Inverse eigenvalue problem for pentadiagonal matrices

Abstract

In this paper, we propose an algorithm for constructing a pentadiagonal matrix with given prescribed three spectra. Sufficient conditions for solvability of the problem are given. We generate an algorithmic procedure to construct the solution matrices and we given a numerical example illustrating the construction algorithm.

Keywords:

• inverse eigenvalue problem
• spectrum
• pentadiagonal matrix

Introduction

The free undamped infinitesimal vibrations, of frequency $\mathit{\omega }$, of a thin straight beam of length $L$ are governed by the Euler–Bernoulli equation(1.1) $$\begin{array}{c}\hfill \frac{d^2}{{\mathit{dx}}^2}(E(x)I(x)\frac{d^{2}u(x)}{{\mathit{dx}}^2})=A(x)\mathit{\rho }(x){\mathit{\omega }}^{2}u(x),0\le x\le L.\end{array}$$(1.1) Here $E(x)$ is Young’s modulus, $\mathit{\rho }(x)$ is the density, $A(x)$ is the cross sectional area at section $x$ and $I(x)$ is the second moment of this area about the axis through the centroid at right angles to the plane of vibration. The problem can be discretized by assuming that the interval $[0,L]$ is divided into $n$ intervals with length $h_i$, and that $\mathit{\rho }(x)$, $E(x)$, $A(x)$ and $I(x)$ has the values ${\mathit{\rho }}_i,E_i,A_i$ and $I_i$ on the subintervals $(x_i,x_{i+1})$, respectively. Figure 1 shows a simple discrete model for the transversely vibrating Euler–Bernoulli beam. The parameters of discrete model can be defined as(1.2) $$\begin{array}{c}\hfill {\ell }_i=h_i,m_i={\mathit{\rho }}_i{A}_i{h}_i,k_i=\frac{E_i{I}_i}{h_i},i=0,1,\dots ,n.\end{array}$$(1.2) Figure 2 shows the configuration around the mass $m_r$ and the discrete system is obtained. For small displacement, the rotations are$$\begin{array}{c}\hfill {\mathit{\theta }}_r=(u_r-u_{r-1})/{\ell }_r,r=0,1,\dots ,n.\end{array}$$The moment ${\mathit{\tau }}_r$ needed to produce a relative rotation of the two rigid rods on either side of $m_r$ is$$\begin{array}{c}\hfill {\mathit{\tau }}_r=k_{r+1}({\mathit{\theta }}_{r+1}-{\mathit{\theta }}_r),r=0,1,\dots ,n-1.\end{array}$$Equilibrium of the rod linking $m_r$ and $m_{r+1}$ yields the shearing forces$$\begin{array}{c}\hfill {\mathit{\varphi }}_r=({\mathit{\tau }}_r-{\mathit{\tau }}_{r+1})/{\ell }_{r+1},r=-1,0,\dots ,n-1.\end{array}$$Balance equation for mass $m_r$ is$$\begin{array}{c}\hfill m_r{u}_r^{\prime \prime }={\mathit{\varphi }}_r-{\mathit{\varphi }}_{r-1},r=-1,0,\dots ,n.\end{array}$$Here ${\mathit{\varphi }}_{-2},{\mathit{\varphi }}_n$ and ${\mathit{\tau }}_{-1},{\mathit{\tau }}_n$ denote external shearing forces and bending moments, respectively, applied to the ends. Suppose that the left hand end is clamped so that$$u_{-1}=0=u_0.$$The governing equations in the matrix form can be written as

Fig. 1 Discrete model of a vibrating beam.[3]

Fig. 2 The configuration around $m_r$.[3]

