lambda}. \end{eqnarray} | (12) |
Write
\begin{equation} \begin{array}{ccc} R^\top A_{11}R\triangleq F = \left[ \begin{array}{ccc} f_{11} & F_{12} & \ \ \ F_{13} \\ F_{21} & F_{22} & \ \ \ F_{23} \\ F_{31} & F_{32} & \ \ \ F_{33} \\ \end{array} \right]&\begin{array}{c} t \\ 2s \\ 2(k+2l) \\ \end{array} \\ \begin{array}{ccc} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t & \ \ \ 2s & \ 2(k+2l) \\ \end{array}& \end{array}, \end{equation} | (13) |
then the equation of (11) is equivalent to
\begin{eqnarray} && F_{12} = F_{21}^\top \Lambda_1, \ F_{21} = F_{12}^\top, \end{eqnarray} | (14) |
\begin{eqnarray} && F_{13} = F_{31}^\top \tilde{\Lambda}_2, \ F_{31} = F_{13}^\top, \end{eqnarray} | (15) |
\begin{eqnarray} && F_{23} = F_{32}^\top \tilde{\Lambda}_2, \ F_{32} = F_{23}^\top \Lambda_1, \end{eqnarray} | (16) |
\begin{eqnarray} && F_{22} = F_{22}^\top \Lambda_1, \end{eqnarray} | (17) |
\begin{eqnarray} && F_{33} = F_{33}^\top \tilde{\Lambda}_2. \end{eqnarray} | (18) |
Because the elements of \Lambda_1, \tilde{\Lambda}_2 are distinct, we can obtain the following relations by Eqs (14)-(18)
\begin{eqnarray} &&F_{12} = 0, \ F_{21} = 0, \ F_{13} = 0, \ F_{31} = 0, \ F_{23} = 0, \ F_{32} = 0, \end{eqnarray} | (19) |
\begin{eqnarray} &&F_{22} = \text{diag} \left\{ \left[\begin{array}{cc} 0 & h_{1} \\ \lambda_1 h_{1} & 0 \\ \end{array}\right], \cdots, \left[\begin{array}{cc} 0 & h_{s} \\ \lambda_{2s-1} h_{s} & 0 \\ \end{array}\right] \right\}, \end{eqnarray} | (20) |
\begin{eqnarray} &&F_{33} = \text{diag}\left\{G_{1}, \cdots, G_{k}, \left[\begin{array}{cc} 0 & G_{k+1} \\ G_{k+1}^\top\tilde{\delta}_{k+1} & 0 \\ \end{array}\right], \cdots, \left[\begin{array}{cc} 0 & G_{k+l} \\ G_{k+l}^\top\tilde{\delta}_{k+2l-1} & 0 \\ \end{array}\right]\right\}, \end{eqnarray} | (21) |
where
\begin{eqnarray*} && G_i = a_iB_i, \ G_{k+j} = a_{k+2j-1}D_1+a_{k+2j}D_2, \ G_{k+j}^\top = G_{k+j}, \\ && B_i = \left[\begin{array}{cc} 1 & \frac{1-\alpha_{i}}{\beta_{i}} \\ -\frac{1-\alpha_{i}}{\beta_{i}} & 1 \\ \end{array}\right], \ D_1 = \left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array}\right], \ D_2 = \left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}\right], \end{eqnarray*} |
and 1\leq i\leq k, 1\leq j\leq l . h_{1}, \cdots, h_{s}, a_{1}, \cdots, a_{k+2l} are arbitrary real numbers. It follows from Eq (12) that
\begin{equation} A_{21} = A_{12}^\top E, \end{equation} | (22) |
where E = R\tilde{\Lambda}R^{-1} .
