Eigenvalue assignment of second-order singular systems by acceleration–velocity–displacement feedback

Abstract

The eigenvalue assignment problem of second-order singular system is investigated by using acceleration–velocity–displacement feedback. The conditions are established to ensure the solvability of partial eigenvalue assignment problem of second-order singular system. The derived results are extended to complete eigenvalue assignment problem of second-order singular system. The presented solvability conditions are easily tested. Then, the methods are given to solve the eigenvalue assignment problem of second-order singular systems. Finally, several examples are given to validate our results and algorithms.

Proof of some lemmas

In this part, we give the proofs of Lemma 2.2, Lemma 2.3, Lemma 2.4 and Lemma 2.5.

1.1 Proof of Lemma 2.2

(1) It is easily seen that a base of ${\mathcal {N}}({\mathcal {E}}_2)$ can be taken as the columns of

$$\begin{aligned} N({\mathcal {E}}_2)=\left[ \begin{array}{cc} -C_{13} &{}\quad -C_{14} \\ 0 &{}\quad 0 \\ 0 &{}\quad 0 \\ I &{}\quad 0 \\ 0 &{}\quad I \\ \end{array} \right] . \end{aligned}$$

Then, we have

$$\begin{aligned} \text{ rank }[{\mathcal {E}}_2, {\mathcal {A}}_2 N({\mathcal {E}}_2), {\mathcal {B}}_2]&=\text{ rank }\left[ \begin{array}{cccccccc} I_{n_1} &{}\quad C_{11} &{} \quad 0 &{}\quad C_{13} &{}\quad C_{14} &{}\quad 0 &{}\quad -K_{13} &{} \quad B_1 \\ 0 &{}\quad 0 &{}\quad I_{n_2} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad -K_{23} &{}\quad B_2 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{} -K_{33} &{}\quad B_3 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad -I_{n_3} &{}\quad 0 &{} \quad B_4 \\ 0 &{}\quad I_{n_1} &{} \quad 0 &{} \quad 0 &{}\quad 0 &{}\quad -C_{14} &{} \quad -C_{13} &{} \quad 0 \\ \end{array} \right] \\&=\text{ rank }\left[ \begin{array}{cccccccc} I_{n_1} &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad I_{n_1} &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad -C_{14} &{} \quad -C_{13} &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad I_{n_2} &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad -K_{23} &{} \quad B_2 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad -K_{33} &{} \quad B_3 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad -I_{n_3} &{} \quad 0 &{} \quad B_4 \\ \end{array} \right] \\&=\text{ rank }\left[ \begin{array}{cccccccc} I_{n_1} &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad I_{n_1} &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad C_{14} &{} \quad C_{13} \\ 0 &{} \quad 0 &{} \quad I_{n_2} &{} \quad 0 &{} \quad 0 &{} \quad B_2 &{} \quad 0 &{} \quad K_{23} \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad B_3 &{} \quad 0&{} \quad K_{33} \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad B_4 &{} \quad I_{n_3} &{} \quad 0 \\ \end{array} \right] \\&=\text{ rank }\left[ \begin{array}{cccccccc} I_{n_1} &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad I_{n_1} &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad -C_{14}B_4 &{} \quad C_{14} &{} \quad C_{13} \\ 0 &{} \quad 0 &{} \quad I_{n_2} &{} \quad 0 &{} \quad 0 &{} \quad B_2 &{} \quad 0 &{} \quad K_{23} \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad B_3 &{} \quad 0&{} \quad K_{33} \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad I_{n_3} &{} \quad 0 \\ \end{array} \right] . \end{aligned}$$

Note that $B_3$ has full row rank. Then $[{\mathcal {E}}_2, {\mathcal {A}}_2 N({\mathcal {E}}_2), {\mathcal {B}}_2]$ has full row rank.

