Page 118 - IJOCTA-15-1

D. Kasinathan et.al. / IJOCTA, Vol.15, No.1, pp.103-122 (2025)

Take b = max{b 0 , b 1 }, for each y ∈ B b . Hence the
ς∈J
n p
¯
¯
p
b ≤ 8 p−1 2 p−1 M M g [E∥φ∥ + L p ] claim.
ϑ
p
¯ p
p
p
p
+ 2 p−1 M [E∥y 1 ∥ + E∥η∥ ] + M (1 + M ) b ∗
ς
g
σ 1
p
p
p p
∗
+ M ς [L (1 + M )] b + M C p ς 2 ¯ ∗
p ˜
p
L G b
ς
ς
f
σ 2
∞ p X
¯ 1 ∗
X
¯ 1
p
L + M
+ M CHς pH−1 ˆ ϑ p L k q L b Step 2: The operator Φ is a contraction mapping.
k
ς
0
k=1 0<ς k <ς Case (i): Let ς ∈ J and y, ¯y ∈ B we have
T
∞ p X o
¯ 2
¯ 2 ∗
X
+ M p L q L b
ς k k
k=1 0<ς k <ς
n p
¯
¯
p
≤ 8 p−1 2 p−1 M M g [E∥φ∥ + L p ]
ϑ
(Φy)(ς)
h
p
p
¯ p
p
p
+ 2 p−1 M [E∥y 1 ∥ + E∥η∥ ] + M (1 + M )
ς g σ 1 ϑ
Z
p
p
p ˜
+ M ς [L (1 + M )] + M C p ς 2 ¯ = T α (ς)[y 1 + η] − g(ς, y ς + z ς , σ 1 (ϑ, τ, y τ + z τ )dτ)
p
p p
L G
ς
ς
f
σ 2
∞ p X 0
X
¯ 1
¯ 1
+ M p L q L ς ϑ
ϑ k k Z Z
k=1 0<ς k <ς + T α (ς − ϑ) f(ϑ, y ϑ + z ϑ , σ 2 (ϑ, τ,
∞ p
X q X 2 i ∗ p pH−1 ˆ o 0 0
¯ 2
p
+ M ς L k L b + M CHς L
k
ς
k=1 0<ς k <ς y τ + z τ )dτ) dϑ
(4) ς
Z
:= b 1 , + T α (ς − ϑ)G(ς, ϑ, y ϑ + z ϑ )dw(ϑ)
0
where ς
Z
H
+ T α (ς − ϑ)σ(ϑ)dB (ϑ)
Q
0
n p
¯
p
¯
b 1 = 8 p−1 2 p−1 M M g [E∥φ∥ + L p ]
ϑ
h
p
¯ p
p
p
p
+ 2 p−1 M [E∥y 1 ∥ + E∥η∥ ] + M (1 + M )
ς
g
σ 1
p
p p
p ˜
p
+ M ς [L (1 + M )] + M C p ς 2 ¯
p
ς f σ 2 ς L G and
∞ p X
X
¯ 1
¯ 1
+ M p L q L
ϑ k k
k=1 0<ς k <ς
∞ p
X q X 2 i ∗
p
¯ 2
+ M ς L k L b (Φ¯y)(ς)
k
k=1 0<ς k <ς
o = T α (ς)[¯y 1 + η]
p
+ M CHς pH−1 ˆ
L .
ς
ϑ
Z
− g(ς, ¯y ς + z ς , σ 1 (ϑ, τ, ¯y τ + z τ )dτ)
Dividing both side of (4) by b and letting as
0
b → ∞, we get
ς ϑ
Z Z

+ T α (ς − ϑ) f(ϑ, ¯y ϑ + z ϑ , σ 2 (ϑ, τ, ¯y τ + z τ )dτ) dϑ
n
p
p
¯ p
p p
p
32 p−1 M (1 + M ) + M ς [L (1 + M )] 0 0
ς
g
f
σ 1
σ 2
ς
∞ p Z
p
p ˜
+ M C p ς 2 ¯ p X L 1 q X L 1 + T α (ς − ϑ)G(ς, ϑ, ¯y ϑ + z ϑ )dw(ϑ)
L G + M
ς ϑ k k
k=1 0<ς k <ς 0
∞ p o Z ς
p X 2 q X 2 p
+ M L L l ≥ 1. H
ϑ k k + T α (ς − ϑ)σ(ϑ)dB (ϑ),
Q
k=1 0<ς k <ς
0
This contradicts with (H 8 ). hence, we have
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