Page 40 - IJOCTA-15-2

Analyzing the Black-Scholes equation with fractional coordinate derivatives using . . .
this case, we have and
n
n
µ 1 E i−1 + µ 2 E − µ 1 E n = v M x−1
u
i+1
i
u X
n
n 2
∥R ∥ h = t h|R | , n = 0, 1, 2, · · · , N t ,
β
0
n−1
n−1
n−1
− µ 1 E − µ 2 E + µ 1 E + 2b E j
i−1 i i+1 n−1 i j=1
(92)
(101)
n−1

X β β
n
l
+ b − b E + E l−1 + R ,
n−l−1 n−l i i i where
l=1
n
n
n
n
E = (E , E , . . . , E M x−1 ), n = 0, 1, 2, · · · , N t ,
1
2
(102)
and
2
n
where R = O(∆t 2−β +h ). Furthermore, the
i
initial and boundary conditions imply that
n
n
n
R = (R , R , . . . , R n ), n = 0, 1, 2, · · · , N t .
( 1 2 M x−1
n
n
E = E M x = 0, n = 0, 1, 2, . . . , N t , (93) (103)
0
n
n
0
E = 0, i = 0, 1, 2, . . . , M x . Since E = E n = 0 and R = R n = 0, we
0
i
0
M x
M x
Therefore
have
n
n−1
µ 1 E n + µ 2 E − µ 1 E n = −µ 1 E i−1 − µ 2 E n−1 Z D
i
i+1
i−1
i
2
n
2
n 2
n
∥E ∥ = |E (x)| dx = ∥E (x)∥ 2,
h L
(104)
n−1
X β β 0
l
+ µ 1 E n−1 + b − b E + E l−1
i+1 n−l−1 n−l i i n = 0, 1, 2, · · · , N t ,
l=1
n
+ R . and
i
(94)
Z D
Similar to the previous section, for n = ∥R ∥ = |R (x)| dx = ∥R (x)∥ 2,
2
n
2
n 2
n
h
L
0, 1, 2, · · · , N t , we define two grid functions as fol- 0 (105)
lows: n = 0, 1, 2, · · · , N t .
( n
E , x ∈ (x j− 1 , x j+ 1 ), j = 1, 2, · · · , M x − 1, By using the Parseval identity, we will have
j
n
E (x) = 2 2
h
h
0, x ∈ (x L , x L + ] ∪ [x R − , x R ], M x−1 +∞
2 2 n 2 X n 2 X n 2
(95) ∥E ∥ = h|E | = D |s | ,
h
j
j
and j=1 j=−∞ (106)
( n n = 0, 1, 2, · · · , N t ,
R , x ∈ (x j− 1 , x j+ 1 ), j = 1, 2, · · · , M x − 1,
j
n
R (x) = 2 2
h
h
0, x ∈ (x L , x L + ] ∪ [x R − , x R ]. and
2 2
(96) M x−1 +∞
X n 2 X n 2
n 2
n
n
Then, E (x) and R (x) have the following Fourier ∥R ∥ = h|R | = D |r | , (107)
j
h
j
series expansion: j=1 j=−∞
n = 0, 1, 2, · · · , N t .
+∞ jπx
n 2i
X
n
E (x) = s e D , n = 0, 1, 2, · · · , N t , Again for x j = x L + jh, for the solution of Eq.
j
j=−∞
(94), we will have
(97)
2πj
n
and E = s e , ω = , (108)
n iωx j
j
+∞ jπx D
n
n 2i
X
R (x) = r e D n = 0, 1, 2, · · · , N t , and
j 2πj
n
n iωx j
j=−∞ R = r e , ω = . (109)
j
(98) D
√
where i = −1, D = x R − x L , and Lemma 5. There is a positive constant η such
( n 1 R D n 2i jπx that
s = E (x)e D dx, n 1
j D 0 (99) |r | ≤ η|r |. (110)
n
n
D dx.
r = 1 R D R (x)e 2i jπx
j D 0
n
2
Proof. Since R = O(∆t 2−β + h ), ∃ η > 0 such
We define the norm ∥.∥ h as i
that
v
2
n
u M x−1 |R | ≤ η(∆t 2−β + h ). (111)
u X i
n
n 2
∥E ∥ h = t h|E | , n = 0, 1, 2, · · · , N t , Then, according to Eq. (107)
j
j=1 √
2
n
(100) ∥R ∥ h ≤ η D(∆t 2−β + h ). (112)
i
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