(1.3) $$\begin{array}{c}\hfill \mathit{\theta }=L^{-1}{N}^{T}u,\end{array}$$(1.3) (1.4) $$\begin{array}{c}\hfill \mathit{\tau }=\hat{K}{N}^T\mathit{\theta },\end{array}$$(1.4) (1.5) $$\begin{array}{c}\hfill \mathit{\varphi }=L^{-1}{N}^T\mathit{\tau }-{\ell }_n^{-1}{\mathit{\tau }}_n{e}_n,\end{array}$$(1.5) (1.6) $$\begin{array}{c}\hfill M{u}^{\prime \prime }=-N\mathit{\varphi }+{\mathit{\varphi }}_n{e}_n.\end{array}$$(1.6) where $u=[u_1,u_2,\cdots ,u_n],\mathit{\theta }=[{\mathit{\theta }}_1,{\mathit{\theta }}_2,\cdots ,{\mathit{\theta }}_n],\mathit{\tau }=[{\mathit{\tau }}_0,{\mathit{\tau }}_1,\cdots ,{\mathit{\tau }}_{n-1}],\mathit{\varphi }=[{\mathit{\varphi }}_0,{\mathit{\varphi }}_1,\cdots ,{\mathit{\varphi }}_{n-1}],\hat{K}=\mathrm{diag}(k_r),L=\mathrm{diag}({\ell }_r),M=\mathrm{diag}(m_r),e_n=[0,0,\cdots ,0,1]$, and$$N=\left(\begin{array}{ccccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill & \hfill \cdots \hfill & \hfill 0\hfill \\ \hfill ·\hfill & \hfill ·\hfill & \hfill ·\hfill & \hfill ·\hfill & \hfill ·\hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill \hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right).$$Equations (1.3(1.3) $$\begin{array}{c}\hfill \mathit{\theta }=L^{-1}{N}^{T}u,\end{array}$$(1.3) ), (1.4(1.4) $$\begin{array}{c}\hfill \mathit{\tau }=\hat{K}{N}^T\mathit{\theta },\end{array}$$(1.4) ), (1.5(1.5) $$\begin{array}{c}\hfill \mathit{\varphi }=L^{-1}{N}^T\mathit{\tau }-{\ell }_n^{-1}{\mathit{\tau }}_n{e}_n,\end{array}$$(1.5) ), (1.6(1.6) $$\begin{array}{c}\hfill M{u}^{\prime \prime }=-N\mathit{\varphi }+{\mathit{\varphi }}_n{e}_n.\end{array}$$(1.6) ) lead to(1.7) $$\begin{array}{c}\hfill M{u}^{\prime \prime }+\mathit{Ku}={\mathit{\varphi }}_n{e}_n+{\ell }_n^{-1}{\mathit{\tau }}_n{\mathit{Ne}}_n,\end{array}$$(1.7) where(1.8) $$\begin{array}{c}\hfill K={\mathit{NL}}^{-1}N\hat{K}{N}^T{L}^{-1}{N}^T.\end{array}$$(1.8) If we seek the solution of the form$$q=u\sin (\mathit{\omega }t+\mathit{\phi }),$$then$$q^{\prime \prime }=-{\mathit{\omega }}^{2}u\sin (\mathit{\omega }t+\mathit{\phi }),$$so that for free vibration Equation (1.7(1.7) $$\begin{array}{c}\hfill M{u}^{\prime \prime }+\mathit{Ku}={\mathit{\varphi }}_n{e}_n+{\ell }_n^{-1}{\mathit{\tau }}_n{\mathit{Ne}}_n,\end{array}$$(1.7) ) leads to(1.9) $$\begin{array}{c}\hfill (K-\mathit{\lambda }M)u=0,\mathit{\lambda }={\mathit{\omega }}^2.\end{array}$$(1.9) where $K$ is a pentadiagonal matrix and $M=\mathrm{diag}(m_r)$. We let $M=D^2$ and $X=\mathit{Du}$ then we obtain the following eigenvalue problem(1.10) $$\begin{array}{c}\hfill \mathit{HX}=\mathit{\lambda }X,\end{array}$$(1.10) where(1.11) $$\begin{array}{c}\hfill H=D^{-1}{\mathit{KD}}^{-1},\end{array}$$(1.11) is a symmetric pentadiagonal matrix as follows(1.12) $$\begin{array}{c}\hfill H=\left(\begin{array}{cccccc}\hfill a_1\hfill & \hfill b_1\hfill & \hfill c_1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill b_1\hfill & \hfill a_2\hfill & \hfill b_2\hfill & \hfill c_2\hfill & \hfill \hfill & \hfill \hfill \\ \hfill c_1\hfill & \hfill b_2\hfill & \hfill a_3\hfill & \hfill b_3\hfill & \hfill c_3\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill c_{n-2}\hfill & \hfill b_{n-1}\hfill & \hfill a_n\hfill \end{array}\right),\end{array}$$(1.12) where(1.13) $$\begin{array}{c}\hfill \{\begin{array}{c}{a}_i=\frac{1}{m_i}[\frac{k_i}{l_i^2}+\frac{k_{i+1}}{l_i^2}+\frac{2k_{i+1}}{l_i{l}_{i+1}}+\frac{k_{i+2}}{l_{i+1}^2}],\hfill \\ b_i=-\frac{1}{{(m_i{m}_{i+1})}^{\frac{1}{2}}}[\frac{k_{i+1}}{l_i{l}_{i+1}}+\frac{k_{i+1}}{l_{i+1}^2}+\frac{k_{i+2}}{l_{i+1}^2}+\frac{k_{i+2}}{l_{i+1}{l}_{i+2}}],\hfill \\ c_i=\frac{1}{{(m_i{m}_{i+2})}^{\frac{1}{2}}}\left[\frac{k_{i+2}}{l_{i+1}{l}_{i+2}}\right].\hfill \end{array}\end{array}$$(1.13) See [1–4]. Thus, the inverse problem for a discrete beam corresponds to an inverse problem for a symmetric pentadiagonal matrix in which the the elements of the first off-diagonals are negative, all other elements being positive. In this paper we consider the problem of constructing a symmetric pentadiagonal matrix of the form (1.12(1.12) $$\begin{array}{c}\hfill H=\left(\begin{array}{cccccc}\hfill a_1\hfill & \hfill b_1\hfill & \hfill c_1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill b_1\hfill & \hfill a_2\hfill & \hfill b_2\hfill & \hfill c_2\hfill & \hfill \hfill & \hfill \hfill \\ \hfill c_1\hfill & \hfill b_2\hfill & \hfill a_3\hfill & \hfill b_3\hfill & \hfill c_3\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill c_{n-2}\hfill & \hfill b_{n-1}\hfill & \hfill a_n\hfill \end{array}\right),\end{array}$$(1.12) ) and by procedure in Appendix 2 we construct masses $m_i$, stiffnesses $k_i$ and lengths $l_i$ of a discrete beam. We propose an algorithm for constructing a symmetric pentadiagonal matrix of the form (1.12(1.12) $$\begin{array}{c}\hfill H=\left(\begin{array}{cccccc}\hfill a_1\hfill & \hfill b_1\hfill & \hfill c_1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill b_1\hfill & \hfill a_2\hfill & \hfill b_2\hfill & \hfill c_2\hfill & \hfill \hfill & \hfill \hfill \\ \hfill c_1\hfill & \hfill b_2\hfill & \hfill a_3\hfill & \hfill b_3\hfill & \hfill c_3\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill c_{n-2}\hfill & \hfill b_{n-1}\hfill & \hfill a_n\hfill \end{array}\right),\end{array}$$(1.12) ) with given three spectra. The classical problem of constructing pentadiagonal matrix $H$ by spectra was solved by Boley and Golub,[3, 5] Singh [4, 6] and Chu et al. [7]. The set of eigenvalues of matrix $H$ are called the spectrum of $H$ and is denoted by $\mathit{\sigma }(H)$. The matrix obtained by knocking the $(m+1)$th row and column of $H$ is denoted by $H_{m+1}$ and the matrix obtained from $H$ by knocking off the $(m+1)$th row and $(m+1)$th column and $(m+2)$th row and $(m+2)$th column is denoted by $H_{m+1,m+2}$, respectively. $\mathit{\sigma }(H_{m+1})$ and $\mathit{\sigma }(H_{m+1,m+2})$ denote the spectra of matrix $H_{m+1}$ and $H_{m+1,m+2},$ respectively. Boley and Golub [5] proposed an inverse eigenvalue problem for a general symmetric matrix with $2p+1$ bands, the case $p=2$, is a pentadiagonal matrix. They construct, a pentadiagonal matrix with given spectral data $\mathit{\sigma }(H)$, $\mathit{\sigma }(H_1)$ and $\mathit{\sigma }(H_{1,2})$. Singh [4] derived a discrete model of beam equation that leads to a pentadiagonal matrix and solved direct and inverse problem of the beam equation. Chu et al. [7] proposed a numerical method for constructing a symmetric pentadiagonal Toeplitz matrix from three largest eigenvalues. Caddemi and Calio [8] presented the exact closed-form solution for the vibration modes and the eigen-value equation of the Euler–Bernoulli beam-column in the presence of an arbitrary number of concentrated open cracks, and they presented the exact closed-form expressions for the vibration modes of the Euler–Bernoulli beam in the presence of multiple concentrated cracks.[9] Gladwell [10] derive a closed form procedure to construct the in-line system of masses and springs with a minimal mass for given overall stiffness from the one spectrum, also he studied the problem of constructing a discrete model of a contilever beam in flexural vibration having a specified two spectrum, and formulate the problem of finding a minimal mass solution for the given length and stiffness, and obtain explicit solutions in simple cases. Using the idea of discrete model [10] Ghanbari et al [11] construct a Rod using one spectrum and minimal mass condition. In this paper, we construct a pentadiagonal matrix $H$ with given spectral data $\mathit{\sigma }(H)={({\mathit{\lambda }}_i)}_1^n$ , $\mathit{\sigma }(H_{m+1})={({\mathit{\mu }}_i)}_1^{n-1}$ and $\mathit{\sigma }(H_{m+1,m+2})={({\mathit{\nu }}_i)}_1^{n-2}$, for some $0\le m\le n-2$. Physically ${({\mathit{\lambda }}_i)}_1^n$ are the eigenvalues corresponding to the end conditions such as fixed-free, ${({\mathit{\mu }}_i)}_1^{n-1}$ are the eigenvalues of discrete system which the mass $(m+1)$th is fixed and ${({\mathit{\nu }}_i)}_1^{n-2}$ are the eigenvalues of discrete system which the masses $(m+1)$th and $(m+2)$th are fixed. Clearly the eigenvalue must interlace; and for simplicity we assume that the interlacing is strict.(1.14) $$\begin{array}{cc}\hfill & {\mathit{\lambda }}_1>{\mathit{\mu }}_1>{\mathit{\lambda }}_2>\cdots >{\mathit{\mu }}_{n-1}>{\mathit{\lambda }}_n\hfill \\ \hfill & {\mathit{\mu }}_1>{\mathit{\nu }}_1>{\mathit{\mu }}_2>\cdots >{\mathit{\nu }}_{n-2}>{\mathit{\mu }}_{n-1}.\hfill \end{array}$$(1.14)  