Theorem 1. Suppose that \Lambda = \mathit{\text{diag}} \{\lambda_{1}, \cdots, \lambda_{p}\} \in {\mathbb{C}}^{p \times p} , X = [x_{1}, \cdots, x_{p}] \in {\mathbb{C}}^{n \times p} , where diagonal elements of \Lambda are all distinct, X is of full column rank p , and both \Lambda and X are closed under complex conjugation in the sense that \lambda_{2i} = \bar{\lambda}_{2i-1} \in {\mathbb{C}} , x_{2i} = \bar{x}_{2i-1} \in {\mathbb{C}}^{n} for i = 1, \cdots, m , and \lambda_{j} \in {\mathbb{R}} , x_{j} \in {\mathbb{R}}^{n} for j = 2m+1, \cdots, p . Rearrange the matrix \Lambda as (4) , and adjust the column vectors of X with corresponding to those of \Lambda . Let \Lambda, X transform into \tilde{\Lambda}, \tilde{X} by (6)-(7) and QR-decomposition of the matrix \tilde{X} be given by (9) . Then the general solution of (2) can be expressed as
\begin{equation} {\mathcal{S}}_{A} = \left\{A \left| A = Q\left[ \begin{array}{cc} R^{-\top}\left[ \begin{array}{ccc} f_{11} & 0 & 0 \\ 0 & F_{22} & 0 \\ 0 & 0 & F_{33} \\ \end{array} \right]R^{-1} & A_{12} \\ A_{12}^\top E & A_{22} \\ \end{array} \right]Q^\top \right. \right\}, \end{equation} | (23) |
where E = R\tilde{\Lambda}R^{-1}, f_{11} is arbitrary real number, A_{12} \in {\mathbb{R}}^{p \times (n-p)}, A_{22}\in {\mathbb{R}}^{(n-p) \times (n-p)} are arbitrary real-valued matrices and F_{22}, F_{33} are given by (20)-(21) .
In order to solve Problem BAP, we need the following lemma.
Lemma 1. [16] Let A, B be two real matrices, and X be an unknown variable matrix. Then
\begin{eqnarray*} && \frac{\partial tr(BX)}{\partial X} = B^\top, \ \frac{\partial tr(X^\top B^\top)}{\partial X} = B^\top, \ \frac{\partial tr(AXBX)}{\partial X} = (BXA+AXB)^\top, \\ && \frac{\partial tr(AX^\top BX^\top)}{\partial X} = BX^\top A+AX^\top B, \ \frac{\partial tr(AXBX^\top)}{\partial X} = AXB+A^\top XB^\top. \end{eqnarray*} |
By Theorem 1 , we can obtain the explicit representation of the solution set {\mathcal{S}}_{A} . It is easy to verify that \mathcal{S}_A is a closed convex subset of {\bf \mathbb{R}}^{n \times n}\times {\bf \mathbb{R}}^{n \times n}. By the best approximation theorem (see Ref. [17]), we know that there exists a unique solution of Problem BAP. In the following we will seek the unique solution \hat{A} in \mathcal{S}_A. For the given matrix \tilde{A} \in {\bf \mathbb{R}}^{n \times n}, write
\begin{equation} \begin{array}{cc} Q^\top \tilde{A}Q = \left[ \begin{array}{cc} \tilde{A}_{11}& \tilde{A}_{12} \\ \tilde{A}_{21}& \tilde{A}_{22} \\ \end{array}\right]&\begin{array}{c} p \\ n-p \\ \end{array} \\ \begin{array}{cc} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p & \ n-p \\ \end{array}& \end{array}, \end{equation} | (24) |
then
\begin{eqnarray*} \|A-\tilde{A}\|^2 & = & \left\|\left[ \begin{array}{cc} R^{- \top}\left[ \begin{array}{ccc} f_{11} & 0 & 0 \\ 0 & F_{22} & 0 \\ 0 & 0 & F_{33} \\ \end{array} \right]R^{-1}-\tilde{A}_{11} & A_{12}-\tilde{A}_{12} \\ A_{12}^\top E-\tilde{A}_{21} & A_{22}-\tilde{A}_{22} \\ \end{array} \right]\right\|^2\\ & = &\left\| R^{- \top}\left[ \begin{array}{ccc} f_{11} & 0 & 0 \\ 0 & F_{22} & 0 \\ 0 & 0 & F_{33} \\ \end{array} \right]R^{-1}-\tilde{A}_{11}\right\|^2\\ &+& \| A_{12}-\tilde{A}_{12}\|^2+\|A_{12}^\top E-\tilde{A}_{21} \|^2+\|A_{22}-\tilde{A}_{22}\|^2. \end{eqnarray*} |
