Inverse Eigenvalue Problem For Mass-Spring-Inerter Systems ¹¹1The work was supported by the National Natural Science Foundation of China under Grants 11925109 and 11688101.

Zhaobo Liu²²2Z. B. Liu, Q. D. Xie and C. Li are with the Key Laboratory of Systems and Control, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, P. R. China. They are also with the School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, P. R. China. Corresponding author: Chanying Li (Email: cyli@amss.ac.cn). , Qida Xie²²footnotemark: 2 , Chanying Li ²²footnotemark: 2

Abstract

This paper has solved the inverse eigenvalue problem for “fixed-free” mass-chain systems with inerters. It is well known that for a spring-mass system wherein the adjacent masses are linked through a spring, the natural frequency assignment can be achieved by choosing appropriate masses and spring stiffnesses if and only if the given positive eigenvalues are distinct. However, when we involve inerters, multiple eigenvalues in the assignment are allowed. In fact, arbitrarily given a set of positive real numbers, we derive a necessary and sufficient condition on the multiplicities of these numbers, which are assigned as the natural frequencies of the concerned mass-spring-inerter system.

1 Introduction

Natural frequency, an inherent attribute of mechanical vibration systems, has attracted wide attention for its importance. In particular, purposefully allocating the natural frequencies to some pre-specified values provides an effective way to induce or evade resonance (see [2], [23]). This naturally raises the inverse eigenvalue problem (IEP), that is, to construct a vibration system whose natural frequencies, or mathematically known as eigenvalues, are given beforehand.

A well-known result on this problem is due to [8] and [17], which is addressed for mass-spring systems. Observe that in such a basic system, the adjacent masses are linked merely by a spring. Therefore, the IEP turns out to be the construction of a Jacobi matrix with its eigenvalues being assigned to a set of specified positive numbers. Borrowing the tools for Jacobi matrices, [8] and [17] assert that the IEP is solvable if and only if the given positive eigenvalues are distinct. Later on, a various of inverse problems on Jacobi matrices and Jacobi operators are investigated as well. But when dampers are taken into consideration, the matrices associated with masses, spring stiffnesses and damping coefficients in mass-spring-damper systems are no longer Jacobi matrices and the quadratic inverse eigenvalue problem (QIEP) is put forth. Nevertheless, most of the literatures on mass-spring-damper systems focus on distinct eigenvalues assignment [3, 13, 1, 16].

Interestingly, the IEP admits multiple eigenvalues, if we introduce a mechanical element called the inerter. This new mechanical device can simulate masses by changing inertance. It was theoretically first studied by [20], completing the analogy between electrical and mechanical networks (see Figure 1.2). Through physical realization, inerters have been applied to many engineering fields such as vibration isolators, landing gears, train suspensions, building vibration control, and so on [6, 22, 14, 21, 11]. In a mass-spring-inerter system, the neighbouring masses are linked by a parallel combination of a spring and an inerter. As a starting point, we restrict our interest in this paper to “fixed-free” systems. The term “fixed-free” means one end of the mass-chain system is attached to the ground while the other end is hanging free, as shown in Figure 1.2.

The free vibration equation of such a mass-spring-inerter system is described by

\boldsymbol{(M+B)\ddot{x}}+\boldsymbol{Kx}=0,

where $\boldsymbol{x}=(x_{1},x_{2},\ldots,x_{n})^{\mathrm{T}}\in\mathbb{R}^{n}$ and

\displaystyle\boldsymbol{M}=\mbox{diag}\{m_{1},m_{2},\ldots,m_{n}\},

(1)

\displaystyle\boldsymbol{K}

\displaystyle=

\displaystyle\begin{bmatrix}k_{1}+k_{2}&-k_{2}&&\\ -k_{2}&k_{2}+k_{3}&-k_{3}&\\ &\ddots&\ddots&\ddots&\\ &&-k_{n-1}&k_{n-1}+k_{n}&-k_{n}\\ &&&-k_{n}&k_{n}\end{bmatrix},

(2)

\displaystyle\boldsymbol{B}

\displaystyle=

\displaystyle\begin{bmatrix}b_{1}+b_{2}&-b_{2}&&\\ -b_{2}&b_{2}+b_{3}&-b_{3}&\\ &\ddots&\ddots&\ddots&\\ &&-b_{n-1}&b_{n-1}+b_{n}&-b_{n}\\ &&&-b_{n}&b_{n}\end{bmatrix}.

(3)

Here, real numbers $m_{j}>0$ , $k_{j}>0$ , $b_{j}\geq 0$ for $j=1,\ldots,n$ stand for the masses, spring stiffnesses and inertances. Unlike mass-spring systems, the well-studied Jacobi matrix theory cannot illuminate the IEP for mass-spring-inerter systems since the inertial matrix in (3) is a tridiagonal matrix. Recently, [10] found that inerters render the multiple eigenvalues possible for a mass-chain system. It showed that the multiplicity $t_{i}$ of a natural frequency $\lambda_{i}$ must fulfill $n\geq 2t_{i}-1$ . Beyond that, little is known for the multiple eigenvalue case.

The purpose of this paper is to solve the IEP for mass-spring-inerter systems, where the eigenvalues are arbitrarily specified to $n$ positive real numbers. We deduce a necessary and sufficient condition for this assignment on the multiplicities of the given numbers. With the proposed critical criterion, the set structure of the given real numbers will be intuitively clear for the natural frequency assignment. Our construction further implies that $m$ masses of the system can be arbitrarily fixed beforehand for the assignment, where $m$ is the amount of the distinct assigned eigenvalues. More precisely, our construction is carried out by only adjusting $n-m$ massess, $n$ spring stiffnesses and $n$ inertances. It degenerates to the claim that, if the pre-specified eigenvalues are all distinct ( $m=n$ ), the IEP can be worked out by recovering $\boldsymbol{K}$ and $\boldsymbol{B}$ , whereas $\boldsymbol{M}$ is fixed arbitrarily. This claim is exactly the main result of [20], which demonstrates an advantage of using inerters in the mass-fixed situation. Unfortunately, not all the natural frequency assignments are realizable by merely adjusting spring stiffnesses and inertances. An example of five-degree-of-freedom system in this paper shows that there exist some restrictive relationships between masses and given eigenvalues.

The organization of this paper is as follows. In Section 2, we state the main result by deducing a necessary and sufficient condition of the IEP for mass-spring-inerter systems, while the proofs are included in Sections 3 and 4. Conclusions are drawn in Section 5.

Refer to caption — Figure 1.1: Mass-spring-inerter system

2 Main Result

The natural frequencies of a mass-spring-inerter system are completely determined by the eigenvalues of matrix pencil $\boldsymbol{K}-\lambda(\boldsymbol{M}+\boldsymbol{B})$ , where $\boldsymbol{M},\boldsymbol{K},\boldsymbol{B}$ are defined by (1), (2) and (3), respectively. So, with a slight abuse of language, we will not distinguish the term “eigenvalues” from the “natural frequencies” in this article. We now raise our problem.

Problem 1. Arbitrarily given a set of real numbers $0<\lambda_{1}\leq\lambda_{2}\leq\cdots\leq\lambda_{n}$ , is it possible to recover matrices $\boldsymbol{M},\boldsymbol{K},\boldsymbol{B}$ in (1), (2) and (3) by choosing $m_{j}>0$ , $k_{j}>0$ and $b_{j}\geq 0$ for $j=1,\ldots,n$ , so that the $n$ eigenvalues of matrix pencil $\boldsymbol{K}-\lambda(\boldsymbol{M}+\boldsymbol{B})$ are exactly $\lambda_{i}$ , $i=1,\ldots,n$ ?

Both [10] and [17] offered a positive answer to Problem 1 for the special case where the eigenvalues are all distinct. But the general situation should involve multiple eigenvalues, which is covered by the following theorem.

Theorem 2.1.

Let $\prod_{i=1}^{m}(\lambda-\lambda_{i})^{t_{i}}$ be a polynomial with $0<\lambda_{1}<\lambda_{2}<\cdots<\lambda_{m}$ and $\sum_{i=1}^{m}{t_{i}}=n$ . Then, there exist some matrices $\boldsymbol{K},\boldsymbol{M},\boldsymbol{B}$ in the forms of (1)–(3) such that

\displaystyle\prod_{i=1}^{m}(\lambda-\lambda_{i})^{t_{i}}\Big{|}\det(\boldsymbol{K}-\lambda(\boldsymbol{M}+\boldsymbol{B}))

(4)

if and only if

\displaystyle t_{i}\leq i,\quad i=1,\ldots,m.

(5)

Remark 2.1.

Theorem 2.1 completely solves Problem 1 by providing the critical criterion (5). As indicated later (see Proposition 4.1 for details), when (5) holds, the recover of the relevant matrices allows a total of $m$ masses being taken arbitrarily, where $m$ is the number of distinct eigenvalues $\lambda_{i}$ given beforehand. Particularly, for $m=n$ , each mass $m_{i},1\leq i\leq n$ can be taken any fixed quantity in advance, as proved in [10]. However, when $m<n$ , it is generally impossible to achieve the natural frequency assignment with all the masses arbitrarily given. Example 2.1 suggests a restrictive relation between the masses and eigenvalues.

Example 2.1.

Let $n=5$ , $t_{1}=t_{2}=1$ , $t_{3}=3$ and $0<\lambda_{1}<\lambda_{2}<\lambda_{3}$ . If there exist some $m_{j}>0$ , $k_{j}>0$ , $b_{j}\geq 0$ , $j=1,2,\ldots,5$ such that $\prod_{i=1}^{3}(\lambda-\lambda_{i})^{t_{i}}\Big{|}\det(\boldsymbol{K}-\lambda(\boldsymbol{M}+\boldsymbol{B}))$ , then

\displaystyle\max_{j\in[2,4]}\frac{m_{j}}{m_{j+1}}>\frac{\lambda_{1}}{8\lambda_{3}}\left(1-\left(\frac{\lambda_{2}}{\lambda_{3}}\right)^{\frac{1}{3}}\right).

(6)

Clearly, (5) holds, but the masses cannot be taken arbitrarily. The proof of (6) is provided in Appendix A.

3 Proof of the necessity of Theorem 2.1.

This section is devoted to proving the necessity of Theorem 2.1, which is relatively easier than the argument for sufficiency. We begin by expressing $\det(\boldsymbol{K}-\lambda(\boldsymbol{M}+\boldsymbol{B}))$ in terms of a recursive sequence of polynomials. First, write

	$\displaystyle\boldsymbol{K}-\lambda(\boldsymbol{M}+\boldsymbol{B})=$
	$\displaystyle\begin{bmatrix}\begin{smallmatrix}k_{1}+k_{2}-\lambda(m_{1}+b_{1}+b_{2})&-k_{2}+\lambda b_{2}&\\ -k_{2}+\lambda b_{2}&k_{2}+k_{3}-\lambda(m_{2}+b_{2}+b_{3})&-k_{3}+\lambda b_{3}\\ &\ddots&\ddots&\ddots\\ &&-k_{n-1}+\lambda b_{n-1}&k_{n-1}+k_{n}-\lambda(m_{n-1}+b_{n-1}+b_{n})&-k_{n}+\lambda b_{n}\\ &&&-k_{n}+\lambda b_{n}&k_{n}-\lambda(m_{n}+b_{n})\\ \end{smallmatrix}\end{bmatrix}.$

For $j=1,\ldots,n$ , let $\boldsymbol{M_{j}}$ , $\boldsymbol{K_{j}}$ and $\boldsymbol{B_{j}}$ be some matrices defined analogously as $\boldsymbol{M},\boldsymbol{K}$ and $\boldsymbol{B}$ in (1)–(3), respectively, but with order $j$ instead of $n$ . Next, denote $f_{j}(\lambda)$ as the determinant of $\boldsymbol{K_{j}}-\lambda(\boldsymbol{M_{j}}+\boldsymbol{B_{j}})$ , $j=1,\ldots,n$ . Let $g_{1}(\lambda)=1$ and $g_{j}(\lambda)$ the leading principal minor of $\boldsymbol{K_{j}}-\lambda(\boldsymbol{M_{j}}+\boldsymbol{B_{j}})$ of order $j-1$ , where $j=2,\ldots,n$ . So, to calculate $\det(\boldsymbol{K}-\lambda(\boldsymbol{M}+\boldsymbol{B}))$ , we only need to treat $f_{n}(\lambda)$ .

Remark 3.1.

For each $j=1,\ldots,n$ , since $\det\boldsymbol{K_{j}}=\prod_{l=1}^{j}k_{l}>0$ , the Gershgorin’s circle theorem indicates that both $\boldsymbol{K_{j}}$ and $\boldsymbol{M_{j}}+\boldsymbol{B_{j}}$ are positive definite matrices and so do their leading principal submatrices. Then, it follows that the roots of $f_{j}(\lambda)$ and $g_{j}(\lambda)$ are all real and positive (see [8, Theorem 1.4.3] ).

To facilitate the subsequent analysis, we introduce the following definition.

Definition 3.1.

Let $f(\lambda)$ and $g(\lambda)$ be two polynomials with degree $s$ , where $s\in\mathbb{N}^{+}$ . Suppose $f(\lambda)$ and $g(\lambda)$ both have $s$ distinct real roots, which are denoted by $\alpha_{1}<\cdots<\alpha_{s}$ and $\beta_{1}<\cdots<\beta_{s}$ , respectively. We say $g(\lambda)\ll f(\lambda)$ , if their leading coefficients are of the same sign and

\beta_{1}<\alpha_{1}<\beta_{2}<\alpha_{2}<\cdots<\beta_{s}<\alpha_{s}.

The proof depends on a simple observation below.

Lemma 3.1.

The polynomials $\{f_{j}(\lambda)\}_{j=1}^{n}$ and $\{g_{j}(\lambda)\}_{j=1}^{n}$ satisfy

\displaystyle\left\{\begin{aligned} &f_{j+1}(\lambda)=(-\lambda m_{j+1})g_{j+1}(\lambda)+(k_{j+1}-\lambda b_{j+1})f_{j}(\lambda)\\ &g_{j+1}(\lambda)=f_{j}(\lambda)+(k_{j+1}-\lambda b_{j+1})g_{j}(\lambda)\end{aligned},\quad j=1,\ldots,n-1\right.

(7)

with $g_{1}(\lambda)=1$ and $f_{1}(\lambda)=k_{1}-\lambda(m_{1}+b_{1})$ .

Proof.

By the definition of $\boldsymbol{K_{1}}-\lambda(\boldsymbol{M_{1}}+\boldsymbol{B_{1}})$ , it is trivial that $f_{1}(\lambda)=k_{1}-\lambda(m_{1}+b_{1})$ . For $j=1,\ldots,n-1$ , expanding the leading principal minor of $\boldsymbol{K_{j+1}}-\lambda(\boldsymbol{M_{j+1}}+\boldsymbol{B_{j+1}})$ of order $j$ by cofactors of the $j$ th row shows $g_{j+1}(\lambda)=f_{j}(\lambda)+(k_{j+1}-\lambda b_{j+1})g_{j}(\lambda)$ . Furthermore, the expansion of $\det(\boldsymbol{K_{j+1}}-\lambda(\boldsymbol{M_{j+1}}+\boldsymbol{B_{j+1}}))$ by cofactors of the $(j+1)$ th row yields

$\displaystyle f_{j+1}(\lambda)$	$\displaystyle=$	$\displaystyle(k_{j+1}-\lambda(m_{j+1}+b_{j+1}))g_{j+1}(\lambda)-(k_{j+1}-\lambda b_{j+1})^{2}g_{j}(\lambda)$
	$\displaystyle=$	$\displaystyle(-\lambda m_{j+1})g_{j+1}(\lambda)+(k_{j+1}-\lambda b_{j+1})f_{j}(\lambda)$
		$\displaystyle+(k_{j+1}-\lambda b_{j+1})^{2}g_{j}(\lambda)-(k_{j+1}-\lambda b_{j+1})^{2}g_{j}(\lambda)$
	$\displaystyle=$	$\displaystyle(-\lambda m_{j+1})g_{j+1}(\lambda)+(k_{j+1}-\lambda b_{j+1})f_{j}(\lambda),$

as desired. $\Box$

Lemma 3.2.

Let $f(\lambda)$ and $g(\lambda)$ be two polynomials that $g(\lambda)\ll f(\lambda)$ , then for any $a,b>0$ , $g(\lambda)\ll ag(\lambda)+bf(\lambda)\ll f(\lambda)$ .

Proof.

Since $g(\lambda)\ll f(\lambda)$ , considering Definition 3.1, we let $\deg(f(\lambda))=\deg(g(\lambda))=s$ for some $s\in\mathbb{N}^{+}$ and let $\{\alpha_{i}\}_{i=1}^{s}$ and $\{\beta_{i}\}_{i=1}^{s}$ be the roots of $f(\lambda)$ and $g(\lambda)$ , respectively. Clearly,

\beta_{1}<\alpha_{1}<\beta_{2}<\alpha_{2}<\cdots<\beta_{s}<\alpha_{s}.

Without loss of generality, assume the leading coefficients of $f(\lambda)$ and $g(\lambda)$ are both positive. Then,

\mbox{sgn}\,((ag(\lambda)+bf(\lambda))(\beta_{i}))=(-1)^{s+1-i}=-(-1)^{s-i}=-\mbox{sgn}\,((ag(\lambda)+bf(\lambda))(\alpha_{i})).