Statement of results

The matrices $H_{m+1}$ and $H_{m+1,m+2}$ are of the form(2.1) $$\begin{array}{c}\hfill H_{m+1}=\left(\begin{array}{ccc}\hfill \mathbf{B}\hfill & \hfill {\mathbf{b}}_m\hfill & \hfill 0\hfill \\ \hfill {\mathbf{b}}_m^T\hfill & \hfill a_{m+2}\hfill & \hfill {\mathbf{c}}_{m+2}^T\hfill \\ \hfill 0\hfill & \hfill {\mathbf{c}}_{m+2}\hfill & \hfill \mathbf{C}\hfill \end{array}\right),H_{m+1,m+2}=\left(\begin{array}{cc}\hfill \mathbf{B}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \mathbf{C}\hfill \end{array}\right)\end{array}$$(2.1) where $\mathbf{B}$ and $\mathbf{C}$ are pentadiagonal matrix of order $m$ and $p$, respectively that $m+p+2=n$, ${\mathbf{b}}_m^T=(0,\dots ,0,c_m)$ and ${\mathbf{c}}_{m+2}^T=(b_{m+2},c_{m+2},0,\dots ,0)$. Express the eigenvalue problem for $H_{m+1}$ in terms of the normalized eigenvectors ${(y_j)}_1^m$ and ${(z_j)}_1^p$ of $\mathbf{B}$ and $\mathbf{C}$, respectively. Thus if $X=[x_1,\dots ,x_m,x_{m+1},x_{m+2},\dots ,x_{n-1}]$ is eigenvector of $H_{m+1}$ then we can write $X$ as follows$$X=\left(\begin{array}{ccc}\hfill Y\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill 1\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill Z\hfill \end{array}\right)\left(\begin{array}{c}\hfill p_1\hfill \\ \hfill ⋮\hfill \\ \hfill p_m\hfill \\ \hfill x_{m+1}\hfill \\ \hfill q_1\hfill \\ \hfill ⋮\hfill \\ \hfill q_p\hfill \end{array}\right),Y=[y_1,y_2,\cdots ,y_m],Z=[z_1,z_2,\cdots ,z_p].$$The eigenvalue problem $H_{m+1}X=\mathit{\lambda }X$ becomes$$\left(\begin{array}{ccc}\hfill \mathbf{B}\hfill & \hfill {\mathbf{b}}_m\hfill & \hfill 0\hfill \\ \hfill {\mathbf{b}}_m^T\hfill & \hfill a_{m+2}\hfill & \hfill {\mathbf{c}}_{m+2}^T\hfill \\ \hfill 0\hfill & \hfill {\mathbf{c}}_{m+2}\hfill & \hfill \mathbf{C}\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill Y\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill 1\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill Z\hfill \end{array}\right)\left(\begin{array}{c}\hfill p_1\hfill \\ \hfill ⋮\hfill \\ \hfill p_m\hfill \\ \hfill x_{m+1}\hfill \\ \hfill q_1\hfill \\ \hfill ⋮\hfill \\ \hfill q_p\hfill \end{array}\right)=\mathit{\lambda }\left(\begin{array}{ccc}\hfill Y\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill 1\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill Z\hfill \end{array}\right)\left(\begin{array}{c}\hfill p_1\hfill \\ \hfill ⋮\hfill \\ \hfill p_m\hfill \\ \hfill x_{m+1}\hfill \\ \hfill q_1\hfill \\ \hfill ⋮\hfill \\ \hfill q_p\hfill \end{array}\right).$$Premultiplying this equation by$$\left(\begin{array}{ccc}\hfill Y^T\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill 1\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill Z^T\hfill \end{array}\right),$$and using orthogonality of matrices $Y$ and $Z$ and $Y^T\mathit{BY}=\mathrm{diag}({\mathit{\sigma }}_i),Z^T\mathit{CZ}=\mathrm{diag}({\mathit{\eta }}_i)$ we obtain$$\left(\begin{array}{ccccccc}\hfill \mathit{\lambda }-{\mathit{\sigma }}_1\hfill & \hfill \hfill & \hfill \hfill & \hfill -s_1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \mathit{\lambda }-{\mathit{\sigma }}_m\hfill & \hfill -s_m\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill -s_1\hfill & \hfill \cdots \hfill & \hfill -s_m\hfill & \hfill \mathit{\lambda }-a_{m+2}\hfill & \hfill -t_1\hfill & \hfill \cdots \hfill & \hfill -t_p\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill -t_1\hfill & \hfill \mathit{\lambda }-{\mathit{\eta }}_1\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill ⋮\hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill -t_p\hfill & \hfill \hfill & \hfill \hfill & \hfill \mathit{\lambda }-{\mathit{\eta }}_p\hfill \end{array}\right)\left(\begin{array}{c}\hfill p_1\hfill \\ \hfill ⋮\hfill \\ \hfill p_m\hfill \\ \hfill x_{m+1}\hfill \\ \hfill q_1\hfill \\ \hfill ⋮\hfill \\ \hfill q_p\hfill \end{array}\right)=0$$where ${\mathit{\sigma }}_i$ and ${\mathit{\eta }}_i$ are eigenvalues of matrices $\mathbf{B}$ and $\mathbf{C}$, in $H_{m+1,m+2}$ respectively and(2.2) $$\begin{array}{c}\hfill \begin{array}{ccc}\hfill s_j=c_m{y}_{m,j}\hfill & \hfill ,\hfill & \hfill t_k=b_{m+2}{z}_{1,k}+c_{m+2}{z}_{2,k}.\hfill \end{array}\end{array}$$(2.2) Thus$$\begin{array}{cc}\hfill & (\mathit{\lambda }-{\mathit{\sigma }}_j)p_j-s_j{x}_{m+1}=0,j=1,2,\dots ,m\hfill \\ \hfill & (\mathit{\lambda }-{\mathit{\eta }}_k)q_k-t_k{x}_{m+1}=0,k=1,2,\dots ,p\hfill \end{array}$$so that$$\begin{array}{c}\hfill \{a_{m+2}-{\mathit{\mu }}_i+\sum _{j=1}^m\frac{s_j^2}{{\mathit{\mu }}_i-{\mathit{\sigma }}_j}+\sum _{k=1}^p\frac{t_k^2}{{\mathit{\mu }}_i-{\mathit{\eta }}_k}\}{x}_{m+1,i}=0,i=1,2,\dots ,n-1.\end{array}$$  Theorem 2.1. If all ${({\mathit{\nu }}_i)}_1^{n-2}$ and ${({\mathit{\mu }}_i)}_1^{n-1}$ are distinct then $x_{m+1,i}\ne 0;i=1,2,\dots ,n-1$.   Proof

