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Page 102 Zhou et al. Intell Robot 2023;3(1):95-112 I http://dx.doi.org/10.20517/ir.2023.05

1) When the sliding mode moves outside the domain Ω, the following two conditions exist:

Case 1: Considering the situation that ( ) > ( ) > 0, then equation (27) can be rewritten as:

2
Õ
4 ( ) = − [ ( ) − ( + 1)] [ ( ) + ( + 1)] . (28)
=1

1

, and ( ) ( ) > ( ), thus ( ) ( ) > ( ),
Since ( ) > ( ) , then ( ) > ( )
( )
we can obtain:
( ) − ( + 1)

= ( ) − ( ) ( ) ( ) + ( )| ( )| [ ( )] + ( )

= [1 − ( ) ( )] ( ) + ( )| ( )| [ ( )] + ( )

≥ [1 − ( ) ( )] ( ) + ( )| ( )| − | ( )| (29)

≥ [1 − ( ) ( )] ( ) + ( ) −

≥ ( ) −
> 0
( ) + ( + 1)

= [1 + ( ) ( )] ( ) − ( )| ( )| [ ( )] − ( )

≥ ( ) + ( )[ 1− ( ) ( ) − | ( )| [ ( )]] − ( )
1−
≥ ( ) ( ) − ( )
( )

≥ ( ) ( ) ( ) − ( ) (30)

≥ ( ) ( ) − ( )

≥ ( ) ( ) − | ( )|

≥ ( ) −
≥ 0.
It can be seen from the above derivation that ( ) − ( + 1) > 0 and ( + 1) + ( ) > 0 are tenable, we
can easy to deduce that:

2
Õ
Δ ( ) = − [ ( ) − ( + 1)] [ ( + 1) + ( )] < 0. (31)
=1

Case 2: Moreover, another situation is that ( ) < − ( ) < 0, similar to the proof for case 1, because
1

( ) < − ( ) and ( ) < − ( ) ,then ( ) | ( )| > ( ),wecangetthat − ( ) | ( )| +
( )

| ( )| < − ( ) + < 0.
Í 2 2
Therefore, Δ ( ) = − ( − ) < 0 holds in this case. Through the analysis of the above knowledge,
=1
2

the system will enter the domain Ω in step, the ( ) − (0) < − ( − ) , then we get ( ) <
2
2
2
∗
∗
∗
∗
2 2
2

2

2

∗
+
(0) − ( − ) < ( ) , available = (0)−( ( ) ) + 1, and ∈ .
∗
∗

2
( − )
2) When the sliding variables enter into the domain Ω, | ( )| ≤ ( ) . To prove | ( + 1)| ≤ ( ) , it is
essential to divide the analyses due to the location of ( ).