Page 34 - IJOCTA-15-2
P. 34

Analyzing the Black-Scholes equation with fractional coordinate derivatives using . . .
                                                                                      2
                Now suppose that g(t) is a derivative func-   Lemma 2. Let g(t) ∈ C [0, t n ] and 0 < β < 1,
            tion. According to the definition of the Caputo   then
                                                                                   ′
            derivative in 17 and by substituting t = t n , we       1    Z  t n  g (s)      1    β
            have                                                   Γ(1 − β)  0  (t n − s) β  ds −  λ  b g(t n )
                                                                                                 0
                                         ′
                  β
                 ∂ g(t)       1    Z  t  g (s)
                       =                      ds                    n−1
                                                                    X
                  ∂t β    Γ(1 − β)  0  (t − s) β                  −      b β   − b β   g(t l ) − b β  g(t 0 )
                                                                                   n−l
                                                                          n−l−1
                                                                                               n−1
                                          ′
                                   Z
                              1      t n  g (s)                     l=1
                       =                        ds
                          Γ(1 − β)  0  (t n − s) β                ≤     1      1 − β  +  4 − β  − (1 + 2 −β )
                                               ′                     Γ(2 − β)   12     2 − β
                                    n Z
                              1    X    t k   g (s)
                       =                            ds.           × max |g (t)|∆t 2−β .
                                                                            ′′
                          Γ(1 − β)          (t n − s) β
                                   k=1  t k−1                       0≤t≤t n
                         ′   g(t n ) − g(t n−1 )
            Considering g =                 + O(∆t),
                                   ∆t                         Proof. See. 35
            in which O represents the big-O notation, we have
                β
               ∂ g(t n )      1      X
                                      n
                       =                 (g(t n ) − g(t n−1 ))    By inserting the n-th time surface in Eq. (14),
                 ∂t β    Γ(1 − β)∆t
                                     k=1                      we have
                           Z
                              t k
               + ∆tO(∆t)        (t n − s) −β ds.
                                                                                 2
                                                                  β
                             t k−1                               ∂ u(x, t n )  ∂ u(x, t n )   ∂u(x, t n )
                                                       (18)                = A           + G
                                                                    ∂t β          ∂x 2          ∂x       (24)
            On the other hand,
                                                                           − Hu(x, t n ) + f(x, t n ).
               Z                      1−β
                 t k               ∆t
                    (t n − s) −β ds =      (n − k + 1) 1−β
                                                                              β
                                   1 − β                                     ∂ u(x, t n )
                                                                  By placing            from Eq. (23) in Eq.
                t k−1
                                                                                ∂t
                                                                                 β
               − (n − k) 1−β  + O(∆t 2−β ).                   (24), we obtain
                                                       (19)
            Thus, from Eq. (18)                                 1    β        n−1   β      β
                                                                               X
                                                                   b u(x, t n ) −   b n−l−1  − b n−l  u(x, t l )
                                                                    0
               β
                                      n
              ∂ g(t n )       1      X                          λ
                      =                   g(t k ) − g(t k−1 )                  l=1
                ∂t β    Γ(2 − β)∆t β                                                2
                                     k=1                            β               ∂ u(x, t n )  ∂u(x, t n )
                                                              − b n−1 u(x, t 0 ) = A  ∂x 2  + G   ∂x
                 (n − k + 1) 1−β  − (n − k) 1−β  + O(∆t 2−β ).
                                                                               − Hu(x, t n ) + f(x, t n ).
                                                       (20)                                              (25)
            Consider the following definitions:                   On the other hand, the BICs are
                      
                      l := n − k,                                         (
                                                                            u(x L , t n ) = λ(t n ),
                                       β
                       λ := Γ(2 − β)∆t ,               (21)                                              (26)
                       β          1−β      1−β                              u(x R , t n ) = ζ(t n ),
                      
                       b := (l + 1)    − (l)   .
                        l                                     and
                Therefore, we will have                                       u(x, 0) = υ(x).            (27)
                β
               ∂ g(t n )  1  n−1  β
                           X
                       =       b (g(t n−l ) − g(t n−l−1 ))
                 ∂t β    λ      l                      (22)   3.2. Space discretization
                            l=0
                        + O(∆t 2−β ).
                                                              On the space domain (x L , x R ), suppose that M x is
                                P
                By rewriting the   , we will have             the number of node points for space discretization
                                                                                                          d
                                                              and h is the space step length defined as h =
                                                                                                         M x
                β


               ∂ g(t n )  1  β        n−1   β      β         where d = x R − x L . In this case, we consider
                                      X
                       =    b g(t n ) −   b      − b
                             0
                 ∂t β     λ                n−l−1    n−l       the node points as X L = x 0 < x 1 < x 2 < · · · <
                                      l=1
                                                                   = x R where for i = 0, 1, 2, · · · , M x , we have

                                                              x M x
                         β                2−β
               × g(t l ) − b n−1 g(t 0 ) + O(∆t  ).           x i = x L + ih. In addition, for the space deriva-
                                                       (23)   tives, we consider the following relations:
                                                           229
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