Page 75 - IJOCTA-15-4

Analysis and analytical solution of incommensurate fuzzy fractional nabla difference systems...
(Here, all vector operations of the same type are and
element-wise). Thus, ⃗x(1) exists and satisfies 1 1
k(u, v, s) = (u + v) + (u − v)s (42)
⃗
⃗
⃗x(1) = ⃗α⃗x 0 ⊕⃗q(W⃗x 0 ⊕⃗p). (36) 2 2
[t] [t]
Therefore, ⃗x(1) is unique because ⃗x 0 is unique. is used to interchange f i and g i if coefficients
We complete the proof by induction. Suppose are negative. We note that k(u, v, 1) = u and
there exists a unique ⃗x(i) for i < N, satisfying k(u, v, −1) = v.
Equation (2). We show ⃗x(N) exists and is unique.
From Equation (34)
[0]
[0]
˜
˜
Now, applying q i to z i (0) ∼ (d i (0), f , ˜g )
N−1 ! i i
⃗ α X Γ(N − s − ⃗α) gives the right-hand side of the NN. Corollary 1
⃗x(N) = ⃗x(s)
Γ(1 − ⃗α) Γ(N − s + 1) plays an important role, yielding
s=0
⃗ ⊕⃗q(W⃗x(N − 1) ⃗ ⊕⃗p). ˜ ˜ [0] [0]
˜
˜
˜
α i
∇ x i (1) = q i (z i (0)) ∼ (d i (0), f , ˜g ),
(37) i i
The right hand side of Equation (37) is a fuzzy
where
combination of ⃗x(0), ⃗x(1), . . . , ⃗x(N−1). Since they
are well-defined fuzzy numbers by the induction ˜ ˜
˜
d i (0) =q i (d i (0)),
hypothesis, ⃗x(N) is well-defined.
Furthermore, suppose ⃗y is another solution of
˜
Equation (2). Then, it satisfies f (r) =d i (0) − q i (d i (0) − f (r)), (43)
˜ [0]
[0]
˜
˜
˜
˜
i i
!
⃗ α N−1 Γ(N − s − ⃗α)
X
⃗y(N) = ⃗y(s)
Γ(1 − ⃗α) Γ(N − s + 1) ˜ [0] ˜ [0] ˜
˜
s=0 ˜ g (r) =q i (d i (0) + ˜g (r)) − d i (0).
i
i
⃗ ⊕⃗q(W⃗y(N − 1) ⃗ ⊕⃗p).
(38)
But y(s) = x(s) for s ≤ N − 1 by the uniqueness
assumption of induction, which leads to Since α i ≥ 0, it follows from Equation (36)
that
!

⃗ α N−1 Γ(N − s − ⃗α)
X
⃗y(N) = ⃗x(s)
Γ(1 − ⃗α) Γ(N − s + 1) [1] [1]
s=0 x i (1) ∼ (d i (1), f , g ), (44)
i
i
⃗ ⊕⃗q(W⃗x(N − 1) ⃗ ⊕⃗p) = ⃗x(N),
(39) where
˜
˜
and it completes the proof. □ d i (1) =α i d i (0) + d i (0),
4. Algorithm for finding fuzzy solution
[1] [0] ˜ [0]
˜
f (r) =α i f (r) + f (r),
A base for constructing an algorithm to compute a i i i (45)
fuzzy solution is Equation (37) and the triple rep-
T
resentation. Suppose ⃗x(0) = [x 1 (0), . . . , x ν (0)] , [1] [0] ˜ [0]
g (r) =α i g (r) + ˜g (r).
i
i
i
is a given fuzzy input. Let Supp(x i (0)) =
[0] [0]
[a i (0), b i (0)], with x i (0) ∼ (d i (0), f , g ) where
i
i
T
[0] [0] Thus, ⃗x(1) = [x 1 (1), . . . , x ν (1)] .
f , g i ∈ U, for i = 1, . . . , ν. Then,
i
z i (0) =w i1 x 1 (0) ⊕ · · · ⊕ w iν x ν (0) + p i
(40)
[0]
[0]
˜
˜
∼(d(0), f , ˜g ) We obtain ⃗x(T) for T > 1 recursively. Sup-
i i [t] [t] [0] [0]
pose x i (t) ∼ (d i (t), f , g ) where f , g ∈ U,
where i i i i
ν for t = 1, . . . , T and i = 1, . . . , ν. We derive a re-
X
˜
d i (0) = w ij d j (0) + p i , cursive formula for each component of the triple.
Define
j=1
ν ⃗z(t − 1) = W⃗x(t − 1) ⃗ ⊕⃗p.
[0]
[0]
X
f ˜ [0] = |w ij |k(f , g , sgn(w ij )), (41) Then,
i j j
j=1
z i (t − 1) = w i1 x 1 (t − 1) ⊕ · · · ⊕ w iν x ν (t − 1) ⊕ p i
ν
[0] X [0] [0] ˜ ˜ [t−1] [t−1]
˜ g = |w ij |k(g , f , sgn(w ij )), ∼ (d(t − 1), f i , ˜g i ),
i j j
j=1 (46)
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