Characterization of the mechanical properties and thermal conductivity of epoxy-silica functionally graded materials

Jaafar Sh. AbdulRazaq; Abdul Kareem F. Hassan; Nuha H. Jasim; Jaafar Sh. AbdulRazaq; Abdul Kareem F. Hassan; Nuha H. Jasim

doi:10.3934/matersci.2023010

AIMS Materials Science

2023, Volume 10, Issue 1: 182-199. doi: 10.3934/matersci.2023010

Previous Article Next Article

Research article Special Issues

Characterization of the mechanical properties and thermal conductivity of epoxy-silica functionally graded materials

1.
Department of Mechanical Engineering, College of Engineering, University of Basrah, Basrah, Iraq
2.
Department of Materials Engineering, College of Engineering, University of Basrah, Basrah, Iraq

Received: 28 August 2022 Revised: 19 December 2022 Accepted: 13 January 2023 Published: 18 January 2023

A functionally graded material (FGM) was prepared using epoxy resin reinforced with silicon dioxide with a particle size of 100 μm and weight percentages of 0, 20, 40, 60, and 80 wt%. In a gravity-molding process using the hand layup technique, specimens with international standard (ASTM)-calculated dimensions were created in a mold of poly(methyl methacrylate), which is also known as acrylic. Tensile, flexural, impact, infrared wave, and thermal conductivity tests, and X-ray diffraction (XRD) were conducted on specimens of the five layers of the FGM. The XRD and infrared spectroscopy demonstrated that the compositions of the silica particles and epoxy had a strong association with their physical structures. The findings of experimental tests indicated that increasing the ratio of silicon dioxide enhanced the mechanical properties, and the increase in modulus of elasticity was directly related to the weight percentage of the reinforcement material. The composite with 80% silica had a 526.88% higher modulus of elasticity than the pure epoxy specimen. Both tensile and flexural strengths of the composite material were maximal when 40 wt% of the particle silicon dioxide was utilized, which were 68.5% and 67.8% higher than those of the neat epoxy, respectively. The test results also revealed that the impact resistance of the FGM increased when the silica proportion increased, with a maximum value of 60 wt% silica particle content, which was an increase of 76.98% compared to pure epoxy. In addition, the thermal properties of epoxy resin improved when SiO₂ was added to the mixture. Thus, the addition of silica filler to composite materials directly proportionally increased their thermal conductivity to the weight ratio of the reinforcement material, which was 32.68–383.66%. FGM composed of up to 80% silica particles had the highest thermal conductivity.

Keywords:

Citation: Jaafar Sh. AbdulRazaq, Abdul Kareem F. Hassan, Nuha H. Jasim. Characterization of the mechanical properties and thermal conductivity of epoxy-silica functionally graded materials[J]. AIMS Materials Science, 2023, 10(1): 182-199. doi: 10.3934/matersci.2023010

Related Papers:

[1]	Lina Liu, Huiting Zhang, Yinlan Chen . The generalized inverse eigenvalue problem of Hamiltonian matrices and its approximation. AIMS Mathematics, 2021, 6(9): 9886-9898. doi: 10.3934/math.2021574
[2]	Shixian Ren, Yu Zhang, Ziqiang Wang . An efficient spectral-Galerkin method for a new Steklov eigenvalue problem in inverse scattering. AIMS Mathematics, 2022, 7(5): 7528-7551. doi: 10.3934/math.2022423
[3]	Yalçın Güldü, Ebru Mişe . On Dirac operator with boundary and transmission conditions depending Herglotz-Nevanlinna type function. AIMS Mathematics, 2021, 6(4): 3686-3702. doi: 10.3934/math.2021219
[4]	Batirkhan Turmetov, Valery Karachik . On solvability of some inverse problems for a nonlocal fourth-order parabolic equation with multiple involution. AIMS Mathematics, 2024, 9(3): 6832-6849. doi: 10.3934/math.2024333
[5]	Wei Ma, Zhenhao Li, Yuxin Zhang . A two-step Ulm-Chebyshev-like Cayley transform method for inverse eigenvalue problems with multiple eigenvalues. AIMS Mathematics, 2024, 9(8): 22986-23011. doi: 10.3934/math.20241117
[6]	Liangkun Xu, Hai Bi . A multigrid discretization scheme of discontinuous Galerkin method for the Steklov-Lamé eigenproblem. AIMS Mathematics, 2023, 8(6): 14207-14231. doi: 10.3934/math.2023727
[7]	Lingling Sun, Hai Bi, Yidu Yang . A posteriori error estimates of mixed discontinuous Galerkin method for a class of Stokes eigenvalue problems. AIMS Mathematics, 2023, 8(9): 21270-21297. doi: 10.3934/math.20231084
[8]	Jia Tang, Yajun Xie . The generalized conjugate direction method for solving quadratic inverse eigenvalue problems over generalized skew Hamiltonian matrices with a submatrix constraint. AIMS Mathematics, 2020, 5(4): 3664-3681. doi: 10.3934/math.2020237
[9]	Phakhinkon Phunphayap, Prapanpong Pongsriiam . Extremal orders and races between palindromes in different bases. AIMS Mathematics, 2022, 7(2): 2237-2254. doi: 10.3934/math.2022127
[10]	Hannah Blasiyus, D. K. Sheena Christy . Two-dimensional array grammars in palindromic languages. AIMS Mathematics, 2024, 9(7): 17305-17318. doi: 10.3934/math.2024841

Abstract

1. Introduction

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])

$\begin{equation*} (\lambda^2 A_1+\lambda A_0+A_1^\top)x = 0, \end{equation*}$

with $A_i \in {\mathbb{R}}^{n \times n}, \ i = 0, 1$ and $A_0^\top = A_0.$ The eigenvalues $\lambda$ , 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])

$\begin{equation} Ax = \lambda A^\top x, \end{equation}$

(1)

with $A \in {\mathbb{R}}^{n \times n}$ is a given matrix, $\lambda\in {\mathbb{C}}$ and nonzero vectors $x\in {\mathbb{C}}^{n}$ are the wanted eigenvalues and eigenvectors of the vibration model. We can obtain $\frac{1}{\lambda} x^\top A^\top = x^\top A$ by transposing the equation (1). Thus, $\lambda$ and $\frac{1}{\lambda}$ 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 $\ast$ -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 $(\Lambda, X)$ in the form

$\begin{equation*} \Lambda = \text{diag} \{\lambda_{1}, \cdots, \lambda_{p}\} \in {\mathbb{C}}^{p \times p}, \end{equation*}$

and

$\begin{equation*} X = [x_{1}, \cdots, x_{p}] \in {\mathbb{C}}^{n \times p}, \end{equation*}$

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 $j = 2m+1, \cdots, p$ , find a real-valued matrix $A$ that satisfy the equation

$\begin{equation} AX = A^\top X\Lambda. \end{equation}$

(2)

Namely, each pair $(\lambda_{t}, x_{t})$ , $t = 1, \cdots, p$ , is an eigenpair of the matrix pencil

$P(\lambda) = A x- \lambda A^\top x.$

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 $\tilde{A} \in {\mathbb{R}}^{n \times n}$ , find $\hat{A} \in {\mathcal{S}}_{A}$ such that

$\begin{equation} \|\hat{A}-\tilde{A}\| = \min\limits_{{A} \in {\mathcal{S}_{A}}} \|A-\tilde{A}\|, \end{equation}$

(3)

where $\| \cdot \|$ is the Frobenius norm, and ${\mathcal{S}}_{A}$ 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 $\Lambda$ and using the QR-decomposition, the expression of the general solution of Problem IPEP is derived. Also, we show that the best approximation solution $\hat{A}$ of Problem BAP is unique and derive an explicit formula for it.