(2) Denote ${\widehat{I}}_{n-n_1}^T=[0_{(n-n_1)\times n_1}, I_{n-n_1}]$. Clearly, $N({\widetilde{M}})={\widehat{I}}_{n-n_1}$. Then we have ${\widetilde{Z}}^T {\widetilde{C}}-{\widetilde{Z}}^T {\widetilde{C}} {\widetilde{I}}_{n_1}{\widetilde{I}}_{n_1}^T={\widetilde{Z}}^T {\widetilde{C}} (I_n-{\widetilde{I}}_{n_1}{\widetilde{I}}_{n_1}^T)={\widetilde{Z}}^T {\widetilde{C}} [0, {\widehat{I}}_{n-n_1}]=0$. Thus,

$$\begin{aligned} \left[ {\widetilde{Z}}^T, -({\widetilde{I}}_{n_1}^T {\widetilde{C}}^T {\widetilde{Z}})^T\right] {\mathcal {E}}_2=[{\widetilde{Z}}^T {\widetilde{I}}_{n_1}, {\widetilde{Z}}^T {\widetilde{C}}-{\widetilde{Z}}^T {\widetilde{C}} {\widetilde{I}}_{n_1}{\widetilde{I}}_{n_1}^T]=0. \end{aligned}$$

Let ${\widetilde{C}}=\left[ {\widetilde{C}}_{1}, {\widetilde{C}}_{2}\right] , \ {\widetilde{C}}_{1}\in {{\mathbb {R}}}^{n\times n_1}$. Then, we have

$$\begin{aligned} \text{ rank }({\mathcal {E}}_2)= & {} \text{ rank }\left[ \begin{array}{ccc} {\widetilde{I}}_{n_1} &{}\quad {\widetilde{C}}_{1} &{} \quad {\widetilde{C}}_{2} \\ 0 &{}\quad I_{n_1} &{}\quad 0 \\ \end{array} \right] \\= & {} \text{ rank }\left[ \begin{array}{ccc} {\widetilde{I}}_{n_1} &{}\quad {\widetilde{C}}_{2} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad I_{n_1} \\ \end{array} \right] = n_1+\text{ rank }[{\widetilde{M}}, {\widetilde{C}} N({\widetilde{M}})]. \end{aligned}$$

Thus, $\dim ({\mathcal {N}}({\mathcal {E}}_2^T))=n-\text{ rank }[{\widetilde{M}}, {\widetilde{C}} N({\widetilde{M}})]$. On the other hand,

$$\begin{aligned} \text{ rank }(\left[ {\widetilde{Z}}^T, -({\widetilde{I}}_{n_1}^T {\widetilde{C}}^T {\widetilde{Z}})^T\right] ^T)=\text{ rank }({\widetilde{Z}})=n-\text{ rank }([{\widetilde{M}}, {\widetilde{C}} N({\widetilde{M}})]). \end{aligned}$$

Thus, $N({\mathcal {E}}_2^T)=\left[ {\widetilde{Z}}^T, -({\widetilde{I}}_{n_1}^T {\widetilde{C}}^T {\widetilde{Z}})^T\right] ^T$.

(3) Observe that

$$\begin{aligned} \text{ rank }[{\mathcal {E}}_2, {\mathcal {B}}_2]&=\text{ rank }\left[ \begin{array}{cccccc} I_{n_1} &{} \quad C_{11} &{} \quad 0 &{} \quad C_{13} &{} \quad C_{14} &{} \quad B_1 \\ 0 &{} \quad 0 &{} \quad I_{n_2} &{} \quad 0 &{} \quad 0 &{} \quad B_2 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad B_3 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad B_4 \\ 0 &{} \quad I_{n_1} &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ \end{array} \right] \\&= \text{ rank }\left[ \begin{array}{cccccc} I_{n_1} &{} \quad 0 &{} \quad C_{13} &{} \quad C_{14} &{} \quad B_1 &{} \quad C_{11}\\ 0 &{} \quad I_{n_2} &{} \quad 0 &{} \quad 0 &{} \quad B_2 &{} \quad 0\\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad B_3 &{} \quad 0\\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad B_4 &{} \quad 0\\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad I_{n_1}\\ \end{array} \right] \\&=\text{ rank }({\widetilde{M}})+\text{ rank }\left[ \begin{array}{ccccc} I_{n_1} &{} \quad 0 &{} \quad C_{13} &{} \quad C_{14} &{} \quad B_1 \\ 0 &{} \quad I_{n_2} &{} \quad 0 &{} \quad 0 &{} \quad B_2 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad B_3 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad B_4 \end{array} \right] \\&= \text{ rank }({\widetilde{M}})+\text{ rank }[{\widetilde{M}}, {\widetilde{C}}N({\widetilde{M}}), {\widetilde{B}}] . \end{aligned}$$

