Page 40 - IJOCTA-15-3

M. Aychluh et.al. / IJOCTA, Vol.15, No.3, pp.407-425 (2025)
Equation (11) has the following form: Since δ ∈ (0, 1), we have:
1 − υ 1 − υ τ υ
m (t) − m 0 = Φ (t, m (t)) ||m|| ≤ ||m 0 || + 1 + ||Φ||
B (υ) B (υ) (1 − υ) Γ (υ)

υ R t υ−1
+ (t − s) Φ (s, m ((s)) ds 1 − υ γ υ υ
B (υ) Γ (υ) 0 + Φ 0 1 + τ .
B (υ) Γ (υ + 1)
Assume that
1 − υ γ υ
− Φ (0, m (0)) 1 + t υ .
B (υ) Γ (υ + 1) ||Φ|| ≤ C Φ (1 + ||m||) .
(12)
5 5 Then,
Define T : C [0, τ] , R → C [0, τ] , R as:
1 − υ γ υ
||m|| ≤ ||m 0 || + 1 + τ υ
B (υ) Γ (υ + 1)

× (Φ 0 + C Φ (1 + ||m||)) .
Let

1 − υ γ υ υ
k = 1 + τ .
1 − υ B (υ) Γ (υ + 1)
T[m] (t) = m 0 + Φ (t, m (t))
B (υ) Then
υ R t υ−1 ||m|| ≤ ||m 0 || + k (Φ 0 + C Φ (1 + ||m||)) .
+ (t − s) Φ (s, m ((s)) ds
B (υ) Γ (υ) 0 By rearranging this inequality, we obtain:
||m 0 || + kΦ 0 + kC Φ

1 − υ γ υ ||m|| ≤ .
− Φ (0, m (0)) 1 + t υ . 1 − kC Φ
B (υ) Γ (υ + 1)
Therefore,
(13)
Since Φ is continuous, and the modified ABC in- ||m 0 || + kΦ 0 + kC Φ
||m|| ≤ M := .
tegral preserves continuity ⇒ T is continuous. 1 − kC Φ
Next, we need to show that all possible solutions This proves uniform boundedness.
to the equation m = δT[m] for δ ∈ [0, 1] are uni-
5
formly bounded in the space C [0, τ], R + . For
δ ∈ [0, 1], the equation m = δT[m] expands to: Let t 1 , t 2 ∈ [0, τ]. From the definition of T,
we have:
1 − υ
m (t) = δm 0 + δ Φ (t, m (t)) T[m] (t 2 ) − T[m] (t 1 ) =
B (υ)
υ R t υ−1 1 − υ
+ 0 (t − s) Φ (s, m ((s)) ds (Φ (t 2 , m (t 2 )) − Φ (t 1 , m (t 1 )))
B (υ) Γ (υ) B (υ)
(14)
υ
1 − υ γ υ + (I 2 − I 1 )
− Φ (0, m (0)) 1 + t υ . B (υ) Γ (υ)
B (υ) Γ (υ + 1)
Taking the supremum norm of both sides, we υ υ υ
+ Φ 0 [t − t ] ,
have: B (υ) Γ (υ + 1) 1 2

1 − υ
||m|| ≤ δ||m 0 || + δ ||Φ (t, m (t)) || where
B (υ) R t 2 υ−1
I 2 − I 1 = (t 2 − s) Φ (s, m ((s)) ds
t 1
υ R t υ−1
+ (t − s) ||Φ (s, m ((s)) ||ds R t 1 υ−1 υ−1
B (υ) Γ (υ) 0 + (t 2 − s) − (t 1 − s) Φ (s, m ((s)) ds.
0
Since Φ is Lipschitz continuous in m and contin-
1 − υ γ υ
+ Φ (0, m (0)) 1 + t υ . uous in t, there exists η Φ such that:
B (υ) Γ (υ + 1)
||Φ(t 2 , m(t 2 )) − Φ(t 1 , m(t 1 ))||
Simplify the integral:
(15)
υ
1 − υ τ
||m|| ≤ δ||m 0 || + δ ||Φ|| + ||Φ|| ≤ η Φ (|t 2 − t 1 | + ||m(t 2 − m(t 1 )||) .
B (υ) B (υ) Γ (υ)
Using mean value theorem, there exists χ ∈
(t 1 , t 2 ) such that
1 − υ γ υ
+ Φ 0 1 + τ υ . υ−1 υ−1 υ−2
B (υ) Γ (υ + 1) | (t 2 − s) −(t 1 − s) | = (υ−1)(χ−s) |t 2 −t 1 |,
412