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Peng et al. Intell Robot 2022;2(3):298­312 I http://dx.doi.org/10.20517/ir.2022.27 Page 304

We use the reciprocally convex approach [27] to deal with the cross item in Equation (18). By using Schur’s
complement, one can get

1 Õ p 2 p 2 2 2 2
( )| + | ( )| − | ( )| < 0.
¤
( ) + 2 ( ) − | ( )| − 2 | (19)

=1, ≠
In the following, we will prove that
(i) with ( ) = 0, the impulsive system Equation (12) is exponentially stable.

Since 2 ( ) = 2 ¤

( ( )+2 ( )) and ( ) is continuous in ∈ [ , +1 ), the integration of Equation (19)
yields

¹ p 1 Õ p
2
2 ( ) − 2 ( ) ≤ 2 [2 | 2 | ( )| ]. (20)
( )| +

=1, ≠
¯
It follows from 2 ( − ) ≤ that

( +1 ) ≤ Θ + 2 ( − +1 ) ( ), (21)
where

Õ p p Õ p p
2
2
Θ = [− ( − 2 ( )| + ( +1 )] + 2 ( )| . (22)
+1 ) + | | ( )| + 2 |
( +1 )| − |
=1 =1, ≠

Then taking Equation (17) and Equation (22) into account, denoting = [ ( ), [− ( +1 ) + ( )]] , we
have

Õ
p p 2
−2 − [ ( ) − ( +1 )]| + Ω ≤ 0. (23)
Θ ≤ −|( )
=1, ≠
Hence, one can conclude that

Õ p p
2
( )| .
( ) ≤ | ( )| + 2 ( 0 − ) ( 0 ) + 2 | 2 (24)
=1, ≠
From Equation (24), we have

Õ p
2 2
( ) ≥ { }| ( )| + | ( )| ,
=1
which implies that

p Õ p
2 2 2 2 ( 0 − )
( )| +
{ }| ( )| ≤ (2 − 1)| | − ( )| + ( 0 ).
=1, ≠
It follows from Ω < 0 that for ≥ 0
2
{ }| ( )| ≤ 2 ( 0 − ) ( 0 ). (25)
For < 0, the system takes the form ¤ = ( ) + ( ), then its corresponding solution is given by

¹
(− + )
( ) = (0) + ( ) . (26)
0