Page 124 - IJOCTA-15-3

Amol D. Khandagale et al. / IJOCTA, Vol.15, No.3, pp.493-502 (2025)
κ
Theorem 1. ( [25, 33]) Let K ⊂ C 1−κ (J , R) By applying D + to either sides of (7), it can
p
be closed, convex, and bounded in nature. Let be deduced from Lemma 1, Lemma 6, and Defi-
the operators ∆ : C 1−κ (J , R) → C 1−κ (J , R) and nition 2 that
Ω : K → C 1−κ (J , R) satisfying: κ τ(1−σ)
D + [φ (s) − f (s, φ (s))] = D + z (s) . (8)
p p
(1) ∆ is D-contraction of nonlinear type,
κ
From (8), and the hypothesis D + [φ − f (·, φ (·))] ∈
(2) Ω is completely continuous and p
(3) u = ∆u + Ωw ⇒ u ∈ K for all w ∈ K. C 1−κ [a, b], we have
1−τ(1−σ) τ(1−σ)
Then, the mapping ∆ + Ω owns a fixed point in DI + z = D + z ∈ C 1−κ [a, b] . (9)
p p
K.
Also, since z ∈ C 1−κ [a, b], by Lemma 3,
Lemma 9. Let a function f : J × R → R 1−τ(1−σ)
I + z ∈ C 1−κ [a, b] . (10)
p
such that, for any φ ∈ C 1−κ [p, q], f (· , φ (·)) ∈
C 1−κ [p, q]. Then, φ ∈ C 1−κ [p, q] is a solution of It follows from equations (9) and (10) that
fractional IVP 1−τ(1−σ) z ∈ C 1
I + 1−κ [a, b] .
p
σ,τ
D + [φ (s) − f (s, φ (s))] = z (s) , s ∈ J ,
p 1−τ(1−σ)
(4) Thus, z and I + z verify the conditions of
p
1−κ
I + [φ (s) − f (s, φ (s))] p + = φ 0 , κ = σ+τ−στ, Lemma 4.
p 1−τ(1−σ)
(5) Next, applying I + to either sides of
p
if and only if φ (s) satisfies the following VIE Equation (8), and using Def. 3, and Lemma 4,
it implies that
κ−1
φ 0 (s − p)
σ,τ
φ (s) = + f (s, φ (s)) D + [φ (s) − f (s, φ (s))] = z (s)
Γ (κ) p
h i
s 1−τ(1−σ) +
1 Z σ−1 I + z (s) (p ) τ(1−σ)−1
p
+ (s − w) z (w) dw. (6) − (s − p) .
Γ (σ) Γ (τ (1 − σ))
p (11)
Proof. Let φ be a solution of (4)–(5). Conse- Since, 1 − κ < 1 − τ (1 − σ), using Lemma 5,
quently, by Lemma 4, we obtain we get
h 1−τ(1−σ) i
σ
σ,τ
I +D + [φ (s) − f (s, φ (s))] = φ (s) − f (s, φ (s)) I + z (s) p + = 0.
p p p
1−κ
+
I + [φ (s) − f (s, φ (s))](p ) Hence, (11) reduces to
− p (s − p) κ−1 σ,τ
Γ (κ) D + [φ (s) − f (s, φ (s))] = z (s) , s ∈ J .
p
σ
= I +z (s) . Now, in order to show that equation (5) also
p
1−κ
Then, holds, apply I + to either sides of (7), then by
p
σ
φ (s) − f (s, φ (s)) = I +z (s) means of Lemma 1 and Lemma 2, we get
p 1−κ 1−κ 1−κ σ
+
1−κ
I + [φ (s) − f (s, φ (s))](p ) I + φ (s) = φ 0 + I + f (s, φ (s)) + I + I +z (s) .
p
p
p
p
+ p (s − p) κ−1
Γ (κ) Since, 1 − κ < 1 − τ (1 − σ) it follows from
κ−1 Lemma 5, when taking the limit as s → p +
φ 0 (s − p)
σ
= I +z (s) + . I + [φ (s) − f (s, φ (s))] p + = φ 0 .
1−κ
p
Γ (κ) p
So, This completes the proof. □
φ 0 (s − p) κ−1
σ
φ (s) = + f (s, φ (s)) + I +z (s)
p
Γ (κ) 3. Main result
(7) Taking into account the following theories, we can
φ 0 (s − p) κ−1 demonstrate the main result:
= + f (s, φ (s))
Γ (κ) Q1: Following weak contraction condition
s holds for the function f (s, ·)
1 Z σ−1
+ (s − w) z (w) dw, s ∈ J . |f (s, v) − f (s, w)| ≤ ψ D (|v − w|) ,
Γ (σ)
p for all s ∈ J and v, w ∈ R, where ψ D (s)
Thus, Equation (6) holds. Now, to demonstrate is a D−function.
the sufficiency. Let φ ∈ C κ [p, q] satisfy Equa- Q2: For every s ∈ J and v ∈ R, there
1−κ
tion (6), which can be expressed as (7). is a continuous function ξ ∈ C 1−κ (J , R),
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