Page 126 - IJOCTA-15-3

Amol D. Khandagale et al. / IJOCTA, Vol.15, No.3, pp.493-502 (2025)
1−κ 1−κ
(s − p) Ωz (s) (s − p) φ (s)
s |φ 0 |
(s − p) 1−κ Z σ−1 κ−1 ≤ + (s − p) 1−κ |f (s, φ (s)) − f (s, 0)|
≤ (s − w) (s − p) ∥ξ∥ ds Γ (κ)
Γ (σ) C 1−κ
p + (s − p) 1−κ |f (s, 0)|
σ
(q − p) β(κ, σ) s
≤ ∥ξ∥ , ∀ s ∈ J . (s − p) 1−κ Z
Γ (σ) C 1−κ + (s − w) σ−1 g (w, z (w)) dw
Γ (σ)
Taking maximum over s, we get p
σ
(q − p) β(κ, σ) |φ 0 |
∥Ωz∥ ≤ ∥ξ∥ . ≤ + (q − p) 1−κ ψ D ∥φ∥ + Λ
C 1−κ Γ (σ) C 1−κ Γ (κ) C 1−κ
σ
This proves the uniform boundedness of the op- (q − p) β(κ, σ)
erator Ω on K. Let s 1 , s 2 ∈ J with s 1 < s 2 . For + Γ (σ) ∥ξ∥ C 1−κ
any z ∈ K, one has |φ 0 |
≤ + (q − p) 1−κ ψ D (η) + Λ
Γ (κ)
σ
(q − p) β(κ, σ)
+ ∥ξ∥
1−κ 1−κ
(s 1 − p) Ωz (s 1 ) − (s 2 − p) Ωz (s 2 ) Γ (σ) C 1−κ

|φ 0 |
s 1
1−κ Z 1−κ
(s 1 − p) ≤ + (q − p) η + Λ
= (s 1 − w) σ−1 g (w, z (w)) dw Γ (κ)
Γ (σ) σ

p
(q − p) β(κ, σ)
+ ∥ξ∥ .
C 1−κ
s 2 Γ (σ)
Z
(s 2 − p) 1−κ σ−1
− (s 2 − w) g (w, z (w)) dw
Γ (σ)
p Applying maximum over s, we get
∥ξ∥ β(κ, σ) |φ 0 | 1−κ
≤ C 1−κ (s 1 − p) 1−κ (s 1 − p) σ+κ−1 ∥φ∥ ≤ + (q − p) η + Λ

Γ (σ) C 1−κ Γ (κ)
σ
(q − p) β(κ, σ)
−(s 2 − p) 1−κ (s 2 − p) σ+κ−1 ∥ξ∥
+
Γ (σ) C 1−κ
∥ξ∥ β(κ, σ)
σ
σ
≤ C 1−κ |(s 1 − p) − (s 2 − p) | . ≤ η.
Γ (σ)
Hence, φ ∈ K. Thus, all the three assumptions of
Thus, there exists a ε > 0, for some η such that the Theorem 1 are hold, which implies that there

|s 2 − s 1 | ≤ ε ⇒ (s 1 − p) 1−κ Ωz (s 1 ) exists φ ∈ K such that ∆φ + Ωφ = φ. Thus, the
operator ∆+Ω has a fixed point in K. Hence, the

−(s 2 − p) 1−κ Ωz (s 2 ) ≤ η,

IVP (1)–(2) owns a solution in C 1−κ (J , R). □
for all s 2 , s 1 ∈ J and z ∈ K. This proves the
equicontinuity of ΩK in C 1−κ (J , R) and hence, Remark 1. For τ = 1 the IVP (1)–(2) reduces
by Arzela–Ascoli theorem, it is compact. to a problem with the fractional derivative opera-
tor of Caputo type. Hence, the results obtained in
the Theorem 2 coincide with the results published
Step III: There exists φ ∈ K such that φ = 32,33
in.
∆φ + Ωz for all z ∈ K.
Let φ ∈ C 1−κ (J , R) and z ∈ K, such that φ = 4. Application
∆φ + Ωz. Using Hypothesis (Q1), we can write
Consider the following IVP:
1−κ
(s − p) φ (s)
1 1 1
,
2 2
1−κ D + [φ (s) − ln(1 + |φ(s)|)] = sin(|φ(s)|),
= (s − p) (∆φ (s) + Ωz (s)) 0 s + 1
2

1−κ 1−κ (18)
≤ (s − p) ∆φ (s) + (s − p) Ωz (s)

1
|φ 0 | s ∈ J = (0, ],
≤ + (s − p) 1−κ f (s, φ (s)) 2

Γ (κ)
1− 3
4 +
s I + [φ (s) − ln(1 + |φ(s)|)] (0 ) = 2, (19)
1−κ Z 0
(s − p)
1
3
1
1
+ (s − w) σ−1 g (w, z (w)) dw Here, σ = , τ = , κ = , p = 0, q = , φ 0 = 2,
Γ (σ)
2 2 4 2

p f (s, φ (s)) = ln(1 + |φ(s)|),
498