# Finite-Time Control of Singular Linear Semi-Markov Jump Systems
## 3 Results
### Stochastically Finite-Time Stabilization
: _It should be pointed out that it is required to assume that \(P_{i}E^{T}+HS_{i}^{T}\) is nonsingular in the proof of Theorem 1. It is assumed that there exists a feasible solution \(\alpha\), in which \(P_{i}\), \(Q_{i}\), \(S_{i}\) and \(Y_{i}\) condition (5) for a prescribed matrix \(H\), but \(P_{i}E^{T}+HS_{i}^{T}\) is singular. In this case, according to [38], the matrix \(\Delta S_{i}\in^{n\times(n-r)}\) with \(\|\Delta S_{i}\|\ll 1\) must exist so that \(P_{i}E^{T}+H(S_{i}+\Delta S_{i})^{T}\) is a nonsingular matrix. Meanwhile, the condition (5) is satisfied for \(\alpha\), \(P_{i}\), \(Q_{i}\), \(S_{i}+\Delta S_{i}\) and \(Y_{i}\)._  
However, the condition (5c) in Theorem 1 is difficult to verify. In the sequel, a numerically tractable condition will be provided in Theorem 2 to get the matrix gains \(K_{i}\) by some strict linear matrix inequalities. **Theorem 2**. : _For given positive scalars \(c_{1}\), \(c_{2}\) and \(T\) with \(c_{1}<c_{2}\) and symmetric positive-definite matrix \(R\), the SSMJS in (1) with \(\omega(t)=0\) is stochastically finite-time stabilizable with respect to \((c_{1},c_{2},T,R)\), if there exist positive scalars \(\alpha\) and \(0<\theta<1\); symmetric positive-definite matrices \(P_{i}\) and \(Z_{i}\); and matrices \(S_{i}\) and \(Y_{i}\) with appropriate dimensions such that (5a), (5b) and the following matrix inequalities are satisfied:_  
\[\theta I<R^{\frac{1}{2}}L^{-1}\bigg{(} \begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix} T^{-1}P_{i}T^{-T} \begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix} + \begin{bmatrix}0&0\\ 0&Z_{i}\end{bmatrix} \bigg{)}L^{-T}R^{\frac{1}{2}}, \tag{25a}\] \[R^{\frac{1}{2}}L^{-1}\bigg{(} \begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix} T^{-1}P_{i}T^{-T} \begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix} + \begin{bmatrix}0&0\\ 0&Z_{i}\end{bmatrix} \bigg{)}L^{-T}R^{\frac{1}{2}}<I,\] (25b) \[c_{1}e^{\alpha T}-\theta c_{2}<0. \tag{25c}\]  
_Furthermore, the control law can be obtained by (6)._Proof of Theorem 2.: First, it follows from (7) and (8) that the matrix \(Q_{i}\) can be also decomposed to  
\[\bar{Q}_{i}=L^{-T}R^{\frac{1}{2}}Q_{i}R^{\frac{1}{2}}L^{-1}= \begin{bmatrix}\bar{Q }_{i11}&\bar{Q}_{i12}\\ \bar{Q}_{i21}&\bar{Q}_{i22}\end{bmatrix} . \tag{26}\]  
Multiplying (5c) by \(T^{T}\) and \(L^{T}\) on the left and right, respectively, gives  
\[Z_{R}^{T}E^{T}Z_{L}^{T}=Z_{R}^{T}E^{T}R^{\frac{1}{2}}Q_{i}R^{\frac{1}{2}}EP_{i} E^{T}Z_{L}^{T},\]  
and further gives  
\[\bar{E}^{T}=\bar{E}^{T}\bar{Q}_{i}\bar{E}\bar{P}_{i}\bar{E}^{T}, \tag{27}\]  
which straightforwardly gives \(\bar{Q}_{i11}\bar{P}_{i11}=I\). With this observation, \(\bar{Q}_{i}\) can be constructed as  
\[\bar{Q}_{i}= \begin{pmatrix}\begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix}P_{i}\begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix}+\begin{bmatrix}0&0\\ 0&Z_{i}\end{bmatrix}\end{pmatrix} ^{-1}, \tag{28}\]  
where \(Z_{i}\in\mathbb{R}^{(n-r)\times(n-r)}\) is symmetric positive-definite. Next we rewrite \(Q_{i}\) as  
\[Q_{i}=R^{-\frac{1}{2}}L^{T} \begin{pmatrix}\begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix}\bar{P}_{i}\begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix}+\begin{bmatrix}0&0\\ 0&Z_{i}\end{bmatrix}\end{pmatrix} ^{-1}LR^{-\frac{1}{2}}, \tag{29}\]  
and its inverse matrix as  
\[Q_{i}^{-1}=R^{\frac{1}{2}}L^{-1} \begin{pmatrix}\begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix}\bar{P}_{i}\begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix}+\begin{bmatrix}0&0\\ 0&Z_{i}\end{bmatrix}\end{pmatrix} L^{-T}R^{\frac{1}{2}}.\]  
From above discussion, to solve the conditions (5c) and (5d), we can set (25a) and (25b), where \(0<\theta<1\) and \(\theta<\lambda_{\min}\big{(}Q_{i}^{-1}\big{)}<\lambda_{\max}\big{(}Q_{i}^{-1} \big{)}<1\). If (25c) holds, we have  
\[\lambda_{\max}\big{(}Q_{i}^{-1}\big{)}c_{1}e^{\pi T}<c_{1}e^{\pi T}<\theta c_{ 2}<\lambda_{\min}\big{(}Q_{i}^{-1}\big{)}c_{2}. \tag{30}\]  
Note that \(\lambda_{\max}(Q_{i})=\frac{1}{\lambda_{\min}\big{(}Q_{i}^{-1}\big{)}}\), \(\lambda_{\min}(Q_{i})=\frac{1}{\lambda_{\max}\big{(}Q_{i}^{-1}\big{)}}\) and (5d) can be guaranteed. **Remark 4**.