Suppose that there exists an $i$ such that ${\mathit{\mu }}_i$ is eigenvalue of $H_{m+1}$ and $x_{m+1,i}$, $(m+1)$th component of eigenvector corresponding to ${\mathit{\mu }}_i$, is zero; thus$$\left(\begin{array}{ccc}\hfill \mathbf{B}-{\mathit{\mu }}_i\hfill & \hfill {\mathbf{b}}_m\hfill & \hfill 0\hfill \\ \hfill {\mathbf{b}}_m^T\hfill & \hfill a_{m+2}-{\mathit{\mu }}_i\hfill & \hfill {\mathbf{c}}_{m+2}^T\hfill \\ \hfill 0\hfill & \hfill {\mathbf{c}}_{m+2}\hfill & \hfill \mathbf{C}-{\mathit{\mu }}_i\hfill \end{array}\right)X=0,$$and(2.3) $$\begin{array}{c}\hfill \begin{array}{c}\hfill (\mathbf{B}-{\mathit{\mu }}_{i}I)\hfill \end{array}\left(\begin{array}{c}\hfill x_{1,i}\hfill \\ \hfill x_{2,i}\hfill \\ \hfill ⋮\hfill \\ \hfill x_{m,i}\hfill \end{array}\right)=0,\begin{array}{c}\hfill (\mathbf{C}-{\mathit{\mu }}_{i}I)\hfill \end{array}\left(\begin{array}{c}\hfill x_{m+2,i}\hfill \\ \hfill x_{m+3,i}\hfill \\ \hfill ⋮\hfill \\ \hfill x_{n-1,i}\hfill \end{array}\right)=0.\end{array}$$(2.3) Since $X$ is eigenvector of $H_{m+1}$ corresponding to ${\mathit{\mu }}_i$, thus $X\ne 0$ and from (2.3) ${\mathit{\mu }}_i$ is eigenvalue of $H_{m+1,m+2}$ which is a contradiction, since eigenvalues of matrices $H_{m+1}$ and $H_{m+1,m+2}$ are distinct. $\square$  

By Theorem 2.1 we have$$a_{m+2}-\mathit{\lambda }+\sum _{j=1}^m\frac{s_j^2}{\mathit{\lambda }-{\mathit{\sigma }}_j}+\sum _{k=1}^p\frac{t_k^2}{\mathit{\lambda }-{\mathit{\eta }}_k}=-\frac{P_{n-1}(\mathit{\lambda })}{P_m(\mathit{\lambda })Q_p(\mathit{\lambda })},$$where $P_{n-1}(\mathit{\lambda }),P_m(\mathit{\lambda })$ and $Q_p(\mathit{\lambda })$ are characteristic polynomials of the matrices $H_{m+1},\mathbf{B}$ and $\mathbf{C}$, respectively. This implies that(2.4) $$\begin{array}{c}\hfill s_j^2=-\frac{P_{n-1}({\mathit{\sigma }}_j)}{P_m^{\prime }({\mathit{\sigma }}_j)Q_p({\mathit{\sigma }}_j)},t_k^2=-\frac{P_{n-1}({\mathit{\eta }}_k)}{P_m({\mathit{\eta }}_k)Q_p^{\prime }({\mathit{\eta }}_k)},\end{array}$$(2.4) where $j=1,2,\dots ,m,k=1,2,\dots ,p.$ Interlace condition (1.14(1.14) $$\begin{array}{cc}\hfill & {\mathit{\lambda }}_1>{\mathit{\mu }}_1>{\mathit{\lambda }}_2>\cdots >{\mathit{\mu }}_{n-1}>{\mathit{\lambda }}_n\hfill \\ \hfill & {\mathit{\mu }}_1>{\mathit{\nu }}_1>{\mathit{\mu }}_2>\cdots >{\mathit{\nu }}_{n-2}>{\mathit{\mu }}_{n-1}.\hfill \end{array}$$(1.14) ) implies that $s_j^2$ and $t_k^2$ in (2.4(2.4) $$\begin{array}{c}\hfill s_j^2=-\frac{P_{n-1}({\mathit{\sigma }}_j)}{P_m^{\prime }({\mathit{\sigma }}_j)Q_p({\mathit{\sigma }}_j)},t_k^2=-\frac{P_{n-1}({\mathit{\eta }}_k)}{P_m({\mathit{\eta }}_k)Q_p^{\prime }({\mathit{\eta }}_k)},\end{array}$$(2.4) ) are positive. Now $c_m^2$ and $y_m$ may be computed from(2.5) $$\begin{array}{c}\hfill c_m^2=\sum _{j=1}^m{s}_j^2,y_m=\frac{\mathbf{s}}{c_m},\end{array}$$(2.5) where $\mathbf{s}=[s_1,\cdots ,s_m]$. From trace formula we can compute $a_{m+1}$ and $a_{m+2}$ as follows(2.6) $$\begin{array}{c}\hfill a_{m+1}=\sum _{i=1}^n{\mathit{\lambda }}_i-\sum _{j=1}^{n-1}{\mathit{\mu }}_j,a_{m+2}=\sum _{j=1}^{n-1}{\mathit{\mu }}_j-\sum _{k=1}^{n-2}{\mathit{\nu }}_k.\end{array}$$(2.6) Now we consider eigenvalue problem for matrix $H$. Thus if $X=[x_1,\dots ,x_m,x_{m+1},x_{m+2},\dots ,x_n]$ is eigenvector of $H$ then we can write $X$ as follows(2.7) $$\begin{array}{c}\hfill X=\left(\begin{array}{cccc}\hfill Y\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill \hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill Z\hfill \end{array}\right)\left(\begin{array}{c}\hfill p_1\hfill \\ \hfill ⋮\hfill \\ \hfill p_m\hfill \\ \hfill x_{m+1}\hfill \\ \hfill x_{m+2}\hfill \\ \hfill q_1\hfill \\ \hfill ⋮\hfill \\ \hfill q_p\hfill \end{array}\right)\end{array}$$(2.7) The eigenvalue problem becomes as follows$$\left(\begin{array}{cccccccc}\hfill \mathit{\lambda }-{\mathit{\sigma }}_1\hfill & \hfill \hfill & \hfill \hfill & \hfill -h_1\hfill & \hfill -s_1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \mathit{\lambda }-{\mathit{\sigma }}_m\hfill & \hfill -h_m\hfill & \hfill -s_m\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill -h_1\hfill & \hfill \cdots \hfill & \hfill -h_m\hfill & \hfill \mathit{\lambda }-a_{m+1}\hfill & \hfill -b_{m+1}\hfill & \hfill -d_1\hfill & \hfill \cdots \hfill & \hfill -d_p\hfill \\ \hfill -s_1\hfill & \hfill \cdots \hfill & \hfill -s_m\hfill & \hfill -b_{m+1}\hfill & \hfill \mathit{\lambda }-a_{m+2}\hfill & \hfill -t_1\hfill & \hfill \cdots \hfill & \hfill -t_p\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill -d_1\hfill & \hfill -t_1\hfill & \hfill \mathit{\lambda }-{\mathit{\eta }}_1\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill -d_p\hfill & \hfill -t_p\hfill & \hfill \hfill & \hfill \hfill & \hfill \mathit{\lambda }-{\mathit{\eta }}_p\hfill \end{array}\right)\left(\begin{array}{c}\hfill p_1\hfill \\ \hfill ⋮\hfill \\ \hfill p_m\hfill \\ \hfill x_{m+1}\hfill \\ \hfill x_{m+2}\hfill \\ \hfill q_1\hfill \\ \hfill ⋮\hfill \\ \hfill q_p\hfill \end{array}\right)=0,$$where(2.8) $$\begin{array}{c}\hfill h_i=c_{m-1}{y}_{m-1,i}+b_m{y}_{m,i},\end{array}$$(2.8) (2.9) $$\begin{array}{c}\hfill d_i=c_{m+1}{z}_{1,i}\end{array}$$(2.9) Thus$$\begin{array}{c}\hfill (\mathit{\lambda }-{\mathit{\sigma }}_j)p_j-h_j{x}_{m+1}-s_j{x}_{m+2}=0,j=1,2,\dots ,m\\ \hfill (\mathit{\lambda }-{\mathit{\eta }}_k)q_k-d_k{x}_{m+1}-t_k{x}_{m+2}=0,k=1,2,\dots ,p,\end{array}$$so that for $\mathit{\lambda }={({\mathit{\lambda }}_i)}_{i=1}^n$ we have(2.10) $$\begin{array}{c}\hfill \{\begin{array}{c}(a_{m+1}-\mathit{\lambda }+\sum _{j=1}^m\frac{h_j^2}{\mathit{\lambda }-{\mathit{\sigma }}_j}+\sum _{k=1}^p\frac{d_k^2}{\mathit{\lambda }-{\mathit{\eta }}^k})x_{m+1}\hfill \\ +(b_{m+1}+\sum _{j=1}^m\frac{h_j{s}_j}{\mathit{\lambda }-{\mathit{\sigma }}_j}+\sum _{k=1}^p\frac{d_k{t}_k}{\mathit{\lambda }-{\mathit{\eta }}^k})x_{m+2}=0,\hfill \\ (b_{m+1}+\sum _{j=1}^m\frac{h_j{s}_j}{\mathit{\lambda }-{\mathit{\sigma }}_j}+\sum _{k=1}^p\frac{d_k{t}_k}{\mathit{\lambda }-{\mathit{\eta }}^k})x_{m+1}\hfill \\ +(a_{m+2}-\mathit{\lambda }+\sum _{j=1}^m\frac{s_j^2}{\mathit{\lambda }-{\mathit{\sigma }}_j}+\sum _{k=1}^p\frac{t_k^2}{\mathit{\lambda }-{\mathit{\eta }}^k})x_{m+2}=0.\hfill \end{array}\end{array}$$(2.10)   Theorem 2.2. If all ${({\mathit{\lambda }}_i)}_1^n$ and ${({\mathit{\nu }}_i)}_1^{n-2}$ are distinct then $(x_{m+1},x_{m+2})\ne 0$.   Proof