2. The solution of Problem IPEP

We first rearrange the matrix $\Lambda$ as

$\begin{equation} \begin{array}{ccc} \Lambda = \left[ \begin{array}{ccc} 1 & 0 & \ \ \ 0 \\ 0 & \Lambda_1 & \ \ \ 0 \\ 0 & 0 & \ \ \ \Lambda_2 \\ \end{array} \right]&\begin{array}{c} t \\ 2s \\ 2(k+2l) \\ \end{array} \\ \begin{array}{ccc} \ \ \ \ \ \ \ \ \ \ \ t & \ 2s & 2(k+2l) \\ \end{array}& \end{array}, \end{equation}$

(4)

where $t+2s+2(k+2l) = p$ , $t = 0 \ or \ 1,$

$\begin{eqnarray*} && \Lambda_1 = \text{diag} \{\lambda_1, \lambda_2, \cdots, \lambda_{2s-1}, \lambda_{2s}\}, \lambda_i \in \mathbb{R}, \ \lambda_{2i-1}^{-1} = \lambda_{2i}, \ 1\leq i\leq s, \\ && \Lambda_2 = \text{diag} \{\delta_1, \cdots, \delta_k, \delta_{k+1}, \delta_{k+2}, \cdots, \delta_{k+2l-1}, \delta_{k+2l}\}, \ \delta_j \in \mathbb{C}^{2 \times 2}, \end{eqnarray*}$

with

$\begin{eqnarray*} &&\delta_j = \left[ \begin{array}{cc} \alpha_j+\beta_{j} i & 0 \\ 0 & \alpha_j-\beta_{j} i \\ \end{array} \right], \ i = \sqrt{-1}, \ 1\leq j\leq k+2l, \\ &&\delta_j^{-1} = \bar{\delta}_j, \ 1 \leq j \leq k, \\ &&\delta_{k+2j-1}^{-1} = \delta_{k+2j}, \ 1 \leq j \leq l, \end{eqnarray*}$

and the adjustment of the column vectors of $X$ corresponds to those of $\Lambda$ .

Define $T_p$ as

$\begin{equation} T_p = \text{diag}\left\{ I_{t+2s}, \frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -i \\ 1 & i \\ \end{array}\right], \cdots, \frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -i \\ 1 & i \\ \end{array}\right]\right\} \in \mathbb{C}^{p \times p}, \end{equation}$

(5)

where $i = \sqrt{-1}$ . It is easy to verify that $T_p^\mathrm{H} T_p = I_p$ . Using this matrix of (5), we obtain

$\begin{equation} \tilde{\Lambda} = T_p^\mathrm{H} \Lambda T_p = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \Lambda_1 & 0 \\ 0 & 0 & \tilde{\Lambda}_2 \\ \end{array} \right], \end{equation}$

(6)

$\begin{equation} \tilde{X} = XT_p = [x_t, \cdots, x_{t+2s}, \sqrt{2}y_{t+2s+1}, \sqrt{2}z_{t+2s+1}, \cdots, \sqrt{2}y_{p-1}, \sqrt{2}z_{p-1}], \end{equation}$

(7)

where

$\begin{equation*} \tilde{\Lambda}_2 = \text{diag} \left\{ \left[\begin{array}{cc} \alpha_1 & \beta_1 \\ -\beta_1 & \alpha_1 \\ \end{array}\right], \cdots, \left[\begin{array}{cc} \alpha_{k+2l} & \beta_{k+2l} \\ -\beta_{k+2l} & \alpha_{k+2l} \\ \end{array}\right] \right\}\triangleq \text{diag} \{\tilde{\delta}_1, \cdots, \tilde{\delta}_{k+2l}\}, \end{equation*}$

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 $(20)-(21)$ .

3. The solution of Problem BAP

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 $(34)$ .

4. A numerical example

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

$\|\hat{A}X-\hat{A}^\top X\Lambda\| = 1.7538\times 10^{-8}.$

5. Concluding remarks

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.

