Page 39 - IJOCTA-15-2

M. M. Parsa, K. Sayevand, H. Jafari, I. Masti / IJOCTA, Vol.15, No.2, pp.225-244 (2025)
which is equivalent to Then

(µ 2 − i (2µ 1 sin(ωh))) ξ 1 r 1 β 0 r−1 β β
X

(75) ξ + ξ r−1 = 2b r−1 ξ + b − b
h i |R| r−l−1 r−l
β 0
= (−µ 2 + 2b ) + i (2µ 1 sin(ωh)) ξ . l=1
0

l
Therefore × ξ + ξ l−1 .

" #
β
(−µ 2 + 2b ) + i (2µ 1 sin(ωh)) 0
1
ξ = 0 ξ . (76) (83)
µ 2 − i (2µ 1 sin(ωh))
It is clear that |R| ≥ 1. Therefore
β
From Eq. (76) and b = 1, we have
0 r−1

0
2
2
2
(2 − µ 2 ) + 4µ sin (ωh) r r−1 ≤ 2b β ξ + X b β − b β
ξ + ξ

0 2
1 2
|ξ | = 1 2 |ξ | . (77) r−1 r−l−1 r−l
2
2
µ + 4µ sin (ωh) l=1
2 1

l
× ξ + ξ l−1 .

Also, from Lemma 3, we have (84)
0 2
1 2
|ξ | ≤ |ξ | . Therefore Let |ξ r−1 | max = max{|ξ , |ξ |, . . . , |ξ r−1 ||}. In
0
1
0
1
|ξ | ≤ |ξ |. (78) this case, we will have
r
|ξ | + |ξ r−1 | max ≤ 2b β |ξ r−1 | max + 2|ξ r−1 | max
r−1
Now, for r ≥ 2, we have r−1 β β
X
× b r−l−1 − b r−l ,
r iωjh
r iω(j+1)h
r iω(j−1)h
µ 1 ξ e + µ 2 ξ e − µ 1 ξ e =
l=1
− µ 1 ξ r−1 iω(j−1)h − µ 2 ξ r−1 iωjh (85)
e
e
or
0 iωjh
+ µ 1 ξ r−1 iω(j+1)h + 2b β ξ e r r−1 r−1
e
r−1 |ξ | + |ξ | max ≤ 2|ξ | max
!
r−1

X β β l iωjh l−1 iωjh X (86)
r−1
+ b − b ξ e + ξ e . β β β
r−l−1 r−l × b + b − b .
r−1 r−l−1 r−l
l=1 l=1
(79)
From Lemma 4, we have
β
r
|ξ | + |ξ r−1 | max ≤ 2|ξ r−1 | max b . (87)
The above equation can be rewritten as: 0
β
h i We know that b = 1. Therefore
r
µ 2 − µ 1 e iωh − e −iωh ξ = 0
r
|ξ | ≤ |ξ r−1 | max . (88)
h i
−µ 2 + µ 1 e iωh − e −iωh ξ r−1
Thus, from Eq. (78) and Eq. (88), it is obvious
that
r−1

β 0 X β β l l−1 r r−1
+ 2b ξ + b − b ξ + ξ , |ξ | ≤ |ξ |, r ≥ 1. (89)
r−1 r−l−1 r−l
l=1 Since this equation holds for every r ≥ 1, it also
(80)
holds for n = 1, 2, . . . , N t , and
n
|ξ | ≤ |ξ n−1 |, n = 1, 2, . . . , N t . (90)
which is equivalent to
Thus, for every n = 1, 2, . . . , N t we have
r
(µ 2 − i (2µ 1 sin(ωh))) ξ =
n
0
|ξ | ≤ |ξ |. (91)
[(−µ 2 + i (2µ 1 sin(ωh))] ξ r−1 + 2b β ξ 0
r−1
(81) In the other words, our proposed method is un-

r−1
X β β conditionally stable for solving the TFB-S model.
l
+ b − b ξ + ξ l−1 .
r−l−1 r−l
l=1
Assuming that R := µ 2 − i (2µ 1 sin(ωh)),
therefore
In what follows, the convergence of the im-

r−1
β
X
β
β
0

r
ξ + ξ r−1 R = 2b r−1 ξ + b r−l−1 − b r−l plicit discrete scheme (41) using the Fourier anal-
l=1 n
ysis method is discussed. We assume that E =
i
l
b n
× ξ + ξ l−1 . |U −U |, where U and U are the approximate
b n
n
n
i
i
i
i
(82) and exact solutions of Eq. (41), respectively. In
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