Question

The aim of transforming the Black-Scholes PDE is of course to find a form where an relatively easy solution exists. Most of the steps seem to be straightforward - please use this reference: http://planetmath.org/encyclopedia/AnalyticSolutionOfBlackScholesPDE.html

...all but one, actually the last one where a convection-diffusion equation is being transformed into the basic diffusion equation.

[In the article you find it here: "Under the new coordinate system (z), we have the relations amongst vector fields ... leading to the following transformation of equation..."]

The u_x-term vanishes by some magic coordinate transformation. When you look at the derivatives they even seems wrong to me because they state that tau=s and then derive delta/delta tau = delta/delta s + some extra term (delta/delta y * -(r-1/2 sigma^2).

I simply don't get it: first how it works and second how they find that kind of transformation.

Answer 1

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:

V_t + (1/2)s²S²V_SS = rV - rSV_S.

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 = e^x, so the log price x follows normal Brownian motion. This yields the equation:

V_t + (1/2)s²(V_xx - V_x) = r(V - V_x).

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(V_xx - V_x - V_tau) = r(V - V_x).

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

u_xx + (something)u_x + (something else)u = u_tau.

(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.