Conflict of interest

The authors declare no conflict of interest.

References

[1]	Jahromi BH, Ajdari A, Nayeb-Hashemi H, et al. (2010) Autofrettage of layered and functionally graded metal-ceramic composite vessels. Compos Struct 92: 1813–1822. https://doi.org/10.1016/J.COMPSTRUCT.2010.01.019 doi: 10.1016/J.COMPSTRUCT.2010.01.019
[2]	Abood AM, Khazal H, Hassan AF (2021) On the determination of first-mode stress intensity factors and T-stress in a continuous functionally graded beam using digital image correlation method. AIMS Mater Sci 9: 56–70. https://doi.org/10.1016/j.matpr.2021.02.768 doi: 10.1016/j.matpr.2021.02.768
[3]	Willert-Porada M (2010) Design and fabrication strategy in the world of functional gradation. Int J Mater Prod Technol 39: 59–71. https://doi.org/10.1504/IJMPT.2010.034260 doi: 10.1504/IJMPT.2010.034260
[4]	Bondioli F, Cannillo V, Fabbri E, et al. (2006) Preparation and characterization of epoxy resins filled with submicron spherical zirconia particles. Polimery 51: 794–798. https://doi.org/10.14314/polimery.2006.794 doi: 10.14314/polimery.2006.794
[5]	Haque ZU, Turner DT (1987) Influence of particulate fillers on the indentation hardness of a glassy cross-linked polymer. J Mater Sci 22: 3379–3384. https://doi.org/10.1007/BF01161208 doi: 10.1007/BF01161208
[6]	Conradi M (2013) Nanosilica-reinforced polymer composites. Mater Tehnol 3: 285–293. Available from: http://www.dlib.si/?URN=URN:NBN:SI:doc-KHZVVWFD
[7]	Phong NT, Gabr MH, Anh LH, et al. (2013) Improved fracture toughness and fatigue life of carbon fiber reinforced epoxy composite due to incorporation of rubber nanoparticles. J Mater Sci 48: 6039–6047. https://doi.org/10.1007/s10853-013-7400-z doi: 10.1007/s10853-013-7400-z
[8]	Schmidt D, Shah D, Giannelis EP (2002) New advances in polymer/layered silicate nanocomposites. Curr Opin Solid State Mater Sci 6: 205–212. https://doi.org/10.1016/S1359-0286(02)00049-9 doi: 10.1016/S1359-0286(02)00049-9
[9]	Shen M, Bever MB (1972) Gradients in polymeric materials. J Mater Sci 7: 741–746. https://doi.org/10.1007/BF00549902 doi: 10.1007/BF00549902
[10]	Woldemariam MH, Belingardi G, Koricho EG, et al. (2019) Effects of nanomaterials and particles on mechanical properties and fracture toughness of composite materials: a short review. AIMS Mater Sci 6: 1191–1212. https://doi.org/10.3934/matersci.2019.6.1191 doi: 10.3934/matersci.2019.6.1191
[11]	Hassan AF, Abood AM, Khalaf HI, et al. (2022) A review of functionally graded materials including their manufacture and applications. Int J Mech Eng 7: 744–755. https://doi.org/10.21203/rs.3.rs-1936049/v1 doi: 10.21203/rs.3.rs-1936049/v1
[12]	Chmielewski M, Pietrzak K (2016) Metal-ceramic functionally graded materials - manufacturing, characterization, application. Bull Polish Acad Sci Tech Sci 64: 151–160. https://doi.org/10.1515/bpasts-2016-0017 doi: 10.1515/bpasts-2016-0017
[13]	Hadj Mostefa A, Merdaci S, Mahmoudi N (2018) An overview of functionally graded materials «FGM». SMSD 267–278. https://doi.org/10.1007/978-3-319-89707-3_30 doi: 10.1007/978-3-319-89707-3_30
[14]	Araki N, Tang DW, Ohtani A (2006) Evaluation of thermophysical properties of functionally graded materials. Int J Thermophys 27: 209–219. https://doi.org/10.1007/S10765-006-0034-5 doi: 10.1007/S10765-006-0034-5
