4 added 1 characters in body

I'm afraid the Planetmath page put my browser into an infinite reload loop, so I can't help you with the formalism there.

I would recommend instead looking at the change of variables in the Wikipedia article. The last time I checked it, it seemed to work.

Edit: Okay, I have a formulation that works. I'll write s for sigma, so the equation is initially:

Vt + (1/2)s2S2VSS = rV - rSVS.

Since S follows a lognormal random walk (in particular the stochastic diff eq governing S involves a logarithmic derivative), it is natural to change to x = log S, or S = ex, so the log price x follows normal Brownian motion. This yields the equation:

Vt + (1/2)s2(Vxx - Vx) = r(V - Vx).

Black-Scholes is a final-value problem, i.e., we know the value of the option at time T, and diffusion works backwards. It is therefore natural to negate the time variable (and multiply by a suitable scalar to make things neater). tau = (1/2)s2(T-t). Then we get:

(1/2)s2(Vxx - Vx - Vtau) = r(V - Vx).

Finally we rescale the value function to remove exponential growth effects. u = eax + b(tau)V for undetermined coefficients a and b. We can substitute, multiply the equation by 2eax+b(tau)/s2, and we get:

uxx + (something)ux + (something else)u = utau.

(something) is a degree one polynomial in a and is independent of b. (something else) is a degree one polynomial in b, so we can choose a and b to kill those terms. This yields the diffusion equation.

Hope that helps.

3 full solution

Edit: Okay, I have a formulation that works. I'll write s for sigma, so the equation is initially:

Vt + (1/2)s2S2VSS = rV - rSVS.

Since S follows a lognormal random walk (in particular the stochastic diff eq governing S involves a logarithmic derivative), it is natural to change to x = log S, or S = ex, so the log price x follows normal Brownian motion. This yields the equation:

Vt + (1/2)s2(Vxx - Vx) = r(V - Vx).

Black-Scholes is a final-value problem, i.e., we know the value of the option at time T, and diffusion works backwards. It is therefore natural to negate the time variable (and multiply by a suitable scalar to make things neater). tau = (1/2)s2(T-t). Then we get:

(1/2)s2(Vxx - Vx - Vtau) = r(V - Vx).

Finally we rescale the value function to remove exponential growth effects. u = eax + b(tau) for undetermined coefficients a and b. We can substitute, multiply the equation by 2eax+b(tau)/s2, and we get:

uxx + (something)ux + (something else)u = utau.

(something) is a degree one polynomial in a and is independent of b. (something else) is a degree one polynomial in b, so we can choose a and b to kill those terms. This yields the diffusion equation.

Hope that helps.

2 Okay, wikipedia uses bizarre characters in URLs.

I'm afraid the Planetmath page put my browser into an infinite reload loop, so I can't help you with the formalism there.

I would recommend instead looking at the change of variables in the Wikipedia article. The last time I checked it, it seemed to work.

1