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Sun et al. Complex Eng Syst 2022;2:17 I http://dx.doi.org/10.20517/ces.2022.48 Page 7 of 15

Recalling the condition (7), we can get

1 ( ∧ , ) | ( ∧ , )| ≤ .
This implies

− ( ∧ , )
( , ≤ ) ≤ | ( ∧ , )| ≤ .
1
We observe that

( , ≤ ) ≤ − ( ∧ , ) .
1
Letting → ∞ yields that ( ∞, ⩽ ) = 0. Hence, ∞, = ∞ a.s. Therefore, we have ∞, = ∞ a.s. This implies
that the unique solution for the -th subsystem (14) will not explode in finite time.
Step 2. This section proves the existence of a unique global solution for SSS (1). Let 0 > 0 be a sufficiently
large integer, such that 0 > | (0)|, where | (0)| is the initial data of the -th subsystem. For any integer
≥ 0, we define the stopping time sequence as follows:

= inf{ ∈ [ , +1 ) : | ( )| ⩾ }.

Clearly, increases as → ∞. For ∈ [ 0 , 1 ), ( ) = 0, using the It ˆ formula, we have

0 0 0
( ∧ )
Λ( ∧ , ( ∧ ), 0 )
∫ ∧ 0

≤ 0 Λ( 0 , ( 0 ), 0 ) + ( , 0 ) , (17)
0
where ( , 0 ) = Λ( , ( ), 0 )+LΛ( , ( ), ( ), 0 ). Letting = 1,accordingtocondition(8)inAssumption
3, we derive that
1 Λ( 1 , ( 1 ), 1 ) ≤ 1 Λ( 1 , ( 1 ), 0 )
∫
1

≤ [ 0 Λ( 0 , ( 0 ), 0 ) + ( , 0 ) ]. (18)
0
For ∈ [ 1 , 2 ), ( ) = 1, we obtain
1 1 1
( ∧ )
Λ( ∧ , ( ∧ ), 1 )
∫ ∧ 1

≤ 1 Λ( 1 , ( 1 ), 1 ) + ( , 1 ) . (19)
1
Combining (18) and (19), it implies that
1 1 1
( ∧ )
Λ( ∧ , ( ∧ ), 1 )
∫ ∫ ∧ 1
1

≤ [ 0 Λ( 0 , ( 0 ), 0 ) + ( , 0 ) ] + ( , 1 ) . (20)
0 1
For ∈ [ −1 , ) and ( ) = −1, we assume that
( ∧ −1 ) Λ( ∧ −1 , ( ∧ −1 ), −1 )

∫ ∧ −1

≤ −1 Λ( −1 , ( −1 ), −1 ) + ( , −1 )
−1
∫
1

≤ ( −1 , 0 ) 0 Λ( 0 , ( 0 ), 0 ) + ( −1 , 0 ) ( , 0 )
0
∫ ∫
2 −1

+ ( −1 , 0 )−1 ( , 1 ) + · · · + ( , −2 )

1 −2
∫ ∧ −1

+ ( , −1 ) . (21)
−1