[15]	El-Galy IM, Saleh BI, Ahmed MH (2019) Functionally graded materials classifications and development trends from industrial point of view. SN Appl Sci 1: 1–23. https://doi.org/10.1007/S42452-019-1413-4 doi: 10.1007/S42452-019-1413-4
[16]	Gasik MM (2010) Functionally Graded Materials: Bulk processing techniques. Int J Mater Prod Technol 39: 20–29. https://doi.org/10.1504/IJMPT.2010.034257 doi: 10.1504/IJMPT.2010.034257
[17]	Park JJ, Yoon KG, Lee JY (2011) Thermal and mechanical properties of epoxy/micro- and nano- mixed silica composites for insulation materials of heavy electric equipment. Trans Electr Electron Mater 12: 98–101. https://doi.org/10.4313/TEEM.2011.12.3.98 doi: 10.4313/TEEM.2011.12.3.98
[18]	Farouq W, Khazal H, Hassan AKF, (2019) Fracture analysis of functionally graded material using digital image correlation technique and extended element-free Galerkin method. Opt Lasers Eng 121: 307–322. https://doi.org/10.1016/j.optlaseng.2019.04.021 doi: 10.1016/j.optlaseng.2019.04.021
[19]	Migliaresi C, Pegoretti A (2002) Fundamentals of polymeric composite materials. Integr Biomater Sci 69–117. https://doi.org/10.1007/0-306-47583-9_3 doi: 10.1007/0-306-47583-9_3
[20]	Linec M, Mušič B (2019) The effects of silica-based fillers on the properties of epoxy molding compounds. Materials (Basel) 12: 1–11. https://doi.org/10.3390/ma12111811 doi: 10.3390/ma12111811
[21]	Al-Ajaj EA, Alguraby AS, Jawad MK (2015) Nano-SiO₂ addition effect on flexural stress and hardness of EP/MWCNT. Eng Tech J 33: 1248–1257. Available from: https://etj.uotechnology.edu.iq/article_116689.html.
[22]	Rihan Y, Abd El-Bary B (2012) Wear resistance and electrical properties of functionally graded epoxy-resin/silica composites. International Atomic Energy Agency (IAEA) 44: 23–27. Available from: https://inis.iaea.org/search/search.aspx?orig_q=RN:44066233.
[23]	Muslim NB, Hamzah AF, Al-kawaz AE, (2018) Study of mechanical properties of wollastonite filled epoxy functionally graded composite. Int J Mech Eng Technol 9: 669–677. Available from: http://repository.uobabylon.edu.iq/papers/publication.aspx?pubid=18509.
[24]	Parihar RS, Setti SG, Sahu RK (2018) Recent advances in the manufacturing processes of functionally graded materials: A review. Sci Eng Compos Mater 25: 309–336. https://doi.org/10.1515/secm-2015-0395 doi: 10.1515/secm-2015-0395
[25]	Sombatsompop N, Wood AK (1997) Measurement of thermal conductivity of polymers using an improved Lee's Disc apparatus. Polym Test 16: 203–223. https://doi.org/10.1016/S0142-9418(96)00043-8 doi: 10.1016/S0142-9418(96)00043-8
[26]	Pandey A, Kattamuri PK, Sastry BS (2021) Measurement of thermal conductivity of sandstone using Lee's apparatus: A case study. Mining Metall Explor 38: 1997–2003. https://doi.org/10.1007/s42461-021-00461-4 doi: 10.1007/s42461-021-00461-4
[27]	Daramola OO, Akintayo O (2017) Mechanical properties of epoxy matrix composites reinforced with green silica particles. Ann Fac Eng Hunedoara-Int J Eng 4: 167–174. Available from: https://www.researchgate.net/publication/327551546.
