Page 91 - IJOCTA-15-1

Existence of mild solution for fuzzy fractional differential equation utilizing the Hilfer-Katugampola . . .
Lemma 2. The Cauchy problem (1) can be rep-
Γ(u)Γ(v) resented as the following integral equation
B(u, v) = .
Γ(u + v)
ϱ
t
ϱ µ−1
Definition 2. 24 The katugampola left-sided frac- e x(t) = e x 0 (t ) + 1 Z t − s ϱ ω−1
tional integral of order m with the lower limit u ϱ µ−1 Γ(µ) Γ(ω) 0 ϱ
of ex ∈ C(U, H) for −∞ < 0 < t < ∞ is defined × s ϱ−1 [Aex(s) + F(s, ex(s))]ds. (4)
by
Proof. We can cite 22 for reference, which
ϱ 1−τ Z t s ϱ−1
ϱ τ will be omitted here.
I + ex(t) = Γ(τ) (t − s ) e x(s)ds,
0
ϱ 1−τ
ϱ
0
Remark 3. 25 The wright function M ex (δ) is
defind by,
where t > b, ϱ > 0 and τ > 0.
The Katugampola fractional integral is defined ∞ t−1
X (−δ)
with respect to an additional parameter ϱ > 0. M ω (δ) = ,
These operators have special properties based on t=1 (t − 1)! ω(1 − ωt)
the value of ϱ.
where ω ∈ (0, 1), δ ∈ [0, +∞). Which satisfies the
+
Remark 2. Specifically, as ϱ → 0 , the equality given below
Katugampola fractional integral converge to the
Hadamard fractional integral, Z ∞ Γ(1 + s)
s
δ M ω (δ)dδ = (1 + ωs) , s ≥ 0,
τ,ϱ 1 Z t t τ−1 ds 0
lim I + ex(t) = log e x(s) ,
ϱ→0 0 Γ(τ) 0 s s where M ω (δ) is a probability density function and
is given as follows:
when the parameter ϱ = 1, they coincide with the
1 1 1
Riemann-Liouville fractional integral, M ω (δ) = δ −1− ω ζ ω (δ − ω ) ≤ 0,
ω
Z t
1 e x(s)
τ,1
I + ex(t) = ds.
0 Γ(τ) 0 (t − s) 1−τ for 0 < ω < 1, δ ∈ [0, +∞).
Definition 3. 12 Let Order τ and type ω sat- Definition 4. A mild solution of the Cauchy
isfy τ ∈ (0, 1] and ω ∈ [0, 1]. The fuzzy Hilfer- problem (1) is defined as the function ex ∈
Katugampola generalized Hukuhara fractional de- C τ,ω (U, H)
rivative, with respect to t with ϱ > 0 of a function
t ∈ C 1−µ,ϱ [0, u], is defined by ϱ Z ϱ ϱ ω−1
t
t t − s
e x(t) = R τ,ω e x 0 +
ϱ ϱ
0
d
ϱ
ϱ τ,ω ϱ ω(1−τ) ϱ−1 ϱ (1−τ)(1−ω) t − s ϱ
D + ex(t) = s e x(t) ϱ−1
0
0
0 I + ds I + × s H ω F(s, ex(s))ds, (5)
ϱ

= ϱ ω(1−τ) ϱ (1−τ)(1−ω) e x(t).
I +
0
0 γ ϱ I +
where
Lemma 1. 4 Assume that the linear operator A
acts as the infinitesimal generator of a semigroup
ϱ ϱ ω−1 ϱ
of class C 0 if and only if t 1+ϱ(µ−1)−ω t t
R τ,ω = I 0 + H ω ,
• The set A has the property of being closed ϱ ϱ ϱ
ϱ Z ∞ ϱ ω
and D(A) = B. t t
H ω = ωδM ω (δ)J δ dδ.
• The resolvent set p(A) of A includes pos- ϱ 0 ϱ
itive real numbers, ∀ α > 0, ϱ
t
Definition 5. 26 As for operator R τ,ω ( ) and
ϱ
1 t ϱ
∥R(β, A)∥ ≤ , H ω ( )
ϱ
β
t ϱ t ϱ
where • R τ,ω ( ) and H ω ( ) are lin-
ϱ
ϱ
∞ q ear, bounded and compact operators, so
Z t>0 t>0
q
R(β, A) = (β I − A) −1 s = e −β t J(t)sdt.
we have
0
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