By Lemma 2.1, there exist nonsingular matrices P, Q, V such that (2.5) holds. Then from above equation we have

$$\begin{aligned} \text{ rank }[{\mathcal {E}}_2, {\mathcal {B}}_2]&= \text{ rank }[{\widetilde{M}}, {\widetilde{C}}N({\widetilde{M}}), {\widetilde{B}}]+\text{ rank }(M)\\&=\text{ rank }[PMQ, (PCQ-PB{\widetilde{F}}_v^T)N({\widetilde{M}}), PBV]+\text{ rank }(M)\\&=\text{ rank }[M, (C-B{\widetilde{F}}_v^T Q^{-1})N(M), B]+\text{ rank }(M)\\&=\text{ rank }[M, C N(M), B]+\text{ rank }(M). \end{aligned}$$

1.2 Proof of Lemma 2.3

(1) Observe that

$$\begin{aligned}&\left[ \begin{array}{cc} I_n &{}\quad 0 \\ \lambda M+C_2 &{}\quad I_n \\ \end{array} \right] (\widetilde{{\mathcal {A}}}_1-\lambda \widetilde{{\mathcal {E}}}_1)\left[ \begin{array}{cc} 0 &{}\quad -I_n \\ I_n &{}\quad -\lambda I_n \\ \end{array} \right] \nonumber \\&\quad =\left[ \begin{array}{cc} I_n &{}\quad 0 \\ 0 &{}\quad \lambda ^2 M+\lambda C+K \\ \end{array} \right] ,\nonumber \\&\qquad \left[ \begin{array}{cc} I_n &{}\quad 0 \\ -\lambda K-C_1 &{}\quad I_n \\ \end{array} \right] (\widetilde{{\mathcal {E}}}_1-\lambda \widetilde{{\mathcal {A}}}_1)\left[ \begin{array}{cc} I_n &{}\quad \lambda I_n \\ 0 &{}\quad I_n \\ \end{array} \right] \nonumber \\&\quad = \left[ \begin{array}{cc} I_n &{}\quad 0 \\ 0 &{} \quad \lambda ^2 K +\lambda C+M \\ \end{array} \right] . \end{aligned}$$

(A1)

Then, by [19], $\widetilde{{\mathcal {A}}}_1-\lambda \widetilde{{\mathcal {E}}}_1$ is a strong linearization of $\lambda ^2\,M+\lambda C+K$.

(2) By (A1), $(-1)^{n(n+1)}|\widetilde{{\mathcal {A}}}_1-\lambda \widetilde{{\mathcal {E}}}_1|=|\lambda ^2\,M+\lambda C+K|$. Thus, $\widetilde{{\mathcal {A}}}_1-\lambda \widetilde{{\mathcal {E}}}_1$ is regular if and only if $\lambda ^2 M+\lambda C+K$ is regular.

(3) It is easily seen that

$$\begin{aligned}&\widetilde{{\mathcal {A}}}_1{\widehat{x}}=\lambda \widetilde{{\mathcal {E}}}_1 {\widehat{x}}\\&\quad \Leftrightarrow \left[ \begin{array}{cc} 0 &{}\quad I \\ -K &{}\quad -C_2 \\ \end{array} \right] \left[ \begin{array}{c} x \\ {\widetilde{x}} \\ \end{array} \right] =\lambda \left[ \begin{array}{cc} I &{}\quad 0 \\ C_1 &{}\quad M \\ \end{array} \right] \left[ \begin{array}{c} x \\ {\widetilde{x}} \\ \end{array} \right] \\&\quad \Leftrightarrow (\lambda ^2 M+\lambda C+K)x=0,\ {\widetilde{x}}=\lambda x. \end{aligned}$$

Thus, result (3) of this lemma holds.