The proof is similar to Theorem 2.1. $\square$  If we denote the coefficient matrix of the system (2.10(2.10) $$\begin{array}{c}\hfill \{\begin{array}{c}(a_{m+1}-\mathit{\lambda }+\sum _{j=1}^m\frac{h_j^2}{\mathit{\lambda }-{\mathit{\sigma }}_j}+\sum _{k=1}^p\frac{d_k^2}{\mathit{\lambda }-{\mathit{\eta }}^k})x_{m+1}\hfill \\ +(b_{m+1}+\sum _{j=1}^m\frac{h_j{s}_j}{\mathit{\lambda }-{\mathit{\sigma }}_j}+\sum _{k=1}^p\frac{d_k{t}_k}{\mathit{\lambda }-{\mathit{\eta }}^k})x_{m+2}=0,\hfill \\ (b_{m+1}+\sum _{j=1}^m\frac{h_j{s}_j}{\mathit{\lambda }-{\mathit{\sigma }}_j}+\sum _{k=1}^p\frac{d_k{t}_k}{\mathit{\lambda }-{\mathit{\eta }}^k})x_{m+1}\hfill \\ +(a_{m+2}-\mathit{\lambda }+\sum _{j=1}^m\frac{s_j^2}{\mathit{\lambda }-{\mathit{\sigma }}_j}+\sum _{k=1}^p\frac{t_k^2}{\mathit{\lambda }-{\mathit{\eta }}^k})x_{m+2}=0.\hfill \end{array}\end{array}$$(2.10) ) by $A(\mathit{\lambda })$, then by Theorem 2.2, $\det A(\mathit{\lambda })=0$ for $\mathit{\lambda }={({\mathit{\lambda }}_i)}_1^n$. Thus(2.11) $$\begin{array}{c}\hfill \begin{array}{c}\hfill \det A(\mathit{\lambda })=\frac{P_n(\mathit{\lambda })}{P_m(\mathit{\lambda })Q_p(\mathit{\lambda })},\hfill \end{array}\end{array}$$(2.11) where $P_n(\mathit{\lambda })$ denote characteristic polynomial of matrix $H$. Thus(2.12) $$\begin{array}{c}\hfill \begin{array}{c}\hfill \det A({\mathit{\lambda }}_i)=0,\mathrm{for}i=1,2,\dots ,n-1.\hfill \end{array}\end{array}$$(2.12) The system of nonlinear equations (2.12(2.12) $$\begin{array}{c}\hfill \begin{array}{c}\hfill \det A({\mathit{\lambda }}_i)=0,\mathrm{for}i=1,2,\dots ,n-1.\hfill \end{array}\end{array}$$(2.12) ) can be solved by Newton’s method [12] for ${(h_j)}_{j=1}^m$, ${(d_k)}_{k=1}^p$ and $b_{m+1}$ . With known $h_j$ and $d_k,$ Equation (2.9(2.9) $$\begin{array}{c}\hfill d_i=c_{m+1}{z}_{1,i}\end{array}$$(2.9) ) implies that(2.13) $$\begin{array}{c}\hfill c_{m+1}^2=\sum _{i=1}^m{d}_i^2,z_1=\frac{\mathbf{d}}{c_{m+1}},\end{array}$$(2.13) where $\mathbf{d}=[d_1,\dots ,d_p]$. Since vectors $y_m$ and $y_{m-1}$ are orthonormal, multiplying the Equation (2.8(2.8) $$\begin{array}{c}\hfill h_i=c_{m-1}{y}_{m-1,i}+b_m{y}_{m,i},\end{array}$$(2.8) ) by $y_m^T$ we find(2.14) $$\begin{array}{c}\hfill b_m=y_m^T\mathbf{h},c_{m-1}=\pm \parallel \mathbf{h}-b_m{y}_m\parallel ,y_{m-1}=\frac{\mathbf{h}-b_m{y}_m}{c_{m-1}},\end{array}$$(2.14) where $\mathbf{h}=[h_1,\dots ,h_m]$ and $\parallel .\parallel$ denote Euclidean norm of a vector. Also vectors $z_1$ and $z_2$ are orthonormal, multiplying the Equation (2.2(2.2) $$\begin{array}{c}\hfill \begin{array}{ccc}\hfill s_j=c_m{y}_{m,j}\hfill & \hfill ,\hfill & \hfill t_k=b_{m+2}{z}_{1,k}+c_{m+2}{z}_{2,k}.\hfill \end{array}\end{array}$$(2.2) ) by $z_1^T$ we obtain(2.15) $$\begin{array}{c}\hfill b_{m+2}=z_1^T\mathbf{t},c_{m+2}=\pm \parallel \mathbf{t}-b_{m+2}{z}_1\parallel ,z_2=\frac{\mathbf{t}-b_{m+2}{z}_1}{c_{m+2}},\end{array}$$(2.15) where $\mathbf{t}=[t_1,\dots ,t_p]$. In this procedure, a Gram-Schmidt process for finding orthonormal vectors $z_1,z_2,y_m,y_{m-1}$ and elements $b_i,c_i$ is necessary. Having $y_m,y_{m-1},z_1$ and $z_2$, matrices $B$ and $C$ in $H_{m+1,m+2}$ can be computed by using the block Lanczos algorithm (see the Appendix 1). The following algorithm lists the steps for constructing the matrix $H$:

Construction Algorithm:

1. Compute $a_{m+1}$ and $a_{m+2}$ from (2.6(2.6) $$\begin{array}{c}\hfill a_{m+1}=\sum _{i=1}^n{\mathit{\lambda }}_i-\sum _{j=1}^{n-1}{\mathit{\mu }}_j,a_{m+2}=\sum _{j=1}^{n-1}{\mathit{\mu }}_j-\sum _{k=1}^{n-2}{\mathit{\nu }}_k.\end{array}$$(2.6) ).

2. Compute $y_m$ and $c_m$ from (2.5(2.5) $$\begin{array}{c}\hfill c_m^2=\sum _{j=1}^m{s}_j^2,y_m=\frac{\mathbf{s}}{c_m},\end{array}$$(2.5) ).

3. Compute $c_{m+1},z_1$ from (2.13(2.13) $$\begin{array}{c}\hfill c_{m+1}^2=\sum _{i=1}^m{d}_i^2,z_1=\frac{\mathbf{d}}{c_{m+1}},\end{array}$$(2.13) ) and $c_{m-1},b_m,y_{m-1}$ from (2.14(2.14) $$\begin{array}{c}\hfill b_m=y_m^T\mathbf{h},c_{m-1}=\pm \parallel \mathbf{h}-b_m{y}_m\parallel ,y_{m-1}=\frac{\mathbf{h}-b_m{y}_m}{c_{m-1}},\end{array}$$(2.14) ).

4. Compute $b_{m+2},c_{m+2},z_2$ from (2.15(2.15) $$\begin{array}{c}\hfill b_{m+2}=z_1^T\mathbf{t},c_{m+2}=\pm \parallel \mathbf{t}-b_{m+2}{z}_1\parallel ,z_2=\frac{\mathbf{t}-b_{m+2}{z}_1}{c_{m+2}},\end{array}$$(2.15) ).

5. Construct matrices $\mathbf{B}$ and $\mathbf{C}$ from Lanczos algorithm (see Appendix 1).

Note that different choices for the square roots for $c_m,c_{m-1},c_{m+1},c_{m+2}$ and vectors $\mathbf{s}$ and $\mathbf{t}$ will lead to different matrices $H$, that is why we have multiple solutions. Summarizing the previous results, we come to the following corollary: Corollary 2.3. Combining conditions of Theorem 2.1, Theorem2.2and condition(1.14(1.14) $$\begin{array}{cc}\hfill & {\mathit{\lambda }}_1>{\mathit{\mu }}_1>{\mathit{\lambda }}_2>\cdots >{\mathit{\mu }}_{n-1}>{\mathit{\lambda }}_n\hfill \\ \hfill & {\mathit{\mu }}_1>{\mathit{\nu }}_1>{\mathit{\mu }}_2>\cdots >{\mathit{\nu }}_{n-2}>{\mathit{\mu }}_{n-1}.\hfill \end{array}$$(1.14) )we can construct pentadiagonal matrix $H$ by three prescribed spectra.   