Therefore, \|A-\tilde{A}\| = \min if and only if
\begin{eqnarray} && \left\|R^{-\top}\left[ \begin{array}{ccc} f_{11} & 0 & 0 \\ 0 & F_{22} & 0 \\ 0 & 0 & F_{33} \\ \end{array} \right]R^{-1}-\tilde{A}_{11}\right\|^2 = \min, \end{eqnarray} | (25) |
\begin{eqnarray} && \| A_{12}-\tilde{A}_{12}\|^2+\|A_{12}^\top E-\tilde{A}_{21} \|^2 = \min, \end{eqnarray} | (26) |
\begin{eqnarray} &&A_{22} = \tilde{A}_{22}. \end{eqnarray} | (27) |
Let
\begin{equation} \begin{array}{c} R^{-1} = \left[ \begin{array}{c} R_1 \\ R_2 \\ R_3 \\ \end{array} \right] \end{array}, \end{equation} | (28) |
then the relation of (25) is equivalent to
\begin{equation} \|R_1^\top f_{11}R_1+ R_2^\top F_{22}R_2 + R_3^\top F_{33} R_3- \tilde{A}_{11} \|^2 = \min. \end{equation} | (29) |
Write
\begin{equation} R_1 = \left[r_{1, t}\right], \ R_2 = \left[ \begin{array}{c} r_{2, 1} \\ \vdots \\ r_{2, 2s}\\ \end{array} \right], \ R_3 = \left[ \begin{array}{c} r_{3, 1} \\ \vdots \\ r_{3, k+2l} \\ \end{array} \right], \end{equation} | (30) |
where r_{1, t} \in {\mathbb{R}}^{t \times p}, r_{2, i} \in {\mathbb{R}}^{1 \times p}, r_{3, j} \in {\mathbb{R}}^{2 \times p}, \ i = 1, \cdots, 2s, \ j = 1, \cdots, k+2l .
Let
\begin{equation} \left\{ \begin{array}{rcl} && J_t = r_{1, t}^\top r_{1, t}, \\ && J_{t+i} = \lambda_{2i-1}r_{2, 2i}^\top r_{2, 2i-1}+r_{2, 2i-1}^\top r_{2, 2i} \ (1 \leq i \leq s), \\ && J_{r+i} = r_{3, i}^\top B_i r_{3, i} \ (1 \leq i \leq k), \\ && J_{r+k+2i-1} = r_{3, k+2i}^\top D_1 \tilde{\delta}_{k+2i-1}r_{3, k+2i-1}+r_{3, k+2i-1}^\top D_1 r_{3, k+2i} \ (1 \leq i \leq l), \\ && J_{r+k+2i} = r_{3, k+2i}^\top D_2 \tilde{\delta}_{k+2i-1}r_{3, k+2i-1}+r_{3, k+2i-1}^\top D_2 r_{3, k+2i} \ (1 \leq i \leq l), \end{array} \right. \end{equation} | (31) |
with r = t+s, q = t+s+k+2l. Then the relation of (29) is equivalent to
\begin{eqnarray*} && g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l}) = \\ &&\|f_{11}J_t+h_1J_{t+1}+\cdots+h_sJ_{r}+a_1J_{r+1}+\cdots+a_{k+2l}J_{q}-\tilde{A}_{11}\|^2 = \text{min}, \end{eqnarray*} |
that is,
\begin{eqnarray*} && g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l}) \\ && = \text{tr} [(f_{11}J_t+h_1J_{t+1}+\cdots+h_sJ_{r}+a_1J_{r+1}+\cdots+a_{k+2l}J_{q}-\tilde{A}_{11})^\top \\ && (f_{11}J_t+h_1J_{t+1}+\cdots+h_sJ_{r}+a_1J_{r+1}+\cdots+a_{k+2l}J_{q}-\tilde{A}_{11})]\\ && = f_{11}^2c_{t, t}+2f_{11}h_1c_{t, t+1}+\cdots+2f_{11}h_sc_{t, r}+2f_{11}a_1c_{t, r+1}+\cdots+2f_{11}a_{k+2l}c_{t, q}-2f_{11}e_t\\ && +h_1^2c_{t+1, t+1}+\cdots+2h_1h_sc_{t+1, r}+2h_1a_1c_{t+1, r+1}+\cdots+2h_1a_{k+2l}c_{t+1, q}-2h_1e_{t+1} \\ && +\cdots \\ && +h_s^2c_{r, r}+2h_sa_1c_{r, r+1}+\cdots+2h_sa_{k+2l}c_{r, q}-2h_se_{r}\\ && +a_1^2c_{r+1, r+1}+\cdots+2a_1a_{k+2l}c_{r+1, q}-2a_1e_{r+1}\\ && +\cdots \\ && +a_{k+2l}^2c_{q, q}-2a_{k+2l}e_{q}+\text{tr} (\tilde{A}_{11}^\top \tilde{A}_{11}), \end{eqnarray*} |
where c_{i, j} = \text{tr} (J_i^\top J_j), e_i = \text{tr} (J_i^\top\tilde{A}_{11}) (i, j = t, \cdots, t+s+k+2l) and c_{i, j} = c_{j, i} .