This implies that for each $1\leq i\leq s$ , there is a root of $ag(\lambda)+bf(\lambda)$ falling in interval $(\beta_{i},\alpha_{i})$ . Observing that the degree of $ag(\lambda)+bf(\lambda)$ is $s$ , the result follows immediately. $\Box$

We present an important property enjoyed by sequence $\{f_{j}(\lambda),g_{j}(\lambda)\}_{j=1}^{n}$ .

Lemma 3.3.

Suppose for some $j\in[1,n-1]$ ,

\displaystyle\frac{(-\lambda)g_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}\ll\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))},

(8)

then $\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}.$ Moreover,
(i) if $b_{j+1}\neq 0$ and $k_{j+1}-\lambda b_{j+1}|\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ , then $(f_{j+1}(\lambda),g_{j+1}(\lambda))=(f_{j}(\lambda),g_{j}(\lambda))(\lambda-\frac{k_{j+1}}{b_{j+1}})$ and $\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}\ll\frac{-f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ ;
(ii) if $b_{j+1}\neq 0$ and $k_{j+1}-\lambda b_{j+1}\nmid\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ , then $(f_{j+1}(\lambda),g_{j+1}(\lambda))=(f_{j}(\lambda),g_{j}(\lambda))$ and $\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j}(\lambda)(k_{j+1}-\lambda b_{j+1})}{(f_{j}(\lambda),g_{j}(\lambda))}$ ;
(iii) if $b_{j+1}=0$ , then $(f_{j+1}(\lambda),g_{j+1}(\lambda))=(f_{j}(\lambda),g_{j}(\lambda))$ and $\frac{(-\lambda)f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}\ll\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}$ .

Proof.

First, according to the definitions of $\{f_{j}(\lambda),g_{j}(\lambda)\}_{j=1}^{n}$ , it is apparent that $\deg(f_{j}(\lambda))=j$ and $\deg(g_{j}(\lambda))=j-1$ for each $j\in[1,n]$ . Let $\deg((f_{j}(\lambda),g_{j}(\lambda)))=j-s_{j}$ for some integer $s_{j}\in[1,j]$ . Recalling (8), denote the roots of $\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ and $\frac{(-\lambda)g_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ by $\alpha_{j,1}<\cdots<\alpha_{j,s_{j}}$ and $0<\beta_{j,1}<\cdots<\beta_{j,s_{j}-1}$ , respectively. These roots fulfill

\displaystyle 0<\alpha_{j,1}<\beta_{j,1}<\cdots<\alpha_{j,s_{j}-1}<\beta_{j,s_{j}-1}<\alpha_{j,s_{j}}.

(9)

In addition, the second equation of (7) implies $(f_{j}(\lambda),g_{j}(\lambda))|(f_{j}(\lambda),g_{j+1}(\lambda))$ . Now, we prove this lemma by discussing three cases.

(i) $b_{j+1}\neq 0$ and $k_{j+1}-\lambda b_{j+1}|\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ . For this case, there exists some $1\leq t_{j}\leq s_{j}$ such that $\alpha_{j,t_{j}}=\frac{k_{j+1}}{b_{j+1}}$ . We shall evaluate the sign of $\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))(k_{j+1}-\lambda b_{j+1})}$ at $\alpha_{j,i},1\leq i\leq s_{j}$ . In fact, according to the second equation of (7),

			$\displaystyle\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))(k_{j+1}-\lambda b_{j+1})}(\alpha_{j,i})$
		$\displaystyle=$	$\displaystyle\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))(k_{j+1}-\lambda b_{j+1})}(\alpha_{j,i})+\frac{g_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}(\alpha_{j,i}),\qquad 1\leq i\leq s_{j}.$

Note that the leading coefficient of $(f_{j}(\lambda),g_{j}(\lambda))$ is positive, the definitions of $f_{j}(\lambda)$ and $g_{j}(\lambda)$ read

\displaystyle\left\{\begin{aligned} &f_{j}(\lambda)=(-1)^{j-s_{j}}(f_{j}(\lambda),g_{j}(\lambda))\prod_{l=1}^{s_{j}}(\alpha_{j,l}-\lambda)\\ &g_{j}(\lambda)=(-1)^{j-s_{j}}(f_{j}(\lambda),g_{j}(\lambda))\prod_{l=1}^{s_{j}-1}(\beta_{j,l}-\lambda)\end{aligned},\right.

(10)

which, together with (9), yields that for $1\leq i\leq s_{j}$ ,

	$\displaystyle\mbox{sgn}\,\left(\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))(k_{j+1}-\lambda b_{j+1})}(\alpha_{j,i})\right)$	$\displaystyle=$	$\displaystyle(-1)^{j-s_{j}}\mbox{sgn}\,\left(\prod_{l=1,l\neq t_{j}}^{s_{j}}(\alpha_{j,l}-\alpha_{j,i})\right)$
		$\displaystyle=$	$\displaystyle\left\{\begin{array}[]{lcl}0,&i\neq t_{j}\\ (-1)^{j-s_{j}+t_{j}-1},&i=t_{j}\end{array}\right.$

and

\displaystyle\mbox{sgn}\,\left(\frac{g_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}(\alpha_{j,i})\right)=(-1)^{j-s_{j}}\mbox{sgn}\,\left(\prod_{l=1}^{s_{j-1}}(\beta_{j,l}-\alpha_{j,i})\right)=(-1)^{j-s_{j}+i-1}.

Therefore,

\mbox{sgn}\,\left(\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))(k_{j+1}-\lambda b_{j+1})}(\alpha_{j,i})\right)=(-1)^{j-s_{j}+i-1}\quad\mbox{for}\quad 1\leq i\leq s_{j}.

This means that for each $i\in[1,s_{j}-1]$ , there exists exactly one root of $\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))(k_{j+1}-\lambda b_{j+1})}$ between $\alpha_{j,i}$ and $\alpha_{j,i+1}$ , and hence $\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))(k_{j+1}-\lambda b_{j+1})}\ll\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ .

Moreover, $(f_{j}(\lambda),g_{j}(\lambda))|(f_{j}(\lambda),g_{j+1}(\lambda))$ , so

(f_{j}(\lambda),g_{j+1}(\lambda))=(f_{j}(\lambda),g_{j}(\lambda))\left(\lambda-\frac{k_{j+1}}{b_{j+1}}\right)

and

\displaystyle\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}=\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))\left(\lambda-\frac{k_{j+1}}{b_{j+1}}\right)}\ll\frac{-f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}=\frac{\left(\frac{k_{j+1}}{b_{j+1}}-\lambda\right)f_{j}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}.

Applying Lemma 3.2 to the first equation of (7), we thus deduce

\displaystyle\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}\ll\frac{\left(\frac{k_{j+1}}{b_{j+1}}-\lambda\right)f_{j}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}=\frac{-f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}.

(12)

Now, $(f_{j+1}(\lambda),g_{j+1}(\lambda))=(f_{j}(\lambda),g_{j+1}(\lambda))$ becasue of $\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}$ , it follows from (12) that

\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}\ll\frac{-f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}.

(ii) $b_{j+1}\neq 0$ and $(k_{j+1}-\lambda b_{j+1})\nmid\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ . We first assume $\frac{k_{j+1}}{b_{j+1}}\in(\alpha_{j,t_{j}},\beta_{j,t_{j}})$ for some $1\leq t_{j}\leq s_{j}-1$ and evaluate the sign of $\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ at points $\frac{k_{j+1}}{b_{j+1}}$ and $\alpha_{j,i},1\leq i\leq s_{j}$ . Since Lemma 3.1 shows

\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}(\alpha_{j,i})=(k_{j+1}-\alpha_{j,i}b_{j+1})\frac{g_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}(\alpha_{j,i}),\quad 1\leq i\leq s_{j},

by (9) and (10),

	$\displaystyle\mbox{sgn}\,\left(\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}(\alpha_{j,i})\right)$	$\displaystyle=$	$\displaystyle(-1)^{j-s_{j}}\mbox{sgn}\,\left((k_{j+1}-\alpha_{j,i}b_{j+1})\prod_{l=1}^{s_{j-1}}(\beta_{j,l}-\alpha_{j,i})\right)$
		$\displaystyle=$	$\displaystyle\left\{\begin{array}[]{lcl}(-1)^{j-s_{j}+i-1},&1\leq i\leq t_{j}\\ (-1)^{j-s_{j}+i},&t_{j}+1\leq i\leq s_{j}\end{array}.\right.$

Similarly, by Lemma 3.1, (9) and (10),

			$\displaystyle\mbox{sgn}\,\left(\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}\left(\frac{k_{j+1}}{b_{j+1}}\right)\right)$
		$\displaystyle=$	$\displaystyle\mbox{sgn}\,\left(\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}\left(\frac{k_{j+1}}{b_{j+1}}\right)+\frac{(k_{j+1}-\lambda b_{j+1})g_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}\left(\frac{k_{j+1}}{b_{j+1}}\right)\right)=(-1)^{j-s_{j}+t_{j}}.$

So, for each $i\in[1,s_{j}-1]$ with $i\neq t_{j}$ , there is a root of $\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ falling in $(\alpha_{j,i},\alpha_{j,i+1})$ , and the rest two roots of $\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ lie in $(\alpha_{j,t_{j}},\frac{k_{j+1}}{b_{j+1}})$ and $(\frac{k_{j+1}}{b_{j+1}},\alpha_{j,t_{j}+1})$ , respectively. This infers $(f_{j}(\lambda),g_{j+1}(\lambda))=(f_{j}(\lambda),g_{j}(\lambda))$ by noting that $(f_{j}(\lambda),g_{j}(\lambda))|(f_{j}(\lambda),g_{j+1}(\lambda))$ . Hence, $\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j}(\lambda)(k_{j+1}-\lambda b_{j+1})}{(f_{j}(\lambda),g_{j+1}(\lambda))}$ and applying Lemma 3.2 to the first equation of (7), one has

\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j}(\lambda)(k_{j+1}-\lambda b_{j+1})}{(f_{j}(\lambda),g_{j+1}(\lambda))}.

Then, $(f_{j+1}(\lambda),g_{j+1}(\lambda))=(f_{j}(\lambda),g_{j+1}(\lambda))=(f_{j}(\lambda),g_{j}(\lambda))$ and consequently

	$\displaystyle\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}$	$\displaystyle=$	$\displaystyle\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}$
		$\displaystyle\ll$	$\displaystyle\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j}(\lambda)(k_{j+1}-\lambda b_{j+1})}{(f_{j}(\lambda),g_{j+1}(\lambda))}=\frac{f_{j}(\lambda)(k_{j+1}-\lambda b_{j+1})}{(f_{j}(\lambda),g_{j}(\lambda))}.$

As for the situations where $\frac{k_{j+1}}{b_{j+1}}\in(0,\alpha_{j,1})$ , $\frac{k_{j+1}}{b_{j+1}}\in\bigcup_{l=1}^{s_{j}-1}[\beta_{j,l},\alpha_{j,l+1})$ and $\frac{k_{j+1}}{b_{j+1}}\in(\alpha_{j,s_{j}},+\infty)$ , an analogous treatment can be employed.

(iii) $b_{j+1}=0$ . We also first calculate the sign of $\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ at the roots of $\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ and $\frac{g_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ . As before,

\displaystyle\left\{\begin{array}[]{ll}\mbox{sgn}\,\left(\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}(\alpha_{j,i})\right)=(-1)^{j-s_{j}+i-1},&1\leq i\leq s_{j}\\ \mbox{sgn}\,\left(\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}(\beta_{j,i})\right)=(-1)^{j-s_{j}+i},&1\leq i\leq s_{j}-1\end{array}.\right.

(16)

Now, $\deg(g_{j+1}(\lambda))=j$ and the leading coefficient of $(f_{j}(\lambda),g_{j}(\lambda))$ is positive, if number $\theta_{j,1}>\alpha_{j,s_{j}}$ is sufficiently large, it is evident that $(f_{j}(\theta_{j,1}),g_{j+1}(\theta_{j,1}))>0$ and

\displaystyle\mbox{sgn}\,\left(\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}(\theta_{j,1})\right)=(-1)^{\deg(g_{j+1}(\lambda))}=(-1)^{j}.

So, each interval in $\{(\alpha_{j,s_{j}},+\infty),(\alpha_{j,i},\beta_{j,i}),i=1,\cdots,s_{j}-1\}$ contains exactly one root of $\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ , which concludes

(f_{j}(\lambda),g_{j+1}(\lambda))=(f_{j}(\lambda),g_{j}(\lambda))\quad\mbox{and}\quad\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}\ll\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}.

Let $\gamma_{j,1}<\cdots<\gamma_{j,s_{j}}$ be the roots of $\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}$ . From the first equation of (7) and (16), it follows that

\displaystyle\left\{\begin{array}[]{ll}\mbox{sgn}\,\left(\frac{f_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}(\alpha_{j,i})\right)=\mbox{sgn}\,\left(\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}(\alpha_{j,i})\right)=(-1)^{j-s_{j}+i},&1\leq i\leq s_{j}\\ \mbox{sgn}\,\left(\frac{f_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}(\gamma_{j,i})\right)=\mbox{sgn}\,\left(\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}(\gamma_{j,i})\right)=(-1)^{j-s_{j}+i},&1\leq i\leq s_{j}\\ \mbox{sgn}\,\left(\frac{f_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}(0)\right)=\mbox{sgn}\,\left(\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}(0)\right)=(-1)^{j-s_{j}}\end{array}.\right.

(20)

Because $\deg(f_{j+1}(\lambda))=j+1$ and the leading coefficient of $(f_{j}(\lambda),g_{j}(\lambda))$ is positive, we can choose a sufficiently large $\theta_{j,2}>\gamma_{s_{j}}>\alpha_{s_{j}}$ such that $(f_{j}(\theta_{j,2}),g_{j+1}(\theta_{j,2}))>0$ and

\displaystyle\mbox{sgn}\,\left(\frac{f_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}(\theta_{j,2})\right)=(-1)^{j+1}.

Recall that $\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}\ll\frac{g_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}$ , so each interval in

\{(0,\alpha_{j,1}),(\gamma_{j,s_{j}},+\infty),(\gamma_{j,i},\alpha_{j,i+1}),i=1,\ldots,s_{j}-1\}

contains exactly one root of $\frac{f_{j+1}(\lambda)}{(f_{j}(\lambda),g_{j+1}(\lambda))}$ . Because of this property,

(f_{j+1}(\lambda),g_{j+1}(\lambda))=(f_{j}(\lambda),g_{j+1}(\lambda))=(f_{j}(\lambda),g_{j}(\lambda)),

and hence $\frac{(-\lambda)f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}\ll\frac{(-\lambda)g_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}$ . $\Box$

It is ready to prove the necessity of Theorem 2.1. To this end, we introduce some notations. Let $f(\lambda)$ be a polynomial whose roots $\{z_{i}\}_{i=1}^{p}$ are all real and $z_{i}<z_{j}$ for every $i<j$ . Denote $\xi(f(\lambda),z_{i})$ as the multiplicity of root $z_{i}$ and for a real number $\alpha$ , define $\zeta(f(\lambda),\alpha)\triangleq\max\{i\in[1,p]:z_{i}<\alpha\}$ .

The proof of the necessity of Theorem 2.1: First, in view of (4), we know that $\lambda_{i},1\leq i\leq m$ are the $m$ distinct roots of $f_{n}(\lambda)$ with multiplicities $t_{i}$ . To proceed the argument, note that Lemma 3.1 gives $\frac{(-\lambda)g_{1}(\lambda)}{(f_{1}(\lambda),g_{1}(\lambda))}\ll\frac{f_{1}(\lambda)}{(f_{1}(\lambda),g_{1}(\lambda))}$ , then by using Lemma 3.3,

\displaystyle\frac{(-\lambda)g_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}\ll\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))},\quad j=1,\ldots,n.

(21)

Particularly, it turns out that all the roots of $\frac{f_{n}(\lambda)}{(f_{n}(\lambda),g_{n}(\lambda))}$ are distinct. So $\xi\left(\frac{f_{n}(\lambda)}{(f_{n}(\lambda),g_{n}(\lambda))},\lambda_{i}\right)\leq 1$ for all $1\leq i\leq m$ and then

\displaystyle\xi\left((f_{n}(\lambda),g_{n}(\lambda)),\lambda_{i}\right)=\xi\left(f_{n}(\lambda),\lambda_{i}\right)-\xi\left(\frac{f_{n}(\lambda)}{(f_{n}(\lambda),g_{n}(\lambda))},\lambda_{i}\right)\geq t_{i}-1.