(4) It is easily seen that (2.9) holds. Observe that

$$\begin{aligned} N({\mathcal {E}}_1)=\left[ \begin{array}{c} 0 \\ N(M) \\ \end{array} \right] ,\ {\mathcal {A}}_1 N({\mathcal {E}}_1)=\left[ \begin{array}{c} N(M) \\ -C_2 N(M) \\ \end{array} \right] . \end{aligned}$$

Then we have

$$\begin{aligned}&\text{ rank }[{\mathcal {E}}_1, {\mathcal {A}}_1 N({\mathcal {E}}_1), {\mathcal {B}}_1]\\&\quad =\text{ rank }\left[ \begin{array}{cccc} I &{}\quad 0 &{}\quad N(M) &{}\quad 0 \\ C_1 &{}\quad M &{}\quad -C_2 N(M) &{} \quad B \\ \end{array} \right] \\&\quad =\text{ rank }\left[ \begin{array}{cccc} I &{} \quad 0 &{}\quad 0 &{} \quad 0 \\ C_2 &{}\quad M &{}\quad -C N(M) &{}\quad B \\ \end{array} \right] \\&\quad =n+\text{ rank }[M, CN(M), B]. \end{aligned}$$

Then, $\text{ rank }[{\mathcal {E}}_1, {\mathcal {A}}_1 N({\mathcal {E}}_1), {\mathcal {B}}_1]=2n$ if and only if $\text{ rank }[M, CN(M), B]=n$.

(5) The second-order system (1.1) can be re-written as

$$\begin{aligned} {\widetilde{{\mathcal {E}}}_1 [x^T, \dot{{\widetilde{x}}}^T]^T=\widetilde{{\mathcal {A}}}_1 [x^T, {\widetilde{x}}^T]^T+{\mathcal {B}}_1 u.} \end{aligned}$$

(A2)

By Result (1) of this lemma, $\widetilde{{\mathcal {A}}}_1-\lambda \widetilde{{\mathcal {E}}}_1$ is a strong linearization of $\lambda ^2 M+\lambda C+K$. Then $ind(\widetilde{{\mathcal {E}}}_1, \widetilde{{\mathcal {A}}}_1)=ind_1(M, C, K)\leqslant 1$. Further from [5][Section 1], we know that for any continuous input function u, system (A2) must have a continuous solution. Observe that systems (1.1) and (A2) are equivalent. Thus, it is ensured that system (1.1) has a continuous solution for a continuous input u, i.e., system (1.1) is impulse-free.

1.3 Proof of Lemma 2.4

(1) It is easily seen that

$$\begin{aligned} |\widetilde{{\mathcal {A}}}_2-\lambda \widetilde{{\mathcal {E}}}_2|&= \left| \begin{array}{cc} -C_2-\lambda M_1 &{}\quad -K-\lambda C_1 \\ I_{n_1} &{}\quad -\lambda {\widetilde{I}}_{n_1}^T \\ \end{array} \right| \\&= \left| \begin{array}{cc} -C_2-\lambda M_1 &{}\quad -\lambda ^2 M-\lambda C-K\\ I_{n_1} &{}\quad 0 \\ \end{array} \right| \\&= (-1)^{n(n_1+1)}\left| \lambda ^2 M+\lambda C+K\right| . \end{aligned}$$

Thus, $\left| \widetilde{{\mathcal {A}}}_2-\lambda \widetilde{{\mathcal {E}}}_2\right| \not \equiv 0\Leftrightarrow \left| \lambda ^2\,M+\lambda C+K\right| \not \equiv 0$. Then result (1) of Lemma 2.4 follows.

(2) Assume that $(\lambda , x)$ is a finite eigenpair of $\lambda ^2\,M+\lambda C+K$ and ${\widetilde{x}}=\lambda {\widetilde{I}}_{n_1}^T x$. Then we see that $\lambda M_1 {\widetilde{x}} +\lambda C_1 x+C_2 {\widetilde{x}}+Kx=0$. Thus $\widetilde{{\mathcal {A}}}_2 [{\widetilde{x}}^T, x^T]^T=\lambda \widetilde{{\mathcal {E}}}_2 [{\widetilde{x}}^T, x^T]^T$, i.e., $(\lambda , [{\widetilde{x}}^T, x^T]^T)$ is a finite eigenpair of $\widetilde{{\mathcal {A}}}_2-\lambda \widetilde{{\mathcal {E}}}_2$.

