Thermal Diffusivity : Conversion of Black-Scholes Equation

Please check the derivitves first - all of my coments for you are in YELLOW,
the problem description in it's entierty is in green. This problem has several
parts:
1) - get the derivitves of u(x,t) and plug them into the PDE, then reduse that
2
PDE to: u = u . - This is the part giving me hell right now!
2 xx
2) - Show that the alpha and beta chosen conform to the specs in the problem
description (that should be pretty much a no-brainer.)

3) - transform the terminal condition for the PDE: V ( X , ) = max ( eX - 1 , 0) into the
form given - I haven't gotten that far, so I'm not sure how difficult that will be?

Please respond in pdf or word format - and PLEASE! - no hand written /
scanned documents!

Problem - 4
V(X , )
Define u (X , ) = , where and are constants yet to be specified.
( X+ )
e
( X+ )
Then: V(X , ) = u (X , ) e .

Now, starting with the PDE for V (X , ):
d V ( X , ) r - - r ( V ( X , ) ).
2 2
d2
V(X , ) = V(X , ) +
d

d 2 dX2 dX 2
and using the chain rule, work out the PDE for u (X , ). Also show that there is a
unique choice of and which simplifies the latter PDE to:
2 d2
u (X , ) = u (X , )
d
- < X < .
d 2 dX2
2 r
Now, by divine inspiration, introduce the ratio = .
2

Show that your choice of and , when expressed in terms of this , reads:
2
-1 -
( - 1) ( + 1) .
2
= =
2 8
Finally, show that the I.C. for V ( X , ) is equivalent to the following I.C. for u (X , ):
[ 0____________________For ( X 0) ]
u ( X , 0) =
X ( + 1) X ( - 1)
2
e - e 2 _____For ( X > 0)


Note that the system describing u ( X , ) is formally identical to the heat equation of
2

an infinite rod with thermal diffusivity k= ! Thus, the volitility of the commodity
2
which underlies our call option is generating a mathmatical diffusion process.

Homework Set # 6 Page - 11 of 19
( X+ )
We define: V(X , ) = u (X , ) e

d V (X , ) r - - r (V ( X , ) )
2 2
d2
Our PDE is: V (X , ) =
d
V(X , ) +
d 2 dX2 dX 2

So we need the following partial derivitives:
d
V(X , ) =
d u ( X , ) e( X + ) + d e( X + ) u ( X , )

d d d
d ( d ( X + ) + d e( X + ) u ( X , )
V X , ) = u (X , ) e
dX dX dX

d d ( d d ( X+ ) + d e( X+ ) u ( X , )
2
V(X , ) = V X , ) = u ( X , ) e
d

dX
2 dX dX dX dX dX

( X+ ) ( X+ )
V(X , ) = u (X , ) e u (X , )
d d
+ e
d d
d ( ( X+ ) ( X+ )
V X , ) =
d (
u X , ) e + e u (X , )
dX dX

d d (
2
( X+ ) ( X+ )
V(X , ) = u X , ) e u ( X , ) = ......
d
+ e
dX
2 dX dX
d d ( ( X+ ) d
u ( X , )
( X+ )
u X , ) e + e
dX dX dX

d d ( ( X+ )
d2

( X + ) ( X + )
u X , ) e = dX2
u (X , ) e

+ u (X , ) e
dX dX

d ( X+ )
u ( X , ) = e
2 ( X+ ) ( X+ ) d u ( X , )
u ( X , ) + e
e
dX dX
( X+ ) ( X+ ) 2 ( X+ ) ( X+ )
V = u e + u e + e u + e u
XX XX X

And now we have (as last time) all of our partial derivitives and our PDE:
I'm not even sure if these derivitives
(u + u)
( X+ )
V = e are correct, so please check!!!

- But I'm positive that this is the PDE
(u + u)
( X+ ) we need to re-substitue the
V = e
X X
derivitves into!
u
( X+ ) 2
V = e + u + u + u
XX XX X


2 2
V = V + V r - - r (V(X , ))
2 XX X 2

Homework Set # 6 Page - 12 of 19
Substituting these into our PDE we get:

e
( X+ )
( )
u + u = ......


2 ( X+ ) 2
( X+ )
u + u + u + u + e
( X+ )
( ) r - - r V
2
e e u + u
2 XX X X 2

( X+ )
Dividing through by e gives us:

2 2
( u + u = ) u + u + 2 u + u + u + u r - - r V
( )
2 XX X X 2 e( X+ )





2 2
( u + u = ) u + u + 2 u + u + u + u r - - r V
( )
2 XX X X 2 e( X+ )

2 r
=
2 2
- -1
( + 1) ( - 1)
2
= =
8 2


( )
2
- 2
= + 2 + 1
8

2 2 2 2
- - 1
= - - = +
8 4 8 2 2

2
2 r 2 2 r 2
- -
2r
2 2 2 2
=
-

-

=

+
1
8 4 8 2 2

2 2 2 2
-4 r 2 r -2 r 1
= - - = +
4 2 8 2 2
8 4 2

2 2
-r r -r 1
= - - = +
2 2 8 2 2
2


2 2 2 2 2 2
u + u = u + u + 2 u + u + u r - u + u r - u - ....
2 XX 2 2 X 2 X X 2 2
r V
( X+ )
e
Homework Set # 6 Page - 13 of 19
2 2 2 2 2 2
2
u + u = u + u + u + u + u r - u + ur - u - ....
2 XX 2 2 X 2 X X 2 2
r V
-r2
2

2

2

2
u + r
- - u = u + u
2
+ u + ........ ( X+ )
XX e
2 2 8 2 2 2
2
2 2 2
r V
u + u r - u + ur - u -
X 2 X X 2 2 ( X+ )
e

2 2 2 2 2
r u r u u 2
u - - - = u + u + u + ........
2 2 8 2 XX 2 2
2
2 2 2
r V
u + u r - u + ur - u -
X 2 X X 2 2 ( X+ )
e




2 2 2 2 2 2
u -
r u
-
r u u
-

u + u
-r + 1 + -r + 1 u + ........
2 2 8
=
2 XX 2 2 2 2 2 2
2

2 2 2 2
-r + 1 u + u r - u + -r ( u r) + u r - -r u + u - ......
2 2 X 2 X X 2 2 2 2 4
2
r V
- r 1 - r2 r 2
+ X+ - -
2 2 2 2 2 8
e
2 2
2 2 2 2 2 2 2 r u u
u -
r u
-
r u u
-

u -
r u
+
u
+
-r + 1 u - X
+
X
+ ....
2 2 8
=
2 XX 2 4 2 2 2 2 4
2 2 2

2 2 2
r ( u r) u r r u u r V
u r - u - + + - -
X X 2

2 2
2
2 4 - r X X - r2 t r t 2 t
+ + - -
2 2 2 2 2 8
e

2
2 2 u r u
u X X ur r V
u = u + + - + -
2 XX 4 2 4 2 - r X X - r2 t r t 2 t
+ + - -
2 2 2 2 2 8
e

This is as far as I could get?


Homework Set # 6 Page - 14 of 19