Skip to content

Turning the Black-Scholes PDE into the heat equation

Question

Show how a change of variables converts the Black-Scholes equation into the heat equation.

Short answer

Set τ=T−t\tau=T-t, x=ln⁡S+(r−12σ2)τx=\ln S+(r-\tfrac12\sigma^2)\tau and V=e−rτu(x,τ)V=e^{-r\tau}u(x,\tau); then uτ=12σ2uxxu_\tau=\tfrac12\sigma^2u_{xx}. The log variable removes the SS and S2S^2 coefficients, shifting xx by the drift absorbs the first-derivative term, and the discount factor cancels the −rV-rV term. Rescaling time by 12σ2\tfrac12\sigma^2 gives the standard uτ=uxxu_\tau=u_{xx}.

Adapted from Joshi, Denson & Downes, Quant Job Interview Questions and Answers, Q6.17, p.190; question and answer rewritten by SelfMadeQuant.com.

See the hint and the full worked solution in the question bank (free account).

Open the worked solution

Lessons behind this question

Start with a free lesson: Options is part of a free topic. Create a free account, no card needed, to read it in full with worked solutions.

Read it free