Numerical examples

Here we illustrate the construction procedure by an example. We consider the Euler–Bernoulli beam equation as follows$$\begin{array}{c}\hfill {(\frac{e^x}{12}{u}^{\prime \prime }(x))}^{\prime \prime }=\mathit{\lambda }{e}^{x}u(x),0⩽x⩽1.\end{array}$$Suppose that $n=8,m=3,p=3,$ then the prescribed three spectra of discrete beam are as follows$$\begin{array}{cc}\hfill \mathit{\sigma }(H)& =\{5088.6035,4071.4626,2756.2258,1525.3776,645.0498,179.9517,\hfill \\ \hfill & 20.8332,0.3219\},\hfill \\ \hfill \mathit{\sigma }(H_4)& =\{4259.9030,4070.9909,1868.4599,1524.4599,329.7466,177.7887,3.1369\},\hfill \\ \hfill \mathit{\sigma }(H_{4,5})& =\{4166.1655,3391.5356,1712.0069,616.5175,281.8484,13.0717\}.\hfill \end{array}$$These eigenvalues satisfy the conditions of Theorem 2.1, Theorem 2.2 and condition (1.14(1.14) $$\begin{array}{cc}\hfill & {\mathit{\lambda }}_1>{\mathit{\mu }}_1>{\mathit{\lambda }}_2>\cdots >{\mathit{\mu }}_{n-1}>{\mathit{\lambda }}_n\hfill \\ \hfill & {\mathit{\mu }}_1>{\mathit{\nu }}_1>{\mathit{\mu }}_2>\cdots >{\mathit{\nu }}_{n-2}>{\mathit{\mu }}_{n-1}.\hfill \end{array}$$(1.14) ). Step 1 of algorithm gives$$a_4=2053.3403,a_5=2053.3403,$$from (2.4(2.4) $$\begin{array}{c}\hfill s_j^2=-\frac{P_{n-1}({\mathit{\sigma }}_j)}{P_m^{\prime }({\mathit{\sigma }}_j)Q_p({\mathit{\sigma }}_j)},t_k^2=-\frac{P_{n-1}({\mathit{\eta }}_k)}{P_m({\mathit{\eta }}_k)Q_p^{\prime }({\mathit{\eta }}_k)},\end{array}$$(2.4) ) we obtain $\mathbf{t}=[-1214.4311,711.7686,-80.4818],\mathbf{s}=[-178.0074,-241.3591,162.9956]$ and by step 2 we find$$c_3=341.3333,y_3=[-0.5215,-0.7071,0.4775].$$Solving the system (2.12(2.12) $$\begin{array}{c}\hfill \begin{array}{c}\hfill \det A({\mathit{\lambda }}_i)=0,\mathrm{for}i=1,2,\dots ,n-1.\hfill \end{array}\end{array}$$(2.12) ) by using Newton’s method, we find$$\begin{array}{cc}\hfill b_4& =-1368.0009,\mathbf{h}=[943.9315,967.3227,-401.5158],\hfill \\ \hfill \mathbf{d}& =[247.2544,-227.2631,61.0344],\hfill \end{array}$$and step 3 gives$$c_4=341.3333,c_2=341.3333,b_3=-1368.0009$$$$z_1=[0.7244,-0.6658,0.1788],y_2=[0.6753,0.0000,0.7375].$$Using step 4, we find$$b_5=-1368.0009,c_5=341.3333,z_2=[-0.6547,-0.5832,0.4809].$$Using the block Lanczos algorithm (Appendix 1), we find$$\begin{array}{cc}\hfill \mathbf{B}=& \left(\begin{array}{ccc}\hfill 2053.34028\hfill & \hfill -1368.0009\hfill & \hfill 341.3333\hfill \\ \hfill -1368.0009\hfill & \hfill 2053.3408\hfill & \hfill -1368.0009\hfill \\ \hfill 341.3333\hfill & \hfill -1368.0009\hfill & \hfill 2053.3403\hfill \end{array}\right),\hfill \\ \hfill \mathbf{C}=& \left(\begin{array}{ccc}\hfill 2053.3403\hfill & \hfill -1368.0009\hfill & \hfill 341.3333\hfill \\ \hfill -1368.0009\hfill & \hfill 1666.5589\hfill & \hfill -641.3060\hfill \\ \hfill 341.3333\hfill & \hfill -641.3060\hfill & \hfill 301.2256\hfill \end{array}\right).\hfill \end{array}$$Thus we constructed a symmetric pentadiagonal matrix $H$ of the form (1.12(1.12) $$\begin{array}{c}\hfill H=\left(\begin{array}{cccccc}\hfill a_1\hfill & \hfill b_1\hfill & \hfill c_1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill b_1\hfill & \hfill a_2\hfill & \hfill b_2\hfill & \hfill c_2\hfill & \hfill \hfill & \hfill \hfill \\ \hfill c_1\hfill & \hfill b_2\hfill & \hfill a_3\hfill & \hfill b_3\hfill & \hfill c_3\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill c_{n-2}\hfill & \hfill b_{n-1}\hfill & \hfill a_n\hfill \end{array}\right),\end{array}$$(1.12) ) as follows$$\begin{array}{cc}\hfill a& =(2053.3403,2053.3403,2053.3403,2053.3403,2053.3403,2053.3403,1666.5589,\hfill \\ \hfill & 301.2256),\hfill \\ \hfill b& =(-1368.0009,-1368.0009,-1368.0009,-1368.0009,-1368.0009,-1368.0009,\hfill \\ \hfill & -641.3060),\hfill \\ \hfill c& =(341.3333,341.3333,341.3333,341.3333,341.3333,341.3333).\hfill \end{array}$$Computing the spectrum of the solution by MATLAB confirms the spectral data of the matrix $H$. With this $H$ and by Appendix 2 we find masses $m_i$, stiffnesses $k_i$ and lengths $l_i$ of discrete beam as follows$$\begin{array}{c}\hfill \begin{array}{c}d=(0.3764,0.4006,0.4265,0.4540,0.4832,0.5144,0.5476,0.5829),m_i=d_i^2,\hfill \\ l_i=c(0.0088208350)\hfill \\ k=c^2(0.00331977086,0.00376179321,0.0042626702,0.00483023802,\hfill \\ 0.0054733768284,0.00620214849,0.00702795496,0.00796371629).\hfill \end{array}\end{array}$$The pentadiagonal matrices $K$ in (1.8(1.8) $$\begin{array}{c}\hfill K={\mathit{NL}}^{-1}N\hat{K}{N}^T{L}^{-1}{N}^T.\end{array}$$(1.8) ) and $H$ in (1.12(1.12) $$\begin{array}{c}\hfill H=\left(\begin{array}{cccccc}\hfill a_1\hfill & \hfill b_1\hfill & \hfill c_1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill b_1\hfill & \hfill a_2\hfill & \hfill b_2\hfill & \hfill c_2\hfill & \hfill \hfill & \hfill \hfill \\ \hfill c_1\hfill & \hfill b_2\hfill & \hfill a_3\hfill & \hfill b_3\hfill & \hfill c_3\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill c_{n-2}\hfill & \hfill b_{n-1}\hfill & \hfill a_n\hfill \end{array}\right),\end{array}$$(1.12) ) are independent of $c$. But for different values of $c$ we construct different discrete beam which have the same spectrum.

For $\mathbf{t}=[1214.4311,711.7686,80.4818],\mathbf{s}=[178.0074,241.3591,162.9956]$ we obtain$$\begin{array}{cc}\hfill a& =(685.3402,3421.3404,2053.3403,2053.3403,2053.3403,3388.1410,601.4182,\hfill \\ \hfill & 31.5656)\hfill \\ \hfill & b=(-1.4662,1209.0707,-175.6775,-260.8086,1205.3432,39.7682,-94.6102),\hfill \\ \hfill & c=(725.3157,1477.2257,341.3333,1301.8410,731.4934,98.7763).\hfill \end{array}$$This pentadiagonal matrix have the same spectrum but for this we cannot construct a discrete beam since by (1.13(1.13) $$\begin{array}{c}\hfill \{\begin{array}{c}{a}_i=\frac{1}{m_i}[\frac{k_i}{l_i^2}+\frac{k_{i+1}}{l_i^2}+\frac{2k_{i+1}}{l_i{l}_{i+1}}+\frac{k_{i+2}}{l_{i+1}^2}],\hfill \\ b_i=-\frac{1}{{(m_i{m}_{i+1})}^{\frac{1}{2}}}[\frac{k_{i+1}}{l_i{l}_{i+1}}+\frac{k_{i+1}}{l_{i+1}^2}+\frac{k_{i+2}}{l_{i+1}^2}+\frac{k_{i+2}}{l_{i+1}{l}_{i+2}}],\hfill \\ c_i=\frac{1}{{(m_i{m}_{i+2})}^{\frac{1}{2}}}\left[\frac{k_{i+2}}{l_{i+1}{l}_{i+2}}\right].\hfill \end{array}\end{array}$$(1.13) ) all $a_i,c_i$ must be positive and all $b_i$ must be negative.

Conclusion

In this paper, we formulated and solved the inverse problem of constructing a pentadiagonal matrix $H$ using three given spectra with interlacing property (1.14(1.14) $$\begin{array}{cc}\hfill & {\mathit{\lambda }}_1>{\mathit{\mu }}_1>{\mathit{\lambda }}_2>\cdots >{\mathit{\mu }}_{n-1}>{\mathit{\lambda }}_n\hfill \\ \hfill & {\mathit{\mu }}_1>{\mathit{\nu }}_1>{\mathit{\mu }}_2>\cdots >{\mathit{\nu }}_{n-2}>{\mathit{\mu }}_{n-1}.\hfill \end{array}$$(1.14) ). This problem will have multiple solutions because in the constructing process we have different choices for $c_m,c_{m-1},c_{m+1},c_{m+2}$ and vectors $\mathbf{s}$ and $\mathbf{t}$. Also for pentadiagonal matrix we construct masses $m_i$, stiffnesses $k_i$ and lengths $l_i$ of corresponding discrete beam. Numerical example shows the accuracy of this method and details of the algorithm confirms the simplicity of the construction algorithm.

