Page 12 - IJOCTA-15-1

P. Kumar / IJOCTA, Vol.15, No.1, pp.1-13 (2025)

Which gives
 γ C
 D I(t) = K 1 (t, I, P, B, E),
t

 γ C

D P(t) = K 2 (t, I, P, B, E), (25) ∥ζ n (t)∥ = ∥G n (t) − G n−1 (t)∥
t
γ
C
 D B(t) = K 3 (t, I, P, B, E), Z t
 t 1
 γ−1
 γ C ∥ζ n (t)∥ =∥ (t − ϑ)
D E(t) = K 4 (t, I, P, B, E), Γ(γ) 0
t
with initial conditions I(0) = I 0 , P(0) = [K(G n−1 (ϑ), ϑ) − K(G n−2 (ϑ), ϑ)] dϑ ∥ .(31)
P 0 , B(0) = B 0 , E(0) = E 0 . Here K 1 , K 2 , K 3 , Then, we get
and K 4 are the singular-type kernels representing
the right-hand side expressions of model (24), re- Z t
spectively. ∥ζ n (t)∥ ≤ 1 (t − ϑ) γ−1
Γ(γ) 0
3.2. Solution existence ∥K(G n−1 (ϑ), ϑ) − K(G n−2 (ϑ), ϑ)∥ dϑ. (32)
Here we check the existence and uniqueness of the
solution with the application of some well-known Now we fix K as a Lipschitzian respect to G, so
mathematical results. In this regard, we consider we have
the following initial value problem (IVP) for rep-
resenting the model (25) Z t
L γ−1
∥ζ n (t)∥ ≤ (t − ϑ)
Γ(γ)
γ C 0
D G(t) = K(G(t), t),
t
(26) ∥G n−1 (ϑ) − G n−2 (ϑ)∥ dϑ. (33)
G(0) = G 0 ,
where
Therefore, we get the following result
G(t) = [I(t), P(t), B(t), E(t)],
G 0 = [I(0), P(0), B(0), E(0)], L Z t
∥ζ n (t)∥ ≤ (t − ϑ) γ−1 ∥ζ n−1 (ϑ)∥ dϑ. (34)

 K 1 (t, I, P, B, E), Γ(γ) 0

K 2 (t, I, P, B, E),

K(G(t), t) = Theorem 3. The given IVP (26) has a unique
 K 3 (t, I, P, B, E),
 solution under the contraction for K.
K 4 (t, I, P, B, E).

Proof. From (34), we have
Let us consider the following Volterra integral
Z t
equation of the IVP (26) L γ−1
∥ζ n (t)∥ ≤ (t − ϑ) ∥ζ n−1 (ϑ)∥ dϑ. (35)
Γ(γ) 0
G(t) = G 0 + 1 R t (t − ϑ) γ−1 K(G(ϑ), ϑ)dϑ.
Γ(γ) 0 Using the value of ∥ζ n−1 (t)∥, we have
(27)
Following iterative approach on the nonlinear ker- γ 2
Lt
nel K, we derive the expression ∥ζ n (t)∥ ≤ ∥ζ n−2 (t)∥ . (36)
Γ(γ + 1)
G n (t) = G 0 + 1 R t (t − ϑ) γ−1 K(G n−1 (ϑ), ϑ)dϑ. Similarly, for ∥ζ n−2 (t)∥
Γ(γ) 0
(28)
We choose G 0 (t) = G 0 . Then, the difference of two Lt γ 3
successive terms is defined by ∥ζ n (t)∥ ≤ ∥ζ n−3 (t)∥ . (37)
Γ(γ + 1)

1 Z t Then, from the successive iterations, we get
G n (t) − G n−1 (t) = (t − ϑ) γ−1
Γ(γ) 0
[K(G n−1 (ϑ), ϑ) − K(G n−2 (ϑ), ϑ)]dϑ. (29) Lt γ n
∥ζ n (t)∥ ≤ ∥ζ 0 (t)∥
Γ(γ + 1)
γ n
We choose ζ n = G n (t) − G n−1 (t). Therefore, we t n
≤ L max ∥G 0 ∥ .(38)
have Γ(γ + 1) t∈[0,T]
n
n P
X If we say G(t) = ζ j (t), then G(t) exists and
G n (t) = ζ j (t). (30) j=0
j=0 continuous.
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