Green's Function Solution

We now use the Green's function for the Diffusion or Heat equation[4], which is the solution to that equation for a point (or delta function) source at point z' at time t'=0

$$G(z-z';t)=\frac{1}{\sqrt{4 \pi t'}}e^{-\frac{(z-z')^2}{4t'}}.$$ (19)

The verification of this Green's function solution is shown in Appendix A. The Green's function shows the Gaussian diffusion of the pointlike input with distance from the input (z-z') increasing as the square root of the time t', as in a random walk.

We can use the Green's function to write the solution for $\tilde{y}(z,t')$ in terms of summing over its input values at points z'on the boundary at the initial time t'=0

 

$$\tilde{y}(z,t')=\int_{-\infty}^{\infty} dz'\tilde{y}(z',0) \frac{1}{\sqrt{4\pi t'}} e^{-\frac{(z-z')^2}{4t'}}.$$ (20)

Putting in the initial condtions at t'=0, where $\tilde{y}(z',0)$ vanishes for negative z', gives

$$\tilde{y}(z,t')=\frac{1}{\sqrt{4\pi t'}} \int_{0}^{\infty} dz... ... 1 \right) \frac{1}{\sqrt{4\pi t'}} e^{-\frac{(z-z')^2}{4t'}}.$$ (21)

To do the integral we change the variable to

 

$$\frac{z'-z}{\sqrt{2t'}}, \quad {\rm and}$$ q = (22)

$$\sqrt{2t'} dq.$$ dz' = (23)

The lower limit on the q integral is now

$$-\frac{z}{\sqrt{2t'}}=-d2,$$ (24)

where substitution gives the dimensionless

$$d2= \frac{\ln{x/c}+(r-v^2/2)(t^*-t)}{v\sqrt{(t^*-t)}}.$$ (25)

In the first term we now complete the square to get a new variable

$$q' = q - \sqrt{2t'} \frac{v^2/2}{(r-v^2/2)}.$$ (26)

The new lower limit on q' in the first term is now -d1 where

$$d1 = \frac{\ln{x/c}+(r+v^2/2)(t^*-t)}{v\sqrt{(t^*-t)}}.$$ (27)

After completing the square on the first term, the exponent simplifies to

$$\ln{x/c}+r(t^*-t)-\frac{1}{2}q'^2.$$ (28)

Both integrals are now related to the Cumulative Distribution Function of the Normal Distribution

$$N(x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^x e^{-\frac{1}{2}t^2}dt.$$ (29)

If we change t to -t in the above integral and invert the limits we get the form of our integrals

$$N(x) = \frac{1}{\sqrt{2 \pi}} \int_{-x}^{\infty} e^{-\frac{1}{2}t^2}dt.$$ (30)

We now have our solution for the canonical $\tilde{y}$

$$\tilde{y}(z,t') = c ( e^{\ln{(x/c)}+r(t^*-t)}N(d1)-N(d2) )$$ (31)

Finally, we use the facts that $\tilde{y} = \hat{y} = y$, and that $e^{\ln{(x/c)}}=x/c$, and the conversion

$$w(x,t)=\tilde{w}(u,t) = e^{-r(t^*-t)} y(u,t),$$ (32)

to get the Black-Scholes solution

 

w(x,t) = x N(d1)-c e^-r(t*-t)N(d2). (33)