analytic solution of Black-Scholes PDE

analytic solution of Black-Scholes PDE

Here we present
an analytical
solution for the

Black-Scholes partial differential equation

,

r

⁢

f

=

∂

⁡

f

∂

⁡

t

+

r

⁢

x

⁢

∂

⁡

f

∂

⁡

x

+

1

2

⁢

σ

2

⁢

x

2

⁢

∂

2

⁡

f

∂

⁡

x

2

,

f

=

f

⁢

(

t

,

x

)

,

(1)

over the domain

0

<

x

<

∞

,

0

≤

t

≤

T

,
with

terminal

condition

f

⁢

(

T

,

x

)

=

ψ

⁢

(

x

)

,
by reducing this parabolic PDE to
the

heat equation

of physics.

We begin by making the substitution:

u

=

e

-

r

⁢

t

⁢

f

,

which is motivated by the fact that it is the portfolio value

discounted by the

interest rate

r

(see the

derivation

of the

Black-Scholes formula

)
that is a

martingale

.
Using the

product rule

on

f

=

e

r

⁢

t

⁢

u

,
we derive the PDE that the function

u

must satisfy:

r

⁢

f

=

r

⁢

e

r

⁢

t

⁢

u

=

r

⁢

e

r

⁢

t

⁢

u

+

e

r

⁢

t

⁢

∂

⁡

u

∂

⁡

t

+

r

⁢

x

⁢

e

r

⁢

t

⁢

∂

⁡

u

∂

⁡

x

+

1

2

⁢

σ

2

⁢

x

2

⁢

e

r

⁢

t

⁢

∂

2

⁡

u

∂

⁡

x

2

;

or simply,

0

=

∂

⁡

u

∂

⁡

t

+

r

⁢

x

⁢

∂

⁡

u

∂

⁡

x

+

1

2

⁢

σ

2

⁢

x

2

⁢

∂

2

⁡

u

∂

⁡

x

2

.

(2)

Next, we make the substitutions:

y

=

log

⁡

x

,

s

=

T

-

t

.

These changes of variables can be motivated
by observing that:

•

The underlying process described by the variable

x

is a geometric Brownian motion
(as explained in the derivation of the Black-Scholes formula itself),
so that

log

⁡

x

describes a

Brownian motion

, possibly
with a drift.
Then

log

⁡

x

should satisfy some sort of diffusion equation
(well-known in physics).

•

The evolution of the system
is backwards from the terminal state of the system. Indeed,
the

boundary condition

is given as a terminal state,
and the

coefficient

of

∂

⁡

u

/

∂

⁡

t

is

positive

in equation (

2

).
(Compare with the standard heat equation,

0

=

-

∂

⁡

u

/

∂

⁡

t

+

∂

⁡

u

/

∂

⁡

x

, which describes a
temperature evolving forwards in time.)
So to get to the heat equation,
we have to use a substitution to reverse time.

Since

∂

⁡

u

∂

⁡

s

=

-

∂

⁡

u

∂

⁡

t

,

∂

⁡

u

∂

⁡

x

=

∂

⁡

u

∂

⁡

y

⁢

d

⁢

y

d

⁢

x

=

1

x

⁢

∂

⁡

u

∂

⁡

y

,

and

∂

2

⁡

u

∂

⁡

x

2

=

∂

∂

⁡

x

⁢

(

1

x

⁢

∂

⁡

u

∂

⁡

y

)

=

-

1

x

2

⁢

∂

⁡

u

∂

⁡

y

+

1

x

2

⁢

∂

2

⁡

u

∂

⁡

y

2

,

substituting in equation (

2

),
we find:

0

=

-

∂

⁡

u

∂

⁡

s

+

(

r

-

1

2

⁢

σ

2

)

⁢

∂

⁡

u

∂

⁡

y

+

1

2

⁢

σ

2

⁢

∂

2

⁡

u

∂

⁡

y

2

.