Conversely, if $(\lambda , [{\widetilde{x}}^T, x^T]^T)$ is a finite eigenpair of $\widetilde{{\mathcal {A}}}_2-\lambda \widetilde{{\mathcal {E}}}_2$, where ${\widetilde{x}}\in {{\mathbb {C}}}^{n_1}$, then from (2.13) we see that $\lambda (M_1 {\widetilde{x}} +C_1 x)+C_2 {\widetilde{x}}+Kx=0$ and ${\widetilde{x}}=\lambda {\widetilde{I}}_{n_1}^T x$. Substituting the second equation into the first equation and using (2.11), (2.12) yield $(\lambda ^2 M+\lambda C+K)x=0$, i.e., $(\lambda , x)$ is an eigenpair of $\lambda ^2 M+\lambda C+K$.

(3) Consider the differential equation

$$\begin{aligned} \widetilde{{\mathcal {E}}}_2 \left[ \begin{array}{c} \dot{{\widetilde{x}}} \\ {\dot{x}} \\ \end{array} \right] \!=\!\widetilde{{\mathcal {A}}}_2\left[ \begin{array}{c} {\widetilde{x}} \\ x \\ \end{array} \right] \!+\!\left[ \begin{array}{c} u \\ 0\\ \end{array} \right] , {\widetilde{x}}\!\in \! {{\mathbb {C}}}^{n_1}, u\!\in \! {{\mathbb {C}}}^n. \end{aligned}$$

(A3)

Note that $ind(\widetilde{{\mathcal {E}}}_2, \widetilde{{\mathcal {A}}}_2)\leqslant 1$. Then for any continuous input function u, there exists a continuous solution $[{\widetilde{x}}^T, x^T]^T$ to (A3) such that $\widetilde{{\mathcal {E}}}_2[{\widetilde{x}}^T, x^T]^T$ is continuously differentiable, i.e., $[{\widetilde{x}}^T, x^T]^T$ satisfies that

$$\begin{aligned}&M_1 \dot{{\widetilde{x}}}+C_1 {\dot{x}}=-C_2 {\widetilde{x}}-K x+u, \end{aligned}$$

(A4)

$$\begin{aligned}&{\widetilde{I}}_{n_1}^T{\dot{x}}={\widetilde{x}}, \end{aligned}$$

(A5)

$$\begin{aligned}&\widetilde{{\mathcal {E}}}_2 [\dot{{\widetilde{x}}}^T, {\dot{x}}^T]^T\ \text{ is } \text{ continuous }. \end{aligned}$$

(A6)

From (A4) and (A5), we see that

$$\begin{aligned} M_1 {\widetilde{I}}_{n_1}^T \ddot{x}+C_1 {\dot{x}}+C_2 {\widetilde{I}}_{n_1}^T {\dot{x}}+Kx=u. \end{aligned}$$

Further by (2.11) and (2.12) we know that x satisfies that

$$\begin{aligned} M \ddot{x}+ C{\dot{x}}+Kx=u. \end{aligned}$$

(A7)

By (A6), (2.11)–(2.13), we know that $M\ddot{x}$, $C{\dot{x}}$ are continuous. Thus, for any continuous u, there exists a continuous solution x to equation (A7) such that $Mx,\ Cx$ are, respectively, twice and once continuously differentiable. Then $ind_2(M, C, K)\leqslant 1$.

1.4 Proof of Lemma 2.5

(1) By Theorem 2.4, Theorem 3.13, Theorem 3.14 in [5], the proof of Theorem 9 in [18] and the first equation in (2.14), we know that there exist nonsingular matrices P, Q, V such that

