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Tan et al. Complex Eng Syst 2023;3:6  I http://dx.doi.org/10.20517/ces.2023.10   Page 13 of 23


               According to Schur complement, it is inferred that (14) is equivalent to Π      < 0 and (15) is equivalent to
                     ∑   
               −Q    +           (ℎ)Q      < 0 .
                        =1

               Case 2. If  ∧   ,    ≠ ∅ and  ∧   ,      ≠ ∅,    ∈  ∧   ,     , denote       =  ∑   ∈ ∧   ,            (ℎ). Since  ∧   ,    ≠ ∅, we know that
                     > 0, thus  ∑             (ℎ)P      can be presented as
                             =1

                       
                    ∑              ∑                           ∑
                               (ℎ)P      =          (ℎ)P      +         (ℎ)P      +          (ℎ)P     
                                   ∧                          ∧
                      =1            ∈    ,                    ∈    ,     ,  ≠                          (27)
                                   ∑                                        ∑
                                                                                             (ℎ)
                                =              (ℎ)P      +         (ℎ)P      − (        (ℎ) +       )  P      .
                                   ∧                                      ∧       −        (ℎ) −      
                                    ∈    ,                                 ∈    ,     ,  ≠  
                                                 ∧         ∑                              ∧
                                           (ℎ)        ) and                  (ℎ)               ,    ≠   , there
               It is obvious that 0 ≤     ≤ 1(   ∈    ,       ∧             = 1. So for ∀   ∈    ,    
                                 −        (ℎ)−                 ∈    ,     −        (ℎ)−     
               is
                         
                     ∑                ∑                   ∑
                                                       (ℎ)
                                 (ℎ)P      =            (             (ℎ)(P      − P      ) +         (ℎ)(P      − P      )).  (28)
                                    ∧       −        (ℎ) −        ∧
                        =1           ∈    ,     ,  ≠       ∈    ,  
                                                        ∑   
               By applying Schur complement and         (ℎ) = −    =1,  ≠            (ℎ), it is deduced that (16), (17) are equivalent
                                                   ∑   
               to Π      < 0 and (18) is equivalent to −Q    +          (ℎ)Q      < 0 .
                                                       =1
               Case 3. If  ∧   ,    = ∅ and  ∧   ,      ≠ ∅,    ∈  ∧   ,     , assume there exists    ≠    and    ∈  ∧   ,     . Denote       =
                       (ℎ) =               (ℎ). Noting that       < 0,  ∑             (ℎ)P      can be presented as
                                                     =1

                       
                    ∑                           ∑                                 ∑
                                                                                                (ℎ)
                               (ℎ)P      =         (ℎ)P      +          (ℎ)P      =         (ℎ)P      −        P      .  (29)
                                               ∧                                ∧        −     
                       =1                      ∈    ,     ,  ≠                   ∈    ,     ,  ≠  
               It is obvious that  ∑  ∧          (ℎ) = −        (ℎ) = −      > 0. So for ∀   ∈  ∧  ,    ≠   , there is
                                 ∈    ,     ,  ≠                                ,    
                   
                ∑                ∑             (ℎ)
                           (ℎ)P      =       [        (ℎ)(P      − P      )] =         (ℎ)(P      − P      ) =               (ℎ)(P      − P      ). (30)
                               ∧        −     
                   =1           ∈    ,     ,  ≠  
                                                        ∑   
               By applying Schur complement and         (ℎ) = −    =1,  ≠            (ℎ) , we konw that (19) is equivalent to Π      < 0.

               In summary, if inequalities (14) − (19) hold, the interconnected semi-Markovian control system (10) with
               partially accessible TRs and dynamic METM is stochastically stable.


               Remark 3. Theorem 1 designed sufficient conditions to ensure (10) is stochastically stable. However, it is dif-
               ficult to directly use this result to acquire the controller gain matrices. Theorem 2 gives a LMI-based sufficient
               criterion for the solvability.


               Theorem 2. For given a positive real number       > 0,       > 0,    > 0,    0 > 0,         > 0 (    = 1, 2, 3, · · · ,    + 3),
                     ∈ (0, 1) and         (ℎ) ∈ [   ,         ], the system (10) with partially accessible TRs and dynamic METM is
                                            
                                                                            
               stochastically stable if there are positive symmetric matrices         > 0,          > 0, Q      > 0, Q    > 0,R    > 0, Ω    > 0,
                              
               and matrices    , S 1  , S 2  , S 3   and S 4   with proper dimensions, such that the linear matrix inequalities hold:
                                
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