Page 38 - IJOCTA-15-2

Analyzing the Black-Scholes equation with fractional coordinate derivatives using . . .
Hereunder, we will show the stability of the Hence, if x i = x L + ih, we may suppose that the
implicit discrete scheme (41) by using the Fourier solution of Eq. (60) has the following form:
b n
n
n
analysis method. We assume that e = |U −U |, n n iωx j 2πj
i
i
i
n
b n
j
where U and U are the approximate and exact e = ξ e , ω = D . (67)
i
i
solutions of Eq. (41), respectively. In this case, Lemma 3. In (60), we have
we have
2
2
2
(2 − µ 2 ) + 4µ sin (ωh) ≤ 1. (68)
1
2
2
2
n
1
2
µ 1 e n + µ 2 e − µ 1 e n = µ + 4µ sin (ωh)
i+1
i
i−1
− µ 1 e n−1 − µ 2 e n−1 + µ 1 e n−1
i−1 i i+1 Proof. By putting µ 2 = 1 + 2Ω + Φ, where
Aλ
Ω = and Φ = Hλ we get
n−1
X
l
0
+ 2b β e + b β − b β n−l e + e l−1 , h 2
i
n−1 i
i
n−l−1
l=1 µ 2 ≥ 1 =⇒ 4µ 2 ≥ 4 =⇒ 4 − 4µ 2 ≤ 0. (69)
(60)
2
2
2
Now if we assume that S = µ + 4µ sin (ωh),
2
1
n
and e = e n = 0. For n = 0, 1, 2, · · · , N t , we then S ≥ 0. Therefore
0
M x
define a grid function as follows: 4 − 4µ 2 + S
4 − 4µ 2 + S ≤ S =⇒ ≤ 1, (70)
S
 or
n
 e , x ∈ (x 1 , x 1 ), 2 2 2
 j j− j+ (2 − µ 2 ) + 4µ sin (ωh)
 2 2 1
 ≤ 1. (71)
 j = 1, 2, · · · , M x − 1, 2 2 2
n
e (x) = µ + 4µ sin (ωh)
1
2



 h h β 1−β
0, x ∈ (x L , x L + ] ∪ [x R − , x R ]. Lemma 4. The coefficients b = (l + 1)

2 2 l
(61) − (l) 1−β apply to the following properties:
β
Because e n = e n , for e n with the period 1. b > 0, l = 0, 1, 2, · · · ,
l
0
j
β
M x
β
β
β
D = x R − x L , a periodic extension can be done. 2. b > b > b > . . . > b β n−1 > b n,
1
2
0
P n−1 β β β β
Then, using the properties of the Fourier series, 3. l=1 (b l−1 − b ) + b n−1 = b = 1.
0
l
n
e (x) can be written as follows:
Proof. The proof is easily done.
+∞ jπx
n 2i
n
X
e (x) = ξ e D , n = 0, 1, 2, · · · , N t , (62) Theorem 3. The numerical scheme (41) for
j
j=−∞ solving the TFB-S model is unconditionally sta-
ble.
√ R jπx
n
1
e (x)e
where i = −1 and ξ = D 0 D n 2i D dx. n n iωjh
j
j
We define the norm ∥.∥ h as Proof. We assume that e = ξ e where i =
√
−1. Therefore, from (60), we have
µ 1 ξ e + µ 2 ξ e − µ 1 ξ e =
v n iω(j−1)h n iωjh n iω(j+1)h
u M x−1
u X
n
n 2
∥e ∥ h = t h|e | , n = 0, 1, 2, · · · , N t , − µ 1 ξ n−1 iω(j−1)h − µ 2 ξ n−1 iωjh
e
e
j
j=1 n−1 iω(j+1)h β 0 iωjh
(63) + µ 1 ξ e + 2b n−1 ξ e
where n−1 β β l iωjh l−1 iωjh
X
n
n
n
e = (e , e , . . . , e n ), n = 0, 1, 2, · · · , N t . + b n−l−1 − b n−l ξ e + ξ e .
1 2 M x−1 l=1
(64) (72)
n
Since e = e n = 0, we have
0 M x Now, for n = 1, we have
D 1 iω(j−1)h 1 iωjh 1 iω(j+1)h
Z
n
2
2
n 2
n
∥e ∥ = |e (x)| dx = ∥e (x)∥ 2, µ 1 ξ e + µ 2 ξ e − µ 1 ξ e =
h
L
0 iωjh
0 iω(j+1)h
0 iω(j−1)h
0 (65) − µ 1 ξ e − µ 2 ξ e + µ 1 ξ e
n = 0, 1, 2, · · · , N t .
β 0 iωjh
+ 2b ξ e .
By using the Parseval identity, we have 0
(73)
Eq. (73) can be written in the following form:
M x−1 +∞
X X h i
n 2
n 2
n 2
∥e ∥ = h|e | = D |ξ | , µ 2 − µ 1 e iωh − e −iωh ξ 1
h j j (66)
j=1 j=−∞ h i (74)
β iωh −iωh 0
n = 0, 1, 2, · · · , N t . = −µ 2 + 2b + µ 1 e − e ξ ,
0
233