(22)

Now, by (21) and Lemma 3.3, for each $j=1,\ldots,n-1$ ,

\displaystyle(f_{j+1}(\lambda),g_{j+1}(\lambda))=\left\{\begin{array}[]{ll}(f_{j}(\lambda),g_{j}(\lambda))(\lambda-\frac{k_{j+1}}{b_{j+1}}),&\mbox{if}~{}b_{j+1}\neq 0,(\lambda-\frac{k_{j+1}}{b_{j+1}})|\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}\\ (f_{j}(\lambda),g_{j}(\lambda)),&\mbox{otherwise}\end{array},\right.

which yields that for all $1\leq i\leq m$ ,

			$\displaystyle\xi\left((f_{j+1}(\lambda),g_{j+1}(\lambda)),\lambda_{i}\right)$		(26)
		$\displaystyle=$	$\displaystyle\left\{\begin{array}[]{ll}\xi\left((f_{j}(\lambda),g_{j}(\lambda)),\lambda_{i}\right)+1,&\mbox{if}~{}b_{j+1}\neq 0,k_{j+1}=\lambda_{i}b_{j+1},(\lambda-\lambda_{i})\|\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}\\ \xi\left((f_{j}(\lambda),g_{j}(\lambda)),\lambda_{i}\right),&\mbox{otherwise}\end{array}.\right.$		(26)

On the other hand, given $j\in[1,n-1]$ and $i\in[1,m]$ , if $b_{j+1}\neq 0,k_{j+1}=\lambda_{i}b_{j+1}$ and $(\lambda-\lambda_{i})|\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ , Lemma 3.3 (i) indicates $\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}\ll\frac{-f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ . Moreover, $\lambda_{i}$ is a root of $\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ , it follows that $\zeta\left(\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))},\lambda_{i}\right)=\zeta\left(\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))},\lambda_{i}\right)+1$ . Otherwise, by virtue of Lemma 3.3, at least one of the following cases will happen:
(i) if $b_{j+1}\neq 0,k_{j+1}=\lambda_{i}b_{j+1}$ and $(k_{j+1}-\lambda b_{j+1})|\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ , then $\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}\ll\frac{-f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))};$
(ii) if $b_{j+1}\neq 0$ and $k_{j+1}-\lambda b_{j+1}\nmid\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}$ , then $\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}\ll\frac{f_{j}(\lambda)(k_{j+1}-\lambda b_{j+1})}{(f_{j}(\lambda),g_{j}(\lambda))}$ ;
(iii) if $b_{j+1}=0$ , then $\frac{(-\lambda)f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))}\ll\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))}$ .
All the above three cases lead to $\zeta\left(\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))},\lambda_{i}\right)\geq\zeta\left(\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))},\lambda_{i}\right)$ . So,

\displaystyle\left\{\begin{array}[]{ll}\zeta\left(\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))},\lambda_{i}\right)=\zeta\left(\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))},\lambda_{i}\right)+1,&\mbox{if case (i) occurs}\\ \zeta\left(\frac{f_{j+1}(\lambda)}{(f_{j+1}(\lambda),g_{j+1}(\lambda))},\lambda_{i}\right)\geq\zeta\left(\frac{f_{j}(\lambda)}{(f_{j}(\lambda),g_{j}(\lambda))},\lambda_{i}\right),&\mbox{otherwise}\end{array}.\right.

(29)

Clearly, $\zeta\left(\frac{f_{1}(\lambda)}{(f_{1}(\lambda),g_{1}(\lambda))},\lambda_{i}\right)\geq\xi\left((f_{1}(\lambda),g_{1}(\lambda)),\lambda_{i}\right)=0$ . By (26) and (29), it can be derived inductively that

\zeta\left(\frac{f_{n}(\lambda)}{(f_{n}(\lambda),g_{n}(\lambda))},\lambda_{i}\right)\geq\xi\left((f_{n}(\lambda),g_{n}(\lambda)),\lambda_{i}\right).

Together with (22), the above inequality shows that for each $1\leq i\leq m$ ,

\displaystyle i-1=\zeta\left(f_{n}(\lambda),\lambda_{i}\right)\geq\zeta\left(\frac{f_{n}(\lambda)}{(f_{n}(\lambda),g_{n}(\lambda))},\lambda_{i}\right)\geq\xi\left((f_{n}(\lambda),g_{n}(\lambda)),\lambda_{i}\right)\geq t_{i}-1,

which completes the proof.

4 Proof of sufficiency of Theorem 2.1.

The sufficiency of Theorem 2.1 is a direct consequence of the following proposition.

Proposition 4.1.

Given $m$ real numbers $M_{1},\ldots,M_{m}>0$ and a polynomial $\prod_{i=1}^{m}(\lambda-\lambda_{i})^{t_{i}}$ with $0<\lambda_{1}<\lambda_{2}<\cdots<\lambda_{m}$ and $\sum_{i=1}^{m}{t_{i}}=n$ , if $t_{i}\leq i$ for each $i=1,\ldots,m$ , then there exist some $m_{j}>0$ , $k_{j}>0$ , $b_{j}\geq 0$ , $j=1,\ldots,n$ and $m$ distinct indices $i_{l},l=1,\ldots,m$ such that $m_{i_{l}}=M_{l}$ and

\displaystyle\prod_{i=1}^{m}(\lambda-\lambda_{i})^{t_{i}}\Big{|}\det(\boldsymbol{K}-\lambda(\boldsymbol{M}+\boldsymbol{B})).

We now begin the construction of the required mass-chain system for Proposition 4.1. That is, to find a sequence of $\{(k_{i},b_{i},m_{i})\}_{i=1}^{n}$ and $m$ indices $i_{l}$ such that $\prod_{i=1}^{m}(\lambda-\lambda_{i})^{t_{i}}|f_{n}(\lambda)$ and $m_{i_{l}}=M_{l}$ , $l=1,\ldots,m$ , where $f_{n}(\lambda)=\det(\boldsymbol{K}-\lambda(\boldsymbol{M}+\boldsymbol{B}))$ . Taking account to [10, Theorem 4], we take $T=\max_{1\leqslant i\leqslant m}t_{i}>1$ . Moreover, denote $q_{j}\triangleq|S_{j}|$ , where

\displaystyle S_{j}\triangleq\{\lambda_{i}:t_{i}\geq j+1\},\quad j=1,\ldots,T-1.

(30)

Evidently, $\sum_{j=1}^{T-1}q_{j}=n-m$ . We reorder the elements of $S_{j}$ by $s_{j}(1)<\cdots<s_{j}(q_{j})$ .

An important observation of Section 3 is that (22) implies every element in $\bigcup_{j=1}^{T-1}S_{j}$ is equal to some $k_{i}/b_{i},i\in[2,n]$ . This could help us to design a rule to determine which $k_{i}/b_{i}\in\bigcup_{j=1}^{T-1}S_{j}$ . It is the key idea of our proof, so we offer an example to elaborate it.

Example 4.1.

Take $n=15,m=8,t_{1}=t_{5}=t_{7}=t_{8}=1,t_{2}=t_{4}=2,t_{3}=3,t_{6}=4$ and $0<\lambda_{1}<\lambda_{2}<\cdots<\lambda_{8}$ in Proposition 4.1. Then, we get sets $S_{j},j=1,2,3$ as shown in Figure 4.4. Note that by (2)–(3), for all $i\in[2,15]$ , $-k_{i}$ and $-b_{i}$ are located in the secondary diagonals of matrices K and B, respectively. We now introduce a $15\times 15$ matrix $\textbf{A}=(a_{ij})$ and assign the elements $a_{i,i+1}$ in the secondary diagonal the values taken from sets $S_{j},j=1,2,3$ (see Figure 4.4). Specifically, for $j=1$ , we treat the first $1+q_{1}=5$ elements of $a_{i,i+1}$ by skipping $a_{5,6}$ and letting $a_{i,i+1}=s_{1}(4-i+1)$ for $i=1,2,3,4$ . So, $a_{1,2}=\lambda_{6},a_{2,3}=\lambda_{4},a_{3,4}=\lambda_{3},a_{4,5}=\lambda_{2}$ , as illustrated in Fig. 3. Repeat this procedure for elements $a_{i,i+1},i=\sum_{l=1}^{j-1}q_{l}+2,\ldots,\sum_{l=1}^{j}q_{l}+2$ with $j=2,3$ . For each $i$ with $a_{i,i+1}\in\bigcup_{j=1}^{3}S_{j}$ , we assign $k_{i+1}/b_{i+1}$ a value equivalent to $a_{i,i+1}$ . As a result, $k_{2}/b_{2}=k_{7}/b_{7}=k_{10}/b_{10}=\lambda_{6}$ , $k_{3}/b_{3}=\lambda_{4}$ , $k_{4}/b_{4}=k_{8}/b_{8}=\lambda_{3}$ , $k_{5}/b_{5}=\lambda_{2}$ .

In general, the rule to determine $k_{i}/b_{i}\in\bigcup_{j=1}^{T-1}S_{j}$ is summarized as follows:

\displaystyle\frac{k_{i}}{b_{i}}=s_{j+1}(q_{j+1}-l+1),\qquad i=1+j+l,\quad j=0,\ldots,T-2,\quad l=1,\ldots,q_{j+1}.

(31)

The proof of Proposition 4.1 thus will be completed in three steps.
Step 1: For each $i=1+j+l$ with $j=0,\ldots,T-2$ and $l=1,\ldots,q_{j+1}$ , we assign $k_{i}/b_{i}$ a value taken from $\bigcup_{j=1}^{T-1}S_{j}$ in the light of (31). So, the cardinal of set $\{i\in[2,n]:\frac{k_{i}}{b_{i}}\not\in\bigcup_{j=1}^{T-1}S_{j}\}$ is $(n-1)-(n-m)=m-1$ .
Step 2: Rewrite the elements of $\{1\}\cup\{i\in[2,n]:\frac{k_{i}}{b_{i}}\not\in\bigcup_{j=1}^{T-1}S_{j}\}$ by $1=i_{1}<\cdots<i_{m}$ . Let $m_{i_{l}}=M_{l}$ for $l=1,\ldots,m$ .
Step 3: Based on the above steps, we compute $(f_{i}(\lambda),g_{i}(\lambda))$ for each $i=1,\ldots,n$ . Then, take some suitable parameters $(m_{i},b_{i},k_{i})$ , so that if $\frac{k_{i+1}}{b_{i+1}}\not\in\bigcup_{j=1}^{T-1}S_{j}$ ,

\displaystyle\left\{\begin{array}[]{l}\frac{f_{i+1}(\lambda)}{(f_{i+1}(\lambda),g_{i+1}(\lambda))}=-\lambda m_{i+1}\frac{g_{i+1}(\lambda)}{(f_{i+1}(\lambda),g_{i+1}(\lambda))}+(k_{i+1}-\lambda b_{i+1})\frac{f_{i}(\lambda)}{(f_{i}(\lambda),g_{i}(\lambda))}\\ \frac{g_{i+1}(\lambda)}{(f_{i+1}(\lambda),g_{i+1}(\lambda))}=\frac{f_{i}(\lambda)}{(f_{i}(\lambda),g_{i}(\lambda))}+(k_{i+1}-\lambda b_{i+1})\frac{g_{i}(\lambda)}{(f_{i}(\lambda),g_{i}(\lambda))}\\ \end{array},\right.

(34)

otherwise, for $\frac{k_{i+1}}{b_{i+1}}\in\bigcup_{j=1}^{T-1}S_{j}$ ,

\displaystyle\left\{\begin{array}[]{l}\frac{f_{i+1}(\lambda)}{(f_{i+1}(\lambda),g_{i+1}(\lambda))}=-\lambda m_{i+1}\frac{g_{i+1}(\lambda)}{(f_{i+1}(\lambda),g_{i+1}(\lambda))}-b_{i+1}\frac{f_{i}(\lambda)}{(f_{i}(\lambda),g_{i}(\lambda))}\\ (\lambda-\frac{k_{i+1}}{b_{i+1}})\frac{g_{i+1}(\lambda)}{(f_{i+1}(\lambda),g_{i+1}(\lambda))}=\frac{f_{i}(\lambda)}{(f_{i}(\lambda),g_{i}(\lambda))}+(k_{i+1}-\lambda b_{i+1})\frac{g_{i}(\lambda)}{(f_{i}(\lambda),g_{i}(\lambda))}\end{array}.\right.

(37)

The construction of $\{\frac{f_{i}(\lambda)}{(f_{i}(\lambda),g_{i}(\lambda))},\frac{g_{i}(\lambda)}{(f_{i}(\lambda),g_{i}(\lambda))}\}_{i=1}^{n}$ will be achieved by an induction method from $n$ to $1$ .

To proceed the proof, we first derive some technical lemmas.

Lemma 4.1.

Let $F(\lambda)=\mu\prod_{i=1}^{p}(\lambda-\alpha_{i})$ and $G(\lambda)=\nu\prod_{i=1}^{p}(\lambda-\beta_{i})$ be two polynomials of degree $p$ that $G(\lambda)\ll F(\lambda)$ , where $|\mu|>|\nu|$ , $\alpha_{1}<\cdots<\alpha_{p}$ and $\beta_{1}<\cdots<\beta_{p}$ . Then, $F(\lambda)-G(\lambda)$ has $p$ real roots $\gamma_{1}<\cdots<\gamma_{p}$ satisfying $\gamma_{i}\in(\alpha_{i},\beta_{i+1})$ for $i=1,\ldots,p-1$ and $\gamma_{p}>\alpha_{p}$ . Moreover, the following two statements hold:
(i) when $p>1$ , for any $\eta\in(0,\min_{1\leqslant i\leqslant p-1}(\alpha_{i+1}-\alpha_{i}))$ , if

\displaystyle\displaystyle\frac{|\nu|}{|\mu|}<\ \frac{\min\{1,\left(\min_{1\leqslant i\leqslant p-1}(\alpha_{i+1}-\alpha_{i})-\eta\right)^{p}\}\min\{1,\eta^{p}\}}{2+2\max_{j\in[1,p]}\left|\prod_{i=1}^{p}(\alpha_{j}-\beta_{i})\right|}

(38)

then

\displaystyle\max_{1\leqslant i\leqslant p}(\gamma_{i}-\alpha_{i})<\eta;

(39)

(ii) when $p=1$ , for any $\eta\in(0,\frac{1}{2})$ , (39) holds provided that

\displaystyle|\nu|<\frac{|\mu|}{2}\min\left\{\frac{\eta}{\alpha_{1}-\beta_{1}},1\right\}.

(40)

Proof.

Note that $G(\lambda)\ll F(\lambda)$ indicates $\mu\nu>0$ and for each $j=1,\ldots,p-1$ ,

\displaystyle(F(\alpha_{j})-G(\alpha_{j}))(F(\beta_{j+1})-G(\beta_{j+1}))=-\mu\nu\prod_{i=1}^{p}(\alpha_{j}-\beta_{i})(\beta_{j+1}-\alpha_{i})<0,

which means $F(\lambda)-G(\lambda)$ has a root $\gamma_{j}$ in $(\alpha_{j},\beta_{j+1})$ . Further, since $\mu\nu>0$ and $|\mu|>|\nu|$ , $\mu(F(\lambda)-G(\lambda))>0$ holds for all sufficiently large $\lambda>\alpha_{p}$ . On the other hand,

\mu(F(\alpha_{p})-G(\alpha_{p}))=-\mu\nu\prod_{i=1}^{p}(\alpha_{p}-\beta_{i})<0,

so $F(\lambda)-G(\lambda)$ has a root $\gamma_{p}$ in $(\alpha_{p},\infty)$ . Clearly, $\gamma_{1}<\cdots<\gamma_{p}$ .

Next, we show statement (i). Observe that $F(\lambda)-G(\lambda)=(\mu-\nu)\prod_{i=1}^{p}(\lambda-\gamma_{i})$ , then for each $j\in[1,p]$ ,

\displaystyle(\mu-\nu)\prod_{i=1}^{p}(\alpha_{j}-\gamma_{i})=F(\alpha_{j})-G(\alpha_{j})=-\nu\prod_{i=1}^{p}(\alpha_{j}-\beta_{i}).

(41)

Let $p>1$ . If (39) fails, denote $l$ as the smallest subscript $i\in[1,p]$ such that $\gamma_{i}-\alpha_{i}\geq\eta$ . Hence,

0<\alpha_{i+1}-\alpha_{i}-\eta\leq\alpha_{i+1}-\gamma_{i}\leq\alpha_{l}-\gamma_{i},\quad i=1,\ldots,l-1.

Moreover, it is clear that $\gamma_{i}-\alpha_{l}\geq\gamma_{l}-\alpha_{l}>0$ for all $i\geq l$ , then by (38) and (41),

			$\displaystyle\left(\min_{1\leqslant i\leqslant p-1}(\alpha_{i+1}-\alpha_{i})-\eta\right)^{l-1}(\gamma_{l}-\alpha_{l})^{p-l+1}$
		$\displaystyle\leq$	$\displaystyle\left\|\prod_{i=1}^{p}(\alpha_{l}-\gamma_{i})\right\|\leq\left\|\frac{\nu/\mu}{1-\nu/\mu}\right\|\max_{j\in[1,p]}\left\|\prod_{i=1}^{p}(\alpha_{j}-\beta_{i})\right\|$
		$\displaystyle<$	$\displaystyle 2\left\|\frac{\nu}{\mu}\right\|\max_{j\in[1,p]}\left\|\prod_{i=1}^{p}(\alpha_{j}-\beta_{i})\right\|<\min\left\{1,\left(\min_{1\leqslant i\leqslant p-1}(\alpha_{i+1}-\alpha_{i})-\eta\right)^{p}\right\}\min\{1,\eta^{p}\}$
		$\displaystyle\leq$	$\displaystyle\left(\min_{1\leqslant i\leqslant p-1}(\alpha_{i+1}-\alpha_{i})-\eta\right)^{l-1}\eta^{p-l+1},$

which contradicts to $\gamma_{l}-\alpha_{l}\geq\eta$ . So, statement (i) is true.