[28]	Sipaut CS, Ahmad N, Adnan R, et al. (2007) Properties and morphology of bulk epoxy composites filled with modified fumed silica-epoxy nanocomposites. J Appl Sci 7: 27–34. https://doi.org/10.1002/jrs.1315 doi: 10.1002/jrs.1315
[29]	Kim D, Chung I, Kim G (2014) Study on mechanical and thermal properties of fiber-reinforced epoxy/hybrid-silica composite. Fibers Polym 14: 2141–2147. https://doi.org/10.1007/s12221-013-2141-9 doi: 10.1007/s12221-013-2141-9
[30]	Misra N, Kapusetti G, Pattanayak DK, et al. (2011) Fabrication and characterization of epoxy/silica functionally graded composite material. Indian J Phys 85: 1393–1404. https://doi.org/10.1007/s12648-011-0161-0 doi: 10.1007/s12648-011-0161-0
[31]	Ahmada T, Mamat O (2012) Tronoh silica sand nanoparticle production and applications design for composites. Defect Diffus Forum 330: 39–47. https://doi.org/10.4028/www.scientific.net/DDF.330.39
[32]	Chen L, Chai S, Liu K, et al. (2012) Enhanced epoxy/silica composites mechanical properties by introducing graphene oxide to the interface. ACS Appl Mater Interfaces 4: 4398–4404. https://doi.org/10.1021/am3010576 doi: 10.1021/am3010576
[33]	Yusra AFI, Khalil HPSA, Davoudpour Y, et al. (2015) Characterization of plant nanofiber-reinforced epoxy composites. BioResources 10: 8268–8280. https://doi.org/10.15376/biores.10.4.8268-8280 doi: 10.15376/biores.10.4.8268-8280
[34]	Vikram Singh V, Jaiswal P, et al. (2021) Mechanical properties of silica fume reinforced epoxy glass microballoons syntactic foams. Mater Today Proc 46: 672–676. https://doi.org/10.1016/j.matpr.2020.11.745 doi: 10.1016/j.matpr.2020.11.745
[35]	Park JJ (2018) Mechanical properties of epoxy/micro-silica and epoxy/micro-alumina composites. Trans Electr Electron Mater 19: 481–485. https://doi.org/10.1007/s42341-018-0074-0 doi: 10.1007/s42341-018-0074-0
[36]	Park J, Park J (2018) Electrical and mechanical strength properties of epoxy/micro silica and alumina composites for power equipment for power equipment. J Korean Inst Electr Electron Mater Eng 31: 496–501. https://doi.org/10.4313/JKEM.2018.31.7.496 doi: 10.4313/JKEM.2018.31.7.496
[37]	Ozcan UE, Karabork F, Yazman S, et al. (2019) Effect of silica/graphene nanohybrid particles on the mechanical properties of epoxy coatings. Arab J Sci Eng 44: 5723–5731. https://doi.org/10.1007/s13369-019-03724-x doi: 10.1007/s13369-019-03724-x
[38]	Alameri I, Oltulu M (2020) Mechanical properties of polymer composites reinforced by silica-based materials of various sizes. Appl Nanosci 10: 4087–4102. https://doi.org/10.1007/s13204-020-01516-6 doi: 10.1007/s13204-020-01516-6
[39]	Poh CL, Mariatti M, Ahmad Fauzi MN, et al. (2014) Tensile, dielectric, and thermal properties of epoxy composites filled with silica, mica, and calcium carbonate. J Mater Sci Mater Electron 25: 2111–2119. https://doi.org/10.1007/s10854-014-1847-9 doi: 10.1007/s10854-014-1847-9

This article has been cited by:

Jiajie Luo, Lina Liu, Sisi Li, Yongxin Yuan, A direct method for the simultaneous updating of finite element mass, damping and stiffness matrices, 2022, 0308-1087, 1, 10.1080/03081087.2022.2092047

Reader Comments

Your name:*

Email:*
© 2023 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)