The block Lanczos algorithm [3]

Suppose that $\mathrm{\Lambda }=\mathrm{diag}({\mathit{\lambda }}_1,{\mathit{\lambda }}_2,\dots ,{\mathit{\lambda }}_n)$ and two vectors $x_1$ and $x_2$ are given. Let $X_1=[x_1,x_2]$, this algorithm constructs a pentadiagonal matrix $\mathbf{J}$ of the form$$\mathbf{J}=\left(\begin{array}{cccc}\hfill A_1\hfill & \hfill -B_1^T\hfill & \hfill \hfill & \hfill \hfill \\ \hfill -B_1\hfill & \hfill A_2\hfill & \hfill -B_2^T\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill -B_{s-1}^T\hfill \\ \hfill \hfill & \hfill \hfill & \hfill -B_{s-1}\hfill & \hfill A_s\hfill \end{array}\right)$$and orthogonal matrix $X=[X_1,X_2,\cdots ,X_s]$ such that(5.1) $$\begin{array}{c}\hfill \mathrm{\Lambda }X=XJ.\end{array}$$(5.1) Here $n=2s$ for some integer $s$, $A_i$ are symmetric matrices of order $2\times 2$ and $B_i$ are upper triangular matrix of order 2. The first $n\times 2$ block of Equation (Appendix 1.1(1.1) $$\begin{array}{c}\hfill \frac{d^2}{{\mathit{dx}}^2}(E(x)I(x)\frac{d^{2}u(x)}{{\mathit{dx}}^2})=A(x)\mathit{\rho }(x){\mathit{\omega }}^{2}u(x),0\le x\le L.\end{array}$$(1.1) ) gives(5.2) $$\begin{array}{c}\hfill \mathrm{\Lambda }{X}_1=X_1{A}_1-X_2{B}_1,\end{array}$$(5.2) since matrix $X$ is orthogonal, premultiplying (Appendix 1.2(1.2) $$\begin{array}{c}\hfill {\ell }_i=h_i,m_i={\mathit{\rho }}_i{A}_i{h}_i,k_i=\frac{E_i{I}_i}{h_i},i=0,1,\dots ,n.\end{array}$$(1.2) ) by $X_1^T$ we obtain(5.3) $$\begin{array}{c}\hfill A_1=X_1^T\mathrm{\Lambda }{X}_1.\end{array}$$(5.3) Equation (Appendix 1.2(1.2) $$\begin{array}{c}\hfill {\ell }_i=h_i,m_i={\mathit{\rho }}_i{A}_i{h}_i,k_i=\frac{E_i{I}_i}{h_i},i=0,1,\dots ,n.\end{array}$$(1.2) ) can be written as(5.4) $$\begin{array}{c}\hfill X_2{B}_1=X_1{A}_1-\mathrm{\Lambda }{X}_1=Z_2.\end{array}$$(5.4) Suppose $X_2=[y_1,y_2],Z_2=[z_1,z_2]$ and$$B_1=\left(\begin{array}{cc}\hfill b_{11}\hfill & \hfill b_{12}\hfill \\ \hfill 0\hfill & \hfill b_{22}\hfill \end{array}\right).$$The first column of Equation (Appendix 1.4(1.4) $$\begin{array}{c}\hfill \mathit{\tau }=\hat{K}{N}^T\mathit{\theta },\end{array}$$(1.4) ) gives$$y_1{b}_{11}=z_1\Rightarrow b_{11}=\pm \parallel z_1\parallel ,y_1=z_1/b_{11}$$and the second column of (Appendix 1.4(1.4) $$\begin{array}{c}\hfill \mathit{\tau }=\hat{K}{N}^T\mathit{\theta },\end{array}$$(1.4) ) gives$$b_{12}{y}_1+b_{22}{y}_2=z_2$$which leads to$$b_{12}=y_1^T{z}_2,b_{22}{y}_2=z_2-b_{12}{y}_1=w_2,$$so that$$b_{22}=\pm \parallel w_2\parallel ,y_2=w_2/b_{22}.$$In this process, a Gram–Schmidt process for finding orthonormal vectors $y_1,y_2$ and matrix $B_1$ is necessary. By repeating the above process we can find matrices $A_i,B_i$ and $\mathbf{J}$. This algorithm can be applied for the $n=2s-1,$ similarly. We have used this case in our numerical example.

Reconstruction of masses $m_i$, stiffnesses $k_i$ and lengths $l_i$ [3]

Suppose that we have computed$$H=D^{-1}\mathit{KD}$$where $D=\mathrm{diag}(d_1,d_2,\cdots ,d_n),m_i=d_i^2$ and $K$ is the pentadiagonal matrix in (1.8(1.8) $$\begin{array}{c}\hfill K={\mathit{NL}}^{-1}N\hat{K}{N}^T{L}^{-1}{N}^T.\end{array}$$(1.8) ). We now untangle $H$ to give masses $m_i$, stiffnesses $k_i$ and lengths $l_i$.

First, we apply external static forces $f_1,f_2$ to masses 1 and 2, and deform the system as shown in Figure 3. For this configuration the static equation is

Fig. 3 Two static forces are needed to deflect all the masses by the same amount.[3]

$$\mathit{Ku}=f=[f_1,-f_2,0,\cdots ,0],u=[1,\cdots ,1]$$i.e.$$\mathit{DHDu}=f$$or$$\mathit{Hd}=[d_1^{-1}{f}_1,-d_2^{-1}{f}_2,0,\cdots ,0],d=[d_1,d_2,\cdots ,d_n].$$Consider this equation. We know $H$ and the last two components $d_{n-1},d_n$. But $H$ is pentadiagonal so that, knowing $d_{n-1},d_n$, we can compute $d_{n-2},\cdots ,d_1$, and find $d_1^{-1}{f}_1,d_2^{-1}{f}_2$ and hence $f_1,f_2$.

Having found the masses $(m_i=d_i^2)$, we find the lengths. We apply a single force $k_1{l}_1^{-1}$ at $m_1$, and deform the system as shown in Figure 4. Now the equation

Fig. 4 One static force will deflect the beam as a straight line.[3]

$${\mathit{Ku}}^0=[k_1{l}_1^{-1},0,\cdots ,0],u^0=[l_1,l_1+l_2,\dots ,l_1+l_2+\dots +l_n],$$yields$$H[d_1{l}_1,d_2(l_1+l_2),\dots ,d_n(l_1+l_2+\dots +l_n)].$$This means that if we invert the equation$$\mathit{Hx}=[1,0,\dots ,0],$$we will find$$d_i(l_1+l_2+\cdots +l_i)={\mathit{cx}}_i,c=d_1^{-1}{k}_1{l}_1^{-1},$$this yields the $l_i$ as follows$$l_1=c\left(\frac{x_1}{d_1}\right),l_i=c(\frac{x_i}{d_i}-\frac{x_{i-1}}{d_{i-1}}),i\ge 2.$$The last step is to find the $k_i$. By Equations (1.8(1.8) $$\begin{array}{c}\hfill K={\mathit{NL}}^{-1}N\hat{K}{N}^T{L}^{-1}{N}^T.\end{array}$$(1.8) ) and (1.11(1.11) $$\begin{array}{c}\hfill H=D^{-1}{\mathit{KD}}^{-1},\end{array}$$(1.11) ) we find$$\hat{K}=N^{-1}{\mathit{LN}}^{-1}{\mathit{DHDN}}^{-T}{\mathit{LN}}^{-T},$$which gives $\hat{K}$ and the stiffnesses $k_i$. From (1.8(1.8) $$\begin{array}{c}\hfill K={\mathit{NL}}^{-1}N\hat{K}{N}^T{L}^{-1}{N}^T.\end{array}$$(1.8) ) and (1.11(1.11) $$\begin{array}{c}\hfill H=D^{-1}{\mathit{KD}}^{-1},\end{array}$$(1.11) ) we note that the matrices $K,H$ are independent of constant $c$.