When $p=1$ , (40) and (41) lead to

\displaystyle\gamma_{1}-\alpha_{1}\leq\left|\frac{\nu/\mu}{1-\nu/\mu}\right|(\alpha_{1}-\beta_{1})<2\left|\frac{\nu}{\mu}\right|(\alpha_{1}-\beta_{1})<\eta.

The statement (ii) is proved. $\Box$

The subsequent parts focus on Step 3 of the construction, whose key idea is to select some appropriate candidates for the roots of $\frac{g_{n}(\lambda)}{(f_{n}(\lambda),g_{n}(\lambda))}$ . We set these roots as $\lambda_{i}+\rho_{i}$ , $i=1,\ldots,m-1$ , where $\rho_{i}=\varepsilon^{(n+1)^{m-i}}$ ,

\displaystyle\varepsilon=\frac{\Delta^{n^{2}+n+1}}{n2^{3(n+1)^{3}}\Lambda^{(n+1)^{2}}},\quad\Delta\triangleq\frac{1}{2}\min\{1,\min_{1\leq i\leq m-1}(\lambda_{i+1}-\lambda_{i})\},\quad\Lambda\triangleq 1+\lambda_{m}.

(42)

Next, define

\displaystyle C_{1}\triangleq\frac{\Delta}{2^{n+1}\Lambda}\quad\mbox{and}\quad C_{2}(j)\triangleq\frac{2^{2(n+1)^{2}}\Lambda^{n+1}}{\Delta^{n}\varepsilon^{(n+1)^{m-j-1}}},\quad j=1,\ldots,m-1,

(43)

as well as

\displaystyle C\triangleq\frac{2^{2n+1}(1+\Lambda^{n})}{C_{1}^{2n^{2}}\rho_{1}^{2n}}\quad\mbox{and}\quad M\triangleq\sum_{k=1}^{m}M_{k}.

(44)

Remark 4.1.

We remark that $\varepsilon<1$ , $\Lambda/\Delta\geq 2$ in (42) and $C>1$ in (44). Moreover, $C_{2}(j),j=1,\ldots,m-1$ defined by (43) satisfy (see Appendix B)

\displaystyle\left\{\begin{array}[]{ll}(1+C_{2}(m-1))^{n}\rho_{m-1}<\frac{\Delta}{2},\\ (1+C_{2}(j))^{n}\rho_{j}<(1+C_{2}(j+1))^{n}\rho_{j+1},&j=1,\ldots,m-2,\quad m>2\\ \frac{n(1+C_{2}(j-1))^{n}}{\Delta}\rho_{j-1}<\frac{1}{4}\left(\frac{\Delta^{n}}{2^{(n+1)^{2}}\Lambda^{n+1}}\right)\rho_{j+1},&j=2,\ldots,m-2,\quad m>3\end{array}.\right.

(48)

The above series of constants are repeatedly used in the next two lemmas (Lemmas 4.2–4.3), whose proofs are contained in Appendix C. Both the two lemmas concern the following polynomials

\displaystyle F(\lambda)=\prod_{i=1}^{p}(\lambda-\alpha_{i})\quad\mbox{and}\quad G(\lambda)=\prod_{i=1}^{p-1}(\lambda-\beta_{i}),\qquad p\in[2,m],

(49)

whose roots $\{\alpha_{i}\}_{i=1}^{p}$ and $\{\beta_{i}\}_{i=1}^{p-1}$ satisfy

\displaystyle\alpha_{p}\in[\lambda_{p},\bar{\lambda})\qquad\mbox{for some number }\bar{\lambda}>\lambda_{p}

(50)

and for each $i=1,\ldots,p-1$ ,

\displaystyle\lambda_{i}+C_{1}^{n}\rho_{i}\leq\alpha_{i}+C_{1}^{n}\rho_{i}<\beta_{i}<\lambda_{i}+(1+C_{2}(i))^{n}\rho_{i}<\lambda_{i+1}.

(51)

Lemma 4.2.

Let $F(\lambda)$ and $G(\lambda)$ be two polynomials defined by (49)–(51) with $\bar{\lambda}=\Lambda$ in (50). For any given constants $\mu,\nu,m^{*}$ satisfying $0<m^{*}<M$ and $-\frac{\mu}{\nu}>CM$ , the following two statements hold.
(i) If $p>2$ , then there exist two monic polynomials $F_{0}(\lambda)$ and $G_{0}(\lambda)$ with distinct roots $\alpha_{1}^{\prime}<\cdots<\alpha_{p-1}^{\prime}$ and $\beta_{1}^{\prime}<\cdots<\beta_{p-2}^{\prime}$ , respectively, satisfying $\alpha_{p-1}^{\prime}\in(\alpha_{p-1},\lambda_{p})$ and for $i\in[1,p-2]$ ,

\displaystyle\alpha_{i}^{\prime}\in(\alpha_{i},\beta_{i}),\qquad C_{1}(\beta_{i}-\alpha_{i})\leq\beta_{i}^{\prime}-\alpha_{i}^{\prime}\leq C_{2}(i)(\beta_{i}-\alpha_{i}).

(52)

In addition, for some numbers $\lambda^{*},b^{*}>0$ and $\mu_{0},\nu_{0}$ with $-\frac{\mu_{0}}{\nu_{0}}>-\frac{\mu}{\nu}-m^{*}$ , $F_{0}(\lambda)$ and $G_{0}(\lambda)$ fulfill

\displaystyle\left\{\begin{array}[]{l}\mu F(\lambda)=-m^{*}\nu\lambda G(\lambda)+b^{*}\mu_{0}(\lambda^{*}-\lambda)F_{0}(\lambda)\\ \nu G(\lambda)=\mu_{0}F_{0}(\lambda)+b^{*}\nu_{0}(\lambda^{*}-\lambda)G_{0}(\lambda)\end{array}.\right.

(55)

(ii) If $p=2$ , then there are some numbers $\lambda^{*},b^{*}>0$ , $\alpha_{1}^{\prime}\in(\alpha_{1},\lambda_{2})$ and $\mu_{0},\nu_{0}$ with $-\frac{\mu_{0}}{\nu_{0}}>-\frac{\mu}{\nu}-m^{*}$ such that polynomials $F_{0}(\lambda)=\lambda-\alpha_{1}^{\prime}$ and $G_{0}(\lambda)=1$ satisfy equation (55).

Lemma 4.3.

Given $\mu,\nu,\lambda^{*}$ with $\lambda_{p}<\lambda^{*}<\Lambda$ and $\frac{\mu}{\nu}<0$ , the following two statements hold.
(i) For polynomials $F(\lambda)$ and $G(\lambda)$ defined by (49)–(51) with $\bar{\lambda}=\lambda^{*}$ in (50), there exist two monic polynomials $F_{0}(\lambda)$ and $G_{0}(\lambda)$ with distinct roots $\alpha_{1}^{\prime}<\cdots<\alpha_{p}^{\prime}$ and $\beta_{1}^{\prime}<\cdots<\beta_{p-1}^{\prime}$ , respectively, such that $\alpha_{p}^{\prime}=\lambda^{*}$ and (52) holds for all $i\in[1,p-1]$ . In addition, for some numbers $m^{*},b^{*}>0$ and $\mu_{0},\nu_{0}$ with $-\frac{\mu_{0}}{\nu_{0}}>-\frac{\lambda_{1}}{\Lambda}\frac{\mu}{\nu}$ , $F_{0}(\lambda)$ and $G_{0}(\lambda)$ fulfill

\displaystyle\left\{\begin{array}[]{l}\mu F(\lambda)=-m^{*}\nu\lambda G(\lambda)-b^{*}\mu_{0}F_{0}(\lambda)\\ \nu(\lambda-\lambda^{*})G(\lambda)=\mu_{0}F_{0}(\lambda)+b^{*}\nu_{0}(\lambda^{*}-\lambda)G_{0}(\lambda)\\ \end{array}.\right.

(58)

(ii) For $F(\lambda)=\lambda-\alpha_{1}$ with $\alpha_{1}<\lambda^{*}$ and $G(\lambda)=1$ , there are some numbers $m^{*},b^{*}>0$ and $\mu_{0},\nu_{0}$ with $-\frac{\mu_{0}}{\nu_{0}}>-\frac{\lambda_{1}}{\Lambda}\frac{\mu}{\nu}$ such that polynomials $F_{0}(\lambda)=\lambda-\lambda^{*}$ and $G_{0}(\lambda)=1$ satisfy equation (58).

Lemma 4.4.

Let $F_{n}(\lambda)=\prod_{i=1}^{m}(\lambda-\lambda_{i})$ and $G_{n}(\lambda)=\prod_{i=1}^{m-1}(\lambda-\lambda_{i}-\rho_{i})$ be two polynomials and $\mu_{n},\nu_{n}$ be two numbers satisfying

\displaystyle-\frac{\mu_{n}}{\nu_{n}}>C\left(\frac{\Lambda}{\lambda_{1}}\right)^{n}\frac{\Lambda M}{\Lambda-\lambda_{1}}.

(59)

Then, there exist some monic polynomials $\{F_{j}(\lambda)\}_{j=1}^{n-1}$ , $\{G_{j}(\lambda)\}_{j=1}^{n-1}$ and some sequences of numbers $\{(\lambda_{j}^{*},b_{j},m_{j})\}_{j=2}^{n}$ , $\{(\mu_{j},\nu_{j})\}_{j=1}^{n-1}$ such that for each $j\in[1,n-1]$ , the following two properties hold:
(i) $\lambda_{j+1}^{*},b_{j+1},m_{j+1}>0$ and $-\frac{\mu_{j}}{\nu_{j}}>C(\frac{\Lambda}{\lambda_{1}})^{j}\frac{\Lambda M}{\Lambda-\lambda_{1}}$ ;
(ii) if $j\in[1,n-1]\setminus\bigcup_{h=0}^{T-2}[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]$ , then

\displaystyle\left\{\begin{array}[]{l}\mu_{j+1}F_{j+1}(\lambda)=-m_{j+1}\nu_{j+1}\lambda G_{j+1}(\lambda)+b_{j+1}\mu_{j}(\lambda_{j+1}^{*}-\lambda)F_{j}(\lambda)\\ \nu_{j+1}G_{j+1}(\lambda)=\mu_{j}F_{j}(\lambda)+b_{j+1}\nu_{j}(\lambda_{j+1}^{*}-\lambda)G_{j}(\lambda)\end{array},\right.

(62)

otherwise, for $j\in\bigcup_{l=0}^{T-2}[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]$ ,

\displaystyle\left\{\begin{array}[]{l}\mu_{j+1}F_{j+1}(\lambda)=-m_{j+1}\nu_{j+1}\lambda G_{j+1}(\lambda)-b_{j+1}\mu_{j}F_{j}(\lambda)\\ \nu_{j+1}(\lambda-\lambda_{j+1}^{*})G_{j+1}(\lambda)=\mu_{j}F_{j}(\lambda)+b_{j+1}\nu_{j}(\lambda_{j+1}^{*}-\lambda)G_{j}(\lambda)\\ \lambda_{j+1}^{*}=s_{l+1}(l+1+\sum_{h=1}^{l+1}q_{h}-j)\end{array},\right.

(66)

where $s_{l+1}(l+1+\sum_{h=1}^{l+1}q_{h}-j)\in S_{l+1}$ and $S_{l+1}$ is defined by (30).

Proof.

For $j=n-1,\ldots,1$ , we construct a series of numbers $\lambda_{j+1}^{*},b_{j+1},m_{j+1}$ , $\mu_{j},\nu_{j}$ and polynomials $F_{j}(\lambda)$ , $G_{j}(\lambda)$ on the basis of $\mu_{j+1},\nu_{j+1}$ , and $F_{j+1}(\lambda)$ , $G_{j+1}(\lambda)$ , according to the following strategies:

if $j\in[1,n-1]\setminus\bigcup_{l=0}^{T-2}[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]$ , we apply Lemma 4.2 with $F(\lambda)=F_{j+1}(\lambda)$ , $G(\lambda)=G_{j+1}(\lambda)$ , $m^{*}=m_{j+1}$ , $\mu=\mu_{j+1}$ , $\nu=\nu_{j+1}$ to obtain $\lambda_{j+1}^{*},b_{j+1}$ , $\mu_{j},\nu_{j}$ and $F_{j}(\lambda)$ , $G_{j}(\lambda)$ , where

\displaystyle m_{j+1}\triangleq\left\{\begin{array}[]{ll}M_{l+2},&j=l+1+\sum_{h=1}^{l+1}q_{h}\,\,\mbox{for}\,\,l<T-2\\ M_{j+1-n+m},&j>T-2+n-m\end{array};\right.

(69)

b)

if $j\in[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]$ for some $l\in[0,T-2]$ , we set $\lambda_{j+1}^{*}=s_{l+1}(l+1+\sum_{h=1}^{l+1}q_{h}-j)$ and apply Lemma 4.3 with $F(\lambda)=F_{j+1}(\lambda)$ , $G(\lambda)=G_{j+1}(\lambda)$ , $\lambda^{*}=\lambda_{j+1}^{*}$ , $\mu=\mu_{j+1}$ , $\nu=\nu_{j+1}$ to obtain $m_{j+1},b_{j+1}$ , $\mu_{j},\nu_{j}$ , and $F_{j}(\lambda)$ , $G_{j}(\lambda)$ .

We shall use the induction method to show that either strategy a) or strategy b) can be implemented for each $j=n-1,\ldots,1$ . First, let $j=n-1$ . Observe that $T\leq m$ , it is easy to compute

(T-2)+\sum_{h=1}^{l+1}q_{h}\leq(m-2)+(n-m)=n-2.

Hence, $n-1\notin\bigcup_{l=0}^{T-2}[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]$ . In addition, since (59) and (69) yield $-\frac{\mu_{n}}{\nu_{n}}>CM$ and $m_{n}\in(0,M)$ , by applying Lemma 4.2(i) with $F(\lambda)=F_{n}(\lambda)$ , $G(\lambda)=G_{n}(\lambda)$ , $m^{*}=m_{n}$ , $\mu=\mu_{n}$ , $\nu=\nu_{n}$ , we can find some numbers $\lambda_{n}^{*},b_{n},\mu_{n-1},\nu_{n-1}$ with $\lambda_{n}^{*},b_{n}>0$ and

-\frac{\mu_{n-1}}{\nu_{n-1}}>-\frac{\mu_{n}}{\nu_{n}}-m_{n}>C\left(\frac{\Lambda}{\lambda_{1}}\right)^{n-1}\frac{\Lambda M}{\Lambda-\lambda_{1}},

and two monic polynomials $F_{n-1}(\lambda)$ , $G_{n-1}(\lambda)$ such that (62) holds. So, strategy a) applies and both (i) and (ii) are true for these $\lambda_{n}^{*},b_{n},\mu_{n-1},\nu_{n-1},F_{n-1}(\lambda),G_{n-1}(\lambda)$ .

Now, assume that we have constructed the required $\{(\lambda_{j}^{*},b_{j},m_{j})\}_{j=n-r+1}^{n}$ , $\{(\mu_{j},\nu_{j})\}_{j=n-r}^{n-1}$ , $\{F_{j}(\lambda)\}_{j=n-r}^{n-1}$ and $\{G_{j}(\lambda)\}_{j=n-r}^{n-1}$ for some $r\in[1,n-2]$ by following either strategy a) or strategy b), so that properties (i) and (ii) hold for $j=n-1,\ldots,n-r$ . Considering Lemmas 4.2 and 4.3, we write $F_{n-j}(\lambda)=\prod_{i=1}^{z_{n-j}}(\lambda-\alpha_{n-j}(i))$ with $\alpha_{n-j}(1)<\cdots<\alpha_{n-j}(z_{n-j})$ and $G_{n-j}(\lambda)=\prod_{i=1}^{z_{n-j}-1}(\lambda-\beta_{n-j}(i))$ with $\beta_{n-j}(1)<\cdots<\beta_{n-j}(z_{n-j}-1)$ , $j=0,1,\ldots,r$ . Here, for each $j=0,\ldots,r-1$ ,

\displaystyle z_{n-j-1}=\left\{\begin{array}[]{ll}z_{n-j}-1,&n-j-1\not\in\cup_{l=0}^{T-2}[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]\\ z_{n-j},&n-j-1\in\cup_{l=0}^{T-2}[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]\\ \end{array}.\right.

(72)

Furthermore, if $z_{n-r}>1$ , then for each $j\in[1,r]$ and $i\in[1,z_{n-r}-1]$ ,

\displaystyle\left\{\begin{array}[]{ll}\alpha_{n-j}(i)\in(\alpha_{n-j+1}(i),\beta_{n-j+1}(i))\\ C_{1}(\beta_{n-j+1}(i)-\alpha_{n-j+1}(i))\leq\beta_{n-j}(i)-\alpha_{n-j}(i)\leq C_{2}(i)(\beta_{n-j+1}(i)-\alpha_{n-j+1}(i))\end{array}.\right.