$$\begin{aligned}&{\widetilde{M}}=P M Q=\left[ \begin{array}{cccc} I_{n_1} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{} \quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{} \quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{} \quad 0 &{}\quad 0 &{}\quad 0 \end{array} \right] ,\ {\widetilde{C}}=P C Q=\left[ \begin{array}{cccc} {\widetilde{C}}_{11} &{}\quad 0 &{} \quad {\widetilde{C}}_{13} &{}\quad {\widetilde{C}}_{14} \\ 0 &{}\quad I_{n_2} &{}\quad 0 &{} \quad 0 \\ {\widetilde{C}}_{31} &{}\quad 0 &{}\quad 0 &{} \quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{array} \right] ,\\&{\widetilde{K}}=P K Q=\left[ \begin{array}{cccc} {\widetilde{K}}_{11} &{}\quad {\widetilde{K}}_{12} &{}\quad {\widetilde{K}}_{13} &{} \quad 0 \\ {\widetilde{K}}_{21} &{}\quad {\widetilde{K}}_{22} &{}\quad {\widetilde{K}}_{23} &{}\quad 0 \\ {\widetilde{K}}_{31} &{}\quad {\widetilde{K}}_{32} &{} \quad {\widetilde{K}}_{33} &{}\quad 0 \\ {\widetilde{K}}_{41} &{}\quad {\widetilde{K}}_{42} &{} \quad {\widetilde{K}}_{43} &{} \quad 0\\ 0 &{}\quad 0 &{} \quad 0 &{} \quad I_{n_3} \end{array} \right] ,\ {\widetilde{B}}=P B V=\left[ \begin{array}{c} {\widetilde{B}}_1 \\ {\widetilde{B}}_2 \\ {\widetilde{B}}_3 \\ {\widetilde{B}}_4 \\ {\widetilde{B}}_5 \end{array} \right] , \end{aligned}$$

where ${\widetilde{K}}_{11},\ {\widetilde{C}}_{11}\in {{\mathbb {R}}}^{n_1\times n_1}$, ${\widetilde{C}}_{31},\ {\widetilde{K}}_{31}\in {{\mathbb {R}}}^{d_1\times n_1}$, ${\widetilde{K}}_{22}\in {{\mathbb {R}}}^{n_2\times n_2}$, ${\widetilde{B}}_1\in {{\mathbb {R}}}^{n_1},\ {\widetilde{B}}_2\in {{\mathbb {R}}}^{n_2}$, ${\widetilde{B}}_3\in {{\mathbb {R}}}^{d_1}$, ${\widetilde{B}}_4\in {{\mathbb {R}}}^{n_3}$. Moreover, matrices ${\widetilde{C}}_{31}$ and $[{\widetilde{B}}_3^T, {\widetilde{B}}_4^T]^T$ have full row rank.

From Lemma 13 in [20] and the second equation in (2.14), we know that $d_1=0$. Then, we have result (1) of Lemma 2.5.

(2) From Result (1) of this lemma, we see that (1.1) can be reduced to

$$\begin{aligned} {{\widetilde{M}} \ddot{{\widetilde{x}}}+{\widetilde{C}}\dot{{\widetilde{x}}}+{\widetilde{K}}{\widetilde{x}}={\widetilde{B}}{\widetilde{u}}, \ {\widetilde{x}}=Q^{-1}x, \ {\widetilde{u}}=V^{-1}u.} \end{aligned}$$

(A8)

Denote

$$\begin{aligned} {{\widetilde{x}}=\left[ \begin{array}{c} {\widetilde{x}}_1 \\ {\widetilde{x}}_2 \\ \end{array} \right] ,\ {\widehat{x}}=\left[ \begin{array}{c} \dot{{\widetilde{x}}}_1 \\ {\widetilde{x}} \\ \end{array} \right] ,\ {\widetilde{x}}_1\in {{\mathbb {C}}}^{n_1}.} \end{aligned}$$

Then (A8) can be rewritten as

$$\begin{aligned} {{\mathcal {E}}_2 \dot{{\widehat{x}}}={\mathcal {A}}_2 {\widehat{x}} +{\mathcal {B}}_2 {\widetilde{u}}.} \end{aligned}$$

(A9)

Then, (1.1) has a continuous solution x such that Mx is twice and Cx is once continuously differentiable if and only if (A9) has a continuous solution ${\widehat{x}}$ such that ${\mathcal {E}}_2 {\widehat{x}}$ is once continuously differentiable. Thus, $ind_2(M, C, K)=ind({\mathcal {E}}_2, {\mathcal {A}}_2)$.