1. Consider the following Black-Scholes PDE for European call:
@V
1
2S2
@2V
@V
rV = 0; (0; 1) (0; T ]; T 0
+
+ (r )S
@t
2
@S2
@S
V (S; t) = 0;
for S = 0;
V (S; t) = S
Ke
r(T t)
; for S
! 1
with suitable initial condition V (S; 0):
With the following transformation
S = Kex; t = T
2
;
2
x
2
V (s; t) = V Ke
; T
2
(
v(x; ) =: K exp
1
(q
2
{
q :=
2r
;
q
:=
2(r
)
;
2
2
) =: v(x; );
and
] }
[
1)x
1
(q
1)2 + qy(x; )
4
the above Black-Scholes PDE becomes the following 1-D heat conduction parabolic PDE:
@y
@2y
x
0;
x
@ = @x2 ; x
2 R;
2
{
}
y(x; 0) = max
exp(
(q
+ 1))
exp(
(q
1)); 0
; x
R
;
2
2
y(x; ) = 0; for1x !
;
1
2
! 1
(
)
y(x; ) = exp
(q
+ 1)x +
(q
+ 1)
for x
:
2
4
Solve the transformed
PDE by the following schemes:
Forward-Euler for time & central difference for space (FTCS) scheme.
Backward-Euler for time & central difference for space (BTCS) scheme.
Crank-Nicolson finite difference scheme
The values of the parameters are T = 1; K = 10; r = 0:06; = 0:3 and = 0.
Continued on the next page
1
2. Consider the following Black-Scholes PDE for European put:
@V
1
@2V
@V
+ 2S2
+ (r )S
rV = 0; (0; ) (0; T ]; T 0
@S2
@S
@t
2
t)
1
r(T
S; for S = 0;
V (S; t) = Ke
V (S; t) = 0; for S ! 1
with suitable initial condition V (S; 0):
Using the transformation given above the Black-Scholes PDE becomes the following problem:
@y
@2y
@
= @x2 ; x 2 R;0;
x
x
y(x; 0) = max
exp(
(q
1))
exp(
(q
+ 1)); 0
; x
R
;
2
2
2
1
1
2
}
{
for x !
y(x; ) = exp
2 (q
1)x + 4 (q
1)
;
;
(
)
y(x; ) = 0 for x ! 1;
Solve the transformed PDE by the following schemes:
Forward-Euler for time & central difference for space (FTCS) scheme.
Backward-Euler for time & central difference for space (BTCS) scheme.
Crank-Nicolson finite difference scheme
The values of the parameters are T = 1; K = 10; r = 0:06; = 0:3 and = 0.
2