(75)

Recall that $\alpha_{n}(i)=\lambda_{i}$ and $\beta_{n}(i)=\lambda_{i}+\rho_{i}$ , (75) implies that for $i\in[1,z_{n-r}-1]$ ,

\displaystyle\beta_{n-r}(i)-\alpha_{n-r}(i)\geq C_{1}^{r}(\beta_{n}(i)-\alpha_{n}(i))=C_{1}^{r}\rho_{i}.

(76)

Moreover, by (75) again,

$\displaystyle\beta_{n-j}(i)-\alpha_{n}(i)$	$\displaystyle=$	$\displaystyle\beta_{n-j}(i)-\alpha_{n-j}(i)+\alpha_{n-j}(i)-\alpha_{n}(i)$
	$\displaystyle\leq$	$\displaystyle C_{2}(i)(\beta_{n-j+1}(i)-\alpha_{n-j+1}(i))+(\beta_{n-j+1}(i)-\alpha_{n}(i))$
	$\displaystyle\leq$	$\displaystyle(1+C_{2}(i))(\beta_{n-j+1}(i)-\alpha_{n}(i)),$

then a straightforward calculation leads to

\displaystyle\alpha_{n}(i)<\alpha_{n-r}(i)<\beta_{n-r}(i)<\alpha_{n}(i)+(1+C_{2}(i))^{r}(\beta_{n}(i)-\alpha_{n}(i)).

(77)

Note that by (48) in Remark 4.1,

(1+C_{2}(i))^{r}(\beta_{n}(i)-\alpha_{n}(i))<(1+C_{2}(i))^{n}\rho_{i}<\Delta,

so by virtue of (76) and (77),

\displaystyle\lambda_{i}+C_{1}^{n}(i)\rho_{i}<\alpha_{n-r}(i)+C_{1}^{n}(i)\rho_{i}<\beta_{n-r}(i)<\lambda_{i}+(1+C_{2}(i))^{n}\rho_{i}<\lambda_{i+1}.

(78)

We remark that no matter strategy a) or b) applies, it always infers

\displaystyle\left\{\begin{array}[]{l}\alpha_{n-r}(z_{n-r})<\beta_{n-r}(z_{n-r})<\lambda_{z_{n-r}+1},\quad z_{n-r+1}-1=z_{n-r}\\ \alpha_{n-r}(z_{n-r})=\lambda_{n-r+1}^{*}<\Lambda,\quad~{}~{}~{}~{}\quad~{}~{}~{}~{}~{}z_{n-r+1}=z_{n-r}\\ \end{array},\right.

and hence

\displaystyle\alpha_{n-r}(z_{n-r})\in[\lambda_{z_{n-r}},\Lambda).

(80)

We now verify that at least one of the strategies a) and b) is valid for $j=n-r-1$ . It is discussed by two cases.

Case 1: $n-r-1\not\in\cup_{l=0}^{T-2}[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]$ . Because of (72), we estimate $z_{n-r}$ directly by

\displaystyle z_{n-r}\geq m-\left(n-2-\sum_{h=1}^{T-1}q_{h}\right)=m-\left(n-2-(n-m)\right)=2.

Now, $-\frac{\mu_{n-r}}{\nu_{n-r}}>C(\frac{\Lambda}{\lambda_{1}})^{n-r}\frac{\Lambda M}{\Lambda-\lambda_{1}}$ , it thus gives $-\frac{\mu_{n-r}}{\nu_{n-r}}>CM$ . So combining (78) and (80), it shows that the assumptions of Lemma 4.2 are fulfilled. Therefore, Lemma 4.2 is applicable and strategy a) works. We thus conclude properties (i) and (ii) for $j=n-r-1$ by Lemma 4.2 and

\displaystyle-\frac{\mu_{n-r-1}}{\nu_{n-r-1}}>-\frac{\mu_{{n-r}}}{\nu_{n-r}}-m_{n-r}>C\left(\frac{\Lambda}{\lambda_{1}}\right)^{n-r}\frac{\Lambda M}{\Lambda-\lambda_{1}}-M>C\left(\frac{\Lambda}{\lambda_{1}}\right)^{n-r-1}\frac{\Lambda M}{\Lambda-\lambda_{1}}.

Case 2: $n-r-1\in[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]$ for some $l\in[0,T-2]$ . If $n-r-1<l+\sum_{h=1}^{l+1}q_{h}$ , then $n-r\in[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]$ , and hence $F_{n-r}(\lambda)$ is constructed by strategy b). In view of Lemma 4.3, the maximal root of $F_{n-r}(\lambda)$ is equivalent to $\lambda_{n-r+1}^{*}$ , which shows

\displaystyle s_{l+1}(l+2+\sum\nolimits_{h=1}^{l+1}q_{h}-n+r)>s_{l+1}(l+1+\sum\nolimits_{h=1}^{l+1}q_{h}-n+r)=\lambda_{n-r+1}^{*}=\alpha_{n-r}(z_{n-r}).

If $n-r-1=l+\sum_{h=1}^{l+1}q_{h}$ , then $F_{n-r}(\lambda)$ is constructed by strategy a). Observe that

\displaystyle z_{n-r}=T-1-(T-l-2)=l+1,

so by (78), $\alpha_{n-r}(z_{n-r})<\lambda_{z_{n-r}+1}=\lambda_{l+2}$ . On the other hand, if we suppose $s_{l+1}(l+2+\sum_{h=1}^{l+1}q_{h}-n+r)=\lambda_{i}$ for some $i\in[1,m]$ , then $i\geq t_{i}\geq l+2$ because of (30). As a consequence,

\displaystyle s_{l+1}(l+2+\sum\nolimits_{h=1}^{l+1}q_{h}-n+r)\geq\lambda_{l+2},

(81)

and hence

\displaystyle s_{l+1}(l+2+\sum\nolimits_{h=1}^{l+1}q_{h}-n+r)>\alpha_{n-r}(z_{n-r}).

(82)

We thus conclude (82) is always true in strategy a). By (80),

\alpha_{n-r}(z_{n-r})\in[\lambda_{z_{n-r}},s_{l+1}(l+2+\sum\nolimits_{h=1}^{l+1}q_{h}-n+r)).

This combining with (78) and (80) verifies the assumptions of Lemma 4.3. Hence, Lemma 4.3 applies and strategy b) can be implemented. So, properties (i) and (ii) hold for $j=n-r-1$ due to Lemma 4.3 and

-\frac{\mu_{n-r-1}}{\nu_{n-r-1}}>-\frac{\mu_{n-r}}{\nu_{n-r}}\frac{\lambda_{1}}{\Lambda}>C\left(\frac{\Lambda}{\lambda_{1}}\right)^{n-r-1}\frac{\Lambda M}{\Lambda-\lambda_{1}}.

The induction is completed. $\Box$

Proof of Proposition 4.1.

Let $\{\mu_{n},\nu_{n},F_{n}(\lambda),G_{n}(\lambda)\}$ be defined in Lemma 4.4, then we can construct a series of numbers $\{(\lambda_{j}^{*},b_{j},m_{j})\}_{j=2}^{n}$ , $\{(\mu_{j},\nu_{j})\}_{j=1}^{n-1}$ and some monic polynomials $\{F_{j}(\lambda)\}_{j=1}^{n-1}$ , $\{G_{j}(\lambda)\}_{j=1}^{n-1}$ . Note that $-\frac{\mu_{1}}{\nu_{1}}>M>M_{1}$ , set

\displaystyle\left\{\begin{array}[]{l}m_{1}=M_{1},\quad b_{1}=-\frac{\mu_{1}}{\nu_{1}}-M_{1},\quad k_{1}=-\alpha_{1}(1)\frac{\mu_{1}}{\nu_{1}}\\ k_{j}=\lambda_{j}^{*}b_{j},\quad j=2,\ldots,n\\ \end{array}.\right.

(85)

We shall see that $\{(k_{j},b_{j},m_{j})\}_{j=1}^{n}$ meet the requirement of Proposition 4.1.

In fact, define a sequence of polynomials $\{D_{j}(\lambda)\}_{j=1}^{n}$ as follows:

\displaystyle\left\{\begin{array}[]{ll}D_{1}(\lambda)=1\\ D_{k+1}(\lambda)=D_{k}(\lambda),&k\in[1,n-1]\setminus\bigcup_{l=0}^{T-2}[l+1+\sum_{h=1}^{l}q_{j},l+\sum_{h=1}^{l+1}q_{j}]\\ D_{k+1}(\lambda)=(\lambda-\lambda_{k+1}^{*})D_{k}(\lambda),&k\in\bigcup_{l=0}^{T-2}[l+1+\sum_{h=1}^{l}q_{j},l+\sum_{h=1}^{l+1}q_{j}]\end{array},\right.

(89)

and let

\displaystyle f_{j}(\lambda)=\frac{\mu_{j}}{\nu_{1}}D_{j}(\lambda)F_{j}(\lambda)\quad\mbox{and}\quad g_{j}(\lambda)=\frac{\nu_{j}}{\nu_{1}}D_{j}(\lambda)G_{j}(\lambda),\quad j=1,\ldots,n.

(90)

Clearly, $f_{n}(\lambda)=\frac{\mu_{n}}{\nu_{1}}D_{n}(\lambda)F_{n}(\lambda)=\frac{\mu_{n}}{\nu_{1}}\prod_{j=1}^{m}(\lambda-\lambda_{j})^{t_{j}}.$ The rest of the proof is to check whether $\{f_{j}(\lambda)\}_{j=1}^{n}$ satisfy the recursive formula in Lemma 3.1.

Firstly, (85) and (90) imply $f_{1}(\lambda)=k_{1}-\lambda(m_{1}+b_{1})$ . Moreover, in view of Lemma 4.4, $G_{1}(\lambda)=1$ , which indicates $g_{1}(\lambda)=1$ . Next, by Lemma 4.4 again, for $j\in[1,n-1]\setminus\bigcup_{l=0}^{T-2}[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]$ ,

\displaystyle\left\{\begin{array}[]{l}f_{j+1}(\lambda)=-m_{j+1}\lambda g_{j+1}(\lambda)+(k_{j+1}-\lambda b_{j+1})f_{j}(\lambda)\\ g_{j+1}(\lambda)=f_{j}(\lambda)+(k_{j+1}-\lambda b_{j+1})g_{j}(\lambda)\end{array},\right.

(93)

and for $j\in\bigcup_{l=0}^{T-2}[l+1+\sum_{h=1}^{l}q_{h},l+\sum_{h=1}^{l+1}q_{h}]$ , we compute

$\displaystyle f_{j+1}(\lambda)$	$\displaystyle=$	$\displaystyle\frac{1}{\nu_{1}}\mu_{j+1}D_{j+1}(\lambda)F_{j+1}(\lambda)=\frac{1}{\nu_{1}}D_{j+1}(\lambda)(-m_{j+1}\nu_{j+1}\lambda G_{j+1}(\lambda)-b_{j+1}\mu_{j}F_{j}(\lambda))$
	$\displaystyle=$	$\displaystyle-m_{j+1}\lambda g_{j+1}(\lambda)-\frac{(\lambda-\lambda_{j+1}^{*})D_{j}(\lambda)}{\nu_{1}}b_{j+1}\mu_{j}F_{j}(\lambda)$
	$\displaystyle=$	$\displaystyle-m_{j+1}\lambda g_{j+1}(\lambda)+(k_{j+1}-\lambda b_{j+1})f_{j}(\lambda),$
$\displaystyle g_{j+1}(\lambda)$	$\displaystyle=$	$\displaystyle\frac{\nu_{j+1}}{\nu_{1}}D_{j+1}(\lambda)G_{j+1}(\lambda)=\frac{D_{j}(\lambda)}{\nu_{1}}\nu_{j+1}(\lambda-\lambda_{j+1}^{*})G_{j+1}(\lambda)$
	$\displaystyle=$	$\displaystyle\frac{D_{j}(\lambda)}{\nu_{1}}(\mu_{j}F_{j}(\lambda)+b_{j+1}\nu_{j}(\lambda_{j+1}^{*}-\lambda)G_{j}(\lambda))=f_{j}(\lambda)+(k_{j+1}-\lambda b_{j+1})g_{j}(\lambda).$

The assertion thus follows. $\Box$

5 Concluding remarks.

The emergence of inerters in engineering brings some new phenomena in the study of inverse problems. Particularly, it enables a mass-chain system to possess multiple eigenvalues. This paper has solved the IEP for the “fixed-free” case, where the real numbers for eigenvalue assignment can be taken arbitrarily positive. Another common situation is that the both ends of the system are fixed at the wall. For such “fixed-fixed” systems, the construction cannot follow readily from the method developed in Section 4. To address this issue, a more detailed analysis is required and it would be our next work.

Appendix A Proof of Example 2.1

For each $j=1,\ldots,5$ , let $F_{j}(\lambda)$ and $G_{j}(\lambda)$ be two monic polynomials such that

\displaystyle\mu_{j}F_{j}(\lambda)=\frac{f_{j}(\lambda)}{\left(f_{j}(\lambda),g_{j}(\lambda)\right)}\quad\mbox{and}\quad\nu_{j}G_{j}(\lambda)=\frac{g_{j}(\lambda)}{\left(f_{j}(\lambda),g_{j}(\lambda)\right)},

where $\mu_{j}$ and $\nu_{j}$ are the leading coefficients of $f_{j}(\lambda)$ and $g_{j}(\lambda)$ , respectively. Apparently, $\mu_{j}\nu_{j}<0$ and

\displaystyle(F_{j}(\lambda),G_{j}(\lambda))=1,\quad j=1,\ldots,5.

(94)

Moreover, denote $\alpha_{j}(1)<\cdots<\alpha_{j}(s_{j})$ and $\beta_{j}(1)<\cdots<\beta_{j}(s_{j}-1)$ as the roots of $F_{j}(\lambda)$ and $G_{j}(\lambda)$ . Define

J\triangleq\{j:k_{j}=\lambda_{3}b_{j},j=2,3,4,5\}\quad\mbox{ and}\quad H\triangleq\{j:F_{j-1}(\lambda_{3})=0,j=2,3,4,5\}.

We first assert that $J\cap H$ cannot contain any adjacent natural numbers. Otherwise, suppose there is a number $i\in\{1,2,3\}$ such that $i+1,i+2\in J\cap H$ . So, $k_{i+1}/b_{i+1}=\lambda_{3}$ and by Lemma 3.3,

\displaystyle\mu_{i+1}F_{i+1}(\lambda)=-m_{i+1}\nu_{i+1}\lambda G_{i+1}(\lambda)-b_{i+1}\mu_{i}F_{i}(\lambda).

(95)

Since the definition of $H$ indicates $F_{i}(\lambda_{3})=F_{i+1}(\lambda_{3})=0$ , then by (95),

m_{i+1}\nu_{i+1}\lambda_{3}G_{i+1}(\lambda_{3})=-b_{i+1}\mu_{i}F_{i}(\lambda_{3})-\mu_{i+1}F_{i+1}(\lambda_{3})=0.

Hence, $(\lambda-\lambda_{3})|(F_{i+1}(\lambda),G_{i+1}(\lambda))$ , which contradicts to (94) that $(F_{i+1}(\lambda),G_{i+1}(\lambda))=1$ .

We in fact have derived $|J\cap H|\leq 2$ , which together with Lemma 3.2 implies $F(\lambda_{3})=0$ and $5\not\in H$ . On the other hand, (22) and (26) in Section 3 infer $|J\cap H|\geq 2$ , and hence $|J\cap H|=2$ . Observe that $J\cap H\not=\{2,3\}$ or $\{3,4\}$ , it then follows that $J\cap H=\{2,4\}$ . So, $\alpha_{1}(1)=\lambda_{3}$ and by Lemma 3.3(i), $s_{2}=1$ and $\alpha_{2}(1)<\lambda_{3}$ . For the roots of $F_{j}(\lambda),j=3,4,5$ , they can be discussed similarly by using Lemma 3.3 repeatedly. Indeed, $3\notin J\cap H$ and $4\in H$ lead to $s_{3}=2$ and $F_{3}(\lambda_{3})=0$ , respectively. So,

\displaystyle\left\{\begin{array}[]{ll}0<\alpha_{3}(1)<\alpha_{2}(1)<\alpha_{3}(2)=\lambda_{3},&b_{3}=0\\ \alpha_{3}(1)<\min\{\alpha_{2}(1),k_{3}/b_{3}\}<\lambda_{3},\quad\mbox{and}\quad\alpha_{3}(2)=\lambda_{3}<k_{3}/b_{3},&b_{3}>0\end{array}.\right.

In addition, $4\in J\cap H$ indicates $s_{4}=2$ and

\displaystyle\alpha_{4}(1)<\alpha_{3}(1)<\alpha_{4}(2)<\alpha_{3}(2)=\lambda_{3}.

