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.


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.

  • $\begingroup$ Thank you Scott - anyway it is the same thing here: x=ln(S/K)+(r-sigma^2/2)*tau - I don't understand how you come to this term and how the actual transformation works. p.s.: you are right, there are some browser problems with this site, IE works well though $\endgroup$ – vonjd Oct 25 '09 at 6:49
  • $\begingroup$ I got this solution from Strang's <i>Computational Science and Engineering</i>, which references <i>The Mathematics of Financial Derivatives</i> by Wilmott, Harrison, and Dewynne. $\endgroup$ – S. Carnahan Oct 27 '09 at 4:16
  • $\begingroup$ Scott: Thank you very much, again! Could you please elaborate on this part: "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. " Could you please give an example for "something" and "something else" so that these terms vanish? $\endgroup$ – vonjd Oct 27 '09 at 18:09
  • $\begingroup$ If you work it out, you get: (something) = 2r/s^2 - 1 - 2a, so you can set a = r/s^2 - 1/2 to make it vanish. (something else) = a^2/s^2 - 2ar/s^2 + a - 2r/s^2 + b, so you can set b = 2r/s^2 + 2ar/s^2 - a - a^2/s^2 to make it vanish. $\endgroup$ – S. Carnahan Oct 28 '09 at 2:36

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