Page 92 - IJOCTA-15-1

R. Hariharan, R. Udhayakumar / IJOCTA, Vol.15, No.1, pp.82-91 (2025)
Theorem 3. If Under the hypotheses (H1) −

ϱ
ϱ(µ−1) 1−µ (H4) are fulfilled, then the Cauchy problem (1)
t Lt ϱ Γ(ϱ(µ − 1) + 1)

R τ,ω has a mild solution on U.

e x ≤ ∥ex∥,

ϱ
(Γ(µ)) 2
Proof. Let we consider B r = {ex ∈

ϱ
t C τ,ω (U, H) : ∥ex∥ ≤ H}, ∀ H > 0. Then B r is

L 1
H ω

e x ≤ ∥ex∥, for all ω ∈ B.
ϱ Γ(ω)

closed, bounded and convex subset of C τ,ω (U, H),
the operator Ξ is defined on B r by
t ϱ t ϱ
• R τ,ω ( ) and H ω ( ) are
ϱ t>0 ϱ t>0
strongly continuous operators, i.e., for all ϱ Z ϱ ϱ ω−1
t
t 1 , t 2 ∈ U, we have t t − s
Ξex(t) =R τ,ω e x 0 +

ϱ ϱ
ϱ 0 ϱ
t t

2 1
R τ,ω e x − R τ,ω t − s

e x → 0, ϱ ϱ

ϱ ϱ
× s ϱ−1 H ω F(s, ex(s))ds, t ∈ U.

ϱ ϱ
ϱ
t t

2 1
ϱ ϱ
H ω e x − H ω 2 1

e x → 0, as t → t .
ϱ ϱ

Next to show that Ξ has a fixed point on B r .
Theorem 1. 27,28 Let B be a Banach space and E Step-1:
is a nonempty, closed, bounded, and convex sub-
Now, we show that Ξex ∈ B r , whenever ex ∈ B r .
set of a Banach space B, such that H : E → E is a
Let we consider ∀H > 0, ∃t ∈ U, such that
compact operator. Then H has at least one fixed
point in E.
∥Ξex(t)∥ > H, (6)
Theorem 2. 29 Let M n (t) be a sequence of func-
tions from [a,b] to R which is uniformly bounded
and equicontinuous. Then, M n (t) has a uniformly In the following, for our convenience, we put
convergent subsequence.
F(s, ex(s)) = F(s).
3. Existence of mild solution
Therefore,
The upcoming findings will utilize the following
assumptions:
(H1) J(t) remains continuous under the uni- ∥Ξex(t)∥ C τ,ω
form operator topology for t > 0, and t ϱ 1−µ
t ϱ

{J(t)} ϱ≥0 is uniformly bounded, there ex- = sup
R τ,ω e x 0

t∈U ϱ ϱ
ists L E > 1 such that
t
ϱ
ϱ
Z t − s ϱ ω−1 t − s ϱ

sup ∥J(t)∥ = L E . + s ϱ−1 H ω F(s)ds
ϱ∈[0,∞) 0 ϱ ϱ
ϱ 1−µ
ϱ
t
t
(H2) For any t ∈ U, the function F : B → H ≤ sup
R τ,ω e x 0
is almost continuous, and for every ex ∈ t∈U ϱ
ϱ
C τ,ω (U, H), the function F : U → H is
Z ϱ ϱ ω−1 ϱ ϱ
t

t − s ϱ−1 t − s
strongly measurable. +
s H ω F(s)ds

ϱ ϱ
+
2
(H3) There exist function q(·) ∈ L (U, R ) and 0
B r := {ex ∈ H : ∥ex∥ ≤ r} ⊂ H, r > 0, and LΓ(ϱ(µ − 1) + 1) L 1 t ϱ(1−µ) ∥m(s)∥
≤ ∥ex 0 ∥ +
ϱ
such that for any ex, h ∈ C τ,ω (U, H) and Γ(µ) 2 ϱ 1−µ ω−1 Γ(ω)
∀ t ∈ [0, a], we have Z t
ϱ
ϱ ω−1 ϱ−1
× (t − s ) s ds

F(t, ex(t)) − F(t, h(t)) ≤ q(t) ex(t) − h(t) .

0
+
2
(H4) There exists a function φ ∈ L (U, R ) ≤ LΓ(ϱ(µ − 1) + 1) ∥ex 0 ∥
such that Γ(µ) 2

ϱ 1+τ(ω−1)
F(t, ex(t)) ≤ φ(t), u ∥m(s)∥ ϱ − 1

+ L 1 B , ω .
∀ex ∈ C τ,ω (U, H) and for almost all t ∈ U. ϱ Γ(ω) ϱ
and there is a constant S φ > 0, such that
Next dividing to both side by H and taking limit
∥φ(t)∥
lim = S φ . as H → ∞, then we have a contradiction to our
H→∞ H
assumption (6).
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