(97)

At last, since $5\notin J\cap H$ , $s_{5}=3$ . By $|J\cap H|=2$ and (26), $(f_{5}(\lambda),g_{5}(\lambda))=(\lambda-\lambda_{3})^{2},$ which immediately gives $F_{5}(\lambda)=\prod_{i=1}^{3}(\lambda-\lambda_{i})$ . So, $\alpha_{5}(1)=\lambda_{1}<\alpha_{5}(2)=\lambda_{2}<\alpha_{5}(3)=\lambda_{3}$ . If $b_{5}>0$ , then (97) infers that $\lambda_{3}<k_{5}/b_{5}$ . Therefore,

\displaystyle\left\{\begin{array}[]{ll}0<\alpha_{5}(1)<\alpha_{4}(1)<\alpha_{5}(2)<\alpha_{4}(2)<\alpha_{5}(3)=\lambda_{3},&b_{5}=0\\ \alpha_{5}(1)<\alpha_{4}(1)<\alpha_{5}(2)<\alpha_{4}(2)<\alpha_{5}(3)=\lambda_{3}<k_{5}/b_{5},&b_{5}>0\end{array}.\right.

We thus summarize

\displaystyle\left\{\begin{array}[]{l}s_{5}=3,\quad s_{4}=s_{3}=2,\quad s_{2}=s_{1}=1\\ \alpha_{2}(s_{2}),\alpha_{4}(s_{4})<\lambda_{3}\quad\mbox{and}\quad\alpha_{1}(s_{1})=\alpha_{3}(s_{3})=\alpha_{5}(s_{5})=\lambda_{3}\\ \alpha_{j+1}(i)<\alpha_{j}(i)<\alpha_{j+1}(i+1),\quad j\in[1,4],\,i\in[1,s_{j+1}-1]\end{array}\right.

(102)

and

\displaystyle\left\{\begin{array}[]{ll}k_{j}/b_{j}=\lambda_{3}\,\,\mbox{and}\,\,b_{j}>0,&j=2,4\\ k_{j}/b_{j}>\lambda_{3}\,\,\mbox{if}\,\,b_{j}>0,&j=3,5\end{array}.\right.

(105)

This means $J=J\cap H=\{2,4\}$ and as a consequence, by Lemma 3.3,

\displaystyle\left\{\begin{array}[]{l}\mu_{j+1}F_{j+1}(\lambda)=-\lambda m_{j+1}\nu_{j+1}G_{j+1}(\lambda)-b_{j+1}\mu_{j}F_{j}(\lambda)\\ \left(\lambda-\frac{k_{j+1}}{b_{j+1}}\right)\nu_{j+1}G_{j+1}(\lambda)=\mu_{j}F_{j}(\lambda)+(k_{j+1}-\lambda b_{j+1})\nu_{j}G_{j}(\lambda)\end{array},\quad j=1,3\right.

(108)

and

\displaystyle\left\{\begin{array}[]{l}\mu_{j+1}F_{j+1}(\lambda)=-\lambda m_{j+1}\nu_{j+1}G_{j+1}(\lambda)+(k_{j+1}-\lambda b_{j+1})\mu_{j}F_{j}(\lambda)\\ \nu_{j+1}G_{j+1}(\lambda)=\mu_{j}F_{j}(\lambda)+(k_{j+1}-\lambda b_{j+1})\nu_{j}G_{j}(\lambda)\end{array},\quad j=2,4.\right.

(111)

With the above properties, we can also present the relationship between the roots of $F_{j}(\lambda)$ and $G_{j+1}(\lambda)$ for $j=2,4$ . In fact, when $j=2,4$ , $s_{j+1}=s_{j}+1$ and by (21), (102) and (105), $\mu_{j}F_{j}(\lambda)\ll-\nu_{j}(\lambda b_{j+1}-k_{j+1})G_{j}(\lambda)$ . Then, (21), Lemma 3.2, (102) and (111) imply

\displaystyle\alpha_{j}(1)<\beta_{j+1}(1)<\alpha_{j}(2)<\cdots<\alpha_{j}(s_{j})<\beta_{j+1}(s_{j+1}-1)<\alpha_{j+1}(s_{j+1})=\lambda_{3},\quad j=2,4.

(112)

Now, we prove (6) by using reduction to absurdity. Suppose

\displaystyle\max_{j\in[2,4]}\frac{m_{j}}{m_{j+1}}\leq\sigma\triangleq\frac{\lambda_{1}}{8\lambda_{3}}\left(1-\left(\frac{\lambda_{2}}{\lambda_{3}}\right)^{\frac{1}{3}}\right).

(113)

First, it is evident that for each $j\in[1,4]$ , (102), (108) and (111) yield

	$\displaystyle m_{j+1}=-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{F_{j+1}(\alpha_{j}(1))}{\alpha_{j}(1)G_{j+1}(\alpha_{j}(1))}$	$\displaystyle=$	$\displaystyle-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{\prod_{i=1}^{s_{j+1}}(\alpha_{j}(1)-\alpha_{j+1}(i))}{\alpha_{j}(1)\prod_{i=1}^{s_{j+1}-1}(\alpha_{j}(1)-\beta_{j+1}(i))}$		(114)
		$\displaystyle\geq$	$\displaystyle-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{\alpha_{j}(1)-\alpha_{j+1}(1)}{\alpha_{j}(1)}.$		(114)

In particular, when $j=1,3$ , (21) further implies

\displaystyle m_{j+1}=-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{F_{j+1}(\alpha_{j}(s_{j}))}{\alpha_{j}(s_{j})G_{j+1}(\alpha_{j}(s_{j}))}=-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{\prod_{i=1}^{s_{j+1}}(\lambda_{3}-\alpha_{j+1}(i))}{\lambda_{3}\prod_{i=1}^{s_{j+1}-1}(\lambda_{3}-\beta_{j+1}(i))}<-\frac{\mu_{j+1}}{\nu_{j+1}}.

(115)

Note that by comparing the leading coefficients of the polynomials in (108)–(111), we assert that for each $j\in[1,4]$ with $b_{j+1}>0$ ,

\displaystyle-\frac{\mu_{j}}{\nu_{j}}=-\frac{\mu_{j+1}+m_{j+1}\nu_{j+1}}{\nu_{j+1}-\mu_{j}}.

(116)

Finally, let us complete the proof by considering the following four cases.

Case 1: $b_{3}b_{5}>0$ . First, we estimate $m_{j}/m_{j+1}$ for $j=2,4$ . Let $\varsigma_{j}=k_{j+1}/b_{j+1}$ , then $\varsigma_{j}>\alpha_{j+1}(s_{j+1})=\lambda_{3}$ by (102)–(105). Therefore, noting that $\nu_{j+1}\mu_{j}>0$ , (21) and (116) lead to

$\displaystyle-\frac{\mu_{j}}{\nu_{j}}$	$\displaystyle=$	$\displaystyle-\frac{\mu_{j+1}+m_{j+1}\nu_{j+1}}{\nu_{j+1}-\mu_{j}}>-\frac{\mu_{j+1}+m_{j+1}\nu_{j+1}}{\nu_{j+1}}=-\frac{\mu_{j+1}}{\nu_{j+1}}\left(1-\frac{F_{j+1}(\varsigma_{j})}{\varsigma_{j}G_{j+1}(\varsigma_{j})}\right)$	(117)
	$\displaystyle=$	$\displaystyle-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{\varsigma_{j}\prod_{i=1}^{s_{j+1}-1}(\varsigma_{j}-\beta_{j+1}(i))-\prod_{i=1}^{s_{j+1}}(\varsigma_{j}-\alpha_{j+1}(i))}{\varsigma_{j}\prod_{i=1}^{s_{j+1}-1}(\varsigma_{j}-\beta_{j+1}(i))}$
	$\displaystyle>$	$\displaystyle-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{\alpha_{j+1}(1)\prod_{i=1}^{s_{j+1}-1}(\varsigma_{j}-\beta_{j+1}(i))}{\varsigma_{j}\prod_{i=1}^{s_{j+1}-1}(\varsigma_{j}-\beta_{j+1}(i))}$
	$\displaystyle=$	$\displaystyle-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{\alpha_{j+1}(1)}{\varsigma_{j}}\geq-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{\lambda_{1}}{\varsigma_{j}},\quad j=2,4.$

So, if $\varsigma_{j}\leq 2\lambda_{3}$ , the above inequality reduces to

\displaystyle-\frac{\mu_{j}}{\nu_{j}}>-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{\lambda_{1}}{2\lambda_{3}},\quad j=2,4.

(118)

We next treat $\varsigma_{2}>2\lambda_{3}$ . Since (21) and (102) imply $\beta_{3}(1)<\alpha_{3}(2)$ and $\lambda_{1}=\alpha_{5}(1)<\alpha_{3}(1)$ , then

\displaystyle(\varsigma_{2}-\lambda_{1})G_{3}(\varsigma_{2})=(\varsigma_{2}-\lambda_{1})(\varsigma_{2}-\beta_{3}(1))>(\varsigma_{2}-\alpha_{3}(1))(\varsigma_{2}-\alpha_{3}(2))=F_{3}(\varsigma_{2}),

which is equivalent to $1-\frac{F_{3}(\varsigma_{2})}{\varsigma_{2}G_{3}(\varsigma_{2})}>\frac{\lambda_{1}}{\varsigma_{2}}.$ Further, by (111) again, $\mu_{2}/\nu_{3}=G_{3}(\varsigma_{2})/F_{2}(\varsigma_{2})$ , so (112) shows

	$\displaystyle-\frac{\mu_{2}}{\nu_{2}}$	$\displaystyle=$	$\displaystyle-\frac{\mu_{3}+m_{3}\nu_{3}}{\nu_{3}-\mu_{2}}=-\frac{\mu_{3}}{\nu_{3}}\frac{1-\frac{F_{3}(\varsigma_{2})}{\varsigma_{2}G_{3}(\varsigma_{2})}}{1-\frac{G_{3}(\varsigma_{2})}{F_{2}(\varsigma_{2})}}>-\frac{\mu_{3}}{\nu_{3}}\frac{\lambda_{1}}{\varsigma_{2}}\frac{F_{2}(\varsigma_{2})}{F_{2}(\varsigma_{2})-G_{3}(\varsigma_{2})}$		(119)
		$\displaystyle=$	$\displaystyle-\frac{\mu_{3}}{\nu_{3}}\frac{\lambda_{1}}{\beta_{3}(1)-\alpha_{2}(1)}\frac{\varsigma_{2}-\alpha_{2}(1)}{\varsigma_{2}}>-\frac{\mu_{3}}{\nu_{3}}\frac{\lambda_{1}}{\lambda_{3}}\left(1-\frac{\lambda_{3}}{2\lambda_{3}}\right)>-\frac{\mu_{3}}{\nu_{3}}\frac{\lambda_{1}}{8\lambda_{3}}.$		(120)

As for $\varsigma_{4}>2\lambda_{3}$ , similar to (119), we compute by (112) that

$\displaystyle-\frac{\mu_{4}}{\nu_{4}}$	$\displaystyle>$	$\displaystyle-\frac{\mu_{5}}{\nu_{5}}\frac{\lambda_{1}}{\varsigma_{4}}\frac{F_{4}(\varsigma_{4})}{F_{4}(\varsigma_{4})-G_{5}(\varsigma_{4})}$	(121)
	$\displaystyle=$	$\displaystyle-\frac{\mu_{5}}{\nu_{5}}\frac{\lambda_{1}}{\varsigma_{4}}\frac{(\varsigma_{4}-\alpha_{4}(1))(\varsigma_{4}-\alpha_{4}(2))}{(\varsigma_{4}-\alpha_{4}(1))(\varsigma_{4}-\alpha_{4}(2))-(\varsigma_{4}-\beta_{5}(1))(\varsigma_{4}-\beta_{5}(2))}$
	$\displaystyle=$	$\displaystyle-\frac{\mu_{5}}{\nu_{5}}\frac{\lambda_{1}}{\varsigma_{4}}\frac{(\varsigma_{4}-\alpha_{4}(1))(\varsigma_{4}-\alpha_{4}(2))}{(\beta_{5}(1)+\beta_{5}(2)-\alpha_{4}(1)-\alpha_{4}(2))\varsigma_{4}+\alpha_{4}(1)\alpha_{4}(2)-\beta_{5}(1)\beta_{5}(2)}$
	$\displaystyle>$	$\displaystyle-\frac{\mu_{5}}{\nu_{5}}\frac{\lambda_{1}}{\beta_{5}(1)+\beta_{5}(2)-\alpha_{4}(1)-\alpha_{4}(2)}\frac{\varsigma_{4}-\alpha_{4}(1)}{\varsigma_{4}}\frac{\varsigma_{4}-\alpha_{4}(2)}{\varsigma_{4}}$
	$\displaystyle>$	$\displaystyle-\frac{\mu_{5}}{\nu_{5}}\frac{\lambda_{1}}{2\lambda_{3}}\left(1-\frac{\lambda_{3}}{2\lambda_{3}}\right)^{2}=-\frac{\mu_{5}}{\nu_{5}}\frac{\lambda_{1}}{8\lambda_{3}}.$

As a result, by (118), (120) and (121), the following inequality always holds:

\displaystyle-\frac{\mu_{j}}{\nu_{j}}>-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{\lambda_{1}}{8\lambda_{3}},\quad j=2,4,

and thus (114) and (115) yield

\displaystyle m_{j}>-\frac{\mu_{j+1}}{\nu_{j+1}}\frac{\lambda_{1}}{8\lambda_{3}}\frac{\alpha_{j-1}(1)-\alpha_{j}(1)}{\alpha_{j-1}(1)}>m_{j+1}\frac{\lambda_{1}}{8\lambda_{3}}\frac{\alpha_{j-1}(1)-\alpha_{j}(1)}{\alpha_{j-1}(1)},\quad j=2,4.

(122)

We proceed to the calculation of $m_{3}/m_{4}$ . Note that $\lambda_{1}<\beta_{4}(1)<\alpha_{4}(2)<\alpha_{3}(2)=\lambda_{3}$ , then analogous to (117), we can prove $-\frac{\mu_{3}}{\nu_{3}}>-\frac{\mu_{4}}{\nu_{4}}\frac{\lambda_{1}}{\lambda_{3}},$ which together with (114) and (115) leads to

\displaystyle m_{3}\geq-\frac{\mu_{4}}{\nu_{4}}\frac{\lambda_{1}}{\lambda_{3}}\frac{\alpha_{2}(1)-\alpha_{3}(1)}{\alpha_{2}(1)}>m_{4}\frac{\lambda_{1}}{\lambda_{3}}\frac{\alpha_{2}(1)-\alpha_{3}(1)}{\alpha_{2}(1)}.

(123)

Now, combining (122) and (123), (113) derives

\displaystyle\frac{\lambda_{1}}{8\lambda_{3}}\frac{\alpha_{j}(1)-\alpha_{j+1}(1)}{\alpha_{j}(1)}\leq\sigma,\quad j=1,2,3.

This is no other than

\displaystyle\frac{\alpha_{j}(1)}{\alpha_{j+1}(1)}\leq\frac{\lambda_{1}}{\lambda_{1}-8\lambda_{3}\sigma},\quad j=1,2,3.

Consequently,

\displaystyle\lambda_{3}=\alpha_{1}(1)\leq\left(\frac{\lambda_{1}}{\lambda_{1}-8\lambda_{3}\sigma}\right)^{3}\alpha_{4}(1)<\left(\frac{\lambda_{1}}{\lambda_{1}-8\lambda_{3}\sigma}\right)^{3}\lambda_{2},

which contradicts to the definition of $\sigma$ in (113).

Case 2: $b_{3}=b_{5}=0$ . Then, (111) infers

\displaystyle m_{3}=-\frac{\mu_{3}}{\nu_{3}}\quad\mbox{and}\quad m_{5}=-\frac{\mu_{5}}{\nu_{5}}.

In this case, $\nu_{5}=\mu_{4}$ and then (111) becomes

\displaystyle\left\{\begin{array}[]{l}\mu_{5}(F_{5}(\lambda)-\lambda G_{5}(\lambda))=k_{5}\nu_{5}F_{4}(\lambda)\\ \mu_{4}(G_{5}(\lambda)-F_{4}(\lambda))=k_{5}\nu_{4}G_{4}(\lambda)\end{array}.\right.

(126)

Since $\deg(F_{5}(\lambda)-\lambda G_{5}(\lambda))=2$ and $\deg(G_{5}(\lambda)-F_{4}(\lambda))=1$ , comparing the leading coefficients of the polynomials in (126), we calculate

\displaystyle-\frac{\mu_{4}}{\nu_{4}}\left(\sum_{i=1}^{2}\beta_{5}(i)-\sum_{i=1}^{2}\alpha_{4}(i)\right)=-\frac{\mu_{5}}{\nu_{5}}\left(\sum_{i=1}^{3}\alpha_{5}(i)-\sum_{i=1}^{2}\beta_{5}(i)\right)=k_{5}.

Consequently, by (21) and (112),

\displaystyle-\frac{\mu_{4}}{\nu_{4}}=-\frac{\mu_{5}}{\nu_{5}}\frac{\sum_{i=1}^{3}\alpha_{5}(i)-\sum_{i=1}^{2}\beta_{5}(i)}{\sum_{i=1}^{2}\beta_{5}(i)-\sum_{i=1}^{2}\alpha_{4}(i)}>-\frac{\mu_{5}}{\nu_{5}}\frac{\alpha_{5}(1)}{2\lambda_{3}}>\frac{\lambda_{1}}{2\lambda_{3}}m_{5},

(127)

and hence

\displaystyle m_{3}=-\frac{\mu_{3}}{\nu_{3}}=-\frac{\mu_{4}+m_{4}\nu_{4}}{\nu_{4}-\mu_{3}}>-\frac{\mu_{4}}{\nu_{4}}-m_{4}=\frac{\lambda_{1}}{2\lambda_{3}}m_{5}-m_{4}.