Consequently,
\begin{eqnarray*} \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial f_{11}}&& = 2f_{11}c_{t, t}+2h_1c_{t, t+1}+\cdots+2h_sc_{t, r}+2a_1c_{t, r+1}\\ &&+\cdots+2a_{k+2l}c_{t, q}-2e_t, \\ \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial h_{1}}&& = 2f_{11}c_{t+1, t}+2h_1c_{t+1, t+1}+\cdots+2h_sc_{t+1, r}+2a_1c_{t+1, r+1}\\ &&+\cdots+2a_{k+2l}c_{t+1, q}-2e_{t+1}, \\ &&\cdots\\ \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial h_{s}}&& = 2f_{11}c_{r, t}+2h_1c_{r, t+1}+\cdots+2h_sc_{r, r}+2a_1c_{r, r+1}\\ &&+\cdots+2a_{k+2l}c_{r, q}-2e_{r}, \\ \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial a_{1}}&& = 2f_{11}c_{r+1, t}+2h_1c_{r+1, t+1}+\cdots+2h_sc_{r+1, r}+2a_1c_{r+1, r+1}\\ &&+\cdots+2a_{k+2l}c_{r+1, q}-2e_{r+1}, \\ &&\cdots\\ \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial a_{k+2l}}&& = 2f_{11}c_{q, t}+2h_1c_{q, t+1}+\cdots+2h_sc_{q, r}+2a_1c_{q, r+1}\\ &&+\cdots+2a_{k+2l}c_{q, q}-2e_{q}. \end{eqnarray*} |
Clearly, g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l}) = \text{min} if and only if
\frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial f_{11}} = 0, \cdots, \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial a_{k+2l}} = 0. |
Therefore,
\begin{equation} \begin{split} &f_{11}c_{t, t}+h_1c_{t, t+1}+\cdots+h_sc_{t, r}+a_1c_{t, r+1}+\cdots+a_{k+2l}c_{t, q} = e_t, \\ &f_{11}c_{t+1, t}+h_1c_{t+1, t+1}+\cdots+h_sc_{t+1, r}+a_1c_{t+1, r+1}+\cdots+a_{k+2l}c_{t+1, q} = e_{t+1}, \\ &\cdots\\ &f_{11}c_{r, t}+h_1c_{r, t+1}+\cdots+h_sc_{r, r}+a_1c_{r, r+1}+\cdots+a_{k+2l}c_{r, q} = e_{r}, \\ &f_{11}c_{r+1, t}+h_1c_{r+1, t+1}+\cdots+h_sc_{r+1, r}+a_1c_{r+1, r+1}+\cdots+a_{k+2l}c_{r+1, q} = e_{r+1}, \\ &\cdots\\ &f_{11}c_{q, t}+h_1c_{q, t+1}+\cdots+h_sc_{q, r}+a_1c_{q, r+1}+\cdots+a_{k+2l}c_{q, q} = e_{q}. \end{split} \end{equation} | (32) |
If let
C = \left[ \begin{array}{ccccccc} c_{t, t}&c_{t, t+1}&\cdots&c_{t, r}&c_{t, r+1}&\cdots&c_{t, q}\\ c_{t+1, t}&c_{t+1, t+1}&\cdots&c_{t+1, r}&c_{t+1, r+1}&\cdots&c_{t+1, q}\\ \vdots&\vdots& &\vdots&\vdots& &\vdots\\ c_{r, t}&c_{r, t+1}&\cdots&c_{r, r}&c_{r, r+1}&\cdots&c_{r, q}\\ c_{r+1, t}&c_{r+1, t+1}&\cdots&c_{r+1, r}&c_{r+1, r+1}&\cdots&c_{r+1, q}\\ \vdots&\vdots& &\vdots&\vdots& &\vdots\\ c_{q, t}&c_{q, t+1}&\cdots&c_{q, r}&c_{q, r+1}&\cdots&c_{q, q}\\ \end{array} \right], \ h = \left[ \begin{array}{c} f_{11}\\ h_1\\ \vdots\\ h_s\\ a_1\\ \vdots\\ a_{k+2l}\\ \end{array} \right], \ e = \left[ \begin{array}{c} e_t\\ e_{t+1}\\ \vdots\\ e_{r}\\ e_{r+1}\\ \vdots\\ e_{q}\\ \end{array} \right], |
where C is symmetric matrix. Then the equation (32) is equivalent to
\begin{equation} C h = e, \end{equation} | (33) |
