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:
<https://planetmath.org/AnalyticSolutionOfBlackScholesPDE>


...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.



Transformation of the Black-Scholes PDE into the diffusion equation - shift of coordinate system
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](http://en.wikipedia.org/wiki/Black-Scholes#Derivation). 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.


