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Page 8 of 15 Sun et al. Complex Eng Syst 2022;2:17 I http://dx.doi.org/10.20517/ces.2022.48

By mathematical induction, for ∈ [ , +1 ) and ( ) = , we have

∫ ∧

Λ( ∧ , ( ∧ ), ) =
( ∧ ) Λ( , ( ), ) + ( , ) . (22)

It follows from (8) and (21) that

( ∧ )
Λ( ∧ , ( ∧ ), )
∫ ∧

≤ Λ( , ( ), −1 ) + ( , )

∫
1

≤ ( , 0 ) 0 Λ( 0 , ( 0 ), 0 ) + ( , 0 ) ( , 0 ) + · · ·

0
∫ ∫ ∫ ∧
−1
2

+ ( , −2 ) + ( , −1 ) + ( , ) . (23)

−2 −1
Because > 1, we obtain from (23) that
∫ ∧
( , 0 )
( ∧ ) [ 0 Λ( 0 , ( 0 ), 0 ) + ( , ) ].
Λ( ∧ , ( ∧ ), ) ≤
0
Similar to the proof stated in Part 1, we can derive
∫ ∧
( , 0 )
Λ( ∧ , ( ∧ ), ) ≤
( ∧ ) [ 1 − ( 1 − 2 ¯ ) ( ( ))
0
∫ ∧

− ( 3 − 4 ¯ − 2 ) | ( )| ],
0
where
( ) ( ∫ )
0
1 = sup 0 Λ( 0 , , 0 ) + 2 ¯ sup ( )
0
[ 0 − , 0 ] [ 0 − , 0 ] 0 −
( ∫ )
0

+ 4 ¯ sup 0 | | ,
[ 0 − , 0 ] 0 −
is finite. Then,

Λ( ∧ , ( ∧ ), ) ≤ 1
( ∧ ) ( , 0 ) . (24)

Recalling condition (7), we obtain

1 ( , 0 ) − ( ∧ )
| ( ∧ )| ≤ .

1
This implies

1 ( , 0 ) − ( ∧ )
( ≤ ) ≤ 1 .

Letting → ∞, we observe that ( ≤ ) = 0 and hence ≥ a.s. We let → ∞ in (24) to obtain
∞ ∞
Λ( , ( ), ( )) ≤ ( , 0 ) 1 − .