(3)

The first

partial derivative

with respect to

y

does not cancel (unless

r

=

1

2

⁢

σ

2

)
because we have not take into account
the drift of the Brownian motion.
To cancel the drift (which is linear in time),
we make the substitutions:

z

=

y

+

(

r

-

1

2

⁢

σ

2

)

⁢

τ

,

τ

=

s

.

Under the new coordinate system

(

z

,

τ

)

, we have the relations
amongst

vector fields

:

∂

∂

⁡

z

=

∂

∂

⁡

y

,

∂

∂

⁡

τ

=

-

(

r

-

1

2

⁢

σ

2

)

⁢

∂

∂

⁡

y

+

∂

∂

⁡

s

,

leading to the following
of equation (

3

):

0

=

-

∂

⁡

u

∂

⁡

τ

-

(

r

-

1

2

⁢

σ

2

)

⁢

∂

⁡

u

∂

⁡

z

+

(

r

-

1

2

⁢

σ

2

)

⁢

∂

⁡

u

∂

⁡

z

+

1

2

⁢

σ

2

⁢

∂

2

⁡

u

∂

⁡

z

2

;

or:

∂

⁡

u

∂

⁡

τ

=

1

2

⁢

σ

2

⁢

∂

2

⁡

u

∂

⁡

z

2

,

u

=

u

⁢

(

τ

,

z

)

,

(4)

which is one form of the diffusion equation.
The domain is on

-

∞

<

z

<

∞

and

0

≤

τ

≤

T

;
the initial condition is to be:

u

⁢

(

0

,

z

)

=

e

-

r

⁢

T

⁢

ψ

⁢

(

e

z

)

:=

u

0

⁢

(

z

)

.

The original function

f

can be recovered
by

f

⁢

(

t

,

x

)

=

e

r

⁢

t

⁢

u

⁢

(

T

-

t

,

log

⁡

x

+

(

r

-

1

2

⁢

σ

2

)

⁢

τ

)

.

The fundamental solution of the PDE (

4

)
is known to be:

G

τ

⁢

(

z

)

=

1

2

⁢

π

⁢

σ

2

⁢

τ

⁢

exp

⁡

(

-

z

2

⁢

σ

2

⁢

τ

)

(derived using the

Fourier transform

);
and the solution

u

with initial condition

u

0

is given by the

convolution

:

u

⁢

(

τ

,

z

)

=

u

0

*

G

τ

⁢

(

z

)

=

e

-

r

⁢

T

2

⁢

π

⁢

σ

2

⁢

τ

⁢

∫

-

∞

∞

ψ

⁢

(

e

ζ

)

⁢

exp

⁡

(

-

(

z

-

ζ

)

2

2

⁢

σ

2

⁢

τ

)

⁢

𝑑

ζ

.

In terms of the original function

f

:

f

⁢

(

t

,

x

)

=

e

-

r

⁢

τ

2

⁢

π

⁢

σ

2

⁢

τ

⁢

∫

-

∞

∞

ψ

⁢

(

e

ζ

)

⁢

exp

⁡

(

-

(

log

⁡

x

+

(

r

-

1

2

⁢

σ

2

)

⁢

τ

-

ζ

)

2

2

⁢

σ

2

⁢

τ

)

⁢

𝑑

ζ

,

(

τ

=

T

-

t

) which agrees with the result derived using probabilistic methods (

http://planetmath.org/BlackScholesFormula

).

Title

analytic solution of Black-Scholes PDE

Canonical name

AnalyticSolutionOfBlackScholesPDE

Date of creation

2013-03-22 16:31:34

Last modified on

2013-03-22 16:31:34

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

6

Author

stevecheng (10074)

Entry type

Derivation

Classification

msc 60H10

Classification

msc 91B28

Related topic

ExampleOfSolvingTheHeatEquation

Related topic

BlackScholesPDE

Related topic

BlackScholesFormula

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