Therefore, by (113),

\sigma^{2}m_{5}\geq m_{3}>\frac{\lambda_{1}}{2\lambda_{3}}m_{5}-m_{4}\geq\frac{\lambda_{1}}{2\lambda_{3}}m_{5}-\sigma m_{5},

which infers $\sigma^{2}+\sigma>\frac{\lambda_{1}}{2\lambda_{3}}$ . This is impossible due to the definition of $\sigma$ in (113).

Case 3: $b_{3}=0$ and $b_{5}>0$ . Then, $m_{3}=-\frac{\mu_{3}}{\nu_{3}}$ and a similar argument as (127) indicates

\displaystyle-\frac{\mu_{2}}{\nu_{2}}=-\frac{\mu_{3}}{\nu_{3}}\frac{\sum_{i=1}^{2}\alpha_{3}(i)-\beta_{3}(1)}{\beta_{3}(1)-\alpha_{2}(1)},

and then by (21) and (112),

	$\displaystyle m_{2}$	$\displaystyle=$	$\displaystyle-\frac{\mu_{2}}{\nu_{2}}\frac{\lambda_{3}-\alpha_{2}(1)}{\lambda_{3}}=-\frac{\mu_{3}}{\nu_{3}}\frac{\lambda_{3}-\alpha_{2}(1)}{\lambda_{3}}\frac{\sum_{i=1}^{2}\alpha_{3}(i)-\beta_{3}(1)}{\beta_{3}(1)-\alpha_{2}(1)}$
		$\displaystyle>$	$\displaystyle-\frac{\mu_{3}}{\nu_{3}}\frac{\sum_{i=1}^{2}\alpha_{3}(i)-\beta_{3}(1)}{\lambda_{3}}>m_{3}\frac{\lambda_{1}}{\lambda_{3}}.$

It contradicts to (113) that $\frac{m_{2}}{m_{3}}\leq\sigma<\frac{\lambda_{1}}{\lambda_{3}}$ .

Case 4: $b_{3}>0$ and $b_{5}=0$ . Then, (127) holds and according to (114),

\displaystyle m_{4}>-\frac{\mu_{4}}{\nu_{4}}\frac{\alpha_{3}(1)-\alpha_{4}(1)}{\alpha_{3}(1)}>m_{5}\frac{\lambda_{1}}{2\lambda_{3}}\frac{\alpha_{3}(1)-\alpha_{4}(1)}{\alpha_{3}(1)}.

By (117)–(120) and (123) in Case 1, we can demonstrate

\displaystyle m_{j}>m_{j+1}\frac{\lambda_{1}}{8\lambda_{3}}\frac{\alpha_{j-1}(1)-\alpha_{j}(1)}{\alpha_{j-1}(1)},\quad j=2,3.

The rest of the proof thus keeps the same as that in Case 1. $\Box$

Appendix B Proof of Remark 4.1

We only prove (48), as the other results are trivial. First, note that $C_{2}(m-1)>1$ and $\Lambda/\Delta\geq 2$ , so

	$\displaystyle(1+C_{2}(m-1))^{n}\rho_{m-1}$	$\displaystyle<$	$\displaystyle 2^{n}C_{2}^{n}(m-1)\rho_{m-1}=2^{n}\left(\frac{2^{2n(n+1)^{2}}\Lambda^{n(n+1)}}{\Delta^{n^{2}}\varepsilon^{n}}\right)\varepsilon^{n+1}$
		$\displaystyle<$	$\displaystyle\frac{2^{2(n+1)^{3}}\Lambda^{(n+1)^{2}}}{2\Delta^{n^{2}+n}}\varepsilon<\frac{\Delta}{2n}<\frac{\Delta}{2}.$

Let $m>2$ . Since $2\varepsilon<1$ , for each $j\in[1,m-2]$ ,

$\displaystyle(1+C_{2}(j))^{n}\rho_{j}$	$\displaystyle<$	$\displaystyle 2^{n}C_{2}^{n}(j)\rho_{j}=2^{n}\left(\frac{2^{2n(n+1)^{2}}\Lambda^{n(n+1)}}{\Delta^{n^{2}}\varepsilon^{n(n+1)^{m-j-1}}}\right)\varepsilon^{(n+1)^{m-j}}$
	$\displaystyle=$	$\displaystyle 2^{n}\left(\frac{2^{2n(n+1)^{2}}\Lambda^{n(n+1)}}{\Delta^{n^{2}}}\right)\left(\varepsilon^{(n+1)^{m-j-2}}\right)^{n+1}$
	$\displaystyle<$	$\displaystyle\frac{2^{2n(n+1)^{2}}\Lambda^{n(n+1)}}{\Delta^{n^{2}}}\varepsilon^{(n+1)^{m-j-2}}=C_{2}^{n}(j+1)\rho_{j+1}<(1+C_{2}(j+1))^{n}\rho_{j+1}.$

When $m>3$ , for each $j\in[2,m-2]$ ,

$\displaystyle\frac{n(1+C_{2}(j-1))^{n}\rho_{j-1}}{\Delta}$	$\displaystyle<$	$\displaystyle\frac{n2^{n}C_{2}^{n}(j-1)\rho_{j-1}}{\Delta}=n2^{n}\left(\frac{2^{2n(n+1)^{2}}\Lambda^{n(n+1)}}{\Delta^{n^{2}+1}}\right)\varepsilon^{(n+1)^{m-j}}$
	$\displaystyle=$	$\displaystyle n2^{n}\left(\frac{2^{2n(n+1)^{2}}\Lambda^{n(n+1)}}{\Delta^{n^{2}+1}}\right)\varepsilon^{(n+1)^{m-j}-(n+1)^{m-j-1}}\rho_{j+1}$
	$\displaystyle<$	$\displaystyle n2^{n}\left(\frac{2^{2n(n+1)^{2}}\Lambda^{n(n+1)}\varepsilon}{\Delta^{n^{2}+1}}\right)\rho_{j+1}$
	$\displaystyle<$	$\displaystyle\frac{\Delta^{n}}{2^{(n+1)^{3}}\Lambda^{n+1}}\rho_{j+1}\leq\frac{1}{4}\left(\frac{\Delta^{n}}{2^{(n+1)^{2}}\Lambda^{n+1}}\right)\rho_{j+1}.$

Hence, (48) follows immediately.

Appendix C Proofs of Lemmas 4.2–4.3

Proof of Lemma 4.2.

Let $p\geq 1$ . We first take a number $\lambda^{*}$ satisfying

\displaystyle\lambda^{*}>\alpha_{p}\quad\mbox{and}\quad-\frac{\mu}{\nu}\frac{F(\lambda^{*})}{\lambda^{*}G(\lambda^{*})}=m^{*}.

(128)

Such $\lambda^{*}$ indeed exists because of (49)–(51), which yield $-\frac{\mu}{\nu}\frac{F(\alpha_{p})}{\alpha_{p}G(\alpha_{p})}=0$ and

\displaystyle\lim_{\lambda\rightarrow+\infty}-\frac{\mu}{\nu}\frac{F(\lambda)}{\lambda G(\lambda)}=-\frac{\mu}{\nu}>CM>m^{*}.

Then, let

\displaystyle\left\{\begin{array}[]{l}F_{0}(\lambda)=\frac{\mu F(\lambda)+m^{*}\nu\lambda G(\lambda)}{(\mu+m^{*}\nu)(\lambda-\lambda^{*})}\\ b^{*}=-\frac{\mu+m^{*}\nu}{\nu}\frac{F_{0}(\lambda^{*})}{G(\lambda^{*})}\\ \mu_{0}=\nu\frac{G(\lambda^{*})}{F_{0}(\lambda^{*})}\\ G_{0}(\lambda)=\frac{\nu G(\lambda)-\mu_{0}F_{0}(\lambda)}{(\nu-\mu_{0})(\lambda-\lambda^{*})}\\ \nu_{0}=\frac{\mu_{0}-\nu}{b^{*}}\end{array},\right.

(134)

we shall show that all the above defined numbers and polynomials fulfill our requirements.

Observe that $\mu\nu<0$ and by (50)–(51),

\alpha_{p}\geq\lambda_{p}>\lambda_{p-1}+(1+C_{2}(p-1))^{n}\rho_{p-1}>\beta_{p-1}>\alpha_{p-1}>\cdots>\beta_{1}>\alpha_{1}>0,

then $\lambda G(\lambda)\ll F(\lambda)$ . As a result, by (128),

\displaystyle-\frac{\mu+m^{*}\nu}{\nu}=-\frac{\mu}{\nu}\left(1-\frac{F(\lambda^{*})}{\lambda^{*}G(\lambda^{*})}\right)>-\frac{\mu}{\nu}\left(1-\frac{\prod_{i=1}^{p}(\lambda^{*}-\alpha_{i})}{(\lambda^{*}-\alpha_{1})\prod_{i=1}^{p-1}(\lambda^{*}-\alpha_{i+1})}\right)=0.

(135)

Since (128) indicates $(\lambda-\lambda^{*})|(\mu F(\lambda)+\lambda m^{*}\nu G(\lambda))$ , (135) means $F_{0}(\lambda)$ is a well-defined monic polynomial of degree $p-1$ . Recall that $\lambda G(\lambda)\ll F(\lambda)$ and $|\mu|>m^{*}|\nu|$ , applying Lemma 4.1 to polynomials $\mu F(\lambda)$ and $-m^{*}\nu\lambda G(\lambda)$ shows

\displaystyle\alpha_{i}^{\prime}\in(\alpha_{i},\beta_{i}),\quad i=1,\ldots,p-1.

(136)

Therefore,

\displaystyle\lambda F_{0}(\lambda)\ll F(\lambda)\quad\mbox{and}\quad F_{0}(\lambda)\ll G(\lambda),

(137)

which give $b^{*}=-\frac{\mu+m^{*}\nu}{\nu}\left(\frac{F_{0}(\lambda^{*})}{G(\lambda^{*})}\right)>0,$

\displaystyle\mu_{0}\nu=\nu^{2}\left(\frac{G(\lambda^{*})}{F_{0}(\lambda^{*})}\right)>0\quad\mbox{and}\quad|\mu_{0}|=|\nu|\left(\frac{G(\lambda^{*})}{F_{0}(\lambda^{*})}\right)<|\nu|.

(138)

Consequently, by (134) and (135),

\displaystyle-\frac{\mu_{0}}{\nu_{0}}=-\frac{\mu+m^{*}\nu}{\nu-\mu_{0}}=-\frac{\mu+m^{*}\nu}{\nu}\frac{1}{1-\mu_{0}/\nu}>-\frac{\mu}{\nu}-m^{*}.

(139)

So far, we have verified that $\lambda^{*},b^{*},\mu_{0}$ and $\nu_{0}$ satisfy the condition of Lemma 4.2. For these numbers, the first equality of (55) follows directly from (134). Considering (136), if the second inequality of (52) holds when $p>2$ , then $F_{0}(\lambda)$ in (134) will be the exact polynomial desired for both statements (i) and (ii).

Next, we check $G_{0}(\lambda)$ in (134). The definition of $\mu_{0}$ infers $(\lambda-\lambda^{*})\big{|}(\nu G(\lambda)-\mu_{0}F_{0}(\lambda))$ , hence (138) implies that $G_{0}(\lambda)$ is a well-defined monic polynomial. We discuss this part by considering two cases.
(i) $p=2$ . In this case, $G_{0}(\lambda)=1$ fulfills the second equality of (55) by (134).
(ii) $p>2$ . Taking account to (137) and (138), we apply Lemma 4.1 to polynomials $\nu G(\lambda)$ and $\mu_{0}F_{0}(\lambda)$ , then

\displaystyle\beta_{i}^{\prime}\in(\beta_{i},\alpha_{i+1}^{\prime}),\quad i=1,\ldots,p-2,

(140)

so the roots of $G_{0}(\lambda)$ are distinct. Further, we can verify the second equality of (55) from (134) again.

Now, it remains to show the second inequality of (52) for $i=1,\ldots,p-2$ when $p>2$ . Combining (136) and (140), it gives $\lambda G_{0}(\lambda)\ll F_{0}(\lambda)$ . Fix an index $i\in[1,p-2]$ , we compute

\displaystyle\prod_{j=1}^{p-1}(\beta_{i}^{\prime}-\alpha_{j}^{\prime})=F_{0}(\beta_{i}^{\prime})=\frac{1}{\mu_{0}}(\nu G(\beta_{i}^{\prime})-b^{*}(\lambda^{*}-\beta_{i}^{\prime})\nu_{0}G_{0}(\beta_{i}^{\prime}))=\frac{F_{0}(\lambda^{*})}{G(\lambda^{*})}\prod_{j=1}^{p-1}(\beta_{i}^{\prime}-\beta_{j}).

(141)

Note that $\frac{F_{0}(\lambda^{*})}{G(\lambda^{*})}>1$ by (137) and $\prod_{j>i+1}\frac{\beta_{i}^{\prime}-\beta_{j}}{\beta_{i}^{\prime}-\alpha_{j}^{\prime}}\geq 1$ by (136) and (140), (141) immediately leads to

\displaystyle\frac{\beta_{i}^{\prime}-\alpha_{i}^{\prime}}{\beta_{i}^{\prime}-\beta_{i}}>\frac{\beta_{i}^{\prime}-\beta_{i+1}}{\beta_{i}^{\prime}-\alpha_{i+1}^{\prime}}\prod_{j<i}\frac{\beta_{i}^{\prime}-\beta_{j}}{\beta_{i}^{\prime}-\alpha_{j}^{\prime}}.

(142)

We are going to employ Lemma 4.1 to estimate term $\frac{\beta_{i}^{\prime}-\beta_{i+1}}{\beta_{i}^{\prime}-\alpha_{i+1}^{\prime}}$ in (142). For this, denote $\eta=\frac{1}{2}\min_{j\in[1,p-1]}(\beta_{j}-\alpha_{j})$ . Recall that $C_{1},\rho_{1}<1$ , then by (51), for each $j\in[1,p-1]$ ,

C_{1}^{n}\rho_{1}\leq\min\{1,C_{1}^{n}\rho_{j}\}<\min\{1,\beta_{j}-\alpha_{j}\},

which means $C_{1}^{n}\rho_{1}/2<\min\{1,\eta\}$ . As a result, by (44),

$\displaystyle\left\|\frac{m^{*}\nu}{\mu}\right\|$	$\displaystyle<$	$\displaystyle\frac{m^{*}}{CM}<\left(\frac{C_{1}^{n}\rho_{1}}{2}\right)^{2n}\frac{1}{2+2\Lambda^{n}}<\frac{\min\{1,\eta^{2p}\}}{2+2\Lambda^{p}}$
	$\displaystyle<$	$\displaystyle\frac{\min\left\{1,\left(\frac{1}{2}\min_{1\leqslant l\leqslant p-1}\left(\alpha_{l+1}-\alpha_{l}\right)\right)^{p}\right\}\min\{1,\eta^{p}\}}{2+2\Lambda^{p}}$
	$\displaystyle<$	$\displaystyle\frac{\min\left\{1,\left(\min_{1\leqslant l\leqslant p-1}(\alpha_{l+1}-\alpha_{l})-\eta\right)^{p}\}\min\{1,\eta^{p}\right\}}{2+2\max_{j\in[1,p]}\left\|\alpha_{j}\prod_{l=1}^{p-1}(\alpha_{j}-\beta_{l})\right\|}.$

So, by applying Lemma 4.1 to polynomials $\mu F(\lambda)$ and $-m^{*}\nu\lambda G(\lambda)$ , we conclude

\displaystyle\max_{j\in[1,p-1]}(\alpha_{j}^{\prime}-\alpha_{j})<\eta=\frac{1}{2}\min_{j\in[1,p-1]}(\beta_{j}-\alpha_{j}).