and the solution of the equation (33) is
\begin{equation} h = C^{-1} e. \end{equation} | (34) |
Substituting (34) into (20)-(21), we can obtain f_{11}, F_{22} and F_{33} explicitly. Similarly, the equation of (26) is equivalent to
\begin{eqnarray*} &&g(A_{12}) = \text{tr} (A_{12}^\top A_{12})+\text{tr} (\tilde{A}_{12}^\top \tilde{A}_{12})-2\text{tr} (A_{12}^\top \tilde{A}_{12})\\ &&+\text{tr} (E^\top A_{12}A_{12}^\top E)+\text{tr} (\tilde{A}_{21}^\top \tilde{A}_{21})-2\text{tr} (E^\top A_{12}\tilde{A}_{21}). \end{eqnarray*} |
Applying Lemma 1, we obtain
\frac{\partial g(A_{12})}{\partial A_{12}} = 2A_{12}-2\tilde{A}_{12}+2EE^\top A_{12}-2E\tilde{A}_{21}^\top, |
setting \frac{\partial g(A_{12})}{\partial A_{12}} = 0 , we obtain
\begin{equation} A_{12} = (I_p+EE^\top)^{-1}(\tilde{A}_{12}+E\tilde{A}_{21}^\top), \end{equation} | (35) |
Theorem 2. Given \tilde{A} \in {\mathbb{R}}^{n \times n} , then the Problem BAP has a unique solution and the unique solution of Problem BAP is
\begin{equation} \hat{A} = Q\left[ \begin{array}{cc} R^{-\top}\left[ \begin{array}{ccc} f_{11} & 0 & 0 \\ 0 & F_{22} & 0 \\ 0 & 0 & F_{33} \\ \end{array} \right]R^{-1} & A_{12} \\ A_{12}^\top E & \tilde{A}_{22} \\ \end{array} \right]Q^\top, \end{equation} | (36) |
where E = R\tilde{\Lambda} R^{-1} , F_{22}, F_{33}, A_{12}, \tilde{A}_{22} are given by (20), (21), (35), (24) and f_{11}, h_{1}, \cdots, h_s, a_1, \cdots, a_{k+2l} are given by (34) .
Based on Theorems 1 and 2 , we can describe an algorithm for solving Problem BAP as follows.
Algorithm 1.
1) Input matrices \Lambda , X and \tilde{A} ;
2) Rearrange \Lambda as (4), and adjust the column vectors of X with corresponding to those of \Lambda ;
3) Form the unitary transformation matrix T_p by (5);
4) Compute real-valued matrices \tilde{\Lambda}, \tilde{X} by (6) and (7);
5) Compute the QR-decomposition of \tilde{X} by (9);
6) F_{12} = 0, F_{21} = 0, F_{13} = 0, F_{31} = 0, F_{23} = 0, F_{32} = 0 by (19) and E = R\tilde{\Lambda} R^{-1} ;
7) Compute \tilde{A}_{ij} = Q_i^\top \tilde{A}Q_j, i, j = 1, 2 ;
8) Compute R^{-1} by (28) to form R_1, R_2, R_3 ;
9) Divide matrices R_1, R_2, R_3 by (30) to form r_{1, t}, r_{2, i}, r_{3, j}, i = 1, \cdots, 2s, j = 1, \cdots, k+2l ;
10) Compute J_i, i = t, \cdots, t+s+k+2l, by (31);
11) Compute c_{i, j} = \mbox{tr} (J_i^\top J_j), e_i = \mbox{tr} (J_i^\top\tilde{A}_{11}), i, j = t, \cdots, t+s+k+2l ;
12) Compute f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l} by (34);
13) Compute F_{22}, F_{33} by (20), (21) and A_{22} = \tilde{A}_{22} ;
14) Compute A_{12} by (35) and A_{21} by (22);
15) Compute the matrix \hat{A} by (36).
