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Page 8 of 14 Ma et al. Complex Eng Syst 2023;3:10 I http://dx.doi.org/10.20517/ces.2023.14

2
where 1 2, 1, and 3 are positive constants, 2 ≤ 1 erf(1). Let 3 = 4 2 , with 4 and 4 being integers
1+ 4
and 0 < 4 < 1, 1 + 4 < 4.
Select a fixed-time sliding mode surface as:

¹

¤ 4 4 ¤ 5 ¤ 5 d (25)
2 = + 21 (d c + d c ) + 22 d c + d c
0
where 21 and 22 are positive constants, 0 < < 1 and > 1, = 4, 5 are positive odd integers.
Theorem 4 For the third-order subsystem (20), if the fixed-time sliding mode surface is chosen as (25) and the
fixed-time controller is designed as (26),

4
¤
2 = ¤ + ¤ + ¤ + ℎ( , ) + 21 (d c 4 + d c )
(26)

¤ 5 ¤ 5 6 6 ˆ
+ 22 d c + d c + 1 d 2 c + 2 d 2 c + 3 erf( 2 ) + 2
thentheslidingmodesurface 2 isfixed-timestable, whichwillconvergeintoasmallregionoforiginwithinsettling

1 1
3
≤ + . ℎ( , ) = 1 erf( ) − 2 erf( ) + 3 erf( ) + 1 | | erf ,
time 2 ¯ ¯

4 2 6 2 (1− ¯ 6 ) 5 2 6 2 ( ¯ 6 −1) 2
in which 1 , 2 are positive constants, and 6 , 6 are positive odd integers satisfying 0 < 6 < 1, 6 > 1.
Proof of Theorem 4 The proof process will be conducted in 3 steps: (1) After the angular error converges
to zero according to Theorem 2, 2 and the auxiliary variable can converge into a small region around the
origin within a fixed time; (2) The error variables , can converge into a small region around the origin
within a fixed time; (3) It should be proved that and do not escape to infinity before the angular error
converges to zero.
1 2
Step 1 Select a positive Lyapunov function 4 = , differentiating it and substituting (24)-(26) yields to
2 2
¤
4 = 2 ¤ 2
1
6 6 ˆ
= 2 − 1 d 2 c − 2 d 2 c − 3 erf( 2 ) + ( 2 − 2 )

1
˜
6 +1 6 +1
≤ − 1 | 2 | − 2 | 2 | − 3 2 erf( 2 ) + 2 2
(27)
1
6 +1 6 +1
˜
≤ − 1 | 2 | − 2 | 2 | − 3 2 tanh( 2 ) + 2 2

1
6 +1 6 +1
˜
≤ − 1 | 2 | − 2 | 2 | − 3 | 2 | + | 2 || 2 |

¯
≤ −2 1 ¯ 6 − 2 2 ¯ 6 + 2
¯ 6
¯ 6
3 3
¯

where ¯ 6 = 6 +1 , ¯ 6 = 6 +1 , 2 = 3 2 with 2 being a positive constant. Using the Lemma 4, the third-order
2 2
n o
1 1
subsystem(14)isfixed-timestable,and 2 willconvergeintoasmallset Δ 4 = 2 | ( 2 ) ≤ min 3 ¯ ¯ 6 , 3 ¯ ¯ 6
1 2 6 2 2 6
, which is determined by
around zero in the fixed time 2
1 1
≤ + (28)

2 ¯ ¯

4 2 6 2 (1 − ¯ 6 ) 5 2 6 2 ( ¯ 6 − 1)
Then, 2 will hold in a small region of origin, which guarantees a real sliding mode surface [23] . Therefore, the
auxiliary variable and its derivative will also converge into the origin along the sliding mode surface [24] .
¤
Step 2 According to (25), when the auxiliary = 0, one has
¤

¤ = − 1 erf( ) + 2 erf( ) − 3 erf( ) − 1 | | erf (29)
3
3