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Zhou et al. Intell Robot 2023;3(1):95-112 I http://dx.doi.org/10.20517/ir.2023.05 Page 105

After ( ) ≥ 1 with ( ) ≥ ( ( ) ) proved, we will discuss 0 < ( ) < 1 with ( ) ≤ ( ( ) ) .

1
Φ ( ) 1− .
According to Lemma 3, we can obtain that | ( + 1)| ≤ ( )

1 1
It can be seen from the above proof that under conditions < Φ ( ) 1− and 0 < < 1, namely,
( )
1
Φ ( ) 1− holds.
0 < ( ) ≤ ( ) , | ( + 1)| ≤ ( )

1 1
Φ ( ) 1− Φ ( ) 1−
When −1 < < 0, ( ) = ( ) = − ( )| | .
1− 1−

( + 1)

Φ ( ) 1−
= ( ) ( ) − ( ) ( ( ) ) − ( ) (42)

1
Φ ( ) 1−
= ( ) ( ) + ( ) ( ( )| |) − ( ).

1
1−
Φ ( ) .
Similar to the above proof process, we can obtain | ( + 1)| ≤ ( )

To this end, when ( ) ∈ Ω, ( + 1) ∈ Ω.

3) After the sliding mode variables enter the determination domain Ω, we will further discuss the bounded
convergence region of the tracking errors. Before proving, it is necessary to introduce an important lemma
about the discrete fast terminal sliding mode surface, as follows.

Lemma 4 [20] : Consider a scalar dynamical system

(43)
( + 1) = ( ) − ( ) + ( )

where > 0, > 0 and 0 < < 1. if | ( )| < , > 0, then the state ( ) is always bounded and there is a
finite step to guarantee
( 1 )

1 1−
| ( )|≤ ( )· , . (44)

When the sliding mode variable enters the domain Ω, combined with the analysis of the fractional order fast
terminal sliding mode surface, we can see

1 1 ( ) + 2 ( ) + 2 [| 1 ( )| ( 1 ( ))] = ( ). (45)
By the definition of GL fractional order operator,

Õ
2
1 1 ( ) + 2 ( ) + (−1) | 1 ( − )| ( 1 ( )) = ( ). (46)

=0
According to Lemma 1 and equation (17), the following expression can be obtained:

2
1 ( + 1) = (1 − 1 ) 1 ( ) + | 1 ( )| ( 1 )( )) + Υ ( ) (47)