Example 1. Consider a 11 -DOF system, where
\tilde{A} = \left[ {\begin{array}{rrrrrrrrrrr} 96.1898 & 18.1847 & 51.3250 & 49.0864 & 13.1973 & 64.9115 & 62.5619 & 81.7628 & 58.7045 & 31.1102 & 26.2212 \\ 0.4634 & 26.3803 & 40.1808 & 48.9253 & 94.2051 & 73.1722 & 78.0227 & 79.4831 & 20.7742 & 92.3380 & 60.2843 \\ 77.4910 & 14.5539 & 7.5967 & 33.7719 & 95.6135 & 64.7746 & 8.1126 & 64.4318 & 30.1246 & 43.0207 & 71.1216 \\ 81.7303 & 13.6069 & 23.9916 & 90.0054 & 57.5209 & 45.0924 & 92.9386 & 37.8609 & 47.0923 & 18.4816 & 22.1747 \\ 86.8695 & 86.9292 & 12.3319 & 36.9247 & 5.9780 & 54.7009 & 77.5713 & 81.1580 & 23.0488 & 90.4881 & 11.7418 \\ 8.4436 & 57.9705 & 18.3908 & 11.1203 & 23.4780 & 29.6321 & 48.6792 & 53.2826 & 84.4309 & 97.9748 & 29.6676 \\ 39.9783 & 54.9860 & 23.9953 & 78.0252 & 35.3159 & 74.4693 & 43.5859 & 35.0727 & 19.4764 & 43.8870 & 31.8778 \\ 25.9870 & 14.4955 & 41.7267 & 38.9739 & 82.1194 & 18.8955 & 44.6784 & 93.9002 & 22.5922 & 11.1119 & 42.4167 \\ 80.0068 & 85.3031 & 4.9654 & 24.1691 & 1.5403 & 68.6775 & 30.6349 & 87.5943 & 17.0708 & 25.8065 & 50.7858 \\ 43.1414 & 62.2055 & 90.2716 & 40.3912 & 4.3024 & 18.3511 & 50.8509 & 55.0156 & 22.7664 & 40.8720 & 8.5516 \\ 91.0648 & 35.0952 & 94.4787 & 9.6455 & 16.8990 & 36.8485 & 51.0772 & 62.2475 & 43.5699 & 59.4896 & 26.2482 \\ \end{array}} \right], |
the measured eigenvalue and eigenvector matrices \Lambda and X are given by
\begin{eqnarray*} &&\Lambda = \mbox{diag} \{1.0000, \ -1.8969, \ -0.5272, \ -0.1131+0.9936i, -0.1131-0.9936i, \\ &&1.9228+2.7256i, \ 1.9228-2.7256i, \ 0.1728-0.2450i, \ 0.1728+0.2450i\}, \end{eqnarray*} |
and
\begin{eqnarray*} &&X = \left[ {\begin{array}{rrrrr} -0.0132& -1.0000& 0.1753& 0.0840 + 0.4722i& 0.0840 - 0.4722i \\ -0.0955& 0.3937& 0.1196& -0.3302 - 0.1892i& -0.3302 + 0.1892i \\ -0.1992& 0.5220& -0.0401& 0.3930 - 0.2908i& 0.3930 + 0.2908i \\ 0.0740& 0.0287& 0.6295& -0.3587 - 0.3507i& -0.3587 + 0.3507i \\ 0.4425& -0.3609& -0.5745& 0.4544 - 0.3119i& 0.4544 + 0.3119i \\ 0.4544& -0.3192& -0.2461& -0.3002 - 0.1267i& -0.3002 + 0.1267i \\ 0.2597& 0.3363& 0.9046& -0.2398 - 0.0134i& -0.2398 + 0.0134i \\ 0.1140& 0.0966& 0.0871& 0.1508 + 0.0275i& 0.1508 - 0.0275i \\ -0.0914& -0.0356& -0.2387& -0.1890 - 0.0492i& -0.1890 + 0.0492i \\ 0.2431& 0.5428& -1.0000& 0.6652 + 0.3348i& 0.6652 - 0.3348i \\ 1.0000& -0.2458& 0.2430& -0.2434 + 0.6061i& -0.2434 - 0.6061i \\ \end{array}} \right. \\ &&\left. {\begin{array}{rrrr} 0.6669 + 0.2418i& 0.6669 - 0.2418i& 0.2556 - 