Page 93 - IJOCTA-15-1

Existence of mild solution for fuzzy fractional differential equation utilizing the Hilfer-Katugampola . . .
Therefore, Ξ is bounded and Ξex ∈ B r .
ϱ 1−µ t ϱ t ϱ 1−µ

t
≤
2 R τ,ω 2 e x 0 − 1

ϱ ϱ ϱ
Step-2: t ϱ

× R τ,ω 1 e x 0
Now, we show that Ξ is continuous. Let {ex i } ⊂ ϱ
B r such that ex m → ex ∈ B r as i → ∞.
Z ϱ 1−µ ϱ ϱ ω−1

t 2 t 2 t − s ϱ−1
2
From hypotheses (H2) and (H4), we can write, +
ϱ ϱ s

t 1
∀t ∈ U, ϱ
t − s ϱ
2
× H ω F(s)ds
ϱ
∥F(t, ex i (t))∥ → ∥F(t, ex(t))∥ as i → ∞,

Z ϱ 1−µ ϱ ϱ ω−1 ϱ 1−µ
t 1 t t − s t

2 2 1
and +
−

ϱ ϱ ϱ
0
ϱ ϱ ω−1 ϱ ϱ
t − s ϱ−1 t − s
1
2
∥F(t, ex i (t)) − F(t, ex(t))∥ ≤ 2m(t) ∀i ∈ N. × s H ω F(s)ds
ϱ ϱ
By Lebesgue dominated convergence theorem, for
Z t 1 ϱ 1−µ ϱ ϱ ω−1

t 1 t − s ϱ−1
1
any t ∈ U, +
s

ϱ ϱ
0
ϱ ϱ ϱ ϱ
t − s t − s
2
1
× H ω − H ω F(s)ds
ϱ ϱ
∥Ξex i (t) − Ξex(t)∥ C τ,ω
ϱ 1−µ
t ∥Ξex(t 2 ) − Ξex(t 1 )∥ C τ,ω ≤ J + I 1 + I 2 + I 3 .
≤ sup
t∈U ϱ J is automatically tends to 0.
t ϱ ϱ ϱ ϱ t 2 ϱ

Z ω−1
Z ϱ 1−µ ϱ ω−1
2

t − s t − s
t 2 t − s ϱ−1
×
s ϱ−1 H ω F i (s)ds I 1 =
s

ϱ ϱ
ϱ ϱ
0 t 1
t t − s ϱ t − s ϱ t − s
ϱ
ϱ
Z ω−1
ϱ ϱ
− s ϱ−1 H ω F(s)ds
× H ω 2 F(s)ds ,

0 ϱ ϱ
ϱ
ϱ
L 1 t ϱ(1−µ) Z t
Z t 1 t ϱ 2 1−µ t − s ϱ ω−1 t ϱ 1−µ
2
ϱ
1
ϱ ω−1 ϱ−1

≤ (t − s ) s I 2 =
−
ϱ 1−µ ω−1 Γ(ω) 0
0 ϱ ϱ ϱ
ϱ
ϱ
ϱ
× ∥F(s, ex i (s)) − F(s, ex(s))∥ds t − s ϱ ω−1 t − s ϱ

× 1 s ϱ−1 H ω 2 F(s)ds ,

L 1 u ϱ(1−µ) Z t ϱ ϱ ω−1 ϱ−1 ϱ ϱ
≤ (t − s ) s
Z ϱ 1−µ ϱ ω−1
ϱ τ(ω−1) Γ(ω) 0
t 1 t 1 t − s ϱ ϱ−1
1
I 3 =
s
× ∥F(s, ex i (s)) − F(s, ex(s))∥ds.
0 ϱ ϱ
ϱ ϱ ϱ ϱ
t − s t − s
1
2
× H ω − H ω F(s)ds
ϱ ϱ
Apply i → ∞, then ∥Ξex i (t)−Ξex(t)∥ → 0. There-
fore we conclude that Ξ is continuous. L 1 t ϱ(1−µ) ∥m(s)∥
ϱ
ϱ ω
I 1 ≤ 2 (t − t ) ,
ϱ τ(ω−1) Γ(ω + 1) 2 1
Step-3:
L 1 ∥m(s)∥ Z t 1 ϱ(1−µ)
ϱ
ϱ ω−1
Now, we show that Ξ is equicontinuous for each I 2 ≤ τ(ω−1) s ϱ−1 t 2 (t − s )
2
t ∈ U, let for any ex ∈ B r and 0 ≤ t 1 ≤ t 2 ≤ u, ϱ Γ(ω) 0

ϱ(1−µ) ϱ ϱ ω−1
− t 1 (t − s ) ds.
1
ϱ 1−µ
t 2
≤ sup ϱ ϱ
∥Ξex(t 2 ) − Ξex(t 1 )∥ C τ,ω Therefore, I 1 → 0 and I 2 → 0 as t → t and from
1
2
t 2 ∈U ϱ ϱ ϱ
the definition (5), we obtain I 3 → 0 as t → t .

ϱ Z t 2 ϱ ϱ ω−1 2 1

t t − s
×
R τ,ω 2 e x 0 + 2 s ϱ−1 Therefore, we say that Ξ is equicontinuous.
ϱ 0 ϱ

ϱ ϱ
ϱ 1−µ
t − s
t
2 1
× H ω F(s)ds
− sup Step-4:
ϱ
t 1 ∈U ϱ

ϱ Z t 1 ϱ ϱ ω−1 Now, we need to prove that t ∈ U, Ξ(t) =

t t − s
×
R τ,ω 1 e x 0 + 1 s ϱ−1 {(Ξex)(t) : t ∈ B r } is relatively compact in
ϱ 0 ϱ

ϱ ϱ
C τ,ω (U, H). Take 0 ≤ t ≤ u then, for every ϵ > 0
t − s
1
× H ω F(s)ds
and α > 0, let t ∈ B r and define the operator Ξ
ϱ
on B r by
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