(143)

Then, $\beta_{i+1}-\alpha_{i+1}^{\prime}\geq\frac{\beta_{i+1}-\alpha_{i+1}}{2}$ , which together with (51) indicates

	$\displaystyle\frac{\beta_{i}^{\prime}-\beta_{i+1}}{\beta_{i}^{\prime}-\alpha_{i+1}^{\prime}}$	$\displaystyle=$	$\displaystyle 1+\frac{\beta_{i+1}-\alpha_{i+1}^{\prime}}{\alpha_{i+1}^{\prime}-\beta_{i}^{\prime}}\geq 1+\frac{\beta_{i+1}-\alpha_{i+1}}{2(\beta_{i+1}-\beta_{i}^{\prime})}$		(144)
		$\displaystyle\geq$	$\displaystyle 1+\frac{C_{1}^{n}\rho_{i+1}}{2(\lambda_{i+1}-\beta_{i}^{\prime})}\geq 1+\frac{\Delta^{n}}{2^{(n+1)^{2}}\Lambda^{n+1}}\rho_{i+1}.$		(144)

Next, we deal with $\prod_{j<i}\frac{\beta_{i}^{\prime}-\beta_{j}}{\beta_{i}^{\prime}-\alpha_{j}^{\prime}}$ in (142) for $i\geq 2$ . As a matter of fact, by (51), for any $j<i$ ,

	$\displaystyle\frac{\beta_{i}^{\prime}-\beta_{j}}{\beta_{i}^{\prime}-\alpha_{j}^{\prime}}$	$\displaystyle>$	$\displaystyle\frac{\beta_{i}^{\prime}-\beta_{j}}{\beta_{i}^{\prime}-\alpha_{j}}=1-\frac{\beta_{j}-\alpha_{j}}{\beta_{i}^{\prime}-\alpha_{j}}$		(145)
		$\displaystyle>$	$\displaystyle 1-\frac{\beta_{j}-\lambda_{j}}{\beta_{i}^{\prime}-\lambda_{j}}>1-\frac{\beta_{j}-\lambda_{j}}{\lambda_{i}-\lambda_{j}}\geq 1-\frac{(1+C_{2}(j))^{n}\rho_{j}}{\Delta}.$		(145)

Now, if $i\geq 2$ , substituting (144) and (145) into (142) yields

$\displaystyle\frac{\beta_{i}^{\prime}-\alpha_{i}^{\prime}}{\beta_{i}^{\prime}-\beta_{i}}$	$\displaystyle>$	$\displaystyle\left(1+\frac{\Delta^{n}}{2^{(n+1)^{2}}\Lambda^{n+1}}\rho_{i+1}\right)\prod_{j<i}\left(1-\frac{(1+C_{2}(j))^{n}\rho_{j}}{\Delta}\right)$	(146)
	$\displaystyle\geq$	$\displaystyle\left(1+\frac{\Delta^{n}}{2^{(n+1)^{2}}\Lambda^{n+1}}\rho_{i+1}\right)\left(1-\frac{(1+C_{2}(i-1))^{n}\rho_{i-1}}{\Delta}\right)^{i-1}$
	$\displaystyle>$	$\displaystyle\left(1+\frac{\Delta^{n}}{2^{(n+1)^{2}}\Lambda^{n+1}}\rho_{i+1}\right)\left(1-\frac{(1+C_{2}(i-1))^{n}\rho_{i-1}}{\Delta}\right)^{n},$

where the second inequality follows from (48) in Remark 4.1. Since the Bernoulli inequality gives

\left(1-\frac{(1+C_{2}(i-1))^{n}\rho_{i-1}}{\Delta}\right)^{n}>1-\frac{n(1+C_{2}(i-1))^{n}\rho_{i-1}}{\Delta},

(146) reduces to

	$\displaystyle\frac{\beta_{i}^{\prime}-\alpha_{i}^{\prime}}{\beta_{i}^{\prime}-\beta_{i}}$	$\displaystyle>$	$\displaystyle\left(1+\frac{\Delta^{n}}{2^{(n+1)^{2}}\Lambda^{n+1}}\rho_{i+1}\right)\left(1-\frac{n(1+C_{2}(i-1))^{n}\rho_{i-1}}{\Delta}\right)$		(147)
		$\displaystyle>$	$\displaystyle 1+\frac{\Delta^{n}}{2^{(n+1)^{2}+1}\Lambda^{n+1}}\rho_{i+1}.$		(147)

As for $i=1$ , it is trivial that $\frac{\beta_{i}^{\prime}-\alpha_{i}^{\prime}}{\beta_{i}^{\prime}-\beta_{i}}>1+\frac{\Delta^{n}}{2^{(n+1)^{2}}\Lambda^{n+1}}\rho_{i+1}.$ So, both the two cases lead to (147). Hence,

\displaystyle\frac{\beta_{i}^{\prime}-\alpha_{i}^{\prime}}{\beta_{i}-\alpha_{i}}<\frac{\beta_{i}^{\prime}-\alpha_{i}^{\prime}}{\beta_{i}-\alpha_{i}^{\prime}}=1+\frac{1}{\frac{\beta_{i}^{\prime}-\alpha_{i}^{\prime}}{\beta_{i}^{\prime}-\beta_{i}}-1}\leq 1+\frac{2^{(n+1)^{2}+1}\Lambda^{n+1}}{\Delta^{n}\rho_{i+1}}<C_{2}(i).

(148)

On the other hand, in view of (143),

\displaystyle\frac{\beta_{i}^{\prime}-\alpha_{i}^{\prime}}{\beta_{i}-\alpha_{i}}>\frac{\beta_{i}-\alpha_{i}^{\prime}}{\beta_{i}-\alpha_{i}}\geq\frac{1}{2}\geq C_{1}.

(149)

Therefore, (52) is a direct result of (148) and (149). $\Box$

Proof of Lemma 4.3.

(i) Let $p\geq 2$ and set

\displaystyle\left\{\begin{array}[]{l}m^{*}=-\frac{\mu}{\nu}\frac{F(\lambda^{*})}{\lambda^{*}G(\lambda^{*})}\\ F_{0}(\lambda)=\frac{\mu F(\lambda)+m^{*}\nu\lambda G(\lambda)}{\mu+m^{*}\nu}\end{array}.\right.

(152)

By (49) and (51), it is clear that $m^{*}>0$ . Observe that

\displaystyle\left|\frac{m^{*}\nu}{\mu}\right|=\frac{\prod_{j=1}^{p}(\lambda^{*}-\alpha_{j})}{\lambda^{*}\prod_{j=1}^{p-1}(\lambda^{*}-\beta_{j})}<\frac{\prod_{j=1}^{p}(\lambda^{*}-\alpha_{j})}{(\lambda^{*}-\alpha_{1})\prod_{j=1}^{p-1}(\lambda^{*}-\alpha_{j+1})}=1,

then $F_{0}(\lambda)$ is a well-defined monic polynomial of $\deg(F_{0}(\lambda))=\deg(F(\lambda))=p.$ Now, (51) indicates $\lambda G(\lambda)\ll F(\lambda)$ , by applying Lemma 4.1 to polynomials $\mu F(\lambda)$ and $-m^{*}\nu\lambda G(\lambda)$ , it follows that the first $p-1$ roots of $F_{0}(\lambda)$ satisfy

\displaystyle\alpha_{j}^{\prime}\in(\alpha_{j},\beta_{j}),\quad j=1,\ldots,p-1.

(153)

Moreover, the definition of $m^{*}$ in (152) shows

(\lambda-\lambda^{*})|(\mu F(\lambda)+m^{*}\nu\lambda G(\lambda)),

which yields $\alpha_{p}^{\prime}=\lambda^{*}.$ Hence, $F(\lambda)\ll F_{0}(\lambda)$ .

Next, let

\displaystyle\left\{\begin{array}[]{l}\mu_{0}=\tau\nu\\ b^{*}=-\frac{\mu+m^{*}\nu}{\mu_{0}}\\ \nu_{0}=\frac{\mu_{0}-\nu}{b^{*}}\\ G_{0}(\lambda)=\frac{\nu(\lambda-\lambda^{*})G(\lambda)-\mu_{0}F_{0}(\lambda)}{b^{*}\nu_{0}(\lambda^{*}-\lambda)}\end{array},\right.

(158)

where

\displaystyle\tau=\left\{\begin{array}[]{l}v_{1},\quad p>2\\ v_{2},\quad p=2\\ \end{array},\right.

(161)

with

$\displaystyle v_{1}$	$\displaystyle=$	$\displaystyle\frac{1}{2}\frac{\min\{1,\left(\min_{1\leqslant l\leqslant p-2}(\beta_{l+1}-\beta_{l})-\eta_{1}\right)^{p-1}\}\min\{1,\eta_{1}^{p-1}\}}{2+2\max_{j\in[1,p-1]}\left\|\prod_{h=1}^{p-1}(\beta_{j}-\alpha_{h}^{\prime})\right\|},$
$\displaystyle v_{2}$	$\displaystyle=$	$\displaystyle\frac{1}{4}\min\left\{\frac{\eta_{2}}{\beta_{1}-\alpha_{1}^{\prime}},1\right\},$
$\displaystyle\eta_{1}$	$\displaystyle=$	$\displaystyle\min\left\{\frac{\min_{1\leqslant l\leqslant p-2}(\beta_{l+1}-\beta_{l})}{4},\min_{1\leqslant l\leqslant p-1}(\beta_{l}-\alpha_{l})\right\},$
$\displaystyle\eta_{2}$	$\displaystyle=$	$\displaystyle\min\left\{\beta_{1}-\alpha_{1},\frac{1}{2}\right\}.$

Therefore,

\displaystyle b^{*}=-\frac{\mu+m^{*}\nu}{\mu_{0}}=-\frac{\mu}{\tau\nu}\left(1-\frac{F(\lambda^{*})}{\lambda^{*}G(\lambda^{*})}\right)>0.

Note that $\tau\in(0,1)$ , then (158) gives $\mu_{0}/\nu\in(0,1)$ . As a result, by $\lambda G(\lambda)\ll F(\lambda)$ ,

\displaystyle-\frac{\mu_{0}}{\nu_{0}}=-\frac{\mu+m^{*}\nu}{\nu-\mu_{0}}>-\frac{\mu}{\nu}\left(1-\frac{F(\lambda^{*})}{\lambda^{*}G(\lambda^{*})}\right)>-\frac{\mu}{\nu}\frac{\lambda^{*}G(\lambda^{*})-(\lambda^{*}-\alpha_{1})G(\lambda^{*})}{\lambda^{*}G(\lambda^{*})}>-\frac{\lambda_{1}}{\Lambda}\frac{\mu}{\nu}>0.

So, $G_{0}(\lambda)$ is a well-defined monic polynomial of degree $p-1$ . Since (153) means that $\frac{F_{0}(\lambda)}{(\lambda-\lambda^{*})}\ll G(\lambda)$ , by applying Lemma 4.1 to polynomials $\nu G(\lambda)$ and $\mu_{0}\frac{F_{0}(\lambda)}{(\lambda-\lambda^{*})}$ , we deduce $\beta_{p-1}^{\prime}>\beta_{p-1}$ and

\displaystyle\beta_{j}^{\prime}\in(\beta_{j},\alpha_{j+1}^{\prime}),\quad j=1,\ldots,p-2.

(162)

Observe that plugging (152) and (158) into (58) immediately shows the validity of (58).

At last, we show the second inequality of (52) for $i=1,\ldots,p-1$ . Fix an index $i\in[1,p-1]$ . By plugging $\lambda=\beta_{i}$ into (58), we obtain $\mu F(\beta_{i})=-b^{*}\mu_{0}F_{0}(\beta_{i}),$ which equals to

\displaystyle\prod_{j=1}^{p}(\beta_{i}-\alpha_{j}^{\prime})=\frac{1}{1-\frac{F(\lambda^{*})}{\lambda^{*}G(\lambda^{*})}}\prod_{j=1}^{p}(\beta_{i}-\alpha_{j}).

(163)

Note that $\frac{\beta_{i}-\alpha_{j}}{\beta_{i}-\alpha_{j}^{\prime}}>1$ for all $j<i$ because of (153). As for $j\in[i+1,p-1]$ , by (51) and (153),

\displaystyle\frac{\beta_{i}-\alpha_{j}}{\beta_{i}-\alpha_{j}^{\prime}}=1-\frac{\alpha_{j}^{\prime}-\alpha_{j}}{\alpha_{j}^{\prime}-\beta_{i}}\geq 1-\frac{\beta_{j}-\lambda_{j}}{\alpha_{j}-\beta_{i}}\geq 1-\frac{\beta_{j}-\lambda_{j}}{\lambda_{i+1}-\beta_{i}},

which together with (48) yields

\displaystyle\frac{\beta_{i}-\alpha_{j}}{\beta_{i}-\alpha_{j}^{\prime}}\geq 1-\frac{\beta_{j}-\lambda_{j}}{(\lambda_{i+1}-\lambda_{i})-(\beta_{i}-\lambda_{i})}\geq 1-\frac{(1+C_{2}(j))^{n}\rho_{j}}{2\Delta-(1+C_{2}(i))^{n}\rho_{i}}>\frac{1}{2}.

Furthermore, since $\alpha_{p}\in[\lambda_{p},\lambda^{*})\subset[\lambda_{p},\Lambda)$ , (42), (48) and (51) lead to

\displaystyle\frac{\beta_{i}-\alpha_{p}}{\beta_{i}-\alpha_{p}^{\prime}}\geq\frac{\lambda_{i+1}-\beta_{i}}{\Lambda}\geq\frac{\Delta-(1+C_{2}^{n}(i))\rho_{i}}{\Lambda}>\frac{\Delta}{2\Lambda}.

Consequently, it follows from (162) and (163) that

\displaystyle\frac{\beta_{i}^{\prime}-\alpha_{i}^{\prime}}{\beta_{i}-\alpha_{i}}>\frac{\beta_{i}-\alpha_{i}^{\prime}}{\beta_{i}-\alpha_{i}}=\frac{1}{1-\frac{F(\lambda^{*})}{\lambda^{*}G(\lambda^{*})}}\prod_{j\not=i}\frac{\beta_{i}-\alpha_{j}}{\beta_{i}-\alpha_{j}^{\prime}}>\frac{1}{2^{n}}\frac{\Delta}{2\Lambda}=C_{1}.

Now, we prove $\beta_{i}^{\prime}-\alpha_{i}^{\prime}\leq C_{2}(i)(\beta_{i}-\alpha_{i})$ for each $i\in[1,p-1]$ . If $p>2$ ,

\displaystyle\left|\frac{\mu_{0}}{\nu}\right|=\tau=v_{1}<\frac{\min\{1,\left(\min_{1\leqslant l\leqslant p-2}(\beta_{l+1}-\beta_{l})-\eta_{1}\right)^{p-1}\}\min\{1,\eta_{1}^{p-1}\}}{2+2\max_{j\in[1,p-1]}\left|\prod_{h=1}^{p-1}(\beta_{j}-\alpha_{h}^{\prime})\right|}.

By applying Lemma 4.1 to polynomials $\nu G(\lambda)$ and $\mu_{0}\frac{F_{0}(\lambda)}{(\lambda-\lambda^{*})}$ , it infers

\displaystyle\max_{i\in[1,p-1]}(\beta_{i}^{\prime}-\beta_{i})<\eta_{1}\leq\min_{1\leqslant i\leqslant p-1}(\beta_{i}-\alpha_{i}).

When $p=2$ ,

\displaystyle\left|\frac{\mu_{0}}{\nu}\right|=\tau<\frac{1}{2}\min\left\{\frac{\eta_{2}}{\beta_{1}-\alpha_{1}^{\prime}},1\right\}

and applying Lemma 4.1 to polynomials $\nu G(\lambda)$ and $\mu_{0}\frac{F_{0}(\lambda)}{(\lambda-\lambda^{*})}$ shows

\displaystyle\beta_{1}^{\prime}-\beta_{1}<\eta_{2}\leq\beta_{1}-\alpha_{1}.

So both cases result in

\displaystyle\max_{i\in[1,p-1]}(\beta_{i}^{\prime}-\beta_{i})<\min_{1\leqslant i\leqslant p-1}(\beta_{i}-\alpha_{i}),

which implies that $\beta_{i}^{\prime}-\alpha_{i}<2(\beta_{i}-\alpha_{i})$ for each $i\in[1,p-1]$ . Then,

\displaystyle\beta_{i}^{\prime}-\alpha_{i}^{\prime}<\beta_{i}^{\prime}-\alpha_{i}<2(\beta_{i}-\alpha_{i})\leq C_{2}(i)(\beta_{i}-\alpha_{i}).

This finishes the proof of statement (i).

(ii) Let $m^{*}=-\frac{\mu}{\nu}\frac{\lambda^{*}-\alpha_{1}}{\lambda^{*}}$ , $\mu_{0}=\frac{1}{2}\nu$ , $b^{*}=-\frac{\mu}{\nu}\frac{2\alpha_{1}}{\lambda^{*}}$ and $\nu_{0}=\frac{1}{4}\frac{\nu^{2}}{\mu}\frac{\lambda^{*}}{\alpha_{1}}$ , we can directly compute that $F_{0}(\lambda)=\lambda-\lambda^{*}$ and $G_{0}(\lambda)=1$ satisfy equation (58). $\Box$

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Inverse Eigenvalue Problem For Mass-Spring-Inerter Systems 111The work was supported by the National Natural Science Foundation of China under Grants 11925109 and 11688101.

Abstract

1 Introduction

2 Main Result

Theorem 2.1.

Remark 2.1.

Example 2.1.

3 Proof of the necessity of Theorem 2.1.

Remark 3.1.

Definition 3.1.

Lemma 3.1.

Proof.

Lemma 3.2.

Proof.

Lemma 3.3.

Proof.

4 Proof of sufficiency of Theorem 2.1.

Proposition 4.1.

Example 4.1.

Lemma 4.1.

Proof.

Remark 4.1.

Lemma 4.2.

Lemma 4.3.

Lemma 4.4.

Proof.

Proof of Proposition 4.1.

5 Concluding remarks.

Appendix A Proof of Example 2.1

Appendix B Proof of Remark 4.1

Appendix C Proofs of Lemmas 4.2–4.3

Proof of Lemma 4.2.

Proof of Lemma 4.3.

References

Inverse Eigenvalue Problem For Mass-Spring-Inerter Systems ¹¹1The work was supported by the National Natural Science Foundation of China under Grants 11925109 and 11688101.