0.1080i& 0.2556 + 0.1080i \\ -0.1172 - 0.0674i& -0.1172 + 0.0674i& -0.5506 - 0.1209i& -0.5506 + 0.1209i \\ 0.5597 - 0.2765i& 0.5597 + 0.2765i& -0.3308 + 0.1936i& -0.3308 - 0.1936i \\ -0.7217 - 0.0566i& -0.7217 + 0.0566i& -0.7306 - 0.2136i& -0.7306 + 0.2136i \\ 0.0909 + 0.0713i& 0.0909 - 0.0713i& 0.5577 + 0.1291i& 0.5577 - 0.1291i \\ 0.1867 + 0.0254i& 0.1867 - 0.0254i& 0.2866 + 0.1427i& 0.2866 - 0.1427i \\ -0.5311 - 0.1165i& -0.5311 + 0.1165i& -0.3873 - 0.1096i& -0.3873 + 0.1096i \\ 0.2624 + 0.0114i& 0.2624 - 0.0114i& -0.6438 + 0.2188i& -0.6438 - 0.2188i \\ -0.0619 - 0.1504i& -0.0619 + 0.1504i& 0.2787 - 0.2166i& 0.2787 + 0.2166i \\ 0.3294 - 0.1718i& 0.3294 + 0.1718i& 0.9333 + 0.0667i& 0.9333 - 0.0667i \\ -0.4812 + 0.5188i& -0.4812 - 0.5188i& 0.6483 - 0.1950i& 0.6483 + 0.1950i \\ \end{array}} \right]. \end{eqnarray*} |
Using Algorithm 1, we obtain the unique solution of Problem BAP as follows:
\begin{eqnarray*} \hat{A} = \left[ {\begin{array}{rrrrrrrrrrr} 34.2563 & 41.7824 & 33.3573 & 33.6298 & 23.8064 & 42.0770 & 50.0641 & 37.5705 & 31.0908 & 48.6169 & 19.0972\\ 18.8561 & 35.2252 & 35.9592 & 44.3502 & 31.9918 & 55.2920 & 55.3052 & 54.3793 & 31.3909 & 60.8345 & 16.9540\\ 29.6359 & 7.6805 & 19.1249 & 17.7183 & 16.7082 & 40.0636 & 18.2916 & 49.9437 & 37.6913 & 15.6027 & 4.9603\\ 58.8782 & 51.4906 & 47.8974 & 35.6985 & 45.6889 & 56.0434 & 53.0908 & 56.5402 & 55.5120 & 38.3447 & 35.8894\\ 33.4087 & 46.9635 & 9.7767 & 41.4215 & 51.4466 & 52.1058 & 65.6724 & 60.1293 & 5.8061 & 62.0139 & 16.5231\\ 31.6580 & 51.2359 & 24.7978 & 65.5567 & 61.7840 & 62.5494 & 58.9363 & 74.7099 & 52.2105 & 55.8532 & 44.3925\\ 19.2961 & 51.2333 & 22.4280 & 56.9340 & 42.6348 & 45.8453 & 56.3729 & 61.5555 & 31.6836 & 67.9525 & 40.2012\\ 41.2796 & 71.3821 & 34.4140 & 33.2817 & 77.4393 & 60.8944 & 32.1411 & 108.5056 & 49.6078 & 19.8351 & 85.7434\\ 64.0890 & 57.6524 & 19.1280 & 25.0394 & 39.0524 & 66.7740 & 20.9023 & 48.8512 & 14.4695 & 18.9284 & 24.8348\\ 37.2550 & 32.3254 & 38.3534 & 59.7358 & 33.5902 & 54.0265 & 50.7770 & 70.2011 & 65.4159 & 58.0720 & 40.0652\\ 28.1301 & 14.7638 & 8.9507 & 20.0963 & 25.5907 & 59.6940 & 30.8558 & 66.8781 & 30.4807 & 23.6107 & 12.9984\\ \end{array}} \right], \end{eqnarray*} |
and
\|\hat{A}X-\hat{A}^\top X\Lambda\| = 8.2431\times 10^{-13}. |
Therefore, the new model \hat{A}X = \hat{A}^\top X\Lambda reproduces the prescribed eigenvalues (the diagonal elements of the matrix \Lambda ) and eigenvectors (the column vectors of the matrix X ).
