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Page 218                           Li et al. Intell Robot 2023;3(2):213-21  I http://dx.doi.org/10.20517/ir.2023.13

               By    > 2   and (8), we get

                                                              2    
                                               2 2  2  + 2     <   − 2  .                               (9)
                                                       10
                                                             2   −   
               By Condition (C ), we obtain
                              2
                                                    2
                                                2 2
                                                       + 2     <    +    − 4  .                        (10)
                                                    10
               By    ⩾ 0,    > 0 and    > 2  , we get
                                                 2                       2    
                                      +    − 4   − (  − 2  ) =    − 2   +    +  > 0,                   (11)
                                               2   −                       − 2  
                                       2    
               which implies    +    − 4   >  2  −    − 2  .

               By (9), (10), and (11), we have

                                                              2    
                                               2 2
                                                   2
                                                       + 2     <   − 2  .                              (12)
                                                   10        2   −   
                             2 2       2    
               Denote   (  ) =       + 2   −  + 2   and    =     . By (12), we know that   (  ) < 0 has a solution for variable
                             10        2  −  
                                                                                             2
                                              2   2                                (  −2  )(1−2     )
                 . By   (  ) < 0, we have Δ = 4 − 4   (−  + 2  ) > 0. By Δ > 0, we get    <  10  . Combining
                                              10  2  −                                 2     2
                                                                                          10
                                  2
               0 ⩽    <  (  −2  )(1−2     )  and    > 2  , we have 2     2  < 1. In summary, we obtain 2     2  < 1,    > 2  ,
                                  10
                            2     2                         10                             10
                               10
                                2
                       (  −2  )(1−2     )
               0 ⩽    <         10  .
                          2      2
                             10
                                                                       2
                                                             (  −2  )(1−2     )
               Sufficiency: By 2     2  < 1,    > 2   and 0 ⩽    <     10  , we get (12). By (12), we have      ∈
                                 10                              2      2
                        √         √                                 10
                −2(  −2  )−    −2(  −2  )+     , where    = 4(   − 2  ) − 4(   − 2  )   [2  (   − 2  ) + 2    ]. From the value range of
                          ,
                                                       2
                                                                    2
                 2(  −2  )   2 10  2(  −2  )   10                  10
                                  2
                   , it can be seen that Condition (C ) and Condition (C ) hold. Therefore, since the real parts of the zero
                                                                2
                                              1
               point of    (  ) are less than −2  , it can be concluded that all eigenvalues of    have negative real parts. Lemma
               1 implies that the system (6) is mean square stable. Therefore, there exists an admissible cooperative control
               strategy    ∈ U, such that for any initial value, the follower can track the leader under the distributed control
               law   .
                                                                           2
                                                                  (  −2  )(1−2     )
               Remark 1 The conditions    > 2  , 2     2  < 1 and 0 ⩽    <  10  stated in Theorem 1 highlight the
                                                10                    2     2
                                                                        10
               influenceofmultiplicativenoisesandboththetransitionrates    and    onthecooperatabilityofthesystem. Itis
               shown that smaller multiplicative noises, lower transition rate   , and higher transition rate    are all favorable
               for the cooperatability of the system. Moreover, the transition rates    and    have lower and upper bounds,
               respectively. What is more, the noises and the system parameters satisfy the corresponding inequality.
               We have the following corollary for the case without measurement noises.
               Corollary 1 Suppose Assumption 1 and    10 = 0 hold. In that case, there exists an admissible cooperative
               control strategy denoted by    ∈ U, such that for any initial value, the follower can track the leader under the
               distributed control law   , if and only if    > 2  .
               5. NUMERICAL SIMULATION
               In this section, we will use a numerical example to demonstrate the effectiveness of our control laws.
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