Mathematical Modeling in Finance with Stochastic Processes

7.2. SOLUTION OF THE BLACK-SCHOLES EQUATION 229

Gather like terms:

uτ=uxx+ [2α+ (k−1)]ux+ [α^2 + (k−1)α−k−β]u.

Chooseα = −(k−1)/2 so that the ux coefficient is 0, and then choose
β=α^2 + (k−1)α−k=−(k+ 1)^2 /4 so theucoefficient is likewise 0. With
this choice, the equation is reduced to

uτ=uxx.

We need to transform the initial condition too. This transformation is

u(x,0) =e−(−(k−1)/2)x−(−(k+1)

(^2) /4)· 0
v(x,0)
=e((k−1)/2)xmax(ex− 1 ,0)
= max

(

e((k+1)/2)x−e((k−1)/2)x, 0

)

.

For future reference, we notice that this function is strictly positive when
the argumentxis strictly positive, that isu 0 (x)>0 whenx >0, otherwise,
u 0 (x) = 0 forx≤0.
We are in the final stage since we are ready to apply the heat-equation
solution representation formula:

u(x,τ) =

1

2

√

πτ

∫∞

−∞

u 0 (s)e−(x−s)

(^2) / 4 τ
ds.
However, first we want to make a change of variable in the integration, by
takingz= (s−x)/

√

2 τ, (and therebydz= (− 1 /

√

2 τ)dx) so that the inte-
gration becomes:

u(x,τ) =

1

√

2 π

∫∞

−∞

u 0

(

z

√

2 τ+x

)

e−z

(^2) / 2
dz.
We may as well only integrate over the domain whereu 0 >0, that is for
z >−x/

√

2 τ. On that domain,u 0 =e((k+1)/2)·(x+z

√ 2 τ

)−e((k−1)/2)·(x+z

√ 2 τ

)so

we are down to:

1

√

2 π

∫∞

−x/

√

2 τ

e

k+1 2 (x+z√ 2 τ)

e−z

(^2) / 2
dz−

1

√

2 π

∫∞

−x/

√

2 τ

e

k− 21 (x+z√ 2 τ)

e−z

(^2) / 2
dz
Call the two integralsI 1 andI 2 respectively.