Transformation of the Black-Scholes PDE into the diffusion equation - shift of coordinate system 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.
 A: 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.