Example 2. (Example 4.1 of [12]) Given \alpha = \cos(\theta) , \beta = \sin(\theta) with \theta = 0.62 and \lambda_1 = 0.2, \lambda_2 = 0.3, \lambda_3 = 0.4 . Let
\begin{equation*} J_0 = \left[ {\begin{array}{rr} 0_2 & \Gamma\\ I_2 & I_2\\ \end{array}} \right], \ J_s = \left[ {\begin{array}{cc} 0_3 & \mbox{diag} \{\lambda_1, \lambda_2, \lambda_3 \}\\ I_3 & 0_3\\ \end{array}} \right], \end{equation*} |
where \Gamma = \left[{\begin{array}{cc} \alpha & -\beta\\ \beta & \alpha\\ \end{array}} \right]. We construct
\begin{equation*} \tilde{A} = \left[ {\begin{array}{rr} J_0 & 0\\ 0 & J_s\\ \end{array}} \right], \end{equation*} |
the measured eigenvalue and eigenvector matrices \Lambda and X are given by
\begin{eqnarray*} \Lambda = \mbox{diag} \{5, 0.2, 0.8139+0.5810i, 0.8139-0.5810i\}, \end{eqnarray*} |
and
\begin{eqnarray*} X = \left[ {\begin{array}{rrrr} -0.4155& 0.6875& -0.2157 - 0.4824i& -0.2157 + 0.4824i\\ -0.4224& -0.3148& -0.3752 + 0.1610i& -0.3752 - 0.1610i\\ -0.0703& -0.6302& -0.5950 - 0.4050i& -0.5950 + 0.4050i\\ -1.0000& -0.4667& 0.2293 - 0.1045i& 0.2293 + 0.1045i\\ 0.2650& 0.3051& -0.2253 + 0.7115i& -0.2253 - 0.7115i\\ 0.9030& -0.2327& 0.4862 - 0.3311i& 0.4862 + 0.3311i\\ -0.6742& 0.3132& 0.5521 - 0.0430i& 0.5521 + 0.0430i\\ 0.6358& 0.1172& -0.0623 - 0.0341i& -0.0623 + 0.0341i\\ -0.4119& -0.2768& 0.1575 + 0.4333i& 0.1575 - 0.4333i\\ -0.2062& 1.0000& -0.1779 - 0.0784i& -0.1779 + 0.0784i\\ \end{array}} \right]. \end{eqnarray*} |
Using Algorithm 1, we obtain the unique solution of Problem BAP as follows:
\begin{equation*} \hat{A} = \left[ {\begin{array}{rrrrrrrrrr} -0.1169& -0.2366& 0.6172& -0.7195& -0.0836& 0.2884& 0.0092& -0.0490& -0.0202& 0.0171\\ -0.0114& -0.0957& 0.1462& 0.6194& 0.3738& -0.1637& 0.1291& -0.0071& 0.0972& 0.1247\\ 0.7607& -0.0497& 0.5803& -0.0346& 0.0979& 0.2959& 0.0937& -0.1060& 0.1323& -0.0339\\ -0.0109& 0.6740& -0.3013& 0.7340& 0.1942& -0.0872& 0.0054& 0.0051& 0.0297& 0.0814\\ 0.1783& 0.2283& 0.2643& 0.0387& 0.0986& -0.3125& -0.0292& 0.2926& -0.0717& -0.0546\\ 0.0953& 0.1027& 0.0360& 0.2668& -0.2418& 0.1206& 0.1406& -0.0551& 0.3071& 0.2097\\ -0.0106& -0.2319& 0.1946& -0.0298& -0.1935& 0.0158& -0.0886& 0.0216& -0.0560& 0.2484\\ 0.1044& 0.1285& 0.1902& 0.2277& 0.6961& 0.1657& 0.0728& -0.0262& -0.0831& -0.0001\\ 0.0906& 0.0021& 0.0764& -0.1264& 0.2144& 0.6703& -0.0850& 0.0764& -0.0104& -0.0149\\ -0.1245& 0.0813& 0.1952& -0.0784& 0.0760& -0.0875& 0.7978& -0.0093& 0.0206& -0.1182\\ \end{array}} \right], \end{equation*} |
and
\|\hat{A}X-\hat{A}^\top X\Lambda\| = 1.7538\times 10^{-8}. |
Therefore, the new model \hat{A}X = \hat{A}^\top X\Lambda reproduces the prescribed eigenvalues (the diagonal elements of the matrix \Lambda ) and eigenvectors (the column vectors of the matrix X ).
In this paper, we have developed a direct method to solve the linear inverse palindromic eigenvalue problem by partitioning the matrix \Lambda and using the QR-decomposition. The explicit best approximation solution is given. The numerical examples show that the proposed method is straightforward and easy to implement.
The authors declare no conflict of interest.
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