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Li et al. Intell Robot 2023;3(2):213-21 I http://dx.doi.org/10.20517/ir.2023.13 Page 217

the distributed control law . This holds if and only if the following conditions are met: 2 2 < 1, > 2 ,
10
2
0 ⩽ < ( −2 )(1−2 ) .
10
2 2
10
Proof: Denote ( ) = 1 ( ) − 0 ( ). By Assumptions 1 and (1)−(4), we get

d ( ) = [ + 10 ( ( ))] 1 ( )d + 10 ( ( )) 10 1 ( )d 10 ( ), (6)

where if 10 ( ( )) = 0, then we denote (1) = and (1) = 0; if 10 ( ( )) = 1, then we denote (2) = +
and (2) = 10.

Denote = diag( (1) ⊕ (1), (2) ⊕ (2)) + diag( (1) ⊕ (1), (2) ⊕ (2)) + ⊗ 1. By the definition
T
of , we have

2 0 0 0 −
= + 2 2 2 +
0 2( + ) 0 −
10

2 −
= 2 2 2 . (7)
2( + ) + −
10
Necessity: If 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 , it implies that the system (6) is
mean square stable. According to Lemma 1, it can be inferred that all eigenvalues of have negative real parts.

− 2 + −
2
Noting that | − | = , we have | − | = ( − 2 ) + ( + −
2
2 2
− − 2( + ) − +
10
2 − )( − 2 ) − 2 − .
2
2 2
2
2 2
10 10
Denote = − 2 and ( ) = + ( + − 2 − ) − 2 − . As all eigenvalues of
2
2 2
2
2 2
2
10 10
have negative real parts, we know that the real parts of the zero point of ( ) are less than −2 . As the real
parts of the zero point of ( ) are less than −2 , by considering the image of the function ( ), the following
two conditions can be inferred.
2
2 2
2
2
2 2
Condition (C ) : (−2 ) = 4 − 2 ( + − 2 − ) − 2 − 10 > 0.
1
10
2 − − + 2
2 2
Condition (C ) : 2 10 + 2 < 0.
2
By Condition (C ), we obtain
1
2
2 2
(2 − )(2 + 2 + ) > 2 . (8)
10
In the following, we discuss the Conditions (C ) and (C ).
1
2
2 2
2
2 2
(1) If 2 + 2 + 2 = 0 holds, then we have (2 − )(2 + 2 + ) = 0. This contradicts the
10 10
inequality (8). Therefore, this situation does not hold.
2 2
(2) If 2 + 2 + 2 > 0 holds, by (8), we get 0 ≤ < 2 .
10
2
2 2
By 2 − − + 10 + 2 < 0, we have + − 2 − 2 > 4 . By + − 2 − 2 > 4 ,
2 2
2 2
2 10 10
2 2
2
2
2 2
2 + 2 + 2 > 0 and 0 ≤ < 2 , we obtain (−2 ) = 4 − 2 ( + − 2 − ) − 2 −
10 10
2
2
2 2
2
2
2
2 2
10 < −4 − 2 − 10 = −4 − (2 + ) < −4 + 2 = 2 ( − 2 ) < 0. This
2
2 2
10
contradicts Condition (C ). Therefore, this situation also is not valid.
1
2 2
(3) If 2 + 2 + 2 < 0, then by ⩾ 0, > 0 and (8), we have > 2 .
10