## 4 Parameter Solutions
### State Feedback Control Law Design  
From (25) and (26), the state feedback control law \(K_{i}\) involves making the system SS. In the following theorem, based on the stability theory and LMI method, we propose a sufficient condition for SS of the system when the transition rates are partially unknown, and the parameter solution of the controller can be derived from the above condition. **Theorem 2**: _System (1) with \(u(t)=K_{i}x(t)\) and partly unknown TRs is stochastically stable if there exists a set of symmetric and positive-definite matrices \(X_{i}\), and a set of matrices \(Y_{i},i\in U^{(i)}\), satisfying:_  
\[\left[ \begin{array}{cc}\epsilon_{1k}^{i}&\Omega_{1k}^{i}\\ *&\Delta_{1k}^{i}\end{array} \right]<0,\forall i\in U_{k}^{(i)}. \tag{28}\]  
\[\left[ \begin{array}{cc}\epsilon_{2k}^{i}&\Omega_{2k}^{i}\\ *&\Delta_{2k}^{i}\end{array} \right]<0,\forall i\in U_{uk}^{(i)}. \tag{29}\]  
\[\left[ \begin{array}{cc}A_{i}X_{i}+X_{i}A_{i}^{T}&X_{i}\\ *&X_{j}\end{array} \right]<0,\forall i\in U_{k}^{(i)},j\
eq i. \tag{30}\]  
\[A_{i}X_{i}+X_{i}A_{i}^{T}+X_{i}\geq 0,\quad\forall i\in U_{uk}^{(i)}. \tag{31}\]  
Moreover, if the above inequations are solvable, the controller gain can be computed from the relation \(K_{i}=Y_{i}X_{i}^{-1}\), where  
\[\epsilon_{1k}^{i}=\left(1+\sum\limits_{j\in U_{k}^{(i)}}\pi_{ij} \right)\left(A_{i}X_{i}+X_{i}A_{i}^{T}\right)+\pi_{ii}X_{i}+Y_{i}^{T}B_{i}^{T} +B_{i}Y_{i}.\] \[\Omega_{1k}^{i}=\left[\sqrt{\pi_{1,i_{1}^{i}}}X_{i},\quad\sqrt{ \pi_{1,i_{2}^{i}}}X_{i_{2}},\quad\cdots,\sqrt{\pi_{1,i_{m-1}^{i}}}X_{i}\right].\] \[\Delta_{1k}^{i}=\text{diag}\Big{(}-X_{i_{1}^{i}},\quad-X_{i_{2}^ {i}},\cdots,\quad-X_{i_{m-1}^{i}}\Big{)}.\] \[\epsilon_{2k}^{i}=\left(1+\sum\limits_{j\in U_{k}^{(i)}}\pi_{ij} \right)\left(A_{i}X_{i}+X_{i}A_{i}^{T}\right)+Y_{i}^{T}B_{i}^{T}+B_{i}Y_{i}.\] \[\Omega_{2k}^{i}=\left[\sqrt{\pi_{1,k_{1}^{i}}}X_{i},\quad\sqrt{ \pi_{1,k_{2}^{i}}}X_{i},\quad\cdots,\sqrt{\pi_{1,k_{m}^{i}}}X_{i}\right].\] \[\Delta_{2k}^{i}=\text{diag}\Big{(}-X_{k_{1}^{i}},\quad-X_{k_{2}^ {i}},\cdots,\quad-X_{k_{m}^{i}}\Big{)}.\ \{l_{1}^{i},l_{2}^{i},\cdots l_{m-1}^{i}\}=\{k_{1}^{i},k_{2}^{i},\cdots k_{m}^ {i}\}\backslash\{i\}.\]  
To prove Theorem 2, we first introduce the following lemma. **Lemma 2** ([44]). : _The unforced system (1) (\(u(t)=0\)) is \(SS\) if and only if there exists a set of symmetric and positive-definite matrices \(X_{i},i\in U\), satisfying:_  
\[A_{i}^{T}P_{i}+P_{i}A_{i}+\sum\limits_{j=1}^{N}\pi_{ij}P_{j}<0 \tag{32}\]  
Lemma 2 shows the SS of the Markov jump systems. System (1) with \(u(t)=K_{i}x(t)\) can be rewritten as \(\dot{x}(t)=(A_{i}+B_{i}K_{i})x(t)\). It can be concluded from Lemma 2 that the SS condition of the system with the feedback controller can be obtained simply by replacing \(A_{i}\) with \(A_{i}+B_{i}K_{i}\). Proof of Theorem 2.: Consider two cases: \(i\in U_{k}^{(i)}\) and \(i\in U_{uk}^{(i)}\). Case 1: \(i\in U_{k}^{(i)}\). Since \(\sum\limits_{j=1}^{N}\pi_{ij}=0\), the SS inequality of the system with \(u(t)=K_{i}x(t)\) can be rewritten as:  
\[(A_{i}+B_{i}K_{i})^{T}P_{i}+P_{i}(A_{i}+B_{i}K_{i})+\sum\limits_{j=1}^{N}\pi_ {ij}P_{j}+\sum\limits_{j=1}^{N}\pi_{ij}\Big{(}A_{i}^{T}P_{i}+P_{i}A_{i}\Big{)}<0 \tag{33}\]  
Denote  
\[\theta_{i}= A_{i}^{T}P_{i}+K_{i}^{T}B_{i}^{T}P_{i}+P_{i}B_{i}K_{i}+P_{i}A_{i}+ \pi_{ii}P_{i}+\sum\limits_{j\in U_{k}^{(i)},j\
eq i}\pi_{ij}P_{j}+\sum\limits _{j\in U_{uk}^{(i)}}\pi_{ij}P_{j}\] \[\epsilon_{1k}^{i}=\left(1+\sum\limits_{j\in U_{k}^{(i)}}\pi_{ij} \right)\left(A_{i}X_{i}+X_{i}A_{i}^{T}\right)+\pi_{ii}X_{i}+Y_{i}^{T}B_{i}^{T} +B_{i}Y_{i}.\]0  
\[\epsilon_{1k}^{i}=\left(1+\sum\limits_{j\in U_{k}^{(i)}}\pi_{ij} \right)\left(A_{i}X_{i}+X_{i}A_{i}^{T}\right)+\pi_{ii}X_{i}+Y_{i}^{T}B_{i}^{T} +B_{i}Y_{i}.\]1 \[\epsilon_{1k}^{i}=\left(1+\sum\limits_{j\in U_{k}^{(i)}}\pi_{ij} \right)\left(A_{i}X_{i}+X_{i}A_{i}^{T}\right)+\pi_{ii}X_{i}+Y_{i}^{T}B_{i}^{T} +B_{i}Y_{i}.\]2  
Thus, \(\theta_{i}<0\) can be guaranteed by  
\[\left\{